DOT/FAA/AR-04/51 

Office of Aviation Research 

Washington, D.C. 20591

THERMOD, an Enhanced 

Thermal Model for 

Determining Aircraft 

Operational Temperatures 

December 2004 

Final Report 

This document is available to the U.S. public 

through the National Technical Information 

Service (NTIS), Springfield, Virginia 22161. 

U.S. Department of Transportation 

Federal Aviation Administration
NOTICE 

This document is disseminated under the sponsorship of the U.S. 

Department of Transportation in the interest of information exchange. The 

United States Government assumes no liability for the contents or use 

thereof. The United States Government does not endorse products or 

manufacturers. Trade or manufacturer's names appear herein solely 

because they are considered essential to the objective of this report. This 

document does not constitute FAA certification policy. Consult your local 

FAA aircraft certification office as to its use. 

This report is available at the Federal Aviation Administration William J. 

Hughes Technical Center's Full-Text Technical Reports page: 

actlibrary.tc.faa.gov in Adobe Acrobat portable document format (PDF). 
Technical Report Documentation Page

1. Report No.

DOT/FAA/AR-04/51

2. Government Accession No. 3. Recipient's Catalog No.

5. Report Date

December 2004 

4. Title and Subtitle

THERMOD, AN ENHANCED THERMAL MODEL FOR DETERMINING 

AIRCRAFT OPERATIONAL TEMPERATURES 6. Performing Organization Code

7. Author(s)

Nathan Govindarajoo, Ph.D, PE.

8. Performing Organization Report No. 

9. Performing Organization Name and Address 10. Work Unit No. (TRAIS) 

1889 Old Dixie Highway, #203 

Vero Beach, FL 32960

11. Contract or Grant No.

13. Type of Report and Period Covered

 Final Report

12. Sponsoring Agency Name and Address

U.S. Department of Transportation 

Federal Aviation Administration 

Office of Aviation Research 

Washington, DC 20591

14. Sponsoring Agency Code

 ACE-120, AGATE

15. Supplementary Notes

The FAA William J. Hughes Technical Center Technical Monitor was Peter Shyprykevich.

16. Abstract

An enhanced version of a thermal analysis computer program called THERMOD was developed for determining the maximum 

operating limit (MOL) temperatures of general aviation aircraft. The project was undertaken under a Federal Aviation 

Administration-AGATE sponsorship. This report is the second of two reports prepared in conjunction with this project. 

Enhancements included program debugging and corrections as well as development of an alternate implicit finite difference 

method, which is used in transient analysis. Numerical validation of THERMOD was undertaken with respect to the finite 

element methods. Good correlation was found. The validated THERMOD can be used to determine the MOL temperatures of a 

typical aircraft that has a low wing configuration. 

17. Key Words 

Maximum operating limit temperature, Thermal analysis, 

Conduction, Convection, Radiation, Steady state, Transient 

state

18. Distribution Statement 

This document is available to the public through the National 

Technical Information Service (NTIS), Springfield, Virginia 

22161.

19. Security Classif. (of this report)

 Unclassified 

20. Security Classif. (of this page)

 Unclassified

21. No. of Pages

 282

22. Price

Form DOT F1700.7 (8-72) Reproduction of completed page authorized 
TABLE OF CONTENTS 

Page 

EXECUTIVE SUMMARY ix 

1. INTRODUCTION 1-1 

1.1 Background 1-1 

1.2 THERMOD Geometry and Input Data 1-3 

1.3 Project Objectives 1-7 

2. DEBUGGING AND PROGRAM CORRECTIONS 2-1 

2.1 BUG 1 2-1 

2.2 BUG 2 2-1 

2.3 BUG 3 2-2 

2.4 Algorithm Development for IBFDM 2-3 

3. NUMERICAL VALIDATION OF THERMOD 3-1 

3.1 Sample Problem 1 3-2 

3.2 Sample Problem 2 3-17 

3.3 Sample Problem 3 3-23 

3.4 Sample Problem 4 3-39 

3.5 Sample Problem 5 3-56 

3.6 Sample Problem 6 3-74 

4. SAMPLE PROBLEM 4-1 

4.1 Input Data 4-1 

4.2 Input File 4-5 

4.3 Output Files 4-7 

5. SUMMARY 5-1 

6. REFERENCES 6-1 

APPENDIX A—THERMOD CODE

iii
LIST OF FIGURES 

Figure Page 

1-1 A Schematic Presentation of THERMOD 1-2 

1-2 Overall Model Geometry and Typical Temperature Profiles at Critical Locations 1-4 

1-3 Input Thermal Properties for THERMOD Model 1-5 

1-4 A Typical Flight Profile During Transient Cooling 1-6 

2-1 Control Volumes for Developing Thermal Equations 2-4 

3-1 Contour Plot of the FEM Output of Sample Problem 1 3-16 

3-2 Contour Plot of the FEM Output of Sample Problem 2 3-22 

3-3 Contour Plot of the FEM Output of Sample Problem 3 3-37 

3-4 Contour Plot of the FEM Output of Sample Problem 4 3-54 

3-5 Contour Plot of the FEM Output of Sample Problem 5 3-72 

3-6 Contour Plot of the FEM Output of Sample Problem 6 3-89 

LIST OF TABLES 

Table Page 

3-1 THERMOD Input File (input.dat) of Sample Problem 1 3-2 

3-2 THERMOD Output File (summary.dat) of Sample Problem 1 3-4 

3-3 THERMOD-Simulated Temperatures (°F) of the Right Wing of Sample 

Problem 1 3-7 

3-4 Finite Element Method Input File of Sample Problem 1 3-8 

3-5 Truncated FEM Output File of Sample Problem 1 3-16 

3-6 Finite Element Method-Simulated Temperatures (°F) of the Right Wing of 

Sample Problem 1 3-17 

3-7 THERMOD Output File (summary.dat) for Sample Problem 2 3-18 

3-8 THERMOD-Simulated Temperatures (°F) of the Right Wing of Sample 

Problem 2 3-21 

3-9 Truncated FEM Output File of Sample Problem 2 3-22 

3-10 Finite Element Method-Simulated Temperatures (°F) of the Right Wing of 

Sample Problem 2 3-23 

iv
3-11 THERMOD Input File (input.dat) of Sample Problem 3 3-24 

3-12 THERMOD Output File (summary.dat) for Sample Problem 3 3-25 

3-13 THERMOD-Simulated Temperatures (°F) of the Right Wing of Sample 

Problem 3 3-29 

3-14 Finite Element Method Input File of Sample Problem 3 3-29 

3-15 Truncated FEM Output File of Sample Problem 3 3-38 

3-16 Finite Element Method-Simulated Temperatures (°F) of the Right Wing of 

Sample Problem 3 3-39 

3-17 THERMOD Input File (input.dat) of Sample Problem 4 3-40 

3-18 THERMOD Output file (summary.dat) for Sample Problem 4 3-41 

3-19 THERMOD-Simulated Temperatures (°F) of the Right Wing of Sample 

Problem 4 3-45 

3-20 Finite Element Method Input File of Sample Problem 4 3-45 

3-21 Truncated FEM Output File of Sample Problem 4 3-55 

3-22 Finite Element Method-Simulated Temperatures (°F) of the Right Wing of 

Sample Problem 4 3-56 

3-23 THERMOD Input File (input.dat) of Sample Problem 5 3-57 

3-24 THERMOD Output File (summary.dat) for Sample Problem 5 3-58 

3-25 THERMOD-Simulated Temperatures (°F) of the Right Wing of Sample 

Problem 5 3-62 

3-26 Finite Element Method Input File of Sample Problem 5 3-63 

3-27 Truncated FEM Output File of Sample Problem 5 3-73 

3-28 Finite Element Method-Simulated Temperatures (°F) of the Right Wing of 

Sample Problem 5 3-74 

3-29 THERMOD Input File (input.dat) of Sample Problem 6 3-74 

3-30 THERMOD Output File (summary.dat) for Sample Problem 6 3-76 

3-31 THERMOD-Simulated Temperatures (°F) of the Right Wing of Sample 

Problem 6 3-79 

v
3-32 Finite Element Method Input File of Sample Problem 6 3-80 

3-33 Truncated FEM Output File of Sample Problem 6 3-90 

3-34 Finite Element Method-Simulated Temperatures (°F) of the Right Wing of 

Sample Problem 6 3-91 

4-1 Temperature and Associated Radiation Data 4-1 

4-2 Thickness of Thermal Elements of the Wing 4-2 

4-3 Thickness of Thermal Elements of the Fuselage Side 4-2 

4-4 Thickness of the Thermal Elements of the Floor 4-2 

4-5 Thickness of the Thermal Elements of the Roof 4-3 

4-6 Overall Dimensions of the Thermal Model 4-3 

4-7 Absorptivity and Emissivity Properties of the Exterior Surfaces of the Wing and 

Fuselage 4-3

4-8 Thermal Properties of Various Solid Materials 4-3 

4-9 Thermal Properties of Air 4-4 

4-10 Miscellaneous Properties 4-4 

4-11 Input File (input.dat) for the Sample Problem 4-6 

4-12 Sample Problem Output Data File (summary.dat) 4-7 

4-13 Sample Problem Output Data File (transient.dat) 4-11 

vi
LIST OF ACRONYMS 

EFFDM Explicit forward finite difference method 

FAA Federal Aviation Administration 

FEM Finite element method 

IBFDM Implicit backward finite difference method 

MOL Maximum operating limit 

SPC 

vii/viii
EXECUTIVE SUMMARY

Composite aircraft structural elements are unfavorably affected by an increase in temperature 

due to exposure to the thermal environment. For design purposes, the synergistic effects of 

extreme ambient temperatures and the accompanying solar radiation should be taken into 

consideration in determining the maximum operating limit (MOL) temperatures experienced by 

the structural elements. Allowable design properties of composite materials may then be 

generated based on these MOL temperatures. THERMOD is a computer model that was 

developed for determining the MOL temperatures for aircraft under various paint schemes. The 

selected low-wing geometry is best suited to general aviation aircraft. 

In determining the MOL temperatures, THERMOD considers the effects of radiation, 

convection, and conduction. Radiation includes direct solar radiation, infrared sky radiation, and 

their reflections from the tarmac. Infrared emission from the surrounding structures, including 

the tarmac, wings and the fuselage and its interaction among these structures, are also 

considered. Convection due to the wind, as well as turbulent convection within the cabin, is also 

simulated, as are one-dimensional conduction through element thickness. The model 

incorporates the effect of fillets at the wing-fuselage junction. Due consideration is also given to 

the greenhouse effect within the cabin, which allows for a realistic modeling of the thermal 

environment within the cabin. The effect of shade underneath the wing is also modeled. 

The soaked temperatures of an aircraft are predicted based on steady-state assumptions, giving 

conservative steady-state temperatures. Because limit loads generally occur in flight conditions, 

a transient (unsteady-state) thermal analysis is used to simulate the thermal conditions while the 

aircraft executes the following maneuvers: taxi, takeoff, climb, and cruise. These maneuvers are 

cooling effects that in addition to intentional cooling of the cabin through opening the door 

simulate a more realistic thermal environment for predicting the MOL temperature. 

THERMOD formulates a total of 67 independent equations within a nonlinear system of 

equations. The nonlinearity is introduced through radiation effects and through convective 

properties with the cabin, which are modeled as temperature-dependent. This system of 

equations is solved using the Newton-Raphson iteration technique. 

This report is the third of four reports prepared in conjunction with this project. The fourth 

report is entitled “THERMOD User’s Manual,” and it will serve as an instructive guide for users 

who wish to use THERMOD for determining aircraft MOL temperatures. The first two reports 

validated THERMOD analyses with test data and studied the sensitivity to input variables. 

The enhanced version of THERMOD was developed for simulating the MOL temperatures of 

general aviation aircraft. Enhancements included program debugging and corrections as well as 

development of an alternate implicit finite difference method. This method for solving the 

transient problem is unconditionally stable over all time and spatial domain as opposed to the 

existing explicit forward finite difference method. Numerical validation of THERMOD was 

undertaken with respect to the finite element methods. Good correlation was found. 

ix/x
1. INTRODUCTION.

An enhanced version of an original thermal analysis computer program called THERMOD [1] 

has been developed for simulating maximum operating limit (MOL) temperatures for general 

aviation aircrafts. This project was undertaken under a Federal Aviation Administration (FAA)-

AGATE sponsorship. 

This report is the third of four reports prepared in conjunction with this project. The fourth 

report, “THERMOD User’s Manual,” serves as a guide for users who wish to use THERMOD 

for determining aircraft MOL temperatures [2]. The first two reports validated THERMOD 

analyses with test data and studied the sensitivity to input variables [3 and 4]. 

To help the reader get acquainted with THERMOD, background information on pertinent 

thermal aspects of the program and input data requirements are presented. This is followed by a 

statement of the project objectives. A more complete description of THERMOD and the 

development of model equations can be found in the original THERMOD document [1]. 

1.1 BACKGROUND.

Structural components strength and sharpness of composite airframe are unfavorably affected by 

an increase in their temperatures due to exposure to the thermal environment. For design 

purposes, the combined effects of extreme ambient temperature and the accompanying solar 

radiation should be taken into consideration in determining the MOL temperatures experienced 

by the structural elements. Design properties of composite materials may then be generated 

based on these MOL temperatures. THERMOD was developed for determining the MOL 

temperatures of low-wing aircraft under various paint schemes. 

In determining the MOL temperatures, THERMOD considers the effects of radiation, 

convection, and conduction. Radiation includes direct solar radiation, infrared sky radiation, and 

their reflections from the tarmac. Infrared emission from the surrounding elements, including 

the tarmac, wings, fuselage, and its interaction among these elements, are also considered. 

Convection due to the wind as well as turbulent convection within the cabin is also simulated, as 

are one-dimensional conduction through element thickness. The model incorporates the effect of 

fillets at the wing-fuselage junction. Due consideration is also given to the greenhouse effect 

within the cabin that allows for a realistic modeling of the thermal environment within the cabin. 

The effect of shade underneath the wing is also modeled. 

In addition to the above factors, the following assumptions are made in THERMOD: (1) 

nonparticipating medium (air); (2) discretized space and time domains; (3) discretized elements 

being isothermal, opaque, diffuse, gray and characterized by uniform radiosity, irradiation, and 

material properties; (4) nonopaque materials such as windows and windshields are considered 

transparent with associated transmissivity values; and (5) constant material properties with 

respect to time (and, hence, temperature). These assumptions are necessary in simplifying the 

complexities involved in a three-dimensional thermal problem being addressed. 

The soaked temperatures of an aircraft are predicted based on steady-state assumptions, giving 

conservative steady-state temperatures. Because limit loads generally occur in flight conditions, 

1-1
a transient (unsteady-state) thermal analysis is used to simulate the thermal conditions while the 

aircraft executes the following maneuvers: taxi, takeoff, climb, and cruise. These maneuvers are 

cooling effects that in addition to intentional cooling of the cabin through opening the door 

simulate realistic thermal environment for predicting the MOL temperature. 

THERMOD formulates a total of 67 independent equations within a nonlinear system of 

equations. The nonlinearity is introduced through radiation effects and through convective 

properties with the cabin, which are modeled as temperature-dependent. This system of 

equations is solved using the Newton-Raphson iteration technique [5]. 

As previously noted, THERMOD considers all three heat transfer mechanisms (convection, 

conduction, and radiation) normally associated with an aircraft parked in the open. Figure 1-1 is 

a schematic representation of a thermal environment showing short wavelength solar radiation, 

long wavelength sky (infrared) radiation, convective heat transfer due to the wind, and onedimensional conduction (through the thickness). THERMOD also considers reflected solar 

energy between the fuselage and wing, diffused solar energy reflected from the tarmac, and 

infrared reflections. The fillets and greenhouse effects are considered as well. Because 

THERMOD simulates a total of 67 independent equations, 67 unknowns are needed. These 

unknowns are 53 temperatures and 14 radiosites. The 53 temperatures (T1 through T53) and 

their locations are noted in figure 1-1. 

FIGURE 1-1. A SCHEMATIC PRESENTATION OF THERMOD 

1-2
Note that not all the unknown temperatures are shown. T2, for example, is located immediately 

below T1 at the skin-core interface (see figure 1-2 for further clarity). Similarly, T27 is located 

immediately below T26 at the insulatory material B-composite floor interface. 

The geometry of the aircraft shown is low wing with sandwich construction for skins. The 

sandwich can be either foam or honeycomb. Nonsandwich skins are also admissible if the 

thickness of the core is assumed to be very small. 

A typical THERMOD analysis begins at the steady-state phase and continues on to the transient 

phase; a phase in which cooling is introduced due to aircraft maneuvers. At the point of 

application of limit or gust load, the temperatures at all 53 locations are noted. Of these 53 

temperatures, 9 temperatures are considered nonstructural. These nonstructural temperatures are 

the three tarmac temperatures (T51, T52, and T53); four insulatory material temperatures (T25, 

T26, T43, and T44); and two transparent material temperatures (T49 and T50), leaving 44 

temperatures to be considered structural. 

THERMOD repeats its analysis over different time periods, as requested. The maximum

temperature over these time periods is then reported as the MOL temperature. This MOL 

temperature may be located at the surface (depending on how dark the paint is), inside the cabin, 

within the floor space, or located anywhere else in the aircraft. 

1.2 THERMOD GEOMETRY AND INPUT DATA.

Figure 1-1 showed a typical cross section of a small general aviation aircraft. A simplified 

rendering of the model is shown in figure 1-2. THERMOD is built on this simplified model. 

Figure 1-2 also shows typical sections at critical locations. In addition to the overall geometry as 

indicated in figure 1-2, a set of input data characterizing each layer, as shown in figure 1-3, is 

also needed. The layer properties include thickness, density, thermal conductivity, and specific 

heat capacity. To consider surface effects, all exposed aircraft surfaces are assigned their 

respective absorptivity and emissivity values. These values are dependent on paint color and 

surface texture. The tarmac is also considered a surface, whose surface and other thermal 

properties are required as well. THERMOD models the greenhouse and fillet effects. 

Transmissivity and fillet color information is needed to consider these two effects. The degree 

of greenhouse effect is also dependent on the percentage of transparent material. This 

information is input in the form of transparent surface area in relation to the overall cabin surface 

area. To simulate convective coefficients, the kinematic viscosity and Prandtl number are 

needed. Because the aircraft is left out in the open, it is subjected to the environment. The 

climatic data of the environment, for each time period, includes ambient temperature, sky 

temperature, solar radiation, and wind speed. 

The final piece of information concerns the cooling that takes place when the aircraft executes 

the following maneuvers: taxi, takeoff, climb, and cruise. This data was furnished through a 

flight profile that is unique to a particular aircraft and is part of the input. The flight profile 

provides information on typical aircraft maneuvering speed with respect to time, from which, the 

time-varying convective coefficient is determined. This coefficient is then used in the 

subsequent transient cooling process. An example flight profile is shown in figure 1-4. 

1-3
L cabin

Wwing

L ledge

L wing

H cabin

H wing Tarmac

d 

b 

a c 

Foam

Foam

Outer Skin

Inner skin

Spar cap

Air space

Spar cap

Inner skin

Outer skin

 Detail a: Wing

Outer skin F

o

a

m

Inner skin

Insulatory

material C Detail b:

Fuselage side

Detail c: Floor 

Insulatory material B

Outer skin

Inner skin

Floor space 

Foam

Inner skin

Outer skin

Foam

Floor

Bellypan

 Detail d: Roof

Outer skin

Inner skin

Insulatory material A

Foam

Thermal model: Top view

Thermal model: Front view

Fuselage

Wing

T1

T2

T3

T4

T5

T6

T7

 T8

T9

T10

 T43 T42 T41 T40 T39

T26

T27

T28

T37

T38

T21

T22

T23

T24

T25

T29

T30

T35

T36

Wcabin

L flg

FIGURE 1-2. OVERALL MODEL GEOMETRY AND TYPICAL TEMPERATURE 

PROFILES AT CRITICAL LOCATIONS 

1-4
1-5

FIGURE 1-3. INPUT THERMAL PROPERTIES FOR THE THERMOD MODEL 
Velocity (mph)

Time (sec)

Ambient wind speed

V 0

V 1 V 2

V 3 V4

V5

V6 V7

V 8 V9

Taxi

Stop

Take-off Run 

Accelerate

Climb 

Accelerate

Cruise

Open door Close

door 

Greenhouse cooling

Limit load application 

0 

 Time

V0=0 mph

V1=10 mph V0- V1=5 sec

V2=10 mph V1- V2=120 sec

V3=0 mph V2- V3=5 sec

V4=0 mph V3- V4=60 sec

V5=75 mph V4- V5=20 sec

V6=110 mph V5- V6=5 sec

V7=110 mph V6- V7=120 sec

V8=190 mph V7- V8=60 sec

V9=190 mph V8-V9=60 sec

Ambient wind

speed=9.5 mph

FIGURE 1-4. A TYPICAL FLIGHT PROFILE DURING TRANSIENT COOLING 

Fuselage is modeled as a box and treated as an enclosure in modeling the greenhouse effect. The 

fuselage roof is treated as a surface with an equivalent transmissivity, allowing solar radiation to 

enter the cabin. The insulatory materials (A), (B), and (C) are simplified representations of cabin 

roof interior, cabin floor, and cabin side interiors. 

A total of 28 distinct layers are considered. Each layer, labeled P1 to P28 in figure 1-3, is 

assigned four properties, i.e., thickness, density, thermal conductivity, and specific heat capacity. 

The tarmac is regarded as a special layer and also characterized by these four properties, as 

indicated in figure 1-3. The eight exposed surfaces are assigned absorptivity (α)or emissivity (ε) 

properties, depending on their surface texture and color. The wing, fuselage, and the interior (B) 

and (C) surfaces are assigned scatter factors (sf). The aircraft is subjected to climatic factors that 

include solar radiation (Qsol), wind (Vwind), ambient temperature (Tamb), and sky temperature 

(Tsky). 

Cooling is assumed to occur as soon as the door is opened. The greenhouse cooling duration is a 

function of the soaked steady-state ambient temperature of the cabin and of the tolerable ambient 

temperature desired before closing the door. This tolerable temperature, arbitrarily selected, is 

1-6
input into THERMOD as ambcabt. The MOL temperature is determined at the point when limit 

load is applied. This moment in time is located at 395 seconds from point V0 , as shown in 

figure 1-4. The ambient wind speed is added to the flight speed to give the total relative wind 

speed. THERMOD uses this relative wind speed at selected times to determine convective 

coefficient, which is a required input in the cooling process. 

1.3 PROJECT OBJECTIVES. 

The objectives of the project were to 

• Debug, identify, and correct the existing errors in THERMOD. 

• develop and validate an alternate transient solution process based on the implicit 

backward finite difference method (IBFDM). The existing transient model in 

THERMOD is based on the explicit forward finite difference method (EFFDM), which is 

not unconditionally stable. The solution to some problems may not converge. An 

IBFDM, however, guarantees convergence by virtue of it being unconditionally stable 

across all time and spatial domain. 

• undertake numerical validation of THERMOD, with respect to the finite element 

methods (FEM). The three modes of heat transfer, i.e., conduction, convection, and 

radiation, were investigated. In addition, both the steady-state analysis and the transient 

analysis were considered. 

• solve a sample aircraft problem. The problem entailed developing input data sets and 

determining the MOL temperature of a typical small aircraft. 

1-7/1-8
2. DEBUGGING AND PROGRAM CORRECTIONS.

Several bugs were detected in the older THERMOD program [1]. This section discusses these 

bugs and presents remedial actions that were taken to correct them. 

2.1 BUG 1.

BUG 1 caused occasional program nonconvergence. The problem was traced to subroutine 

updradio. This subroutine updates radiosity at each time step of the transient analysis. 

Radiosity is defined as the total energy leaving a surface, which includes both infrared reflection 

and emission. There are a total of 14 radiosity functions in THERMOD. Two of these radiosity 

functions were found to be in error. The infrared emission terms in these functions were 

inadvertently left out. The subroutine is called from either subroutine trnsien1 or trnsien2, 

depending on whether the EFFDM or IBFDM is selected. Subroutine updradio in turn calls 

subroutines ludcmp and lubksb. 

The two radiosities affected were J12 and J14. In both functions, the infrared emission effect of 

the tarmac were inadvertently left out, which are shown below: 

• function J12 : sig*t(51)**4/((1-emista)/emista) 

• function J14 : sig*t(53)**4/((1-emista)/emista) 

These errors have been fixed by including the terms in the appropriate locations. Simulation 

runs of sample problems were found to converge.

2.2 BUG 2.

BUG 2 occasionally produced negative temperatures in some of the 53 temperatures that defined 

THERMOD, which is not necessarily an error. In the solution process, the nonlinear solution 

routine determines the temperature values that satisfy user input tolerances for temperatures or 

thermal functions. As long as one of the tolerances are satisfied, the solution is said to have 

converged, regardless of whether there are any negative temperatures found as part of the 

solution. Mathematically, the solution is correct. 

The variable-based convergence criterion must satisfy the following condition: 

 ( )

tolx

abs

j ≤

∑ (2-1)

X

X X

j

j

j j

i

i i −

∑

=

=

+

+

67

1

67

1

1

1

2-1
where: 

j ij

j j

X X

X X i

The previous value of the unknownvariable

The current value of the unknownvariable 1

=

= +

The function-based convergence criterion must satisfy the following condition: 

( )

tolf

fn

abs fn fn

j

j

j

j j

i

i i

≤

−

∑

∑

=

=

+

+

67

1

67

1

1

1 (2-2)

where: 

j j

j j

fn fn

fn fn

i

i

The previous value of thefunction

The current value of the function 1

=

= +

In the older version of THERMOD, when either of these two criteria were satisfied, the solution 

was completed. Because of this approach, both the tolerances were not satisfied at the same

time. In the current version of THERMOD, both criteria must be satisfied simultaneously before 

the program exits. This is enforced through the statement 

“if(errf.le.tolf.and.errx.le.tolx) goto 20” 

This dual enforcement removed the problem of negative temperatures for sample problems 

considered. 

2.3 BUG 3.

BUG 3 prevented solution convergence for certain laminate thickness or when too small a time 

step was used. This problem occurs only in the transient phase of THERMOD. This problem

was caused by the EFFDM. This method is not unconditionally stable and is the only method 

found in the original THERMOD program. An alternate finite difference method IBFDM, was 

developed to solve this problem. Simulation studies demonstrated that IBFDM has effectively 

eliminated this problem. 

2-2
The IBFDM is unconditionally stable over all spatial and time domain [6]. This stability is 

attributed to the fact that, unlike EFFDM, IBFDM considers the transient problem as a system of 

equations to be solved in a simultaneous manner. This imparts stability to the solution process 

and allows the use of less refined spatial discretization and larger time steps. The advantages of 

IBFDM over EFFDM come, however, at the expense of more complex programming and 

bookkeeping procedures. In THERMOD, for example, the development of the IBFDM 

necessitated the rearrangement of the transient equations, their assembly into a set of

simultaneous equations, their nonlinear solution using the Newton-Raphson iteration technique 

[5], and the retrieval of the solved transient temperatures. These solution steps require a 

substantial increase in computing time. However, this increase in computing time may be offset 

by the use of less refined spatial discretization and larger time steps, exploiting the inherent 

stability of IBFDM. 

The following section discusses the development of the algorithm for IBFDM followed by an 

example problem substantiating IBFDM with respect to the existing EFFDM. In addition, an 

example problem will be presented that will illustrate the inherent stability of IBFDM over 

EFFDM. 

2.4 ALGORITHM DEVELOPMENT FOR IBFDM. 

IBFDM in THERMOD can be conceptually viewed as a procedure that recognizes the fact that 

the temperature of a point in space at a certain instance in time is dependent on the temperatures 

of all other points that are directly associated with that point. The temperature of each of these 

other points in turn has its own dependence on temperatures and other points associated with it, 

and so on. One then notices a chain-like link being formed among all transient temperatures of a 

system. A unique solution for this link exists for every time step, and only a system of 

simultaneous equations will provide this solution. THERMOD sets up these equations, 

assembles them, solves them, and retrieves the transient temperatures for each time step 

requested. 

The governing equation of the transient heat transfer process is based on the formula 

E E E (2-3) in out & − & = ∆ &

Where = rate of incoming energy; = rate of outgoing energy; and Ein

& Eout& ∆E& = rate of energy 

gain or energy lost. A positive ∆E& indicates energy is gained, while a negative∆E& indicates 

energy is lost. A zero ∆E& implies a steady-state system where there is no gain or lost of energy 

in the system. Note that equation 2-3 does not include the effect of energy generation, which is 

certainly absent in the case of an aircraft parked in the open. 

The implementation of the IBFDM transient programming rests on setting up equation 2-3 for 

each one of the control volumes defined in the system and solving the resulting set of 

simultaneous equations. 

2-3
The procedure can be demonstrated by developing the equation for control volume 3 shown in 

figure 2-1. This control volume may be viewed to be within the wing of an aircraft where the 

surface effects are absent. The only mode of heat transfer in this region is conduction. Only 

through thickness, one-dimensional conduction is assumed. Along-wing span and along-wing 

width conduction were neglected in THERMOD to produce conservative temperatures. 

T3 

T4 

Layer 2: k2, ρ2, c2 

d3 Control volume 3 

d4 Control volume 4 

y2

d2 Control volume 2 

T1 

T2 Layer 1: k1, ρ1, c1 

d1 

y1 

Surface 

Control volume 1 

Area, A 

FIGURE 2-1. CONTROL VOLUMES FOR DEVELOPING THERMAL EQUATIONS 

In developing the equations, it is assumed that the reader has a basic understanding of the 

principles of heat transfer. For an overview of the mathematical derivation of the three modes of 

heat transfer, i.e., conduction, convection, and radiation, the reader is referred to reference 1. 

2-4
From figure 2-1, for control volume 3 with cross-sectional area A 

A

y

T T

E k

i i

in a

1

1

3

1

2

+ + − & = where ( ) 2

1

1 d2 d3 y = +

1

2

3

1

2

2

2 3 ( ) 2 2

k k

d d

ka d d

+

+ =

 A

y

T i i

out b

2

1

4

1

3

+ +

& T

E k − = where ( ) 2

1

2 3 4 y = d + d

2

2

4

1

2

3

3 4 ( ) 2 2

k k

d d

b d k d

+

+ =

t

T T

E mc

i i

∆

− ∆ =

+

3

1

3

1 &

where: m = mass of the control volume = ρ1 A d3 

ρ1 = density of the control volume 

 c1 = specific heat capacity of the control volume 

∆t = time step 

In the above equations, i+1 T represents new temperature, which is an unknown quantity, and i T

represents the previous temperature, which is a known quantity. 

Substituting the above expressions for , , and Ein

& Eout& ∆E& into equation 2-3 gives 

t

T T A Ad c

y

T T A k

y

T T k

i i i i

b

i i

a ∆

− = − − − + + + + +

3

1

3

1 3 1

2

1

4

1

3

1

1

3

1

2 ρ

Eliminating A, and rearranging, one obtains: 

(2-4) i i i ia T a T a T a4T3

1

3 4

1

2 3

1

1 2 + + = + + +

where: 

1

1

y

k

a a =

















∆ = − − − t

d c

y

k

y

k

a a b 1 3 1

1 2

2

ρ

2-5
2

3

y

k

a b =

t

d c

a

∆ = − 1 3 1

4

ρ

Note that in equation 2-4, a1, a2, a3, a4, and i T are known quantities. The unknowns are T , 

, and T . It is observed that equation 2-4 is implicit, in that there are more than one 

unknown. To solve this equation, one needs to create another equation representing the control 

volume adjacent to the current one. This equation would contain other unknown temperatures. 

Thus, one needs to develop a series of equations representing all control volumes that were used 

to define the thermal system, before a unique solution for the set of new temperatures, 

1

2

i+

1

3

i+ T 1

4

i+

i+1 T could 

be found. 

The difference between IBFDM and EFFDM can be observed by studying the derivation of 

equation 2-4. Had and been defined in terms of the previous temperatures, the only 

unknown in equation 2-4 would have been T , and this would have rendered the equation 

explicit. Because and were expressed in terms of the new temperatures, which are 

unknown quantities, equation 2-4 was rendered implicit. 

Ein

&

Ein

&

Eout

&

Eout

&

1

3

i+

In THERMOD, a total of 39 control volumes were created, resulting in 117 unknown 

temperatures, i+1 T . Note that each control volume was designed to have three temperatures, as 

illustrated in the derivation of the equations above. Twenty-seven of the control volumes were 

within the structure where the only mode of heat transfer was by means of conduction, as was 

illustrated in the above sample equations. The remaining twelve control volumes represented 

surface elements where convection and radiation effects, in addition to conduction effects, were 

considered. The 117 equations were assembled and subsequently solved using the nonlinear 

Newton-Raphson iteration procedure. A description of this procedure is found in references 3 

and 4. Note that the nonlinear solution procedure is required for two reasons. First, infrared 

radiation, a capability modeled in THERMOD, follows the Stefan-Boltzman Law, which is 

based on a nonlinear formulation, where radiation R = σ T4

. In this formulation, the 

temperature, T, is an unknown quantity. Second, the convection coefficient within the cabin was 

modeled to be dependent on temperatures, which again, are an unknown quantity. 

The transient solution is initiated by treating the temperatures obtained from the steady-state 

analysis as the initial set of previous temperatures. The solved new temperatures are then treated 

as the previous temperatures in the next time step. The analysis is repeated until a solution 

satisfying a required set of tolerance in temperatures and functions is obtained. Once solved, the 

117 temperatures are retrieved, from which only the 53 temperatures defining THERMOD are 

reported. (See figures 1-1 and 1-2 to locate these temperatures.) 

2-6
As indicated, THERMOD is a thermal system with 67 unknowns, where 53 of the unknowns are 

temperatures and 14 are radiosity. Radiosity is defined as the radiative flux that leaves a surface, 

and it includes both the reflected irradiation as well as emission [6]. Because THERMOD 

depends on temperatures, it follows that radiosity will change with time in a transient 

environment. This change in radiosity is implemented in THERMOD by updating its value 

using the new temperatures determined at the end of each time step. 

The various subroutines, their definitions, and the program flow chart addressing IBFDM are 

contained in appendix A. 

The development of IBFDM required extensive rearrangement of the transient equations, 

substantial bookkeeping, a nonlinear solution of the resulting simultaneous equations, and data 

retrieval. These extra efforts were expected to increase the computing resources. This was 

indeed the case. The amount of computing time spent was vastly different for the previous two 

problems. The EFFDM solution process took less than 1 minute, while the IBFDM took about 

90 minutes. 

As was previously noted, the primary purpose of IBFDM was to overcome probable instability 

problems that may arise using EFFDM. The substantial increase in the computing effort of 

IBFDM, however, seems to be a worthwhile price to pay, given that a potential stability problem

was solved and satisfactory results are expected. It is also noted that because IBFDM is 

inherently stable regardless of the time step duration, the computing effort may be substantially 

reduced by increasing the time step until accuracy of the solution is not affected. 

2-7/2-8
3. NUMERICAL VALIDATION OF THERMOD.

This section presents the numerical validation of THERMOD with respect to FEMs. A total of

eight problems were formulated and solved using the THERMOD program and then the FEM 

programs. The results from both methods of analysis were then compared. 

It was beyond the scope of this project to develop a complete FEM thermal model of an aircraft. 

The FEM sample problems were, thus, kept as simple as possible and were yet able to capture 

the three fundamental modes of heat transfer mechanisms, i.e., conduction, convection, and 

radiation. The problems, where possible, addressed each transport mechanism individually first, 

then progressed into more complex thermal systems involving a combination of these 

mechanisms. 

In developing sample input files for THERMOD, all 15 sets of data defining the entire thermal 

problem of an aircraft must be generated and used in all sample problems. This is because, 

unlike the FEM methods, THERMOD was not developed as a general-purpose thermal program. 

It was designed to solve a very specific thermal problem associated with an aircraft subjected to 

the thermal environment. While the solution process in itself is rather complex, the inputs to the 

program were designed to be simple and straightforward. Because of this constraint, aircraft 

problems in its entirety will be solved in THERMOD during this validation exercise. The FEM 

methods, however, are not subject to this constraint. 

Of the eight validation problems solved, six involved steady-state analysis. The other two 

involved both steady-state and transient analyses. The FEM models of the six steady-state 

problems were analyzed using UAI/NASTRAN software. The FEM models of the two 

combined steady-state and transient problems were analyzed using MSC/NASTRAN software. 

The primary reason for using two different FEM packages was that the thermal module of the 

UAI/NASTRAN package was found to be deficient in the transient analysis. It lacked a timedependent convective coefficient, which is needed in analyzing transient cooling, where air is 

forced over the aircraft while maneuvering according to various flight profile information. The 

MSC/NASTRAN thermal module was found to have this capability. 

To have a better understanding of the sample problems presented in this section, it is helpful to 

be familiar with the THERMOD User’s Manual [2], which gives a complete description of input 

data sets, variable definitions, formats, and other general information. Only information directly 

relevant to the validation problems will be discussed in detail in this section, while the 

discussion of other data will be kept to a minimum. 

It is assumed that the reader have a working knowledge of the FEM methods, sufficient to 

decipher NASTRAN input data decks. The relevant FEM information pertaining to NASTRAN 

can be found in the following documents: UAI/NASTRAN User’s Reference Manual [7], 

UAI/NASTRAN User’s Guide [8], MSC/NASTRAN Quick Reference Guide [9], 

MSC/NASTRAN User’s Guide [10], MSC/NASTRAN Command Reference Guide [11], and 

MSC/NASTRAN Thermal Analysis User’s Guide [12]. 

3-1
3.1 SAMPLE PROBLEM 1.

The purpose of sample problem 1 was to illustrate the conduction mechanism of the heat transfer 

process. Table 3-1 shows the THERMOD input file, input.dat, of sample problem 1. As 

previously noted, a total of 15 sets of data must always be used to complete a successful run of 

THERMOD. These sets of data define a thermal problem of a complete aircraft. 

TABLE 3-1. THERMOD INPUT FILE (input.dat) OF SAMPLE PROBLEM 1 

198e-6,0.704 

1, 0.020,125,0.300,0.30 

2, 0.375,4.4,0.021,0.24 

3, 0.020,125,0.300,0.30 

4, 0.350,125,0.300,0.30 

5, 10.70,0.067,0.016,0.24 

6, 0.350,125,0.300,0.30 

7, 0.020,125,0.300,0.30 

8, 0.375,4.4,0.021,0.24 

9, 0.020,125,0.300,0.30 

10,0.030,125,0.300,0.30 

11,0.375,4.4,0.021,0.24 

12,0.020,125,0.300,0.30 

13,0.250,10.,0.020,0.24 

14,48.00,0.067,0.016,0.24 

15,0.020,125,0.300,0.30 

16,0.375,4.4,0.021,0.24 

17,0.020,125,0.300,0.30 

18,0.250,10.,0.020,0.24 

19,30.00,0.067,0.016,0.24 

20,5.000,10.,0.020,0.24 

21,0.020,125,0.300,0.30 

22,0.375,4.4,0.021,0.24 

23,0.020,125,0.300,0.30 

24,2.500,0.067,0.016,0.24 

25,0.020,125,0.300,0.30 

26,0.375,4.4,0.021,0.24 

27,0.020,125,0.300,0.30 

28,0.250,130,0.800,0.20 

3.5,34.0,2.5,5.5,22.0,11.0,3.0,4.0,30,24,28 

0.70,0.9,0.9,2.0,150,0.80,0.22 

0.3,0.9,0.3,0.9,0.3,0.9,0.1,0.9 

3,100 

0.9,0.9,0.9,0.9 

0.95,0.47,0.3,0.8,0.95,0.47,0.95,0.47 

9 

0,10,10,0,0,75,110,110,190,190 

5,120,5,60,20,5,120,60,60 

3-2
TABLE 3-1. THERMOD INPUT FILE (input.dat) OF SAMPLE PROBLEM 1 (Continued) 

7 

60 570.67 459.67 14 330 10 10 0.36 

75 573.67 459.67 14 355 10 10 0.31 

90 578.67 459.67 14 355 10 12 0.33 

75 581.67 459.67 14 330 10 10 0.31 

60 582.67 459.67 14 291 10 10 0.36 

45 583.67 459.67 14 231 10 10 0.44 

30 582.67 459.67 14 160 10 10 0.57 

2,100,180,1 

10,1.0,1.0 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10 

For comparison purposes with the FEM method, the focus was on the right wing of the aircraft. 

(The right wing is indicated by RHS in figure 1-1.) As illustrated in the input.dat file of 

table 3-1, the right wing was modeled as follows: 

• Layer 1: 0.02-in.-thick top outer fiber glass skin 

• Layer 2: 0.375-in.-thick top foam core 

• Layer 3: 0.02-in.-thick top inner fiber glass skin 

• Layer 4: 0.35-in.-thick top fiber glass spar cap 

• Layer 5: 10.70-in.-deep air space between the top and bottom spar caps 

• Layer 6: 0.35-in.-thick bottom fiber glass spar cap 

• Layer 7: 0.02-in.-thick bottom inner fiber glass skin 

• Layer 8: 0.375-in.-thick bottom foam core 

• Layer 9: 0.02-in.-thick bottom outer fiber glass skin 

The summary.dat file of table 3-2 summarizes the results of the THERMOD analysis. As 

observed in table 3-2, a total of seven time periods were involved in the THERMOD analysis. 

Steady-state temperature results obtained for time period 3 was chosen for comparison purposes 

with the FEM method. A total of 10 temperature locations were identified on the right wing, as 

illustrated in figure 1-1. These 10 temperatures from the top to the bottom are T1, T2.....T10. 

Their values were extracted from table 3-2 and summarized in table 3-3. The results indicated 

that the top surface of the wing, because of its exposure to the sun, was hotter than the lower 

surface, which was more sheltered. There was a significant difference in temperature between 

the upper cap (T5) and the lower cap (T6). This was attributed to the 10.70 in. of air (which has 

low thermal conductivity) that separates them. 

3-3
TABLE 3-2. THERMOD OUTPUT FILE (summary.dat) OF SAMPLE PROBLEM 1 

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ 

THERMOD: A THERMAL PROGRAM FOR AIRCRAFTS 

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ 

 

 

 

 SUMMARY OF STEADY-STATE AND TRANSIENT ANALYSIS 

 

 

 

 (Temperatures are shown for all time periods) 

 

 

 TEMPERATURES AT THE END OF STEADY STATE ANALYSIS 

 

 TEMPERATURES IN DEGREES FAHRENHEIT 

 1 2 3 4 5 6 7 

 

 1 124.0 132.5 136.1 134.0 126.4 115.2 102.8 

 2 124.0 132.5 136.1 134.0 126.4 115.2 102.8 

 3 123.7 132.2 135.8 133.7 126.2 115.3 103.0 

 4 123.7 132.2 135.8 133.7 126.2 115.3 103.0 

 5 123.7 132.2 135.8 133.7 126.2 115.3 103.0 

 6 114.4 120.2 124.7 124.6 121.1 116.3 110.5 

 7 114.3 120.2 124.7 124.6 121.1 116.4 110.5 

 8 114.3 120.2 124.7 124.6 121.1 116.4 110.5 

 9 114.1 119.9 124.4 124.3 121.0 116.4 110.7 

 10 114.1 119.9 124.4 124.3 121.0 116.4 110.7 

 11 122.1 131.4 135.9 132.9 124.8 113.5 101.4 

 12 122.1 131.4 135.9 132.9 124.8 113.5 101.4 

 13 121.9 131.1 135.6 132.7 124.7 113.6 101.6 

 14 121.9 131.1 135.6 132.7 124.7 113.6 101.6 

 15 121.9 131.1 135.6 132.7 124.7 113.6 101.6 

 16 114.3 120.2 124.7 124.6 121.1 116.3 110.5 

 17 114.3 120.2 124.7 124.6 121.1 116.3 110.5 

 18 114.3 120.2 124.7 124.6 121.1 116.3 110.5 

 19 114.1 119.9 124.4 124.3 121.0 116.4 110.7 

 20 114.1 119.9 124.4 124.3 121.0 116.4 110.7 

 21 205.8 225.9 231.9 222.2 200.7 169.4 134.2 

 22 205.8 225.8 231.8 222.1 200.7 169.5 134.2 

 23 207.4 216.5 217.2 213.6 202.4 180.2 143.1 

 24 207.4 216.4 217.2 213.6 202.4 180.2 143.1 

 25 208.4 209.9 206.9 207.6 203.6 187.7 149.4 

 26 198.9 198.2 195.5 196.8 195.4 183.4 146.2 

 27 150.5 153.0 154.4 155.5 153.6 146.5 127.6 

 28 150.4 153.0 154.4 155.4 153.6 146.5 127.6 

3-4
TABLE 3-2. THERMOD OUTPUT FILE (summary.dat) OF SAMPLE 

PROBLEM 1 (Continued) 

 29 147.0 149.8 151.5 152.5 150.6 143.9 126.3 

 30 147.0 149.8 151.5 152.5 150.6 143.9 126.3 

 31 131.8 135.6 138.7 139.6 137.5 132.4 120.5 

 32 131.8 135.6 138.7 139.6 137.5 132.4 120.5 

 33 131.8 135.6 138.7 139.6 137.5 132.4 120.5 

 34 131.8 135.6 138.7 139.6 137.5 132.4 120.5 

 35 116.7 121.5 125.8 126.6 124.5 120.8 114.7 

 36 116.7 121.5 125.8 126.6 124.4 120.8 114.7 

 37 113.2 118.3 122.9 123.7 121.5 118.2 113.3 

 38 113.2 118.3 122.9 123.7 121.4 118.2 113.3 

 39 199.5 179.7 148.8 180.0 196.6 194.7 173.9 

 40 199.5 179.7 148.9 180.1 196.6 194.6 173.9 

 41 197.6 186.6 168.7 186.2 194.5 188.4 162.3 

 42 197.6 186.6 168.8 186.2 194.5 188.4 162.3 

 43 196.2 191.4 182.7 190.5 193.1 184.1 154.2 

 44 191.8 188.5 182.6 187.8 189.2 180.4 151.1 

 45 169.9 170.8 168.7 171.5 170.2 161.4 138.9 

 46 169.8 170.7 168.7 171.5 170.0 161.3 138.8 

 47 138.6 145.3 148.8 148.2 142.9 134.1 121.3 

 48 138.4 145.2 148.7 148.0 142.7 133.9 121.2 

 49 145.6 148.6 149.7 150.4 147.7 139.7 123.0 

 50 147.6 150.5 151.3 152.1 149.5 141.3 123.9 

 51 113.6 120.8 125.1 123.7 118.1 110.0 100.6 

 52 108.2 115.1 119.2 118.0 112.7 105.1 96.2 

 53 113.6 120.8 125.1 123.7 118.1 110.0 100.6 

 

Maximum Structural Temperature for Period 1 = 207.37 Occurring 

at Location 24 

Maximum Structural Temperature for Period 2 = 225.86 Occurring 

at Location 21 

Maximum Structural Temperature for Period 3 = 231.88 Occurring 

at Location 21 

Maximum Structural Temperature for Period 4 = 222.15 Occurring 

at Location 21 

Maximum Structural Temperature for Period 5 = 202.43 Occurring 

at Location 24 

Maximum Structural Temperature for Period 6 = 194.68 Occurring 

at Location 39 

Maximum Structural Temperature for Period 7 = 173.92 Occurring 

at Location 39 

 

Maximum Structural Temperature Over All 7 Periods =231.88 

Occurring at Location 21 at Period 3 

 

 

3-5
TABLE 3-2. THERMOD OUTPUT FILE (summary.dat) OF SAMPLE 

PROBLEM 1 (Continued) 

 TEMPERATURES AT THE END OF TRANSIENT ANALYSIS 

 

 TEMPERATURES IN DEGREES FAHRENHEIT 

 1 2 3 4 5 6 7 

 

 1 112.9 116.8 121.7 123.8 123.4 122.5 119.8 

 2 113.2 117.3 122.2 124.1 123.5 122.3 119.2 

 3 122.9 131.1 134.8 133.0 126.0 115.7 104.1 

 4 123.2 131.5 135.2 133.3 126.1 115.6 103.7 

 5 123.2 131.6 135.2 133.3 126.0 115.4 103.5 

 6 114.6 120.5 125.0 124.9 121.3 116.5 110.6 

 7 114.3 120.1 124.6 124.6 121.3 116.7 111.0 

 8 114.3 120.0 124.5 124.6 121.3 116.9 111.3 

 9 112.7 116.6 121.6 123.8 123.6 123.1 120.9 

 10 112.7 116.4 121.4 123.7 123.7 123.3 121.2 

 11 112.7 116.8 121.7 123.8 123.3 122.4 119.7 

 12 113.0 117.2 122.1 124.0 123.3 122.1 119.1 

 13 121.3 130.1 134.6 132.1 124.6 114.2 102.9 

 14 121.5 130.5 135.0 132.3 124.6 114.0 102.4 

 15 121.5 130.6 135.0 132.3 124.6 113.8 102.2 

 16 114.5 120.4 124.9 124.8 121.2 116.4 110.5 

 17 114.3 120.1 124.6 124.6 121.2 116.6 111.0 

 18 114.2 120.0 124.5 124.6 121.3 116.8 111.2 

 19 112.7 116.6 121.6 123.8 123.6 123.1 120.9 

 20 112.7 116.4 121.4 123.7 123.7 123.3 121.2 

 21 125.4 131.3 136.7 137.5 134.8 130.9 124.7 

 22 127.4 133.3 138.7 139.3 136.4 132.3 125.2 

 23 181.2 187.5 192.2 187.4 180.5 171.4 140.3 

 24 183.1 189.2 193.8 188.9 182.0 172.9 140.9 

 25 200.7 202.6 205.1 200.1 195.7 189.2 147.8 

 26 185.7 185.5 187.8 185.4 184.4 181.0 145.5 

 27 150.7 153.2 154.6 155.7 153.8 146.7 127.7 

 28 149.5 152.1 153.7 154.7 152.8 145.8 127.3 

 29 146.5 149.3 151.2 152.1 150.2 143.7 126.3 

 30 145.7 148.6 150.4 151.4 149.5 143.1 126.0 

 31 131.5 135.1 138.2 139.3 137.7 133.2 121.9 

 32 131.5 135.1 138.2 139.3 137.7 133.2 121.9 

 33 131.5 135.1 138.2 139.3 137.7 133.2 121.9 

 34 131.5 135.1 138.2 139.3 137.7 133.2 121.9 

 35 117.2 121.6 125.9 127.3 125.8 123.2 117.7 

 36 116.4 120.7 125.2 126.5 125.2 122.7 117.5 

 37 112.1 115.7 120.6 123.1 123.4 123.4 121.5 

 38 112.0 115.5 120.5 123.0 123.3 123.4 121.6 

 39 121.2 122.3 123.8 129.2 131.3 131.5 127.8 

 40 123.2 124.0 125.1 130.7 133.0 133.1 128.7 

3-6
TABLE 3-2. THERMOD OUTPUT FILE (summary.dat) OF SAMPLE 

PROBLEM 1 (Continued) 

 41 172.1 166.9 160.2 168.7 173.2 172.1 150.2 

 42 173.6 168.3 161.6 170.0 174.5 173.3 150.8 

 43 183.1 179.5 176.6 179.9 182.2 181.5 152.3 

 44 179.2 177.0 176.6 177.6 178.7 178.0 149.3 

 45 162.3 161.7 161.5 163.9 164.5 162.3 141.2 

 46 160.8 160.3 160.1 162.6 163.2 161.0 140.6 

 47 121.5 123.0 125.1 129.8 131.5 131.4 127.2 

 48 120.0 121.6 123.8 128.6 130.3 130.3 126.7 

 49 123.6 126.8 131.0 132.5 132.0 130.4 123.1 

 50 127.8 130.8 134.8 136.0 135.3 133.2 123.9 

 51 169.1 187.2 193.0 184.0 164.8 137.5 109.0 

 52 157.9 174.6 180.0 172.0 154.5 129.7 103.8 

 53 169.1 187.2 193.0 184.0 164.8 137.5 109.0 

 

Maximum Structural Temperature for Period 1 = 183.07 Occurring 

at Location 24 

Maximum Structural Temperature for Period 2 = 189.17 Occurring 

at Location 24 

Maximum Structural Temperature for Period 3 = 193.83 Occurring 

at Location 24 

Maximum Structural Temperature for Period 4 = 188.86 Occurring 

at Location 24 

Maximum Structural Temperature for Period 5 = 181.98 Occurring 

at Location 24 

Maximum Structural Temperature for Period 6 = 173.32 Occurring 

at Location 42 

Maximum Structural Temperature for Period 7 = 150.75 Occurring 

at Location 42 

 

Maximum Structural Temperature Over All 7 Periods =193.83 

Occurring at Location 24 at Period 3 

TABLE 3-3. THERMOD-SIMULATED TEMPERATURES (°F) OF THE RIGHT WING OF 

SAMPLE PROBLEM 1 

T1 T2 T3 T4 T5 T6 T7 T8 T9 T10

136.1 136.1 135.8 135.8 135.8 124.7 124.7 124.7 124.4 124.4

An FEM model of the wing was developed based on the information used in the THERMOD 

model. One-half of the wing was modeled as a 17-ft-long structure with layer properties as 

indicated above. A total of 153 CQUAD4 elements, each 0.001 ft thick, was employed, resulting 

in a total of 324 grid points (nodes). Material properties, representing conductivity, were 

assigned using MAT4 cards. PSHELL cards were used to assign the thickness and material 

properties for each of the element. 

3-7
Boundary conditions were imposed along the outer nodes of the model. Recall that the objective 

of this sample problem was to validate the conductive aspect of THERMOD. The FEM 

boundary conditions were then set such that all top surface nodes assumed the value of T1 

(136.1°F), and all bottom surface nodes assumed the value of T10 (124.4°F), whose values were 

obtained from the earlier THERMOD analysis. The outer left and right most edges of the wing 

were assigned boundary temperatures of 124.4°F, thus completely encapsulating the FEM 

model. The temperature profile across the wing depth is now entirely due to conduction. Table 

3-4 illustrates the FEM input file. 

TABLE 3-4. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE PROBLEM 1 

ID D:\Natha,MSC/N 

APP HEAT 

SOL 1 

TIME 10000 

CEND 

 TITLE = CONDUCTION IN RECTANGULAR ADIABATIC PLATE (t=0.001ft) 

 SUBTITLE = 10.7 in gap filled by air; k=0.016BTU/hr/ft/degF 

 SPC = 1 

 THERMAL= ALL 

 FLUX = ALL 

 SPCF = ALL 

BEGIN BULK 

$ *************************************************************************** 

$ Written by : MSC/NASTRAN for Windows 

$ Version : 6.00 

$ Translator : UAI/NASTRAN 

$ From Model : D:\Thermod_Validation\FEM\conduction_only.MOD 

$ Date : Mon Mar 06 11:27:09 2000 

$ *************************************************************************** 

$ 

PARAM,K6ROT,100. 

PARAM,MAXRATIO,1.E+8 

CORD2C 1 0 0. 0. 0. 0. 0. 1.+MSC/NC1

+MSC/NC1 1. 0. 1. 

CORD2S 2 0 0. 0. 0. 0. 0. 1.+MSC/NC2

+MSC/NC2 1. 0. 1. 

$ MSC/NASTRAN for Windows Constraint Set 1 : steady-state_bound-temp 

SPC 1 1 1 124.4 

SPC 1 20 1 124.4 

SPC 1 21 1 124.4 

SPC 1 22 1 124.4 

SPC 1 23 1 124.4 

SPC 1 24 1 124.4 

SPC 1 25 1 124.4 

SPC 1 26 1 124.4 

SPC 1 27 1 124.4 

SPC 1 28 1 124.4 

SPC 1 29 1 124.4 

SPC 1 30 1 124.4 

SPC 1 31 1 124.4 

SPC 1 32 1 124.4 

SPC 1 33 1 124.4 

SPC 1 34 1 124.4 

SPC 1 35 1 124.4 

SPC 1 36 1 124.4 

SPC 1 37 1 124.4 

3-8
TABLE 3-4. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 1 (Continued) 

SPC 1 55 1 124.4 

SPC 1 56 1 124.4 

SPC 1 73 1 124.4 

SPC 1 109 1 124.4 

SPC 1 128 1 124.4 

SPC 1 145 1 124.4 

SPC 1 146 1 124.4 

SPC 1 164 1 124.4 

SPC 1 181 1 124.4 

SPC 1 182 1 124.4 

SPC 1 199 1 124.4 

SPC 1 200 1 124.4 

SPC 1 201 1 136.1 

SPC 1 202 1 136.1 

SPC 1 203 1 136.1 

SPC 1 204 1 136.1 

SPC 1 205 1 136.1 

SPC 1 206 1 136.1 

SPC 1 207 1 136.1 

SPC 1 208 1 136.1 

SPC 1 209 1 136.1 

SPC 1 210 1 136.1 

SPC 1 211 1 136.1 

SPC 1 212 1 136.1 

SPC 1 213 1 136.1 

SPC 1 214 1 136.1 

SPC 1 215 1 136.1 

SPC 1 216 1 136.1 

SPC 1 235 1 124.4 

SPC 1 253 1 124.4 

SPC 1 254 1 124.4 

SPC 1 307 1 124.4 

SPC 1 308 1 124.4 

$ MSC/NASTRAN for Windows Property 1 : 0.02 in glass_outer_layer 

PSHELL 1 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 2 : 3/8 in foam 

PSHELL 2 2 0.001 2 2 0. 

$ MSC/NASTRAN for Windows Property 3 : 0.02 in inner glass 

PSHELL 3 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 4 : 0.350 in cap 

PSHELL 4 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 5 : 10.70 in air space 

PSHELL 5 3 0.001 3 3 0. 

$ MSC/NASTRAN for Windows Property 6 : 0.350 in cap 

PSHELL 6 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 7 : 0.02 in inner glass 

PSHELL 7 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 8 : 3/8 in foam 

PSHELL 8 2 0.001 2 2 0. 

$ MSC/NASTRAN for Windows Property 9 : 0.02 in glass_outer_layer 

PSHELL 9 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Material 1 : glass 

MAT4 1 0.3 1. 

$ MSC/NASTRAN for Windows Material 2 : foam 

MAT4 2 0.021 1. 

$ MSC/NASTRAN for Windows Material 3 : air 

MAT4 3 0.016 1. 

3-9
TABLE 3-4. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 1 (Continued) 

GRID 1 0 0. 0. 0. 0 

GRID 3 0 1.1.667E-3 0. 0 

GRID 5 0 3.1.667E-3 0. 0 

GRID 7 0 5.1.667E-3 0. 0 

GRID 9 0 7.1.667E-3 0. 0 

GRID 10 0 8.1.667E-3 0. 0 

GRID 11 0 9.1.667E-3 0. 0 

GRID 12 0 10.1.667E-3 0. 0 

GRID 14 0 12.1.667E-3 0. 0 

GRID 15 0 13.1.667E-3 0. 0 

GRID 18 0 16.1.667E-3 0. 0 

GRID 20 0 17. 0. 0. 0 

GRID 21 0 16. 0. 0. 0 

GRID 22 0 15. 0. 0. 0 

GRID 23 0 14. 0. 0. 0 

GRID 24 0 13. 0. 0. 0 

GRID 25 0 12. 0. 0. 0 

GRID 26 0 11. 0. 0. 0 

GRID 27 0 10. 0. 0. 0 

GRID 28 0 9. 0. 0. 0 

GRID 29 0 8. 0. 0. 0 

GRID 30 0 7. 0. 0. 0 

GRID 31 0 6. 0. 0. 0 

GRID 32 0 5. 0. 0. 0 

GRID 33 0 4. 0. 0. 0 

GRID 34 0 3. 0. 0. 0 

GRID 35 0 2. 0. 0. 0 

GRID 36 0 1. 0. 0. 0 

GRID 37 0 0.1.667E-3 0. 0 

GRID 41 0 3.0.032917 0. 0 

GRID 44 0 6.0.032917 0. 0 

GRID 45 0 7.0.032917 0. 0 

GRID 47 0 9.0.032917 0. 0 

GRID 48 0 10.0.032917 0. 0 

GRID 50 0 12.0.032917 0. 0 

GRID 51 0 13.0.032917 0. 0 

GRID 52 0 14.0.032917 0. 0 

GRID 53 0 15.0.032917 0. 0 

GRID 54 0 16.0.032917 0. 0 

GRID 55 0 17.0.032917 0. 0 

GRID 56 0 17.1.667E-3 0. 0 

GRID 58 0 15.1.667E-3 0. 0 

GRID 59 0 14.1.667E-3 0. 0 

GRID 62 0 11.1.667E-3 0. 0 

GRID 67 0 6.1.667E-3 0. 0 

GRID 69 0 4.1.667E-3 0. 0 

GRID 71 0 2.1.667E-3 0. 0 

GRID 73 0 0.0.032917 0. 0 

GRID 75 0 1.0.034583 0. 0 

GRID 76 0 2.0.034583 0. 0 

GRID 77 0 3.0.034583 0. 0 

GRID 78 0 4.0.034583 0. 0 

GRID 79 0 5.0.034583 0. 0 

GRID 80 0 6.0.034583 0. 0 

GRID 83 0 9.0.034583 0. 0 

GRID 85 0 11.0.034583 0. 0 

GRID 87 0 13.0.034583 0. 0 

3-10
TABLE 3-4. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 1 (Continued) 

GRID 88 0 14.0.034583 0. 0 

GRID 98 0 11.0.032917 0. 0 

GRID 101 0 8.0.032917 0. 0 

GRID 104 0 5.0.032917 0. 0 

GRID 105 0 4.0.032917 0. 0 

GRID 107 0 2.0.032917 0. 0 

GRID 108 0 1.0.032917 0. 0 

GRID 109 0 0.0.034583 0. 0 

GRID 112 0 2. 0.06375 0. 0 

GRID 115 0 5. 0.06375 0. 0 

GRID 116 0 6. 0.06375 0. 0 

GRID 118 0 8. 0.06375 0. 0 

GRID 120 0 10. 0.06375 0. 0 

GRID 121 0 11. 0.06375 0. 0 

GRID 122 0 12. 0.06375 0. 0 

GRID 123 0 13. 0.06375 0. 0 

GRID 124 0 14. 0.06375 0. 0 

GRID 125 0 15. 0.06375 0. 0 

GRID 128 0 17.0.034583 0. 0 

GRID 129 0 16.0.034583 0. 0 

GRID 130 0 15.0.034583 0. 0 

GRID 133 0 12.0.034583 0. 0 

GRID 135 0 10.0.034583 0. 0 

GRID 137 0 8.0.034583 0. 0 

GRID 138 0 7.0.034583 0. 0 

GRID 145 0 0. 0.06375 0. 0 

GRID 146 0 0. 0.95542 0. 0 

GRID 147 0 1. 0.95542 0. 0 

GRID 150 0 4. 0.95542 0. 0 

GRID 152 0 6. 0.95542 0. 0 

GRID 160 0 14. 0.95542 0. 0 

GRID 161 0 15. 0.95542 0. 0 

GRID 164 0 17. 0.06375 0. 0 

GRID 165 0 16. 0.06375 0. 0 

GRID 172 0 9. 0.06375 0. 0 

GRID 174 0 7. 0.06375 0. 0 

GRID 177 0 4. 0.06375 0. 0 

GRID 178 0 3. 0.06375 0. 0 

GRID 180 0 1. 0.06375 0. 0 

GRID 181 0 0. 1.01917 0. 0 

GRID 182 0 0. 1.0175 0. 0 

GRID 183 0 1. 1.0175 0. 0 

GRID 184 0 2. 1.0175 0. 0 

GRID 188 0 6. 1.0175 0. 0 

GRID 193 0 11. 1.0175 0. 0 

GRID 195 0 13. 1.0175 0. 0 

GRID 196 0 14. 1.0175 0. 0 

GRID 197 0 15. 1.0175 0. 0 

GRID 198 0 16. 1.0175 0. 0 

GRID 199 0 17. 1.0175 0. 0 

GRID 200 0 17. 1.01917 0. 0 

GRID 201 0 16. 1.01917 0. 0 

GRID 202 0 15. 1.01917 0. 0 

GRID 203 0 14. 1.01917 0. 0 

GRID 204 0 13. 1.01917 0. 0 

GRID 205 0 12. 1.01917 0. 0 

GRID 206 0 11. 1.01917 0. 0 

3-11
TABLE 3-4. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 1 (Continued) 

GRID 207 0 10. 1.01917 0. 0 

GRID 208 0 9. 1.01917 0. 0 

GRID 209 0 8. 1.01917 0. 0 

GRID 210 0 7. 1.01917 0. 0 

GRID 211 0 6. 1.01917 0. 0 

GRID 212 0 5. 1.01917 0. 0 

GRID 213 0 4. 1.01917 0. 0 

GRID 214 0 3. 1.01917 0. 0 

GRID 215 0 2. 1.01917 0. 0 

GRID 216 0 1. 1.01917 0. 0 

GRID 219 0 1. 0.98625 0. 0 

GRID 220 0 2. 0.98625 0. 0 

GRID 221 0 3. 0.98625 0. 0 

GRID 222 0 4. 0.98625 0. 0 

GRID 223 0 5. 0.98625 0. 0 

GRID 226 0 8. 0.98625 0. 0 

GRID 227 0 9. 0.98625 0. 0 

GRID 228 0 10. 0.98625 0. 0 

GRID 229 0 11. 0.98625 0. 0 

GRID 231 0 13. 0.98625 0. 0 

GRID 232 0 14. 0.98625 0. 0 

GRID 235 0 17. 0.98625 0. 0 

GRID 241 0 12. 1.0175 0. 0 

GRID 243 0 10. 1.0175 0. 0 

GRID 244 0 9. 1.0175 0. 0 

GRID 245 0 8. 1.0175 0. 0 

GRID 246 0 7. 1.0175 0. 0 

GRID 248 0 5. 1.0175 0. 0 

GRID 249 0 4. 1.0175 0. 0 

GRID 250 0 3. 1.0175 0. 0 

GRID 253 0 0. 0.98625 0. 0 

GRID 254 0 0. 0.98458 0. 0 

GRID 257 0 3. 0.98458 0. 0 

GRID 258 0 4. 0.98458 0. 0 

GRID 259 0 5. 0.98458 0. 0 

GRID 260 0 6. 0.98458 0. 0 

GRID 264 0 10. 0.98458 0. 0 

GRID 267 0 13. 0.98458 0. 0 

GRID 268 0 14. 0.98458 0. 0 

GRID 269 0 15. 0.98458 0. 0 

GRID 270 0 16. 0.98458 0. 0 

GRID 273 0 16. 0.98625 0. 0 

GRID 274 0 15. 0.98625 0. 0 

GRID 277 0 12. 0.98625 0. 0 

GRID 282 0 7. 0.98625 0. 0 

GRID 283 0 6. 0.98625 0. 0 

GRID 292 0 2. 0.95542 0. 0 

GRID 293 0 3. 0.95542 0. 0 

GRID 295 0 5. 0.95542 0. 0 

GRID 297 0 7. 0.95542 0. 0 

GRID 298 0 8. 0.95542 0. 0 

GRID 299 0 9. 0.95542 0. 0 

GRID 300 0 10. 0.95542 0. 0 

GRID 301 0 11. 0.95542 0. 0 

GRID 302 0 12. 0.95542 0. 0 

GRID 303 0 13. 0.95542 0. 0 

GRID 306 0 16. 0.95542 0. 0 

3-12
TABLE 3-4. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 1 (Continued) 

GRID 307 0 17. 0.95542 0. 0 

GRID 308 0 17. 0.98458 0. 0 

GRID 313 0 12. 0.98458 0. 0 

GRID 314 0 11. 0.98458 0. 0 

GRID 316 0 9. 0.98458 0. 0 

GRID 317 0 8. 0.98458 0. 0 

GRID 318 0 7. 0.98458 0. 0 

GRID 323 0 2. 0.98458 0. 0 

GRID 324 0 1. 0.98458 0. 0 

CQUAD4 1 9 1 37 3 36 

CQUAD4 2 9 36 3 71 35 

CQUAD4 3 9 35 71 5 34 

CQUAD4 4 9 34 5 69 33 

CQUAD4 5 9 33 69 7 32 

CQUAD4 6 9 32 7 67 31 

CQUAD4 7 9 31 67 9 30 

CQUAD4 8 9 30 9 10 29 

CQUAD4 9 9 29 10 11 28 

CQUAD4 10 9 28 11 12 27 

CQUAD4 11 9 27 12 62 26 

CQUAD4 12 9 26 62 14 25 

CQUAD4 13 9 25 14 15 24 

CQUAD4 14 9 24 15 59 23 

CQUAD4 15 9 23 59 58 22 

CQUAD4 16 9 22 58 18 21 

CQUAD4 17 9 21 18 56 20 

CQUAD4 18 8 37 73 108 3 

CQUAD4 19 8 3 108 107 71 

CQUAD4 20 8 71 107 41 5 

CQUAD4 21 8 5 41 105 69 

CQUAD4 22 8 69 105 104 7 

CQUAD4 23 8 7 104 44 67 

CQUAD4 24 8 67 44 45 9 

CQUAD4 25 8 9 45 101 10 

CQUAD4 26 8 10 101 47 11 

CQUAD4 27 8 11 47 48 12 

CQUAD4 28 8 12 48 98 62 

CQUAD4 29 8 62 98 50 14 

CQUAD4 30 8 14 50 51 15 

CQUAD4 31 8 15 51 52 59 

CQUAD4 32 8 59 52 53 58 

CQUAD4 33 8 58 53 54 18 

CQUAD4 34 8 18 54 55 56 

CQUAD4 35 7 73 109 75 108 

CQUAD4 36 7 108 75 76 107 

CQUAD4 37 7 107 76 77 41 

CQUAD4 38 7 41 77 78 105 

CQUAD4 39 7 105 78 79 104 

CQUAD4 40 7 104 79 80 44 

CQUAD4 41 7 44 80 138 45 

CQUAD4 42 7 45 138 137 101 

CQUAD4 43 7 101 137 83 47 

CQUAD4 44 7 47 83 135 48 

CQUAD4 45 7 48 135 85 98 

CQUAD4 46 7 98 85 133 50 

CQUAD4 47 7 50 133 87 51 

CQUAD4 48 7 51 87 88 52 

3-13
TABLE 3-4. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 1 (Continued) 

CQUAD4 49 7 52 88 130 53 

CQUAD4 50 7 53 130 129 54 

CQUAD4 51 7 54 129 128 55 

CQUAD4 52 6 109 145 180 75 

CQUAD4 53 6 75 180 112 76 

CQUAD4 54 6 76 112 178 77 

CQUAD4 55 6 77 178 177 78 

CQUAD4 56 6 78 177 115 79 

CQUAD4 57 6 79 115 116 80 

CQUAD4 58 6 80 116 174 138 

CQUAD4 59 6 138 174 118 137 

CQUAD4 60 6 137 118 172 83 

CQUAD4 61 6 83 172 120 135 

CQUAD4 62 6 135 120 121 85 

CQUAD4 63 6 85 121 122 133 

CQUAD4 64 6 133 122 123 87 

CQUAD4 65 6 87 123 124 88 

CQUAD4 66 6 88 124 125 130 

CQUAD4 67 6 130 125 165 129 

CQUAD4 68 6 129 165 164 128 

CQUAD4 69 5 145 146 147 180 

CQUAD4 70 5 180 147 292 112 

CQUAD4 71 5 112 292 293 178 

CQUAD4 72 5 178 293 150 177 

CQUAD4 73 5 177 150 295 115 

CQUAD4 74 5 115 295 152 116 

CQUAD4 75 5 116 152 297 174 

CQUAD4 76 5 174 297 298 118 

CQUAD4 77 5 118 298 299 172 

CQUAD4 78 5 172 299 300 120 

CQUAD4 79 5 120 300 301 121 

CQUAD4 80 5 121 301 302 122 

CQUAD4 81 5 122 302 303 123 

CQUAD4 82 5 123 303 160 124 

CQUAD4 83 5 124 160 161 125 

CQUAD4 84 5 125 161 306 165 

CQUAD4 85 5 165 306 307 164 

CQUAD4 86 1 182 181 216 183 

CQUAD4 87 1 183 216 215 184 

CQUAD4 88 1 184 215 214 250 

CQUAD4 89 1 250 214 213 249 

CQUAD4 90 1 249 213 212 248 

CQUAD4 91 1 248 212 211 188 

CQUAD4 92 1 188 211 210 246 

CQUAD4 93 1 246 210 209 245 

CQUAD4 94 1 245 209 208 244 

CQUAD4 95 1 244 208 207 243 

CQUAD4 96 1 243 207 206 193 

CQUAD4 97 1 193 206 205 241 

CQUAD4 98 1 241 205 204 195 

CQUAD4 99 1 195 204 203 196 

CQUAD4 100 1 196 203 202 197 

CQUAD4 101 1 197 202 201 198 

CQUAD4 102 1 198 201 200 199 

CQUAD4 103 2 253 182 183 219 

CQUAD4 104 2 219 183 184 220 

CQUAD4 105 8 220 184 250 221 

3-14
TABLE 3-4. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 1 (Continued) 

CQUAD4 106 2 221 250 249 222 

CQUAD4 107 2 222 249 248 223 

CQUAD4 108 2 223 248 188 283 

CQUAD4 109 2 283 188 246 282 

CQUAD4 110 2 282 246 245 226 

CQUAD4 111 2 226 245 244 227 

CQUAD4 112 2 227 244 243 228 

CQUAD4 113 2 228 243 193 229 

CQUAD4 114 2 229 193 241 277 

CQUAD4 115 2 277 241 195 231 

CQUAD4 116 2 231 195 196 232 

CQUAD4 117 2 232 196 197 274 

CQUAD4 118 2 274 197 198 273 

CQUAD4 119 2 273 198 199 235 

CQUAD4 120 3 254 253 219 324 

CQUAD4 121 3 324 219 220 323 

CQUAD4 122 3 323 220 221 257 

CQUAD4 123 3 257 221 222 258 

CQUAD4 124 3 258 222 223 259 

CQUAD4 125 3 259 223 283 260 

CQUAD4 126 3 260 283 282 318 

CQUAD4 127 3 318 282 226 317 

CQUAD4 128 3 317 226 227 316 

CQUAD4 129 3 316 227 228 264 

CQUAD4 130 3 264 228 229 314 

CQUAD4 131 3 314 229 277 313 

CQUAD4 132 3 313 277 231 267 

CQUAD4 133 3 267 231 232 268 

CQUAD4 134 3 268 232 274 269 

CQUAD4 135 3 269 274 273 270 

CQUAD4 136 3 270 273 235 308 

CQUAD4 137 4 146 254 324 147 

CQUAD4 138 4 147 324 323 292 

CQUAD4 139 4 292 323 257 293 

CQUAD4 140 4 293 257 258 150 

CQUAD4 141 4 150 258 259 295 

CQUAD4 142 4 295 259 260 152 

CQUAD4 143 4 152 260 318 297 

CQUAD4 144 4 297 318 317 298 

CQUAD4 145 4 298 317 316 299 

CQUAD4 146 4 299 316 264 300 

CQUAD4 147 4 300 264 314 301 

CQUAD4 148 4 301 314 313 302 

CQUAD4 149 4 302 313 267 303 

CQUAD4 150 4 303 267 268 160 

CQUAD4 151 4 160 268 269 161 

CQUAD4 152 4 161 269 270 306 

CQUAD4 153 4 306 270 308 307 

ENDDATA 

The results of the one-half FEM wing model analysis are presented in figure 3-1 in the form of a 

contour plot. The contours along the outer nodes are in agreement with the imposed boundary 

conditions. The contours of the inner layers appear to be indicative of the temperature profile of 

the THERMOD analysis. 

3-15
FIGURE 3-1. CONTOUR PLOT OF THE FEM OUTPUT OF SAMPLE PROBLEM 1 

The truncated FEM output file is shown in table 3-5. The nodes, from top to bottom of the wing, 

located at the mid-span of the model are indicated by nodes 209, 245, 226, 317, 298, 118, 137, 

101, 10, and 29. The temperatures of these nodes were extracted from this file and are 

summarized in table 3-6. Mid-span was chosen because it was sufficiently far away from any 

edge effects that might be present. 

TABLE 3-5. TRUNCATED FEM OUTPUT FILE OF SAMPLE PROBLEM 1 

1 CONDUCTION IN RECTANGULAR ADIABATIC PLATE (T=0.001FT) MARCH 6, 2000 UAI/NASTRAN VERSION 20.1 PAGE 

16 

 10.7 IN GAP FILLED BY AIR; K=0.016BTU/HR/FT/DEGF 

0 

 

 T E M P E R A T U R E V E C T O R 

 

 POINT ID. TYPE ID VALUE ID+1 VALUE ID+2 VALUE ID+3 VALUE ID+4 VALUE ID+5 VALUE 

 1 TEMP 1.244000E+02 

 3 TEMP 1.244011E+02 

 5 TEMP 1.244011E+02 

 7 TEMP 1.244011E+02 

 9 TEMP 1.244011E+02 1.244011E+02 1.244011E+02 1.244011E+02 

 14 TEMP 1.244011E+02 1.244011E+02 

 18 TEMP 1.244011E+02 

 20 TEMP 1.244000E+02 1.244000E+02 1.244000E+02 1.244000E+02 1.244000E+02 1.244000E+02 

 26 TEMP 1.244000E+02 1.244000E+02 1.244000E+02 1.244000E+02 1.244000E+02 1.244000E+02 

 32 TEMP 1.244000E+02 1.244000E+02 1.244000E+02 1.244000E+02 1.244000E+02 1.244000E+02 

 41 TEMP 1.246966E+02 

 44 TEMP 1.246966E+02 1.246966E+02 

 47 TEMP 1.246966E+02 1.246966E+02 

 50 TEMP 1.246966E+02 1.246966E+02 1.246966E+02 1.246961E+02 1.246825E+02 1.244000E+02 

 56 TEMP 1.244000E+02 

 58 TEMP 1.244011E+02 1.244011E+02 

 62 TEMP 1.244011E+02 

 67 TEMP 1.244011E+02 

 69 TEMP 1.244011E+02 

 71 TEMP 1.244011E+02 

 73 TEMP 1.244000E+02 

 75 TEMP 1.246836E+02 1.246972E+02 1.246977E+02 1.246977E+02 1.246977E+02 1.246977E+02 

 83 TEMP 1.246977E+02 

 85 TEMP 1.246977E+02 

 87 TEMP 1.246977E+02 1.246977E+02 

3-16
TABLE 3-5. TRUNCATED FEM OUTPUT FILE OF SAMPLE PROBLEM 1 (Continued) 

 98 TEMP 1.246966E+02 

 101 TEMP 1.246966E+02 

 104 TEMP 1.246966E+02 1.246966E+02 

 107 TEMP 1.246961E+02 1.246825E+02 1.244000E+02 

 112 TEMP 1.247165E+02 

 115 TEMP 1.247170E+02 1.247170E+02 

 118 TEMP 1.247170E+02 

 120 TEMP 1.247170E+02 1.247170E+02 1.247170E+02 1.247170E+02 1.247170E+02 1.247165E+02 

 128 TEMP 1.244000E+02 1.246836E+02 1.246972E+02 

 133 TEMP 1.246977E+02 

 135 TEMP 1.246977E+02 

 137 TEMP 1.246977E+02 1.246977E+02 

 145 TEMP 1.244000E+02 1.244000E+02 1.355065E+02 

 150 TEMP 1.357830E+02 

 152 TEMP 1.357830E+02 

 160 TEMP 1.357829E+02 1.357764E+02 

 164 TEMP 1.244000E+02 1.247021E+02 

 172 TEMP 1.247170E+02 

 174 TEMP 1.247170E+02 

 177 TEMP 1.247170E+02 1.247170E+02 

 180 TEMP 1.247021E+02 1.244000E+02 1.244000E+02 1.360979E+02 1.360989E+02 

 188 TEMP 1.360989E+02 

 193 TEMP 1.360989E+02 

 195 TEMP 1.360989E+02 1.360989E+02 1.360989E+02 1.360979E+02 1.244000E+02 1.244000E+02 

 201 TEMP 1.361000E+02 1.361000E+02 1.361000E+02 1.361000E+02 1.361000E+02 1.361000E+02 

1 CONDUCTION IN RECTANGULAR ADIABATIC PLATE (T=0.001FT) MARCH 6, 2000 UAI/NASTRAN VERSION 20.1 PAGE 

17 

 10.7 IN GAP FILLED BY AIR; K=0.016BTU/HR/FT/DEGF 

0 

 

 T E M P E R A T U R E V E C T O R 

 

 POINT ID. TYPE ID VALUE ID+1 VALUE ID+2 VALUE ID+3 VALUE ID+4 VALUE ID+5 VALUE 

 207 TEMP 1.361000E+02 1.361000E+02 1.361000E+02 1.361000E+02 1.361000E+02 1.361000E+02 

 213 TEMP 1.361000E+02 1.361000E+02 1.361000E+02 1.361000E+02 

 219 TEMP 1.355396E+02 1.357971E+02 1.358033E+02 1.358034E+02 1.358034E+02 

 226 TEMP 1.358034E+02 1.358034E+02 1.358034E+02 1.358034E+02 

 231 TEMP 1.358034E+02 1.358033E+02 

 235 TEMP 1.244000E+02 

 241 TEMP 1.360989E+02 

 243 TEMP 1.360989E+02 1.360989E+02 1.360989E+02 1.360989E+02 

 248 TEMP 1.360989E+02 1.360989E+02 1.360989E+02 

 253 TEMP 1.244000E+02 1.244000E+02 

 257 TEMP 1.358022E+02 1.358023E+02 1.358023E+02 1.358023E+02 

 264 TEMP 1.358023E+02 

 267 TEMP 1.358023E+02 1.358022E+02 1.357960E+02 1.355375E+02 

 273 TEMP 1.355396E+02 1.357971E+02 

 277 TEMP 1.358034E+02 

 282 TEMP 1.358034E+02 1.358034E+02 

 292 TEMP 1.357764E+02 1.357829E+02 

 295 TEMP 1.357830E+02 

 297 TEMP 1.357830E+02 1.357830E+02 1.357830E+02 1.357830E+02 1.357830E+02 1.357830E+02 

 303 TEMP 1.357830E+02 

 306 TEMP 1.355065E+02 1.244000E+02 1.244000E+02 

 313 TEMP 1.358023E+02 1.358023E+02 

 316 TEMP 1.358023E+02 1.358023E+02 1.358023E+02 

 323 TEMP 1.357960E+02 1.355375E+02 

 

TABLE 3-6. FINITE ELEMENT METHOD-SIMULATED TEMPERATURES (°F) OF THE 

RIGHT WING OF SAMPLE PROBLEM 1 

T1 T2 T3 T4 T5 T6 T7 T8 T9 T10

136.1 136.1 135.8 135.8 135.8 124.7 124.7 124.7 124.4 124.4

When compared, the inner temperatures of tables 3-3 and 3-6 show excellent agreement, thus 

validating the conduction capability of the THERMOD program. 

3.2 SAMPLE PROBLEM 2.

Sample problem 2 is similar to sample problem 1, except the 10.7-in.-deep air space between the 

spar caps was replaced with a fuel-filled space. The input.dat file is altered by changing the 

conductivity of layer 5 from 0.016 (Btu/hr)/(ft°F), that of air, to 0.081 (Btu/hr)/(ft°F), estimated 

to be that of oil [6]. All other thermal properties of the layer were kept the same as in sample 

problem 1. 

3-17
Temperatures from the THERMOD run is shown in table 3-7, from which the steady-state 

temperatures T1, T2.....T10 (top to bottom of wing) were extracted and tabulated in table 3-8 for 

time period 3. 

TABLE 3-7. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE PROBLEM 2 

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ 

THERMOD: A THERMAL PROGRAM FOR AIRCRAFTS 

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ 

 

 

 

 SUMMARY OF STEADY-STATE AND TRANSIENT ANALYSIS 

 

 

 

 (Temperatures are shown for all time periods) 

 

 

 TEMPERATURES AT THE END OF STEADY STATE ANALYSIS 

 

 TEMPERATURES IN DEGREES FAHRENHEIT 

 1 2 3 4 5 6 7 

 

 1 123.7 132.2 135.9 133.8 126.3 115.2 102.9 

 2 123.7 132.2 135.9 133.8 126.3 115.2 103.0 

 3 122.7 131.0 134.7 132.8 125.7 115.4 103.8 

 4 122.7 131.0 134.7 132.8 125.7 115.4 103.8 

 5 122.7 130.9 134.6 132.7 125.7 115.4 103.8 

 6 115.4 121.5 125.9 125.6 121.7 116.2 109.7 

 7 115.3 121.4 125.8 125.5 121.6 116.2 109.8 

 8 115.3 121.4 125.8 125.5 121.6 116.2 109.8 

 9 114.3 120.2 124.6 124.6 121.1 116.4 110.6 

 10 114.3 120.2 124.6 124.6 121.1 116.4 110.6 

 11 122.0 131.2 135.6 132.8 124.7 113.6 101.6 

 12 122.0 131.2 135.6 132.8 124.7 113.6 101.6 

 13 121.1 130.0 134.5 131.9 124.3 113.8 102.5 

 14 121.1 130.0 134.5 131.9 124.3 113.8 102.5 

 15 121.1 129.9 134.4 131.8 124.3 113.9 102.6 

 16 115.1 121.4 125.9 125.5 121.5 116.0 109.5 

 17 115.1 121.3 125.8 125.4 121.4 116.0 109.6 

 18 115.1 121.3 125.8 125.4 121.4 116.0 109.6 

 19 114.3 120.1 124.6 124.5 121.1 116.3 110.5 

 20 114.3 120.1 124.6 124.5 121.1 116.3 110.5 

 21 205.8 225.9 231.9 222.2 200.7 169.4 134.2 

 22 205.8 225.8 231.8 222.1 200.7 169.5 134.2 

 23 207.4 216.5 217.2 213.6 202.4 180.2 143.1 

 24 207.4 216.4 217.2 213.5 202.4 180.2 143.1 

 25 208.4 209.9 206.9 207.6 203.6 187.7 149.4 

3-18
TABLE 3-7. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE 

PROBLEM 2 (Continued) 

 26 198.9 198.2 195.5 196.8 195.4 183.4 146.2 

 27 150.5 153.0 154.4 155.5 153.6 146.5 127.6 

 28 150.4 153.0 154.4 155.4 153.6 146.5 127.6 

 29 147.0 149.8 151.5 152.5 150.6 143.9 126.3 

 30 147.0 149.8 151.5 152.5 150.6 143.9 126.3 

 31 131.8 135.6 138.7 139.6 137.5 132.4 120.5 

 32 131.8 135.6 138.7 139.6 137.5 132.4 120.5 

 33 131.8 135.6 138.7 139.6 137.5 132.4 120.5 

 34 131.8 135.6 138.7 139.6 137.5 132.4 120.5 

 35 116.7 121.5 125.8 126.6 124.5 120.8 114.7 

 36 116.7 121.5 125.8 126.6 124.4 120.8 114.7 

 37 113.2 118.3 122.9 123.7 121.5 118.2 113.3 

 38 113.2 118.3 122.9 123.7 121.4 118.2 113.3 

 39 199.5 179.7 148.8 180.0 196.6 194.7 173.9 

 40 199.5 179.7 148.9 180.1 196.6 194.6 173.9 

 41 197.6 186.6 168.7 186.2 194.5 188.4 162.3 

 42 197.6 186.6 168.8 186.2 194.5 188.4 162.3 

 43 196.2 191.4 182.7 190.5 193.1 184.1 154.2 

 44 191.8 188.5 182.6 187.8 189.2 180.4 151.1 

 45 169.9 170.8 168.7 171.5 170.1 161.4 138.9 

 46 169.8 170.7 168.7 171.5 170.0 161.3 138.8 

 47 138.6 145.3 148.8 148.2 142.9 134.1 121.3 

 48 138.4 145.2 148.7 148.0 142.7 133.9 121.2 

 49 145.6 148.6 149.7 150.4 147.7 139.7 123.0 

 50 147.6 150.5 151.3 152.1 149.5 141.3 123.9 

 51 113.6 120.9 125.1 123.7 118.1 110.0 100.6 

 52 108.2 115.1 119.2 118.0 112.7 105.1 96.2 

 53 113.6 120.9 125.1 123.7 118.1 110.0 100.6 

 

Maximum Structural Temperature for Period 1 = 207.37 Occurring 

at Location 24 

Maximum Structural Temperature for Period 2 = 225.86 Occurring 

at Location 21 

Maximum Structural Temperature for Period 3 = 231.88 Occurring 

at Location 21 

Maximum Structural Temperature for Period 4 = 222.15 Occurring 

at Location 21 

Maximum Structural Temperature for Period 5 = 202.43 Occurring 

at Location 24 

Maximum Structural Temperature for Period 6 = 194.68 Occurring 

at Location 39 

Maximum Structural Temperature for Period 7 = 173.93 Occurring 

at Location 39 

 

Maximum Structural Temperature Over All 7 Periods =231.88 

Occurring at Location 21 at Period 3 

 

3-19
TABLE 3-7. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE 

PROBLEM 2 (Continued) 

 

 TEMPERATURES AT THE END OF TRANSIENT ANALYSIS 

 

 TEMPERATURES IN DEGREES FAHRENHEIT 

 1 2 3 4 5 6 7 

 

 1 112.8 116.8 121.7 123.8 123.4 122.5 119.8 

 2 113.2 117.3 122.1 124.1 123.5 122.3 119.3 

 3 122.0 129.9 133.7 132.1 125.5 115.9 104.9 

 4 122.3 130.3 134.1 132.3 125.6 115.7 104.5 

 5 122.3 130.3 134.1 132.3 125.5 115.5 104.3 

 6 115.5 121.7 126.1 125.8 121.9 116.4 109.8 

 7 115.3 121.3 125.7 125.5 121.8 116.6 110.2 

 8 115.2 121.2 125.6 125.5 121.9 116.7 110.5 

 9 112.8 116.7 121.6 123.8 123.6 123.1 120.8 

 10 112.7 116.5 121.5 123.8 123.7 123.3 121.2 

 11 112.7 116.7 121.6 123.7 123.3 122.4 119.7 

 12 113.0 117.2 122.1 124.0 123.3 122.1 119.2 

 13 120.5 129.1 133.6 131.3 124.2 114.4 103.8 

 14 120.7 129.4 133.9 131.5 124.3 114.2 103.3 

 15 120.8 129.4 133.9 131.5 124.2 114.1 103.1 

 16 115.3 121.6 126.0 125.6 121.6 116.1 109.6 

 17 115.0 121.2 125.7 125.4 121.6 116.3 110.1 

 18 115.0 121.1 125.6 125.4 121.7 116.5 110.4 

 19 112.8 116.6 121.6 123.8 123.6 123.1 120.8 

 20 112.7 116.5 121.5 123.8 123.7 123.3 121.2 

 21 125.4 131.3 136.7 137.5 134.8 130.9 124.7 

 22 127.4 133.3 138.7 139.3 136.4 132.3 125.2 

 23 181.2 187.5 192.2 187.4 180.5 171.4 140.3 

 24 183.1 189.2 193.8 188.9 182.0 172.9 140.9 

 25 200.7 202.6 205.1 200.1 195.7 189.2 147.8 

 26 185.7 185.5 187.8 185.4 184.4 181.0 145.5 

 27 150.7 153.2 154.6 155.7 153.8 146.7 127.7 

 28 149.5 152.1 153.7 154.7 152.8 145.8 127.3 

 29 146.5 149.3 151.2 152.1 150.2 143.7 126.3 

 30 145.7 148.6 150.4 151.4 149.5 143.1 126.0 

 31 131.5 135.1 138.2 139.3 137.7 133.2 121.9 

 32 131.5 135.1 138.2 139.3 137.7 133.2 121.9 

 33 131.5 135.1 138.2 139.3 137.7 133.2 121.9 

 34 131.5 135.1 138.2 139.3 137.7 133.2 121.9 

 35 117.2 121.6 125.9 127.3 125.8 123.2 117.7 

 36 116.4 120.7 125.2 126.5 125.2 122.7 117.5 

 37 112.1 115.7 120.6 123.1 123.4 123.4 121.5 

 38 112.0 115.5 120.5 123.0 123.3 123.4 121.6 

 39 121.2 122.3 123.8 129.2 131.3 131.5 127.8 

 40 123.2 124.0 125.1 130.7 133.0 133.1 128.7 

 41 172.1 166.9 160.2 168.7 173.2 172.1 150.2 

3-20
TABLE 3-7. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE 

PROBLEM 2 (Continued) 

 42 173.6 168.3 161.5 170.0 174.5 173.3 150.8 

 43 183.1 179.5 176.6 179.9 182.2 181.5 152.3 

 44 179.2 177.0 176.6 177.6 178.7 178.0 149.3 

 45 162.3 161.7 161.5 163.9 164.5 162.3 141.2 

 46 160.8 160.3 160.1 162.6 163.2 161.0 140.6 

 47 121.5 123.0 125.1 129.8 131.5 131.4 127.2 

 48 120.0 121.6 123.8 128.6 130.3 130.3 126.7 

 49 123.6 126.8 131.0 132.5 132.0 130.4 123.1 

 50 127.8 130.8 134.8 136.0 135.3 133.2 123.9 

 51 169.1 187.2 193.0 184.0 164.8 137.5 109.0 

 52 157.8 174.5 180.0 172.0 154.5 129.7 103.8 

 53 169.1 187.2 193.0 184.0 164.8 137.5 109.0 

 

Maximum Structural Temperature for Period 1 = 183.07 Occurring 

at Location 24 

Maximum Structural Temperature for Period 2 = 189.17 Occurring 

at Location 24 

Maximum Structural Temperature for Period 3 = 193.83 Occurring 

at Location 24 

Maximum Structural Temperature for Period 4 = 188.85 Occurring 

at Location 24 

Maximum Structural Temperature for Period 5 = 181.98 Occurring 

at Location 24 

Maximum Structural Temperature for Period 6 = 173.32 Occurring 

at Location 42 

Maximum Structural Temperature for Period 7 = 150.76 Occurring 

at Location 42 

 

Maximum Structural Temperature Over All 7 Periods =193.83 

Occurring at Location 24 at Period 3 

TABLE 3-8. THERMOD-SIMULATED TEMPERATURES (°F) OF THE RIGHT WING OF 

SAMPLE PROBLEM 2 

T1 T2 T3 T4 T5 T6 T7 T8 T9 T10

135.9 135.9 134.7 134.7 134.6 125.9 125.8 125.8 124.6 124.6

The results of the one-half FEM wing model analysis are presented in figure 3-2 in the form of a 

contour plot. The contours along the outer nodes are in agreement with the imposed boundary 

conditions while the contours of the inner layers appear to be indicative of the temperature 

profile of the THERMOD analysis. 

3-21
FIGURE 3-2. CONTOUR PLOT OF THE FEM OUTPUT OF SAMPLE PROBLEM 2 

The truncated FEM output file is shown in table 3-9. The nodes, from top to bottom of the wing, 

located at mid-span of the model are indicated by nodes 209, 245, 226, 317, 298, 118, 137, 101, 

10, and 29. The temperatures of these nodes were extracted from this file and are reproduced in 

table 3-10. Mid-span was chosen because it was sufficiently far away from any edge effects that 

might be present. 

TABLE 3-9. TRUNCATED FEM OUTPUT FILE OF SAMPLE PROBLEM 2 

1 CONDUCTION IN RECTANGULAR ADIABATIC PLATE (T=0.001FT) MARCH 6, 2000 UAI/NASTRAN VERSION 20.1 PAGE 

16 

 10.7 IN GAP FILLED BY OIL; K=0.081BTU/HR/FT/DEGF 

0 

 

 T E M P E R A T U R E V E C T O R 

 

 POINT ID. TYPE ID VALUE ID+1 VALUE ID+2 VALUE ID+3 VALUE ID+4 VALUE ID+5 VALUE 

 1 TEMP 1.246000E+02 

 3 TEMP 1.246039E+02 

 5 TEMP 1.246044E+02 

 7 TEMP 1.246044E+02 

 9 TEMP 1.246044E+02 1.246044E+02 1.246044E+02 1.246044E+02 

 14 TEMP 1.246044E+02 1.246044E+02 

 18 TEMP 1.246039E+02 

 20 TEMP 1.246000E+02 1.246000E+02 1.246000E+02 1.246000E+02 1.246000E+02 1.246000E+02 

 26 TEMP 1.246000E+02 1.246000E+02 1.246000E+02 1.246000E+02 1.246000E+02 1.246000E+02 

 32 TEMP 1.246000E+02 1.246000E+02 1.246000E+02 1.246000E+02 1.246000E+02 1.246000E+02 

 41 TEMP 1.257877E+02 

 44 TEMP 1.257885E+02 1.257885E+02 

 47 TEMP 1.257885E+02 1.257885E+02 

 50 TEMP 1.257885E+02 1.257885E+02 1.257877E+02 1.257777E+02 1.256599E+02 1.246000E+02 

 56 TEMP 1.246000E+02 

 58 TEMP 1.246044E+02 1.246044E+02 

 62 TEMP 1.246044E+02 

 67 TEMP 1.246044E+02 

 69 TEMP 1.246044E+02 

 71 TEMP 1.246044E+02 

 73 TEMP 1.246000E+02 

 75 TEMP 1.256639E+02 1.257821E+02 1.257921E+02 1.257929E+02 1.257929E+02 1.257929E+02 

 83 TEMP 1.257929E+02 

 85 TEMP 1.257929E+02 

 87 TEMP 1.257929E+02 1.257921E+02 

3-22
TABLE 3-9. TRUNCATED FEM OUTPUT FILE OF SAMPLE PROBLEM 2 (Continued) 

 98 TEMP 1.257885E+02 

 101 TEMP 1.257885E+02 

 104 TEMP 1.257885E+02 1.257885E+02 

 107 TEMP 1.257777E+02 1.256599E+02 1.246000E+02 

 112 TEMP 1.258588E+02 

 115 TEMP 1.258703E+02 1.258703E+02 

 118 TEMP 1.258703E+02 

 120 TEMP 1.258703E+02 1.258703E+02 1.258703E+02 1.258702E+02 1.258694E+02 1.258588E+02 

 128 TEMP 1.246000E+02 1.256639E+02 1.257821E+02 

 133 TEMP 1.257929E+02 

 135 TEMP 1.257929E+02 

 137 TEMP 1.257929E+02 1.257929E+02 

 145 TEMP 1.246000E+02 1.246000E+02 1.340483E+02 

 150 TEMP 1.346296E+02 

 152 TEMP 1.346297E+02 

 160 TEMP 1.346277E+02 1.345958E+02 

 164 TEMP 1.246000E+02 1.257333E+02 

 172 TEMP 1.258703E+02 

 174 TEMP 1.258703E+02 

 177 TEMP 1.258702E+02 1.258694E+02 

 180 TEMP 1.257333E+02 1.246000E+02 1.246000E+02 1.358935E+02 1.358954E+02 

 188 TEMP 1.358956E+02 

 193 TEMP 1.358956E+02 

 195 TEMP 1.358956E+02 1.358956E+02 1.358954E+02 1.358935E+02 1.246000E+02 1.246000E+02 

 201 TEMP 1.359000E+02 1.359000E+02 1.359000E+02 1.359000E+02 1.359000E+02 1.359000E+02 

1 CONDUCTION IN RECTANGULAR ADIABATIC PLATE (T=0.001FT) MARCH 6, 2000 UAI/NASTRAN VERSION 20.1 PAGE 

17 

 10.7 IN GAP FILLED BY OIL; K=0.081BTU/HR/FT/DEGF 

0 

 

 T E M P E R A T U R E V E C T O R 

 

 POINT ID. TYPE ID VALUE ID+1 VALUE ID+2 VALUE ID+3 VALUE ID+4 VALUE ID+5 VALUE 

 207 TEMP 1.359000E+02 1.359000E+02 1.359000E+02 1.359000E+02 1.359000E+02 1.359000E+02 

 213 TEMP 1.359000E+02 1.359000E+02 1.359000E+02 1.359000E+02 

 219 TEMP 1.341632E+02 1.346795E+02 1.347096E+02 1.347113E+02 1.347115E+02 

 226 TEMP 1.347115E+02 1.347115E+02 1.347115E+02 1.347115E+02 

 231 TEMP 1.347113E+02 1.347096E+02 

 235 TEMP 1.246000E+02 

 241 TEMP 1.358956E+02 

 243 TEMP 1.358956E+02 1.358956E+02 1.358956E+02 1.358956E+02 

 248 TEMP 1.358956E+02 1.358956E+02 1.358956E+02 

 253 TEMP 1.246000E+02 1.246000E+02 

 257 TEMP 1.347052E+02 1.347069E+02 1.347070E+02 1.347070E+02 

 264 TEMP 1.347070E+02 

 267 TEMP 1.347069E+02 1.347052E+02 1.346750E+02 1.341568E+02 

 273 TEMP 1.341632E+02 1.346795E+02 

 277 TEMP 1.347115E+02 

 282 TEMP 1.347115E+02 1.347115E+02 

 292 TEMP 1.345958E+02 1.346277E+02 

 295 TEMP 1.346297E+02 

 297 TEMP 1.346297E+02 1.346297E+02 1.346297E+02 1.346297E+02 1.346297E+02 1.346297E+02 

 303 TEMP 1.346296E+02 

 306 TEMP 1.340483E+02 1.246000E+02 1.246000E+02 

 313 TEMP 1.347070E+02 1.347070E+02 

 316 TEMP 1.347070E+02 1.347070E+02 1.347070E+02 

 323 TEMP 1.346750E+02 1.341568E+02 

 

When compared, the inner temperatures of tables 3-8 and 3-10 show excellent agreement, again 

validating the conduction capability of the THERMOD program. 

TABLE 3-10. FINITE ELEMENT METHOD-SIMULATED TEMPERATURES (°F) OF THE 

RIGHT WING OF SAMPLE PROBLEM 2 

T1 T2 T3 T4 T5 T6 T7 T8 T9 T10

135.9 135.9 134.7 134.7 134.6 125.9 125.8 125.8 124.6 124.6

3.3 SAMPLE PROBLEM 3.

Sample problem 3, isolated and studied the effects of surface convection. Convection was 

introduced into the system by surface airflow. Depending on the ambient and surface 

temperatures, heat may either be added or removed from the system. 

3-23
As in sample problems 1 and 2, a complete aircraft problem was solved using the THERMOD 

program. A particular item of interest was that the absorptivity and emissivity of all external 

surfaces, i.e., the fuselage, wing, and tarmac surfaces, were assigned 1.0 and 0.0 values, 

respectively. Such surfaces absorb all heat and do not emit any energy (no radiation). The only 

way for surface heat transfer to take place was through conduction and convection. Table 3-11 

shows the THERMOD input file for this sample problem. 

TABLE 3-11. THERMOD INPUT FILE (input.dat) OF SAMPLE PROBLEM 3 

198e-6,0.704 

1, 0.020,125,0.300,0.30 

2, 0.375,4.4,0.021,0.24 

3, 0.020,125,0.300,0.30 

4, 0.350,125,0.300,0.30 

5, 10.70,0.067,0.016,0.24 

6, 0.350,125,0.300,0.30 

7, 0.020,125,0.300,0.30 

8, 0.375,4.4,0.021,0.24 

9, 0.020,125,0.300,0.30 

10,0.030,125,0.300,0.30 

11,0.375,4.4,0.021,0.24 

12,0.020,125,0.300,0.30 

13,0.250,10.,0.020,0.24 

14,48.00,0.067,0.016,0.24 

15,0.020,125,0.300,0.30 

16,0.375,4.4,0.021,0.24 

17,0.020,125,0.300,0.30 

18,0.250,10.,0.020,0.24 

19,30.00,0.067,0.016,0.24 

20,5.000,10.,0.020,0.24 

21,0.020,125,0.300,0.30 

22,0.375,4.4,0.021,0.24 

23,0.020,125,0.300,0.30 

24,2.500,0.067,0.016,0.24 

25,0.020,125,0.300,0.30 

26,0.375,4.4,0.021,0.24 

27,0.020,125,0.300,0.30 

28,0.250,130,0.800,0.20 

3.5,34.0,2.5,5.5,22.0,11.0,3.0,4.0,30,24,28 

0.70,1.0,0.0,2.0,150,0.80,0.22 

0.3,0.9,0.3,0.9,0.3,0.9,0.1,0.9 

3,100 

0.9,0.9,0.9,0.9 

1.0,0.0,1.0,0.0,1.0,0.0,1.0,0.0 

9 

0,10,10,0,0,75,110,110,190,190 

5,120,5,60,20,5,120,60,60 

7 

60 570.67 459.67 14 330 10 10 0.36 

3-24
TABLE 3-11. THERMOD INPUT FILE (input.dat) OF SAMPLE PROBLEM 3 (Continued) 

75 573.67 459.67 14 355 10 10 0.31 

90 578.67 459.67 14 355 10 12 0.33 

75 581.67 459.67 14 330 10 10 0.31 

60 582.67 459.67 14 291 10 10 0.36 

45 583.67 459.67 14 231 10 10 0.44 

30 582.67 459.67 14 160 10 10 0.57 

2,100,180,1 

10,1.0,1.0 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10 

Temperatures from the THERMOD run is shown in table 3-12, from which the steady-state 

temperatures T1, T2.....T10 (top to bottom of wing) for time period 3 were extracted and 

tabulated in table 3-13. 

TABLE 3-12. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE PROBLEM 3 

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ 

 THERMOD: A THERMAL PROGRAM FOR AIRCRAFTS 

 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ 

 

 

 

 SUMMARY OF STEADY-STATE AND TRANSIENT ANALYSIS 

 

 

 

 (Temperatures are shown for all time periods) 

 

 

 TEMPERATURES AT THE END OF STEADY STATE ANALYSIS 

 

 TEMPERATURES IN DEGREES FAHRENHEIT 

 1 2 3 4 5 6 7 

 

 1 321.1 366.1 380.0 356.4 308.3 244.1 181.8 

 2 321.1 366.1 380.0 356.3 308.3 244.1 181.8 

 3 315.9 359.8 373.5 350.5 303.7 241.1 180.3 

 4 315.8 359.8 373.4 350.5 303.6 241.1 180.3 

 5 315.5 359.4 373.0 350.1 303.3 240.9 180.2 

 6 119.2 123.9 129.2 131.2 130.3 128.7 125.3 

 7 118.9 123.5 128.8 130.8 130.0 128.5 125.2 

 8 118.9 123.5 128.8 130.8 129.9 128.5 125.2 

3-25
TABLE 3-12. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE 

PROBLEM 3 (Continued) 

 9 113.6 117.2 122.3 124.9 125.3 125.5 123.7 

 10 113.6 117.1 122.3 124.9 125.3 125.5 123.7 

 11 321.1 366.1 380.0 356.4 308.3 244.1 181.8 

 12 321.1 366.1 380.0 356.3 308.3 244.1 181.8 

 13 315.9 359.8 373.5 350.5 303.7 241.1 180.3 

 14 315.8 359.8 373.4 350.5 303.6 241.1 180.3 

 15 315.5 359.4 373.0 350.1 303.3 240.9 180.2 

 16 119.2 123.9 129.2 131.2 130.3 128.7 125.3 

 17 118.9 123.5 128.8 130.8 130.0 128.5 125.2 

 18 118.9 123.5 128.8 130.8 129.9 128.5 125.2 

 19 113.6 117.2 122.3 124.9 125.3 125.5 123.7 

 20 113.6 117.1 122.3 124.9 125.3 125.5 123.7 

 21 277.2 308.5 317.8 302.4 269.0 222.0 171.4 

 22 277.1 308.3 317.6 302.2 268.9 221.9 171.4 

 23 244.4 257.4 258.7 253.9 238.6 210.2 166.7 

 24 244.3 257.2 258.5 253.7 238.5 210.2 166.7 

 25 221.4 221.6 217.3 219.9 217.3 202.0 163.4 

 26 194.1 189.2 184.0 189.0 192.2 184.4 150.1 

 27 147.9 147.4 147.9 151.8 153.7 150.8 135.0 

 28 147.9 147.4 147.9 151.8 153.7 150.8 135.0 

 29 144.6 144.4 145.3 149.1 151.0 148.4 134.0 

 30 144.6 144.4 145.3 149.1 151.0 148.4 134.0 

 31 130.2 131.3 134.0 137.5 138.9 137.9 129.2 

 32 130.2 131.3 134.0 137.5 138.9 137.9 129.2 

 33 130.2 131.3 134.0 137.5 138.9 137.9 129.2 

 34 130.2 131.3 134.0 137.5 138.9 137.9 129.2 

 35 115.7 118.3 122.7 125.8 126.9 127.4 124.5 

 36 115.7 118.3 122.7 125.8 126.9 127.4 124.5 

 37 112.4 115.3 120.1 123.1 124.2 125.0 123.5 

 38 112.4 115.3 120.1 123.1 124.2 125.0 123.5 

 39 210.7 173.9 128.8 177.2 210.2 219.4 200.8 

 40 210.7 174.0 129.0 177.3 210.2 219.3 200.6 

 41 200.6 178.8 152.2 180.6 199.7 201.5 178.1 

 42 200.6 178.8 152.3 180.6 199.6 201.4 178.0 

 43 193.6 182.2 168.6 183.0 192.3 189.0 162.3 

 44 186.0 177.5 168.6 178.6 185.6 182.2 156.6 

 45 161.4 156.6 152.3 160.0 165.0 163.1 145.6 

 46 161.3 156.5 152.2 159.9 164.9 163.0 145.5 

 47 126.1 126.8 129.0 133.4 135.6 135.7 129.8 

 48 125.9 126.6 128.8 133.2 135.4 135.6 129.7 

 49 126.2 124.5 123.5 127.4 129.7 127.1 114.5 

 50 128.7 126.7 125.5 129.6 132.0 129.2 115.7 

 51 188.8 207.4 215.7 208.8 191.6 168.5 144.8 

 52 183.9 201.5 209.6 203.3 187.3 165.7 143.4 

 53 188.8 207.4 215.7 208.8 191.6 168.5 144.8 

 

3-26
TABLE 3-12. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE 

PROBLEM 3 (Continued) 

Maximum Structural Temperature for Period 1 = 321.12 Occurring 

at Location 1 

Maximum Structural Temperature for Period 2 = 366.11 Occurring 

at Location 1 

Maximum Structural Temperature for Period 3 = 380.01 Occurring 

at Location 1 

Maximum Structural Temperature for Period 4 = 356.36 Occurring 

at Location 1 

Maximum Structural Temperature for Period 5 = 308.29 Occurring 

at Location 1 

Maximum Structural Temperature for Period 6 = 244.09 Occurring 

at Location 1 

Maximum Structural Temperature for Period 7 = 200.77 Occurring 

at Location 39 

 

Maximum Structural Temperature Over All 7 Periods =380.01 

Occurring at Location 1 at Period 3 

 

 

 TEMPERATURES AT THE END OF TRANSIENT ANALYSIS 

 

 TEMPERATURES IN DEGREES FAHRENHEIT 

 1 2 3 4 5 6 7 

 

 1 129.1 135.7 141.4 142.2 138.9 134.3 128.1 

 2 134.7 142.5 148.5 148.5 143.9 137.6 129.6 

 3 301.7 342.8 355.8 334.6 291.1 233.0 176.4 

 4 306.4 348.5 361.8 340.0 295.3 235.7 177.7 

 5 306.1 348.2 361.4 339.7 295.1 235.5 177.6 

 6 124.1 129.7 135.2 136.6 134.5 131.5 126.7 

 7 118.8 123.3 128.7 130.7 129.9 128.4 125.2 

 8 118.6 123.1 128.4 130.4 129.7 128.3 125.1 

 9 111.4 114.5 119.6 122.5 123.4 124.3 123.1 

 10 111.2 114.3 119.3 122.3 123.2 124.1 123.1 

 11 129.1 135.7 141.4 142.2 138.9 134.3 128.1 

 12 134.7 142.5 148.5 148.5 143.9 137.6 129.6 

 13 301.7 342.8 355.8 334.6 291.1 233.0 176.4 

 14 306.4 348.5 361.8 340.0 295.3 235.7 177.7 

 15 306.1 348.2 361.4 339.7 295.1 235.5 177.6 

 16 124.1 129.7 135.2 136.6 134.5 131.5 126.7 

 17 118.8 123.3 128.7 130.7 129.9 128.4 125.2 

 18 118.6 123.1 128.4 130.4 129.7 128.3 125.1 

 19 111.4 114.5 119.6 122.5 123.4 124.3 123.1 

 20 111.2 114.3 119.3 122.3 123.2 124.1 123.1 

 21 130.8 137.4 142.9 143.7 140.4 135.8 128.8 

 22 133.5 140.4 145.8 146.4 142.7 137.7 129.7 

 23 203.9 216.1 216.0 215.1 202.6 188.7 155.1 

3-27
TABLE 3-12. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE 

PROBLEM 3 (Continued) 

 24 205.8 217.8 217.4 216.7 204.1 190.2 155.8 

 25 210.8 216.5 210.1 213.7 205.9 197.6 157.6 

 26 186.1 186.3 181.1 186.4 185.3 182.1 149.5 

 27 148.1 147.5 148.0 151.9 153.9 150.9 135.1 

 28 147.0 146.6 147.2 151.1 153.0 150.2 134.8 

 29 144.2 144.1 145.1 148.9 150.7 148.2 133.9 

 30 143.4 143.4 144.4 148.2 150.0 147.6 133.6 

 31 129.7 131.0 133.7 137.1 138.6 137.6 129.1 

 32 129.7 131.0 133.7 137.1 138.6 137.6 129.1 

 33 129.7 131.0 133.7 137.1 138.6 137.6 129.1 

 34 129.7 131.0 133.7 137.1 138.6 137.6 129.1 

 35 116.0 118.5 122.9 126.0 127.2 127.6 124.6 

 36 115.2 117.8 122.3 125.4 126.5 127.0 124.4 

 37 111.3 114.3 119.2 122.2 123.2 124.2 123.1 

 38 111.2 114.1 119.1 122.1 123.1 124.1 123.1 

 39 119.2 119.2 120.3 126.8 130.2 131.8 129.2 

 40 121.4 120.9 121.3 128.3 132.0 133.7 130.5 

 41 174.2 164.5 149.6 167.4 176.6 178.9 159.7 

 42 175.8 166.0 150.9 168.8 178.0 180.2 160.4 

 43 185.3 179.9 167.2 180.8 185.2 186.1 160.0 

 44 178.6 175.6 167.2 176.9 179.3 179.9 154.7 

 45 159.1 156.3 150.9 159.8 163.2 163.9 146.6 

 46 157.5 154.8 149.6 158.4 161.9 162.6 145.9 

 47 119.0 119.6 121.3 127.1 129.9 131.4 128.6 

 48 117.6 118.3 120.3 125.9 128.8 130.2 127.9 

 49 116.9 118.7 121.3 124.6 125.7 125.5 119.8 

 50 120.3 121.9 123.7 127.3 128.2 127.8 120.4 

 51 306.1 348.1 361.3 339.6 295.0 235.5 177.6 

 52 288.9 327.5 340.0 320.5 279.9 225.7 172.8 

 53 306.1 348.1 361.3 339.6 295.0 235.5 177.6 

 

Maximum Structural Temperature for Period 1 = 306.45 Occurring 

at Location 4 

Maximum Structural Temperature for Period 2 = 348.51 Occurring 

at Location 4 

Maximum Structural Temperature for Period 3 = 361.78 Occurring 

at Location 4 

Maximum Structural Temperature for Period 4 = 339.99 Occurring 

at Location 4 

Maximum Structural Temperature for Period 5 = 295.35 Occurring 

at Location 4 

Maximum Structural Temperature for Period 6 = 235.71 Occurring 

at Location 4 

Maximum Structural Temperature for Period 7 = 177.71 Occurring 

at Location 4 

Maximum Structural Temperature Over All 7 Periods =361.78 

Occurring at Location 4 at Period 3 

3-28
TABLE 3-13. THERMOD-SIMULATED TEMPERATURES (°F) OF THE RIGHT WING OF 

SAMPLE PROBLEM 3 

T1 T2 T3 T4 T5 T6 T7 T8 T9 T10

380.0 380.0 373.5 373.4 373.0 129.2 128.8 128.8 122.3 122.3

As in sample problems 1 and 2, a one-half wing structure was modeled using the 

UAI/NASTRAN FEM program. Boundary conditions were imposed based on the THERMODsimulated data for period 3. The wing top was constrained to 380.0°F, and the wing sides were 

constrained to 124.4°F, which happens to be the highest ambient temperature (time period 6 of 

table 3-11). Because the ratio of the wing length to its depth is sufficiently high, edge effects 

due to the wing sides (and the boundary temperature used) are insignificant, as the FEM results 

show. The bottom wing surface was free of any constraint, except that it was subjected to 

convection, with a film coefficient h = 1.3433 (Btu/hr)/(ft2

°F). This film coefficient was 

calculated internally by the THERMOD program and is based on the theory of airflow on a flat 

surface and a wind speed of 14 ft/sec. 

The FEM input file is shown in table 3-14. The boundary conditions were imposed by the SPC 

cards. An ambient temperature of 119.0°F (pertaining to period 3 of the THERMOD input) was 

applied through the scalar point, SPOINT, bulk data. The boundary elements, CHBDY, through 

which convection is realized, were modeled using the LINE elements connecting the nodes at the 

wing bottom. Property cards PHBDY and material cards MAT4 simulated the film coefficient 

and other element properties. 

TABLE 3-14. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE PROBLEM 3 

ID D:\Natha,MSC/N 

APP HEAT 

SOL 1 

TIME 10000 

CEND 

 TITLE = CONDUCTION & CONVECTION IN RECT. ADIABATIC PLATE (t=0.001ft) 

 SUBTITLE = 10.7 in air gap; k=0.016BTU/hr/ft/degF; h=1.3433BTU/hr/ft^2/degF 

 SPC = 1 

 THERMAL= ALL 

 FLUX = ALL 

 SPCF = ALL 

BEGIN BULK 

$ *************************************************************************** 

$ Written by : MSC/NASTRAN for Windows 

$ Version : 6.00 

$ Translator : UAI/NASTRAN 

$ From Model : D:\Thermod_Validation\FEM\conduction&convection.MOD 

$ Date : Tue Mar 07 08:27:39 2000 

$ *************************************************************************** 

$ 

PARAM,K6ROT,100. 

PARAM,MAXRATIO,1.E+8 

CORD2C 1 0 0. 0. 0. 0. 0. 1.+MSC/NC1

+MSC/NC1 1. 0. 1. 

CORD2S 2 0 0. 0. 0. 0. 0. 1.+MSC/NC2

+MSC/NC2 1. 0. 1. 

3-29
TABLE 3-14. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 3 (Continued) 

$ MSC/NASTRAN for Windows Constraint Set 1 : bound_temp&ambient constr 

SPC 1 37 1 124.4 

SPC 1 55 1 124.4 

SPC 1 56 1 124.4 

SPC 1 73 1 124.4 

SPC 1 109 1 124.4 

SPC 1 128 1 124.4 

SPC 1 145 1 124.4 

SPC 1 146 1 124.4 

SPC 1 164 1 124.4 

SPC 1 181 1 124.4 

SPC 1 182 1 124.4 

SPC 1 199 1 124.4 

SPC 1 200 1 124.4 

SPC 1 201 1 380.0 

SPC 1 202 1 380.0 

SPC 1 203 1 380.0 

SPC 1 204 1 380.0 

SPC 1 205 1 380.0 

SPC 1 206 1 380.0 

SPC 1 207 1 380.0 

SPC 1 208 1 380.0 

SPC 1 209 1 380.0 

SPC 1 210 1 380.0 

SPC 1 211 1 380.0 

SPC 1 212 1 380.0 

SPC 1 213 1 380.0 

SPC 1 214 1 380.0 

SPC 1 215 1 380.0 

SPC 1 216 1 380.0 

SPC 1 235 1 124.4 

SPC 1 253 1 124.4 

SPC 1 254 1 124.4 

SPC 1 307 1 124.4 

SPC 1 308 1 124.4 

spc,1,500,1,119.0 

spoint,500 

chbdy,1000,100,line,1,36,,,,500,500 

chbdy,1001,100,line,36,35,,,,500,500 

chbdy,1002,100,line,35,34,,,,500,500 

chbdy,1003,100,line,34,33,,,,500,500 

chbdy,1004,100,line,33,32,,,,500,500 

chbdy,1005,100,line,32,31,,,,500,500 

chbdy,1006,100,line,31,30,,,,500,500 

chbdy,1007,100,line,30,29,,,,500,500 

chbdy,1008,100,line,29,28,,,,500,500 

chbdy,1009,100,line,28,27,,,,500,500 

chbdy,1010,100,line,27,26,,,,500,500 

chbdy,1011,100,line,26,25,,,,500,500 

chbdy,1012,100,line,25,24,,,,500,500 

chbdy,1013,100,line,24,23,,,,500,500 

chbdy,1014,100,line,23,22,,,,500,500 

chbdy,1015,100,line,22,21,,,,500,500 

chbdy,1016,100,line,21,20,,,,500,500 

phbdy,100,200,0.001 

mat4,200,1.3433 

$mat4,200,0.0001 

3-30
TABLE 3-14. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 3 (Continued) 

$ 

$ MSC/NASTRAN for Windows Property 1 : 0.02 in glass_outer_layer 

PSHELL 1 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 2 : 3/8 in foam 

PSHELL 2 2 0.001 2 2 0. 

$ MSC/NASTRAN for Windows Property 3 : 0.02 in inner glass 

PSHELL 3 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 4 : 0.350 in cap 

PSHELL 4 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 5 : 10.70 in air space 

PSHELL 5 3 0.001 3 3 0. 

$ MSC/NASTRAN for Windows Property 6 : 0.350 in cap 

PSHELL 6 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 7 : 0.02 in inner glass 

PSHELL 7 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 8 : 3/8 in foam 

PSHELL 8 2 0.001 2 2 0. 

$ MSC/NASTRAN for Windows Property 9 : 0.02 in glass_outer_layer 

PSHELL 9 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 10 : 10.70 in oil 

PSHELL 10 4 0.001 4 4 0. 

$ MSC/NASTRAN for Windows Material 1 : glass 

MAT4 1 0.3 1. 

$ MSC/NASTRAN for Windows Material 2 : foam 

MAT4 2 0.021 1. 

$ MSC/NASTRAN for Windows Material 3 : air 

MAT4 3 0.016 1. 

$ MSC/NASTRAN for Windows Material 4 : oil 

MAT4 4 0.081 1. 

GRID 1 0 0. 0. 0. 0 

GRID 3 0 1.1.667E-3 0. 0 

GRID 5 0 3.1.667E-3 0. 0 

GRID 7 0 5.1.667E-3 0. 0 

GRID 9 0 7.1.667E-3 0. 0 

GRID 10 0 8.1.667E-3 0. 0 

GRID 11 0 9.1.667E-3 0. 0 

GRID 12 0 10.1.667E-3 0. 0 

GRID 14 0 12.1.667E-3 0. 0 

GRID 15 0 13.1.667E-3 0. 0 

GRID 18 0 16.1.667E-3 0. 0 

GRID 20 0 17. 0. 0. 0 

GRID 21 0 16. 0. 0. 0 

GRID 22 0 15. 0. 0. 0 

GRID 23 0 14. 0. 0. 0 

GRID 24 0 13. 0. 0. 0 

GRID 25 0 12. 0. 0. 0 

GRID 26 0 11. 0. 0. 0 

GRID 27 0 10. 0. 0. 0 

GRID 28 0 9. 0. 0. 0 

GRID 29 0 8. 0. 0. 0 

GRID 30 0 7. 0. 0. 0 

GRID 31 0 6. 0. 0. 0 

GRID 32 0 5. 0. 0. 0 

GRID 33 0 4. 0. 0. 0 

GRID 34 0 3. 0. 0. 0 

GRID 35 0 2. 0. 0. 0 

GRID 36 0 1. 0. 0. 0 

3-31
TABLE 3-14. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 3 (Continued) 

GRID 37 0 0.1.667E-3 0. 0 

GRID 41 0 3.0.032917 0. 0 

GRID 44 0 6.0.032917 0. 0 

GRID 45 0 7.0.032917 0. 0 

GRID 47 0 9.0.032917 0. 0 

GRID 48 0 10.0.032917 0. 0 

GRID 50 0 12.0.032917 0. 0 

GRID 51 0 13.0.032917 0. 0 

GRID 52 0 14.0.032917 0. 0 

GRID 53 0 15.0.032917 0. 0 

GRID 54 0 16.0.032917 0. 0 

GRID 55 0 17.0.032917 0. 0 

GRID 56 0 17.1.667E-3 0. 0 

GRID 58 0 15.1.667E-3 0. 0 

GRID 59 0 14.1.667E-3 0. 0 

GRID 62 0 11.1.667E-3 0. 0 

GRID 67 0 6.1.667E-3 0. 0 

GRID 69 0 4.1.667E-3 0. 0 

GRID 71 0 2.1.667E-3 0. 0 

GRID 73 0 0.0.032917 0. 0 

GRID 75 0 1.0.034583 0. 0 

GRID 76 0 2.0.034583 0. 0 

GRID 77 0 3.0.034583 0. 0 

GRID 78 0 4.0.034583 0. 0 

GRID 79 0 5.0.034583 0. 0 

GRID 80 0 6.0.034583 0. 0 

GRID 83 0 9.0.034583 0. 0 

GRID 85 0 11.0.034583 0. 0 

GRID 87 0 13.0.034583 0. 0 

GRID 88 0 14.0.034583 0. 0 

GRID 98 0 11.0.032917 0. 0 

GRID 101 0 8.0.032917 0. 0 

GRID 104 0 5.0.032917 0. 0 

GRID 105 0 4.0.032917 0. 0 

GRID 107 0 2.0.032917 0. 0 

GRID 108 0 1.0.032917 0. 0 

GRID 109 0 0.0.034583 0. 0 

GRID 112 0 2. 0.06375 0. 0 

GRID 115 0 5. 0.06375 0. 0 

GRID 116 0 6. 0.06375 0. 0 

GRID 118 0 8. 0.06375 0. 0 

GRID 120 0 10. 0.06375 0. 0 

GRID 121 0 11. 0.06375 0. 0 

GRID 122 0 12. 0.06375 0. 0 

GRID 123 0 13. 0.06375 0. 0 

GRID 124 0 14. 0.06375 0. 0 

GRID 125 0 15. 0.06375 0. 0 

GRID 128 0 17.0.034583 0. 0 

GRID 129 0 16.0.034583 0. 0 

GRID 130 0 15.0.034583 0. 0 

GRID 133 0 12.0.034583 0. 0 

GRID 135 0 10.0.034583 0. 0 

GRID 137 0 8.0.034583 0. 0 

GRID 138 0 7.0.034583 0. 0 

GRID 145 0 0. 0.06375 0. 0 

GRID 146 0 0. 0.95542 0. 0 

GRID 147 0 1. 0.95542 0. 0 

3-32
TABLE 3-14. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 3 (Continued) 

GRID 150 0 4. 0.95542 0. 0 

GRID 152 0 6. 0.95542 0. 0 

GRID 160 0 14. 0.95542 0. 0 

GRID 161 0 15. 0.95542 0. 0 

GRID 164 0 17. 0.06375 0. 0 

GRID 165 0 16. 0.06375 0. 0 

GRID 172 0 9. 0.06375 0. 0 

GRID 174 0 7. 0.06375 0. 0 

GRID 177 0 4. 0.06375 0. 0 

GRID 178 0 3. 0.06375 0. 0 

GRID 180 0 1. 0.06375 0. 0 

GRID 181 0 0. 1.01917 0. 0 

GRID 182 0 0. 1.0175 0. 0 

GRID 183 0 1. 1.0175 0. 0 

GRID 184 0 2. 1.0175 0. 0 

GRID 188 0 6. 1.0175 0. 0 

GRID 193 0 11. 1.0175 0. 0 

GRID 195 0 13. 1.0175 0. 0 

GRID 196 0 14. 1.0175 0. 0 

GRID 197 0 15. 1.0175 0. 0 

GRID 198 0 16. 1.0175 0. 0 

GRID 199 0 17. 1.0175 0. 0 

GRID 200 0 17. 1.01917 0. 0 

GRID 201 0 16. 1.01917 0. 0 

GRID 202 0 15. 1.01917 0. 0 

GRID 203 0 14. 1.01917 0. 0 

GRID 204 0 13. 1.01917 0. 0 

GRID 205 0 12. 1.01917 0. 0 

GRID 206 0 11. 1.01917 0. 0 

GRID 207 0 10. 1.01917 0. 0 

GRID 208 0 9. 1.01917 0. 0 

GRID 209 0 8. 1.01917 0. 0 

GRID 210 0 7. 1.01917 0. 0 

GRID 211 0 6. 1.01917 0. 0 

GRID 212 0 5. 1.01917 0. 0 

GRID 213 0 4. 1.01917 0. 0 

GRID 214 0 3. 1.01917 0. 0 

GRID 215 0 2. 1.01917 0. 0 

GRID 216 0 1. 1.01917 0. 0 

GRID 219 0 1. 0.98625 0. 0 

GRID 220 0 2. 0.98625 0. 0 

GRID 221 0 3. 0.98625 0. 0 

GRID 222 0 4. 0.98625 0. 0 

GRID 223 0 5. 0.98625 0. 0 

GRID 226 0 8. 0.98625 0. 0 

GRID 227 0 9. 0.98625 0. 0 

GRID 228 0 10. 0.98625 0. 0 

GRID 229 0 11. 0.98625 0. 0 

GRID 231 0 13. 0.98625 0. 0 

GRID 232 0 14. 0.98625 0. 0 

GRID 235 0 17. 0.98625 0. 0 

GRID 241 0 12. 1.0175 0. 0 

GRID 243 0 10. 1.0175 0. 0 

GRID 244 0 9. 1.0175 0. 0 

GRID 245 0 8. 1.0175 0. 0 

GRID 246 0 7. 1.0175 0. 0 

GRID 248 0 5. 1.0175 0. 0 

3-33
TABLE 3-14. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 3 (Continued) 

GRID 249 0 4. 1.0175 0. 0 

GRID 250 0 3. 1.0175 0. 0 

GRID 253 0 0. 0.98625 0. 0 

GRID 254 0 0. 0.98458 0. 0 

GRID 257 0 3. 0.98458 0. 0 

GRID 258 0 4. 0.98458 0. 0 

GRID 259 0 5. 0.98458 0. 0 

GRID 260 0 6. 0.98458 0. 0 

GRID 264 0 10. 0.98458 0. 0 

GRID 267 0 13. 0.98458 0. 0 

GRID 268 0 14. 0.98458 0. 0 

GRID 269 0 15. 0.98458 0. 0 

GRID 270 0 16. 0.98458 0. 0 

GRID 273 0 16. 0.98625 0. 0 

GRID 274 0 15. 0.98625 0. 0 

GRID 277 0 12. 0.98625 0. 0 

GRID 282 0 7. 0.98625 0. 0 

GRID 283 0 6. 0.98625 0. 0 

GRID 292 0 2. 0.95542 0. 0 

GRID 293 0 3. 0.95542 0. 0 

GRID 295 0 5. 0.95542 0. 0 

GRID 297 0 7. 0.95542 0. 0 

GRID 298 0 8. 0.95542 0. 0 

GRID 299 0 9. 0.95542 0. 0 

GRID 300 0 10. 0.95542 0. 0 

GRID 301 0 11. 0.95542 0. 0 

GRID 302 0 12. 0.95542 0. 0 

GRID 303 0 13. 0.95542 0. 0 

GRID 306 0 16. 0.95542 0. 0 

GRID 307 0 17. 0.95542 0. 0 

GRID 308 0 17. 0.98458 0. 0 

GRID 313 0 12. 0.98458 0. 0 

GRID 314 0 11. 0.98458 0. 0 

GRID 316 0 9. 0.98458 0. 0 

GRID 317 0 8. 0.98458 0. 0 

GRID 318 0 7. 0.98458 0. 0 

GRID 323 0 2. 0.98458 0. 0 

GRID 324 0 1. 0.98458 0. 0 

CQUAD4 1 9 1 37 3 36 

CQUAD4 2 9 36 3 71 35 

CQUAD4 3 9 35 71 5 34 

CQUAD4 4 9 34 5 69 33 

CQUAD4 5 9 33 69 7 32 

CQUAD4 6 9 32 7 67 31 

CQUAD4 7 9 31 67 9 30 

CQUAD4 8 9 30 9 10 29 

CQUAD4 9 9 29 10 11 28 

CQUAD4 10 9 28 11 12 27 

CQUAD4 11 9 27 12 62 26 

CQUAD4 12 9 26 62 14 25 

CQUAD4 13 9 25 14 15 24 

CQUAD4 14 9 24 15 59 23 

CQUAD4 15 9 23 59 58 22 

CQUAD4 16 9 22 58 18 21 

CQUAD4 17 9 21 18 56 20 

CQUAD4 18 8 37 73 108 3 

CQUAD4 19 8 3 108 107 71 

3-34
TABLE 3-14. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 3 (Continued) 

CQUAD4 20 8 71 107 41 5 

CQUAD4 21 8 5 41 105 69 

CQUAD4 22 8 69 105 104 7 

CQUAD4 23 8 7 104 44 67 

CQUAD4 24 8 67 44 45 9 

CQUAD4 25 8 9 45 101 10 

CQUAD4 26 8 10 101 47 11 

CQUAD4 27 8 11 47 48 12 

CQUAD4 28 8 12 48 98 62 

CQUAD4 29 8 62 98 50 14 

CQUAD4 30 8 14 50 51 15 

CQUAD4 31 8 15 51 52 59 

CQUAD4 32 8 59 52 53 58 

CQUAD4 33 8 58 53 54 18 

CQUAD4 34 8 18 54 55 56 

CQUAD4 35 7 73 109 75 108 

CQUAD4 36 7 108 75 76 107 

CQUAD4 37 7 107 76 77 41 

CQUAD4 38 7 41 77 78 105 

CQUAD4 39 7 105 78 79 104 

CQUAD4 40 7 104 79 80 44 

CQUAD4 41 7 44 80 138 45 

CQUAD4 42 7 45 138 137 101 

CQUAD4 43 7 101 137 83 47 

CQUAD4 44 7 47 83 135 48 

CQUAD4 45 7 48 135 85 98 

CQUAD4 46 7 98 85 133 50 

CQUAD4 47 7 50 133 87 51 

CQUAD4 48 7 51 87 88 52 

CQUAD4 49 7 52 88 130 53 

CQUAD4 50 7 53 130 129 54 

CQUAD4 51 7 54 129 128 55 

CQUAD4 52 6 109 145 180 75 

CQUAD4 53 6 75 180 112 76 

CQUAD4 54 6 76 112 178 77 

CQUAD4 55 6 77 178 177 78 

CQUAD4 56 6 78 177 115 79 

CQUAD4 57 6 79 115 116 80 

CQUAD4 58 6 80 116 174 138 

CQUAD4 59 6 138 174 118 137 

CQUAD4 60 6 137 118 172 83 

CQUAD4 61 6 83 172 120 135 

CQUAD4 62 6 135 120 121 85 

CQUAD4 63 6 85 121 122 133 

CQUAD4 64 6 133 122 123 87 

CQUAD4 65 6 87 123 124 88 

CQUAD4 66 6 88 124 125 130 

CQUAD4 67 6 130 125 165 129 

CQUAD4 68 6 129 165 164 128 

CQUAD4 69 5 145 146 147 180 

CQUAD4 70 5 180 147 292 112 

CQUAD4 71 5 112 292 293 178 

CQUAD4 72 5 178 293 150 177 

CQUAD4 73 5 177 150 295 115 

CQUAD4 74 5 115 295 152 116 

CQUAD4 75 5 116 152 297 174 

CQUAD4 76 5 174 297 298 118 

3-35
TABLE 3-14. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 3 (Continued) 

CQUAD4 77 5 118 298 299 172 

CQUAD4 78 5 172 299 300 120 

CQUAD4 79 5 120 300 301 121 

CQUAD4 80 5 121 301 302 122 

CQUAD4 81 5 122 302 303 123 

CQUAD4 82 5 123 303 160 124 

CQUAD4 83 5 124 160 161 125 

CQUAD4 84 5 125 161 306 165 

CQUAD4 85 5 165 306 307 164 

CQUAD4 86 1 182 181 216 183 

CQUAD4 87 1 183 216 215 184 

CQUAD4 88 1 184 215 214 250 

CQUAD4 89 1 250 214 213 249 

CQUAD4 90 1 249 213 212 248 

CQUAD4 91 1 248 212 211 188 

CQUAD4 92 1 188 211 210 246 

CQUAD4 93 1 246 210 209 245 

CQUAD4 94 1 245 209 208 244 

CQUAD4 95 1 244 208 207 243 

CQUAD4 96 1 243 207 206 193 

CQUAD4 97 1 193 206 205 241 

CQUAD4 98 1 241 205 204 195 

CQUAD4 99 1 195 204 203 196 

CQUAD4 100 1 196 203 202 197 

CQUAD4 101 1 197 202 201 198 

CQUAD4 102 1 198 201 200 199 

CQUAD4 103 2 253 182 183 219 

CQUAD4 104 2 219 183 184 220 

CQUAD4 105 8 220 184 250 221 

CQUAD4 106 2 221 250 249 222 

CQUAD4 107 2 222 249 248 223 

CQUAD4 108 2 223 248 188 283 

CQUAD4 109 2 283 188 246 282 

CQUAD4 110 2 282 246 245 226 

CQUAD4 111 2 226 245 244 227 

CQUAD4 112 2 227 244 243 228 

CQUAD4 113 2 228 243 193 229 

CQUAD4 114 2 229 193 241 277 

CQUAD4 115 2 277 241 195 231 

CQUAD4 116 2 231 195 196 232 

CQUAD4 117 2 232 196 197 274 

CQUAD4 118 2 274 197 198 273 

CQUAD4 119 2 273 198 199 235 

CQUAD4 120 3 254 253 219 324 

CQUAD4 121 3 324 219 220 323 

CQUAD4 122 3 323 220 221 257 

CQUAD4 123 3 257 221 222 258 

CQUAD4 124 3 258 222 223 259 

CQUAD4 125 3 259 223 283 260 

CQUAD4 126 3 260 283 282 318 

CQUAD4 127 3 318 282 226 317 

CQUAD4 128 3 317 226 227 316 

CQUAD4 129 3 316 227 228 264 

CQUAD4 130 3 264 228 229 314 

CQUAD4 131 3 314 229 277 313 

CQUAD4 132 3 313 277 231 267 

CQUAD4 133 3 267 231 232 268 

3-36
TABLE 3-14. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 3 (Continued) 

CQUAD4 134 3 268 232 274 269 

CQUAD4 135 3 269 274 273 270 

CQUAD4 136 3 270 273 235 308 

CQUAD4 137 4 146 254 324 147 

CQUAD4 138 4 147 324 323 292 

CQUAD4 139 4 292 323 257 293 

CQUAD4 140 4 293 257 258 150 

CQUAD4 141 4 150 258 259 295 

CQUAD4 142 4 295 259 260 152 

CQUAD4 143 4 152 260 318 297 

CQUAD4 144 4 297 318 317 298 

CQUAD4 145 4 298 317 316 299 

CQUAD4 146 4 299 316 264 300 

CQUAD4 147 4 300 264 314 301 

CQUAD4 148 4 301 314 313 302 

CQUAD4 149 4 302 313 267 303 

CQUAD4 150 4 303 267 268 160 

CQUAD4 151 4 160 268 269 161 

CQUAD4 152 4 161 269 270 306 

CQUAD4 153 4 306 270 308 307 

ENDDATA 

The results of the one-half FEM wing model analysis are presented in figure 3-3 in the form of a 

contour plot. The contours along the outer nodes are in agreement with the imposed boundary 

conditions, while the contours of the inner layers appear to be indicative of the temperature 

profile of the THERMOD analysis. Also note the surface boundary conditions enforced on the 

top and sides of the wing. 

FIGURE 3-3. CONTOUR PLOT OF THE FEM OUTPUT OF SAMPLE PROBLEM 3 

3-37
The truncated FEM output file is shown in table 3-15. The nodes, from top to bottom of the 

wing, located at mid-span of the model are indicated by nodes 209, 245, 226, 317, 298, 118, 137, 

101, 10, and 29. The temperatures of these nodes were extracted from this file and are 

reproduced in table 3-16. Mid-span was chosen because it was sufficiently far away from any 

edge effects that might be present. 

TABLE 3-15. TRUNCATED FEM OUTPUT FILE OF SAMPLE PROBLEM 3 

1 CONDUCTION & CONVECTION IN RECT. ADIABATIC PLATE (T=0.001FT) AUGUST 15, 2000 UAI/NASTRAN VERSION 20.1 PAGE 

16 

 10.7 IN AIR GAP; K=0.016BTU/HR/FT/DEGF; H=1.3433BTU/HR/FT^2/DEGF 

0 

 

 T E M P E R A T U R E V E C T O R 

 

 POINT ID. TYPE ID VALUE ID+1 VALUE ID+2 VALUE ID+3 VALUE ID+4 VALUE ID+5 VALUE 

 1 TEMP 1.243670E+02 

 3 TEMP 1.215624E+02 

 5 TEMP 1.222429E+02 

 7 TEMP 1.222786E+02 

 9 TEMP 1.222806E+02 1.222807E+02 1.222807E+02 1.222806E+02 

 14 TEMP 1.222786E+02 1.222896E+02 

 18 TEMP 1.215624E+02 

 20 TEMP 1.243670E+02 1.215388E+02 1.224165E+02 1.222185E+02 1.222653E+02 1.222543E+02 

 26 TEMP 1.222569E+02 1.222563E+02 1.222564E+02 1.222564E+02 1.222563E+02 1.222569E+02 

 32 TEMP 1.222543E+02 1.222653E+02 1.222185E+02 1.224165E+02 1.215388E+02 1.244000E+02 

 41 TEMP 1.287578E+02 

 44 TEMP 1.287905E+02 1.287900E+02 

 47 TEMP 1.287901E+02 1.287900E+02 

 50 TEMP 1.287883E+02 1.287975E+02 1.287578E+02 1.289130E+02 1.278674E+02 1.244000E+02 

 56 TEMP 1.244000E+02 

 58 TEMP 1.224406E+02 1.222429E+02 

 62 TEMP 1.222812E+02 

 67 TEMP 1.222812E+02 

 69 TEMP 1.222896E+02 

 71 TEMP 1.224406E+02 

 73 TEMP 1.244000E+02 

 75 TEMP 1.278909E+02 1.289371E+02 1.287821E+02 1.288218E+02 1.288126E+02 1.288148E+02 

 83 TEMP 1.288144E+02 

 85 TEMP 1.288148E+02 

 87 TEMP 1.288218E+02 1.287821E+02 

 98 TEMP 1.287905E+02 

 101 TEMP 1.287901E+02 

 104 TEMP 1.287883E+02 1.287975E+02 

 107 TEMP 1.289130E+02 1.278674E+02 1.244000E+02 

 112 TEMP 1.293606E+02 

 115 TEMP 1.292379E+02 1.292401E+02 

 118 TEMP 1.292397E+02 

 120 TEMP 1.292395E+02 1.292401E+02 1.292379E+02 1.292470E+02 1.292076E+02 1.293606E+02 

 128 TEMP 1.244000E+02 1.278909E+02 1.289371E+02 

 133 TEMP 1.288126E+02 

 135 TEMP 1.288143E+02 

 137 TEMP 1.288144E+02 1.288143E+02 

 145 TEMP 1.244000E+02 1.244000E+02 3.669625E+02 

 150 TEMP 3.730168E+02 

 152 TEMP 3.730167E+02 

 160 TEMP 3.730125E+02 3.728751E+02 

 164 TEMP 1.244000E+02 1.283041E+02 

 172 TEMP 1.292397E+02 

 174 TEMP 1.292395E+02 

 177 TEMP 1.292470E+02 1.292076E+02 

 180 TEMP 1.283041E+02 1.244000E+02 1.244000E+02 3.799533E+02 3.799752E+02 

 188 TEMP 3.799756E+02 

 193 TEMP 3.799756E+02 

 195 TEMP 3.799756E+02 3.799756E+02 3.799752E+02 3.799533E+02 1.244000E+02 1.244000E+02 

 201 TEMP 3.800000E+02 3.800000E+02 3.800000E+02 3.800000E+02 3.800000E+02 3.800000E+02 

1 CONDUCTION & CONVECTION IN RECT. ADIABATIC PLATE (T=0.001FT) AUGUST 15, 2000 UAI/NASTRAN VERSION 20.1 PAGE 

17 

 10.7 IN AIR GAP; K=0.016BTU/HR/FT/DEGF; H=1.3433BTU/HR/FT^2/DEGF 

0 

 

 T E M P E R A T U R E V E C T O R 

 

 POINT ID. TYPE ID VALUE ID+1 VALUE ID+2 VALUE ID+3 VALUE ID+4 VALUE ID+5 VALUE 

 207 TEMP 3.800000E+02 3.800000E+02 3.800000E+02 3.800000E+02 3.800000E+02 3.800000E+02 

 213 TEMP 3.800000E+02 3.800000E+02 3.800000E+02 3.800000E+02 

 219 TEMP 3.676886E+02 3.733311E+02 3.734622E+02 3.734663E+02 3.734662E+02 

 226 TEMP 3.734662E+02 3.734662E+02 3.734662E+02 3.734663E+02 

 231 TEMP 3.734663E+02 3.734622E+02 

 235 TEMP 1.244000E+02 

 241 TEMP 3.799756E+02 

 243 TEMP 3.799756E+02 3.799756E+02 3.799756E+02 3.799756E+02 

 248 TEMP 3.799756E+02 3.799756E+02 3.799756E+02 

 253 TEMP 1.244000E+02 1.244000E+02 

 257 TEMP 3.734379E+02 3.734420E+02 3.734419E+02 3.734419E+02 

 264 TEMP 3.734419E+02 

3-38
TABLE 3-15. TRUNCATED FEM OUTPUT FILE OF SAMPLE PROBLEM 3 (Continued) 

 267 TEMP 3.734420E+02 3.734379E+02 3.733062E+02 3.676435E+02 

 273 TEMP 3.676886E+02 3.733311E+02 

 277 TEMP 3.734662E+02 

 282 TEMP 3.734662E+02 3.734663E+02 

 292 TEMP 3.728751E+02 3.730125E+02 

 295 TEMP 3.730167E+02 

 297 TEMP 3.730167E+02 3.730167E+02 3.730167E+02 3.730167E+02 3.730167E+02 3.730167E+02 

 303 TEMP 3.730168E+02 

 306 TEMP 3.669625E+02 1.244000E+02 1.244000E+02 

 313 TEMP 3.734419E+02 3.734419E+02 

 316 TEMP 3.734419E+02 3.734419E+02 3.734419E+02 

 323 TEMP 3.733062E+02 3.676435E+02 

 500 TEMP 1.190000E+02 

When compared, the temperatures of tables 3-13 and 3-16 show excellent agreement, validating 

the conduction and convection capabilities of the THERMOD program. 

3.4 SAMPLE PROBLEM 4.

The primary purpose of sample problem 4 was to study the effects of solar radiation on an 

aircraft. As in sample problems 1 and 2, THERMOD was set up to solve a complete aircraft

problem, while the FEM method was confined to the wing model. 

• The direct solar radiation rays are in the visible regions of the energy spectrum, have a 

fixed incident angle, are of short wavelengths, and are characterized by high energy 

levels. 

• The infrared radiation falls in the nonvisible region of the energy spectrum. It is of long 

wavelength and characterized by low energy level. 

In addition, all external surfaces, except the wing surface, were assigned absorptivity and 

emissivity values of 1.0 and 0.0, respectively. The values for the wing surface were 0.9 and 0.0, 

respectively. A 0.0 emissivity value prevents the emission of infrared energy, thereby removing 

its influence on the thermal system. For the same reason, the sky radiation was assigned an 

absolute temperature of 0.0. The conduction and convection heat transfer modes, however, were 

left intact, as in sample problem 3. The THERMOD input file, input.dat, is illustrated in 

table 3-17. 

TABLE 3-16. FINITE ELEMENT METHOD-SIMULATED TEMPERATURES (°F) OF THE 

RIGHT WING OF SAMPLE PROBLEM 3 

T1 T2 T3 T4 T5 T6 T7 T8 T9 T10

380.0 380.0 373.5 373.4 373.0 129.2 128.8 128.8 122.3 122.3

An aircraft parked in the open is subjected to three types of radiation: direct solar radiation, 

diffused solar radiation, and infrared radiation. 

• The diffused solar radiation is also in the visible region of the energy spectrum but does 

not have a fixed incident angle. It is diffused in nature. 

• The direct solar radiation type is explored in sample problem 4. 

3-39
TABLE 3-17. THERMOD INPUT FILE (input.dat) OF SAMPLE PROBLEM 4 

198e-6,0.704 

1, 0.020,125,0.300,0.30 

2, 0.375,4.4,0.021,0.24 

3, 0.020,125,0.300,0.30 

4, 0.350,125,0.300,0.30 

5, 10.70,0.067,0.016,0.24 

6, 0.350,125,0.300,0.30 

7, 0.020,125,0.300,0.30 

8, 0.375,4.4,0.021,0.24 

9, 0.020,125,0.300,0.30 

10,0.030,125,0.300,0.30 

11,0.375,4.4,0.021,0.24 

12,0.020,125,0.300,0.30 

13,0.250,10.,0.020,0.24 

14,48.00,0.067,0.016,0.24 

15,0.020,125,0.300,0.30 

16,0.375,4.4,0.021,0.24 

17,0.020,125,0.300,0.30 

18,0.250,10.,0.020,0.24 

19,30.00,0.067,0.016,0.24 

20,5.000,10.,0.020,0.24 

21,0.020,125,0.300,0.30 

22,0.375,4.4,0.021,0.24 

23,0.020,125,0.300,0.30 

24,2.500,0.067,0.016,0.24 

25,0.020,125,0.300,0.30 

26,0.375,4.4,0.021,0.24 

27,0.020,125,0.300,0.30 

28,0.250,130,0.800,0.20 

3.5,34.0,2.5,5.5,22.0,11.0,3.0,4.0,30,24,28 

0.70,1.0,0.0,2.0,150,0.80,0.22 

0.3,0.9,0.3,0.9,0.3,0.9,0.1,0.9 

3,100 

0.9,0.9,0.0,0.0 

1.0,0.0,0.9,0.0,1.0,0.0,1.0,0.0 

9 

0,10,10,0,0,75,110,110,190,190 

5,120,5,60,20,5,120,60,60 

7 

60 570.67 450.00 14 330 10 10 0.36 

75 573.67 0.0000 14 355 10 10 0.31 

90 578.67 0.0000 14 355 10 12 0.33 

75 581.67 0.0000 14 330 10 10 0.31 

60 582.67 0.0000 14 291 10 10 0.36 

45 583.67 0.0000 14 231 10 10 0.44 

30 582.67 0.0000 14 160 10 10 0.57 

3-40
TABLE 3-17. THERMOD INPUT FILE (input.dat) OF SAMPLE PROBLEM 4 (Continued) 

2,100,180,1 

10,1.0,1.0 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10 

TABLE 3-18. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE PROBLEM 4 

Temperatures from the THERMOD run is shown in table 3-18, from which the steady-state 

temperatures T1, T2.....T10 (top to bottom of wing) for time period 3 were extracted and 

tabulated in table 3-19. 

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ 

THERMOD: A THERMAL PROGRAM FOR AIRCRAFTS 

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ 

 

 

 

 SUMMARY OF STEADY-STATE AND TRANSIENT ANALYSIS 

 

 

 

 (Temperatures are shown for all time periods) 

 

 

 TEMPERATURES AT THE END OF STEADY STATE ANALYSIS 

 

 TEMPERATURES IN DEGREES FAHRENHEIT 

 1 2 3 4 5 6 7 

 

 1 300.1 340.9 353.9 332.9 289.8 232.1 175.9 

 2 300.1 340.9 353.9 332.9 289.7 232.1 175.9 

 3 295.4 335.2 348.0 327.6 285.6 229.4 174.6 

 4 295.4 335.2 348.0 327.6 285.6 229.4 174.6 

 5 295.1 334.8 347.6 327.3 285.3 229.2 174.5 

 6 118.4 122.9 128.2 130.3 129.5 128.2 125.1 

 7 118.1 122.5 127.8 129.9 129.3 128.1 125.0 

 8 118.1 122.5 127.8 129.9 129.3 128.1 125.0 

 9 113.4 116.9 122.0 124.6 125.1 125.4 123.7 

 10 113.4 116.8 121.9 124.6 125.1 125.3 123.7 

 11 300.1 340.9 353.9 332.9 289.8 232.1 175.9 

 12 300.1 340.9 353.9 332.9 289.7 232.1 175.9 

 13 295.4 335.2 348.0 327.6 285.6 229.4 174.6 

 14 295.4 335.2 348.0 327.6 285.6 229.4 174.6 

3-41
TABLE 3-18. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE 

PROBLEM 4 (Continued) 

 15 295.1 334.8 347.6 327.3 285.3 229.2 174.5 

 16 118.4 122.9 128.2 130.3 129.5 128.2 125.1 

 17 118.1 122.5 127.8 129.9 129.3 128.1 125.0 

 18 118.1 122.5 127.8 129.9 129.3 128.1 125.0 

 19 113.4 116.9 122.0 124.6 125.1 125.4 123.7 

 20 113.4 116.8 121.9 124.6 125.1 125.3 123.7 

 21 277.3 306.8 316.0 300.7 267.3 220.2 169.5 

 22 277.2 306.6 315.8 300.5 267.2 220.2 169.5 

 23 244.7 251.8 252.7 248.1 232.9 204.4 160.4 

 24 244.6 251.6 252.5 247.9 232.8 204.3 160.4 

 25 221.9 213.2 208.4 211.3 208.8 193.3 154.0 

 26 194.0 172.3 166.4 172.1 175.6 167.8 132.5 

 27 147.9 139.9 140.0 144.2 146.4 143.5 127.2 

 28 147.9 139.9 140.0 144.2 146.4 143.4 127.2 

 29 144.6 137.6 138.2 142.2 144.3 141.7 126.8 

 30 144.6 137.5 138.2 142.2 144.3 141.7 126.8 

 31 130.1 127.4 129.9 133.5 135.1 134.1 125.2 

 32 130.1 127.4 129.9 133.5 135.1 134.1 125.2 

 33 130.1 127.4 129.9 133.5 135.1 134.1 125.2 

 34 130.1 127.4 129.9 133.5 135.1 134.1 125.2 

 35 115.7 117.3 121.7 124.8 126.0 126.5 123.5 

 36 115.7 117.3 121.7 124.8 126.0 126.5 123.5 

 37 112.4 115.0 119.8 122.8 123.9 124.7 123.2 

 38 112.4 115.0 119.8 122.8 123.9 124.7 123.2 

 39 219.3 175.9 125.8 178.9 215.2 225.3 205.4 

 40 219.2 175.9 125.9 178.9 215.1 225.2 205.2 

 41 205.0 171.8 142.0 173.4 194.6 197.1 173.1 

 42 204.9 171.8 142.1 173.4 194.5 197.0 173.0 

 43 194.9 168.9 153.4 169.6 180.1 177.3 150.4 

 44 186.3 163.3 153.4 164.4 172.1 169.1 143.4 

 45 161.6 147.1 142.1 150.5 156.0 154.3 136.7 

 46 161.4 147.0 142.0 150.4 155.9 154.2 136.7 

 47 126.1 123.9 125.9 130.5 132.9 133.1 127.1 

 48 125.9 123.8 125.8 130.4 132.8 133.0 127.1 

 49 124.3 92.5 91.1 95.6 98.2 95.5 81.8 

 50 126.8 95.1 93.4 98.1 100.9 97.9 83.4 

 51 188.8 207.4 215.7 208.8 191.6 168.5 144.8 

 52 183.9 201.5 209.6 203.3 187.3 165.7 143.4 

 53 188.8 207.4 215.7 208.8 191.6 168.5 144.8 

 

Maximum Structural Temperature for Period 1 = 300.11 Occurring 

at Location 1 

Maximum Structural Temperature for Period 2 = 340.91 Occurring 

at Location 1 

Maximum Structural Temperature for Period 3 = 353.91 Occurring 

at Location 1 

3-42
TABLE 3-18. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE 

PROBLEM 4 (Continued) 

Maximum Structural Temperature for Period 4 = 332.93 Occurring 

at Location 1 

Maximum Structural Temperature for Period 5 = 289.76 Occurring 

at Location 1 

Maximum Structural Temperature for Period 6 = 232.09 Occurring 

at Location 1 

Maximum Structural Temperature for Period 7 = 205.42 Occurring 

at Location 39 

 

Maximum Structural Temperature Over All 7 Periods =353.91 

Occurring at Location 1 at Period 3 

 

 

 TEMPERATURES AT THE END OF TRANSIENT ANALYSIS 

 

 TEMPERATURES IN DEGREES FAHRENHEIT 

 1 2 3 4 5 6 7 

 

 1 127.3 133.5 139.2 140.1 137.3 133.3 127.6 

 2 132.4 139.6 145.5 145.8 141.8 136.2 129.0 

 3 282.6 319.9 332.1 313.4 274.3 222.1 171.0 

 4 286.9 325.1 337.5 318.2 278.1 224.5 172.2 

 5 286.6 324.7 337.2 317.9 277.9 224.4 172.2 

 6 122.7 128.1 133.6 135.1 133.4 130.7 126.3 

 7 118.0 122.4 127.7 129.8 129.2 128.0 125.0 

 8 117.8 122.2 127.5 129.6 129.0 127.9 124.9 

 9 111.4 114.5 119.5 122.4 123.4 124.2 123.1 

 10 111.2 114.2 119.3 122.2 123.2 124.1 123.1 

 11 127.3 133.5 139.2 140.1 137.3 133.3 127.6 

 12 132.4 139.6 145.5 145.8 141.8 136.2 129.0 

 13 282.6 319.9 332.1 313.4 274.3 222.1 171.0 

 14 286.9 325.1 337.5 318.2 278.1 224.5 172.2 

 15 286.6 324.7 337.2 317.9 277.9 224.4 172.2 

 16 122.7 128.1 133.6 135.1 133.4 130.7 126.3 

 17 118.0 122.4 127.7 129.8 129.2 128.0 125.0 

 18 117.8 122.2 127.5 129.6 129.0 127.9 124.9 

 19 111.4 114.5 119.5 122.4 123.4 124.2 123.1 

 20 111.2 114.2 119.3 122.2 123.2 124.1 123.1 

 21 130.8 137.1 142.6 143.4 140.3 135.5 128.5 

 22 133.6 139.9 145.2 145.9 142.5 137.2 129.2 

 23 204.0 207.7 207.2 206.6 198.8 180.2 146.1 

 24 205.9 209.0 208.1 207.7 200.0 181.2 146.3 

 25 210.6 196.6 189.5 193.7 192.8 177.7 137.0 

 26 186.0 169.5 163.5 169.5 173.2 165.7 132.1 

 27 148.1 140.0 140.1 144.3 146.5 143.5 127.2 

 28 147.0 139.3 139.6 143.7 145.8 143.0 127.1 

 29 144.2 137.3 138.0 142.1 144.1 141.6 126.8 

3-43
TABLE 3-18. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE 

PROBLEM 4 (Continued) 

 30 143.4 136.8 137.5 141.6 143.6 141.1 126.7 

 31 129.7 127.1 129.7 133.3 134.9 133.9 125.1 

 32 129.7 127.1 129.7 133.3 134.9 133.9 125.1 

 33 129.7 127.1 129.7 133.3 134.9 133.9 125.1 

 34 129.7 127.1 129.7 133.3 134.9 133.9 125.1 

 35 116.0 117.5 121.9 125.0 126.2 126.6 123.6 

 36 115.2 116.9 121.4 124.5 125.6 126.2 123.5 

 37 111.3 114.2 119.2 122.2 123.2 124.2 123.0 

 38 111.2 114.1 119.1 122.1 123.1 124.1 123.0 

 39 119.9 119.2 119.9 126.8 130.5 132.1 129.4 

 40 122.1 120.6 120.6 128.0 132.2 133.8 130.5 

 41 176.5 157.0 140.1 159.8 172.7 173.0 153.5 

 42 178.1 158.2 141.0 160.8 173.8 174.0 153.9 

 43 186.5 166.5 152.0 167.4 177.3 174.3 148.1 

 44 178.8 161.5 152.0 162.8 170.1 166.9 141.7 

 45 159.6 147.2 141.0 150.7 157.1 155.7 138.3 

 46 158.0 146.1 140.1 149.7 156.0 154.7 137.9 

 47 119.5 119.2 120.6 126.7 129.9 131.3 128.4 

 48 118.1 118.2 119.9 125.8 128.9 130.4 127.9 

 49 116.2 106.1 108.7 112.2 113.7 113.2 107.3 

 50 119.5 107.0 108.8 112.6 114.3 113.3 105.7 

 51 306.1 348.1 361.3 339.6 295.0 235.5 177.6 

 52 288.9 327.5 340.0 320.5 279.9 225.7 172.8 

 53 306.1 348.1 361.3 339.6 295.0 235.5 177.6 

 

Maximum Structural Temperature for Period 1 = 286.91 Occurring 

at Location 4 

Maximum Structural Temperature for Period 2 = 325.06 Occurring 

at Location 4 

Maximum Structural Temperature for Period 3 = 337.51 Occurring 

at Location 4 

Maximum Structural Temperature for Period 4 = 318.20 Occurring 

at Location 4 

Maximum Structural Temperature for Period 5 = 278.12 Occurring 

at Location 4 

Maximum Structural Temperature for Period 6 = 224.54 Occurring 

at Location 4 

Maximum Structural Temperature for Period 7 = 172.24 Occurring 

at Location 4 

 

Maximum Structural Temperature Over All 7 Periods =337.51 

Occurring at Location 4 at Period 3 

3-44
TABLE 3-19. THERMOD-SIMULATED TEMPERATURES (°F) OF THE RIGHT WING OF 

SAMPLE PROBLEM 4 

As in sample problems 1 through 3, a one-half wing structure was modeled using the 

UAI/NASTRAN FEM program. Two types of loads were applied to the model. These loads, 

based on period 3 of the THERMOD input file, were direct solar radiation of 355 Btu/hr/ft2 and 

convection with a film coefficient of h = 1.3433 (Btu/hr)/(ft2

°F). The film coefficient was 

calculated internally by the THERMOD program and was based on the theory of airflow on a 

flat surface and a wind speed of 14 ft/sec. The radiation load was applied perpendicular to the 

top surface only, while both the top and bottom surfaces were subjected to convection forces. 

The ambient temperature was maintained at 119°F (578.67°R). 

The FEM input file is shown in table 3-20. The ambient temperature of 119°F was imposed 

through the SPC case control card. The ambient temperature boundary condition was scalar in 

nature and implemented through the SPOINT bulk data. The boundary elements, CHBDY, 

through which solar radiation and convection effects were applied, were modeled using the 

LINE boundary shape type, connecting the boundary nodes along the wing top and wing bottom.

The SPACE designation in the CHBDY boundary element indicates that this element emits 

infrared radiation to a black body, in which case, no VIEW bulk data were required. Property 

cards PHBDY and the material cards MAT4, define absorptivity, emissivity, film coefficient, 

and conductivity values. Because of the involvement of infrared radiation, through emission, 

this sample problem was rendered nonlinear. This required the Stefan-Boltzman constant and an 

initial set of temperatures, which were implemented through the PARAM and TEMPERATURE 

statements, respectively. 

T1 T2 T3 T4 T5 T6 T7 T8 T9 T10

353.9 353.9 348.0 348.0 347.6 128.2 127.8 127.8 122.0 121.9

TABLE 3-20. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE PROBLEM 4 

ID D:\Natha,MSC/N 

APP HEAT 

SOL 3 

TIME 10000 

CEND 

 TITLE = direct_radiation_conduction_noemission_convection (t=0.001ft) 

 SUBTITLE = 10.7 in gap filled by air; k=0.016BTU/hr/ft/degF 

 load=1 

 SPC = 1 

 temperature(estimate)=2 

 THERMAL= ALL 

 FLUX = ALL 

 SPCF = ALL 

BEGIN BULK 

$ *************************************************************************** 

$ Written by : MSC/NASTRAN for Windows 

$ Version : 6.00 

$ Translator : UAI/NASTRAN 

$ From Model : D:\Thermod_Validation\FEM\conduction_only.MOD 

$ Date : Mon Mar 06 11:27:09 2000 

$ *************************************************************************** 

3-45
TABLE 3-20. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 4 (Continued) 

$ 

PARAM,K6ROT,100. 

PARAM,MAXRATIO,1.E+8 

param,sigma,1.714e-9 

param,tabs,0.0 

param,maxit,10 

CORD2C 1 0 0. 0. 0. 0. 0. 1.+MSC/NC1

+MSC/NC1 1. 0. 1. 

CORD2S 2 0 0. 0. 0. 0. 0. 1.+MSC/NC2

+MSC/NC2 1. 0. 1. 

$ 

$ MSC/NASTRAN for Windows Property 1 : 0.02 in glass_outer_layer 

PSHELL 1 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 2 : 3/8 in foam 

PSHELL 2 2 0.001 2 2 0. 

$ MSC/NASTRAN for Windows Property 3 : 0.02 in inner glass 

PSHELL 3 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 4 : 0.350 in cap 

PSHELL 4 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 5 : 10.70 in air space 

PSHELL 5 3 0.001 3 3 0. 

$ MSC/NASTRAN for Windows Property 6 : 0.350 in cap 

PSHELL 6 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 7 : 0.02 in inner glass 

PSHELL 7 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 8 : 3/8 in foam 

PSHELL 8 2 0.001 2 2 0. 

$ MSC/NASTRAN for Windows Property 9 : 0.02 in glass_outer_layer 

PSHELL 9 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Material 1 : glass 

MAT4 1 0.3 1. 

$ MSC/NASTRAN for Windows Material 2 : foam 

MAT4 2 0.021 1. 

$ MSC/NASTRAN for Windows Material 3 : air 

MAT4 3 0.016 1. 

$ 

chbdy,1001,100,line,200,201,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1002,100,line,201,202,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1003,100,line,202,203,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1004,100,line,203,204,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1005,100,line,204,205,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1006,100,line,205,206,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1007,100,line,206,207,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1008,100,line,207,208,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1009,100,line,208,209,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1010,100,line,209,210,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1011,100,line,210,211,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1012,100,line,211,212,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1013,100,line,212,213,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1014,100,line,213,214,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1015,100,line,214,215,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1016,100,line,215,216,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1017,100,line,216,181,,,space,500,500,,,0.0,1.0,0.0 

phbdy,100,200,0.001,0.0,0.9 

mat4,200,1.3433 

spoint,500 

spc,1,500,1,578.67 

3-46
TABLE 3-20. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 4 (Continued) 

$ 

$ Direct radiation load from a distant source: load case 1 

qvect,1,355.0,0.0,-1.0,0.0,1001,thru,1017 

$ 

$ view ids 

$view,1,yes,yes,1,1,0.0 

$ 

chbdy,2001,300,line,20,21,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2002,300,line,21,22,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2003,300,line,22,23,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2004,300,line,23,24,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2005,300,line,24,25,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2006,300,line,25,26,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2007,300,line,26,27,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2008,300,line,27,28,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2009,300,line,28,29,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2010,300,line,29,30,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2011,300,line,30,31,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2012,300,line,31,32,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2013,300,line,32,33,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2014,300,line,33,34,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2015,300,line,34,35,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2016,300,line,35,36,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2017,300,line,36,1,,,space,500,500,,,0.0,-1.0,0.0 

phbdy,300,400,0.001,0.0,0.9 

mat4,400,1.3433 

$ 

temp,2,1,578.67,3,578.67,5,578.67 

temp,2,7,578.67,9,578.67,10,578.67 

temp,2,11,578.67,12,578.67,14,578.67 

temp,2,15,578.67,18,578.67,20,578.67 

temp,2,21,578.67,22,578.67,23,578.67 

temp,2,24,578.67,25,578.67,26,578.67 

temp,2,27,578.67,28,578.67,29,578.67 

temp,2,30,578.67,31,578.67,32,578.67 

temp,2,33,578.67,34,578.67,35,578.67 

temp,2,36,578.67,37,578.67,41,578.67 

temp,2,44,578.67,45,578.67,47,578.67 

temp,2,48,578.67,50,578.67,51,578.67 

temp,2,52,578.67,53,578.67,54,578.67 

temp,2,55,578.67,56,578.67,58,578.67 

temp,2,59,578.67,62,578.67,67,578.67 

temp,2,69,578.67,71,578.67,73,578.67 

temp,2,75,578.67,76,578.67,77,578.67 

temp,2,78,578.67,79,578.67,80,578.67 

temp,2,83,578.67,85,578.67,87,578.67 

temp,2,88,578.67,98,578.67,101,578.67 

temp,2,104,578.67,105,578.67,107,578.67 

temp,2,108,578.67,109,578.67,112,578.67 

temp,2,115,578.67,116,578.67,118,578.67 

temp,2,120,578.67,121,578.67,122,578.67 

temp,2,123,578.67,124,578.67,125,578.67 

temp,2,128,578.67,129,578.67,130,578.67 

temp,2,133,578.67,135,578.67,137,578.67 

temp,2,138,578.67,145,578.67,146,578.67 

temp,2,147,578.67,150,578.67,152,578.67 

temp,2,160,578.67,161,578.67,164,578.67 

3-47
TABLE 3-20. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 4 (Continued) 

temp,2,165,578.67,172,578.67,174,578.67 

temp,2,177,578.67,178,578.67,180,578.67 

temp,2,181,578.67,182,578.67,183,578.67 

temp,2,184,578.67,188,578.67,193,578.67 

temp,2,195,578.67,196,578.67,197,578.67 

temp,2,198,578.67,199,578.67,200,578.67 

temp,2,201,578.67,202,578.67,203,578.67 

temp,2,204,578.67,205,578.67,206,578.67 

temp,2,207,578.67,208,578.67,209,578.67 

temp,2,210,578.67,211,578.67,212,578.67 

temp,2,213,578.67,214,578.67,215,578.67 

temp,2,216,578.67,219,578.67,220,578.67 

temp,2,221,578.67,222,578.67,223,578.67 

temp,2,226,578.67,227,578.67,228,578.67 

temp,2,229,578.67,231,578.67,232,578.67 

temp,2,235,578.67,241,578.67,243,578.67 

temp,2,248,578.67,249,578.67,250,578.67 

temp,2,253,578.67,254,578.67,257,578.67 

temp,2,258,578.67,259,578.67,260,578.67 

temp,2,264,578.67,267,578.67,268,578.67 

temp,2,269,578.67,270,578.67,273,578.67 

temp,2,274,578.67,277,578.67,282,578.67 

temp,2,283,578.67,292,578.67,293,578.67 

temp,2,295,578.67,297,578.67,298,578.67 

temp,2,299,578.67,300,578.67,301,578.67 

temp,2,302,578.67,303,578.67,306,578.67 

temp,2,307,578.67,308,578.67,313,578.67 

temp,2,314,578.67,316,578.67,317,578.67 

temp,2,318,578.67,323,578.67,324,578.67 

temp,2,500,578.67 

$ 

GRID 1 0 0. 0. 0. 0 

GRID 3 0 1.1.667E-3 0. 0 

GRID 5 0 3.1.667E-3 0. 0 

GRID 7 0 5.1.667E-3 0. 0 

GRID 9 0 7.1.667E-3 0. 0 

GRID 10 0 8.1.667E-3 0. 0 

GRID 11 0 9.1.667E-3 0. 0 

GRID 12 0 10.1.667E-3 0. 0 

GRID 14 0 12.1.667E-3 0. 0 

GRID 15 0 13.1.667E-3 0. 0 

GRID 18 0 16.1.667E-3 0. 0 

GRID 20 0 17. 0. 0. 0 

GRID 21 0 16. 0. 0. 0 

GRID 22 0 15. 0. 0. 0 

GRID 23 0 14. 0. 0. 0 

GRID 24 0 13. 0. 0. 0 

GRID 25 0 12. 0. 0. 0 

GRID 26 0 11. 0. 0. 0 

GRID 27 0 10. 0. 0. 0 

GRID 28 0 9. 0. 0. 0 

GRID 29 0 8. 0. 0. 0 

GRID 30 0 7. 0. 0. 0 

GRID 31 0 6. 0. 0. 0 

GRID 32 0 5. 0. 0. 0 

GRID 33 0 4. 0. 0. 0 

temp,2,244,578.67,245,578.67,246,578.67 

3-48
TABLE 3-20. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 4 (Continued) 

GRID 34 0 3. 0. 0. 0 

GRID 35 0 2. 0. 0. 0 

GRID 36 0 1. 0. 0. 0 

GRID 37 0 0.1.667E-3 0. 0 

GRID 41 0 3.0.032917 0. 0 

GRID 44 0 6.0.032917 0. 0 

GRID 45 0 7.0.032917 0. 0 

GRID 47 0 9.0.032917 0. 0 

GRID 48 0 10.0.032917 0. 0 

GRID 50 0 12.0.032917 0. 0 

GRID 51 0 13.0.032917 0. 0 

GRID 52 0 14.0.032917 0. 0 

GRID 53 0 15.0.032917 0. 0 

GRID 54 0 16.0.032917 0. 0 

GRID 55 0 17.0.032917 0. 0 

GRID 56 0 17.1.667E-3 0. 0 

GRID 58 0 15.1.667E-3 0. 0 

GRID 59 0 14.1.667E-3 0. 0 

GRID 62 0 11.1.667E-3 0. 0 

GRID 67 0 6.1.667E-3 0. 0 

GRID 69 0 4.1.667E-3 0. 0 

GRID 71 0 2.1.667E-3 0. 0 

GRID 73 0 0.0.032917 0. 0 

GRID 75 0 1.0.034583 0. 0 

GRID 76 0 2.0.034583 0. 0 

GRID 77 0 3.0.034583 0. 0 

GRID 78 0 4.0.034583 0. 0 

GRID 79 0 5.0.034583 0. 0 

GRID 80 0 6.0.034583 0. 0 

GRID 83 0 9.0.034583 0. 0 

GRID 85 0 11.0.034583 0. 0 

GRID 87 0 13.0.034583 0. 0 

GRID 88 0 14.0.034583 0. 0 

GRID 98 0 11.0.032917 0. 0 

GRID 101 0 8.0.032917 0. 0 

GRID 104 0 5.0.032917 0. 0 

GRID 105 0 4.0.032917 0. 0 

GRID 107 0 2.0.032917 0. 0 

GRID 108 0 1.0.032917 0. 0 

GRID 109 0 0.0.034583 0. 0 

GRID 112 0 2. 0.06375 0. 0 

GRID 115 0 5. 0.06375 0. 0 

GRID 116 0 6. 0.06375 0. 0 

GRID 118 0 8. 0.06375 0. 0 

GRID 120 0 10. 0.06375 0. 0 

GRID 121 0 11. 0.06375 0. 0 

GRID 122 0 12. 0.06375 0. 0 

GRID 123 0 13. 0.06375 0. 0 

GRID 124 0 14. 0.06375 0. 0 

GRID 125 0 15. 0.06375 0. 0 

GRID 128 0 17.0.034583 0. 0 

GRID 129 0 16.0.034583 0. 0 

GRID 130 0 15.0.034583 0. 0 

GRID 133 0 12.0.034583 0. 0 

GRID 135 0 10.0.034583 0. 0 

GRID 137 0 8.0.034583 0. 0 

GRID 138 0 7.0.034583 0. 0 

3-49
TABLE 3-20. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 4 (Continued) 

GRID 145 0 0. 0.06375 0. 0 

GRID 146 0 0. 0.95542 0. 0 

GRID 147 0 1. 0.95542 0. 0 

GRID 150 0 4. 0.95542 0. 0 

GRID 152 0 6. 0.95542 0. 0 

GRID 160 0 14. 0.95542 0. 0 

GRID 161 0 15. 0.95542 0. 0 

GRID 164 0 17. 0.06375 0. 0 

GRID 165 0 16. 0.06375 0. 0 

GRID 172 0 9. 0.06375 0. 0 

GRID 174 0 7. 0.06375 0. 0 

GRID 177 0 4. 0.06375 0. 0 

GRID 178 0 3. 0.06375 0. 0 

GRID 180 0 1. 0.06375 0. 0 

GRID 181 0 0. 1.01917 0. 0 

GRID 182 0 0. 1.0175 0. 0 

GRID 183 0 1. 1.0175 0. 0 

GRID 184 0 2. 1.0175 0. 0 

GRID 188 0 6. 1.0175 0. 0 

GRID 193 0 11. 1.0175 0. 0 

GRID 195 0 13. 1.0175 0. 0 

GRID 196 0 14. 1.0175 0. 0 

GRID 197 0 15. 1.0175 0. 0 

GRID 198 0 16. 1.0175 0. 0 

GRID 199 0 17. 1.0175 0. 0 

GRID 200 0 17. 1.01917 0. 0 

GRID 201 0 16. 1.01917 0. 0 

GRID 202 0 15. 1.01917 0. 0 

GRID 203 0 14. 1.01917 0. 0 

GRID 204 0 13. 1.01917 0. 0 

GRID 205 0 12. 1.01917 0. 0 

GRID 206 0 11. 1.01917 0. 0 

GRID 207 0 10. 1.01917 0. 0 

GRID 208 0 9. 1.01917 0. 0 

GRID 209 0 8. 1.01917 0. 0 

GRID 210 0 7. 1.01917 0. 0 

GRID 211 0 6. 1.01917 0. 0 

GRID 212 0 5. 1.01917 0. 0 

GRID 213 0 4. 1.01917 0. 0 

GRID 214 0 3. 1.01917 0. 0 

GRID 215 0 2. 1.01917 0. 0 

GRID 216 0 1. 1.01917 0. 0 

GRID 219 0 1. 0.98625 0. 0 

GRID 220 0 2. 0.98625 0. 0 

GRID 221 0 3. 0.98625 0. 0 

GRID 222 0 4. 0.98625 0. 0 

GRID 223 0 5. 0.98625 0. 0 

GRID 226 0 8. 0.98625 0. 0 

GRID 227 0 9. 0.98625 0. 0 

GRID 228 0 10. 0.98625 0. 0 

GRID 229 0 11. 0.98625 0. 0 

GRID 231 0 13. 0.98625 0. 0 

GRID 232 0 14. 0.98625 0. 0 

GRID 235 0 17. 0.98625 0. 0 

GRID 241 0 12. 1.0175 0. 0 

GRID 243 0 10. 1.0175 0. 0 

GRID 244 0 9. 1.0175 0. 0 

3-50
TABLE 3-20. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 4 (Continued) 

GRID 245 0 8. 1.0175 0. 0 

GRID 246 0 7. 1.0175 0. 0 

GRID 248 0 5. 1.0175 0. 0 

GRID 249 0 4. 1.0175 0. 0 

GRID 250 0 3. 1.0175 0. 0 

GRID 253 0 0. 0.98625 0. 0 

GRID 254 0 0. 0.98458 0. 0 

GRID 257 0 3. 0.98458 0. 0 

GRID 258 0 4. 0.98458 0. 0 

GRID 259 0 5. 0.98458 0. 0 

GRID 260 0 6. 0.98458 0. 0 

GRID 264 0 10. 0.98458 0. 0 

GRID 267 0 13. 0.98458 0. 0 

GRID 268 0 14. 0.98458 0. 0 

GRID 269 0 15. 0.98458 0. 0 

GRID 270 0 16. 0.98458 0. 0 

GRID 273 0 16. 0.98625 0. 0 

GRID 274 0 15. 0.98625 0. 0 

GRID 277 0 12. 0.98625 0. 0 

GRID 282 0 7. 0.98625 0. 0 

GRID 283 0 6. 0.98625 0. 0 

GRID 292 0 2. 0.95542 0. 0 

GRID 293 0 3. 0.95542 0. 0 

GRID 295 0 5. 0.95542 0. 0 

GRID 297 0 7. 0.95542 0. 0 

GRID 298 0 8. 0.95542 0. 0 

GRID 299 0 9. 0.95542 0. 0 

GRID 300 0 10. 0.95542 0. 0 

GRID 301 0 11. 0.95542 0. 0 

GRID 302 0 12. 0.95542 0. 0 

GRID 303 0 13. 0.95542 0. 0 

GRID 306 0 16. 0.95542 0. 0 

GRID 307 0 17. 0.95542 0. 0 

GRID 308 0 17. 0.98458 0. 0 

GRID 313 0 12. 0.98458 0. 0 

GRID 314 0 11. 0.98458 0. 0 

GRID 316 0 9. 0.98458 0. 0 

GRID 317 0 8. 0.98458 0. 0 

GRID 318 0 7. 0.98458 0. 0 

GRID 323 0 2. 0.98458 0. 0 

GRID 324 0 1. 0.98458 0. 0 

CQUAD4 1 9 1 37 3 36 

CQUAD4 2 9 36 3 71 35 

CQUAD4 3 9 35 71 5 34 

CQUAD4 4 9 34 5 69 33 

CQUAD4 5 9 33 69 7 32 

CQUAD4 6 9 32 7 67 31 

CQUAD4 7 9 31 67 9 30 

CQUAD4 8 9 30 9 10 29 

CQUAD4 9 9 29 10 11 28 

CQUAD4 10 9 28 11 12 27 

CQUAD4 11 9 27 12 62 26 

CQUAD4 12 9 26 62 14 25 

CQUAD4 13 9 25 14 15 24 

CQUAD4 14 9 24 15 59 23 

CQUAD4 15 9 23 59 58 22 

CQUAD4 16 9 22 58 18 21 

3-51
TABLE 3-20. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 4 (Continued) 

CQUAD4 17 9 21 18 56 20 

CQUAD4 18 8 37 73 108 3 

CQUAD4 19 8 3 108 107 71 

CQUAD4 20 8 71 107 41 5 

CQUAD4 21 8 5 41 105 69 

CQUAD4 22 8 69 105 104 7 

CQUAD4 23 8 7 104 44 67 

CQUAD4 24 8 67 44 45 9 

CQUAD4 25 8 9 45 101 10 

CQUAD4 26 8 10 101 47 11 

CQUAD4 27 8 11 47 48 12 

CQUAD4 28 8 12 48 98 62 

CQUAD4 29 8 62 98 50 14 

CQUAD4 30 8 14 50 51 15 

CQUAD4 31 8 15 51 52 59 

CQUAD4 32 8 59 52 53 58 

CQUAD4 33 8 58 53 54 18 

CQUAD4 34 8 18 54 55 56 

CQUAD4 35 7 73 109 75 108 

CQUAD4 36 7 108 75 76 107 

CQUAD4 37 7 107 76 77 41 

CQUAD4 38 7 41 77 78 105 

CQUAD4 39 7 105 78 79 104 

CQUAD4 40 7 104 79 80 44 

CQUAD4 41 7 44 80 138 45 

CQUAD4 42 7 45 138 137 101 

CQUAD4 43 7 101 137 83 47 

CQUAD4 44 7 47 83 135 48 

CQUAD4 45 7 48 135 85 98 

CQUAD4 46 7 98 85 133 50 

CQUAD4 47 7 50 133 87 51 

CQUAD4 48 7 51 87 88 52 

CQUAD4 49 7 52 88 130 53 

CQUAD4 50 7 53 130 129 54 

CQUAD4 51 7 54 129 128 55 

CQUAD4 52 6 109 145 180 75 

CQUAD4 53 6 75 180 112 76 

CQUAD4 54 6 76 112 178 77 

CQUAD4 55 6 77 178 177 78 

CQUAD4 56 6 78 177 115 79 

CQUAD4 57 6 79 115 116 80 

CQUAD4 58 6 80 116 174 138 

CQUAD4 59 6 138 174 118 137 

CQUAD4 60 6 137 118 172 83 

CQUAD4 61 6 83 172 120 135 

CQUAD4 62 6 135 120 121 85 

CQUAD4 63 6 85 121 122 133 

CQUAD4 64 6 133 122 123 87 

CQUAD4 65 6 87 123 124 88 

CQUAD4 66 6 88 124 125 130 

CQUAD4 67 6 130 125 165 129 

CQUAD4 68 6 129 165 164 128 

CQUAD4 69 5 145 146 147 180 

CQUAD4 70 5 180 147 292 112 

CQUAD4 71 5 112 292 293 178 

CQUAD4 72 5 178 293 150 177 

CQUAD4 73 5 177 150 295 115 

3-52
TABLE 3-20. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 4 (Continued) 

CQUAD4 74 5 115 295 152 116 

CQUAD4 75 5 116 152 297 174 

CQUAD4 76 5 174 297 298 118 

CQUAD4 77 5 118 298 299 172 

CQUAD4 78 5 172 299 300 120 

CQUAD4 79 5 120 300 301 121 

CQUAD4 80 5 121 301 302 122 

CQUAD4 81 5 122 302 303 123 

CQUAD4 82 5 123 303 160 124 

CQUAD4 83 5 124 160 161 125 

CQUAD4 84 5 125 161 306 165 

CQUAD4 85 5 165 306 307 164 

CQUAD4 86 1 182 181 216 183 

CQUAD4 87 1 183 216 215 184 

CQUAD4 88 1 184 215 214 250 

CQUAD4 89 1 250 214 213 249 

CQUAD4 90 1 249 213 212 248 

CQUAD4 91 1 248 212 211 188 

CQUAD4 92 1 188 211 210 246 

CQUAD4 93 1 246 210 209 245 

CQUAD4 94 1 245 209 208 244 

CQUAD4 95 1 244 208 207 243 

CQUAD4 96 1 243 207 206 193 

CQUAD4 97 1 193 206 205 241 

CQUAD4 98 1 241 205 204 195 

CQUAD4 99 1 195 204 203 196 

CQUAD4 100 1 196 203 202 197 

CQUAD4 101 1 197 202 201 198 

CQUAD4 102 1 198 201 200 199 

CQUAD4 103 2 253 182 183 219 

CQUAD4 104 2 219 183 184 220 

CQUAD4 105 8 220 184 250 221 

CQUAD4 106 2 221 250 249 222 

CQUAD4 107 2 222 249 248 223 

CQUAD4 108 2 223 248 188 283 

CQUAD4 109 2 283 188 246 282 

CQUAD4 110 2 282 246 245 226 

CQUAD4 111 2 226 245 244 227 

CQUAD4 112 2 227 244 243 228 

CQUAD4 113 2 228 243 193 229 

CQUAD4 114 2 229 193 241 277 

CQUAD4 115 2 277 241 195 231 

CQUAD4 116 2 231 195 196 232 

CQUAD4 117 2 232 196 197 274 

CQUAD4 118 2 274 197 198 273 

CQUAD4 119 2 273 198 199 235 

CQUAD4 120 3 254 253 219 324 

CQUAD4 121 3 324 219 220 323 

CQUAD4 122 3 323 220 221 257 

CQUAD4 123 3 257 221 222 258 

CQUAD4 124 3 258 222 223 259 

CQUAD4 125 3 259 223 283 260 

CQUAD4 126 3 260 283 282 318 

CQUAD4 127 3 318 282 226 317 

CQUAD4 128 3 317 226 227 316 

CQUAD4 129 3 316 227 228 264 

CQUAD4 130 3 264 228 229 314 

3-53
TABLE 3-20. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 4 (Continued) 

CQUAD4 131 3 314 229 277 313 

CQUAD4 132 3 313 277 231 267 

CQUAD4 133 3 267 231 232 268 

CQUAD4 134 3 268 232 274 269 

CQUAD4 135 3 269 274 273 270 

CQUAD4 136 3 270 273 235 308 

CQUAD4 137 4 146 254 324 147 

CQUAD4 138 4 147 324 323 292 

CQUAD4 139 4 292 323 257 293 

CQUAD4 140 4 293 257 258 150 

CQUAD4 141 4 150 258 259 295 

CQUAD4 142 4 295 259 260 152 

CQUAD4 143 4 152 260 318 297 

CQUAD4 144 4 297 318 317 298 

CQUAD4 145 4 298 317 316 299 

CQUAD4 146 4 299 316 264 300 

CQUAD4 147 4 300 264 314 301 

CQUAD4 148 4 301 314 313 302 

CQUAD4 149 4 302 313 267 303 

CQUAD4 150 4 303 267 268 160 

CQUAD4 151 4 160 268 269 161 

CQUAD4 152 4 161 269 270 306 

CQUAD4 153 4 306 270 308 307 

ENDDATA 

The results of the one-half FEM wing model analysis are presented in figure 3-4 in the form of a 

contour plot. The contours appear to be indicative of the temperature profile of the THERMOD 

analysis. The top surface, which was subjected to direct solar radiation, is certainly much hotter 

than the bottom surface. 

Note that the temperatures are in absolute terms (°R). 

FIGURE 3-4. CONTOUR PLOT OF THE FEM OUTPUT OF SAMPLE PROBLEM 4 

3-54
The truncated FEM output file is shown in table 3-21. The nodes, from top to bottom of the 

wing, located at mid-span of the model are indicated by nodes 209, 245, 226, 317, 298, 118, 137, 

101, 10, and 29. The temperatures of these nodes were extracted from this file and are 

reproduced in table 3-22. Mid-span was chosen because it was sufficiently far away from any 

edge effects that might be present. Note that temperatures in table 3-22 were reduced from

absolute temperatures (°R) to °F.

TABLE 3-21. TRUNCATED FEM OUTPUT FILE OF SAMPLE PROBLEM 4 

1 DIRECT_RADIATION_CONDUCTION_NOEMISSION_CONVECTION (T=0.001FT MARCH 15, 2000 UAI/NASTRAN VERSION 20.1 PAGE 

18 

 10.7 IN GAP FILLED BY AIR; K=0.016BTU/HR/FT/DEGF 

0 

 

 T E M P E R A T U R E V E C T O R 

 

 POINT ID. TYPE ID VALUE ID+1 VALUE ID+2 VALUE ID+3 VALUE ID+4 VALUE ID+5 VALUE 

 1 TEMP 5.816010E+02 

 3 TEMP 5.816228E+02 

 5 TEMP 5.816228E+02 

 7 TEMP 5.816228E+02 

 9 TEMP 5.816228E+02 5.816228E+02 5.816228E+02 5.816228E+02 

 14 TEMP 5.816228E+02 5.816228E+02 

 18 TEMP 5.816228E+02 

 20 TEMP 5.816010E+02 5.816010E+02 5.816010E+02 5.816010E+02 5.816010E+02 5.816010E+02 

 26 TEMP 5.816010E+02 5.816010E+02 5.816010E+02 5.816010E+02 5.816010E+02 5.816010E+02 

 32 TEMP 5.816010E+02 5.816010E+02 5.816010E+02 5.816010E+02 5.816010E+02 5.816228E+02 

 41 TEMP 5.874816E+02 

 44 TEMP 5.874816E+02 5.874816E+02 

 47 TEMP 5.874816E+02 5.874816E+02 

 50 TEMP 5.874816E+02 5.874816E+02 5.874816E+02 5.874816E+02 5.874816E+02 5.874816E+02 

 56 TEMP 5.816228E+02 

 58 TEMP 5.816228E+02 5.816228E+02 

 62 TEMP 5.816228E+02 

 67 TEMP 5.816228E+02 

 69 TEMP 5.816228E+02 

 71 TEMP 5.816228E+02 

 73 TEMP 5.874816E+02 

 75 TEMP 5.875035E+02 5.875035E+02 5.875035E+02 5.875035E+02 5.875035E+02 5.875035E+02 

 83 TEMP 5.875035E+02 

 85 TEMP 5.875035E+02 

 87 TEMP 5.875035E+02 5.875035E+02 

 98 TEMP 5.874816E+02 

 101 TEMP 5.874816E+02 

 104 TEMP 5.874816E+02 5.874816E+02 

 107 TEMP 5.874816E+02 5.874816E+02 5.875035E+02 

 112 TEMP 5.878863E+02 

 115 TEMP 5.878863E+02 5.878863E+02 

 118 TEMP 5.878863E+02 

 120 TEMP 5.878863E+02 5.878863E+02 5.878863E+02 5.878863E+02 5.878863E+02 5.878863E+02 

 128 TEMP 5.875035E+02 5.875035E+02 5.875035E+02 

 133 TEMP 5.875035E+02 

 135 TEMP 5.875035E+02 

 137 TEMP 5.875035E+02 5.875035E+02 

 145 TEMP 5.878863E+02 8.073008E+02 8.073008E+02 

 150 TEMP 8.073008E+02 

 152 TEMP 8.073008E+02 

 160 TEMP 8.073008E+02 8.073008E+02 

 164 TEMP 5.878863E+02 5.878863E+02 

 172 TEMP 5.878863E+02 

 174 TEMP 5.878863E+02 

 177 TEMP 5.878863E+02 5.878863E+02 

 180 TEMP 5.878863E+02 8.135862E+02 8.135643E+02 8.135643E+02 8.135643E+02 

 188 TEMP 8.135643E+02 

 193 TEMP 8.135643E+02 

 195 TEMP 8.135643E+02 8.135643E+02 8.135643E+02 8.135643E+02 8.135643E+02 8.135862E+02 

 201 TEMP 8.135862E+02 8.135862E+02 8.135862E+02 8.135862E+02 8.135862E+02 8.135862E+02 

1 DIRECT_RADIATION_CONDUCTION_NOEMISSION_CONVECTION (T=0.001FT MARCH 15, 2000 UAI/NASTRAN VERSION 20.1 PAGE 

19 

 10.7 IN GAP FILLED BY AIR; K=0.016BTU/HR/FT/DEGF 

0 

 

 T E M P E R A T U R E V E C T O R 

 

 POINT ID. TYPE ID VALUE ID+1 VALUE ID+2 VALUE ID+3 VALUE ID+4 VALUE ID+5 VALUE 

 207 TEMP 8.135862E+02 8.135862E+02 8.135862E+02 8.135862E+02 8.135862E+02 8.135862E+02 

 213 TEMP 8.135862E+02 8.135862E+02 8.135862E+02 8.135862E+02 

 219 TEMP 8.077054E+02 8.077054E+02 8.077054E+02 8.077054E+02 8.077054E+02 

 226 TEMP 8.077054E+02 8.077054E+02 8.077054E+02 8.077054E+02 

 231 TEMP 8.077054E+02 8.077054E+02 

 235 TEMP 8.077054E+02 

 241 TEMP 8.135643E+02 

 243 TEMP 8.135643E+02 8.135643E+02 8.135643E+02 8.135643E+02 

 248 TEMP 8.135643E+02 8.135643E+02 8.135643E+02 

 253 TEMP 8.077054E+02 8.076835E+02 

 257 TEMP 8.076835E+02 8.076835E+02 8.076835E+02 8.076835E+02 

 264 TEMP 8.076835E+02 

 267 TEMP 8.076835E+02 8.076835E+02 8.076835E+02 8.076835E+02 

 273 TEMP 8.077054E+02 8.077054E+02 

3-55
TABLE 3-21. TRUNCATED FEM OUTPUT FILE OF SAMPLE PROBLEM 4 (Continued) 

 277 TEMP 8.077054E+02 

 282 TEMP 8.077054E+02 8.077054E+02 

 292 TEMP 8.073008E+02 8.073008E+02 

 295 TEMP 8.073008E+02 

 297 TEMP 8.073008E+02 8.073008E+02 8.073008E+02 8.073008E+02 8.073008E+02 8.073008E+02 

 303 TEMP 8.073008E+02 

 306 TEMP 8.073008E+02 8.073008E+02 8.076835E+02 

 313 TEMP 8.076835E+02 8.076835E+02 

 316 TEMP 8.076835E+02 8.076835E+02 8.076835E+02 

 323 TEMP 8.076835E+02 8.076835E+02 

 500 TEMP 5.786700E+02 

 

When compared, the temperatures of tables 3-19 and 3-22 show excellent agreement, validating 

the radiation, conduction, and convection capabilities of the THERMOD program. 

TABLE 3-22. FINITE ELEMENT METHOD-SIMULATED TEMPERATURES (°F) OF THE 

RIGHT WING OF SAMPLE PROBLEM 4 

T1 T2 T3 T4 T5 T6 T7 T8 T9 T10

353.9 353.9 348.0 348.0 347.6 128.2 127.8 127.8 122.0 121.9

3.5 SAMPLE PROBLEM 5.

In sample problem 4, the combined effects of conduction, convection, and direct solar radiation 

were studied. The wing surfaces, in addition to other external surfaces however, were assigned 

0.0 emissivity values, meaning that the surfaces were rendered incapable of emitting infrared 

radiation. The purpose of sample problem 5 was to study and validate the effects of infrared 

emission. This was done by removing the emission restriction on the wing surfaces of sample 

problem 4, by changing it from 0.0 to 0.9. This action was expected to result in lower wing 

temperatures than in sample problem 4. As in sample problems 1 through 4, THERMOD was set 

up to solve a complete aircraft problem, while the FEM method was confined to the wing model. 

To ensure consistency with the FEM wing model, the wing in the THERMOD model was 

isolated from infrared radiation effects from the fuselage and tarmac surfaces by assigning 

emissivity values of 0.0. As in sample problem 4, the sky radiation was assigned an absolute 

temperature of 0.0 to preclude infrared radiation from the sky. The conduction and convection 

heat transfer modes, however, were left intact, as in sample problem 4. 

Despite the above setup, particularly the efforts to isolate the wing, a preliminary comparative 

study of the THERMOD and FEM results showed significant differences in wing temperatures. 

This discrepancy was resolved by further isolating the wing from the tarmac effects by 

increasing the height of the wing from the tarmac to 100.0 ft from its current 2.5 ft. In addition, 

the infrared reflection effects from the fuselage surface were reduced by decreasing the cabin 

height from 4 ft to 0.1 ft. 

Despite these additions, temperature differences still existed between the THERMOD and FEM 

wing models, particularly on elements on the wing top surface. The temperatures on the wing 

bottom surface were found to be in excellent agreement, however. This led to the believe that 

the fillet effect, which has been modeled as an integral part of THERMOD, could be a plausible 

cause of this problem. The fillet effect is defined as the effect found in the fuselage-wing 

3-56
junctions, usually covered by fairing, that serves to increase the local temperatures because the 

surface is concave. A concave surface is viewed as a partial cavity, and in thermal processes, it 

is the nature of cavities to increase temperatures regardless of the paint color. In THERMOD, 

the fillet was modeled as a quadrant. Its temperature-increasing effect was confined to direct 

solar radiation only. This explains why only the top wing surface temperatures were different in 

the THERMOD and FEM models. 

The fillet effect was removed from the wing top by assigning a 0.0 value in the second field of 

input data set 6, in essence eliminating the fillet effect. See the THERMOD User’s Reference 

Manual for further details on the input requirements for fillet effects. With the removal of the 

fillet effect, there was excellent agreement in the temperatures of the THERMOD and FEM wing 

models. The THERMOD input file, input.dat, is illustrated in table 3-23. 

TABLE 3-23. THERMOD INPUT FILE (input.dat) OF SAMPLE PROBLEM 5 

198e-6,0.704 

1, 0.020,125,0.300,0.30 

2, 0.375,4.4,0.021,0.24 

3, 0.020,125,0.300,0.30 

4, 0.350,125,0.300,0.30 

5, 10.70,0.067,0.016,0.24 

6, 0.350,125,0.300,0.30 

7, 0.020,125,0.300,0.30 

8, 0.375,4.4,0.021,0.24 

9, 0.020,125,0.300,0.30 

10,0.030,125,0.300,0.30 

11,0.375,4.4,0.021,0.24 

12,0.020,125,0.300,0.30 

13,0.250,10.,0.020,0.24 

14,48.00,0.067,0.016,0.24 

15,0.020,125,0.300,0.30 

16,0.375,4.4,0.021,0.24 

17,0.020,125,0.300,0.30 

18,0.250,10.,0.020,0.24 

19,30.00,0.067,0.016,0.24 

20,5.000,10.,0.020,0.24 

21,0.020,125,0.300,0.30 

22,0.375,4.4,0.021,0.24 

23,0.020,125,0.300,0.30 

24,2.500,0.067,0.016,0.24 

25,0.020,125,0.300,0.30 

26,0.375,4.4,0.021,0.24 

27,0.020,125,0.300,0.30 

3-57
TABLE 3-23. THERMOD INPUT FILE (input.dat) OF SAMPLE PROBLEM 5 (Continued) 

28,0.250,130,0.800,0.20 

3.5,34.0,100.0,5.5,22.0,11.0,0.1,4.0,30,24,28 

0.70,1.0,0.0,2.0,150,0.80,0.22 

0.3,0.9,0.3,0.9,0.3,0.9,0.1,0.9 

3,0.0 

0.9,0.9,0.0,0.0 

1.0,0.0,0.9,0.9,1.0,0.0,1.0,0.0 

9 

0,10,10,0,0,75,110,110,190,190 

5,120,5,60,20,5,120,60,60 

7 

60 570.67 0.0000 14 330 10 10 0.36 

75 573.67 0.0000 14 355 10 10 0.31 

90 578.67 0.0000 14 355 10 12 0.33 

75 581.67 0.0000 14 330 10 10 0.31 

60 582.67 0.0000 14 291 10 10 0.36 

45 583.67 0.0000 14 231 10 10 0.44 

30 582.67 0.0000 14 160 10 10 0.57 

1,0.02,180,10 

10,1.0,1.0 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10 

Temperatures from the THERMOD run is shown in table 3-24, from which the steady-state 

temperatures T1, T2.....T10 (top to bottom of wing) for time period 3 were extracted and 

tabulated in table 3-25. 

TABLE 3-24. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE PROBLEM 5 

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ 

THERMOD: A THERMAL PROGRAM FOR AIRCRAFTS 

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ 

 

 

 

 SUMMARY OF STEADY-STATE AND TRANSIENT ANALYSIS 

 

 

 

 (Temperatures are shown for all time periods) 

 

 

 TEMPERATURES AT THE END OF STEADY STATE ANALYSIS 

3-58
TABLE 3-24. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE 

PROBLEM 5 (Continued) 

 

 TEMPERATURES IN DEGREES FAHRENHEIT 

 1 2 3 4 5 6 7 

 

 1 146.4 166.2 172.3 162.4 141.2 111.0 79.4 

 2 146.4 166.2 172.3 162.4 141.2 111.0 79.4 

 3 143.7 163.1 169.1 159.5 138.8 109.4 78.6 

 4 143.7 163.1 169.1 159.5 138.8 109.4 78.6 

 5 143.5 162.9 168.9 159.3 138.7 109.3 78.5 

 6 43.4 45.9 49.1 50.6 50.5 50.0 48.3 

 7 43.2 45.7 48.9 50.4 50.3 49.9 48.3 

 8 43.2 45.7 48.9 50.4 50.3 49.9 48.3 

 9 40.5 42.5 45.7 47.5 47.9 48.3 47.5 

 10 40.5 42.5 45.7 47.5 47.9 48.3 47.5 

 11 146.4 166.2 172.3 162.4 141.2 111.0 79.4 

 12 146.4 166.2 172.3 162.4 141.2 111.0 79.4 

 13 143.7 163.1 169.1 159.5 138.8 109.4 78.6 

 14 143.7 163.1 169.1 159.5 138.8 109.4 78.6 

 15 143.5 162.9 168.9 159.3 138.7 109.3 78.5 

 16 43.4 45.9 49.1 50.6 50.5 50.0 48.3 

 17 43.2 45.7 48.9 50.4 50.3 49.9 48.3 

 18 43.2 45.7 48.9 50.4 50.3 49.9 48.3 

 19 40.5 42.5 45.7 47.5 47.9 48.3 47.5 

 20 40.5 42.5 45.7 47.5 47.9 48.3 47.5 

 21 273.7 307.6 319.4 301.1 265.1 215.9 166.4 

 22 273.5 307.4 319.2 300.9 265.0 215.8 166.4 

 23 232.5 254.2 264.3 249.5 225.7 189.9 149.9 

 24 232.4 254.0 264.1 249.3 225.5 189.8 149.9 

 25 203.6 216.8 225.7 213.3 198.0 171.7 138.4 

 26 161.0 169.1 177.3 167.7 158.3 140.2 115.4 

 27 133.2 138.5 144.9 142.3 138.7 131.2 119.6 

 28 133.2 138.5 144.9 142.3 138.7 131.2 119.6 

 29 131.2 136.3 142.6 140.5 137.3 130.5 119.9 

 30 131.2 136.3 142.6 140.5 137.3 130.5 119.9 

 31 122.5 126.7 132.4 132.5 131.1 127.7 121.2 

 32 122.5 126.7 132.4 132.5 131.1 127.7 121.2 

 33 122.5 126.7 132.4 132.5 131.1 127.7 121.2 

 34 122.5 126.7 132.4 132.5 131.1 127.7 121.2 

 35 113.8 117.1 122.3 124.6 125.0 124.9 122.6 

 36 113.8 117.1 122.3 124.6 125.0 124.9 122.6 

 37 111.8 114.9 120.0 122.8 123.6 124.3 122.9 

 38 111.8 114.9 120.0 122.8 123.6 124.3 122.9 

 39 214.6 176.2 128.7 178.9 212.9 221.3 202.5 

 40 214.4 176.2 128.8 178.8 212.8 221.1 202.3 

 41 189.0 172.6 151.6 173.4 187.0 183.5 163.2 

 42 188.9 172.6 151.7 173.3 186.9 183.4 163.0 

 43 171.1 170.1 167.7 169.5 168.8 157.0 135.6 

3-59
TABLE 3-24. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE 

PROBLEM 5 (Continued) 

 44 161.2 164.1 167.7 163.8 160.0 148.1 129.4 

 45 144.7 147.6 151.7 150.1 147.8 140.2 127.3 

 46 144.7 147.5 151.6 150.0 147.8 140.2 127.3 

 47 121.1 124.1 128.8 130.4 130.4 128.9 124.3 

 48 121.0 123.9 128.7 130.3 130.3 128.8 124.3 

 49 88.4 92.9 98.4 95.3 91.9 84.7 74.4 

 50 90.9 95.5 101.1 97.8 94.1 86.4 75.6 

 51 188.8 207.4 215.7 208.8 191.6 168.5 144.8 

 52 183.9 201.5 209.6 203.3 187.3 165.7 143.4 

 53 188.8 207.4 215.7 208.8 191.6 168.5 144.8 

 

Maximum Structural Temperature for Period 1 = 273.69 Occurring 

at Location 21 

Maximum Structural Temperature for Period 2 = 307.55 Occurring 

at Location 21 

Maximum Structural Temperature for Period 3 = 319.43 Occurring 

at Location 21 

Maximum Structural Temperature for Period 4 = 301.07 Occurring 

at Location 21 

Maximum Structural Temperature for Period 5 = 265.15 Occurring 

at Location 21 

Maximum Structural Temperature for Period 6 = 221.30 Occurring 

at Location 39 

Maximum Structural Temperature for Period 7 = 202.48 Occurring 

at Location 39 

 

Maximum Structural Temperature Over All 7 Periods =319.43 

Occurring at Location 21 at Period 3 

 

 

 TEMPERATURES AT THE END OF TRANSIENT ANALYSIS 

 

 TEMPERATURES IN DEGREES FAHRENHEIT 

 1 2 3 4 5 6 7 

 

 1 114.4 119.2 124.3 126.0 124.8 122.8 119.0 

 2 115.2 120.3 125.5 126.9 125.2 122.4 117.9 

 3 141.2 159.5 165.4 156.6 137.5 110.5 81.8 

 4 142.0 160.7 166.7 157.5 137.9 110.2 80.7 

 5 142.7 161.7 167.7 158.3 138.1 109.8 79.6 

 6 45.1 47.7 51.0 52.4 52.4 52.0 50.2 

 7 46.7 49.2 52.5 54.1 54.1 53.7 52.0 

 8 48.4 50.9 54.3 55.9 55.9 55.6 53.9 

 9 103.2 106.0 110.8 113.6 114.5 115.4 114.4 

 10 104.8 107.7 112.5 115.4 116.3 117.3 116.3 

 11 114.4 119.2 124.3 126.0 124.8 122.8 119.0 

 12 115.2 120.3 125.5 126.9 125.2 122.4 117.9 

3-60
TABLE 3-24. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE 

PROBLEM 5 (Continued) 

 13 141.2 159.5 165.4 156.6 137.5 110.5 81.8 

 14 142.0 160.7 166.7 157.5 137.9 110.2 80.7 

 15 142.7 161.7 167.7 158.3 138.1 109.8 79.6 

 16 45.1 47.7 51.0 52.4 52.4 52.0 50.2 

 17 46.7 49.2 52.5 54.1 54.1 53.7 52.0 

 18 48.4 50.9 54.3 55.9 55.9 55.6 53.9 

 19 103.2 106.0 110.8 113.6 114.5 115.4 114.4 

 20 104.8 107.7 112.5 115.4 116.3 117.3 116.3 

 21 124.8 130.4 136.0 137.2 135.0 131.8 126.7 

 22 126.6 132.4 138.1 139.0 136.5 132.7 126.9 

 23 180.2 193.0 200.8 192.9 180.0 159.6 135.1 

 24 181.6 194.5 202.3 194.2 181.1 160.2 135.2 

 25 175.0 185.3 192.9 184.5 173.4 154.2 129.7 

 26 150.4 157.7 164.8 158.5 151.3 137.2 117.3 

 27 133.3 138.6 145.0 142.4 138.7 131.2 119.6 

 28 132.8 138.0 144.4 141.9 138.5 131.1 119.6 

 29 131.2 136.3 142.6 140.5 137.3 130.5 119.9 

 30 131.2 136.3 142.6 140.5 137.3 130.5 119.9 

 31 122.5 126.6 132.3 132.5 131.1 127.7 121.2 

 32 122.5 126.6 132.3 132.5 131.1 127.7 121.2 

 33 122.5 126.6 132.3 132.5 131.1 127.7 121.2 

 34 122.5 126.6 132.3 132.5 131.1 127.7 121.2 

 35 113.7 117.0 122.1 124.5 125.0 124.9 122.6 

 36 113.7 116.9 122.0 124.5 125.0 124.9 122.6 

 37 111.2 114.2 119.2 122.2 123.1 124.1 123.0 

 38 111.1 114.1 119.1 122.1 123.1 124.0 123.0 

 39 116.0 117.1 119.8 124.8 127.3 128.6 126.6 

 40 117.3 118.2 120.5 125.8 128.3 129.4 127.0 

 41 153.0 149.4 144.5 152.4 156.1 151.2 137.7 

 42 154.0 150.4 145.5 153.2 156.8 151.7 137.7 

 43 153.1 155.0 156.3 156.8 155.0 145.8 129.8 

 44 149.5 152.6 156.3 154.7 151.8 142.8 128.0 

 45 141.2 143.0 145.5 146.4 145.5 139.4 127.8 

 46 140.3 142.1 144.5 145.5 144.8 139.0 127.7 

 47 116.4 117.7 120.5 125.3 127.5 128.5 126.3 

 48 115.6 116.9 119.8 124.6 126.9 128.1 126.3 

 49 105.5 108.7 113.7 115.5 115.7 115.1 112.5 

 50 106.3 109.7 114.9 116.3 116.1 114.9 111.5 

 51 306.1 348.1 361.3 339.6 295.0 235.5 177.6 

 52 288.9 327.5 340.0 320.5 279.9 225.7 172.8 

 53 306.1 348.1 361.3 339.6 295.0 235.5 177.6 

 

Maximum Structural Temperature for Period 1 = 181.56 Occurring 

at Location 24 

Maximum Structural Temperature for Period 2 = 194.50 Occurring 

at Location 24 

3-61
TABLE 3-24. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE 

PROBLEM 5 (Continued) 

Maximum Structural Temperature for Period 3 = 202.33 Occurring 

at Location 24 

Maximum Structural Temperature for Period 4 = 194.16 Occurring 

at Location 24 

Maximum Structural Temperature for Period 5 = 181.07 Occurring 

at Location 24 

Maximum Structural Temperature for Period 6 = 160.21 Occurring 

at Location 24 

Maximum Structural Temperature for Period 7 = 137.74 Occurring 

at Location 42 

 

Maximum Structural Temperature Over All 7 Periods =202.33 

Occurring at Location 24 at Period 3 

TABLE 3-25. THERMOD-SIMULATED TEMPERATURES (°F) OF THE RIGHT WING OF 

SAMPLE PROBLEM 5 

T1 T2 T3 T4 T5 T6 T7 T8 T9 T10 

172.3 172.3 169.1 169.1 168.9 49.1 48.9 48.9 45.7 45.7

As in sample problems 1 through 4, a one-half wing structure was modeled using the 

UAI/NASTRAN FEM program. Two types of loads were applied to the model. These loads, 

based on period 3 of the THERMOD input file, were direct solar radiation of 355 Btu/hr/ft

convection with a film coefficient of h = 1.3433 (Btu/hr)/(ft2

°F). This film coefficient was 

calculated internally by the THERMOD program and was based on the theory of airflow on a 

flat surface and a wind speed of 14 ft/sec. The radiation load was applied perpendicular to the 

top surface only, while both the top and bottom surfaces were subjected to convection forces. 

The ambient temperature was maintained at 119°F (578.67°R). Unlike sample problem 4, the 

wings were assigned a nonzero emissivity value of 0.9, allowing the wings to emit infrared 

radiation, thereby, reducing the wing temperatures. 

The FEM input file is shown in table 3-26. The ambient temperature of 119°F was imposed 

through the SPC case control card. The ambient temperature boundary condition was scalar in 

nature and was implemented through the SPOINT bulk data. The boundary elements, CHBDY, 

through which solar radiation and convection effects were applied, were modeled using the 

LINE boundary shape type, connecting the boundary nodes along the wing top and bottom. The 

SPACE designation in the CHBDY boundary element indicates that this element emits infrared 

radiation to a black body, in which case, no VIEW bulk data was required. Property cards 

PHBDY and the material cards MAT4, define absorptivity, emissivity, film coefficient, and 

conductivity values. Note that unlike sample problem 4 where wings were prevented from

emitting radiation, an emissivity value of 0.9 was assigned to the wing in sample problems, as 

indicated in the PHBDY card. Because of the involvement of infrared radiation, through 

emission, this sample problem was rendered nonlinear. This required the Stefan-Boltzman 

constant, σ, and an initial set of temperatures, which were implemented through the PARAM and 

TEMPERATURE statements, respectively. 

2

 and 

3-62
TABLE 3-26. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE PROBLEM 5 

ID D:\Natha,MSC/N 

APP HEAT 

SOL 3 

TIME 10000 

CEND 

 TITLE = direct_radiation_conduction_emission_convection (t=0.001ft) 

 SUBTITLE = 10.7 in gap filled by air; k=0.016BTU/hr/ft/degF 

 load=1 

 SPC = 1 

 temperature(estimate)=2 

 THERMAL= ALL 

 FLUX = ALL 

 SPCF = ALL 

BEGIN BULK 

$ *************************************************************************** 

$ Written by : MSC/NASTRAN for Windows 

$ Version : 6.00 

$ Translator : UAI/NASTRAN 

$ From Model : D:\Thermod_Validation\FEM\conduction_only.MOD 

$ Date : Mon Mar 06 11:27:09 2000 

$ *************************************************************************** 

$ 

PARAM,K6ROT,100. 

PARAM,MAXRATIO,1.E+8 

param,sigma,1.714e-9 

param,tabs,0.0 

param,maxit,10 

CORD2C 1 0 0. 0. 0. 0. 0. 1.+MSC/NC1

+MSC/NC1 1. 0. 1. 

CORD2S 2 0 0. 0. 0. 0. 0. 1.+MSC/NC2

+MSC/NC2 1. 0. 1. 

$ 

$ MSC/NASTRAN for Windows Property 1 : 0.02 in glass_outer_layer 

PSHELL 1 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 2 : 3/8 in foam 

PSHELL 2 2 0.001 2 2 0. 

$ MSC/NASTRAN for Windows Property 3 : 0.02 in inner glass 

PSHELL 3 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 4 : 0.350 in cap 

PSHELL 4 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 5 : 10.70 in air space 

PSHELL 5 3 0.001 3 3 0. 

$ MSC/NASTRAN for Windows Property 6 : 0.350 in cap 

PSHELL 6 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 7 : 0.02 in inner glass 

PSHELL 7 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 8 : 3/8 in foam 

PSHELL 8 2 0.001 2 2 0. 

$ MSC/NASTRAN for Windows Property 9 : 0.02 in glass_outer_layer 

PSHELL 9 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Material 1 : glass 

MAT4 1 0.3 1. 

$ MSC/NASTRAN for Windows Material 2 : foam 

MAT4 2 0.021 1. 

3-63
TABLE 3-26. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 5 (Continued) 

$ MSC/NASTRAN for Windows Material 3 : air 

MAT4 3 0.016 1. 

$ 

chbdy,1001,100,line,200,201,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1002,100,line,201,202,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1003,100,line,202,203,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1004,100,line,203,204,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1005,100,line,204,205,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1006,100,line,205,206,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1007,100,line,206,207,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1008,100,line,207,208,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1009,100,line,208,209,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1010,100,line,209,210,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1011,100,line,210,211,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1012,100,line,211,212,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1013,100,line,212,213,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1014,100,line,213,214,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1015,100,line,214,215,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1016,100,line,215,216,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1017,100,line,216,181,,,space,500,500,,,0.0,1.0,0.0 

phbdy,100,200,0.001,0.9,0.9 

mat4,200,1.3433 

spoint,500 

spc,1,500,1,578.67 

$ 

$ Direct radiation load from a distant source: load case 1 

qvect,1,355.0,0.0,-1.0,0.0,1001,thru,1017 

$ 

$ view ids 

$view,1,yes,yes,1,1,0.0 

$ 

chbdy,2001,300,line,20,21,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2002,300,line,21,22,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2003,300,line,22,23,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2004,300,line,23,24,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2005,300,line,24,25,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2006,300,line,25,26,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2007,300,line,26,27,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2008,300,line,27,28,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2009,300,line,28,29,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2010,300,line,29,30,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2011,300,line,30,31,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2012,300,line,31,32,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2013,300,line,32,33,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2014,300,line,33,34,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2015,300,line,34,35,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2016,300,line,35,36,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2017,300,line,36,1,,,space,500,500,,,0.0,-1.0,0.0 

phbdy,300,400,0.001,0.9,0.9 

mat4,400,1.3433 

$ 

temp,2,1,578.67,3,578.67,5,578.67 

temp,2,7,578.67,9,578.67,10,578.67 

temp,2,11,578.67,12,578.67,14,578.67 

temp,2,15,578.67,18,578.67,20,578.67 

temp,2,21,578.67,22,578.67,23,578.67 

temp,2,24,578.67,25,578.67,26,578.67 

3-64
TABLE 3-26. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 5 (Continued) 

temp,2,27,578.67,28,578.67,29,578.67 

temp,2,30,578.67,31,578.67,32,578.67 

temp,2,33,578.67,34,578.67,35,578.67 

temp,2,36,578.67,37,578.67,41,578.67 

temp,2,44,578.67,45,578.67,47,578.67 

temp,2,48,578.67,50,578.67,51,578.67 

temp,2,52,578.67,53,578.67,54,578.67 

temp,2,55,578.67,56,578.67,58,578.67 

temp,2,59,578.67,62,578.67,67,578.67 

temp,2,69,578.67,71,578.67,73,578.67 

temp,2,75,578.67,76,578.67,77,578.67 

temp,2,78,578.67,79,578.67,80,578.67 

temp,2,83,578.67,85,578.67,87,578.67 

temp,2,88,578.67,98,578.67,101,578.67 

temp,2,104,578.67,105,578.67,107,578.67 

temp,2,108,578.67,109,578.67,112,578.67 

temp,2,115,578.67,116,578.67,118,578.67 

temp,2,120,578.67,121,578.67,122,578.67 

temp,2,123,578.67,124,578.67,125,578.67 

temp,2,128,578.67,129,578.67,130,578.67 

temp,2,133,578.67,135,578.67,137,578.67 

temp,2,138,578.67,145,578.67,146,578.67 

temp,2,147,578.67,150,578.67,152,578.67 

temp,2,160,578.67,161,578.67,164,578.67 

temp,2,165,578.67,172,578.67,174,578.67 

temp,2,177,578.67,178,578.67,180,578.67 

temp,2,181,578.67,182,578.67,183,578.67 

temp,2,184,578.67,188,578.67,193,578.67 

temp,2,195,578.67,196,578.67,197,578.67 

temp,2,198,578.67,199,578.67,200,578.67 

temp,2,201,578.67,202,578.67,203,578.67 

temp,2,204,578.67,205,578.67,206,578.67 

temp,2,207,578.67,208,578.67,209,578.67 

temp,2,210,578.67,211,578.67,212,578.67 

temp,2,213,578.67,214,578.67,215,578.67 

temp,2,216,578.67,219,578.67,220,578.67 

temp,2,221,578.67,222,578.67,223,578.67 

temp,2,226,578.67,227,578.67,228,578.67 

temp,2,229,578.67,231,578.67,232,578.67 

temp,2,235,578.67,241,578.67,243,578.67 

temp,2,244,578.67,245,578.67,246,578.67 

temp,2,248,578.67,249,578.67,250,578.67 

temp,2,253,578.67,254,578.67,257,578.67 

temp,2,258,578.67,259,578.67,260,578.67 

temp,2,264,578.67,267,578.67,268,578.67 

temp,2,269,578.67,270,578.67,273,578.67 

temp,2,274,578.67,277,578.67,282,578.67 

temp,2,283,578.67,292,578.67,293,578.67 

temp,2,295,578.67,297,578.67,298,578.67 

temp,2,299,578.67,300,578.67,301,578.67 

temp,2,302,578.67,303,578.67,306,578.67 

temp,2,307,578.67,308,578.67,313,578.67 

temp,2,314,578.67,316,578.67,317,578.67 

temp,2,318,578.67,323,578.67,324,578.67 

temp,2,500,578.67 

$ 

3-65
TABLE 3-26. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 5 (Continued) 

GRID 1 0 0. 0. 0. 0 

GRID 3 0 1.1.667E-3 0. 0 

GRID 5 0 3.1.667E-3 0. 0 

GRID 7 0 5.1.667E-3 0. 0 

GRID 9 0 7.1.667E-3 0. 0 

GRID 10 0 8.1.667E-3 0. 0 

GRID 11 0 9.1.667E-3 0. 0 

GRID 12 0 10.1.667E-3 0. 0 

GRID 14 0 12.1.667E-3 0. 0 

GRID 15 0 13.1.667E-3 0. 0 

GRID 18 0 16.1.667E-3 0. 0 

GRID 20 0 17. 0. 0. 0 

GRID 21 0 16. 0. 0. 0 

GRID 22 0 15. 0. 0. 0 

GRID 23 0 14. 0. 0. 0 

GRID 24 0 13. 0. 0. 0 

GRID 25 0 12. 0. 0. 0 

GRID 26 0 11. 0. 0. 0 

GRID 27 0 10. 0. 0. 0 

GRID 28 0 9. 0. 0. 0 

GRID 29 0 8. 0. 0. 0 

GRID 30 0 7. 0. 0. 0 

GRID 31 0 6. 0. 0. 0 

GRID 32 0 5. 0. 0. 0 

GRID 33 0 4. 0. 0. 0 

GRID 34 0 3. 0. 0. 0 

GRID 35 0 2. 0. 0. 0 

GRID 36 0 1. 0. 0. 0 

GRID 37 0 0.1.667E-3 0. 0 

GRID 41 0 3.0.032917 0. 0 

GRID 44 0 6.0.032917 0. 0 

GRID 45 0 7.0.032917 0. 0 

GRID 47 0 9.0.032917 0. 0 

GRID 48 0 10.0.032917 0. 0 

GRID 50 0 12.0.032917 0. 0 

GRID 51 0 13.0.032917 0. 0 

GRID 52 0 14.0.032917 0. 0 

GRID 53 0 15.0.032917 0. 0 

GRID 54 0 16.0.032917 0. 0 

GRID 55 0 17.0.032917 0. 0 

GRID 56 0 17.1.667E-3 0. 0 

GRID 58 0 15.1.667E-3 0. 0 

GRID 59 0 14.1.667E-3 0. 0 

GRID 62 0 11.1.667E-3 0. 0 

GRID 67 0 6.1.667E-3 0. 0 

GRID 69 0 4.1.667E-3 0. 0 

GRID 71 0 2.1.667E-3 0. 0 

GRID 73 0 0.0.032917 0. 0 

GRID 75 0 1.0.034583 0. 0 

GRID 76 0 2.0.034583 0. 0 

GRID 77 0 3.0.034583 0. 0 

GRID 78 0 4.0.034583 0. 0 

GRID 79 0 5.0.034583 0. 0 

GRID 80 0 6.0.034583 0. 0 

GRID 83 0 9.0.034583 0. 0 

GRID 85 0 11.0.034583 0. 0 

GRID 87 0 13.0.034583 0. 0 

3-66
TABLE 3-26. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 5 (Continued) 

GRID 88 0 14.0.034583 0. 0 

GRID 98 0 11.0.032917 0. 0 

GRID 101 0 8.0.032917 0. 0 

GRID 104 0 5.0.032917 0. 0 

GRID 105 0 4.0.032917 0. 0 

GRID 107 0 2.0.032917 0. 0 

GRID 108 0 1.0.032917 0. 0 

GRID 109 0 0.0.034583 0. 0 

GRID 112 0 2. 0.06375 0. 0 

GRID 115 0 5. 0.06375 0. 0 

GRID 116 0 6. 0.06375 0. 0 

GRID 118 0 8. 0.06375 0. 0 

GRID 120 0 10. 0.06375 0. 0 

GRID 121 0 11. 0.06375 0. 0 

GRID 122 0 12. 0.06375 0. 0 

GRID 123 0 13. 0.06375 0. 0 

GRID 124 0 14. 0.06375 0. 0 

GRID 125 0 15. 0.06375 0. 0 

GRID 128 0 17.0.034583 0. 0 

GRID 129 0 16.0.034583 0. 0 

GRID 130 0 15.0.034583 0. 0 

GRID 133 0 12.0.034583 0. 0 

GRID 135 0 10.0.034583 0. 0 

GRID 137 0 8.0.034583 0. 0 

GRID 138 0 7.0.034583 0. 0 

GRID 145 0 0. 0.06375 0. 0 

GRID 146 0 0. 0.95542 0. 0 

GRID 147 0 1. 0.95542 0. 0 

GRID 150 0 4. 0.95542 0. 0 

GRID 152 0 6. 0.95542 0. 0 

GRID 160 0 14. 0.95542 0. 0 

GRID 161 0 15. 0.95542 0. 0 

GRID 164 0 17. 0.06375 0. 0 

GRID 165 0 16. 0.06375 0. 0 

GRID 172 0 9. 0.06375 0. 0 

GRID 174 0 7. 0.06375 0. 0 

GRID 177 0 4. 0.06375 0. 0 

GRID 178 0 3. 0.06375 0. 0 

GRID 180 0 1. 0.06375 0. 0 

GRID 181 0 0. 1.01917 0. 0 

GRID 182 0 0. 1.0175 0. 0 

GRID 183 0 1. 1.0175 0. 0 

GRID 184 0 2. 1.0175 0. 0 

GRID 188 0 6. 1.0175 0. 0 

GRID 193 0 11. 1.0175 0. 0 

GRID 195 0 13. 1.0175 0. 0 

GRID 196 0 14. 1.0175 0. 0 

GRID 197 0 15. 1.0175 0. 0 

GRID 198 0 16. 1.0175 0. 0 

GRID 199 0 17. 1.0175 0. 0 

GRID 200 0 17. 1.01917 0. 0 

GRID 201 0 16. 1.01917 0. 0 

GRID 202 0 15. 1.01917 0. 0 

GRID 203 0 14. 1.01917 0. 0 

GRID 204 0 13. 1.01917 0. 0 

GRID 205 0 12. 1.01917 0. 0 

GRID 206 0 11. 1.01917 0. 0 

3-67
TABLE 3-26. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 5 (Continued) 

GRID 207 0 10. 1.01917 0. 0 

GRID 208 0 9. 1.01917 0. 0 

GRID 209 0 8. 1.01917 0. 0 

GRID 210 0 7. 1.01917 0. 0 

GRID 211 0 6. 1.01917 0. 0 

GRID 212 0 5. 1.01917 0. 0 

GRID 213 0 4. 1.01917 0. 0 

GRID 214 0 3. 1.01917 0. 0 

GRID 215 0 2. 1.01917 0. 0 

GRID 216 0 1. 1.01917 0. 0 

GRID 219 0 1. 0.98625 0. 0 

GRID 220 0 2. 0.98625 0. 0 

GRID 221 0 3. 0.98625 0. 0 

GRID 222 0 4. 0.98625 0. 0 

GRID 223 0 5. 0.98625 0. 0 

GRID 226 0 8. 0.98625 0. 0 

GRID 227 0 9. 0.98625 0. 0 

GRID 228 0 10. 0.98625 0. 0 

GRID 229 0 11. 0.98625 0. 0 

GRID 231 0 13. 0.98625 0. 0 

GRID 232 0 14. 0.98625 0. 0 

GRID 235 0 17. 0.98625 0. 0 

GRID 241 0 12. 1.0175 0. 0 

GRID 243 0 10. 1.0175 0. 0 

GRID 244 0 9. 1.0175 0. 0 

GRID 245 0 8. 1.0175 0. 0 

GRID 246 0 7. 1.0175 0. 0 

GRID 248 0 5. 1.0175 0. 0 

GRID 249 0 4. 1.0175 0. 0 

GRID 250 0 3. 1.0175 0. 0 

GRID 253 0 0. 0.98625 0. 0 

GRID 254 0 0. 0.98458 0. 0 

GRID 257 0 3. 0.98458 0. 0 

GRID 258 0 4. 0.98458 0. 0 

GRID 259 0 5. 0.98458 0. 0 

GRID 260 0 6. 0.98458 0. 0 

GRID 264 0 10. 0.98458 0. 0 

GRID 267 0 13. 0.98458 0. 0 

GRID 268 0 14. 0.98458 0. 0 

GRID 269 0 15. 0.98458 0. 0 

GRID 270 0 16. 0.98458 0. 0 

GRID 273 0 16. 0.98625 0. 0 

GRID 274 0 15. 0.98625 0. 0 

GRID 277 0 12. 0.98625 0. 0 

GRID 282 0 7. 0.98625 0. 0 

GRID 283 0 6. 0.98625 0. 0 

GRID 292 0 2. 0.95542 0. 0 

GRID 293 0 3. 0.95542 0. 0 

GRID 295 0 5. 0.95542 0. 0 

GRID 297 0 7. 0.95542 0. 0 

GRID 298 0 8. 0.95542 0. 0 

GRID 299 0 9. 0.95542 0. 0 

GRID 300 0 10. 0.95542 0. 0 

GRID 301 0 11. 0.95542 0. 0 

GRID 302 0 12. 0.95542 0. 0 

GRID 303 0 13. 0.95542 0. 0 

GRID 306 0 16. 0.95542 0. 0 

3-68
TABLE 3-26. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 5 (Continued) 

GRID 307 0 17. 0.95542 0. 0 

GRID 308 0 17. 0.98458 0. 0 

GRID 313 0 12. 0.98458 0. 0 

GRID 314 0 11. 0.98458 0. 0 

GRID 316 0 9. 0.98458 0. 0 

GRID 317 0 8. 0.98458 0. 0 

GRID 318 0 7. 0.98458 0. 0 

GRID 323 0 2. 0.98458 0. 0 

GRID 324 0 1. 0.98458 0. 0 

CQUAD4 1 9 1 37 3 36 

CQUAD4 2 9 36 3 71 35 

CQUAD4 3 9 35 71 5 34 

CQUAD4 4 9 34 5 69 33 

CQUAD4 5 9 33 69 7 32 

CQUAD4 6 9 32 7 67 31 

CQUAD4 7 9 31 67 9 30 

CQUAD4 8 9 30 9 10 29 

CQUAD4 9 9 29 10 11 28 

CQUAD4 10 9 28 11 12 27 

CQUAD4 11 9 27 12 62 26 

CQUAD4 12 9 26 62 14 25 

CQUAD4 13 9 25 14 15 24 

CQUAD4 14 9 24 15 59 23 

CQUAD4 15 9 23 59 58 22 

CQUAD4 16 9 22 58 18 21 

CQUAD4 17 9 21 18 56 20 

CQUAD4 18 8 37 73 108 3 

CQUAD4 19 8 3 108 107 71 

CQUAD4 20 8 71 107 41 5 

CQUAD4 21 8 5 41 105 69 

CQUAD4 22 8 69 105 104 7 

CQUAD4 23 8 7 104 44 67 

CQUAD4 24 8 67 44 45 9 

CQUAD4 25 8 9 45 101 10 

CQUAD4 26 8 10 101 47 11 

CQUAD4 27 8 11 47 48 12 

CQUAD4 28 8 12 48 98 62 

CQUAD4 29 8 62 98 50 14 

CQUAD4 30 8 14 50 51 15 

CQUAD4 31 8 15 51 52 59 

CQUAD4 32 8 59 52 53 58 

CQUAD4 33 8 58 53 54 18 

CQUAD4 34 8 18 54 55 56 

CQUAD4 35 7 73 109 75 108 

CQUAD4 36 7 108 75 76 107 

CQUAD4 37 7 107 76 77 41 

CQUAD4 38 7 41 77 78 105 

CQUAD4 39 7 105 78 79 104 

CQUAD4 40 7 104 79 80 44 

CQUAD4 41 7 44 80 138 45 

CQUAD4 42 7 45 138 137 101 

CQUAD4 43 7 101 137 83 47 

CQUAD4 44 7 47 83 135 48 

CQUAD4 45 7 48 135 85 98 

CQUAD4 46 7 98 85 133 50 

CQUAD4 47 7 50 133 87 51 

CQUAD4 48 7 51 87 88 52 

3-69
TABLE 3-26. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 5 (Continued) 

CQUAD4 49 7 52 88 130 53 

CQUAD4 50 7 53 130 129 54 

CQUAD4 51 7 54 129 128 55 

CQUAD4 52 6 109 145 180 75 

CQUAD4 53 6 75 180 112 76 

CQUAD4 54 6 76 112 178 77 

CQUAD4 55 6 77 178 177 78 

CQUAD4 56 6 78 177 115 79 

CQUAD4 57 6 79 115 116 80 

CQUAD4 58 6 80 116 174 138 

CQUAD4 59 6 138 174 118 137 

CQUAD4 60 6 137 118 172 83 

CQUAD4 61 6 83 172 120 135 

CQUAD4 62 6 135 120 121 85 

CQUAD4 63 6 85 121 122 133 

CQUAD4 64 6 133 122 123 87 

CQUAD4 65 6 87 123 124 88 

CQUAD4 66 6 88 124 125 130 

CQUAD4 67 6 130 125 165 129 

CQUAD4 68 6 129 165 164 128 

CQUAD4 69 5 145 146 147 180 

CQUAD4 70 5 180 147 292 112 

CQUAD4 71 5 112 292 293 178 

CQUAD4 72 5 178 293 150 177 

CQUAD4 73 5 177 150 295 115 

CQUAD4 74 5 115 295 152 116 

CQUAD4 75 5 116 152 297 174 

CQUAD4 76 5 174 297 298 118 

CQUAD4 77 5 118 298 299 172 

CQUAD4 78 5 172 299 300 120 

CQUAD4 79 5 120 300 301 121 

CQUAD4 80 5 121 301 302 122 

CQUAD4 81 5 122 302 303 123 

CQUAD4 82 5 123 303 160 124 

CQUAD4 83 5 124 160 161 125 

CQUAD4 84 5 125 161 306 165 

CQUAD4 85 5 165 306 307 164 

CQUAD4 86 1 182 181 216 183 

CQUAD4 87 1 183 216 215 184 

CQUAD4 88 1 184 215 214 250 

CQUAD4 89 1 250 214 213 249 

CQUAD4 90 1 249 213 212 248 

CQUAD4 91 1 248 212 211 188 

CQUAD4 92 1 188 211 210 246 

CQUAD4 93 1 246 210 209 245 

CQUAD4 94 1 245 209 208 244 

CQUAD4 95 1 244 208 207 243 

CQUAD4 96 1 243 207 206 193 

CQUAD4 97 1 193 206 205 241 

CQUAD4 98 1 241 205 204 195 

CQUAD4 99 1 195 204 203 196 

CQUAD4 100 1 196 203 202 197 

CQUAD4 101 1 197 202 201 198 

CQUAD4 102 1 198 201 200 199 

CQUAD4 103 2 253 182 183 219 

CQUAD4 104 2 219 183 184 220 

CQUAD4 105 8 220 184 250 221 

3-70
TABLE 3-26. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 5 (Continued) 

CQUAD4 106 2 221 250 249 222 

CQUAD4 107 2 222 249 248 223 

CQUAD4 108 2 223 248 188 283 

CQUAD4 109 2 283 188 246 282 

CQUAD4 110 2 282 246 245 226 

CQUAD4 111 2 226 245 244 227 

CQUAD4 112 2 227 244 243 228 

CQUAD4 113 2 228 243 193 229 

CQUAD4 114 2 229 193 241 277 

CQUAD4 115 2 277 241 195 231 

CQUAD4 116 2 231 195 196 232 

CQUAD4 117 2 232 196 197 274 

CQUAD4 118 2 274 197 198 273 

CQUAD4 119 2 273 198 199 235 

CQUAD4 120 3 254 253 219 324 

CQUAD4 121 3 324 219 220 323 

CQUAD4 122 3 323 220 221 257 

CQUAD4 123 3 257 221 222 258 

CQUAD4 124 3 258 222 223 259 

CQUAD4 125 3 259 223 283 260 

CQUAD4 126 3 260 283 282 318 

CQUAD4 127 3 318 282 226 317 

CQUAD4 128 3 317 226 227 316 

CQUAD4 129 3 316 227 228 264 

CQUAD4 130 3 264 228 229 314 

CQUAD4 131 3 314 229 277 313 

CQUAD4 132 3 313 277 231 267 

CQUAD4 133 3 267 231 232 268 

CQUAD4 134 3 268 232 274 269 

CQUAD4 135 3 269 274 273 270 

CQUAD4 136 3 270 273 235 308 

CQUAD4 137 4 146 254 324 147 

CQUAD4 138 4 147 324 323 292 

CQUAD4 139 4 292 323 257 293 

CQUAD4 140 4 293 257 258 150 

CQUAD4 141 4 150 258 259 295 

CQUAD4 142 4 295 259 260 152 

CQUAD4 143 4 152 260 318 297 

CQUAD4 144 4 297 318 317 298 

CQUAD4 145 4 298 317 316 299 

CQUAD4 146 4 299 316 264 300 

CQUAD4 147 4 300 264 314 301 

CQUAD4 148 4 301 314 313 302 

CQUAD4 149 4 302 313 267 303 

CQUAD4 150 4 303 267 268 160 

CQUAD4 151 4 160 268 269 161 

CQUAD4 152 4 161 269 270 306 

CQUAD4 153 4 306 270 308 307 

ENDDATA 

The results of the one-half FEM wing model analysis are presented in figure 3-5 in the form of a 

contour plot. The contours appear to be indicative of the temperature profile of the THERMOD 

analysis. The top surface, which was subjected to direct solar radiation, is certainly much hotter 

than the bottom surface. The temperatures are significantly lower than those in figure 3-4 of 

sample problem 4 because in sample problem 5 the wing surfaces were modeled to emit infrared 

radiation energy. 

3-71
Note that the temperatures are in absolute terms (°R). 

FIGURE 3-5. CONTOUR PLOT OF THE FEM OUTPUT OF SAMPLE PROBLEM 5 

The truncated FEM output file is shown in table 3-27. The nodes, from top to bottom of the 

wing, located at mid-span of the model, are indicated by nodes 209, 245, 226, 317, 298, 118, 

137, 101, 10, and 29. The temperatures of these nodes were extracted from this file and are 

reproduced in table 3-28. Mid-span was chosen because it was sufficiently far away from any 

edge effects that might be present. Note that the temperatures in table 3-28 were reduced from

absolute temperatures (°R) to °F.

When compared, the temperatures of tables 3-25 and 3-28 show excellent agreement, validating 

the radiation, conduction and convection, capabilities of the THERMOD program, including 

infrared emission. 

3-72
TABLE 3-27. TRUNCATED FEM OUTPUT FILE OF SAMPLE PROBLEM 5 

1 DIRECT_RADIATION_CONDUCTION_EMISSION_CONVECTION (T=0.001FT) AUGUST 17, 2000 UAI/NASTRAN VERSION 20.1 PAGE 

18 

 10.7 IN GAP FILLED BY AIR; K=0.016BTU/HR/FT/DEGF 

0 

 

 T E M P E R A T U R E V E C T O R 

 

 POINT ID. TYPE ID VALUE ID+1 VALUE ID+2 VALUE ID+3 VALUE ID+4 VALUE ID+5 VALUE 

 1 TEMP 5.055055E+02 

 3 TEMP 5.055174E+02 

 5 TEMP 5.055174E+02 

 7 TEMP 5.055174E+02 

 9 TEMP 5.055174E+02 5.055174E+02 5.055174E+02 5.055174E+02 

 14 TEMP 5.055174E+02 5.055174E+02 

 18 TEMP 5.055174E+02 

 20 TEMP 5.055055E+02 5.055055E+02 5.055055E+02 5.055055E+02 5.055055E+02 5.055055E+02 

 26 TEMP 5.055055E+02 5.055055E+02 5.055055E+02 5.055055E+02 5.055055E+02 5.055055E+02 

 32 TEMP 5.055055E+02 5.055055E+02 5.055055E+02 5.055055E+02 5.055055E+02 5.055174E+02 

 44 TEMP 5.087106E+02 5.087106E+02 

 47 TEMP 5.087106E+02 5.087106E+02 

 50 TEMP 5.087106E+02 5.087106E+02 5.087106E+02 5.087106E+02 5.087106E+02 5.087106E+02 

 56 TEMP 5.055174E+02 

 58 TEMP 5.055174E+02 5.055174E+02 

 62 TEMP 5.055174E+02 

 67 TEMP 5.055174E+02 

 69 TEMP 5.055174E+02 

 71 TEMP 5.055174E+02 

 73 TEMP 5.087106E+02 

 75 TEMP 5.087225E+02 5.087225E+02 5.087225E+02 5.087225E+02 5.087225E+02 5.087225E+02 

 83 TEMP 5.087225E+02 

 85 TEMP 5.087225E+02 

 87 TEMP 5.087225E+02 5.087225E+02 

 98 TEMP 5.087106E+02 

 101 TEMP 5.087106E+02 

 104 TEMP 5.087106E+02 5.087106E+02 

 107 TEMP 5.087106E+02 5.087106E+02 5.087225E+02 

 112 TEMP 5.089311E+02 

 115 TEMP 5.089311E+02 5.089311E+02 

 118 TEMP 5.089311E+02 

 120 TEMP 5.089311E+02 5.089311E+02 5.089311E+02 5.089311E+02 5.089311E+02 5.089311E+02 

 128 TEMP 5.087225E+02 5.087225E+02 5.087225E+02 

 133 TEMP 5.087225E+02 

 135 TEMP 5.087225E+02 

 137 TEMP 5.087225E+02 5.087225E+02 

 145 TEMP 5.089311E+02 6.285157E+02 6.285157E+02 

 150 TEMP 6.285157E+02 

 152 TEMP 6.285157E+02 

 160 TEMP 6.285157E+02 6.285157E+02 

 164 TEMP 5.089311E+02 5.089311E+02 

 172 TEMP 5.089311E+02 

 174 TEMP 5.089311E+02 

 177 TEMP 5.089311E+02 5.089311E+02 

 180 TEMP 5.089311E+02 6.319413E+02 6.319294E+02 6.319294E+02 6.319294E+02 

 188 TEMP 6.319294E+02 

 193 TEMP 6.319294E+02 

 195 TEMP 6.319294E+02 6.319294E+02 6.319294E+02 6.319294E+02 6.319294E+02 6.319413E+02 

 201 TEMP 6.319413E+02 6.319413E+02 6.319413E+02 6.319413E+02 6.319413E+02 6.319413E+02 

1 DIRECT_RADIATION_CONDUCTION_EMISSION_CONVECTION (T=0.001FT) AUGUST 17, 2000 UAI/NASTRAN VERSION 20.1 PAGE 

19 

0 

 

 T E M P E R A T U R E V E C T O R 

 

 POINT ID. TYPE ID VALUE ID+1 VALUE ID+2 VALUE ID+3 VALUE ID+4 VALUE ID+5 VALUE 

 207 TEMP 6.319413E+02 6.319413E+02 6.319413E+02 6.319413E+02 6.319413E+02 6.319413E+02 

 213 TEMP 6.319413E+02 6.319413E+02 6.319413E+02 6.319413E+02 

 219 TEMP 6.287362E+02 6.287362E+02 6.287362E+02 6.287362E+02 6.287362E+02 

 226 TEMP 6.287362E+02 6.287362E+02 6.287362E+02 6.287362E+02 

 231 TEMP 6.287362E+02 6.287362E+02 

 235 TEMP 6.287362E+02 

 241 TEMP 6.319294E+02 

 243 TEMP 6.319294E+02 6.319294E+02 6.319294E+02 6.319294E+02 

 248 TEMP 6.319294E+02 6.319294E+02 6.319294E+02 

 253 TEMP 6.287362E+02 6.287242E+02 

 257 TEMP 6.287242E+02 6.287242E+02 6.287242E+02 6.287242E+02 

 264 TEMP 6.287242E+02 

 267 TEMP 6.287242E+02 6.287242E+02 6.287242E+02 6.287242E+02 

 273 TEMP 6.287362E+02 6.287362E+02 

 277 TEMP 6.287362E+02 

 282 TEMP 6.287362E+02 6.287362E+02 

 292 TEMP 6.285157E+02 6.285157E+02 

 295 TEMP 6.285157E+02 

 297 TEMP 6.285157E+02 6.285157E+02 6.285157E+02 6.285157E+02 6.285157E+02 6.285157E+02 

 303 TEMP 6.285157E+02 

 306 TEMP 6.285157E+02 6.285157E+02 6.287242E+02 

 313 TEMP 6.287242E+02 6.287242E+02 

 316 TEMP 6.287242E+02 6.287242E+02 6.287242E+02 

 323 TEMP 6.287242E+02 6.287242E+02 

 500 TEMP 5.786700E+02 

 

 41 TEMP 5.087106E+02 

 10.7 IN GAP FILLED BY AIR; K=0.016BTU/HR/FT/DEGF 

3-73
TABLE 3-28. FINITE ELEMENT METHOD-SIMULATED TEMPERATURES (°F) OF THE 

RIGHT WING OF SAMPLE PROBLEM 5 

T1 T2 T3 T4 T5 T6 T7 T8 T9 T10

172.3 172.3 169.1 169.1 168.9 49.3 49.1 49.0 45.9 45.8

3.6 SAMPLE PROBLEM 6.

In sample problem 5, the effects of infrared emission, in addition to the combined effects of

conduction, convection, and direct solar radiation, were studied. The effects of infrared emission 

were carried out by removing the emission restriction on the wing surfaces of sample problem 4, 

by changing it from 0.0 to 0.9. The only external source of heat originated from direct solar 

radiation. Sample problem 6 incorporated the effects of external infrared radiation from the sky, 

in addition to the direct solar radiation. As in sample problems 1 through 5, THERMOD was set 

up to solve a complete aircraft problem, while the FEM method was confined to the wing model. 

To ensure consistency with the FEM wing model, the wing in the THERMOD model was 

isolated from infrared radiation effects from the fuselage and tarmac surfaces by assigning 

emissivity values of 0.0. Unlike sample problem 5, the sky radiation was assigned an absolute 

temperature of 459.67°R, so as to incorporate the infrared radiation from the sky. The 

conduction and convection heat transfer modes were left intact, as in sample problem 5. 

As in sample problem 5, the wings were further isolated from the tarmac effects by arbitrarily 

placing the wing at a height of 100.0 ft from the tarmac. In addition, the infrared reflection 

effects from the fuselage surface were reduced by maintaining a cabin height of 0.1 ft. 

As in sample problem 5, the fillet effect was removed from the wing top by assigning a 0.0 value 

in the second field of input data set 6. See the THERMOD User’s Reference Manual for further 

details on the input requirements for fillet effects. The THERMOD input file, input.dat, is 

illustrated in table 3-29. 

TABLE 3-29. THERMOD INPUT FILE (input.dat) OF SAMPLE PROBLEM 6 

198e-6,0.704 

1, 0.020,125,0.300,0.30 

2, 0.375,4.4,0.021,0.24 

3, 0.020,125,0.300,0.30 

4, 0.350,125,0.300,0.30 

6, 0.350,125,0.300,0.30 

8, 0.375,4.4,0.021,0.24 

10,0.030,125,0.300,0.30 

11,0.375,4.4,0.021,0.24 

12,0.020,125,0.300,0.30 

5, 10.70,0.067,0.016,0.24 

7, 0.020,125,0.300,0.30 

9, 0.020,125,0.300,0.30 

3-74
TABLE 3-29. THERMOD INPUT FILE (input.dat) OF SAMPLE 

PROBLEM 6 (Continued) 

13,0.250,10.,0.020,0.24 

14,48.00,0.067,0.016,0.24 

15,0.020,125,0.300,0.30 

16,0.375,4.4,0.021,0.24 

17,0.020,125,0.300,0.30 

18,0.250,10.,0.020,0.24 

19,30.00,0.067,0.016,0.24 

20,5.000,10.,0.020,0.24 

21,0.020,125,0.300,0.30 

22,0.375,4.4,0.021,0.24 

23,0.020,125,0.300,0.30 

24,2.500,0.067,0.016,0.24 

25,0.020,125,0.300,0.30 

26,0.375,4.4,0.021,0.24 

27,0.020,125,0.300,0.30 

28,0.250,130,0.800,0.20 

3.5,34.0,100.0,5.5,22.0,11.0,0.1,4.0,30,24,28 

0.70,1.0,0.0,2.0,150,0.80,0.22 

0.3,0.9,0.3,0.9,0.3,0.9,0.1,0.9 

3,0.0 

0.9,0.9,0.0,0.0 

1.0,0.0,0.9,0.9,1.0,0.0,1.0,0.0 

9 

0,10,10,0,0,75,110,110,190,190 

5,120,5,60,20,5,120,60,60 

7 

60 570.67 459.67 14 330 10 10 0.36 

75 573.67 459.67 14 355 10 10 0.31 

90 578.67 459.67 14 355 10 12 0.33 

75 581.67 459.67 14 330 10 10 0.31 

60 582.67 459.67 14 291 10 10 0.36 

45 583.67 459.67 14 231 10 10 0.44 

30 582.67 459.67 14 160 10 10 0.57 

1,0.02,180,10 

10,1.0,1.0 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10 

Temperatures from the THERMOD run is shown in table 3-30, from which the steady-state 

temperatures T1, T2.....T10 (top to bottom of wing) for time period 3 were extracted and 

tabulated in table 3-31. 

3-75
TABLE 3-30. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE PROBLEM 6 

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ 

THERMOD: A THERMAL PROGRAM FOR AIRCRAFTS 

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ 

 

 

 

 SUMMARY OF STEADY-STATE AND TRANSIENT ANALYSIS 

 (Temperatures are shown for all time periods) 

 

 

 TEMPERATURES AT THE END OF STEADY STATE ANALYSIS 

 

 3 169.0 187.2 192.8 183.8 164.3 136.8 107.9 

 5 168.8 187.0 192.7 183.6 164.2 136.7 107.9 

 6 88.8 91.1 94.1 95.4 95.2 94.8 93.3 

 7 88.7 91.0 93.9 95.2 95.1 94.8 93.3 

 8 88.7 90.9 93.9 95.2 95.1 94.7 93.2 

 9 86.5 88.4 91.2 92.9 93.3 93.6 92.9 

 10 86.5 88.4 91.2 92.8 93.3 93.6 92.9 

 11 171.1 189.7 195.5 186.1 166.2 137.9 108.3 

 12 171.1 189.7 195.5 186.1 166.2 137.9 108.3 

 13 168.9 187.2 192.8 183.8 164.3 136.8 107.9 

 14 168.9 187.1 192.8 183.8 164.3 136.8 107.9 

 15 168.8 187.0 192.7 183.6 164.2 136.7 107.9 

 16 88.8 91.1 94.1 95.4 95.2 94.8 93.3 

 17 88.7 91.0 93.9 95.2 95.1 94.8 93.3 

 19 86.5 88.4 91.2 92.9 93.3 93.6 92.9 

 22 276.5 310.3 322.1 303.8 268.1 219.1 169.9 

 23 242.5 264.0 274.0 259.4 236.0 200.8 161.5 

 24 242.4 263.8 273.8 259.2 235.8 200.7 161.5 

 25 218.6 231.4 240.1 228.1 213.4 187.9 155.7 

 26 185.0 192.9 200.7 191.5 182.5 165.1 141.4 

 27 143.9 149.0 155.3 152.9 149.4 142.2 131.2 

 29 140.9 145.9 152.0 150.1 147.1 140.6 130.4 

 30 140.9 145.9 152.0 150.1 147.1 140.6 130.4 

 

 

 

 TEMPERATURES IN DEGREES FAHRENHEIT 

 1 2 3 4 5 6 7 

 

 1 171.1 189.7 195.5 186.1 166.2 137.9 108.3 

 2 171.1 189.7 195.5 186.1 166.2 137.9 108.3 

 4 168.9 187.1 192.8 183.8 164.3 136.8 107.9 

 18 88.7 90.9 93.9 95.2 95.1 94.7 93.2 

 20 86.5 88.4 91.2 92.8 93.3 93.6 92.9 

 21 276.7 310.4 322.3 304.0 268.2 219.2 169.9 

 28 143.9 149.0 155.3 152.9 149.4 142.2 131.2 

3-76
TABLE 3-30. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE 

PROBLEM 6 (Continued) 

 31 128.1 132.2 137.8 138.0 136.7 133.5 127.2 

 32 128.1 132.2 137.8 138.0 136.7 133.5 127.2 

 33 128.1 132.2 137.8 138.0 136.7 133.5 127.2 

 34 128.1 132.2 137.8 138.0 136.7 133.5 127.2 

 35 115.2 118.5 123.6 125.9 126.4 126.3 124.0 

 36 115.2 118.5 123.6 125.9 126.4 126.3 124.0 

 37 112.3 115.3 120.4 123.2 124.0 124.7 123.3 

 39 218.4 180.1 132.7 182.8 216.8 225.1 206.3 

 43 190.5 190.1 188.2 189.5 188.3 176.4 155.1 

 44 181.2 184.4 188.2 184.1 180.0 168.0 149.3 

 47 125.1 128.2 132.9 134.5 134.5 132.8 128.3 

 51 188.8 207.4 215.7 208.8 191.6 168.5 144.8 

 53 188.8 207.4 215.7 208.8 191.6 168.5 144.8 

 

Maximum Structural Temperature for Period 1 = 276.67 Occurring 

at Location 21 

 

Maximum Structural Temperature Over All 7 Periods =322.29 

Occurring at Location 21 at Period 3 

 38 112.2 115.3 120.4 123.2 124.0 124.7 123.3 

 40 218.4 180.2 132.9 182.9 216.7 225.0 206.2 

 41 202.0 186.0 165.4 186.7 200.0 196.5 176.2 

 42 202.0 186.0 165.5 186.8 200.0 196.4 176.1 

 45 158.2 161.3 165.5 163.7 161.3 153.5 140.6 

 46 158.0 161.2 165.4 163.6 161.2 153.5 140.6 

 48 124.9 128.0 132.7 134.3 134.3 132.7 128.2 

 49 123.9 128.0 133.1 130.2 127.0 120.3 110.7 

 50 126.1 130.5 135.7 132.6 129.1 121.8 111.6 

 52 183.9 201.5 209.6 203.3 187.3 165.7 143.4 

Maximum Structural Temperature for Period 2 = 310.45 Occurring 

at Location 21 

Maximum Structural Temperature for Period 3 = 322.29 Occurring 

at Location 21 

Maximum Structural Temperature for Period 4 = 304.01 Occurring 

at Location 21 

Maximum Structural Temperature for Period 5 = 268.19 Occurring 

at Location 21 

Maximum Structural Temperature for Period 6 = 225.14 Occurring 

at Location 39 

Maximum Structural Temperature for Period 7 = 206.35 Occurring 

at Location 39 

3-77
TABLE 3-30. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE 

PROBLEM 6 (Continued) 

 TEMPERATURES AT THE END OF TRANSIENT ANALYSIS 

 

 TEMPERATURES IN DEGREES FAHRENHEIT 

 1 2 3 4 5 6 7 

 1 117.0 121.7 126.8 128.5 127.3 125.4 121.6 

 2 118.4 123.4 128.6 130.0 128.3 125.7 121.2 

 3 164.7 181.8 187.5 179.2 161.2 135.7 109.0 

 4 166.1 183.7 189.3 180.7 162.3 136.0 108.7 

 5 167.4 185.3 191.0 182.1 163.1 136.2 108.4 

 6 89.4 91.7 94.7 96.0 95.9 95.4 93.9 

 7 89.8 92.1 95.1 96.5 96.4 96.1 94.6 

 8 90.4 92.7 95.8 97.2 97.1 96.8 95.3 

 9 108.2 111.1 115.8 118.6 119.6 120.5 119.5 

 10 108.8 111.6 116.4 119.3 120.2 121.2 120.2 

 11 117.0 121.7 126.8 128.5 127.3 125.4 121.6 

 12 118.4 123.4 128.6 130.0 128.3 125.7 121.2 

 13 164.7 181.8 187.5 179.2 161.2 135.7 109.0 

 14 166.1 183.6 189.3 180.7 162.3 136.0 108.7 

 15 167.4 185.3 191.0 182.1 163.1 136.2 108.4 

 16 89.4 91.7 94.7 96.0 95.9 95.4 93.9 

 17 89.8 92.1 95.1 96.5 96.4 96.1 94.6 

 18 90.4 92.7 95.8 97.2 97.1 96.8 95.3 

 19 108.2 111.1 115.8 118.6 119.6 120.5 119.5 

 20 108.8 111.6 116.4 119.3 120.2 121.2 120.2 

 21 125.0 130.7 136.2 137.4 135.3 132.0 126.9 

 22 127.1 132.9 138.5 139.5 137.0 133.2 127.5 

 23 189.3 201.7 208.5 201.7 189.3 169.4 145.5 

 24 191.0 203.5 210.4 203.4 190.7 170.4 146.0 

 25 188.7 198.6 205.0 197.8 187.4 168.9 145.2 

 26 167.6 174.4 180.2 175.2 168.6 155.4 137.2 

 27 144.1 149.2 155.5 153.0 149.5 142.3 131.1 

 28 143.2 148.3 154.5 152.3 148.9 142.0 131.1 

 29 140.9 145.9 152.0 150.2 147.1 140.6 130.5 

 30 140.9 145.8 151.9 150.1 147.0 140.6 130.4 

 31 127.8 131.9 137.5 137.8 136.6 133.5 127.2 

 32 127.8 131.9 137.5 137.8 136.6 133.5 127.2 

 33 127.8 131.9 137.5 137.8 136.6 133.5 127.2 

 34 127.8 131.9 137.5 137.8 136.6 133.5 127.2 

 35 114.7 117.9 123.1 125.6 126.1 126.3 124.0 

 36 114.6 117.8 123.0 125.5 126.1 126.3 124.0 

 37 111.2 114.2 119.2 122.2 123.2 124.1 123.1 

 38 111.1 114.1 119.1 122.1 123.1 124.1 123.0 

 39 116.3 117.5 120.0 125.2 127.6 128.9 126.9 

 40 117.9 118.9 121.2 126.4 128.9 130.0 127.7 

 41 163.4 159.9 154.8 162.9 166.6 161.7 148.3 

3-78
TABLE 3-30. THERMOD OUTPUT FILE (summary.dat) FOR SAMPLE 

PROBLEM 6 (Continued) 

 42 164.8 161.4 156.2 164.2 167.7 162.5 148.8 

 43 166.2 168.2 169.0 170.1 168.3 159.2 143.7 

 44 162.8 166.0 169.0 168.0 165.3 156.4 142.0 

 45 152.2 154.1 156.2 157.5 156.6 150.4 139.0 

 46 150.9 152.7 154.8 156.3 155.6 149.6 138.6 

 47 117.0 118.3 121.2 125.9 128.1 129.1 127.0 

 48 115.9 117.2 120.0 124.9 127.2 128.4 126.6 

 49 113.1 116.3 121.1 123.1 123.3 122.7 120.0 

 50 115.2 118.6 123.5 125.0 124.9 123.6 120.3 

 51 306.1 348.1 361.3 339.6 295.0 235.5 177.6 

 52 288.9 327.5 340.0 320.5 279.9 225.7 172.8 

 53 306.1 348.1 361.3 339.6 295.0 235.5 177.6 

 

Maximum Structural Temperature for Period 1 = 191.02 Occurring 

at Location 24 

Maximum Structural Temperature for Period 2 = 203.55 Occurring 

at Location 24 

Maximum Structural Temperature for Period 3 = 210.37 Occurring 

at Location 24 

Maximum Structural Temperature for Period 4 = 203.36 Occurring 

at Location 24 

Maximum Structural Temperature for Period 5 = 190.72 Occurring 

at Location 24 

Maximum Structural Temperature for Period 6 = 170.41 Occurring 

at Location 24 

Maximum Structural Temperature for Period 7 = 148.79 Occurring 

at Location 42 

 

Maximum Structural Temperature Over All 7 Periods =210.37 

Occurring at Location 24 at Period 3 

TABLE 3-31. THERMOD-SIMULATED TEMPERATURES (°F) OF THE RIGHT WING OF 

SAMPLE PROBLEM 6 

T1 T2 T3 T4 T5 T6 T7 T8 T9 T10

195.5 195.5 192.8 192.8 192.7 94.1 93.9 93.9 91.2 91.2

As in the previous sample problems, a one-half wing structure was modeled using the 

UAI/NASTRAN FEM program. Instead of just two types of loads that were applied as in 

sample problem 5, three types of loads were applied to the model in sample problem 6. These 

loads, based on period 3 of the THERMOD input file, were direct solar radiation of 355 

Btu/hr/ft2

, convection with a film coefficient of h = 1.3433 (Btu/hr)/(ft2°F), and infrared sky 

radiation, based on a sky temperature of 0°F (459.67°R). The film coefficient was calculated 

internally by the THERMOD program and was based on the theory of airflow on a flat surface 

and a wind speed of 14 ft/sec. The solar radiation load was applied perpendicular to the top 

surface only, while both the top and bottom surfaces were subjected to the infrared sky radiation. 

Both the top and bottom surfaces were subjected to the convection loads. The ambient 

3-79
temperature was maintained at 119°F (578.67°R). As in sample problem 5, the wings were 

assigned a nonzero emissivity value of 0.9, allowing the wings to emit and absorb infrared 

radiation. The effect of infrared radiation of the tarmac was neglected. 

The FEM input file is shown in table 3-32. The ambient temperature of 119°F was imposed 

through the SPC case control card. This ambient temperature boundary condition was scalar in 

nature and was implemented through the SPOINT bulk data. The boundary elements, CHBDY, 

through which solar radiation and convection effects were applied, were modeled using the 

LINE boundary shape type, connecting the boundary nodes along the wing top and bottom. The 

SPACE designation in the CHBDY boundary element indicates that this element emits infrared 

radiation to a black body, in which case, no VIEW bulk data was required. Property cards 

PHBDY and the material cards MAT4, define absorptivity, emissivity, film coefficient, and 

conductivity values. Note that as in problem 5, the wing was assigned an emissivity of 0.9, as 

indicated in the PHBDY card. Load case 10 combines all three load cases: solar radiation on the 

top surface (indicated by QVECT) and the infrared sky radiation on the top and bottom surfaces 

(indicated by QBDY1). Because of the involvement of infrared radiation, through emission, this 

sample problem was rendered nonlinear. This required the Stefan-Boltzman constant, σ, and an 

initial set of temperatures, which were implemented through the PARAM and TEMPERATURE 

statements, respectively. 

TABLE 3-32. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE PROBLEM 6 

ID D:\Natha,MSC/N 

APP HEAT 

SOL 3 

TIME 10000 

CEND 

 TITLE = dradiation_infrd_conduction_emission_convection (t=0.001ft) 

 SUBTITLE = 10.7 in gap filled by air; k=0.016BTU/hr/ft/degF 

 load = 10 

 SPC = 1 

 temperature(estimate)=2 

 THERMAL= ALL 

 FLUX = ALL 

 SPCF = ALL 

BEGIN BULK 

$ *************************************************************************** 

$ Written by : MSC/NASTRAN for Windows 

$ Version : 6.00 

$ Translator : UAI/NASTRAN 

$ From Model : D:\Thermod_Validation\FEM\conduction_only.MOD 

$ Date : Mon Mar 06 11:27:09 2000 

$ *************************************************************************** 

$ 

PARAM,K6ROT,100. 

PARAM,MAXRATIO,1.E+8 

param,sigma,1.714e-9 

param,tabs,0.0 

param,maxit,10 

CORD2C 1 0 0. 0. 0. 0. 0. 

1.+MSC/NC1 

3-80
TABLE 3-32. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 6 (Continued) 

+MSC/NC1 1. 0. 1. 

CORD2S 2 0 0. 0. 0. 0. 0. 

1.+MSC/NC2 

+MSC/NC2 1. 0. 1. 

$ 

$ MSC/NASTRAN for Windows Property 1 : 0.02 in glass_outer_layer 

PSHELL 1 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 2 : 3/8 in foam 

PSHELL 2 2 0.001 2 2 0. 

$ MSC/NASTRAN for Windows Property 3 : 0.02 in inner glass 

PSHELL 3 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 4 : 0.350 in cap 

PSHELL 4 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 5 : 10.70 in air space 

PSHELL 5 3 0.001 3 3 0. 

$ MSC/NASTRAN for Windows Property 6 : 0.350 in cap 

PSHELL 6 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 7 : 0.02 in inner glass 

PSHELL 7 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Property 8 : 3/8 in foam 

PSHELL 8 2 0.001 2 2 0. 

$ MSC/NASTRAN for Windows Property 9 : 0.02 in glass_outer_layer 

PSHELL 9 1 0.001 1 1 0. 

$ MSC/NASTRAN for Windows Material 1 : glass 

MAT4 1 0.3 1. 

$ MSC/NASTRAN for Windows Material 2 : foam 

MAT4 2 0.021 1. 

$ MSC/NASTRAN for Windows Material 3 : air 

MAT4 3 0.016 1. 

$ 

chbdy,1001,100,line,200,201,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1002,100,line,201,202,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1003,100,line,202,203,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1004,100,line,203,204,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1005,100,line,204,205,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1006,100,line,205,206,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1008,100,line,207,208,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1009,100,line,208,209,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1010,100,line,209,210,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1011,100,line,210,211,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1012,100,line,211,212,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1013,100,line,212,213,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1014,100,line,213,214,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1015,100,line,214,215,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1016,100,line,215,216,,,space,500,500,,,0.0,1.0,0.0 

chbdy,1017,100,line,216,181,,,space,500,500,,,0.0,1.0,0.0 

phbdy,100,200,0.001,0.9,0.9 

mat4,200,1.3433 

spoint,500 

spc,1,500,1,578.67 

$ 

$ Direct radiation load from a distant source: load case 1: 

qvect,1,355.0,0.0,-1.0,0.0,1001,thru,1017 

$ 

$ Infrared radiation from the sky: load case 2 (applied to the top) 

$ and load case 3 (applied to the bottom) 

chbdy,1007,100,line,206,207,,,space,500,500,,,0.0,1.0,0.0 

3-81
TABLE 3-32. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 6 (Continued) 

$ Q0=sigma*(T^4)=1.714e-9*(459.67)^4 

$ =76.5236BTU/hr/ft^2: assuming sky temp.=0degF=459.67 degR. 

Qbdy1,2,76.5236,1001,thru,1017 

qbdy1,3,76.5236,2001,thru,2017 

$ combine the above three load cases 1, 2 and 3 into load case 10: 

load,10,1.0,1.0,1,1.0,2,1.0,3 

$ 

chbdy,2001,300,line,20,21,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2002,300,line,21,22,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2003,300,line,22,23,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2004,300,line,23,24,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2005,300,line,24,25,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2006,300,line,25,26,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2007,300,line,26,27,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2008,300,line,27,28,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2009,300,line,28,29,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2010,300,line,29,30,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2011,300,line,30,31,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2012,300,line,31,32,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2013,300,line,32,33,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2014,300,line,33,34,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2015,300,line,34,35,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2016,300,line,35,36,,,space,500,500,,,0.0,-1.0,0.0 

chbdy,2017,300,line,36,1,,,space,500,500,,,0.0,-1.0,0.0 

phbdy,300,400,0.001,0.9,0.9 

mat4,400,1.3433 

$ 

temp,2,1,578.67,3,578.67,5,578.67 

temp,2,7,578.67,9,578.67,10,578.67 

temp,2,11,578.67,12,578.67,14,578.67 

temp,2,15,578.67,18,578.67,20,578.67 

temp,2,21,578.67,22,578.67,23,578.67 

temp,2,24,578.67,25,578.67,26,578.67 

temp,2,27,578.67,28,578.67,29,578.67 

temp,2,30,578.67,31,578.67,32,578.67 

temp,2,33,578.67,34,578.67,35,578.67 

temp,2,36,578.67,37,578.67,41,578.67 

temp,2,44,578.67,45,578.67,47,578.67 

temp,2,48,578.67,50,578.67,51,578.67 

temp,2,52,578.67,53,578.67,54,578.67 

temp,2,55,578.67,56,578.67,58,578.67 

temp,2,59,578.67,62,578.67,67,578.67 

temp,2,69,578.67,71,578.67,73,578.67 

temp,2,75,578.67,76,578.67,77,578.67 

temp,2,78,578.67,79,578.67,80,578.67 

temp,2,83,578.67,85,578.67,87,578.67 

temp,2,88,578.67,98,578.67,101,578.67 

temp,2,104,578.67,105,578.67,107,578.67 

temp,2,108,578.67,109,578.67,112,578.67 

temp,2,115,578.67,116,578.67,118,578.67 

temp,2,120,578.67,121,578.67,122,578.67 

temp,2,123,578.67,124,578.67,125,578.67 

temp,2,128,578.67,129,578.67,130,578.67 

temp,2,133,578.67,135,578.67,137,578.67 

temp,2,138,578.67,145,578.67,146,578.67 

temp,2,147,578.67,150,578.67,152,578.67 

temp,2,160,578.67,161,578.67,164,578.67 

3-82
TABLE 3-32. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 6 (Continued) 

temp,2,165,578.67,172,578.67,174,578.67 

temp,2,177,578.67,178,578.67,180,578.67 

temp,2,181,578.67,182,578.67,183,578.67 

temp,2,184,578.67,188,578.67,193,578.67 

temp,2,195,578.67,196,578.67,197,578.67 

temp,2,198,578.67,199,578.67,200,578.67 

temp,2,201,578.67,202,578.67,203,578.67 

temp,2,204,578.67,205,578.67,206,578.67 

temp,2,207,578.67,208,578.67,209,578.67 

temp,2,210,578.67,211,578.67,212,578.67 

temp,2,213,578.67,214,578.67,215,578.67 

temp,2,216,578.67,219,578.67,220,578.67 

temp,2,221,578.67,222,578.67,223,578.67 

temp,2,226,578.67,227,578.67,228,578.67 

temp,2,229,578.67,231,578.67,232,578.67 

temp,2,235,578.67,241,578.67,243,578.67 

temp,2,244,578.67,245,578.67,246,578.67 

temp,2,248,578.67,249,578.67,250,578.67 

temp,2,253,578.67,254,578.67,257,578.67 

temp,2,258,578.67,259,578.67,260,578.67 

temp,2,264,578.67,267,578.67,268,578.67 

temp,2,269,578.67,270,578.67,273,578.67 

temp,2,274,578.67,277,578.67,282,578.67 

temp,2,283,578.67,292,578.67,293,578.67 

temp,2,295,578.67,297,578.67,298,578.67 

temp,2,299,578.67,300,578.67,301,578.67 

temp,2,302,578.67,303,578.67,306,578.67 

temp,2,307,578.67,308,578.67,313,578.67 

temp,2,314,578.67,316,578.67,317,578.67 

temp,2,318,578.67,323,578.67,324,578.67 

temp,2,500,578.67 

$ 

GRID 1 0 0. 0. 0. 0 

GRID 3 0 1.1.667E-3 0. 0 

GRID 5 0 3.1.667E-3 0. 0 

GRID 7 0 5.1.667E-3 0. 0 

GRID 9 0 7.1.667E-3 0. 0 

GRID 10 0 8.1.667E-3 0. 0 

GRID 11 0 9.1.667E-3 0. 0 

GRID 12 0 10.1.667E-3 0. 0 

GRID 14 0 12.1.667E-3 0. 0 

GRID 15 0 13.1.667E-3 0. 0 

GRID 18 0 16.1.667E-3 0. 0 

GRID 20 0 17. 0. 0. 0 

GRID 21 0 16. 0. 0. 0 

GRID 22 0 15. 0. 0. 0 

GRID 23 0 14. 0. 0. 0 

GRID 24 0 13. 0. 0. 0 

GRID 25 0 12. 0. 0. 0 

GRID 26 0 11. 0. 0. 0 

GRID 27 0 10. 0. 0. 0 

GRID 28 0 9. 0. 0. 0 

GRID 29 0 8. 0. 0. 0 

GRID 30 0 7. 0. 0. 0 

GRID 31 0 6. 0. 0. 0 

GRID 32 0 5. 0. 0. 0 

GRID 33 0 4. 0. 0. 0 

3-83
TABLE 3-32. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 6 (Continued) 

GRID 34 0 3. 0. 0. 0 

GRID 35 0 2. 0. 0. 0 

GRID 36 0 1. 0. 0. 0 

GRID 37 0 0.1.667E-3 0. 0 

GRID 41 0 3.0.032917 0. 0 

GRID 44 0 6.0.032917 0. 0 

GRID 45 0 7.0.032917 0. 0 

GRID 47 0 9.0.032917 0. 0 

GRID 48 0 10.0.032917 0. 0 

GRID 50 0 12.0.032917 0. 0 

GRID 51 0 13.0.032917 0. 0 

GRID 52 0 14.0.032917 0. 0 

GRID 53 0 15.0.032917 0. 0 

GRID 54 0 16.0.032917 0. 0 

GRID 55 0 17.0.032917 0. 0 

GRID 56 0 17.1.667E-3 0. 0 

GRID 58 0 15.1.667E-3 0. 0 

GRID 59 0 14.1.667E-3 0. 0 

GRID 62 0 11.1.667E-3 0. 0 

GRID 67 0 6.1.667E-3 0. 0 

GRID 69 0 4.1.667E-3 0. 0 

GRID 71 0 2.1.667E-3 0. 0 

GRID 73 0 0.0.032917 0. 0 

GRID 75 0 1.0.034583 0. 0 

GRID 76 0 2.0.034583 0. 0 

GRID 77 0 3.0.034583 0. 0 

GRID 78 0 4.0.034583 0. 0 

GRID 79 0 5.0.034583 0. 0 

GRID 80 0 6.0.034583 0. 0 

GRID 83 0 9.0.034583 0. 0 

GRID 85 0 11.0.034583 0. 0 

GRID 87 0 13.0.034583 0. 0 

GRID 88 0 14.0.034583 0. 0 

GRID 98 0 11.0.032917 0. 0 

GRID 101 0 8.0.032917 0. 0 

GRID 104 0 5.0.032917 0. 0 

GRID 105 0 4.0.032917 0. 0 

GRID 107 0 2.0.032917 0. 0 

GRID 108 0 1.0.032917 0. 0 

GRID 109 0 0.0.034583 0. 0 

GRID 112 0 2. 0.06375 0. 0 

GRID 115 0 5. 0.06375 0. 0 

GRID 116 0 6. 0.06375 0. 0 

GRID 118 0 8. 0.06375 0. 0 

GRID 120 0 10. 0.06375 0. 0 

GRID 121 0 11. 0.06375 0. 0 

GRID 122 0 12. 0.06375 0. 0 

GRID 123 0 13. 0.06375 0. 0 

GRID 124 0 14. 0.06375 0. 0 

GRID 125 0 15. 0.06375 0. 0 

GRID 128 0 17.0.034583 0. 0 

GRID 129 0 16.0.034583 0. 0 

GRID 130 0 15.0.034583 0. 0 

GRID 133 0 12.0.034583 0. 0 

GRID 135 0 10.0.034583 0. 0 

GRID 137 0 8.0.034583 0. 0 

GRID 138 0 7.0.034583 0. 0 

3-84
TABLE 3-32. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 6 (Continued) 

GRID 145 0 0. 0.06375 0. 0 

GRID 146 0 0. 0.95542 0. 0 

GRID 147 0 1. 0.95542 0. 0 

GRID 150 0 4. 0.95542 0. 0 

GRID 152 0 6. 0.95542 0. 0 

GRID 160 0 14. 0.95542 0. 0 

GRID 161 0 15. 0.95542 0. 0 

GRID 164 0 17. 0.06375 0. 0 

GRID 165 0 16. 0.06375 0. 0 

GRID 172 0 9. 0.06375 0. 0 

GRID 174 0 7. 0.06375 0. 0 

GRID 177 0 4. 0.06375 0. 0 

GRID 178 0 3. 0.06375 0. 0 

GRID 180 0 1. 0.06375 0. 0 

GRID 181 0 0. 1.01917 0. 0 

GRID 182 0 0. 1.0175 0. 0 

GRID 183 0 1. 1.0175 0. 0 

GRID 184 0 2. 1.0175 0. 0 

GRID 188 0 6. 1.0175 0. 0 

GRID 193 0 11. 1.0175 0. 0 

GRID 195 0 13. 1.0175 0. 0 

GRID 196 0 14. 1.0175 0. 0 

GRID 197 0 15. 1.0175 0. 0 

GRID 198 0 16. 1.0175 0. 0 

GRID 199 0 17. 1.0175 0. 0 

GRID 200 0 17. 1.01917 0. 0 

GRID 201 0 16. 1.01917 0. 0 

GRID 202 0 15. 1.01917 0. 0 

GRID 203 0 14. 1.01917 0. 0 

GRID 204 0 13. 1.01917 0. 0 

GRID 205 0 12. 1.01917 0. 0 

GRID 206 0 11. 1.01917 0. 0 

GRID 207 0 10. 1.01917 0. 0 

GRID 208 0 9. 1.01917 0. 0 

GRID 209 0 8. 1.01917 0. 0 

GRID 210 0 7. 1.01917 0. 0 

GRID 211 0 6. 1.01917 0. 0 

GRID 212 0 5. 1.01917 0. 0 

GRID 213 0 4. 1.01917 0. 0 

GRID 214 0 3. 1.01917 0. 0 

GRID 215 0 2. 1.01917 0. 0 

GRID 216 0 1. 1.01917 0. 0 

GRID 219 0 1. 0.98625 0. 0 

GRID 220 0 2. 0.98625 0. 0 

GRID 221 0 3. 0.98625 0. 0 

GRID 222 0 4. 0.98625 0. 0 

GRID 223 0 5. 0.98625 0. 0 

GRID 226 0 8. 0.98625 0. 0 

GRID 227 0 9. 0.98625 0. 0 

GRID 228 0 10. 0.98625 0. 0 

GRID 229 0 11. 0.98625 0. 0 

GRID 231 0 13. 0.98625 0. 0 

GRID 232 0 14. 0.98625 0. 0 

GRID 235 0 17. 0.98625 0. 0 

GRID 241 0 12. 1.0175 0. 0 

GRID 243 0 10. 1.0175 0. 0 

GRID 244 0 9. 1.0175 0. 0 

3-85
TABLE 3-32. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 6 (Continued) 

GRID 245 0 8. 1.0175 0. 0 

GRID 246 0 7. 1.0175 0. 0 

GRID 248 0 5. 1.0175 0. 0 

GRID 249 0 4. 1.0175 0. 0 

GRID 250 0 3. 1.0175 0. 0 

GRID 253 0 0. 0.98625 0. 0 

GRID 254 0 0. 0.98458 0. 0 

GRID 257 0 3. 0.98458 0. 0 

GRID 258 0 4. 0.98458 0. 0 

GRID 259 0 5. 0.98458 0. 0 

GRID 260 0 6. 0.98458 0. 0 

GRID 264 0 10. 0.98458 0. 0 

GRID 267 0 13. 0.98458 0. 0 

GRID 268 0 14. 0.98458 0. 0 

GRID 269 0 15. 0.98458 0. 0 

GRID 270 0 16. 0.98458 0. 0 

GRID 273 0 16. 0.98625 0. 0 

GRID 274 0 15. 0.98625 0. 0 

GRID 277 0 12. 0.98625 0. 0 

GRID 282 0 7. 0.98625 0. 0 

GRID 283 0 6. 0.98625 0. 0 

GRID 292 0 2. 0.95542 0. 0 

GRID 293 0 3. 0.95542 0. 0 

GRID 295 0 5. 0.95542 0. 0 

GRID 297 0 7. 0.95542 0. 0 

GRID 298 0 8. 0.95542 0. 0 

GRID 299 0 9. 0.95542 0. 0 

GRID 300 0 10. 0.95542 0. 0 

GRID 301 0 11. 0.95542 0. 0 

GRID 302 0 12. 0.95542 0. 0 

GRID 303 0 13. 0.95542 0. 0 

GRID 306 0 16. 0.95542 0. 0 

GRID 307 0 17. 0.95542 0. 0 

GRID 308 0 17. 0.98458 0. 0 

GRID 313 0 12. 0.98458 0. 0 

GRID 314 0 11. 0.98458 0. 0 

GRID 316 0 9. 0.98458 0. 0 

GRID 317 0 8. 0.98458 0. 0 

GRID 318 0 7. 0.98458 0. 0 

GRID 323 0 2. 0.98458 0. 0 

GRID 324 0 1. 0.98458 0. 0 

CQUAD4 1 9 1 37 3 36 

CQUAD4 2 9 36 3 71 35 

CQUAD4 3 9 35 71 5 34 

CQUAD4 4 9 34 5 69 33 

CQUAD4 5 9 33 69 7 32 

CQUAD4 6 9 32 7 67 31 

CQUAD4 7 9 31 67 9 30 

CQUAD4 8 9 30 9 10 29 

CQUAD4 9 9 29 10 11 28 

CQUAD4 10 9 28 11 12 27 

CQUAD4 11 9 27 12 62 26 

CQUAD4 12 9 26 62 14 25 

CQUAD4 13 9 25 14 15 24 

CQUAD4 14 9 24 15 59 23 

CQUAD4 15 9 23 59 58 22 

CQUAD4 16 9 22 58 18 21 

3-86
TABLE 3-32. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 6 (Continued) 

CQUAD4 17 9 21 18 56 20 

CQUAD4 18 8 37 73 108 3 

CQUAD4 19 8 3 108 107 71 

CQUAD4 20 8 71 107 41 5 

CQUAD4 21 8 5 41 105 69 

CQUAD4 22 8 69 105 104 7 

CQUAD4 23 8 7 104 44 67 

CQUAD4 24 8 67 44 45 9 

CQUAD4 25 8 9 45 101 10 

CQUAD4 26 8 10 101 47 11 

CQUAD4 27 8 11 47 48 12 

CQUAD4 28 8 12 48 98 62 

CQUAD4 29 8 62 98 50 14 

CQUAD4 30 8 14 50 51 15 

CQUAD4 31 8 15 51 52 59 

CQUAD4 32 8 59 52 53 58 

CQUAD4 33 8 58 53 54 18 

CQUAD4 34 8 18 54 55 56 

CQUAD4 35 7 73 109 75 108 

CQUAD4 36 7 108 75 76 107 

CQUAD4 37 7 107 76 77 41 

CQUAD4 38 7 41 77 78 105 

CQUAD4 39 7 105 78 79 104 

CQUAD4 40 7 104 79 80 44 

CQUAD4 41 7 44 80 138 45 

CQUAD4 42 7 45 138 137 101 

CQUAD4 43 7 101 137 83 47 

CQUAD4 44 7 47 83 135 48 

CQUAD4 45 7 48 135 85 98 

CQUAD4 46 7 98 85 133 50 

CQUAD4 47 7 50 133 87 51 

CQUAD4 48 7 51 87 88 52 

CQUAD4 49 7 52 88 130 53 

CQUAD4 50 7 53 130 129 54 

CQUAD4 51 7 54 129 128 55 

CQUAD4 52 6 109 145 180 75 

CQUAD4 53 6 75 180 112 76 

CQUAD4 54 6 76 112 178 77 

CQUAD4 55 6 77 178 177 78 

CQUAD4 56 6 78 177 115 79 

CQUAD4 57 6 79 115 116 80 

CQUAD4 58 6 80 116 174 138 

CQUAD4 59 6 138 174 118 137 

CQUAD4 60 6 137 118 172 83 

CQUAD4 61 6 83 172 120 135 

CQUAD4 62 6 135 120 121 85 

CQUAD4 63 6 85 121 122 133 

CQUAD4 64 6 133 122 123 87 

CQUAD4 65 6 87 123 124 88 

CQUAD4 66 6 88 124 125 130 

CQUAD4 67 6 130 125 165 129 

CQUAD4 68 6 129 165 164 128 

CQUAD4 69 5 145 146 147 180 

CQUAD4 70 5 180 147 292 112 

CQUAD4 71 5 112 292 293 178 

CQUAD4 72 5 178 293 150 177 

CQUAD4 73 5 177 150 295 115 

3-87
TABLE 3-32. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 6 (Continued) 

CQUAD4 74 5 115 295 152 116 

CQUAD4 75 5 116 152 297 174 

CQUAD4 76 5 174 297 298 118 

CQUAD4 77 5 118 298 299 172 

CQUAD4 78 5 172 299 300 120 

CQUAD4 79 5 120 300 301 121 

CQUAD4 80 5 121 301 302 122 

CQUAD4 81 5 122 302 303 123 

CQUAD4 82 5 123 303 160 124 

CQUAD4 83 5 124 160 161 125 

CQUAD4 84 5 125 161 306 165 

CQUAD4 85 5 165 306 307 164 

CQUAD4 86 1 182 181 216 183 

CQUAD4 87 1 183 216 215 184 

CQUAD4 88 1 184 215 214 250 

CQUAD4 89 1 250 214 213 249 

CQUAD4 90 1 249 213 212 248 

CQUAD4 91 1 248 212 211 188 

CQUAD4 92 1 188 211 210 246 

CQUAD4 93 1 246 210 209 245 

CQUAD4 94 1 245 209 208 244 

CQUAD4 95 1 244 208 207 243 

CQUAD4 96 1 243 207 206 193 

CQUAD4 97 1 193 206 205 241 

CQUAD4 98 1 241 205 204 195 

CQUAD4 99 1 195 204 203 196 

CQUAD4 100 1 196 203 202 197 

CQUAD4 101 1 197 202 201 198 

CQUAD4 102 1 198 201 200 199 

CQUAD4 103 2 253 182 183 219 

CQUAD4 104 2 219 183 184 220 

CQUAD4 105 8 220 184 250 221 

CQUAD4 106 2 221 250 249 222 

CQUAD4 107 2 222 249 248 223 

CQUAD4 108 2 223 248 188 283 

CQUAD4 109 2 283 188 246 282 

CQUAD4 110 2 282 246 245 226 

CQUAD4 111 2 226 245 244 227 

CQUAD4 112 2 227 244 243 228 

CQUAD4 113 2 228 243 193 229 

CQUAD4 114 2 229 193 241 277 

CQUAD4 115 2 277 241 195 231 

CQUAD4 116 2 231 195 196 232 

CQUAD4 117 2 232 196 197 274 

CQUAD4 118 2 274 197 198 273 

CQUAD4 119 2 273 198 199 235 

CQUAD4 120 3 254 253 219 324 

CQUAD4 121 3 324 219 220 323 

CQUAD4 122 3 323 220 221 257 

CQUAD4 123 3 257 221 222 258 

CQUAD4 124 3 258 222 223 259 

CQUAD4 125 3 259 223 283 260 

CQUAD4 126 3 260 283 282 318 

CQUAD4 127 3 318 282 226 317 

CQUAD4 128 3 317 226 227 316 

CQUAD4 129 3 316 227 228 264 

CQUAD4 130 3 264 228 229 314 

3-88
TABLE 3-32. FINITE ELEMENT METHOD INPUT FILE OF SAMPLE 

PROBLEM 6 (Continued) 

CQUAD4 131 3 314 229 277 313 

CQUAD4 132 3 313 277 231 267 

CQUAD4 133 3 267 231 232 268 

CQUAD4 134 3 268 232 274 269 

CQUAD4 135 3 269 274 273 270 

CQUAD4 136 3 270 273 235 308 

CQUAD4 137 4 146 254 324 147 

CQUAD4 138 4 147 324 323 292 

CQUAD4 139 4 292 323 257 293 

CQUAD4 140 4 293 257 258 150 

CQUAD4 141 4 150 258 259 295 

CQUAD4 142 4 295 259 260 152 

CQUAD4 143 4 152 260 318 297 

CQUAD4 144 4 297 318 317 298 

CQUAD4 145 4 298 317 316 299 

CQUAD4 146 4 299 316 264 300 

CQUAD4 147 4 300 264 314 301 

CQUAD4 148 4 301 314 313 302 

CQUAD4 149 4 302 313 267 303 

CQUAD4 150 4 303 267 268 160 

CQUAD4 151 4 160 268 269 161 

CQUAD4 152 4 161 269 270 306 

CQUAD4 153 4 306 270 308 307 

ENDDATA 

The results of the one-half FEM wing model analysis are presented in figure 3-6 in the form of a 

contour plot. The contours appear to be indicative of the temperature profile of the THERMOD 

analysis. The top surface, which was subjected to both direct solar radiation and infrared sky 

radiation, is certainly much hotter than the bottom surface, which was subjected to the infrared 

sky radiation only. The temperatures are higher than those found in figure 3-5 of sample 

problem 5 because of the inclusion of both the direct solar radiation and infrared sky radiation. 

Note that the temperatures are in absolute terms (°R). 

FIGURE 3-6. CONTOUR PLOT OF THE FEM OUTPUT OF SAMPLE PROBLEM 6 

3-89
The truncated FEM output file is shown in table 3-33. The nodes, from top to bottom of the 

wing, located at mid-span of the model are indicated by nodes 209, 245, 226, 317, 298, 118, 137, 

101, 10, and 29. The temperatures of these nodes were extracted from this file and are 

reproduced in table 3-34. Mid-span was chosen because it was sufficiently far away from any 

edge effects that might be present. Note that the temperatures in table 3-34 were reduced from

absolute temperatures (°R) to °F.

TABLE 3-33. TRUNCATED FEM OUTPUT FILE OF SAMPLE PROBLEM 6 

1 DRADIATION_INFRD_CONDUCTION_EMISSION_CONVECTION (T=0.001FT) AUGUST 21, 2000 UAI/NASTRAN VERSION 20.1 PAGE 

18 

 10.7 IN GAP FILLED BY AIR; K=0.016BTU/HR/FT/DEGF 

0 

 

 T E M P E R A T U R E V E C T O R 

 

 POINT ID. TYPE ID VALUE ID+1 VALUE ID+2 VALUE ID+3 VALUE ID+4 VALUE ID+5 VALUE 

 1 TEMP 5.397126E+02 

 5 TEMP 5.397238E+02 

 7 TEMP 5.397238E+02 

 9 TEMP 5.397238E+02 5.397238E+02 5.397238E+02 5.397238E+02 

 14 TEMP 5.397238E+02 5.397238E+02 

 18 TEMP 5.397238E+02 

 20 TEMP 5.397126E+02 5.397126E+02 5.397126E+02 5.397126E+02 5.397126E+02 5.397126E+02 

 26 TEMP 5.397126E+02 5.397126E+02 5.397126E+02 5.397126E+02 5.397126E+02 5.397126E+02 

 32 TEMP 5.397126E+02 5.397126E+02 5.397126E+02 5.397126E+02 5.397126E+02 5.397238E+02 

 41 TEMP 5.427075E+02 

 44 TEMP 5.427075E+02 5.427075E+02 

 47 TEMP 5.427075E+02 5.427075E+02 

 56 TEMP 5.397238E+02 

 62 TEMP 5.397238E+02 

 67 TEMP 5.397238E+02 

 69 TEMP 5.397238E+02 

 71 TEMP 5.397238E+02 

 118 TEMP 5.429136E+02 

 253 TEMP 6.548631E+02 6.548519E+02 

 3 TEMP 5.397238E+02 

 50 TEMP 5.427075E+02 5.427075E+02 5.427075E+02 5.427075E+02 5.427075E+02 5.427075E+02 

 58 TEMP 5.397238E+02 5.397238E+02 

 73 TEMP 5.427075E+02 

 75 TEMP 5.427186E+02 5.427186E+02 5.427186E+02 5.427186E+02 5.427186E+02 5.427186E+02 

 83 TEMP 5.427186E+02 

 85 TEMP 5.427186E+02 

 87 TEMP 5.427186E+02 5.427186E+02 

 98 TEMP 5.427075E+02 

 101 TEMP 5.427075E+02 

 104 TEMP 5.427075E+02 5.427075E+02 

 107 TEMP 5.427075E+02 5.427075E+02 5.427186E+02 

 112 TEMP 5.429136E+02 

 115 TEMP 5.429136E+02 5.429136E+02 

 120 TEMP 5.429136E+02 5.429136E+02 5.429136E+02 5.429136E+02 5.429136E+02 5.429136E+02 

 128 TEMP 5.427186E+02 5.427186E+02 5.427186E+02 

 133 TEMP 5.427186E+02 

 135 TEMP 5.427186E+02 

 137 TEMP 5.427186E+02 5.427186E+02 

 145 TEMP 5.429136E+02 6.546570E+02 6.546570E+02 

 150 TEMP 6.546570E+02 

 152 TEMP 6.546570E+02 

 160 TEMP 6.546570E+02 6.546570E+02 

 164 TEMP 5.429136E+02 5.429136E+02 

 172 TEMP 5.429136E+02 

 174 TEMP 5.429136E+02 

 177 TEMP 5.429136E+02 5.429136E+02 

 180 TEMP 5.429136E+02 6.578580E+02 6.578469E+02 6.578469E+02 6.578469E+02 

 188 TEMP 6.578469E+02 

 193 TEMP 6.578469E+02 

 195 TEMP 6.578469E+02 6.578469E+02 6.578469E+02 6.578469E+02 6.578469E+02 6.578580E+02 

 201 TEMP 6.578580E+02 6.578580E+02 6.578580E+02 6.578580E+02 6.578580E+02 6.578580E+02 

1 DRADIATION_INFRD_CONDUCTION_EMISSION_CONVECTION (T=0.001FT) AUGUST 21, 2000 UAI/NASTRAN VERSION 20.1 PAGE 

19 

 10.7 IN GAP FILLED BY AIR; K=0.016BTU/HR/FT/DEGF 

0 

 

 T E M P E R A T U R E V E C T O R 

 

 POINT ID. TYPE ID VALUE ID+1 VALUE ID+2 VALUE ID+3 VALUE ID+4 VALUE ID+5 VALUE 

 207 TEMP 6.578580E+02 6.578580E+02 6.578580E+02 6.578580E+02 6.578580E+02 6.578580E+02 

 213 TEMP 6.578580E+02 6.578580E+02 6.578580E+02 6.578580E+02 

 219 TEMP 6.548631E+02 6.548631E+02 6.548631E+02 6.548631E+02 6.548631E+02 

 226 TEMP 6.548631E+02 6.548631E+02 6.548631E+02 6.548631E+02 

 231 TEMP 6.548631E+02 6.548631E+02 

 235 TEMP 6.548631E+02 

 241 TEMP 6.578469E+02 

 243 TEMP 6.578469E+02 6.578469E+02 6.578469E+02 6.578469E+02 

 248 TEMP 6.578469E+02 6.578469E+02 6.578469E+02 

 257 TEMP 6.548519E+02 6.548519E+02 6.548519E+02 6.548519E+02 

 264 TEMP 6.548519E+02 

 267 TEMP 6.548519E+02 6.548519E+02 6.548519E+02 6.548519E+02 

 273 TEMP 6.548631E+02 6.548631E+02 

3-90
TABLE 3-33. TRUNCATED FEM OUTPUT FILE OF SAMPLE PROBLEM 6 (Continued) 

 277 TEMP 6.548631E+02 

 282 TEMP 6.548631E+02 6.548631E+02 

 292 TEMP 6.546570E+02 6.546570E+02 

 295 TEMP 6.546570E+02 

 297 TEMP 6.546570E+02 6.546570E+02 6.546570E+02 6.546570E+02 6.546570E+02 6.546570E+02 

 303 TEMP 6.546570E+02 

 306 TEMP 6.546570E+02 6.546570E+02 6.548519E+02 

 313 TEMP 6.548519E+02 6.548519E+02 

 316 TEMP 6.548519E+02 6.548519E+02 6.548519E+02 

 323 TEMP 6.548519E+02 6.548519E+02 

 500 TEMP 5.786700E+02 

When compared, the temperatures of tables 3-31 and 3-34 show good agreement at the wing top. 

However, there is a significant difference in the bottom wing temperatures. This is due to the 

fact that the wing bottom in the FEM model was subjected to the sky radiation only, while the 

wing bottom in the THERMOD model was subjected to both the sky radiation and the tarmac 

radiation. In addition, the THERMOD program used control volumes and view factors in 

modeling the radiation effects, while the sample FEM model radiated heat to space, whereby the 

temperature-increasing effects of multiple reflections within an enclosure was neglected. These 

are the reasons why the wing bottom temperatures of the FEM model are lower than the 

THERMOD model, as comparison of tables 3-31 and 3-34 shows. In any case, sample problem

6 validates the radiation, conduction, and convection capabilities of THERMOD program, 

including infrared emission and absorption. 

TABLE 3-34. FINITE ELEMENT METHOD-SIMULATED TEMPERATURES (°F) OF THE 

RIGHT WING OF SAMPLE PROBLEM 6 

T1 T2 T3 T4 T5 T6 T7 T8 T9 T10 

198.2 198.2 195.2 195.2 195.0 83.2 83.0 83.0 80.0 80.0

3-91/3-92
4. SAMPLE PROBLEM.

This section presents a complete thermal solution for an aircraft. The objective was to determine 

the MOL temperature of the aircraft painted with a particular paint scheme and subjected to a 

given thermal environment. The solution involved both the steady-state and the transient 

analysis procedures. The transient analysis will be dependent on the type of flight profile 

information provided. Hypothetical values were used for all required thermal input data. 

4.1 INPUT DATA. 

THERMOD requires one input file, input.dat. Once invoked (by double-clicking THERMOD), 

THERMOD automatically reads in the input.dat file and reads out two output files. These output 

files are summary.dat and transient.dat. The summary.dat file summarizes the temperature 

results and prints the maximum operating temperatures for steady state and transient state. The 

transient.dat file prints the transient temperatures with respect to time. 

A complete description of the input file, input variables, their definitions, and formats is located 

in the THERMOD User’s Manual [2]. Refer to figures 1-1 through 1-4 for a graphical 

description of the problem. 

The environmental data for the sample problem is based on table 4-1. An analysis was 

performed for each hour noted in the table. This is important because the maximum ambient 

temperature could lag the maximum solar radiation, as indicated in the table. To account for 

infrared sky radiation, an effective sky temperature of 0°F was assumed. This temperature can 

vary from about -50°F under a cold, clear sky to +50°F under warm, cloudy conditions [6]. 

Because a desert environment is likely to be simulated during an extremely hot day, it is 

reasonable to assume a cold, clear sky that would allow maximum solar radiation to penetrate 

through the atmosphere. Thus, an effective sky temperature of 0°F was assumed. 

Hour 

Ambient Temperature 

(°F) 

Solar Radiation 

(Btu/hr/f

Numerical values for the geometric and thermal properties for the hypothetical problem are 

presented in the following. 

TABLE 4-1. TEMPERATURE AND ASSOCIATED RADIATION DATA 

Angle* 

(degrees) 

Wind 

(ft/sec) t

2

) 

1100 60 106 10 300

1200 75 109 11 325

1300 90 114 14 325

1400 75 117 13 300

1500 60 120 12 261

1600 45 119 12 201

1700 30 118 10 130

*Angle refers to the sun’s incident angle at different times. To determine the infrared radiation from the sky, an effective 

sky temperature of 0°F was assumed. 

4-1
The dimensional schematics of the model are given in figure 1-2. The dimensions and thermal 

property values of various elements of the hypothetical problem are given in tables 4-2 through 

4-10. 

Thermal Element 

TABLE 4-2. THICKNESS OF THERMAL ELEMENTS OF THE WING 

Thickness (in) 

Outer skin 

Inner skin 

0.30 

Air space 

0.30 

Inner skin 

Outer skin 

0.03 

Foam 0.375 

0.02 

Spar cap 

12.0 

Spar cap 

0.02 

Foam 0.375 

0.03 

Use figure 1-2 (Detail a: Wing) in conjunction with this table. 

TABLE 4-3. THICKNESS OF THERMAL ELEMENTS OF THE FUSELAGE SIDE 

Thermal Element Thickness (in) 

0.03 

Foam 0.375

Inner skin 0.02 

Insulatory material C 

Outer skin 

0.25 

Use figure 1-2 (Detail b: Fuselage side) in conjunction with this table. 

TABLE 4-4. THICKNESS OF THE THERMAL ELEMENTS OF THE FLOOR 

Thermal Element Thickness (in) 

5.0 

Outer skin 

Inner skin 

3.0 

Inner skin 

Outer skin 

Insulatory material B 

0.03 

Foam 0.375 

0.02 

Floor space 

0.02 

Foam 0.375 

0.03 

Use figure 1-2 (Detail c: Floor) in conjunction with this table. The insulatory material B is a symbolic 

representation of the seat, the carpet, the middle console, and other possible items, all combined and 

treated as one material. This assumption is necessary for THERMOD models conduction in one 

dimension only (through-thickness). In addition, THERMOD considers all discretized elements as 

being isothermal, characterized by uniform radiosity, irradiation, and material properties. The thickness 

of insulatory material B is regarded as the weighted average (in terms of surface area) of the items just

described. 

4-2
TABLE 4-5. THICKNESS OF THE THERMAL ELEMENTS OF THE ROOF 

Thermal Element Thickness (in) 

Outer skin 

Inner skin 0.02 

Insulatory material A 0.25 

0.03 

Foam 0.375 

Use figure 1-2 (Detail d: Roof) in conjunction with this table. 

TABLE 4-6. OVERALL DIMENSIONS OF THE THERMAL MODEL 

Dimension Length (ft)

Lwing 35.0

Wwing 3.5

Hwing 2.8

cabin 11.0 

cabin 4.0 

Hcabin 2.5

Lflg 22.0

Lledge 5.5

L

W

Greenhouse effect within the cabin is modeled with the following dimensions: 

Wcabin = 4.0 ft; Hcabin = 2.5 ft; Lcabin = 11.0 ft.

TABLE 4-7. ABSORPTIVITY AND EMISSIVITY PROPERTIES OF THE EXTERIOR 

SURFACES OF THE WING AND FUSELAGE 

Thermal Element Absorptivity Emissivity 

Wing 0.45 0.80

Fuselage sides 0.65 0.90 

Fuselage top 0.50 0.85 

Fuselage bottom 0.60 0.85 

TABLE 4-8. THERMAL PROPERTIES OF VARIOUS SOLID MATERIALS 

Thermal Element

Density 

(lb/ft3

) 

Conductivity 

(Btu/hr/ft/°F) 

Specific Heat Capacity 

(Btu/lb/°F) 

Divinycell HT70 foam [13] 4.4 0.021 0.24 

E-Glass Composite [14] 125 0.30 0.30 

Fuselage Interior Material 

(A, B, C as in figure 1-1) 10 0.020 0.24

Plexiglas [6] 130 0.80 0.20 

Tarmac [6] 150 0.80 0.22 

Note: The numbers in brackets refer to the references from which the corresponding data were obtained. 

4-3
TABLE 4-9. THERMAL PROPERTIES OF AIR 

Thermal 

Element

Density 

(lb/ft3

) 

Conductivity

(Btu/hr/ft/°F) 

Specific Heat 

Capacity 

(Btu/lb/°F)

Kinematic Viscosity 

(ft2

/sec) 

Prandtl 

Number 

Air [6] 0.067 0.016 0.24 198e-6 0.704 

Note: The number in brackets refer to the references from which the corresponding data were obtained. Air properties are based 

on a temperature of 150°F. 

TABLE 4-10. MISCELLANEOUS PROPERTIES 

Item Values

Tarmac surface: asphalt [6] Absorptivity = 0.9; emissivity = 0.9 

Tarmac depth 2.0 ft of concrete 

Insulatory material A and C [15] Absorptivity = 0.5; emissivity = 0.9 

Insulatory material B [15] Absorptivity = 0.6; emissivity = 0.9 

Absorptivity = 0.2; emissivity = 0.9; transmissivity = 0.70; 

thickness = 0.25 in. 

Cabin scatter factor 

Wing = 0.9; fuselage = 0.9 

Plexiglas [6 and 16] 

Sides = 0.9; floor = 0.9 

Exterior scatter factor [6] 

Note: The numbers in brackets refer to the references from which the corresponding data were obtained.

Note that the E-glass values in table 4-8 are based on typical 7781 epoxy-based glass composite 

[14]. Materials A and C could represent some typical composite panels made of felt core, 

sandwiched between layers of glass cloth. The insulatory material B is a symbolic 

representation of the seat, carpet, and other possible items, all combined and treated as one 

material. This assumption is necessary for THERMOD models conduction in one dimension 

only (through-thickness). In addition, THERMOD considers all discretized elements as being 

isothermal, characterized by uniform radiosity, irradiation, and material properties. The thermal 

properties of insulatory materials A, B, and C are simply assumed as shown. 

The thermal values used for Plexiglas in table 4-8 are typical values for this material [6]. The 

tarmac base material is conservatively assumed to be concrete, allowing for substantial 

conduction of heat from the unsheltered tarmac region to the sheltered tarmac region underneath 

the wing. The tarmac, however, is assumed to be paved with an asphalt surface, thereby, 

absorbing the most heat from the sun (solar absorptivity = 0.9; see table 4-10) and emitting the 

most infrared radiation (emissivity = 0.9; see table 4-10) than would a typical concrete-paved 

tarmac. 

In table 4-10, a tarmac depth of 2.0 ft was assumed: judged thick enough to conduct substantial 

heat from the unsheltered tarmac to the sheltered tarmac underneath the wing. The absorptivity 

of the insulatory materials A, B, and C may vary from 0.1 to 0.7 [15], depending on the texture 

and its composition. Their emissivities are generally high [15], therefore, a value of 0.9 was 

4-4
assumed for all of them. Depending on the thickness and tint of the Plexiglas, its transmissivity 

may vary. FAA Title 14 Code of Federal Regulation 23.775 (d) stipulates that the transmissivity 

of the windshield and side windows must not be less than 0.70 when the pilot is seated in the 

normal flight position. Based on this stipulation, a transmissivity value of 0.70 was assumed. A 

Plexiglas thickness of 0.25 in. was assumed in the analysis. 

The scatter factor for the cabin and exterior (wings and fuselage) surfaces are as shown in 

table 4-10. The scatter factor is defined as the fraction of the reflected solar radiation that is 

diffuse; the rest is specular (focus). Generally, the cabin does not have highly polished surfaces, 

therefore, a 0.9 scatter factor is justified. The exterior surface is, generally, not highly polished 

and non-mirror-like. Reference 6 states that for most engineering applications, a dull surface 

may be assumed. A scatter factor of 0.9, which produces a dull surface, is assumed for the 

exterior surface. This means 10% of the reflected solar radiation is specular. 

Cooling takes place when the door is opened and the full convective forces that have always 

been present outside (due to wind, see table 4-1) is assumed to be applied within the cabin until 

an ambient cabin temperature of 140°F is reached. This ambient temperature is calculated by 

taking the weighted average surface temperatures of T25, T26, T43, T44, and T50 at each time 

step of the transient phase. When the ambient temperature reaches 140°F, the door is closed, and 

the aircraft begins taxiing as indicated in figure 1-4. 

Based on the above information, THERMOD determines the steady-state temperature of the 

model. The results are then used as a starting base in a second phase of analysis, in which 

cooling is introduced. This phase of the analysis is called the transient analysis. To undertake 

this analysis, the flight profile information of figure 1-3 is used. The flight profile provides 

information on relative wind velocity, which is then used to determine the instantaneous 

convective coefficient. The higher the coefficient the sooner the temperature reaches a steadystate during the transient cooling phase. At the end of the transient analysis (at the point of 

application of the limit or gust load), the 44 structural temperatures, are compared to each other, 

and the maximum of these structural temperatures is then used as the MOL temperature for a 

given paint scheme. It should be emphasized that THERMOD was developed in this manner so 

that no special remedial measures need to be taken to address temperatures in heat-soaked deep 

locations. These and all other temperatures are addressed as individual unknowns within a 

system of nonlinear equations, and the maximum of these unknowns (structural temperatures) is 

reported as the MOL temperature for a given paint scheme. 

4.2 INPUT FILE. 

The input file, input.dat, for the sample problem is presented in table 4-11. A total of 15 sets of 

data are required to complete the model description. Most of these data sets were developed 

based on the information provided above. Some information, not described above, was also 

included in these sets, which can be found in the THERMOD User’s Manual [2]. 

4-5
TABLE 4-11. INPUT FILE (input.dat) FOR THE SAMPLE PROBLEM 

198e-6,0.704 

1, 0.030,125,0.300,0.30 

2, 0.375,4.4,0.021,0.24 

3, 0.020,125,0.300,0.30 

4, 0.300,125,0.300,0.30 

5, 12.0,0.067,0.016,0.24 

6, 0.300,125,0.300,0.30 

7, 0.020,125,0.300,0.30 

8, 0.375,4.4,0.021,0.24 

9, 0.030,125,0.300,0.30 

10,0.030,125,0.300,0.30 

11,0.375,4.4,0.021,0.24 

12,0.020,125,0.300,0.30 

13,0.250,10.,0.020,0.24 

14,48.00,0.067,0.016,0.24 

15,0.030,125,0.300,0.30 

16,0.375,4.4,0.021,0.24 

17,0.020,125,0.300,0.30 

18,0.250,10.,0.020,0.24 

19,30.00,0.067,0.016,0.24 

20,5.000,10.,0.020,0.24 

21,0.030,125,0.300,0.30 

22,0.375,4.4,0.021,0.24 

23,0.020,125,0.300,0.30 

24,3.00,0.067,0.016,0.24 

25,0.020,125,0.300,0.30 

26,0.375,4.4,0.021,0.24 

27,0.030,125,0.300,0.30 

28,0.250,130,0.800,0.20 

3.5,35.0,2.8,5.5,22.0,11.0,2.5,4.0,30,24,28 

0.70,0.9,0.9,2.0,150,0.80,0.22 

0.5,0.9,0.6,0.9,0.5,0.9,0.2,0.9 

3,100 

0.9,0.9,0.9,0.9 

0.65,0.90,0.45,0.80,0.50,0.85,0.60,0.85 

9 

0,10,10,0,0,75,110,110,190,190 

5,120,5,60,20,5,120,60,60 

7 

60 565.67 459.67 10 300 10 10 0.36 

75 568.67 459.67 11 325 10 10 0.31 

90 573.67 459.67 14 325 10 12 0.33 

75 576.67 459.67 13 300 10 10 0.31 

60 579.67 459.67 12 261 10 10 0.36 

45 578.67 459.67 12 201 10 10 0.44 

30 577.67 459.67 10 130 10 10 0.57 

1,0.02,140,100 

10,1.0,1.0 

4-6
TABLE 4-11. INPUT FILE (input.dat) FOR THE SAMPLE PROBLEM (Continued) 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 

10,10,10,10,10,10,10 

4.3 OUTPUT FILES. 

Two output files were generated: summary.dat and transient.dat. The summary.dat file 

summarizes the temperature results and prints the MOL temperature for steady state and 

transient state, as shown in table 4-12. The MOL temperature at the end of the steady-sate 

analysis was 174.1°F. The MOL temperature at the end of the transient analysis was 151.0°F. 

Thus, the transient analysis has reduced the maximum soaked temperature by 23.1°F. The final 

MOL temperature for this aircraft was 151.0°F. 

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ 

The transient.dat file prints the transient temperatures with respect to time, as shown in 

table 4-13. The variations of all 53 temperatures with respect to time are presented. 

TABLE 4-12. SAMPLE PROBLEM OUTPUT DATA FILE (summary.dat) 

THERMOD: A THERMAL PROGRAM FOR AIRCRAFTS 

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ 

 

 

 

 SUMMARY OF STEADY-STATE AND TRANSIENT ANALYSIS 

 

 

 

 (Temperatures are shown for all time periods) 

 

 

 TEMPERATURES AT THE END OF STEADY STATE ANALYSIS 

 

 TEMPERATURES IN DEGREES FAHRENHEIT 

 1 2 3 4 5 6 7 

 

 1 138.1 149.1 150.4 146.9 137.1 119.1 100.3 

 2 138.1 149.1 150.4 146.9 137.1 119.1 100.3 

 3 137.5 148.4 149.7 146.3 136.6 118.9 100.4 

 4 137.5 148.4 149.7 146.3 136.6 118.9 100.4 

 5 137.5 148.4 149.7 146.3 136.6 118.9 100.4 

 6 112.1 117.9 120.3 120.6 118.4 111.4 103.8 

 7 112.0 117.9 120.3 120.5 118.4 111.4 103.8 

 8 112.0 117.9 120.3 120.5 118.3 111.4 103.8 

 9 111.4 117.2 119.6 119.9 117.9 111.2 103.9 

4-7
TABLE 4-12. SAMPLE PROBLEM OUTPUT DATA FILE (summary.dat) (Continued) 

 10 111.4 117.2 119.6 119.9 117.9 111.2 103.9 

 11 134.1 145.9 148.3 144.1 133.8 116.2 98.1 

 12 134.1 145.9 148.3 144.1 133.8 116.2 98.1 

 13 133.6 145.2 147.7 143.5 133.5 116.1 98.2 

 14 133.6 145.2 147.7 143.5 133.5 116.1 98.2 

 15 133.6 145.2 147.6 143.5 133.4 116.1 98.2 

 16 111.9 117.8 120.3 120.5 118.3 111.3 103.7 

 17 111.9 117.8 120.2 120.5 118.3 111.3 103.7 

 18 111.9 117.8 120.2 120.5 118.3 111.3 103.7 

 19 111.4 117.1 119.6 119.9 117.9 111.2 103.9 

 20 111.4 117.1 119.6 119.9 117.9 111.2 103.9 

 21 135.8 145.0 144.3 141.2 132.7 115.4 92.8 

 22 135.9 145.1 144.4 141.3 132.8 115.6 92.8 

 24 166.3 169.5 166.0 164.4 160.1 143.2 109.1 

 25 187.5 186.5 181.0 180.5 179.1 162.5 120.4 

 26 195.0 193.5 188.7 187.4 186.5 170.6 125.1 

 27 148.4 150.7 149.8 149.4 147.9 137.0 110.9 

 28 148.4 150.7 149.8 149.3 147.9 137.0 110.9 

 46 160.4 158.5 152.1 155.3 156.6 146.8 121.2 

 23 166.2 169.4 165.9 164.3 160.0 143.1 109.0 

 29 145.1 147.6 147.0 146.6 145.1 134.6 109.9 

 30 145.1 147.6 147.0 146.6 145.1 134.6 109.9 

 31 127.6 131.5 132.4 132.3 130.6 122.0 104.5 

 32 127.6 131.5 132.4 132.3 130.6 122.0 104.5 

 33 127.6 131.5 132.4 132.3 130.6 122.0 104.5 

 34 127.6 131.5 132.4 132.3 130.6 122.0 104.5 

 35 110.1 115.5 117.8 118.1 116.2 109.4 99.2 

 36 110.1 115.5 117.8 118.1 116.2 109.4 99.2 

 37 106.8 112.4 115.0 115.4 113.4 107.0 98.2 

 38 106.8 112.4 115.0 115.3 113.4 107.0 98.2 

 39 159.2 147.2 126.8 143.9 153.5 147.2 130.2 

 40 159.3 147.3 127.0 144.0 153.6 147.3 130.2 

 41 174.0 166.3 152.3 161.8 167.1 157.2 129.8 

 42 174.1 166.4 152.4 161.8 167.2 157.2 129.8 

 43 184.4 179.7 170.2 174.3 176.7 164.2 129.6 

 44 187.2 180.7 170.1 175.4 179.5 169.1 135.0 

 45 160.5 158.6 152.2 155.4 156.8 146.9 121.3 

 47 122.3 126.9 126.4 126.7 124.1 115.0 101.6 

 48 122.1 126.7 126.3 126.6 123.9 114.9 101.4 

 49 149.8 150.9 146.4 147.1 146.2 134.9 113.9 

 50 151.7 152.6 148.0 148.7 147.8 136.3 114.4 

 51 109.7 116.7 118.1 117.3 113.0 102.9 90.6 

 52 104.2 111.0 112.9 112.1 108.0 98.6 86.5 

 53 109.7 116.7 118.1 117.3 113.0 102.9 90.6 

 

Maximum Structural Temperature for Period 1 = 174.10 Occurring 

at Location 42 

4-8
TABLE 4-12. SAMPLE PROBLEM OUTPUT DATA FILE (summary.dat) (Continued) 

Maximum Structural Temperature for Period 2 = 169.52 Occurring 

at Location 24 

Maximum Structural Temperature for Period 3 = 165.98 Occurring 

at Location 24 

Maximum Structural Temperature for Period 4 = 164.41 Occurring 

at Location 24 

Maximum Structural Temperature for Period 5 = 167.20 Occurring 

at Location 42 

Maximum Structural Temperature for Period 6 = 157.23 Occurring 

at Location 42 

Maximum Structural Temperature for Period 7 = 130.19 Occurring 

at Location 39 

 

Maximum Structural Temperature Over All 7 Periods =174.10 

Occurring at Location 42 at Period 1 

 

 

 TEMPERATURES AT THE END OF TRANSIENT ANALYSIS 

 

 TEMPERATURES IN DEGREES FAHRENHEIT 

 1 2 3 4 5 6 7 

 

 1 108.5 112.3 117.2 119.5 121.4 118.9 116.5 

 2 109.3 113.3 118.1 120.3 121.8 118.9 116.0 

 3 135.1 145.5 147.0 144.1 135.3 118.9 101.7 

 4 135.9 146.4 147.9 144.8 135.7 118.9 101.2 

 5 136.5 147.2 148.6 145.4 136.0 118.9 100.9 

 6 112.1 117.9 120.3 120.6 118.4 111.4 104.4 

 7 111.9 117.8 120.2 120.5 118.4 111.6 104.6 

 8 111.8 117.6 120.1 120.5 118.5 111.8 104.9 

 9 107.4 110.9 115.6 118.2 120.4 118.4 116.5 

 10 107.3 110.7 115.5 118.1 120.4 118.6 116.9 

 11 108.4 112.2 117.2 119.5 121.3 118.8 116.4 

 12 109.2 113.2 118.1 120.2 121.6 118.7 115.9 

 13 131.7 142.5 145.2 141.7 132.4 116.2 99.7 

 14 132.4 143.4 146.0 142.4 132.7 116.1 99.2 

 15 132.9 144.1 146.6 142.9 132.9 116.1 98.8 

 16 111.9 117.8 120.3 120.5 118.3 111.3 104.3 

 17 111.8 117.7 120.2 120.4 118.3 111.5 104.5 

 18 111.7 117.5 120.0 120.4 118.4 111.7 104.9 

 19 107.4 110.9 115.6 118.2 120.4 118.4 116.5 

 20 107.3 110.7 115.5 118.1 120.4 118.6 116.9 

 21 108.6 112.6 117.5 119.5 121.0 118.3 115.3 

 22 109.5 113.4 118.2 120.2 121.5 118.7 115.3 

 23 135.3 138.6 140.9 139.5 136.8 132.6 114.4 

 24 136.4 139.7 141.8 140.4 137.6 133.3 114.5 

 25 148.7 150.0 150.5 149.3 147.4 143.0 119.3 

 26 153.3 154.1 154.9 153.5 152.3 147.9 122.5 

4-9
TABLE 4-12. SAMPLE PROBLEM OUTPUT DATA FILE (summary.dat) (Continued) 

 27 148.8 151.0 150.1 149.6 148.2 137.2 110.9 

 28 147.8 150.1 149.3 148.8 147.4 136.6 110.8 

 29 145.1 147.7 147.1 146.7 145.2 134.6 110.0 

 30 145.1 147.6 147.0 146.6 145.1 134.6 110.0 

 31 128.1 131.7 132.9 133.3 132.4 124.5 108.4 

 32 128.1 131.7 132.9 133.3 132.4 124.5 108.4 

 33 128.1 131.7 132.9 133.3 132.4 124.5 108.4 

 34 128.1 131.7 132.9 133.3 132.4 124.5 108.4 

 35 111.1 115.8 118.9 120.0 119.6 114.4 106.8 

 36 111.0 115.6 118.8 120.0 119.6 114.5 107.0 

 37 107.1 110.6 115.3 117.8 119.9 117.8 115.7 

 38 107.0 110.5 115.2 117.7 119.9 118.0 116.0 

 39 109.1 111.4 114.9 118.6 121.7 120.2 117.9 

 40 110.0 112.2 115.5 119.2 122.3 120.8 118.1 

 41 137.3 136.0 133.9 137.5 139.3 137.7 123.5 

 42 138.3 136.9 134.7 138.2 140.0 138.3 123.7 

 43 144.9 144.1 142.7 144.6 145.6 143.0 124.5 

 44 147.5 145.2 142.7 145.7 148.0 146.6 128.2 

 45 136.2 135.6 134.5 137.1 138.6 137.5 123.2 

 46 135.1 134.6 133.8 136.4 137.9 136.8 122.9 

 47 109.9 112.1 115.5 119.2 122.2 120.7 118.0 

 49 112.2 115.2 119.2 121.4 123.3 121.1 117.0 

 50 114.3 117.3 121.0 123.0 124.7 122.3 117.2 

 51 166.3 183.2 182.1 174.9 157.7 128.5 98.0 

 52 155.1 170.7 170.4 163.9 148.2 121.6 93.1 

 53 166.3 183.2 182.1 174.9 157.7 128.5 98.0 

 

Maximum Structural Temperature for Period 1 = 148.77 Occurring 

at Location 27 

Maximum Structural Temperature for Period 2 = 151.00 Occurring 

at Location 27 

Maximum Structural Temperature for Period 3 = 150.07 Occurring 

at Location 27 

Maximum Structural Temperature for Period 4 = 149.62 Occurring 

at Location 27 

Maximum Structural Temperature for Period 5 = 148.17 Occurring 

at Location 27 

Maximum Structural Temperature for Period 7 = 123.67 Occurring 

at Location 42 

 

Maximum Structural Temperature Over All 7 Periods =151.00 

Occurring at Location 27 at Period 2 

 48 109.0 111.4 114.9 118.6 121.7 120.2 117.8 

Maximum Structural Temperature for Period 6 = 138.33 Occurring 

at Location 42 

The MOL temperature at the end of the steady-sate analysis was 174.1°F. The MOL temperature at the end of the 

transient analysis was 151.0°F. Thus, the final MOL temperature for this aircraft was 151.0°F. 

4-10
TABLE 4-13. SAMPLE PROBLEM OUTPUT DATA FILE (transient.dat) 

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ 

 

 

 TRANSIENT TEMPERATURES FOR PERIOD: 1 

 COOLING WHILE DOOR IS OPENED: 

 

 2.0000 138.1 138.1 137.5 137.5 . . . 109.7 104.2 109.7 

 4.0000 138.1 138.1 137.6 137.5 . . . 109.7 104.2 109.7 

 6.0000 138.1 138.1 137.6 137.5 . . . 109.7 104.2 109.7 

 8.0000 138.1 138.1 137.6 137.5 . . . 109.7 104.2 109.7 

 10.0000 138.1 138.1 137.6 137.5 . . . 109.7 104.2 109.7 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 386.0000 138.1 138.1 137.6 137.5 . . . 109.7 104.2 109.7 

 388.0000 138.1 138.1 137.6 137.5 . . . 109.7 104.2 109.7 

 390.0000 138.1 138.1 137.6 137.5 . . . 109.7 104.2 109.7 

 COOLING DURING AIRCRAFT MANEUVERS: 

 time(sec) 1 2 3 4 . . . 51 52 53 

 394.0000 138.0 138.1 137.6 137.5 . . . 145.5 136.3 145.5 

 396.0000 137.9 138.0 137.6 137.5 . . . 157.2 146.8 157.2 

 398.0000 137.8 137.9 137.6 137.5 . . . 162.3 151.4 162.3 

 400.0000 137.6 137.8 137.6 137.5 . . . 164.6 153.5 164.6 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 840.0000 108.5 109.3 135.2 135.9 . . . 166.3 155.1 166.3 

 842.0000 108.5 109.3 135.1 135.9 . . . 166.3 155.1 166.3 

 844.0000 108.5 109.3 135.1 135.9 . . . 166.3 155.1 166.3 

 846.0000 108.5 109.3 135.1 135.9 . . . 166.3 155.1 166.3 

 

 

 TRANSIENT TEMPERATURES FOR PERIOD: 2 

 

 COOLING WHILE DOOR IS OPENED: 

 

 time(sec) 1 2 3 4 . . . 51 52 53 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 

 COOLING DURING AIRCRAFT MANEUVERS: 

 

 time(sec) 1 2 3 4 . . . 51 52 53 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 

THERMOD: A THERMAL PROGRAM FOR AIRCRAFTS 

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ 

 

 time(sec) 1 2 3 4 . . . 51 52 53 

 0.0000 138.1 138.1 137.5 137.5 . . . 109.7 104.2 109.7 

 

 

 392.0000 138.1 138.1 137.6 137.5 . . . 119.1 112.6 119.0 

 . . . . . . . . . . . 

4-11
TABLE 4-13. SAMPLE PROBLEM OUTPUT DATA FILE (transient.dat) (Continued) 

 

 TRANSIENT TEMPERATURES FOR PERIOD: 3 

 

 time(sec) 1 2 3 4 . . . 51 52 53 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 

 COOLING DURING AIRCRAFT MANEUVERS: 

 

 time(sec) 1 2 3 4 . . . 51 52 53 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 

 TRANSIENT TEMPERATURES FOR PERIOD: 4 

 

 COOLING WHILE DOOR IS OPENED: 

 

 time(sec) 1 2 3 4 . . . 51 52 53 

 6.0000 146.9 146.9 146.3 146.3 . . . 117.3 112.1 117.3 

 8.0000 146.9 146.9 146.3 146.3 . . . 117.3 112.1 117.3 

 10.0000 146.9 146.9 146.3 146.3 . . . 117.3 112.1 117.3 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 342.0000 146.9 146.9 146.3 146.3 . . . 117.3 112.1 117.3 

 

 COOLING DURING AIRCRAFT MANEUVERS: 

 

 time(sec) 1 2 3 4 . . . 51 52 53 

 346.0000 146.9 146.9 146.3 146.3 . . . 143.5 135.7 143.5 

 348.0000 146.8 146.8 146.3 146.3 . . . 161.1 151.4 161.1 

 350.0000 146.6 146.7 146.3 146.3 . . . 168.8 158.4 168.8 

 352.0000 146.5 146.6 146.3 146.3 . . . 172.2 161.5 172.2 

 354.0000 146.4 146.5 146.3 146.3 . . . 173.7 162.8 173.7 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 790.0000 119.5 120.3 144.2 144.9 145.5 120.6 120.5 174.9 163.9 174.9 

 794.0000 119.5 120.3 144.1 144.9 145.4 120.6 120.5 174.9 163.9 174.9 

 

 COOLING WHILE DOOR IS OPENED: 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 0.0000 146.9 146.9 146.3 146.3 . . . 117.3 112.1 117.3 

 2.0000 146.9 146.9 146.3 146.3 . . . 117.3 112.1 117.3 

 4.0000 146.9 146.9 146.3 146.3 . . . 117.3 112.1 117.3 

 . . . . . . . . . . . 

 336.0000 146.9 146.9 146.3 146.3 . . . 117.3 112.1 117.3 

 338.0000 146.9 146.9 146.3 146.3 . . . 117.3 112.1 117.3 

 340.0000 146.9 146.9 146.3 146.3 . . . 117.3 112.1 117.3 

 344.0000 146.9 146.9 146.3 146.3 . . . 117.3 112.1 117.3 

 792.0000 119.5 120.3 144.1 144.9 145.5 120.6 120.5 174.9 163.9 174.9 

 796.0000 119.5 120.3 144.1 144.8 145.4 120.6 120.5 174.9 163.9 174.9 

 798.0000 119.5 120.3 144.1 144.8 145.4 120.6 120.5 174.9 163.9 174.9 

4-12
TABLE 4-13. SAMPLE PROBLEM OUTPUT DATA FILE (transient.dat) (Continued) 

 

 TRANSIENT TEMPERATURES FOR PERIOD: 5 

 

 COOLING WHILE DOOR IS OPENED: 

 time(sec) 1 2 3 4 . . . 51 52 53 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 

 time(sec) 1 2 3 4 . . . 51 52 53 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 

 COOLING WHILE DOOR IS OPENED: 

 

 . . . . . . . . . . . 

 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 time(sec) 1 2 3 4 . . . 51 52 53 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 

 . . . . . . . . . . . 

 

 COOLING DURING AIRCRAFT MANEUVERS: 

 . . . . . . . . . . . 

 

 TRANSIENT TEMPERATURES FOR PERIOD: 6 

 

 time(sec) 1 2 3 4 . . . 51 52 53 

 . . . . . . . . . . . 

 . . . . . . . . . . . 

 

 COOLING DURING AIRCRAFT MANEUVERS: 

 time(sec) 1 2 3 4 . . . 51 52 53 

 

TRANSIENT TEMPERATURES FOR PERIOD: 7 

 

 COOLING WHILE DOOR IS OPENED: 

 

 . . . . . . . . . . . 

 

 COOLING DURING AIRCRAFT MANEUVERS: 

 

 time(sec) 1 2 3 4 . . . 51 52 53 

 . . . . . . . . . . . 

Note: This file was truncated. Transient data was written for each time period for all 53 temperatures. Interior 

cooling took place when the door was opened. Once cabin ambient temperature of 140°F was reached, the door 

was closed, and the aircraft was subjected to further cooling while maneuvering according to the input flight profile. 

4-13/4-14
5. SUMMARY.

A thermal analysis was performed on a hypothetical aircraft. The wing was painted with a paint 

color containing the following thermal properties: absorptivity = 0.45 and emissivity = 0.80. 

The fuselage was painted as follows: (1) sides: absorptivity = 0.65 and emissivity = 0.90, (2) 

top: absorptivity = 0.50 and emissivity = 0.85, and (3) bottom: absorptivity = 0.60 and 

emissivity = 0.85. 

A few important observations from tables 4-12 and 4-13 should be noted. 

1. A total of 53 temperatures were simulated, out of which 44 temperatures were considered 

structural and 9 were nonstructural. The structural temperatures were T1-T24, T27-T38, 

T39-T42, and T45-T48. The nonstructural temperatures were 

b. Plexiglas temperatures (T49 and T50); and 

2. Transient temperature results were simulated based on the flight profile information in 

figure 1-4. 

4. In some cases, the transient temperatures of table 4-12 are higher than the corresponding 

steady-state temperatures, even though the transient phase is thought to have a cooling 

effect on the plane. Generally, during the transient phase, the aircraft is expected to cool 

Table 4-12 shows all 53 steady-state temperatures for different times of the day. For example, at 

1100 hours (time period 1), temperature T1 is 138.1°F, and at 1300 hours (time period 2), it is 

149.1°F. The hottest structural temperature is T42 (174.1°F), occurring at 1100 hours (time 

period 1). If only steady-state results were used in the Federal Aviation Administration 

certification, the maximum operating limit (MOL) temperature for this aircraft should at least be 

175°F. However, THERMOD was modeled to consider transient analysis, and as a result, the 

benefits associated with this cooling phase should be considered. 

Table 4-12 also shows the results of the transient analysis. The transient analysis was performed 

using the flight profile of figure 1-4 and the steady-state temperatures of table 4-12 as the 

starting base. In general, the transient temperatures of table 4-12 are lower than steady-state 

temperatures. The hottest transient structural temperature is T27 (151.0°F), occurring at 1200 

hours (time period 2). This temperature is practically the same as its steady-state counterpart 

(T27 = 150.7°F). This is due to the fact that T27 was located underneath the 5-inch-thick 

insulatory material (material B, representing the seats, carpet, and other related items), which 

acts as an effective barrier to thermal changes above it. Note that in contrast, surfaces that are 

exposed to convective forces (such as those surfaces represented by T1, T11, T25, T26, T39, 

T43, T44, and T49) undergo rapid cooling. 

a. insulatory material temperatures (T25, T26, T43, and T44); 

c. tarmac temperatures (T51, T52, and T53). 

3. To obtain conservative results, THERMOD assumed the ground to possess thermal 

characteristics similar to that of the tarmac throughout the entire flight profile, even while 

the plane was in the air. 

5-1 
down due to convective heat transfer from the air flowing over (it until equilibrium is 

achieved). 

Thus, in some cases, when the user has set the time of the day as early morning or late 

evenings, and depending on the user’s other ‘weather condition’ inputs, it’s sometimes 

possible that the airplane’s exterior is cooler (due to relatively cooler steady state 

temperatures) than the relatively warmer surroundings (which are warmer due to the user 

specified inputs of solar radiation, sun angle etc.). And thus, for calculating transient 

phase temperatures, when THERMOD moves this relatively warmer air over the airplane, 

it raises the temperature on the airplane’s exterior. 

For the purpose of computing the transient phase temperatures, THERMOD does not take 

airplane altitude into account, but instead, compensates by changing the air flow 

temperature over the airplane to equal the ambient temperatures prescribed for that flight 

profile. The convective heat transfer coefficients vary according to the time and speed 

set for each flight segment by the user. The user also sets the ambient conditions when 

they specify the “time of the day”, “sun inclination angle”, “relative wind speed” and 

“solar radiation” etc. This means, that the airplane would in fact never leave the ground, 

but its flight profile such as shown in figure 1-4 would be simulated by changing the 

ambient conditions and the convective coefficients for each specified flight segment. 

5. The tarmac temperatures (T51, T52, and T53) show significant increase during the 

transient phase. In the steady-state phase, these temperatures were lower because they 

represent regions that were directly under the wing, which provided some measure of 

shade from direct sunlight. However, because the plane is in motion, as in the transient 

phase, these temperatures were essentially the exposed tarmac temperatures. 

6. For a given paint scheme, THERMOD simulated the temperatures at each time period. 

The 44 structural temperatures in the transient phase were compared with each other over 

seven time periods (1100, 1200, 1300, 1400, 1500, 1600, and 1700 hours), and the 

maximum structural temperature obtained from this comparison was used as the MOL 

temperature. This MOL temperature may be located at the surface (depending on how 

dark the paint is), inside the cabin, within the floor space, or elsewhere. It is the 

maximum structural temperature experienced anywhere in the aircraft. 

From this project, it was observed that this hypothetical aircraft with paint absorptivity and 

emissivity values as noted above had a MOL temperature of 174.1°F at the end of the steady-sate 

analysis. The MOL temperature at the end of the transient analysis was 151.0°F. Thus, the 

transient analysis reduced the maximum soaked temperature by 23.1°F. The final MOL 

temperature for this aircraft was 151.0°F. This temperature occurred on the floor surface 

underneath the insulatory material (material B of figure 1-1) covering the floor. 

5-2 
6. REFERENCES.

1. Govindarajoo, N., “THERMOD: A Thermal Model for Predicting Aircraft MOL 

Temperatures,” Report No. REK0002A.DOC, Cirrus Design Corporation, Duluth, MN, 

October 1996. 

2. Govindarajoo, N., “THERMOD, An Enhanced Thermal Model for Determining Aircraft

Operational Temperatures: THERMOD User’s Manual,” FAA report DOT/FAA/AR04/52, December 2004. 

3. 

9. MScNASTRAN 70.5, “Quick Reference Guide,” MacNeal-Schwendler Corp., 815 

Colorado Boulevard, Los Angeles, CA, 1998. 

11. MScNASTRAN 70.5, “Command Reference Guide,” MacNeal-Schwendler Corp., 815 

Colorado Boulevard, Los Angeles, CA, 1998. 

Miller, L. Scott, Waltner, J., Mazmudar, T., Merchant, M., and Tomblin, J., “THERMOD 

Composite Airframe Temperature Prediction Tool Evaluation, Validation, and 

Enhancement With Initial Steady-State Temperature Data,” FAA report DOT/FAA/AR04/30, September 2004.

4. Miller, L.S. “Impact of Aircraft Operation on Composite Airframe Temperatures,” 

AGATE-WP3.3-033051-122, October 2001 (also available at http://www.niar.wichita. 

edu/agate. 

5. Press, W.H. et al., “Numerical Recipes in FORTRAN: The Art of Scientific 

Computing,” second edition, Cambridge Univ. Press, New York, NY, 1994. 

6. Incropera, F.P. and Dewitt, D.P., “Fundamentals of Heat and Mass Transfer,” third 

edition, John Wiley & Sons, New York, NY, 1990. 

7. UAI/NASTRAN 20.1, “User’s Reference Manual,” Universal Analytics, Inc., 3625 Del 

Almo Blvd., Suite 370, Torrance, CA, 1999. 

8. UAI/NASTRAN 20.1, “User’s Guide,” Universal Analytics, Inc., 3625 Del Almo Blvd., 

Suite 370, Torrance, CA, 1999. 

10. MScNASTRAN 70.5, “User’s Guide,” MacNeal-Schwendler Corp., 815 Colorado 

Boulevard, Los Angeles, CA,1998. 

12. MScNASTRAN 68, “Thermal Analysis User’s Guide,” MacNeal-Schwendler Corp., 815 

Colorado Boulevard, Los Angeles, CA, 1994.

13. Technical Brochure, Divinycell International Inc., DIAD Group, 315 Seahawk Drive, 

DeSoto, TX 75115. 

14. Reinhart, T.J., Chairman, “Engineered Materials Handbook Vol. 1, Composites,” ASM 

International, OH, 1987. 

6-1
15. Gubareff, G.G. et al., “Thermal Radiation Properties Survey: A Review of Literature,” 

second edition, Honeywell Research Center, Minneapolis, MN, 1960. 

16. Gebhart, B., “Heat Transfer,” second edition, McGraw-Hill Book Co., 1971. 

6-2
APPENDIX A—THERMOD CODE

THERMOD is a FORTRAN-based program developed using the DIGITAL VISUAL 

FORTRAN 6.0 compiler, an IBM-compatible Pentium II PC, and Windows 98. THERMOD is

noninteractive and runs on either the DOS or Windows platform. A computer with 64 MB or 

more of RAM and with at least 500 MB of hard disk space is recommended. A typical analysis 

may take from as little as 30 seconds to a few hours, depending on the whether the explicit 

forward finite difference method (EFFDM) or the implicit backward finite difference method 

(IBFDM), respectively, is selected. It also depends on the time step used in the transient 

analysis. 

THERMOD requires one input file, input.dat. Once invoked (by double-clicking THERMOD), 

THERMOD automatically reads in the input.dat file and reads out two output files, which are 

summary.dat and transient.dat. The summary.dat file summarizes the temperature results and 

prints the maximum operating temperatures for steady state and transient state. The transient.dat 

file prints the transient temperatures with respect to time. Details on these files and instructions 

on how to use THERMOD can be found in THERMOD User’s Manual [A-1].

THERMOD simulates the service temperatures of an aircraft by setting up and solving a set of

67 equations in a nonlinear system of equations to produce 53 steady-state temperatures and 14 

steady-state radiosity. The solution is performed using the Newton-Raphson iteration technique 

[A-2]. This is followed by a transient analysis, based on either the EFFDM or the IBFDM. 

Figure A-1 shows a flowchart of THERMOD. The following paragraphs give a brief description 

of THERMOD, and section A.1 lists the definitions of the routines. Section A.2 presents the 

THERMOD code in its entity. 

MAIN is the main controlling program that calls subroutine READIN. READIN reads in the 

input file containing the relevant information to run THERMOD. MAIN next calls the 

SOLUTION subroutine, which controls both the steady-state and transient analyses.

SOLUTION performs this function by calling 11 subroutines as shown in figure A-1. 

THERMOD was programmed to simulate the temperatures of an aircraft for a particular paint

scheme chosen. The subroutines VIEWFACT, HCONVECT, QREFCAB, TARMMOD,

INFRQSKY, and SBFILLET are called by SOLUTION only once. SOLUTION also calls 

subroutines QREFFW, QREFTW, NEWTRAPH, and TRANSIEN. 

The steady-state analysis phase is controlled by the NEWTRAPH subroutine, which calls 

THERMODL, LUDCMP, and LUBKSB subroutines. The THERMODL subroutine forms the 

67 equations needed in THERMODL. This is the heart of the steady-state analysis.

THERMODL also calls JACOBIAN to formulate the Jacobians needed in the Newton-Raphson

Iteration solution procedure [A-2]. Having formulated the equations using the THERMODL and 

JACOBIAN subroutines, NEWTRAPH next calls the LUDCMP subroutine, which decomposes

the Jacobian matrix. This is followed by another call by NEWTRAPH to LUBKSB where

backsubstitution is performed to find the solution for the unknowns. 

The transient analysis phase is controlled either by the TRNSIEN1 or TRNSIEN2 subroutines, 

where the EFFDM method is controlled by TRNSIEN1 and the IBFDM method is controlled by 

TRNSIEN2. TRNSIEN1 calls LHSWNG1, RHSWNG1, FUSLGR1, FUSLGRF1, TARMAC1, 

A-1
UPRADIO, and HCOOLCON subroutines. These subroutines together with LUDCMP and 

LUBKSB complete the EFFDM transient analysis. The inclusion of the UPRADIO subroutine, 

which updates the radiosity for each time increment, is necessary because without it, numerical

overflow or other numerical difficulties will result. TRNSIEN2 calls LHSWNG2, RHSWNG2, 

FUSLGR2, FUSLGRF2, TARMAC2, UPRADIO, and HCOOLCON and NEWRAPH2 

subroutines. The UPDRADIO calls LUDCMP and LUBKSB to update the radiosity. The 

NEWRAPH2 subroutine calls LHSWNG2, RHSWNG2, FUSLGR2, FUSLGRF2, TARMAC2, 

LUDCMP, and LUBKSB to complete the IBFDM finite difference analysis. 

A.1 DEFINITION OF SUBROUTINES.

MAIN The main controlling program.

READIN Reads in the input data. 

SOLUTION Controls both the steady-state and transient analyses. 

VIEWFACT The controlling routine for determining view factors for aligned parallel 

rectangles and for perpendicular rectangles with a common edge. 

VIEWFC Determines the view factors for aligned parallel rectangles, and is the 

controlling routine for determining the view factors for perpendicular 

rectangles with a common edge. 

VFACFIJ Determines the view factors for perpendicular rectangles with a common 

edge. 

HCONVECT Determines the steady-state convective coefficients for the fuselage, wing, and 

tarmac. 

QREFCAB Determines the intercepted solar radiation in the cabin, and is the controlling 

outline for determining the diffuse solar radiation that is absorbed in the cabin

by treating the cabin as a partial cavity. 

DIFFSOLA Determines the diffuse solar radiation that is absorbed in a six-sided 

rectangular enclosure by treating it as a partial cavity in which multiple 

reflections and absorptions are considered. 

TARMMODL Determines the infrared emission from the tarmac as well as the tarmac 

temperature.

INFRQSKY Determines the infrared emission from the sky. 

SBFILLET Determines the view factors of fillets by assuming them to be a quadrant.

QREFFW Determines the intercepted solar radiation at the fuselage-wing junction, and 

is the controlling routine for determining the diffuse solar radiation that is 

absorbed at this junction by treating the junction as a partial cavity. 

A-2
QREFTW Determines the intercepted solar radiation at the tarmac-wing bottom region,

and is the controlling routine for determining the diffuse solar radiation that is

absorbed in this region by treating the region as a partial cavity. 

DIFSLWNG Determines the diffuse solar radiation that is absorbed at the tarmac-wing 

bottom region by representing the region as a six-sided rectangular enclosure 

and by treating it as a partial cavity in which multiple reflections and 

absorptions are considered. The region is divided into three smaller 

rectangular boxes, representing the LHS, RHS, and the MIDDLE of the wing. 

NEWTRAPH The controlling routine for the steady-state analysis based on the NewtonRaphson iteration technique. 

THERMODL Formulates the 67 heat transfer equations. 

LUDCMP Decomposes a matrix by the LU decomposition technique. 

LUBKSB Determines the solution of the unknown by backward substitution. 

JACOBIAN Determines the Jacobian for the 67 equations as a requirement of the NewtonRaphson iteration technique. 

TRNSIEN1 The controlling routine for the EFFDM transient analysis. 

LHSWNG1 Performs EFFDM transient analysis of the left wing (the wing exposed to the 

solar radiation). 

RHSWNG1 Performs EFFDM transient analysis of the right wing (the wing that is under 

the shade of the fuselage). 

FUSLGLR1 Performs EFFDM transient analysis of the fuselage from left to right. 

FUSLGRF1 Performs EFFDM transient analysis of the fuselage from top to bottom

(including the Plexiglas cover). 

TARMAC1 Performs EFFDM transient analysis of the tarmac. 

UPRADIO Updates radiosity (all 14) at each time increment of the transient analysis. 

HCOOLCON Determines the convective coefficient at each time increment for the fuselage 

surface and the wing surface based on a given flight profile. 

TRNSIEN2 The controlling routine for the IBFDM transient analysis. 

LHSWNG2 Performs IBFDM transient analysis of the left wing (the wing exposed to the 

solar radiation). 

RHSWNG2 Performs IBFDM transient analysis of the right wing (the wing that is under 

the shade of the fuselage). 

A-3
FUSLGLR2 Performs IBFDM transient analysis of the fuselage from left to right. 

FUSLGRF2 Performs IBFDM transient analysis of the fuselage from top to bottom 

(including the Plexiglas cover). 

TARMAC2 Performs IBFDM transient analysis of the tarmac. 

NEWRAPH2 The controlling subroutine for the nonlinear solution of the IBFDM transient 

problem. The nonlinear solution is provided by the Newton-Raphson iteration 

technique. 

Solid lines represent the original THERMOD program. The dotted lines represent additions to the 

original THERMOD program.

FIGURE A-1. FLOWCHART OF THE THERMOD MODEL 

A-4
A-2. THERMOD PROGRAM CODE. 

program main 

c This is the main controlling program. 

c This program determines the service temperatures of an aircraft. 

c There are 67 equations with 67 unknowns to be solved in a non-linear

c system of equations. 

c 

c Program developed by Nathan Govindarajoo, PhD., PE., Consulting Engineer. 

c Program belongs to FAA and AGATE. 

c 

 write(6,*)'$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$' 

 write(6,*)'WELCOME TO THERMOD: A THERMAL PROGRAM FOR AIRCRAFTS' 

 write(6,*)'$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$' 

 write(6,*)' ' 

write(6,*)'INPUT DATA BEING READ' 

 call readin 

 write(6,*)'INPUT DATA SUCCESSFULLY READ' 

 write(6,*)' ' 

write(6,*)'SOLUTION IN PROGRESS' 

 call solution 

 write(6,*)' ' 

 write(6,*)'THERMOD SUCCESSFULLY COMPLETED' 

 end 

c 

 subroutine readin 

c purpose: reads in the data 

c calls: none

c called from: main 

 common/c1/n,ntrial,tolt,tolfn 

 common/c2/t(200),fn(200),jacob(200,200) 

 common/c3/th(100),k(100) 

 common/c4/widthwng,lengtwng,hwng,locwng,lplane,lfuselg,hfuselg, 

 & wfuselg 

 common/c5/trnsm,absorbta,emista,tarmd,densta,conducta,shcta 

 common/c6/absorbf,emisf,absorbw,emisw,absorbfl,emisfl,absorbfi, 

 &emisfi,absorbpg,emispg,absorftp,emisrftp,absorfbt,emisrfbt, 

 &absorbfb,emisfb 

 common/c7/nperiods,envtdata(100,20) 

 common/c8/icode,nconstrn,idtemp(100),tempcons(100) 

 common/c9/scatffl,scatffi,scatfw,scatff 

 common/c10/aircondc,thdiffus,visckine,prandtl 

 common/c11/icodef11,percent,nparts 

 common/c12/a1,a2,a3,a4,a5,a6(100),a7(100) 

 common/c13/nintervl,vel(100),timeintv(100) 

 common/c14/thicknes(100),density(100),conductv(100),sheatcap(100)

 common/c15/ibegabsf,iendabsf,ibegemsf,iendemsf,ibegabsw,iendabsw, 

 & ibegemsw,iendemsw 

 common/c16/ifindiff,deltime,amcabtmp,intvprnt 

 common/store1/absorbff,emisff,absorbww,emisww,absrftpp,emirftpp, 

 & absrbfbb,emisfbb 

A-5
 real k,lengtwng,locwng,lplane,lfuselg 

 parameter(small=0.00001,smallth=0.0001,glassk=1.0) 

 open(unit=8,file='input.dat',status='old') 

c 

c Variables:

c absorbf=absorbtivity of exterior fuselage side 

c emisf=emissivity of exterior fuselage side 

c absorbfl=absorbtivity of floor of cabin (fuselage interior) 

c emisfl=emissivity of floor of cabin (fuselage interior) 

c absorbfi=absorbtivity of side of cabin (fuselage interior) 

c emisfi=emissivity of side of cabin (fuselage interior) 

c absorbpg=absorbtivity of plexiglass 

c emispg=emissivity of plexiglass 

c absorftp=absorbtivity of the top of roof top of fuselage exterior 

c emisrftp=emissivity of the top of roof top of fuselage exterior 

c absorfbt=absorbtivity of the bottom of roof top in fuselage interior 

c emisrfbt=emissivity of the bottom of roof top in fuselage interior 

c absorbfb=absorptivity of the bottom of fuselage exterior (ie above tarmac) 

c emisfb=emissivity of the bottom of fuselage exterior (ie above tarmac) 

c n=number of equations=67 

c ifindiff=inidicator for the type of finite difference method in transient analysis: 

c =1 means explicit finite difference method 

c =2 means implicit finite difference method 

c ntrial=maximum number of iterations allowed in the non-linear solution

c tolt=temperature tolerance in the solution 

c tolfn=function tolerance in the solution 

c t()=temperature (deg F or Rankine) 

c fn()=equations/or functions 

c jacob(,)=Jacobian 

c th()=element thickness (ft) 

c k()=element conductivity ((BTU/hr/ft/deg F) 

c widthwng=wing width (ft) 

c lengtwng=total wing length (tip-to-tip) (ft) 

c hwng=wing height from ground (ft) 

c locwng=leading edge wing location from the fuselage front (ft) 

c lplane=entire plane length (ft) 

c lfuselg=fuselage length (ft) 

c hfuselg=height of fuselage (ft) 

c wfuselg=width of fuselage (ft) 

c trnsm=transmissivity of plexiglass 

c absorbta=tarmac absorbtivity 

c emista=tarmac emissivity

c tarmd=tarmac depth (overall) (ft) 

c densta=tarmac density (overall) (lb/ft^3) 

c conducta=tarmac conductivity (overall) (BTU/hr/ft/deg F) 

c shcta=tarmac specific heat capacity (overall) (BTU/lbm/deg F) 

c nperiods=number of time intervals (say from the morning till the evening) 

c envtdata(,1)=sun's incident angle (degrees) 

c envtdata(,2)=ambient temperature (Rankine) 

c envtdata(,3)=effective sky temperature (Rankine) 

c envtdata(,4)=wind speed (ft/s) 

A-6
c envtdata(,5)=total (both direct and diffused) solar radiation (BTU/hr/sq.ft) 

c envtdata(,6)=area (sq. ft) of plexiglass perpendicular to the sun, & seen

c by the sun 

c envtdata(,7)=area (sq. ft) of plexiglass horizontal to the sun, & seen 

c by the sun 

c envtdata(,8)=fraction of sun that gets into the fuselage (in terms of 

c the projected area)

c icode=1, means, only one analysis is done knowing the paint scheme

c icode=2, means, analysis is done for all paint schemes, satisfying 

c some temperature requirements 

c nconstrn=number of temperatures that must be satisfied above 

c idtemp()=which temperature 

c tempcons()=the value of the temperature requirement to be satisfied 

c scatffl=fraction of the light reflected from the floor 

c that is diffused 

c scatffi=fraction of the light reflected from the interior fuselage side 

c that is diffused 

c scatfw=fraction of the light reflected from the wing that is diffused 

c scatff=fraction of the light reflected from the exterior fuselage side 

c that is diffused 

c aircondc=conductivity of air (BTU/hr/ft/deg F) 

c thdiffus=air thermal diffusivity (ft^2/sec) 

c visckine=air kinematic viscosity (ft^2/sec) 

c prandtl=Prandtl number 

c icodef11=1: the fillet is completely painted with paint that is the same as 

c that used on the wing 

c icodef11=2: the fillet is completely painted with paint that is the same as 

c that used on the fuselage 

c icodef11=3: the fuselage/wing paint boundary falls on the fillet; need 

c to assign the percentage of the fillet occupied by the wing paint 

c percent=percentage when icodef11=3 

c nparts=determines absorptivity and emissivity increments (1/nparts) 

c a1=total area (sq. ft) of the plexiglass 

c a2=total area (sq. ft) of the fuselage top half 

c a3=a1/(a1+a2)*lfuselg*wfuselg 

c a4=a2/(a1+a2)*lfuselg*wfuselg 

c a5()=area (sq. ft) of plexiglass that intercepts diffuse reflected solar 

c rays from the tarmac

c a6()=same as envtdata(,6) 

c a7()=same as envtdata(,7) 

c nintervl=number of intervals in the flight profile used in cooling phase 

c vel()=speed of aircraft during cooling phase (mph) 

c timeintv()=time interval during each portion of the flight profile (sec) 

c thicknes()=element thickness (in) 

c density()=element density(lb/ft^3) 

c conductv()=element thermal conductivity (BTU/hr/ft/deg F) 

c sheatcap()=element specific heat capacity (BTU/lbm/deg F) 

c ibegabsf=fuselage absorbtivity pointer on the beginning location of 

c the paint matrix 

c iendabsf=fuselage absorbtivity pointer on the ending location of 

c the paint matrix 

A-7
c ibegemsf=fuselage emissivity pointer on the beginning location of 

c the paint matrix 

c iendemsf=fuselage emissivity pointer on the ending location of 

c the paint matrix 

c ibegabsw=wing absorbtivity pointer on the beginning location of 

c the paint matrix 

c iendabsw=wing absorbtivity pointer on the ending location of 

c the paint matrix 

c ibegemsw=wing emissivity pointer on the beginning location of 

c the paint matrix 

c iendemsw=wing emissivity pointer on the ending location of

c the paint matrix 

c deltime=time increment in the transient phase (sec) 

c amcabtmp=ambient cabin temperature that is tolerable before egressing

c and taxing (deg F) 

c intvprnt=print interval in the transient phase 

c 

 read(8,*)visckine,prandtl 

 do 10 i=1,28 

c j is number of the element: it is just a dummy; not used anywhere 

 read(8,*)j,thicknes(i),density(i),conductv(i),sheatcap(i) 

 10 thicknes(i)=thicknes(i)/12. 

 read(8,*)widthwng,lengtwng,hwng,locwng,lplane,lfuselg,hfuselg, 

 & wfuselg,a1,a2,a5 

 read(8,*)trnsm,absorbta,emista,tarmd,densta,conducta,shcta 

 read(8,*)absorbfl,emisfl,absorbfi,emisfi,absorfbt,emisrfbt, 

 & absorbpg,emispg 

 if(absorbfl.eq.1.0)absorbfl=1.-small 

 if(absorbfl.eq.0.0)absorbfl=small 

 if(emisfl.eq.1.0)emisfl=1.-small 

 if(emisfl.eq.0.0)emisfl=small 

 if(absorbfi.eq.1.0)absorbfi=1.-small 

 if(absorbfi.eq.0.0)absorbfi=small 

 if(emisfi.eq.1.0)emisfi=1.-small 

 if(emisfi.eq.0.0)emisfi=small 

 if(absorfbt.eq.1.0)absorfbt=1.-small 

 if(absorfbt.eq.0.0)absorfbt=small 

 if(emisrfbt.eq.1.0)emisrfbt=1.-small 

 if(emisrfbt.eq.0.0)emisrfbt=small 

 if(absorbpg.eq.1.0)absorbpg=1.-small 

 if(absorbpg.eq.0.0)absorbpg=small 

 if(emispg.eq.1.0)emispg=1.-small 

 if(emispg.eq.0.0)emispg=small 

 read(8,*)icodef11,percent 

 read(8,*)scatffl,scatffi,scatfw,scatff 

c read(8,*)icode 

 icode=1 

 if(icode.eq.1)then 

 read(8,*)absorbf,emisf,absorbw,emisw,absorftp,emisrftp, 

 & absorbfb,emisfb 

 if(absorbf.eq.1.0)absorbf=1.-small 

A-8
 if(absorbf.eq.0.0)absorbf=small 

 if(emisf.eq.1.0)emisf=1.-small 

 if(emisf.eq.0.0)emisf=small 

 if(absorbw.eq.1.0)absorbw=1.-small 

 if(absorbw.eq.0.0)absorbw=small 

 if(emisw.eq.1.0)emisw=1.-small 

 if(emisw.eq.0.0)emisw=small 

 if(absorftp.eq.1.0)absorftp=1.-small 

 if(absorftp.eq.0.0)absorftp=small 

 if(emisrftp.eq.1.0)emisrftp=1.-small 

 if(emisrftp.eq.0.0)emisrftp=small 

 if(absorbfb.eq.1.0)absorbfb=1.-small 

 if(absorbfb.eq.0.0)absorbfb=small 

 if(emisfb.eq.1.0)emisfb=1.-small 

 if(emisfb.eq.0.0)emisfb=small 

 elseif(icode.eq.2)then 

 read(8,*)nparts 

 read(8,*)ibegabsf,iendabsf 

 read(8,*)ibegemsf,iendemsf 

 read(8,*)ibegabsw,iendabsw 

 read(8,*)ibegemsw,iendemsw 

 read(8,*)nconstrn 

 read(8,*)(idtemp(i),tempcons(i),i=1,nconstrn) 

 endif 

 read(8,*)nintervl 

 read(8,*)(vel(i),i=1,nintervl+1) 

 read(8,*)(timeintv(i),i=1,nintervl) 

 read(8,*)nperiods 

 do 20 i=1,nperiods 

 20 read(8,*)(envtdata(i,j),j=1,8) 

 n=67 

 read(8,*)ifindiff,deltime,amcabtmp,intvprnt 

 amcabtmp=amcabtmp+459.67 

 read(8,*)ntrial,tolt,tolfn 

c initial trial for the aircraft temperatures and radiosities 

 read(8,*)(t(i),i=1,n) 

c because some of these properties will be destroyed depending on 

c icode, they will be stored for later retrieval 

 absorbff=absorbf 

 emisff=emisf 

 absorbww=absorbw 

 emisww=emisw 

 absrftpp=absorftp 

 emirftpp=emisrftp 

 absrbfbb=absorbfb 

 emisfbb=emisfb 

c 

c reference some of the values as follows: 

c 

c air conductivity & thermal diffusivity within the cabin 

c conductv(14) same as conductv(19) 

A-9
 aircondc=conductv(19) 

 thdiffus=conductv(19)/(density(19)*sheatcap(19))*(1./3600) 

c bec. the model was developed piece-meal, need to reorder things 

c as follows (for internal representation): 

c first thickness: 

 th(1)=thicknes(1) 

 th(2)=thicknes(2) 

 th(3)=thicknes(3) 

 th(4)=thicknes(4) 

 th(29)=thicknes(5) 

 th(30)=thicknes(6) 

 th(5)=thicknes(7) 

 th(6)=thicknes(8) 

 th(7)=thicknes(9) 

 th(8)=thicknes(10) 

 th(9)=thicknes(11) 

 th(10)=thicknes(12)

 th(23)=thicknes(13)

 th(11)=thicknes(14)

 th(27)=thicknes(15)

 th(26)=thicknes(16)

 th(25)=thicknes(17)

 th(28)=thicknes(18)

 th(24)=thicknes(19)

 th(21)=thicknes(20)

 th(20)=thicknes(21)

 th(19)=thicknes(22)

 th(18)=thicknes(23)

 th(31)=thicknes(24)

 th(14)=thicknes(25)

 th(13)=thicknes(26)

 th(12)=thicknes(27)

 th(22)=thicknes(28)

c spar inside the cabin is assumed not to be in direct contact with the 

c floor or belly-pan. its temperature is taken to be the average of the 

c floor bottom and top of belly-pan. because of this, and the fact that 

c floor bottom temp. is usually higher than the belly-pan top, no bulk 

c or turbulent air motion exists. also radiation effects 

c will be assumed to be negligible; therefore, temperatures: 

c t(31), t(32), t(33) and t(32) will all be assumed to be the same. 

c ie equal to the average of the floor bottom and top of belly-pan. 

c therefore the floor space (space between the floor bottom and belly-pan 

c top is assumed=th(31), and all others (th(17); th(16); th(32); and 

c th(15) is assumed very small;= smallth. numerically ok. 

 th(17)=smallth 

 th(16)=smallth 

 th(32)=smallth 

 th(15)=smallth 

c now conductivity: 

 k(1)=conductv(1) 

 k(2)=conductv(2) 

A-10
 k(3)=conductv(3) 

 k(4)=conductv(4) 

 k(29)=conductv(5) 

 k(30)=conductv(6) 

 k(5)=conductv(7) 

 k(6)=conductv(8) 

 k(7)=conductv(9) 

 k(8)=conductv(10) 

 k(9)=conductv(11) 

 k(10)=conductv(12) 

 k(23)=conductv(13) 

 k(11)=conductv(14) 

 k(27)=conductv(15) 

 k(26)=conductv(16) 

 k(25)=conductv(17) 

 k(28)=conductv(18) 

 k(24)=conductv(19) 

 k(21)=conductv(20) 

 k(20)=conductv(21) 

 k(19)=conductv(22) 

 k(18)=conductv(23) 

 k(31)=conductv(24) 

 k(14)=conductv(25) 

 k(13)=conductv(26) 

 k(12)=conductv(27) 

 k(22)=conductv(28) 

c because of the statements made above, k(17); k(16); k(32) and 

c k(15) are taken as somewhat higher than glass (1.0); too high, say 1500 

c will cause numerical overflow during the solution process:

c this and the above thicknesses (smallth) will make k()/th() very large: 

c thereby making the resistance th()/k() negligible, which is what we want. 

 k(17)=glassk 

 k(16)=glassk 

 k(32)=glassk 

 k(15)=glassk 

 close(8)

 return 

 end 

c 

 subroutine solution 

c purpose: controls the solution of the problem 

c calls: viewfact,hconvect,qrefcab,tarmmodl,infrqsky,sbfillet,qreffw, 

c qreftw,newtraph,transien, 

c called from: main 

 common/c1/n,ntrial,tolt,tolfn 

 common/c2/t(200),fn(200),jacob(200,200) 

 common/c4/widthwng,lengtwng,hwng,locwng,lplane,lfuselg,hfuselg,

 & wfuselg 

 common/c5/trnsm,absorbta,emista,tarmd,densta,conducta,shcta 

 common/c6/absorbf,emisf,absorbw,emisw,absorbfl,emisfl,absorbfi, 

 &emisfi,absorbpg,emispg,absorftp,emisrftp,absorfbt,emisrfbt, 

A-11
 &absorbfb,emisfb 

 common/c7/nperiods,envtdata(100,20) 

 common/c8/icode,nconstrn,idtemp(100),tempcons(100) 

 common/c11/icodef11,percent,nparts 

 common/c12/a1,a2,a3,a4,a5,a6(100),a7(100) 

 common/c15/ibegabsf,iendabsf,ibegemsf,iendemsf,ibegabsw,iendabsw, 

 & ibegemsw,iendemsw 

 common/c16/ifindiff,deltime,amcabtmp,intvprnt 

 common/convect/hf(100),hw(100),ht(100) 

 common/tarmacqt/qqtarm(100),tttarm(100) 

 common/qskyinfr/qqsky(100) 

 common/fillet/f11w,f11f 

 common/result/temp(200),temptrn(200) 

 dimension tinitial(100) 

 dimension tempall(100,100),itemp1(2,100),tempall1(100), 

 &location(100),tempp(100),tempp1(100) 

 dimension itemp1tr(2,100),temptr(100,100),temptr1(100), 

 &locattr(100) 

 character*1 indicat(150,150)/22500*'N'/ 

 character*1 indicat1(150,150)/22500*'N'/ 

 parameter(small=0.00001) 

 parameter(tarmleng=100.0) 

 real lfuselg,lplane 

 open(unit=9,file='summary.dat',status='unknown') 

 open(unit=10,file='transient.dat',status='unknown') 

c open(unit=12,file='outconv',status='unknown') 

c open(unit=13,file='odiffsol',status='unknown') 

c open(unit=14,file='out2.dat',status='unknown') 

c open(unit=15,file='out3.dat',status='unknown') 

c open(unit=16,file='transin3.out',status='unknown') 

c open(unit=17,file='transin4.out',status='unknown') 

c Variables:

c hf=fuselage convective coefficient 

c hw=wing convective coefficient 

c ht=tarmac convective coefficient 

c qqtarm=tarmac emission 

c tttarm=tarmac temperature 

c qqsky=sky emission 

c f11w=fillet view factor (for wing) 

c f11f=fillet view factor (for fuselage)

c tinitial=initial guessed temperatures 

c indicat=indicator for the paint matrix 

c tarmleng=tarmac length for finding hw 

c 

c store the initial guessed temperatures 

c 

 do 80 i=1,n 

 80 tinitial(i)=t(i) 

c 

c store the various areas relating the fuselage roof and 

c the plexiglass: 

A-12
c 

 a3=a1/(a1+a2)*lfuselg*wfuselg 

 a4=a2/(a1+a2)*lfuselg*wfuselg 

 do 85 j=1,nperiods 

 a6(j)=envtdata(j,6) 

 85 a7(j)=envtdata(j,7) 

c 

c find view factors 

 call viewfact 

c find convection coefficients (hf) for the fuselage (4 sides) &

c for the left wing (hw) (top & bottom) and the 

c right wing (hw) (top & bottom). 

c hf for all four fuselage sides are the same 

c hw for all the wing sides are the same 

c 

c find hf for the fuselage:

 call hconvect(lplane,hf) 

c find hw for the wings 

 call hconvect(widthwng,hw) 

c now find ht for the tarmac 

 call hconvect(tarmleng,ht) 

c find reflected solar rays (qrefl) within the cabin 

 call qrefcab 

c find tarmac temperature and qtarm (tarmac radiation) 

 call tarmmodl(absorbta,emista,qqtarm,tttarm) 

c find the infrared sky radiation 

 call infrqsky(qqsky) 

c find the modification factors for absorbtivity & emissivity due to fillets

 call sbfillet(icodef11,percent,f11w,f11f) 

 if(icode.eq.2)del=1./nparts 

c 

c for the option chosen determine the temperatures; or paint scheme for 

c a given set of temperatures chosen

c 

c icode=1: just one run with temperature output for a given 

c set of paint scheme: 

c 

 write(10,*)'$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$' 

 write(10,*)'THERMOD: A THERMAL PROGRAM FOR AIRCRAFTS'

 write(10,*)'$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$' 

 if(icode.eq.1)then 

 itotitr=0 

 absinfrf=emisf*(1./(1.-f11f*(1.-emisf))) 

 eemisf=emisf*(1.-f11f)/(1.-f11f*(1.-emisf)) 

 rff=absinfrf/eemisf 

 call qreffw(absorbf,absorbw) 

 call qreftw(absorbw) 

 absinfrw=emisw*(1./(1.-f11w*(1.-emisw))) 

 eemisw=emisw*(1.-f11w)/(1.-f11w*(1.-emisw)) 

 rfw=absinfrw/eemisw 

c write(6,*)'absorbf,emisf,absorbw,emisw' 

A-13
c write(6,*)absorbf,emisf,absorbw,emisw 

 do 10 i=1,nperiods 

 write(6,*)' ' 

write(6,*)'STEADY STATE ANALYSIS BEGUN FOR PERIOD',i 

 itotitr=itotitr+1 

 do 90 itemp=1,n 

 90 t(itemp)=tinitial(itemp) 

 theta=envtdata(i,1) 

 ta=envtdata(i,2) 

 qsky=qqsky(i) 

 qdir=envtdata(i,5) 

 hhf=hf(i) 

 hhw=hw(i) 

 hht=ht(i) 

 qtarm=qqtarm(i) 

 ttarm=tttarm(i) 

c find airplane temperature 

 call newtraph(i,kkk,theta,ta,qsky,hhf,hhw,hht,qdir, 

 & qtarm,ttarm,absorbf,emisf,absorbw,emisw, 

 & eemisf,eemisw,rff,rfw) 

 95 format(1x,i1,1x,i4,70(1x,f4.0)) 

 write(6,*)'STEADY STATE ANALYSIS COMPLETED FOR PERIOD',i 

c write(6,95)i,kkk,(temp(j),j=1,n) 

 itemp1(1,i)=i 

c itemp1(2,i)=kkk 

 do 97 j=1,n 

 97 tempall(j,i)=temp(j) 

write(6,*)'TRANSIENT ANALYSIS BEGUN FOR PERIOD',i 

 if(ifindiff.eq.1)call trnsien1(itotitr,i,theta,ta,qsky,

 & hhf,hhw,hht,qdir,qtarm,ttarm,absorbf,emisf,absorbw, 

 & emisw,eemisf,eemisw,rff,rfw) 

 if(ifindiff.eq.2)call trnsien2(itotitr,i,theta,ta,qsky,

 & hhf,hhw,hht,qdir,qtarm,ttarm,absorbf,emisf,absorbw, 

 & emisw,eemisf,eemisw,rff,rfw) 

 write(6,*)'TRANSIENT ANALYSIS COMPLETED FOR PERIOD',i 

 itemp1tr(1,i)=i 

 do 197 j=1,n 

 197 temptr(j,i)=temptrn(j) 

 10 continue 

c find the maximum temperatures 

 write(9,*)'$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$' 

 write(9,*)'THERMOD: A THERMAL PROGRAM FOR AIRCRAFTS' 

 write(9,*)'$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$' 

 write(9,*)' ' 

 write(9,*)' ' 

 write(9,*)' ' 

write(9,*)'SUMMARY OF STEADY-STATE AND TRANSIENT ANALYSIS' 

 write(9,*)' ' 

 write(9,*)' ' 

 write(9,*)' ' 

c write(9,*)'absorbf,emisf,absorbw,emisw' 

A-14
c write(9,*)absorbf,emisf,absorbw,emisw 

 tempall2=0.0 

 temptr2=0.0 

 do 120 i=1,nperiods 

 tempall1(i)=0.0 

 temptr1(i)=0.0 

c Temperatures T25, T26, T43, T44 (all insulatory surfaces); T51, T52 and 

c T53 (tarmac surfaces); and T49 and T50 (plexiglass material) are 

c considered non-structural. 

 do 125 j=1,48 

 if(j.ne.25.and.j.ne.26.and.j.ne.43.and.j.ne.44)then 

 if(tempall(j,i).gt.tempall1(i))then 

 tempall1(i)=tempall(j,i) 

 location(i)=j

 endif 

 endif 

 if(j.ne.25.and.j.ne.26.and.j.ne.43.and.j.ne.44)then 

 if(temptr(j,i).gt.temptr1(i))then 

 temptr1(i)=temptr(j,i) 

 locattr(i)=j 

 endif 

 endif 

 125 continue 

 if(tempall1(i).gt.tempall2)then 

 tempall2=tempall1(i) 

 locfinal=location(i) 

 iperiodf=i 

 endif 

 if(temptr1(i).gt.temptr2)then 

 temptr2=temptr1(i) 

 locfintr=locattr(i) 

 iperiotr=i 

 endif 

 120 continue 

c print the temperatures 

c do 100 j=1,2

c 100 write(9,101)(itemp1(j,i),i=1,nperiods) 

c steady-state output 

 write(9,*)'(Temperatures are shown for all time periods)' 

 write(9,*)' ' 

 write(9,*)' ' 

write(9,*)'TEMPERATURES AT THE END OF STEADY STATE ANALYSIS' 

 write(9,*)' ' 

 write(9,250) 

 250 format(8x,'TEMPERATURES IN DEGREES FAHRENHEIT') 

 write(9,101)(itemp1(1,i),i=1,nperiods) 

 101 format(4x,100(1x,i5)) 

 write(9,*)' ' 

 do 99 j=1,53

 99 write(9,98)j,(tempall(j,i),i=1,nperiods) 

 98 format(1x,i3,100(1x,f5.1)) 

A-15
 write(9,*)'' 

 do 130 i=1,nperiods 

c write(6,*)'maximum structural temperature for period',i,'=', 

c & tempall1(i),' occurring at location',location(i) 

 write(9,135)i,tempall1(i),location(i) 

 135 format('Maximum Structural Temperature for Period',1x,i3,1x, 

 & '=',1x,f6.2,1x,'Occurring at Location',1x,i3) 

 130 continue 

c write(6,*)'maximum structural temperature over all periods=', 

c & tempall2,' occurring at location ',locfinal,' and at ',

c & 'period ',iperiodf 

 write(9,*)' ' 

 write(9,136)nperiods,tempall2,locfinal,iperiodf 

 136 format('Maximum Structural Temperature Over All',1x,i3, 

 & 1x,'Periods',1x,'=',f6.2,1x,'Occurring at Location',1x,i3, 

 & 1x,'at Period',1x,i3) 

c transient analysis output 

 write(9,*)' ' 

 write(9,*)' ' 

write(9,*)'TEMPERATURES AT THE END OF TRANSIENT ANALYSIS' 

 write(9,*)' ' 

 write(9,250) 

 write(9,101)(itemp1tr(1,i),i=1,nperiods) 

 write(9,*)' ' 

 do 299 j=1,53

 299 write(9,98)j,(temptr(j,i),i=1,nperiods) 

 write(9,*)'' 

 do 1130 i=1,nperiods 

c write(6,*)'maximum structural temperature for period',i,'=', 

c & temptr1(i),' occurring at location',locattr(i) 

 write(9,137)i,temptr1(i),locattr(i) 

 137 format('Maximum Structural Temperature for Period',1x,i3,1x, 

 & '=',1x,f6.2,1x,'Occurring at Location',1x,i3) 

 1130 continue 

c write(6,*)'maximum structural temperature over all',nperiods, 

c 'periods=', 

c & temptr2,' occurring at location ',locfintr,' and at ',

c & 'period ',iperiotr 

 write(9,*)' ' 

 write(9,138)nperiods,temptr2,locfintr,iperiotr 

 138 format('Maximum Structural Temperature Over All',1x,i3, 

 & 1x,'Periods',1x,'=',f6.2,1x,'Occurring at Location',1x,i3, 

 & 1x,'at Period',1x,i3) 

c 

c 

c 

c icode=2: this is the "mother" of all iterations. for a given paint 

c on the fuselage, this option iterates for all paint combination on the 

c wing. this process is repeated for the next fuselage paint. note 

c that the fuselage paint are the same for the sides, top and bottom. 

c this is how it has been programmed. can be modified if requested. 

A-16
c this option outputs a matrix of paint schemes that are permissible for 

c the entire plane without violating a set of temperature 

c requirements requested by the user. 

c 

 elseif(icode.eq.2)then 

 itotitr=0 

 do 440 iabsorbf=ibegabsf,iendabsf

 absorbf=(iabsorbf-1)*del 

 if(iabsorbf.eq.1)absorbf=small 

 if(iabsorbf.eq.nparts+1)absorbf=1.-small 

 absorftp=absorbf 

 absorbfb=absorbf 

 do 460 iemisf=ibegemsf,iendemsf 

 isumm=(iabsorbf-1)*(nparts+1)+iemisf 

 emisf=(iemisf-1)*del 

 if(iemisf.eq.1)emisf=small 

 if(iemisf.eq.nparts+1)emisf=1.-small 

 emisrftp=emisf 

 emisfb=emisf 

 absinfrf=emisf*(1./(1.-f11f*(1.-emisf))) 

 eemisf=emisf*(1.-f11f)/(1.-f11f*(1.-emisf)) 

 rff=absinfrf/eemisf 

 do 470 iabsorbw=ibegabsw,iendabsw 

 absorbw=(iabsorbw-1)*del 

 if(iabsorbw.eq.1)absorbw=small 

 if(iabsorbw.eq.nparts+1)absorbw=1.-small 

 call qreffw(absorbf,absorbw) 

 call qreftw(absorbw) 

 do 450 iemisw=ibegemsw,iendemsw 

 isum=(iabsorbw-1)*(nparts+1)+iemisw 

 emisw=(iemisw-1)*del 

 if(iemisw.eq.1)emisw=small 

 if(iemisw.eq.nparts+1)emisw=1.-small 

 absinfrw=emisw*(1./(1.-f11w*(1.-emisw))) 

 eemisw=emisw*(1.-f11w)/(1.-f11w*(1.-emisw)) 

 rfw=absinfrw/eemisw 

 itot=0 

 itot1=0 

 do 425 iconstrn=1,nconstrn 

 425 tempp(iconstrn)=0.0

 do 426 iconstrn=1,nconstrn 

 426 tempp1(iconstrn)=0.0

 do 410 i=1,nperiods 

 itotitr=itotitr+1 

 do 490 itemp=1,n 

 490 t(itemp)=tinitial(itemp) 

 theta=envtdata(i,1) 

 ta=envtdata(i,2) 

 qsky=qqsky(i) 

 qdir=envtdata(i,5) 

 hhf=hf(i) 

A-17
 hhw=hw(i) 

 hht=ht(i) 

 qtarm=qqtarm(i) 

 ttarm=tttarm(i) 

c find airplane temperature 

 call newtraph(i,kkk,theta,ta,qsky,hhf,hhw,hht,qdir, 

 & qtarm,ttarm,absorbf,emisf,absorbw,emisw, 

 & eemisf,eemisw,rff,rfw) 

 write(9,*)'STEADY STATE ANALYSIS' 

write(9,*)'Total # of Analysis (per period)=',itotitr 

 write(9,*)'absorbf,emisf,absorbw,emisw,period,iteratns' 

 write(9,*)absorbf,emisf,absorbw,emisw,i,kkk 

 write(9,96)(ii,ii=1,n) 

 96 format(7x,70(1x,i4)) 

 write(9,95)i,kkk,(temp(ii),ii=1,n) 

 write(6,*)'STEADY STATE ANALYSIS' 

write(6,*)'Total # of Analysis (per period)=',itotitr 

 write(6,*)'absorbf,emisf,absorbw,emisw,period,iteratns' 

 write(6,*)absorbf,emisf,absorbw,emisw,i,kkk 

 write(6,96)(ii,ii=1,n) 

 write(6,95)i,kkk,(temp(ii),ii=1,n) 

 do 430 iconstrn=1,nconstrn 

 if(temp(idtemp(iconstrn)).gt.tempp(iconstrn)) 

 & tempp(iconstrn)=temp(idtemp(iconstrn)) 

 430 continue 

 if(ifindiff.eq.1)call trnsien1(itotitr,i,theta,ta,qsky,

 & hhf,hhw,hht,qdir,qtarm,ttarm,absorbf,emisf,absorbw, 

 & emisw,eemisf,eemisw,rff,rfw) 

 if(ifindiff.eq.2)call trnsien2(itotitr,i,theta,ta,qsky,

 & hhf,hhw,hht,qdir,qtarm,ttarm,absorbf,emisf,absorbw, 

 & emisw,eemisf,eemisw,rff,rfw) 

 write(6,*)'TRANSIENT ANALYSIS' 

 write(9,*)'TRANSIENT ANALYSIS' 

 write(6,95)i,kkk,(temptrn(ii),ii=1,n) 

 write(9,95)i,kkk,(temptrn(ii),ii=1,n) 

 do 431 iconstrn=1,nconstrn 

 if(temptrn(idtemp(iconstrn)).gt.tempp1(iconstrn)) 

 & tempp1(iconstrn)=temptrn(idtemp(iconstrn)) 

 431 continue 

 410 continue

 do 435 iconstrn=1,nconstrn 

 435 if(tempp(iconstrn).le.tempcons(iconstrn)) 

 & itot=itot+1 

 if(itot.eq.nconstrn)then 

 indicat(isumm,isum)='Y' 

 write(9,*)'passed the steady-state temp. constraints' 

 write(9,*)'absorbf,emisf,absorbw,emisw' 

 write(9,*)absorbf,emisf,absorbw,emisw 

 write(9,96)(ii,ii=1,n) 

 write(9,199)(temp(i),i=1,67) 

 199 format(7x,70(1x,f4.0)) 

A-18
 endif 

 do 436 iconstrn=1,nconstrn 

 436 if(tempp1(iconstrn).le.tempcons(iconstrn)) 

 & itot1=itot1+1 

 if(itot1.eq.nconstrn)then

 indicat1(isumm,isum)='Y' 

 write(9,*)'passed the transient temp. constraints' 

 write(9,*)'absorbf,emisf,absorbw,emisw' 

 write(9,*)absorbf,emisf,absorbw,emisw 

 write(9,96)(ii,ii=1,n) 

 write(9,199)(temptrn(i),i=1,67) 

 endif 

 450 continue 

 470 continue 

 460 continue 

 440 continue 

 nmatrix=(nparts+1)**2 

 write(9,*)'STEADY STATE ANALYSIS ONLY'

 do 420 isumm=1,nmatrix 

 write(9,495)(indicat(isumm,isum),isum=1,nmatrix) 

 420 continue 

 write(9,*)'AT THE END OF TRANSIENT ANALYSIS' 

 do 421 isumm=1,nmatrix 

 write(9,495)(indicat1(isumm,isum),isum=1,nmatrix) 

 421 continue 

 495 format(150a1) 

 endif 

 close(9)

 close(10) 

c close(12) 

c close(13) 

c close(14) 

c close(15) 

c close(16) 

c close(17) 

 return 

 end 

c 

c 

 subroutine hconvect(l,h) 

c purpose: determines the convective coefficient 

c calls: none

c called from: solution 

 common/c7/nperiods,envtdata(100,20) 

 common/c10/k,thdiffus,visckine,prandtl 

 real l,llamn,nusllamn,k 

 dimension h(100) 

 parameter(rellamn=5.e5) 

c 

c Variables:

c h: convective coefficient 

A-19
c l=length 

c nusllamn=Nusselt number in the laminar zone 

c rellamn=Reynold's number in the laminar zone 

c 

 do 10 i=1,nperiods 

 v=envtdata(i,4) 

 rel=v*l/visckine 

 if(rel.le.rellamn)then 

 nusllamn=0.664*rel**(1./2)*prandtl**(1./3) 

 h(i)=k*nusllamn/l 

c write(12,*)'rel,nusllamn,h',rel,nusllamn,h(i) 

 elseif(rel.gt.rellamn)then 

 llamn=rellamn*visckine/v

 nusllamn=0.664*rellamn**(1./2)*prandtl**(1./3) 

 hlamn=k*nusllamn/llamn 

 hturb=0.0296*k/(l-llamn)*prandtl**(1./3)*(v/visckine)**(4./5) 

 & *(5./4)*(l**(4./5)-llamn**(4./5)) 

 h(i)=1./l*(llamn*hlamn+(l-llamn)*hturb) 

c write(12,*)'rel,llamn,hlamn,hturb,h', 

c & rel,llamn,hlamn,hturb,h(i) 

 endif 

 10 continue

 end 

c 

c 

 subroutine viewfact 

c purpose: determines the view factor (controlling routine) 

c calls: viewfc 

c called from: solution 

 common/c4/widthwng,lengtwng,hwng,locwng,lplane,lfuselg,hfuselg,

 & wfuselg 

 common/vf1/fbwa(6,6),fbwb(6,6) 

 common/vf2/fa11,fa12,fa13,fa14,fa15,fa16,fa21,fa22,fa23,fa24, 

 & fa25,fa26,fa31,fa32,fa33,fa34,fa35,fa36,fa41,fa42, 

 & fa43,fa44,fa45,fa46,fa51,fa52,fa53,fa54,fa55,fa56, 

 & fa61,fa62,fa63,fa64,fa65,fa66 

 common/vf4/fa57,fa68 

 common/vf5/fc11,fc12,fc13,fc14,fc15,fc16,fc21,fc22,fc23,fc24, 

 & fc25,fc26,fc31,fc32,fc33,fc34,fc35,fc36,fc41,fc42, 

 & fc43,fc44,fc45,fc46,fc51,fc52,fc53,fc54,fc55,fc56, 

 & fc61,fc62,fc63,fc64,fc65,fc66 

 dimension f(6,6) 

 real lengtwng,l,locwng,lfuselg,lplane 

c 

c Variables:

c fbwa(,)=view factors for the wing bottom (LHS & RHS) 

c fbwb(,)=view factors for the wing bottom (CENTER) 

c fa11....fa66=view factors for the top fuselage-wing junction

c fa57,fa68=view factors for face 5 & 7; and 6 & 8, where 7 is 

c the fuselage face front of the wing and 8 is the

c fuselage face aft of the wing. 

A-20
cfc11....fc66=view factors for the fuselage inside (cabin) 

c 

c view factors for wingbottom 

c LHS & RHS: 

 x=widthwng 

 y=(lengtwng-wfuselg)/2. 

 z=hwng 

 call viewfc(x,y,z,fbwa) 

c CENTER (below fuselage): 

 x=widthwng 

 y=wfuselg 

 z=hwng 

 call viewfc(x,y,z,fbwb) 

c 

c view factors for the top fuselage-wing junction 

 x=widthwng 

 y=(lengtwng-wfuselg)/2 

 l=hfuselg 

 call viewfc(x,y,l,f) 

 fa11=f(1,1) 

 fa12=f(1,2) 

 fa13=f(1,3) 

 fa14=f(1,4) 

 fa15=f(1,5) 

 fa16=f(1,6) 

 fa21=f(2,1) 

 fa22=f(2,2) 

 fa23=f(2,3) 

 fa24=f(2,4) 

 fa25=f(2,5) 

 fa26=f(2,6) 

 fa31=f(3,1) 

 fa32=f(3,2) 

 fa33=f(3,3) 

 fa34=f(3,4) 

 fa35=f(3,5) 

 fa36=f(3,6) 

 fa41=f(4,1) 

 fa42=f(4,2) 

 fa43=f(4,3) 

 fa44=f(4,4) 

 fa45=f(4,5) 

 fa46=f(4,6) 

 fa51=f(5,1) 

 fa52=f(5,2) 

 fa53=f(5,3) 

 fa54=f(5,4) 

 fa55=f(5,5) 

 fa56=f(5,6) 

 fa61=f(6,1) 

 fa62=f(6,2) 

A-21
 fa63=f(6,3) 

 fa64=f(6,4) 

 fa65=f(6,5) 

 fa66=f(6,6) 

c view factors for face 5 & 7; and 6 & 8, where 7 is the fuselage face 

c front of the wing and 8 is the fuselage face aft of the wing. 

c find fa57: 

 x=hfuselg 

 y=(lengtwng-wfuselg)/2 

 l=locwng

 call viewfc(x,y,l,f) 

 fa57=f(1,2) 

c write(9,*)'fa57=',fa57 

c find fa68 

 x=hfuselg 

 y=(lengtwng-wfuselg)/2 

 l=lplane-locwng-widthwng 

 call viewfc(x,y,l,f) 

 fa68=f(1,2) 

c write(9,*)'fa68=',fa68 

c 

c find fij's for the inside of the fuselage. 

c that is find fc11..fc16; fc21..fc26; fc31..fc36; fc41..fc46 

c fc51..fc56; fc61..fc66. note fc11,fc22,fc33,fc44,fc55 & fc66

c are zeroes. 

 x=lfuselg 

 y=wfuselg 

 l=hfuselg 

 call viewfc(x,y,l,f) 

 fc11=f(1,1) 

 fc12=f(1,2) 

 fc13=f(1,3) 

 fc14=f(1,4) 

 fc15=f(1,5) 

 fc16=f(1,6) 

 fc21=f(2,1) 

 fc22=f(2,2) 

 fc23=f(2,3) 

 fc24=f(2,4) 

 fc25=f(2,5) 

 fc26=f(2,6) 

 fc31=f(3,1) 

 fc32=f(3,2) 

 fc33=f(3,3) 

 fc34=f(3,4) 

 fc35=f(3,5) 

 fc36=f(3,6) 

 fc41=f(4,1) 

 fc42=f(4,2) 

 fc43=f(4,3) 

 fc44=f(4,4) 

A-22
 fc45=f(4,5) 

 fc46=f(4,6) 

 fc51=f(5,1) 

 fc52=f(5,2) 

 fc53=f(5,3) 

 fc54=f(5,4) 

 fc55=f(5,5) 

 fc56=f(5,6) 

 fc61=f(6,1) 

 fc62=f(6,2) 

 fc63=f(6,3) 

 fc64=f(6,4) 

 fc65=f(6,5) 

 fc66=f(6,6) 

 return 

 end 

c 

 subroutine viewfc(x,y,l,f) 

c purpose: determines the view factors for: aligned parallel rectangles 

c and for: perpendicular rectangles with a common edge 

c calls: vfacfij 

c called from: viewfact 

 dimension f(6,6) 

 parameter(pi=3.141593) 

 real l 

 a1=x*y 

 a2=x*l 

 a3=a1 

 a4=a2 

 a5=l*y 

 a6=a5 

c find fij for aligned parallel rectangles (fig 13.4) Frank P. Incropera 

c "fundamentals of heat and mass transfer" 

 do 10 i=1,3 

 if(i.eq.1)then

 xb=x/l 

 yb=y/l 

 elseif(i.eq.2)then 

 xb=x/y

 yb=l/y 

 elseif(i.eq.3)then 

 xb=y/x

 yb=l/x 

 endif 

 a=alog((((1+xb**2)*(1+yb**2))/(1+xb**2+yb**2))**.5) 

 b=xb*(1+yb**2)**.5*atan(xb/(1+yb**2)**.5) 

 c=yb*(1+xb**2)**.5*atan(yb/(1+xb**2)**.5) 

 d=xb*atan(xb) 

 e=yb*atan(yb) 

c if(i.eq.1)write(9,*)'a,b,c,d,e',a,b,c,d,e 

 fij=2/(pi*xb*yb)*(a+b+c-d-e) 

A-23
 if(i.eq.1)f(1,3)=fij 

 if(i.eq.2)f(2,4)=fij 

 if(i.eq.3)f(5,6)=fij 

 10 continue

 f(3,1)=f(1,3) 

 f(4,2)=f(2,4) 

 f(6,5)=f(5,6) 

do 20 i=1,6 

 20 f(i,i)=0. 

c find fij for perpendicular rectangles with a common edge 

c (fig 13.6) Frank P. Incropera 

c "fundamentals of heat and mass transfer" 

c 

c go about x-axis 

 w=y/x 

 h=l/x 

 call vfacfij(w,h,fij) 

 f(1,2)=fij 

 f(3,4)=f(1,2) 

 f(2,1)=f(1,2)*a1/a2 

 f(4,3)=f(2,1) 

 call vfacfij(h,w,fij) 

 f(2,3)=fij 

 f(4,1)=f(2,3) 

 f(3,2)=f(2,3)*a2/a3 

 f(1,4)=f(3,2) 

c go about y-axis 

 w=x/y 

 h=l/y 

 call vfacfij(w,h,fij) 

 f(1,5)=fij 

 f(3,6)=f(1,5) 

 f(5,1)=f(1,5)*a1/a5 

 f(6,3)=f(5,1) 

 call vfacfij(h,w,fij) 

 f(5,3)=fij 

 f(6,1)=f(5,3) 

 f(3,5)=f(5,3)*a5/a3 

 f(1,6)=f(3,5) 

c go about l-axis 

 w=x/l 

 h=y/l 

 call vfacfij(w,h,fij) 

 f(4,5)=fij 

 f(2,6)=f(4,5) 

 f(5,4)=f(4,5)*a4/a5 

 f(6,2)=f(5,4) 

 call vfacfij(h,w,fij) 

 f(5,2)=fij 

 f(6,4)=f(5,2) 

A-24
 f(2,5)=f(5,2)*a5/a2 

 f(4,6)=f(2,5) 

c do 50 i=1,6 

c 50 write(9,*)(f(i,j),j=1,6) 

 return 

 end 

c 

 subroutine vfacfij(w,h,fij) 

c purpose: find fij for perpendicular rectangles with a common edge 

c (fig 13.6) Frank P. Incropera 

c "fundamentals of heat and mass transfer" 

c calls: none

c called from: viewfc 

 parameter(pi=3.141593) 

 a=w*atan(1/w) 

 b=h*atan(1/h) 

 c=(h**2+w**2)**.5*atan(1/(h**2+w**2)**.5) 

 d=((1+w**2)*(1+h**2))/(1+w**2+h**2) 

 e=((w**2*(1+w**2+h**2))/((1+w**2)*(w**2+h**2)))**w**2 

 f=((h**2*(1+h**2+w**2))/((1+h**2)*(h**2+w**2)))**h**2 

c write(9,10)a,b,c,d,e,f 

c 10 format(6f10.5) 

 fij=1/(pi*w)*(a+b-c+.25*alog(d*e*f)) 

 return 

 end 

c 

c 

 subroutine newtraph(iperiod,k,theta,ta,qsky,hhf,hhw,hht,qdir, 

 & qtarm,ttarm,absorbf,emisf,absorbw,emisw, 

 & eemisf,eemisw,rff,rfw) 

c purpose: this subroutine calls: linear solvers; ludcmp and lubksb. 

c given an initial guess x for a root in n dimensions, take 

c ntrial Newton-Raphson steps to improve the root. stop if the 

c root converges in either summed absolute variable increments tolx or summed 

c absolute function values tolf. 

c calls: thermodl,ludcmp,lubksb 

c called from: solution 

 common/c1/n,ntrial,tolx,tolf 

 common/c2/x(200),fvec(200),jacob(200,200) 

 common/sc1/indx(200),p(200) 

 common/result/temp(200),temptrn(200) 

c 

c Variables:

c ipower=exponent of 10. to ensure solution converges in positive numbers 

c 

 ipower=1 

 50 do 14 k=1,ntrial 

c user subroutine supplies function values at x in fvec and 

c Jacobian matrix in jacob. 

 call thermodl(iperiod,k,theta,ta,qsky,hhf,hhw,hht,qdir, 

 & qtarm,ttarm,absorbf,emisf,absorbw,emisw, 

A-25
 & eemisf,eemisw,rff,rfw) 

c check function convergence 

 errf=0. 

do 11 i=1,n 

 errf=errf+abs(fvec(i)) 

 11 continue 

c if(errf.le.tolf) goto 20 

c right-hand side of linear equations 

do 12 i=1,n 

 p(i)=-fvec(i) 

 12 continue 

c solve linear eqns. using LU decomposition 

 call ludcmp 

c forward and back substitution 

 call lubksb 

c check root convergence; and update solution 

 errx=0. 

do 13 i=1,n 

 errx=errx+abs(p(i)) 

 x(i)=x(i)+p(i) 

 13 continue 

 write(14,*)'k(trial#),errx,errf',k,errx,errf 

if(errf.le.tolf.and.errx.le.tolx) goto 20 

 14 continue

 20 loopindc=0 

 do 30 i=1,n 

 if(x(i).lt.0.0)then 

 write(6,*)'i=',i,x(i) 

 x(i)=10**ipower 

 loopindc=1

 endif 

 30 continue

 if(loopindc.eq.1)then

 ipower=ipower+1 

 goto 50 

 endif 

c rearrange all temperatures and radiosity in a neat manner as in 

c Fig 1. of the report 

 temp(1)=x(1) 

 temp(2)=x(2) 

 temp(3)=x(3) 

 temp(4)=x(4) 

 temp(5)=x(62) 

 temp(6)=x(63) 

 temp(7)=x(5) 

 temp(8)=x(6) 

 temp(9)=x(7) 

 temp(10)=x(8) 

 temp(11)=x(9) 

 temp(12)=x(10) 

 temp(13)=x(11) 

A-26
 temp(14)=x(12) 

 temp(15)=x(64) 

 temp(16)=x(65) 

 temp(17)=x(13) 

 temp(18)=x(14) 

 temp(19)=x(15) 

 temp(20)=x(16) 

 temp(21)=x(59) 

 temp(22)=x(58) 

 temp(23)=x(57) 

 temp(24)=x(60) 

 temp(25)=x(61) 

 temp(26)=x(35) 

 temp(27)=x(34) 

 temp(28)=x(33) 

 temp(29)=x(32) 

 temp(30)=x(31) 

 temp(31)=(x(31)+x(28))/2. 

 temp(32)=(x(31)+x(28))/2. 

 temp(33)=(x(31)+x(28))/2. 

 temp(34)=(x(31)+x(28))/2. 

 temp(35)=x(28) 

 temp(36)=x(27) 

 temp(37)=x(26) 

 temp(38)=x(25) 

 temp(39)=x(17) 

 temp(40)=x(18) 

 temp(41)=x(19) 

 temp(42)=x(20) 

 temp(43)=x(38) 

 temp(44)=x(39) 

 temp(45)=x(21) 

 temp(46)=x(22) 

 temp(47)=x(23) 

 temp(48)=x(24) 

 temp(49)=x(37) 

 temp(50)=x(36) 

 temp(51)=x(51) 

 temp(52)=x(52) 

 temp(53)=x(53) 

 temp(54)=x(40) 

 temp(55)=x(41) 

 temp(56)=x(42) 

 temp(57)=x(43) 

 temp(58)=x(44) 

 temp(59)=x(45) 

 temp(60)=x(46) 

 temp(61)=x(47) 

 temp(62)=x(48) 

 temp(63)=x(49) 

 temp(64)=x(50) 

A-27
 temp(65)=x(54) 

 temp(66)=x(55) 

 temp(67)=x(56) 

 do 100 i=1,53 

 100 temp(i)=temp(i)-459.67 

 return 

 end 

c 

 subroutine thermodl(iperiod,ik,theta,ta,qsky,hhf,hhw,hht,qdir, 

 & qtarm,ttarm,absorbf,emisf,absorbw,emisw, 

 & eemisf,eemisw,rff,rfw) 

c purpose: calculates and return the function value for each function fn 

c and the jacobian values jacob; all at each new t values for each 

c new newton-raphson trials. core of the thermal model 

c called from: newtraph

c calls:jacobian 

 common/c1/n,ntrial,tolt,tolfn 

 common/c2/t(200),fn(200),jacob(200,200) 

 common/c3/th(100),k(100) 

 common/c4/widthwng,lengtwng,hwng,locwng,lplane,lfuselg,hfuselg,

 & wfuselg 

 common/c5/trnsm,absorbta,emista,tarmd,densta,conducta,shcta 

 common/c6/absorbff,emisff,absorbww,emisww,absorbfl,emisfl,absorbfi 

 &,emisfi,absorbpg,emispg,absorftp,emisrftp,absorfbt,emisrfbt 

 &,absorbfb,emisfb 

 common/c7/nperiods,envtdata(100,20) 

 common/c10/aircondc,thdiffus,visckine,prandtl 

 common/c12/a1,a2,a3,a4,a5,a6(100),a7(100) 

 common/reflecq/qreflin(6,100),qscabsin(6,100),qreflfwr(6,100), 

 &qreflfwl(6,100),qscabfwr(6,100),qscabfwl(6,100),qabstwa1(6,100), 

 &qabstwa2(6,100),qabstwb(6,100) 

 common/vf1/fbwa(6,6),fbwb(6,6) 

 common/vf2/fa11,fa12,fa13,fa14,fa15,fa16,fa21,fa22,fa23,fa24, 

 & fa25,fa26,fa31,fa32,fa33,fa34,fa35,fa36,fa41,fa42, 

 & fa43,fa44,fa45,fa46,fa51,fa52,fa53,fa54,fa55,fa56, 

 & fa61,fa62,fa63,fa64,fa65,fa66 

 common/vf4/fa57,fa68 

 common/vf5/fc11,fc12,fc13,fc14,fc15,fc16,fc21,fc22,fc23,fc24, 

 & fc25,fc26,fc31,fc32,fc33,fc34,fc35,fc36,fc41,fc42, 

 & fc43,fc44,fc45,fc46,fc51,fc52,fc53,fc54,fc55,fc56, 

 & fc61,fc62,fc63,fc64,fc65,fc66 

 common/facttarm/fact 

 dimension jj(20) 

 real jj,jacob,k,nusselt1,nusselt2 

 real lengtwng,locwng,lfuselg 

 parameter(fpi=3.141593/180.) 

 parameter(sig=0.1714e-8) 

 parameter(gravity=32.2) 

 data fact/0.3/ 

c 

c Variables:

A-28
c nusselt1=Nusselt number inside the fuselage (vertical) 

c nusselt2=Nusselt number inside the fuselage (horizontal) 

c sig=Stefan-Boltzmann constant 

c fact=allows for modification of temperature of the tarmac shade 

c underneath the wing 

c qreflin(,)=intercepted solar radiation (both direct and specular) 

c in the cabin 

c qscabsin(,)=absorbed diffuse solar radiation in the cabin 

c qreflfwr(,)=intercepted solar radiation (both direct and specular) 

c by right wing 

c qreflfwl(,)=intercepted solar radiation (both direct and specular) 

c by left wing 

c qscabfwr(,)=absorbed diffuse solar radiation by the right wing 

c qscabfwl(,)=absorbed diffuse solar radiation by the left wing

c qabstwa1(,)=intercepted solar radiation (direct)

c by the tarmac underneath the wing (LHS) 

c qabstwa2(,)=intercepted solar radiation (direct)

c by the tarmac underneath the wing (RHS) 

c qabstwb(,)=intercepted solar radiation (direct) 

c by the tarmac underneath the wing(CENTER) 

c 

c input t(from 40 to 50) into jj(1 to 11): jj=radiosities 

 do 10 i=1,11 

 10 jj(i)=t(39+i) 

c input t(from 54 to 56) into jj(12 to 14): jj=radiosities 

 do 15 i=1,3 

 15 jj(11+i)=t(53+i) 

c note t(51) to t(53) are for temperatures of tarmac under the wing 

c 

c find the values for each functions, fn, using the current t values 

c 

c left wing (conduction): Eqns. 1,2,3,4,5,6,62&63 

c 

 fn(1)=-k(1)/th(1)*(t(1)-t(2))+k(2)/th(2)*(t(2)-t(3)) 

 fn(2)=-k(2)/th(2)*(t(2)-t(3))+k(3)/th(3)*(t(3)-t(4)) 

 fn(3)=-k(3)/th(3)*(t(3)-t(4))+k(4)/th(4)*(t(4)-t(62)) 

 fn(4)=-k(4)/th(4)*(t(4)-t(62))+k(29)/th(29)*(t(62)-t(63)) 

 fn(62)=-k(29)/th(29)*(t(62)-t(63))+k(30)/th(30)*(t(63)-t(5)) 

 fn(63)=-k(30)/th(30)*(t(63)-t(5))+k(5)/th(5)*(t(5)-t(6)) 

 fn(5)=-k(5)/th(5)*(t(5)-t(6))+k(6)/th(6)*(t(6)-t(7)) 

 fn(6)=-k(6)/th(6)*(t(6)-t(7))+k(7)/th(7)*(t(7)-t(8)) 

c 

c right wing (conduction): Eqns. 7,8,9,10,11,12,64&65 

c 

 fn(7)=-k(1)/th(1)*(t(9)-t(10))+k(2)/th(2)*(t(10)-t(11)) 

 fn(8)=-k(2)/th(2)*(t(10)-t(11))+k(3)/th(3)*(t(11)-t(12)) 

 fn(9)=-k(3)/th(3)*(t(11)-t(12))+k(4)/th(4)*(t(12)-t(64)) 

 fn(10)=-k(4)/th(4)*(t(12)-t(64))+k(29)/th(29)*(t(64)-t(65)) 

 fn(64)=-k(29)/th(29)*(t(64)-t(65))+k(30)/th(30)*(t(65)-t(13)) 

 fn(65)=-k(30)/th(30)*(t(65)-t(13))+k(5)/th(5)*(t(13)-t(14)) 

 fn(11)=-k(5)/th(5)*(t(13)-t(14))+k(6)/th(6)*(t(14)-t(15)) 

A-29
 fn(12)=-k(6)/th(6)*(t(14)-t(15))+k(7)/th(7)*(t(15)-t(16)) 

c 

c left fuselage (conduction). Eqns 13,14,17 

 do 40 i=13,14 

 fn(i)=-k(i-5)*((t(i+4)-t(i+5))/th(i-5)) 

 & +k(i-4)*((t(i+5)-t(i+6))/th(i-4)) 

 40 continue

c Eqn. 17 recently modified 

 fn(17)=-k(10)/th(10)*(t(19)-t(20)) 

 & +k(23)/th(23)*(t(20)-t(38)) 

c right fuselage (conduction). Eqns 15,16,18 

 ii=0 

 do 50 i=15,16 

 fn(i)=-k(10-ii)*((t(i+6)-t(i+7))/th(10-ii))

 & +k(9-ii)*((t(i+7)-t(i+8))/th(9-ii)) 

 ii=ii+1 

 50 continue

c Eqn. 18 recently modified 

 fn(18)=-k(23)/th(23)*(t(39)-t(21)) 

 & +k(10)/th(10)*(t(21)-t(22)) 

c 

c Conduction Eqns. 32,33,34,35,36,37,38,39,40,66&67. These eqns model 

c the spar-box and floor section within the cabin. Green house effect 

c is part of this addition. This effect is considered through 

c JJ8 to JJ11 around the four sides of the cabin interior 

c 

c Eqns. 32,33&34: 

c 

 do 53 j=1,3 

 fn(31+j)=-k(12+j)/th(12+j)*(t(26+j)-t(25+j)) 

 & +k(11+j)/th(11+j)*(t(25+j)-t(24+j)) 

 53 continue

c 

c Eqns. 37,38,39&40: 

c 

 do 55 j=1,4 

 fn(36+j)=-k(17+j)/th(17+j)*(t(31+j)-t(30+j)) 

 & +k(16+j)/th(16+j)*(t(30+j)-t(29+j)) 

 55 continue

c 

c Eqns. 35,36,66&67: 

c 

 fn(35)=-k(32)/th(32)*(t(67)-t(29))+k(15)/th(15)*(t(29)-t(28)) 

 fn(36)=-k(31)/th(31)*(t(66)-t(67))+k(32)/th(32)*(t(67)-t(29)) 

 fn(66)=-k(16)/th(16)*(t(30)-t(66))+k(31)/th(31)*(t(66)-t(67)) 

 fn(67)=-k(17)/th(17)*(t(31)-t(30))+k(16)/th(16)*(t(30)-t(66)) 

c 

c wing bottom, facing tarmac. there are a total of 12 eqns. to consider 

c within three sections of the wing: LHS(section: 1); RHS (section 2) 

c MIDDLE (section 3, ie underneath the fuselage). these eqns. are: 

c LHS(section 1): fn19(t8),fn20(j2),fn51(t51),fn54(j12); 

A-30
c RHS(section 2): fn21(t16),fn22(j4),fn53(t53),fn56(j14); 

c MIDDLE(section 3): fn31(t25),fn46(j7),fn52(t52),fn55(j13). 

c simple two-dimensional conduction from the unsheltered tarmac 

c surface to the sheltered surface is considered. a tarmac 

c thickness tarmd is assumed. also the bottom of the sheltered 

c tarmac is considered adiabatic (insulated) so that no q 

c is assumed to sip into the ground (being conservative). 

c also for the depth, tarmd, the temperature of the unsheltered 

c tarmac is assumed constant. note, qtarm and ttarm are known, 

c for they are unsheltered tarmac temperatures. 

c 

 qskytarm=qsky+0.5*qtarm+0.5*(1.-emista)*qsky 

 rlwng=(lengtwng-wfuselg)/2. 

c 

c LHS(section 1): fn19(t8),fn20(j2),fn51(t51),fn54(j12): 

c 

 qnetsky=(sig*t(8)**4-jj(2))/((1.-emisw)/emisw) 

 rad32skt=jj(14)*fbwa(4,1)+qskytarm*fbwa(4,2)+ 

 & qskytarm*fbwa(4,5)+qskytarm*fbwa(4,6) 

 rad12skt=jj(13)*fbwb(4,1)+rad32skt*fbwb(4,2)+ 

 & qskytarm*fbwb(4,5)+qskytarm*fbwb(4,6) 

 fn(19)=qabstwa1(3,iperiod)-qnetsky-hhw*(t(8)-ta)+ 

 & k(7)/th(7)*(t(7)-t(8)) 

 fn(20)=-qnetsky+ 

 & (jj(2)-jj(12))/(1./fbwa(3,1))+ 

 & (jj(2)-rad12skt)/(1./fbwa(3,2))+ 

 & (jj(2)-qskytarm)/(1./fbwa(3,4))+ 

 & (jj(2)-qskytarm)/(1./fbwa(3,5))+ 

 & (jj(2)-qskytarm)/(1./fbwa(3,6)) 

 qnetsky=(sig*t(51)**4-jj(12))/((1.-emista)/emista) 

 fn(51)=(qabstwa1(1,iperiod)-qnetsky-hht*(t(51)-ta))* 

 & rlwng*widthwng+ 

 & 2.*(conducta/(fact*widthwng)*(ttarm-t(51)))*tarmd*rlwng+ 

 & conducta/(fact*widthwng)*(ttarm-t(51))*tarmd*widthwng- 

 & conducta/(.5*wfuselg+.5*rlwng)*(t(51)-t(52))*tarmd*widthwng

 rad32sk=qsky*fbwa(4,2)+jj(4)*fbwa(4,3)+qsky*fbwa(4,5)+ 

 & qsky*fbwa(4,6) 

 rad12sk=rad32sk*fbwb(4,2)+jj(7)*fbwb(4,3)+ 

 & qsky*fbwb(4,5)+qsky*fbwb(4,6) 

 fn(54)=-qnetsky+ 

 & (jj(12)-rad12sk)/(1./fbwa(1,2))+ 

 & (jj(12)-jj(2))/(1./fbwa(1,3))+ 

 & (jj(12)-qsky)/(1./fbwa(1,4))+ 

 & (jj(12)-qsky)/(1./fbwa(1,5))+ 

 & (jj(12)-qsky)/(1./fbwa(1,6)) 

c 

c RHS(section 2): fn21(t16),fn22(j4),fn53(t53),fn56(j14): 

c 

 qnetsky=(sig*t(16)**4-jj(4))/((1.-emisw)/emisw) 

 rad34skt=jj(12)*fbwa(2,1)+qskytarm*fbwa(2,4)

 & +qskytarm*fbwa(2,5)+qskytarm*fbwa(2,6) 

A-31
 rad24skt=jj(13)*fbwb(2,1)+rad34skt*fbwb(2,4) 

 & +qskytarm*fbwb(2,5)+qskytarm*fbwb(2,6) 

 fn(21)=qabstwa2(3,iperiod)-qnetsky-hhw*(t(16)-ta)+ 

 & k(7)/th(7)*(t(15)-t(16)) 

 fn(22)=-qnetsky+ 

 & (jj(4)-jj(14))/(1./fbwa(3,1))+ 

 & (jj(4)-qskytarm)/(1./fbwa(3,2))+ 

 & (jj(4)-rad24skt)/(1./fbwa(3,4))+ 

 & (jj(4)-qskytarm)/(1./fbwa(3,5))+ 

 & (jj(4)-qskytarm)/(1./fbwa(3,6)) 

 qnetsky=(sig*t(53)**4-jj(14))/((1.-emista)/emista) 

 fn(53)=(qabstwa2(1,iperiod)-qnetsky-hht*(t(53)-ta))* 

 & rlwng*widthwng+ 

 & 2.*(conducta/(fact*widthwng)*(ttarm-t(53)))*tarmd*rlwng+ 

 & conducta/(fact*widthwng)*(ttarm-t(53))*tarmd*widthwng- 

 & conducta/(.5*wfuselg+.5*rlwng)*(t(53)-t(52))*tarmd*widthwng

 rad34sk=jj(2)*fbwa(2,3)+qsky*fbwa(2,4)+qsky*fbwa(2,5)+ 

 & qsky*fbwa(2,6) 

 rad24sk=jj(7)*fbwb(2,3)+rad34sk*fbwb(2,4)+ 

 & qsky*fbwb(2,5)+qsky*fbwb(2,6) 

 fn(56)=-qnetsky+ 

 & (jj(14)-qsky)/(1./fbwa(1,2))+ 

 & (jj(14)-jj(4))/(1./fbwa(1,3))+ 

 & (jj(14)-rad24sk)/(1./fbwa(1,4))+ 

 & (jj(14)-qsky)/(1./fbwa(1,5))+ 

 & (jj(14)-qsky)/(1./fbwa(1,6)) 

c 

c MIDDLE(section 3): fn31(t25),fn46(j7),fn52(t52),fn55(j13). 

c 

 qnetsky=(sig*t(25)**4-jj(7))/((1.-emisfb)/emisfb) 

 fn(31)=qabstwb(3,iperiod)-qnetsky-hhf*(t(25)-ta)+ 

 & k(12)/th(12)*(t(26)-t(25)) 

 fn(46)=-qnetsky+ 

 & (jj(7)-jj(13))/(1./fbwb(3,1))+ 

 & (jj(7)-rad32skt)/(1./fbwb(3,2))+ 

 & (jj(7)-rad34skt)/(1./fbwb(3,4))+ 

 & (jj(7)-qskytarm)/(1./fbwb(3,5))+ 

 & (jj(7)-qskytarm)/(1./fbwb(3,6)) 

 qnetsky=(sig*t(52)**4-jj(13))/((1.-emista)/emista) 

 fn(52)=(qabstwb(1,iperiod)-qnetsky-hht*(t(52)-ta))* 

 & wfuselg*widthwng+ 

 & 2.*(conducta/(fact*widthwng)*(ttarm-t(52)))*tarmd*wfuselg+

 & conducta/(.5*wfuselg+.5*rlwng)*(t(51)-t(52))*tarmd*widthwng+ 

 & conducta/(.5*wfuselg+.5*rlwng)*(t(53)-t(52))*tarmd*widthwng

 fn(55)=-qnetsky+ 

 & (jj(13)-rad32sk)/(1./fbwb(1,2))+ 

 & (jj(13)-jj(7))/(1./fbwb(1,3))+ 

 & (jj(13)-rad34sk)/(1./fbwb(1,4))+ 

 & (jj(13)-qsky)/(1./fbwb(1,5))+ 

 & (jj(13)-qsky)/(1./fbwb(1,6)) 

c 

A-32
c equations for the fuselage-wing junction facing the sky 

c there are a total of 8 eqns. (4 for left half and 4 for right half)

c consider the wings first (a total of 4 eqns: 2 on each side) 

c eqns. 23, 24, 25 & 26. 

c 

c when ii=1 (eqn. 23 & 24), left hand side; when ii=2 (eqn. 25 & 26), 

c right hand side 

 do 80 ii=1,2 

 qnetsky=(sig*t(8*(ii-1)+1)**4-rfw*jj(1+(ii-1)*2))/ 

 & ((1.-rfw*eemisw)/eemisw) 

 if(ii.eq.1)fn(23+(ii-1)*2)=absorbw*qreflfwl(1,iperiod)* 

 & sin(fpi*theta)+qscabfwl(1,iperiod)-qnetsky-hhw* 

 & (t(8*(ii-1)+1)-ta)-k(1)*((t(8*(ii-1)+1)-t(8*(ii-1)+2))/th(1)) 

 if(ii.eq.2)fn(23+(ii-1)*2)=absorbw*qreflfwr(1,iperiod)* 

 & sin(fpi*theta)+qscabfwr(1,iperiod)-qnetsky-hhw* 

 & (t(8*(ii-1)+1)-ta)-k(1)*((t(8*(ii-1)+1)-t(8*(ii-1)+2))/th(1)) 

 fn(24+(ii-1)*2)=-qnetsky 

 & +(((jj(1+(ii-1)*2)-jj(5+(ii-1)))/(1./fa12)) 

 & +((jj(1+(ii-1)*2)-qsky)/(1./fa13)) 

 & +((jj(1+(ii-1)*2)-qsky)/(1./fa14)) 

 & +((jj(1+(ii-1)*2)-qsky-fa57*(emisf*sig*t(17+(ii-1)*7) 

 & **4+(1-emisf)*(qsky+.5*qtarm+.5*(1.-emista)*qsky)))/(1./fa15)) 

 & +((jj(1+(ii-1)*2)-qsky-fa68*(emisf*sig*t(17+(ii-1)*7) 

 & **4+(1-emisf)*(qsky+.5*qtarm+.5*(1.-emista)*qsky)))/(1./fa16))) 

 80 continue

c now consider the 4 eqns. of the above related to the fuselage. 

c 2 eqns. on the left hand side (eqns. 27 & 28) and 2 eqns. on

c the right hand side (eqns. 29 & 30). 

c 

c when ii=1, left hand side; when ii=2, right hand side 

 do 90 ii=1,2 

 qnetsky=(sig*t(17+(ii-1)*7)**4-rff*jj(5+(ii-1)))/ 

 & ((1.-rff*eemisf)/eemisf) 

 if(ii.eq.1)fn(27+(ii-1)*2)=absorbf*qreflfwl(2,iperiod)* 

 & cos(fpi*theta)+qscabfwl(2,iperiod)-qnetsky-hhf* 

 & (t(17+(ii-1)*7)-ta)+ 

 & (-1)**ii*k(8)*((t(17+(ii-1)*6)-t(18+(ii-1)*6))/th(8)) 

 if(ii.eq.2)fn(27+(ii-1)*2)=qscabfwr(2,iperiod)-qnetsky-hhf* 

 & (t(17+(ii-1)*7)-ta)+ 

 & (-1)**ii*k(8)*((t(17+(ii-1)*6)-t(18+(ii-1)*6))/th(8)) 

 fn(28+(ii-1)*2)=-qnetsky 

 & +(jj(5+(ii-1))-jj(1+(ii-1)*2))/(1/fa21) 

 & +(jj(5+(ii-1))-qsky)/(1/fa23) 

 & +(jj(5+(ii-1))-qsky-.5*qtarm-.5*(1.-emista)*qsky)/(1/fa24) 

 & +(jj(5+(ii-1))-qsky-.5*qtarm-.5*(1.-emista)*qsky)/(1/fa25) 

 & +(jj(5+(ii-1))-qsky-.5*qtarm-.5*(1.-emista)*qsky)/(1/fa26) 

 90 continue

c Recent modification has added a total of 9 eqns within the cabin. 

c These eqns. are : 4 temperatures within the cabin (Eqn. 41,42,44,45); 

c 1 temperature outside plexiglass (Eqn. 43); and 4 JJ's within 

c the cabin (Eqns. 47,48,49,50). 

A-33
c 

c In order to consider the effects of the roof cover, five 

c equations (eqn. 57, 58, 59, 60 & 61) have been added, giving temperatures 

c t57, t58, t59, t60 & t61. note that in these equations, areas are 

c kept to indicate their comparative influence. 

c also note that eqns. 42 & 43 have been modified. 

c surface 4 (combined roof and glass bottom) is treated as one 

c unit having a uniform temperature, tpgrf; uniform radiosity JJ9;and 

c uniform emissivity; in order to obtain a uniform qnet within the cabin.

c 

c equation 58 (conduction): 

 fn(58)=k(27)/th(27)*(t(59)-t(58))*a2 

 & -k(26)/th(26)*(t(58)-t(57))*a2 

c equation 57 (conduction): 

 fn(57)=k(26)/th(26)*(t(58)-t(57))*a2 

 & -k(25)/th(25)*(t(57)-t(60))*a2 

c equation 60 (conduction): 

 fn(60)=k(25)/th(25)*(t(57)-t(60))*a2 

 & -k(28)/th(28)*(t(60)-t(61))*a2 

c 

c top of plexiglass and roof top facing the sky & tarmac. 

c eqn 43 (t37(plexiglass) & eqn 59 (t59 (roof top)) 

c 

 fn(43)=-emispg*sig*t(37)**4*a1 

 & -hhf*(t(37)-ta)*a1 

 & -k(22)/th(22)*(t(37)-t(36))*a1 

 & +emispg*qsky*a1 

 & +emispg*.5*(1.-emista)*qsky*a5 

 & +emispg*.5*qtarm*a5 

 & +.5*absorbpg*qdir*cos(fpi*theta)*a6(iperiod) 

 & +.5*absorbpg*qdir*sin(fpi*theta)*a7(iperiod) 

 & +.5*absorbpg*.5*(1.-absorbta)*qdir*sin(fpi*theta)*a5 

 & +.5*absorbpg*qreflin(4,iperiod)*sin(fpi*theta)*a3 

 & +.5*absorbpg*qscabsin(4,iperiod)*lfuselg*wfuselg 

 & *a3/(absorbpg*a3+absorfbt*a4) 

c 

 fn(59)=-emisrftp*sig*t(59)**4*a2

 & -hhf*(t(59)-ta)*a2 

 & -k(27)/th(27)*(t(59)-t(58))*a2 

 & +emisrftp*qsky*a2 

 & +absorftp*qdir*sin(fpi*theta)*a2 

c 

c find the weighted average temperature of the ends, tend 

 aa1=lfuselg*wfuselg 

 aa2=a1 

 aa3=lfuselg*hfuselg 

 aa4=aa3 

 aa5=a2 

 atot=aa1+aa2+aa3+aa4+aa5 

 cc1=aa1/atot 

 cc2=aa2/atot 

A-34
 cc3=aa3/atot 

 cc4=aa4/atot 

 cc5=aa5/atot 

 tend=cc1*t(35)+cc2*t(36)+cc3*t(38)+cc4*t(39)+cc5*t(61) 

c 

c for eqns: 42 (bottom of plexiglass facing the cabin floor below); 

c 41 (floor of cabin inside (surface B)); 44 (left hand side of cabin 

c inside (surface C)); and 45 (right hand side of cabin inside

c (surface A)): the effective convection coefficient must be found. 

c these coeffs. are coupled with the surface temperatures. therefore 

c need to read in thermal diffusivity (thdiffus); kinematic viscosity 

c (visckine); gravity (gravity); and Prandtl # (prandtl). 

c from these, Rayleigh number (rayleigh); and Nusselt # (nusselt) 

c must be found within the system of equation. keff=k * nusselt. 

c 

c find the weighted plexiglass and roof bottom temperatures: 

 c1=a1/(a1+a2) 

 c2=a2/(a1+a2) 

 tpgrf=c1*t(36)+c2*t(61) 

c 

 rayleig1=gravity*1./((tpgrf+t(35))/2.)*(t(35)-tpgrf)*th(24)**3/ 

 & (thdiffus*visckine) 

 if(t(35).lt.0.0.or.tpgrf.lt.0.0)then 

 nusselt1=1.0 

 else 

 if(rayleig1.le.3e5)then 

 nusselt1=1.0 

 elseif(rayleig1.gt.3e5.and.rayleig1.le.7e9)then 

 nusselt1=0.069*rayleig1**(1./3.)*prandtl**.074 

 elseif(rayleig1.gt.7e9)then 

 nusselt1=0.069*7e9**(1./3.)*prandtl**.074 

c this is to ensure a reasonable nusselt1 number. we do not want 

c a runaway solution during the Newton-Raphson solution of the 

c system of non-linear equations. 

c write(9,*)'Rayleigh number exceeded 7e9 for vertical' 

 endif 

 endif 

c write(9,*)'vertical',iperiod,ik,t(35),tpgrf,rayleig1,nusselt1 

 rayleig2=gravity*1./((t(38)+t(39))/2.)*abs(t(38)-t(39))* 

 & th(11)**3/(thdiffus*visckine) 

 if(t(38).lt.0.0.or.t(39).lt.0.0)then 

 nusselt2=1.0 

 else 

 if(rayleig2.le.(1e3*(0.2+prandtl)/prandtl))then 

 nusselt2=1.0 

 elseif(rayleig2.gt.(1e3*(0.2+prandtl)/prandtl).and.rayleig2.lt.

 & 1.0e9)then 

 nusselt2=0.18*((prandtl/(0.2+prandtl))*rayleig2)**0.29 

 elseif(rayleig2.gt.1.0e9)then 

 nusselt2=0.18*((prandtl/(0.2+prandtl))*1.0e9)**0.29 

c this is to ensure a reasonable nusselt1 number. we do not want 

A-35
c a runaway solution during the Newton-Raphson solution of the 

c system of non-linear equations. 

c write(9,*)'Rayleigh number exceeded 1e9 for horizontal' 

 endif 

 endif 

c write(9,*)'horizontal',iperiod,ik,t(38),t(39),rayleig2,nusselt2 

c 

c bottom of plexiglass facing the cabin floor below. 

c eqns: eqn 42 (t36);& eqn 48 (JJ9) 

c the plexiglass and roof bottom are treated as "one" unit, 

c (surface 4), for determining radiosity JJ9 and qnet 

 emispgrf=a1/(a1+a2)*emispg+a2/(a1+a2)*emisrfbt 

 qnet=(sig*tpgrf**4-jj(9))/((1.-emispgrf)/emispgrf) 

 qpg=emispg*sig*t(36)**4-emispg/(a3*emispg+a4*emisrfbt)* 

 & (a3*emispg*sig*t(36)**4+a4*emisrfbt*sig*t(61)**4-qnet*(a3+a4)) 

 fn(42)=-qpg*a3 

 & -nusselt1*k(24)/th(24)*(tpgrf-t(35))*a3 

 & +k(22)/th(22)*(t(37)-t(36))*a1 

 & +.5*absorbpg*qdir*cos(fpi*theta)*a6(iperiod) 

 & +.5*absorbpg*qdir*sin(fpi*theta)*a7(iperiod) 

 & +.5*absorbpg*.5*(1.-absorbta)*qdir*sin(fpi*theta)*a5 

 & +.5*absorbpg*qreflin(4,iperiod)*sin(fpi*theta)*a3 

 & +.5*absorbpg*qscabsin(4,iperiod)*lfuselg*wfuselg 

 & *a3/(absorbpg*a3+absorfbt*a4) 

c fn48: 

 fn(48)=-qnet+(jj(9)-jj(11))/(1./fc31)+(jj(9)-jj(10))/(1./fc32)

 & +(jj(9)-jj(8))/(1./fc34)+(jj(9)-sig*tend**4)/(1./fc35) 

 & +(jj(9)-sig*tend**4)/(1./fc36) 

c 

c equation 61 (t61) (roof bottom surface): 

 qrfbt=emisrfbt*sig*t(61)**4-emisrfbt/(a3*emispg+a4*emisrfbt)* 

 & (a3*emispg*sig*t(36)**4+a4*emisrfbt*sig*t(61)**4-qnet*(a3+a4)) 

 fn(61)=-qrfbt*a4 

 & -nusselt1*k(24)/th(24)*(tpgrf-t(35))*a4 

 & +k(28)/th(28)*(t(60)-t(61))*a2 

 & +absorfbt*qreflin(4,iperiod)*sin(fpi*theta)*a4 

 & +absorfbt*qscabsin(4,iperiod)*lfuselg*wfuselg 

 & *a4/(absorbpg*a3+absorfbt*a4) 

c 

c left hand side of cabin inside (surface C) 

c eqn 44 (t38);& eqn 47 (JJ8) 

 qnet=(sig*t(38)**4-jj(8))/((1.-emisfi)/emisfi) 

 fn(44)=-qnet-nusselt2*k(11)/th(11)*(t(38)-t(39)) 

 & +k(23)/th(23)*(t(20)-t(38)) 

 & +absorbfi*qreflin(3,iperiod)*cos(fpi*theta) 

 & +qscabsin(3,iperiod) 

 fn(47)=-qnet+(jj(8)-jj(11))/(1./fc41)+(jj(8)-jj(10))/(1./fc42)

 & +(jj(8)-jj(9))/(1./fc43)+(jj(8)-sig*tend**4)/(1./fc45) 

 & +(jj(8)-sig*tend**4)/(1./fc46) 

c right hand side of cabin inside (surface A) 

c eqn 45 (t39);& eqn 49 (JJ10) 

A-36
 qnet=(sig*t(39)**4-jj(10))/((1.-emisfi)/emisfi) 

 fn(45)=-qnet-k(23)/th(23)*(t(39)-t(21)) 

 & +nusselt2*k(11)/th(11)*(t(38)-t(39)) 

 & +absorbfi*qreflin(1,iperiod)*cos(fpi*theta) 

 & +qscabsin(1,iperiod) 

 fn(49)=-qnet+(jj(10)-jj(11))/(1./fc21)+(jj(10)-jj(9))/(1./fc23)

 & +(jj(10)-jj(8))/(1./fc24)+(jj(10)-sig*tend**4)/(1./fc25) 

 & +(jj(10)-sig*tend**4)/(1./fc26) 

c floor of cabin inside (surface B) 

c eqn 41 (t35):& eqn 50 (JJ11) 

 qnet=(sig*t(35)**4-jj(11))/((1.-emisfl)/emisfl) 

 fn(41)=-qnet-k(21)/th(21)*(t(35)-t(34)) 

 & +nusselt1*k(24)/th(24)*(tpgrf-t(35)) 

 & +absorbfl*qreflin(2,iperiod)*sin(fpi*theta) 

 & +qscabsin(2,iperiod) 

 fn(50)=-qnet+(jj(11)-jj(10))/(1./fc12)+(jj(11)-jj(9))/(1./fc13)

 & +(jj(11)-jj(8))/(1./fc14)+(jj(11)-sig*tend**4)/(1./fc15) 

 & +(jj(11)-sig*tend**4)/(1./fc16) 

c 

c now find the jacobian values for the current t values 

c 

 call jacobian(hhf,hhw,hht,emisf,emisw,eemisf,eemisw,rff, 

 & rfw,nusselt1,nusselt2,c1,c2,cc1,cc2,cc3, 

 & cc4,cc5,tend,tpgrf,emispgrf,rlwng) 

c 

c just a dummy for using ik, (otherwise compiler warning) 

c ik maybe necessary for Rayleigh check, in case during debugging 

 dummyik=ik 

 return 

 end 

c 

 subroutine jacobian(hhf,hhw,hht,emisf,emisw,eemisf,eemisw,rff, 

 & rfw,nusselt1,nusselt2,c1,c2,cc1,cc2,cc3, 

 & cc4,cc5,tend,tpgrf,emispgrf,rlwng) 

c purpose: calculates and return the jacobian values 

c at each new t values 

c called from: thermodl

c calls:none

 common/c1/n,ntrial,tolt,tolfn 

 common/c2/t(200),fn(200),jacob(200,200) 

 common/c3/th(100),k(100) 

 common/c4/widthwng,lengtwng,hwng,locwng,lplane,lfuselg,hfuselg,

 & wfuselg 

 common/c5/trnsm,absorbta,emista,tarmd,densta,conducta,shcta 

 common/c6/absorbff,emisff,absorbww,emisww,absorbfl,emisfl,absorbfi 

 &,emisfi,absorbpg,emispg,absorftp,emisrftp,absorfbt,emisrfbt 

 &,absorbfb,emisfb 

 common/c7/nperiods,envtdata(100,20) 

 common/c10/aircondc,thdiffus,visckine,prandtl 

 common/c12/a1,a2,a3,a4,a5,a6(100),a7(100) 

 common/vf1/fbwa(6,6),fbwb(6,6) 

A-37
 common/vf2/fa11,fa12,fa13,fa14,fa15,fa16,fa21,fa22,fa23,fa24, 

 & fa25,fa26,fa31,fa32,fa33,fa34,fa35,fa36,fa41,fa42, 

 & fa43,fa44,fa45,fa46,fa51,fa52,fa53,fa54,fa55,fa56, 

 & fa61,fa62,fa63,fa64,fa65,fa66 

 common/vf4/fa57,fa68 

 common/vf5/fc11,fc12,fc13,fc14,fc15,fc16,fc21,fc22,fc23,fc24, 

 & fc25,fc26,fc31,fc32,fc33,fc34,fc35,fc36,fc41,fc42, 

 & fc43,fc44,fc45,fc46,fc51,fc52,fc53,fc54,fc55,fc56, 

 & fc61,fc62,fc63,fc64,fc65,fc66 

 common/facttarm/fact 

 real jacob,k,nusselt1,nusselt2 

 parameter(sig=0.1714e-8) 

 parameter(gravity=32.2) 

c 

c initialize: (most jacobian coefficients are zero) 

 do 110 i=1,n 

do 110 j=1,n

 jacob(i,j)=0. 

 110 continue 

c 

c left-wing conduction (eqn. 1 & 2) 

c 

 do 120 i=1,2 

 jacob(i,i)=-k(i)/th(i) 

 jacob(i,i+1)=k(i)/th(i)+k(i+1)/th(i+1) 

 jacob(i,i+2)=-k(i+1)/th(i+1) 

 120 continue 

c 

c left-wing conduction (eqn. 5 & 6) 

c 

 do 125 i=5,6 

 jacob(i,i)=-k(i)/th(i) 

 jacob(i,i+1)=k(i)/th(i)+k(i+1)/th(i+1) 

 jacob(i,i+2)=-k(i+1)/th(i+1) 

 125 continue 

c 

c left-wing conduction (eqn. 3,4,62&63) 

c 

 jacob(3,3)=-k(3)/th(3) 

 jacob(3,4)=k(3)/th(3)+k(4)/th(4) 

 jacob(3,62)=-k(4)/th(4) 

c 

 jacob(4,4)=-k(4)/th(4) 

 jacob(4,62)=k(4)/th(4)+k(29)/th(29) 

 jacob(4,63)=-k(29)/th(29) 

c 

 jacob(62,5)=-k(30)/th(30) 

 jacob(62,62)=-k(29)/th(29) 

 jacob(62,63)=k(29)/th(29)+k(30)/th(30) 

c 

 jacob(63,5)=k(30)/th(30)+k(5)/th(5) 

A-38
 jacob(63,6)=-k(5)/th(5) 

 jacob(63,63)=-k(30)/th(30) 

c 

c right-wing conduction (eqn. 7 & 8) 

c 

 do 130 i=1,2 

 jacob(i+6,i+8)=-k(i)/th(i) 

 jacob(i+6,i+9)=k(i)/th(i)+k(i+1)/th(i+1) 

 jacob(i+6,i+10)=-k(i+1)/th(i+1) 

 130 continue 

c 

c right-wing conduction (eqn. 11 & 12) 

c 

 do 135 i=5,6 

 jacob(i+6,i+8)=-k(i)/th(i) 

 jacob(i+6,i+9)=k(i)/th(i)+k(i+1)/th(i+1) 

 jacob(i+6,i+10)=-k(i+1)/th(i+1) 

 135 continue 

c 

c right-wing conduction (eqn. 9,10,64&65) 

c 

 jacob(9,11)=-k(3)/th(3) 

 jacob(9,12)=k(3)/th(3)+k(4)/th(4) 

 jacob(9,64)=-k(4)/th(4) 

c 

 jacob(10,12)=-k(4)/th(4) 

 jacob(10,64)=k(4)/th(4)+k(29)/th(29) 

 jacob(10,65)=-k(29)/th(29) 

c 

 jacob(64,13)=-k(30)/th(30) 

 jacob(64,64)=-k(29)/th(29) 

 jacob(64,65)=k(29)/th(29)+k(30)/th(30) 

c 

 jacob(65,13)=k(30)/th(30)+k(5)/th(5) 

 jacob(65,14)=-k(5)/th(5) 

 jacob(65,65)=-k(30)/th(30) 

c 

c left fuselage conduction (eqn. 13 to 14) 

 do 140 i=1,2 

 jacob(i+12,i+16)=-k(i+7)/th(i+7) 

 jacob(i+12,i+17)=k(i+7)/th(i+7)+k(i+8)/th(i+8) 

 jacob(i+12,i+18)=-k(i+8)/th(i+8) 

 140 continue 

c add eqn. 17, recently modified 

 jacob(17,19)=-k(10)/th(10) 

 jacob(17,20)=k(10)/th(10)+k(23)/th(23) 

 jacob(17,38)=-k(23)/th(23) 

c right fuselage conduction (eqn. 15 to 16) 

 do 150 i=1,2 

 jacob(i+14,i+20)=-k(11-i)/th(11-i) 

 jacob(i+14,i+21)=k(11-i)/th(11-i)+k(10-i)/th(10-i)

A-39
 jacob(i+14,i+22)=-k(10-i)/th(10-i) 

 150 continue 

c add eqn. 18, recently modified 

 jacob(18,21)=k(23)/th(23)+k(10)/th(10) 

 jacob(18,22)=-k(10)/th(10) 

 jacob(18,39)=-k(23)/th(23) 

c fuselage-wing junction

c wing portion (i=1 implies left hand side. i=2 implies right hand side) 

c (eqns. 23 to 26) 

 do 180 i=1,2 

 jacob(23+(i-1)*2,1+(i-1)*8)=-(4*sig*t(1+(i-1)*8)**3)/ 

 & ((1-rfw*eemisw)/eemisw)-hhw-k(1)/th(1) 

 jacob(23+(i-1)*2,2+(i-1)*8)=k(1)/th(1) 

 jacob(23+(i-1)*2,40+(i-1)*2)=1.*rfw/((1.-rfw*eemisw)/eemisw) 

 jacob(24+(i-1)*2,1+(i-1)*8)=-(4*sig*t(1+(i-1)*8)**3)/ 

 & ((1-rfw*eemisw)/eemisw) 

 jacob(24+(i-1)*2,17+(i-1)*7)=-(4*fa57*emisf*sig*t(17+(i-1)*7)**3 

 & /(1/fa15))-(4*fa68*emisf*sig*t(17+(i-1)*7)**3/(1/fa16)) 

 jacob(24+(i-1)*2,40+(i-1)*2)=1.*rfw/((1.-rfw*eemisw)/eemisw) 

 & +(1/(1/fa12)+1/(1/fa13)+1/(1/fa14)+1/(1/fa15)+1/ 

 & (1/fa16)) 

 jacob(24+(i-1)*2,44+(i-1))=-1/(1/fa12) 

 180 continue 

c fuselage portion (i=1 implies left hand side. i=2 implies right hand side) 

c (eqns. 27 to 30) 

 do 190 i=1,2 

 jacob(27+(i-1)*2,17+(i-1)*7)=-(4*sig*t(17+(i-1)*7)**3)/ 

 & ((1-rff*eemisf)/eemisf)-hhf-k(8)/th(8) 

 jacob(27+(i-1)*2,18+(i-1)*5)=k(8)/th(8) 

 jacob(27+(i-1)*2,44+(i-1))=1.*rff/((1.-rff*eemisf)/eemisf) 

 jacob(28+(i-1)*2,17+(i-1)*7)=-(4*sig*t(17+(i-1)*7)**3)/ 

 & ((1-rff*eemisf)/eemisf) 

 jacob(28+(i-1)*2,40+(i-1)*2)=-1/(1/fa21) 

 jacob(28+(i-1)*2,44+(i-1))=1.*rff/((1.-rff*eemisf)/eemisf) 

 & +(1/(1/fa21)+1/(1/fa23)+1/(1/fa24)+1/(1/fa25)+1/ 

 & (1/fa26)) 

 190 continue 

c 

c Eqns. 32...40; & 66,67 for conduction in the middle section of fuselage

c (through: bottom skin cover; through the spar; through the 

c the upper skin of the spar box cap) 

c 

c Eqns. 32,33 & 34: 

c 

 do 195 i=1,3 

 jacob(31+i,24+i)=-k(11+i)/th(11+i) 

 jacob(31+i,25+i)=k(12+i)/th(12+i)+k(11+i)/th(11+i) 

 jacob(31+i,26+i)=-k(12+i)/th(12+i) 

 195 continue 

c 

c Eqns. 37,38,39 & 40: 

A-40
c 

 do 200 i=6,9 

 jacob(31+i,24+i)=-k(11+i)/th(11+i) 

 jacob(31+i,25+i)=k(12+i)/th(12+i)+k(11+i)/th(11+i) 

 jacob(31+i,26+i)=-k(12+i)/th(12+i) 

 200 continue 

c 

c Eqns. 35,36,66&67: 

c 

 jacob(35,28)=-k(15)/th(15) 

 jacob(35,29)=k(32)/th(32)+k(15)/th(15) 

 jacob(35,67)=-k(32)/th(32) 

c 

 jacob(36,29)=-k(32)/th(32) 

 jacob(36,66)=-k(31)/th(31) 

 jacob(36,67)=k(31)/th(31)+k(32)/th(32) 

c 

 jacob(66,30)=-k(16)/th(16) 

 jacob(66,66)=k(16)/th(16)+k(31)/th(31) 

 jacob(66,67)=-k(31)/th(31) 

c 

 jacob(67,30)=k(17)/th(17)+k(16)/th(16) 

 jacob(67,31)=-k(17)/th(17) 

 jacob(67,66)=-k(16)/th(16) 

c 

c add jacobian 43 for top of plexiglass: 

 jacob(43,36)=k(22)/th(22)*a1 

 jacob(43,37)=-4*emispg*sig*t(37)**3*a1 

 & -hhf*a1 

 & -k(22)/th(22)*a1 

c add jacobian 59 for top of roof cover: 

 jacob(59,58)=k(27)/th(27)*a2 

 jacob(59,59)=-4*emisrftp*sig*t(59)**3*a2 

 & -hhf*a2 

 & -k(27)/th(27)*a2 

c 

c jacobian 57, 58 & 60 for internal conduction of roof cover: 

c jacobian(57; 57, 58, 60): 

 jacob(57,57)=-k(26)/th(26)*a2 

 & -k(25)/th(25)*a2 

 jacob(57,58)=k(26)/th(26)*a2 

 jacob(57,60)=k(25)/th(25)*a2 

c jacobian(58: 57,58,59): 

 jacob(58,57)=k(26)/th(26)*a2 

 jacob(58,58)=-k(27)/th(27)*a2 

 & -k(26)/th(26)*a2 

 jacob(58,59)=k(27)/th(27)*a2 

c jacobian(60: 57, 60, 61) 

 jacob(60,57)=k(25)/th(25)*a2 

 jacob(60,60)=-k(25)/th(25)*a2 

 & -k(28)/th(28)*a2 

A-41
 jacob(60,61)=k(28)/th(28)*a2 

c 

c for eqns: 42 (bottom of plexiglass facing the cabin floor below); 

c 41 (floor of cabin inside (surface B)); 44 (left hand side of cabin 

c inside (surface C)); and 45 (right hand side of cabin inside

c (surface A)): the effective convection coefficent must be found. 

c these coeffs. are coupled with the surface temperatures. therefore 

c need to read in thermal diffusivity (thdiffus); kinematic viscosity 

c (visckine); gravity (gravity); and Prandtl # (prandtl). 

c from these, Rayleigh number (rayleigh); and Nusselt # (nusselt) 

c must be found within the system of equation. keff=k * nusselt. 

c 

c 

c add jacobians 42, 61 & 48 (J9) for bottom of plexiglass 

c and the bottom of roof cover (all inside cabin) 

c jacobian 42: 

 if(nusselt1.eq.1.0)then 

 jacob(42,35)=nusselt1*k(24)/th(24)*a3 

 jacob(42,36)=-(4*emispg*sig*t(36)**3- 

 & emispg/(a3*emispg+a4*emisrfbt)* 

 & (4*a3*emispg*sig*t(36)**3-(a3+a4)/((1.-emispgrf)/emispgrf)* 

 & (4*sig*tpgrf**3*c1)))*a3 

 & -nusselt1*k(24)/th(24)*c1*a3 

 & -k(22)/th(22)*a1 

 jacob(42,37)=k(22)/th(22)*a1 

 jacob(42,48)=emispg/(a3*emispg+a4*emisrfbt)* 

 & (a3+a4)/((1.-emispgrf)/emispgrf)*a3 

 jacob(42,61)=-(-emispg/(a3*emispg+a4*emisrfbt)* 

 & (4*a4*emisrfbt*sig*t(61)**3-(a3+a4)/((1.-emispgrf)/emispgrf)* 

 & (4*sig*tpgrf**3*c2)))*a3 

 & -nusselt1*k(24)/th(24)*c2*a3 

 else 

 value1=-0.069*(gravity/(.5*thdiffus*visckine))**(1./3.) 

 & *prandtl**0.074*th(24) 

 partl35=(t(35)+tpgrf)**(-1./3.)*(4./3.)*(t(35)-tpgrf)**(1./3.) 

 & -(t(35)-tpgrf)**(4./3.)*(1./3.)*(t(35)+tpgrf)**(-4./3.) 

 facnusel=-value1*partl35 

 jacob(42,35)=facnusel*k(24)/th(24)*a3 

 partl36=-(t(35)+tpgrf)**(-1./3.)*(4./3.)*(t(35)-tpgrf)**(1./3.) 

 & -(t(35)-tpgrf)**(4./3.)*(1./3.)*(t(35)+tpgrf)**(-4./3.) 

 facnusel=value1*partl36 

 jacob(42,36)=-(4*emispg*sig*t(36)**3- 

 & emispg/(a3*emispg+a4*emisrfbt)* 

 & (4*a3*emispg*sig*t(36)**3-(a3+a4)/((1.-emispgrf)/emispgrf)* 

 & (4*sig*tpgrf**3*c1)))*a3 

 & -facnusel*k(24)/th(24)*c1*a3 

 & -k(22)/th(22)*a1 

 jacob(42,37)=k(22)/th(22)*a1 

 jacob(42,48)=emispg/(a3*emispg+a4*emisrfbt)* 

 & (a3+a4)/((1.-emispgrf)/emispgrf)*a3 

 partl61=-(t(35)+tpgrf)**(-1./3.)*(4./3.)*(t(35)-tpgrf)**(1./3.) 

A-42
 & -(t(35)-tpgrf)**(4./3.)*(1./3.)*(t(35)+tpgrf)**(-4./3.) 

 facnusel=value1*partl61 

 jacob(42,61)=-(-emispg/(a3*emispg+a4*emisrfbt)* 

 & (4*a4*emisrfbt*sig*t(61)**3-(a3+a4)/((1.-emispgrf)/emispgrf)* 

 & (4*sig*tpgrf**3*c2)))*a3 

 & -facnusel*k(24)/th(24)*c2*a3 

 endif 

c 

c jacobian 61: 

 if(nusselt1.eq.1.0)then 

 jacob(61,35)=nusselt1*k(24)/th(24)*a4 

 jacob(61,36)=-(-emisrfbt/(a3*emispg+a4*emisrfbt)* 

 & (4*a3*emispg*sig*t(36)**3-(a3+a4)/((1.-emispgrf)/emispgrf)* 

 & (4*sig*tpgrf**3*c1)))*a4 

 & -nusselt1*k(24)/th(24)*c1*a4 

 jacob(61,48)=emisrfbt/(a3*emispg+a4*emisrfbt)* 

 & (a3+a4)/((1.-emispgrf)/emispgrf)*a4 

 jacob(61,60)=k(28)/th(28)*a2 

 jacob(61,61)=-(4*emisrfbt*sig*t(61)**3- 

 & emisrfbt/(a3*emispg+a4*emisrfbt)* 

 & (4*a4*emisrfbt*sig*t(61)**3-(a3+a4)/((1.-emispgrf)/emispgrf)* 

 & (4*sig*tpgrf**3*c2)))*a4 

 & -nusselt1*k(24)/th(24)*c2*a4 

 & -k(28)/th(28)*a2 

 else 

 facnusel=-value1*partl35 

 jacob(61,35)=facnusel*k(24)/th(24)*a4 

 facnusel=value1*partl36 

 jacob(61,36)=-(-emisrfbt/(a3*emispg+a4*emisrfbt)* 

 & (4*a3*emispg*sig*t(36)**3-(a3+a4)/((1.-emispgrf)/emispgrf)* 

 & (4*sig*tpgrf**3*c1)))*a4 

 & -facnusel*k(24)/th(24)*c1*a4 

 jacob(61,48)=emisrfbt/(a3*emispg+a4*emisrfbt)* 

 & (a3+a4)/((1.-emispgrf)/emispgrf)*a4 

 jacob(61,60)=k(28)/th(28)*a2 

 facnusel=value1*partl61 

 jacob(61,61)=-(4*emisrfbt*sig*t(61)**3- 

 & emisrfbt/(a3*emispg+a4*emisrfbt)* 

 & (4*a4*emisrfbt*sig*t(61)**3-(a3+a4)/((1.-emispgrf)/emispgrf)* 

 & (4*sig*tpgrf**3*c2)))*a4 

 & -facnusel*k(24)/th(24)*c2*a4 

 & -k(28)/th(28)*a2 

 endif 

c 

c jacobian 48: 

 jacob(48,35)=-4*sig*tend**3*cc1/(1./fc35) 

 & -4*sig*tend**3*cc1/(1./fc36) 

 jacob(48,36)=-4*sig*(tpgrf)**3*c1/

 & ((1.-emispgrf)/emispgrf) 

 & -4*sig*tend**3*cc2/(1./fc35) 

 & -4*sig*tend**3*cc2/(1./fc36) 

A-43
 jacob(48,38)=-4*sig*tend**3*cc3/(1./fc35) 

 & -4*sig*tend**3*cc3/(1./fc36) 

 jacob(48,39)=-4*sig*tend**3*cc4/(1./fc35) 

 & -4*sig*tend**3*cc4/(1./fc36) 

 jacob(48,61)=-4*sig*(tpgrf)**3*c2/

 & ((1.-emispgrf)/emispgrf) 

 & -4*sig*tend**3*cc5/(1./fc35) 

 & -4*sig*tend**3*cc5/(1./fc36) 

 jacob(48,47)=-1./(1/fc34) 

 jacob(48,48)=1./((1.-emispgrf)/emispgrf) 

 & +(1/(1/fc31)+1/(1/fc32)+1/(1/fc34)+1/(1/fc35)+1/ 

 & (1/fc36)) 

 jacob(48,49)=-1./(1/fc32) 

 jacob(48,50)=-1./(1/fc31) 

c add eqns. 44 & 47 (left hand side of cabin interior (surface C)) 

c jacobian 44: 

 if(nusselt2.eq.1.0)then 

 jacob(44,20)=k(23)/th(23) 

 jacob(44,38)=-4*sig*t(38)**3/((1.-emisfi)/emisfi) 

 & -nusselt2*k(11)/th(11)-k(23)/th(23) 

 jacob(44,39)=nusselt2*k(11)/th(11) 

 jacob(44,47)=1./((1.-emisfi)/emisfi) 

 else 

 jacob(44,20)=k(23)/th(23) 

 value2=0.18*(prandtl/(0.2+prandtl)*gravity/(.5*thdiffus*visckine) 

 & *th(11)**3)**0.29 

 ft38t391=(t(38)+t(39))**(-0.29)*(t(38)-t(39)) 

 & *0.29*(abs(t(38)-t(39)))**(-0.71)+(abs(t(38)-t(39)))**0.29 

 & *((t(38)-t(39))*(-0.29)*(t(38)+t(39))**(-1.29) 

 & +(t(38)+t(39))**(-0.29)) 

 facnusel=value2*ft38t391 

 jacob(44,38)=-4*sig*t(38)**3/((1.-emisfi)/emisfi) 

 & -facnusel*k(11)/th(11)-k(23)/th(23) 

 ft38t392=(t(38)+t(39))**(-0.29)*(t(38)-t(39)) 

 & *(-0.29)*(abs(t(38)-t(39)))**(-0.71)+(abs(t(38)-t(39)))**0.29 

 & *((t(38)-t(39))*(-0.29)*(t(38)+t(39))**(-1.29) 

 & +(-1.)*(t(38)+t(39))**(-0.29)) 

 facnusel=-value2*ft38t392 

 jacob(44,39)=facnusel*k(11)/th(11) 

 jacob(44,47)=1./((1.-emisfi)/emisfi) 

 endif 

c 

c jacobian 47: 

 jacob(47,35)=-4*sig*tend**3*cc1/(1./fc45) 

 & -4*sig*tend**3*cc1/(1./fc46) 

 jacob(47,36)=-4*sig*tend**3*cc2/(1./fc45) 

 & -4*sig*tend**3*cc2/(1./fc46) 

 jacob(47,38)=-4*sig*t(38)**3/((1.-emisfi)/emisfi) 

 & -4*sig*tend**3*cc3/(1./fc45) 

 & -4*sig*tend**3*cc3/(1./fc46) 

 jacob(47,39)=-4*sig*tend**3*cc4/(1./fc45) 

A-44
 & -4*sig*tend**3*cc4/(1./fc46) 

 jacob(47,61)=-4*sig*tend**3*cc5/(1./fc45) 

 & -4*sig*tend**3*cc5/(1./fc46) 

 jacob(47,47)=1./((1.-emisfi)/emisfi) 

 & +(1/(1/fc41)+1/(1/fc42)+1/(1/fc43)+1/(1/fc45)+1/ 

 & (1/fc46)) 

 jacob(47,48)=-1./(1/fc43) 

 jacob(47,49)=-1./(1/fc42) 

 jacob(47,50)=-1./(1/fc41) 

c add eqns. 45 & 49 (left hand side of cabin interior (surface C)) 

c jacobian 45: 

 if(nusselt2.eq.1.0)then 

 jacob(45,21)=k(23)/th(23) 

 jacob(45,38)=nusselt2*k(11)/th(11) 

 jacob(45,39)=-4*sig*t(39)**3/((1.-emisfi)/emisfi) 

 & -nusselt2*k(11)/th(11)-k(23)/th(23) 

 jacob(45,49)=1./((1.-emisfi)/emisfi) 

 else 

 jacob(45,21)=k(23)/th(23) 

 facnusel=value2*ft38t391 

 jacob(45,38)=facnusel*k(11)/th(11) 

 facnusel=-value2*ft38t392 

 jacob(45,39)=-4*sig*t(39)**3/((1.-emisfi)/emisfi) 

 & -facnusel*k(11)/th(11)-k(23)/th(23) 

 jacob(45,49)=1./((1.-emisfi)/emisfi) 

 endif 

c 

c 

c jacobian 49: 

 jacob(49,35)=-4*sig*tend**3*cc1/(1./fc25) 

 & -4*sig*tend**3*cc1/(1./fc26) 

 jacob(49,36)=-4*sig*tend**3*cc2/(1./fc25) 

 & -4*sig*tend**3*cc2/(1./fc26) 

 jacob(49,38)=-4*sig*tend**3*cc3/(1./fc25) 

 & -4*sig*tend**3*cc3/(1./fc26) 

 jacob(49,39)=-4*sig*t(39)**3/((1.-emisfi)/emisfi) 

 & -4*sig*tend**3*cc4/(1./fc25) 

 & -4*sig*tend**3*cc4/(1./fc26) 

 jacob(49,61)=-4*sig*tend**3*cc5/(1./fc25) 

 & -4*sig*tend**3*cc5/(1./fc26) 

 jacob(49,47)=-1./(1/fc24) 

 jacob(49,48)=-1./(1/fc23) 

 jacob(49,49)=1./((1.-emisfi)/emisfi) 

 & +(1/(1/fc21)+1/(1/fc23)+1/(1/fc24)+1/(1/fc25)+1/ 

 & (1/fc26)) 

 jacob(49,50)=-1./(1/fc21) 

c add eqns. 41 & 50 (floor (top of spar cap) of cabin interior 

c (surface C)) 

c jacobian 41: 

 if(nusselt1.eq.1.0)then 

 jacob(41,34)=k(21)/th(21) 

A-45
 jacob(41,35)=-4*sig*t(35)**3/((1.-emisfl)/emisfl) 

 & -nusselt1*k(24)/th(24)-k(21)/th(21) 

 jacob(41,36)=nusselt1*k(24)/th(24)*c1 

 jacob(41,50)=1./((1.-emisfl)/emisfl) 

 jacob(41,61)=nusselt1*k(24)/th(24)*c2 

 else 

 jacob(41,34)=k(21)/th(21) 

 facnusel=-value1*partl35 

 jacob(41,35)=-4*sig*t(35)**3/((1.-emisfl)/emisfl) 

 & -facnusel*k(24)/th(24)-k(21)/th(21) 

 facnusel=value1*partl36 

 jacob(41,36)=facnusel*k(24)/th(24)*c1 

 jacob(41,50)=1./((1.-emisfl)/emisfl) 

 facnusel=value1*partl61 

 jacob(41,61)=facnusel*k(24)/th(24)*c2 

 endif 

c 

c jacobian 50: 

 jacob(50,35)=-4*sig*t(35)**3/((1.-emisfl)/emisfl) 

 & -4*sig*tend**3*cc1/(1./fc15) 

 & -4*sig*tend**3*cc1/(1./fc16) 

 jacob(50,36)=-4*sig*tend**3*cc2/(1./fc15) 

 & -4*sig*tend**3*cc2/(1./fc16) 

 jacob(50,38)=-4*sig*tend**3*cc3/(1./fc15) 

 & -4*sig*tend**3*cc3/(1./fc16) 

 jacob(50,39)=-4*sig*tend**3*cc4/(1./fc15) 

 & -4*sig*tend**3*cc4/(1./fc16) 

 jacob(50,61)=-4*sig*tend**3*cc5/(1./fc15) 

 & -4*sig*tend**3*cc5/(1./fc16) 

 jacob(50,47)=-1./(1/fc14) 

 jacob(50,48)=-1./(1/fc13) 

 jacob(50,49)=-1./(1/fc12) 

 jacob(50,50)=1./((1.-emisfl)/emisfl) 

 & +(1/(1/fc12)+1/(1/fc13)+1/(1/fc14)+1/(1/fc15)+1/ 

 & (1/fc16)) 

c 

c 

c wing bottom, facing tarmac. there are a total of 12 eqns. to consider 

c within three sections of the wing: LHS(section: 1); RHS (section 2) 

c MIDDLE (section 3, ie underneath the fuselage). these eqns. are: 

c LHS(section 1): fn19(t8),fn20(j2),fn51(t51),fn54(j12); 

c RHS(section 2): fn21(t16),fn22(j4),fn53(t53),fn56(j14); 

c MIDDLE(section 3): fn31(t25),fn46(j7),fn52(t52),fn55(j13). 

c simple two-dimensional conduction from the unsheltered tarmac 

c surface to the sheltered surface is considered. a tarmac 

c thickness tarmd is assumed. also the bottom of the sheltered 

c tarmac is considered adiabatic (insulated) so that no q 

c is assumed to sip into the ground (being conservative). 

c also for the depth, tarmd, the temperature of the unsheltered 

c tarmac is assumed constant. note, qtarm and ttarm are known, 

c for they are unsheltered tarmac temperatures. 

A-46
c 

c 

c LHS(section 1): jacobian: 19;20;51&54; 

 jacob(19,7)=k(7)/th(7) 

 jacob(19,8)=-(4*sig*t(8)**3)/((1-emisw)/emisw) 

 & -hhw-k(7)/th(7) 

 jacob(19,41)=1./((1.-emisw)/emisw) 

 jacob(20,8)=-(4*sig*t(8)**3)/((1-emisw)/emisw) 

 jacob(20,41)=1./((1.-emisw)/emisw) 

 & +1/(1/fbwa(3,1))+1/(1/fbwa(3,2))+1/(1/fbwa(3,4)) 

 & +1/(1/fbwa(3,5))+1/(1/fbwa(3,6)) 

 jacob(20,54)=-1/(1/fbwa(3,1)) 

 jacob(20,55)=-fbwb(4,1)/(1/fbwa(3,2)) 

 jacob(20,56)=-fbwa(4,1)*fbwb(4,2)/(1/fbwa(3,2)) 

 jacob(51,51)=(-4*sig*t(51)**3/((1.-emista)/emista)-hht) 

 & *rlwng*widthwng 

 & -2*conducta/(fact*widthwng)*tarmd*rlwng 

 & -conducta/(fact*widthwng)*tarmd*widthwng 

 & -conducta/(0.5*wfuselg+0.5*rlwng)*tarmd*widthwng 

 jacob(51,52)=conducta/(0.5*wfuselg+0.5*rlwng)*tarmd*widthwng 

 jacob(51,54)=1./((1.-emista)/emista)*rlwng*widthwng 

 jacob(54,51)=-4*sig*t(51)**3/((1.-emista)/emista) 

 jacob(54,43)=-fbwa(4,3)*fbwb(4,2)/(1/fbwa(1,2)) 

 jacob(54,46)=-fbwb(4,3)/(1/fbwa(1,2)) 

 jacob(54,54)=1/((1-emista)/emista) 

 & +1/(1/fbwa(1,2))+1/(1/fbwa(1,3))+1/(1/fbwa(1,4)) 

 & +1/(1/fbwa(1,5))+1/(1/fbwa(1,6)) 

 jacob(54,41)=-1/(1/fbwa(1,3)) 

c 

c RHS(section 2): jacobian: 21;22;53&56; 

 jacob(21,15)=k(7)/th(7) 

 jacob(21,16)=-(4*sig*t(16)**3)/((1-emisw)/emisw) 

 & -hhw-k(7)/th(7) 

 jacob(21,43)=1./((1.-emisw)/emisw) 

 jacob(22,16)=-(4*sig*t(16)**3)/((1-emisw)/emisw) 

 jacob(22,43)=1./((1.-emisw)/emisw) 

 & +1/(1/fbwa(3,1))+1/(1/fbwa(3,2))+1/(1/fbwa(3,4)) 

 & +1/(1/fbwa(3,5))+1/(1/fbwa(3,6)) 

 jacob(22,54)=-fbwa(2,1)*fbwb(2,4)/(1/fbwa(3,4)) 

 jacob(22,55)=-fbwb(2,1)/(1/fbwa(3,4)) 

 jacob(22,56)=-1/(1/fbwa(3,1)) 

 jacob(53,52)=conducta/(0.5*wfuselg+0.5*rlwng)*tarmd*widthwng 

 jacob(53,53)=(-4*sig*t(53)**3/((1.-emista)/emista)-hht) 

 & *rlwng*widthwng 

 & -2*conducta/(fact*widthwng)*tarmd*rlwng 

 & -conducta/(fact*widthwng)*tarmd*widthwng 

 & -conducta/(0.5*wfuselg+0.5*rlwng)*tarmd*widthwng 

 jacob(53,56)=1./((1.-emista)/emista)*rlwng*widthwng 

 jacob(56,53)=-4*sig*t(53)**3/((1.-emista)/emista) 

 jacob(56,41)=-fbwa(2,3)*fbwb(2,4)/(1/fbwa(1,4)) 

 jacob(56,43)=-1/(1/fbwa(1,3)) 

A-47
 jacob(56,46)=-fbwb(2,3)/(1/fbwa(1,4)) 

 jacob(56,56)=1/((1-emista)/emista) 

 & +1/(1/fbwa(1,2))+1/(1/fbwa(1,3))+1/(1/fbwa(1,4)) 

 & +1/(1/fbwa(1,5))+1/(1/fbwa(1,6)) 

c 

c MIDDLE(section 3): jacobian: 31;46;52&55: 

c 

 jacob(31,25)=-(4*sig*t(25)**3)/((1-emisfb)/emisfb) 

 & -hhf-k(12)/th(12) 

 jacob(31,26)=k(12)/th(12) 

 jacob(31,46)=1./((1.-emisfb)/emisfb) 

 jacob(46,25)=-(4*sig*t(25)**3)/((1-emisfb)/emisfb) 

 jacob(46,46)=1./((1.-emisfb)/emisfb) 

 & +1/(1/fbwb(3,1))+1/(1/fbwb(3,2))+1/(1/fbwb(3,4)) 

 & +1/(1/fbwb(3,5))+1/(1/fbwb(3,6)) 

 jacob(46,54)=-fbwa(2,1)/(1/fbwb(3,4)) 

 jacob(46,55)=-1/(1/fbwb(3,1)) 

 jacob(46,56)=-fbwa(4,1)/(1/fbwb(3,2)) 

 jacob(52,51)=conducta/(0.5*wfuselg+0.5*rlwng)*tarmd*widthwng 

 jacob(52,52)=(-4*sig*t(52)**3/((1.-emista)/emista)-hht) 

 & *wfuselg*widthwng 

 & -2*conducta/(fact*widthwng)*tarmd*wfuselg 

 & -conducta/(0.5*wfuselg+0.5*rlwng)*tarmd*widthwng 

 & -conducta/(0.5*wfuselg+0.5*rlwng)*tarmd*widthwng 

 jacob(52,53)=conducta/(0.5*wfuselg+0.5*rlwng)*tarmd*widthwng 

 jacob(52,55)=1./((1.-emista)/emista)*wfuselg*widthwng 

 jacob(55,52)=-4*sig*t(52)**3/((1.-emista)/emista) 

 jacob(55,41)=-fbwa(2,3)/(1/fbwb(1,4)) 

 jacob(55,43)=-fbwa(4,3)/(1/fbwb(1,2)) 

 jacob(55,46)=-1/(1/fbwb(1,3)) 

 jacob(55,55)=1/((1-emista)/emista) 

 & +1/(1/fbwb(1,2))+1/(1/fbwb(1,3))+1/(1/fbwb(1,4)) 

 & +1/(1/fbwb(1,5))+1/(1/fbwb(1,6)) 

 return 

 end 

c 

 subroutine tarmmodl(absorbta,emista,qtarm,t) 

c purpose: determine the tarmac emission and tarmac temperature 

c called from: solution 

c calls: none

 common/c7/nperiods,envtdata(100,20) 

 common/convect/hf(100),hw(100),ht(100) 

 parameter(fpi=3.141593/180.) 

 parameter(sig=0.1714e-8) 

 parameter(ntrial=1000,tolt=0.0001,tolf=0.0001) 

 real jacob 

 dimension qtarm(100),t(100) 

c assume no heat loss through the ground 

c (ie ground is adiabatic/insulated). 

 do 20 i=1,nperiods 

 theta=envtdata(i,1) 

A-48
 ta=envtdata(i,2) 

 tsky=envtdata(i,3) 

 qdir=envtdata(i,5) 

 h=ht(i) 

 t(i)=100.

 do 10 k=1,ntrial 

 f=absorbta*qdir*sin(fpi*theta)-emista*sig*(t(i)**4-tsky**4) 

 & -h*(t(i)-ta) 

 errf=abs(f) 

 if(errf.le.tolf)then 

 qtarm(i)=emista*sig*t(i)**4 

 goto 20 

 endif 

 p=-f 

 jacob=-4*emista*sig*t(i)**3-h 

 delt=p/jacob

 t(i)=t(i)+delt 

 errt=abs(delt) 

 if(errt.le.tolt)then 

 qtarm(i)=emista*sig*t(i)**4 

 goto 20 

 endif 

 10 continue

 20 continue

 return 

 end 

c 

 subroutine infrqsky(qqsky) 

c purpose: determines the sky radiation 

c called from: solution 

c calls: none

 common/c7/nperiods,envtdata(100,20) 

 dimension qqsky(100) 

 parameter(sig=0.1714e-8) 

 do 10 i=1,nperiods 

 tsky=envtdata(i,3) 

 10 qqsky(i)=sig*tsky**4 

 return 

 end 

c 

 subroutine ludcmp 

 common/c1/n,ntrial,tolx,tolf 

 common/c2/x(200),fvec(200),jacob(200,200) 

 common/sc1/indx(200),p(200) 

 real jacob 

 parameter(tiny=1.0e-20) 

c purpose: given jacob(1:n,1:n), with physical dimension 100 by 100, 

c this routine replaces it by the LU decomposition of a rowwise 

c permutation of itself. jacob is also output, arranged as in eqn(2.3.14) 

c of Numerical Recipes in Fortran (William H. press et.al). 

c indx(1:n) is an output vector that records the row permutation 

A-49
c affected by the partial pivot; d is output as plus minus 1 depending on

c whether the number of row interchanges was even or odd, respectively. 

c this routine is used in combination with lubksb to solve linear equations 

c or invert a matrix. note: tiny=a small number (for ill-conditioned pivot) 

c called from: newtraph, updradio 

c calls: none

 dimension vv(200) 

c vv stores the implicit scaling of each row 

c no row interchanges yet 

 d=1. 

c loop over rows to get the implicit scaling information 

 do 12 i=1,n 

 aamax=0. 

do 11 j=1,n 

 if(abs(jacob(i,j)).gt.aamax)aamax=abs(jacob(i,j)) 

 11 continue 

if(aamax.eq.0)pause 'singular matrix in ludcmp' 

c no nonzero largest element 

c save the scaling 

 vv(i)=1./aamax 

 12 continue

c this is the loop over columns of Crout's method; this is 

c equation (2.3.12), except for i=j 

 do 19 j=1,n 

do 14 i=1,j-1

 sum=jacob(i,j) 

 do 13 k=1,i-1 

 sum=sum-jacob(i,k)*jacob(k,j) 

 13 continue 

 jacob(i,j)=sum 

 14 continue 

c initialize for the search for largest pivot element

 aamax=0. 

c this is i=j of eqn.(2.3.12) and i=j+1...n of eqn.(2.3.13) 

do 16 i=j,n 

 sum=jacob(i,j) 

 do 15 k=1,j-1 

 sum=sum-jacob(i,k)*jacob(k,j) 

 15 continue 

 jacob(i,j)=sum 

c figure of merit for the pivot; is it better than the best so far?

 dum=vv(i)*abs(sum) 

 if (dum.ge.aamax)then

 imax=i 

 aamax=dum 

 endif 

 16 continue 

c do we need to interchange rows? yes, do so 

 if(j.ne.imax)then 

 do 17 k=1,n 

 dum=jacob(imax,k) 

A-50
 jacob(imax,k)=jacob(j,k) 

 jacob(j,k)=dum 

 17 continue 

c ...and change the parity of d; also interchange the scale factor 

 d=-d 

 vv(imax)=vv(j) 

 endif 

 indx(j)=imax 

 if(jacob(j,j).eq.0)jacob(j,j)=tiny 

c if the pivot element is zero the matrix is singular (at least 

c to the precision of the algorithm). for some applications on singular 

c matrices, it is desirable to substitute tiny for zero. 

c now, finally divide by the pivot element 

 if(j.ne.n)then

 dum=1./jacob(j,j) 

 do 18 i=j+1,n 

 jacob(i,j)=jacob(i,j)*dum

 18 continue 

 endif 

c go back for the next column in the reduction 

 19 continue

 return 

 end 

c 

 subroutine lubksb 

 common/c1/n,ntrial,tolx,tolf 

 common/c2/x(200),fvec(200),jacob(200,200) 

 common/sc1/indx(200),b(200) 

 real jacob 

c purpose: solves the set of n linear equations a.x=b. 

c here jacob is input, not as the 

c matrix jacob but rather as its LU decomposition, determined by the routine 

c ludcmp. indx is input as the permutation vector returned by ludcmp. 

c b(1:n) is input as the right-hand side vector b, and returns with the 

c solution vector. jacob and index are not modified by this routine and can 

c be left in place for successive calls with different right-hand sides b. 

c this routine takes into account the possibility that b will begin with many 

c zero elements, so it is efficient for use in matrix inversion (if needed). 

c called from: newtraph,updradio 

c calls: none

 ii=0 

c when ii is set to a positive value, it will become the index 

c of the first nonvanishing element of b. we now do the forward substitution, 

c eqn. (2.3.6). the only new wrinkle is to unscramble the permutation as we 

c go. 

 do 12 i=1,n 

 ll=indx(i) 

 sum=b(ll) 

 b(ll)=b(i) 

 if(ii.ne.0)then

 do 11 j=ii,i-1 

A-51
 sum=sum-jacob(i,j)*b(j) 

 11 continue 

 elseif(sum.ne.0)then 

c a nonzero element was encountered, so from now on we will have 

c to do the sums in the loop above. 

 ii=i 

 endif 

 b(i)=sum 

 12 continue

c now we do the backsubstitution, eqn. (2.3.7) 

 do 14 i=n,1,-1 

 sum=b(i) 

do 13 j=i+1,n 

 sum=sum-jacob(i,j)*b(j)

 13 continue 

c store a component of the solution vector x. 

 b(i)=sum/jacob(i,i) 

 14 continue

 return 

 end 

subroutine qrefcab 

c purpose: determines the intercepted solar radiation and the absorbed 

c diffuse radiation in the cabin. cabin treated as a partial 

c cavity 

c called from: solution 

c calls: diffsola 

 common/c4/widthwng,lengtwng,hwng,locwng,lplane,lfuselg,hfuselg,

 & wfuselg 

 common/c5/trnsm,absorbta,emista,tarmd,densta,conducta,shcta 

 common/c6/absorbff,emisff,absorbww,emisww,absorbfl,emisfl,absorbfi 

 &,emisfi,absorbpg,emispg,absorftp,emisrftp,absorfbt,emisrfbt 

 &,absorbfb,emisfb 

 common/c7/nperiods,envtdata(100,20) 

 common/c9/scatffl,scatffi,scatfw,scatff 

 common/c12/a1,a2,a3,a4,a5,a6(100),a7(100) 

 common/reflecq/qreflin(6,100),qscabsin(6,100),qreflfwr(6,100), 

 &qreflfwl(6,100),qscabfwr(6,100),qscabfwl(6,100),qabstwa1(6,100), 

 &qabstwa2(6,100),qabstwb(6,100) 

 common/vf5/fc11,fc12,fc13,fc14,fc15,fc16,fc21,fc22,fc23,fc24, 

 & fc25,fc26,fc31,fc32,fc33,fc34,fc35,fc36,fc41,fc42, 

 & fc43,fc44,fc45,fc46,fc51,fc52,fc53,fc54,fc55,fc56, 

 & fc61,fc62,fc63,fc64,fc65,fc66 

 dimension qscat(6,100) 

 dimension f(6,6),absorb(6),ttrnsm(6,100) 

 parameter(fpi=3.141593/180.) 

 real lfuselg 

c 

c Variables:

c qreflin(,)=intercepted solar radiation (both direct and specular) 

c in the cabin 

A-52
c qscabsin(,)=absorbed diffuse solar radiation in the cabin 

c 

c initialize

 do 9 j=1,nperiods 

do 9 i=1,6 

 ttrnsm(i,j)=0.0 

 qreflin(i,j)=0.0 

 9 qscat(i,j)=0.0 

c find qrefl for surface A (right hand side of cabin interior) 

c right hand side means as you seated normally and face the wind shield

 do 10 i=1,nperiods 

 theta=fpi*envtdata(i,1) 

 qdir=envtdata(i,5) 

 fopndirs=envtdata(i,8) 

 if(theta.ge.0.0.and.theta.le.atan(hfuselg/(2*wfuselg)))then 

 qreflin(1,i)=qdir*trnsm*fopndirs 

 qscat(1,i)=scatffi*(1.-absorbfi)*qreflin(1,i)*cos(theta) 

 elseif(theta.gt.atan(hfuselg/(2*wfuselg)).and.theta.lt. 

 & atan(hfuselg/wfuselg))then 

 qreflin(1,i)=qdir*trnsm*fopndirs*(1.+(1.-scatffl)*(1.-absorbfl)

 & *(1.-scatffi)**2*(1.-absorbfi)**2) 

 qscat(1,i)=scatffi*(1.-absorbfi)*qreflin(1,i)*cos(theta) 

 elseif(theta.eq.atan(hfuselg/wfuselg))then 

 qreflin(1,i)=qdir*trnsm*fopndirs 

 qscat(1,i)=scatffi*(1.-absorbfi)*qreflin(1,i)*cos(theta) 

 elseif(theta.gt.atan(hfuselg/wfuselg).and.theta.le.(fpi*90)) 

 & then 

 qreflin(1,i)=qdir*trnsm*fopndirs*(1.+(1.-scatffl)*(1.-absorbfl))

 qscat(1,i)=scatffi*(1.-absorbfi)*qreflin(1,i)*cos(theta) 

 endif 

 10 continue 

c find qrefl for surface B (floor of cabin interior) 

 do 20 i=1,nperiods 

 theta=fpi*envtdata(i,1) 

 qdir=envtdata(i,5) 

 fopndirs=envtdata(i,8) 

 if(theta.ge.0.0.and.theta.le.atan(hfuselg/(2*wfuselg)))then 

 qreflin(2,i)=qdir*trnsm*fopndirs*(1.-scatffi)**2* 

 & (1.-absorbfi)**2 

 qscat(2,i)=scatffl*(1.-absorbfl)*qreflin(2,i)*sin(theta) 

 elseif(theta.gt.atan(hfuselg/(2*wfuselg)).and.theta.lt. 

 & atan(hfuselg/wfuselg))then 

 qreflin(2,i)=qdir*trnsm*fopndirs*((1.-scatffi)*(1.-absorbfi) 

 & +(1.-scatffi)**2*(1.-absorbfi)**2) 

 qscat(2,i)=scatffl*(1.-absorbfl)*qreflin(2,i)*sin(theta) 

 elseif(theta.eq.atan(hfuselg/wfuselg))then 

 qreflin(2,i)=qdir*trnsm*fopndirs*(1.-scatffi)*(1.-absorbfi) 

 qscat(2,i)=scatffl*(1.-absorbfl)*qreflin(2,i)*sin(theta) 

 elseif(theta.gt.atan(hfuselg/wfuselg).and.theta.le.(fpi*90)) 

 & then 

 qreflin(2,i)=qdir*trnsm*fopndirs*(1.+(1.-scatffi)*(1.-absorbfi))

A-53
 qscat(2,i)=scatffl*(1.-absorbfl)*qreflin(2,i)*sin(theta) 

 endif 

 20 continue 

c find qrefl for surface C (left hand side of cabin interior) 

 do 30 i=1,nperiods 

 theta=fpi*envtdata(i,1) 

 qdir=envtdata(i,5) 

 fopndirs=envtdata(i,8) 

 if(theta.ge.0.0.and.theta.le.atan(hfuselg/(2*wfuselg)))then 

 qreflin(3,i)=qdir*trnsm*fopndirs*(1.-scatffi)*(1.-absorbfi) 

 qscat(3,i)=scatffi*(1.-absorbfi)*qreflin(3,i)*cos(theta) 

 elseif(theta.gt.atan(hfuselg/(2*wfuselg)).and.theta.lt. 

 & atan(hfuselg/wfuselg))then 

 qreflin(3,i)=qdir*trnsm*fopndirs*((1.-scatffi)*(1.-absorbfi)+ 

 & (1.-scatffl)*(1.-absorbfl)*(1.-scatffi)*(1.-absorbfi))

 qscat(3,i)=scatffi*(1.-absorbfi)*qreflin(3,i)*cos(theta) 

 elseif(theta.ge.atan(hfuselg/wfuselg).and.theta.lt. 

 & atan(2*hfuselg/wfuselg))then 

 qreflin(3,i)=qdir*trnsm*fopndirs*(1.-scatffl)*(1.-absorbfl) 

 & *(1.-scatffi)*(1.-absorbfi) 

 qscat(3,i)=scatffi*(1.-absorbfi)*qreflin(3,i)*cos(theta) 

 elseif(theta.ge.atan(2*hfuselg/wfuselg).and.theta.le.(fpi*90)) 

 & then 

 qreflin(3,i)=0.0 

 qscat(3,i)=0.0

 endif 

 30 continue 

c find qrefl for plexiglass surface on the top of cabin interior 

 do 40 i=1,nperiods 

 theta=fpi*envtdata(i,1) 

 qdir=envtdata(i,5) 

 fopndirs=envtdata(i,8) 

 if(theta.ge.0.0.and.theta.le.atan(hfuselg/(2*wfuselg)))then 

 qreflin(4,i)=qdir*trnsm*fopndirs*(1.-scatffl)*(1.-absorbfl) 

 & *(1.-scatffi)**4*(1.-absorbfi)**4 

 qscat(4,i)=0.0

 elseif(theta.gt.atan(hfuselg/(2*wfuselg)).and.theta.lt. 

 & atan(hfuselg/wfuselg))then 

 qreflin(4,i)=qdir*trnsm*fopndirs*((1.-scatffl)*(1.-absorbfl)* 

 & (1.-scatffi)**2*(1.-absorbfi)**2+ 

 & (1.-scatffl)*(1.-absorbfl)*(1.-scatffi)**3*(1.-absorbfi)**3) 

 qscat(4,i)=0.0

 elseif(theta.eq.atan(hfuselg/wfuselg))then 

 qreflin(4,i)=qdir*trnsm*fopndirs*(1.-scatffl)*(1.-absorbfl)* 

 & (1.-scatffi)**2*(1.-absorbfi)**2 

 qscat(4,i)=0.0

 elseif(theta.gt.atan(hfuselg/wfuselg).and.theta.lt. 

 & atan(2*hfuselg/wfuselg))then 

 qreflin(4,i)=qdir*trnsm*fopndirs*((1.-scatffl)*(1.-absorbfl)* 

 & (1.-scatffi)*(1.-absorbfi)+ 

 & (1.-scatffl)*(1.-absorbfl)*(1.-scatffi)**2*(1.-absorbfi)**2) 

A-54
 qscat(4,i)=0.0

 elseif(theta.eq.atan(2*hfuselg/wfuselg))then 

 qreflin(4,i)=qdir*trnsm*fopndirs*(1.-scatffi)*(1.-absorbfi) 

 & *(1.-scatffl)*(1.-absorbfl) 

 qscat(4,i)=0.0

 elseif(theta.gt.atan(2*hfuselg/wfuselg).and.theta.le.(fpi*90)) 

 & then 

 qreflin(4,i)=qdir*trnsm*fopndirs*((1.-scatffl)*(1.-absorbfl)+ 

 & (1.-scatffi)*(1.-absorbfi)*(1.-scatffl)*(1.-absorbfl))

 qscat(4,i)=0.0

 endif 

 40 continue 

c 

c qscat(4,i) is modified to consider direct solar rays reflected from 

c the tarmac as diffused solar rays, which subsequently hit the glass 

 do 50 i=1,nperiods 

 theta=fpi*envtdata(i,1) 

 qdir=envtdata(i,5) 

 fopndifs=a5/(a1+a2) 

 50 qscat(4,i)=trnsm*fopndifs*.5*(1.-absorbta)*qdir*sin(theta) 

c 

c now find the total scattered energy absorbed at surface A(RHS-inside) 

c , surface B(floor-bottom), surface C(LHS-inside) 

c and the plexiglass for each period. note: surface A=1; surface B=2; 

c surface C=3; and inside plexiglass surface=4. Note: surfaces 

c 5 & 6 are assumed to be ideal black bodies. they absorb all scattered

c rays coming from other surfaces and reflect none (ie qscat=0.0). note 

c that no direct rays actually falls onto these 2 surfaces 

c because of the way the plane is positioned (ie the fuselage isc perpendicular to the rays) 

 f(1,1)=fc22 

 f(1,2)=fc21 

 f(1,3)=fc24 

 f(1,4)=fc23 

 f(1,5)=fc25 

 f(1,6)=fc26 

 f(2,1)=fc12 

 f(2,2)=fc11 

 f(2,3)=fc14 

 f(2,4)=fc13 

 f(2,5)=fc15 

 f(2,6)=fc16 

 f(3,1)=fc42 

 f(3,2)=fc41 

 f(3,3)=fc44 

 f(3,4)=fc43 

 f(3,5)=fc45 

 f(3,6)=fc46 

 f(4,1)=fc32 

 f(4,2)=fc31 

 f(4,3)=fc34 

A-55
 f(4,4)=fc33 

 f(4,5)=fc35 

 f(4,6)=fc36 

 f(5,1)=fc52 

 f(5,2)=fc51 

 f(5,3)=fc54 

 f(5,4)=fc53 

 f(5,5)=fc55 

 f(5,6)=fc56 

 f(6,1)=fc62 

 f(6,2)=fc61 

 f(6,3)=fc64 

 f(6,4)=fc63 

 f(6,5)=fc65 

 f(6,6)=fc66 

 absorb(1)=absorbfi 

 absorb(2)=absorbfl 

 absorb(3)=absorbfi 

 absorb(4)=a1/(a1+a2)*absorbpg+a2/(a1+a2)*absorfbt 

 absorb(5)=1.0 

 absorb(6)=1.0 

 do 60 i=1,nperiods 

 fopndifs=a1/(a1+a2) 

c note: fopndifs=fraction open that allows diffused solar ray to get 

c out of the cabin. this value is the same as the fraction of 

c infrared sky radiation entering the cabin. 

 60 ttrnsm(4,i)=trnsm*fopndifs 

 call diffsola(f,absorb,ttrnsm,qscat,qscabsin) 

 return 

 end 

c 

c 

 subroutine diffsola(f,absorb,trnsm,e,qscabs)

c purpose: determines the diffuse solar radiation that is absorbed by the 

c surfaces. multiple absorption and reflections considered. 

c called from: qrefcab and qreffw 

c calls: none

 common/c7/nperiods,envtdata(100,20) 

 dimension f(6,6),absorb(6),trnsm(6,100),e(6,100),qscabs(6,100), 

 & sumdq(6),edum(6) 

 real intcepq(6) 

 parameter(qismall=1e-23,toldq=0.00001) 

c 

 do 150 j=1,nperiods 

c initialize 

do 20 i=1,6 

 edum(i)=e(i,j) 

 20 qscabs(i,j)=qismall 

do 110 iter=1,1000 

 do 40 i=1,6

 intcepq(i)=0.0 

A-56
 do 40 ii=1,6 

 40 intcepq(i)=intcepq(i)+f(i,ii)*edum(ii) 

 do 50 i=1,6

 sumdq(i)=absorb(i)*intcepq(i) 

 qscabs(i,j)=qscabs(i,j)+sumdq(i) 

 edum(i)=(1.-absorb(i)-trnsm(i,j))*intcepq(i) 

 if(edum(i).lt.0.0)edum(i)=0.0 

 50 continue 

 if(sumdq(1)/qscabs(1,j).le.toldq.and. 

 & sumdq(2)/qscabs(2,j).le.toldq.and. 

 & sumdq(3)/qscabs(3,j).le.toldq.and. 

 & sumdq(4)/qscabs(4,j).le.toldq.and. 

 & sumdq(5)/qscabs(5,j).le.toldq.and. 

 & sumdq(6)/qscabs(6,j).le.toldq)goto 150 

 110 continue 

 150 continue 

c do 170 j=1,nperiods 

c write(13,*)'period,e(i,j)' 

c write(13,*)'j=',j,(e(i,j),i=1,6) 

c write(13,*)'period,qscabs(i,j)' 

c 170 write(13,*)'j=',j,(qscabs(i,j),i=1,6) 

 return 

 end 

c 

 subroutine qreffw(absorbf,absorbw) 

c purpose: determines the intercepted solar radiation and the absorbed 

c diffuse radiation at the fuselage/wing junction. 

c called from: solution 

c calls: diffsola 

 common/c4/widthwng,lengtwng,hwng,locwng,lplane,lfuselg,hfuselg,

 & wfuselg 

 common/c5/trnsm,absorbta,emista,tarmd,densta,conducta,shcta 

 common/vf4/fa57,fa68 

 common/c7/nperiods,envtdata(100,20) 

 common/c9/scatffl,scatffi,scatfw,scatff 

 common/reflecq/qreflin(6,100),qscabsin(6,100),qreflfwr(6,100), 

 &qreflfwl(6,100),qscabfwr(6,100),qscabfwl(6,100),qabstwa1(6,100), 

 &qabstwa2(6,100),qabstwb(6,100) 

 common/vf2/fa11,fa12,fa13,fa14,fa15,fa16,fa21,fa22,fa23,fa24, 

 & fa25,fa26,fa31,fa32,fa33,fa34,fa35,fa36,fa41,fa42, 

 & fa43,fa44,fa45,fa46,fa51,fa52,fa53,fa54,fa55,fa56, 

 & fa61,fa62,fa63,fa64,fa65,fa66 

 common/fillet/f11w,f11f 

 dimension qscatfwr(6,100),qscatfwl(6,100) 

 dimension f(6,6),absorb(6),ttrnsm(6,100),tempab(6,100) 

 real lengtwng 

 parameter(fpi=3.141593/180.) 

c 

c Variables:

c qreflfwr(,)=intercepted solar radiation (both direct and specular) 

c by right wing 

A-57
c qreflfwl(,)=intercepted solar radiation (both direct and specular) 

c by left wing 

c qscabfwr(,)=absorbed diffuse solar radiation by the right wing 

c qscabfwl(,)=absorbed diffuse solar radiation by the left wing

c 

c initialize 

do 10 j=1,nperiods 

 do 10 i=1,6

 ttrnsm(i,j)=0.0 

 qscatfwr(i,j)=0.0 

 qscatfwl(i,j)=0.0 

 qreflfwr(i,j)=0.0 

 qreflfwl(i,j)=0.0 

 qscabfwr(i,j)=0.0 

 10 qscabfwl(i,j)=0.0 

c reference f(i,j)'s 

 f(1,1)=fa11 

 f(1,2)=fa12 

 f(1,3)=fa13 

 f(1,4)=fa14 

 f(1,5)=fa15 

 f(1,6)=fa16 

 f(2,1)=fa21 

 f(2,2)=fa22 

 f(2,3)=fa23 

 f(2,4)=fa24 

 f(2,5)=fa25 

 f(2,6)=fa26 

 f(3,1)=fa31 

 f(3,2)=fa32 

 f(3,3)=fa33 

 f(3,4)=fa34 

 f(3,5)=fa35 

 f(3,6)=fa36 

 f(4,1)=fa41 

 f(4,2)=fa42 

 f(4,3)=fa43 

 f(4,4)=fa44 

 f(4,5)=fa45 

 f(4,6)=fa46 

 f(5,1)=fa51 

 f(5,2)=fa52 

 f(5,3)=fa53 

 f(5,4)=fa54 

 f(5,5)=fa55 

 f(5,6)=fa56 

 f(6,1)=fa61 

 f(6,2)=fa62 

 f(6,3)=fa63 

 f(6,4)=fa64 

 f(6,5)=fa65 

A-58
 f(6,6)=fa66 

 absorb(3)=0.0 

 absorb(4)=0.0 

 absorb(5)=0.0 

 absorb(6)=0.0 

do 15 j=1,nperiods 

 do 15 i=3,6

 15 ttrnsm(i,j)=1.0 

 fdiffsw=1./(1.-f11w*(1.-absorbw)) 

 fdiffsf=1./(1.-f11f*(1.-absorbf)) 

 absorb(1)=absorbw*fdiffsw 

 absorb(2)=absorbf*fdiffsf 

c find qreflfwr, qreflfwl, qscabfwr and qscabfwl: 

c 

c LHS of fuselage: 

c 

c diffused solar interaction between fuselage/wing only 

c 

 do 20 i=1,nperiods 

 theta=fpi*envtdata(i,1) 

 qdir=envtdata(i,5) 

 qreflfwl(1,i)=(1.+(1.-scatff)*(1.-absorbf))*qdir 

 qscatfwl(1,i)=scatfw*(1.-absorbw)*qreflfwl(1,i)*sin(theta) 

 qreflfwl(2,i)=(1.+(1.-scatfw)*(1.-absorbw))*qdir 

 qscatfwl(2,i)=scatff*(1.-absorbf)*qreflfwl(2,i)*cos(theta) 

 20 continue

 call diffsola(f,absorb,ttrnsm,qscatfwl,qscabfwl) 

c store temporarily 

c write(13,*)'Partial diffuse solar fuselage/wing only-LHS' 

 do 25 j=1,nperiods 

c write(13,*)(qscabfwl(i,j),i=1,6) 

 do 25 i=1,6 

 25 tempab(i,j)=qscabfwl(i,j) 

c 

c now consider diffused solar reflection from the tarmac subsequently 

c absorbed by the fuselage/wing 

c 

c reinitialize

 do 125 j=1,nperiods 

 do 125 i=1,6 

 125 qscatfwl(i,j)=0.0 

 do 135 j=1,nperiods 

 theta=fpi*envtdata(j,1) 

 qdir=envtdata(j,5) 

 qtemp=.5*(1.-absorbta)*qdir*sin(theta) 

 qscatfwl(4,j)=qtemp 

 qscatfwl(5,j)=qtemp 

 135 qscatfwl(6,j)=qtemp 

 f(1,4)=0.0

 f(1,5)=0.0

 f(1,6)=0.0

A-59
 call diffsola(f,absorb,ttrnsm,qscatfwl,qscabfwl) 

c write(13,*)'Partial diffuse solar reflected from tarmac-LHS' 

 do 145 j=1,nperiods 

c write(13,*)(qscabfwl(i,j),i=1,6) 

 do 145 i=1,6 

 145 tempab(i,j)=tempab(i,j)+qscabfwl(i,j)

c replace f(1,4),f(1,5),f(1,6): 

 f(1,4)=fa14 

 f(1,5)=fa15 

 f(1,6)=fa16 

c 

c now consider diffused reflection from the fuselage (both direct ray 

c as well as from the tarmac) that fall onto surface 5 & 6 

c 

c reinitialize

 do 126 j=1,nperiods 

 do 126 i=1,6 

 126 qscatfwl(i,j)=0.0 

 do 136 j=1,nperiods 

 theta=fpi*envtdata(j,1) 

 qdir=envtdata(j,5) 

 qscatfwl(5,j)=fa57*((1.-absorbf)*(.5*(1.-absorbta)*qdir* 

 & sin(theta))+scatff*(1.-absorbf)*qdir*cos(theta)) 

 qscatfwl(6,j)=fa68*((1.-absorbf)*(.5*(1.-absorbta)*qdir* 

 & sin(theta))+scatff*(1.-absorbf)*qdir*cos(theta)) 

 136 continue 

 f(2,5)=0.0

 f(2,6)=0.0

 call diffsola(f,absorb,ttrnsm,qscatfwl,qscabfwl) 

 write(13,*)'Partial diffuse solar reflected from the fuselage', 

 & '(both direct ray as well as from the tarmac) that', 

 & 'fall onto surface 5 & 6 -LHS' 

 do 146 j=1,nperiods 

 write(13,*)(qscabfwl(i,j),i=1,6) 

 do 146 i=1,6 

 146 qscabfwl(i,j)=tempab(i,j)+qscabfwl(i,j) 

c replace f(2,5),f(2,6): 

 f(2,5)=fa25 

 f(2,6)=fa26 

c 

c RHS of fuselage: 

c 

c diffused solar interaction between fuselage/wing only 

c 

 do 30 i=1,nperiods 

 theta=fpi*envtdata(i,1) 

 qdir=envtdata(i,5) 

 if(theta.ge.atan(hfuselg/(0.5*lengtwng)))then 

 qreflfwr(1,i)=qdir 

 qscatfwr(1,i)=scatfw*(1.-absorbw)*qreflfwr(1,i)*sin(theta) 

 endif 

A-60
 qreflfwr(2,i)=0.0 

 qscatfwr(2,i)=0.0 

 30 continue

 call diffsola(f,absorb,ttrnsm,qscatfwr,qscabfwr) 

c store temporarily 

 write(13,*)'Partial diffuse solar fuselage/wing only-RHS' 

 do 325 j=1,nperiods 

 write(13,*)(qscabfwr(i,j),i=1,6) 

 do 325 i=1,6 

 325 tempab(i,j)=qscabfwr(i,j) 

c 

c now consider diffused solar reflection from the tarmac subsequently 

c absorbed by the fuselage/wing 

c 

c reinitialize

 do 225 j=1,nperiods 

 do 225 i=1,6 

 225 qscatfwr(i,j)=0.0 

 do 235 j=1,nperiods 

 theta=fpi*envtdata(j,1) 

 qdir=envtdata(j,5) 

 qtemp=.5*(1.-absorbta)*qdir*sin(theta) 

 qscatfwr(4,j)=qtemp 

 qscatfwr(5,j)=qtemp 

 235 qscatfwr(6,j)=qtemp 

 f(1,4)=0.0

 f(1,5)=0.0

 f(1,6)=0.0

 call diffsola(f,absorb,ttrnsm,qscatfwr,qscabfwr) 

 write(13,*)'Partial diffuse solar reflected from tarmac-RHS' 

 do 245 j=1,nperiods 

 write(13,*)(qscabfwr(i,j),i=1,6) 

 do 245 i=1,6 

 245 tempab(i,j)=tempab(i,j)+qscabfwr(i,j)

c replace f(1,4),f(1,5),f(1,6) 

 f(1,4)=fa14 

 f(1,5)=fa15 

 f(1,6)=fa16 

c 

c now consider diffused reflection from the fuselage (originating 

c from the tarmac) that fall onto surface 5 & 6 

c 

c reinitialize

 do 226 j=1,nperiods 

 do 226 i=1,6 

 226 qscatfwr(i,j)=0.0 

 do 236 j=1,nperiods 

 theta=fpi*envtdata(j,1) 

 qdir=envtdata(j,5) 

 qscatfwr(5,j)=fa57*(1.-absorbf)*(.5*(1.-absorbta)*qdir* 

 & sin(theta)) 

A-61
 qscatfwr(6,j)=fa68*(1.-absorbf)*(.5*(1.-absorbta)*qdir* 

 & sin(theta)) 

 236 continue 

 f(2,5)=0.0

 f(2,6)=0.0

 call diffsola(f,absorb,ttrnsm,qscatfwr,qscabfwr) 

 write(13,*)'Partial diffuse solar reflected from the fuselage', 

 & '(originating from the tarmac) that', 

 & 'fall onto surface 5 & 6 -LHS' 

 do 246 j=1,nperiods 

 write(13,*)(qscabfwr(i,j),i=1,6) 

 do 246 i=1,6 

 246 qscabfwr(i,j)=tempab(i,j)+qscabfwr(i,j) 

 return 

 end 

c 

 subroutine qreftw(absorbw) 

c purpose: determines the intercepted solar radiation and the absorbed 

c diffuse radiation between the wing bottom and the tarmac. 

c called from: solution 

c calls: difslwng 

 common/c5/trnsm,absorbta,emista,tarmd,densta,conducta,shcta 

 common/c6/absorbff,emisff,absorbww,emisww,absorbfl,emisfl,absorbfi 

 &,emisfi,absorbpg,emispg,absorftp,emisrftp,absorfbt,emisrfbt 

 &,absorbfb,emisfb 

 common/c7/nperiods,envtdata(100,20) 

 common/vf1/fbwaa(6,6),fbwbb(6,6) 

 common/reflecq/qreflin(6,100),qscabsin(6,100),qreflfwr(6,100), 

 &qreflfwl(6,100),qscabfwr(6,100),qscabfwl(6,100),qabstwa1(6,100), 

 &qabstwa2(6,100),qabstwb(6,100) 

 dimension fbwa1(6,6),fbwa2(6,6),fbwb(6,6), 

 &e1(6,100),e2(6,100),e3(6,100) 

 dimension absorb1(6),absorb2(6),absorb3(6),trnsm1(6), 

 &trnsm2(6),trnsm3(6) 

 parameter(fpi=3.141593/180.) 

c 

c Variables:

c qabstwa1(,)=intercepted solar radiation (direct)

c by the tarmac underneath the wing (LHS) 

c qabstwa2(,)=intercepted solar radiation (direct)

c by the tarmac underneath the wing (RHS) 

c qabstwb(,)=intercepted solar radiation (direct) 

c by the tarmac underneath the wing(CENTER) 

c 

c initialize the diffused solar energy entering each surface 

 do 10 j=1,nperiods 

 do 10 i=1,6 

 e1(i,j)=0.0 

 e2(i,j)=0.0 

 10 e3(i,j)=0.0 

c initialize/reference absorb, trnsm & fbw 

A-62
 do 20 i=1,6 

 absorb1(i)=0.0 

 absorb2(i)=0.0 

 absorb3(i)=0.0 

 trnsm1(i)=1.0 

 trnsm2(i)=1.0 

 trnsm3(i)=1.0 

do 20 j=1,6 

 fbwa1(i,j)=fbwaa(i,j) 

 fbwa2(i,j)=fbwaa(i,j) 

 fbwb(i,j)=fbwbb(i,j) 

 20 continue

c find diffused solar energy entering each surface, e(i,j)'s 

 do 30 i=1,nperiods 

 theta=fpi*envtdata(i,1) 

 qdir=envtdata(i,5) 

 qqscat=0.5*(1.-absorbta)*qdir*sin(theta) 

 e1(4,i)=qqscat 

 e1(5,i)=qqscat 

 e1(6,i)=qqscat 

 e2(2,i)=qqscat 

 e2(5,i)=qqscat 

 e2(6,i)=qqscat 

 e3(5,i)=qqscat 

 e3(6,i)=qqscat 

 30 continue

c input absorb 

 absorb1(1)=absorbta 

 absorb2(1)=absorbta 

 absorb3(1)=absorbta 

c input trnsm 

 trnsm1(1)=0.0 

 trnsm1(3)=0.0 

 trnsm2(1)=0.0 

 trnsm2(3)=0.0 

 trnsm3(1)=0.0 

 trnsm3(3)=0.0 

c some of the reflected diffused solar radiation does not affect

c surface 1 right away: because f14=f41*a4/a1, therefore if f41=0, f14 

c should also = 0. 

c fbwa1(4,1)=0.0 

c fbwa1(5,1)=0.0 

c fbwa1(6,1)=0.0 

c fbwa2(2,1)=0.0 

c fbwa2(5,1)=0.0 

c fbwa2(6,1)=0.0 

c fbwb(5,1)=0.0 

c fbwb(6,1)=0.0 

 fbwa1(1,4)=0.0 

 fbwa1(1,5)=0.0 

 fbwa1(1,6)=0.0 

A-63
 fbwa2(1,2)=0.0 

 fbwa2(1,5)=0.0 

 fbwa2(1,6)=0.0 

 fbwb(1,5)=0.0 

 fbwb(1,6)=0.0 

c input absorb that changes 

 absorb1(3)=absorbw 

 absorb2(3)=absorbw 

 absorb3(3)=absorbfb 

 call difslwng(fbwa1,fbwa2,fbwb,absorb1,absorb2,absorb3, 

 &trnsm1,trnsm2,trnsm3,e1,e2,e3,qabstwa1,qabstwa2,qabstwb) 

 return 

 end 

c 

 subroutine difslwng(fbwa1,fbwa2,fbwb,absorb1,absorb2,absorb3, 

 &trnsm1,trnsm2,trnsm3,e1,e2,e3,qabstwa1,qabstwa2,qabstwb) 

c purpose: determines the diffuse solar radiation that is absorbed by the 

c wing bottom and tarmac surfaces. multiple absorption and

c reflections considered. region divided into 3 portions 

c called from: qreftw 

c calls: none

 common/c7/nperiods,envtdata(100,20) 

 dimension fbwa1(6,6),fbwa2(6,6),fbwb(6,6), 

 & absorb1(6),absorb2(6),absorb3(6),

 & trnsm1(6),trnsm2(6),trnsm3(6), 

 & e1(6,100),e2(6,100),e3(6,100), 

 & qabstwa1(6,100),qabstwa2(6,100),qabstwb(6,100) 

 dimension sumdq1(6),sumdq2(6),sumdq3(6), 

 & edum1(6),edum2(6),edum3(6), 

 & trnsmtd1(6),trnsmtd2(6),trnsmtd3(6) 

 real intcepq1(6),intcepq2(6),intcepq3(6) 

 parameter(qismall=1e-23,toldq=0.00001) 

c 

 do 150 j=1,nperiods 

c initialize 

do 20 i=1,6 

 edum1(i)=e1(i,j) 

 edum2(i)=e2(i,j) 

 edum3(i)=e3(i,j) 

 qabstwa1(i,j)=qismall 

 qabstwa2(i,j)=qismall 

 qabstwb(i,j)=qismall 

 20 continue 

 igo1=0 

 igo2=0 

 igo3=0 

do 110 iter=1,1000 

 do 40 i=1,6

 intcepq1(i)=0.0 

 intcepq2(i)=0.0 

 intcepq3(i)=0.0 

A-64
 do 40 ii=1,6 

 intcepq1(i)=intcepq1(i)+fbwa1(i,ii)*edum1(ii) 

 intcepq2(i)=intcepq2(i)+fbwa2(i,ii)*edum2(ii) 

 intcepq3(i)=intcepq3(i)+fbwb(i,ii)*edum3(ii) 

 40 continue 

 trnsmtd1(2)=trnsm1(2)*intcepq1(2) 

 trnsmtd2(4)=trnsm2(4)*intcepq2(4) 

 trnsmtd3(4)=trnsm3(4)*intcepq3(4) 

 trnsmtd3(2)=trnsm3(2)*intcepq3(2) 

c 

 do 50 i=1,6

c LHS (Section 1): 

 if(igo1.eq.0)then 

 sumdq1(i)=absorb1(i)*intcepq1(i)

 qabstwa1(i,j)=qabstwa1(i,j)+sumdq1(i) 

 if(i.eq.2)then 

 edum1(2)=trnsmtd3(4) 

 else 

 edum1(i)=(1.-absorb1(i)-trnsm1(i))*intcepq1(i) 

 endif 

 endif 

c RHS (section 2): 

 if(igo2.eq.0)then 

 sumdq2(i)=absorb2(i)*intcepq2(i)

 qabstwa2(i,j)=qabstwa2(i,j)+sumdq2(i) 

 if(i.eq.4)then 

 edum2(4)=trnsmtd3(2) 

 else 

 edum2(i)=(1.-absorb2(i)-trnsm2(i))*intcepq2(i) 

 endif 

 endif 

c CENTER (underneath the fuselage; Section 3):

 if(igo3.eq.0)then 

 sumdq3(i)=absorb3(i)*intcepq3(i)

 qabstwb(i,j)=qabstwb(i,j)+sumdq3(i) 

 if(i.eq.2)then 

 edum3(2)=trnsmtd2(4) 

 elseif(i.eq.4)then 

 edum3(4)=trnsmtd1(2) 

 else 

 edum3(i)=(1.-absorb3(i)-trnsm3(i))*intcepq3(i) 

 endif 

 endif 

 50 continue 

c LHS (Section 1): 

 if(sumdq1(1)/qabstwa1(1,j).le.toldq.and. 

 & sumdq1(2)/qabstwa1(2,j).le.toldq.and. 

 & sumdq1(3)/qabstwa1(3,j).le.toldq.and. 

 & sumdq1(4)/qabstwa1(4,j).le.toldq.and. 

 & sumdq1(5)/qabstwa1(5,j).le.toldq.and. 

 & sumdq1(6)/qabstwa1(6,j).le.toldq)igo1=1 

A-65
c RHS (section 2): 

 if(sumdq2(1)/qabstwa2(1,j).le.toldq.and. 

 & sumdq2(2)/qabstwa2(2,j).le.toldq.and. 

 & sumdq2(3)/qabstwa2(3,j).le.toldq.and. 

 & sumdq2(4)/qabstwa2(4,j).le.toldq.and. 

 & sumdq2(5)/qabstwa2(5,j).le.toldq.and. 

 & sumdq2(6)/qabstwa2(6,j).le.toldq)igo2=1 

c CENTER (underneath the fuselage; Section 3):

 if(sumdq3(1)/qabstwb(1,j).le.toldq.and. 

 & sumdq3(2)/qabstwb(2,j).le.toldq.and. 

 & sumdq3(3)/qabstwb(3,j).le.toldq.and. 

 & sumdq3(4)/qabstwb(4,j).le.toldq.and. 

 & sumdq3(5)/qabstwb(5,j).le.toldq.and. 

 & sumdq3(6)/qabstwb(6,j).le.toldq)igo3=1 

 if(igo1.eq.1.and.igo2.eq.1.and.igo3.eq.1)goto 150 

 110 continue 

 150 continue 

c 

 write(13,*)'LHS (Section 1): ' 

 do 170 j=1,nperiods 

 write(13,*)'period,e1(i,j)' 

 write(13,*)'j=',j,(e1(i,j),i=1,6) 

 write(13,*)'period,qabstwa1(i,j)' 

 170 write(13,*)'j=',j,(qabstwa1(i,j),i=1,6) 

 write(13,*)'RHS (section 2):' 

 do 180 j=1,nperiods 

 write(13,*)'period,e2(i,j)' 

 write(13,*)'j=',j,(e2(i,j),i=1,6) 

 write(13,*)'period,qabstwa2(i,j)' 

 180 write(13,*)'j=',j,(qabstwa2(i,j),i=1,6) 

 write(13,*)'CENTER (underneath the fuselage; Section 3): ' 

 do 190 j=1,nperiods 

 write(13,*)'period,e3(i,j)' 

 write(13,*)'j=',j,(e3(i,j),i=1,6) 

 write(13,*)'period,qabstwb(i,j)' 

 190 write(13,*)'j=',j,(qabstwb(i,j),i=1,6) 

 return 

 end 

c 

 subroutine sbfillet(icodef11,percent,f11w,f11f) 

c purpose: determines the view factors for a portion of a quadrant surface to 

c be used for determining the effective absorbtivities 

c and emissivities 

c called from: solution 

c calls: none

 real l 

 parameter(fpi=3.141592654/180, r=1.0,l=1000.) 

c 

c Variables:

c icodef11=1: the normal scenario where the wing paint extend beyond 

c the fillet-fuselage junction. the wing will 

A-66
c be assigned an (infrared & diffused solar) absorbtivity, 

c and emissivity modified by the fillet factor, 

c the fuselage will have f11f=0.0, implying that the flat 

c absorbtivity and emissivity (normal) will be used throughout 

c the fuselage. 

c 

c icodef11=2: the abnormal scenario where the fuselage paint extend 

c beyond the fillet-wing junction, and into the wing to 

c some extend. the fuselage will 

c be assigned an (infrared & diffused solar) absorbtivity, 

c and emissivity modified by the fillet factor, whereas the 

c wing will have f11w=0.0, implying that the flat absorbtivity 

c and emissivity (normal) will be used throughout the wing. 

c 

c icodef11=3: the wing-paint/fuselage paint junction is within the fillet 

c region. you need the percentage of the fillet occupied by the 

c wing paint (percentage range from 0 to 100%). 

c 

c r &l=just dummies 

c 

 if(icodef11.eq.1)percent=100. 

 if(icodef11.eq.2)percent=0.0 

 theta=.9*percent 

c wing region (region a):

c initialize: 

 f11=0.0 

 if(theta.eq.0)goto 10 

 w1=theta*fpi*r 

 a1=w1*l 

 w2=2.*r*sin(fpi*theta/2.) 

 a2=w2*l 

 f21=1. 

 f12=a2/a1*f21 

 f11=1.-f12 

 10 f11w=f11 

c write(9,*)'wing region, theta, f11w',theta,f11w 

c fuselage region (region b): 

c initialize 

 f11=0.0 

 if(theta.eq.90.)goto 20

 theta1=90.-theta 

 w1=theta1*fpi*r 

 a1=w1*l 

 w2=2.*r*sin(fpi*theta1/2.) 

 a2=w2*l 

 f21=1. 

 f12=a2/a1*f21 

 f11=1.-f12 

 20 f11f=f11

c write(9,*)'fuselage region, theta, f11f',theta,f11f 

 return 

A-67
 end 

subroutine trnsien1(itotitr,iperiod,theta,ta,qsky, 

 & hhfo,hhwo,hhto,qdir,qtarm,ttarm,absorbf,emisf,absorbw, 

 & emisw,eemisf,eemisw,rff,rfw) 

c purpose: controls the transient analysis 

c called from: solution 

c calls: lhswng1, rhswng1, fuslglr1, fuslgrf1, tarmac1, updradio 

 common/c1/n,ntrial,tolx,tolf 

 common/c4/widthwng,lengtwng,hwng,locwng,lplane,lfuselg,hfuselg,

 & wfuselg 

 common/c7/nperiods,envtdata(100,20) 

 common/c12/a1,a2,a3,a4,a5,a6(100),a7(100) 

 common/c13/nintervl,vel(100),timeintv(100) 

 common/c16/ifindiff,deltime,amcabtmp,intvprnt 

 common/temp1/t1(1000) 

 common/temp2/t2(1000) 

 common/temp3/t3(1000) 

 common/temp4/t4(1000),tpg1,tpg2,tpg3 

 common/temp5/t5(1000) 

 common/radiosit/radiosj(14) 

 common/data1/ddeltime 

 common/result/temp(200),temptrn(200) 

 dimension tvel(100),tottime(100) 

 real lfuselg,lplane 

 parameter(fhhc=1.0) 

 parameter(sunf=0.5,sunf2=0.5) 

 parameter(temptoln=0.000001) 

c 

c Variables:

c t1()...t5()=temperatures 

c radiosj()=updated radiosity 

c deltime=time increment in transient analysis (sec) 

c temp=steady state temperature 

c temptrn=transient state temperature 

c amcabtmp=ambient cab temperature (Rankine) 

c fhhc=factor for convection in cabin

c temptoln=difference between the current ambient cabin temp and the 

c previous one: to prevent infinite loop 

c intvprnt=intervals for printing purposes 

c sunf=assume there is a fifty percent chance that a spot will be under 

c direct sun throughout the cooling phase. 

c sunf2=only fifty percent of the sun gets into the cabin. or take it 

c the average between nothing getting through to all of the direct rays 

c getting through. 

c 

 if(itotitr.eq.1)then 

 ddeltime=deltime/3600 

c convert the velocity from mph to ft/sec 

do 50 i=1,nintervl+1 

 50 tvel(i)=vel(i)*5280./3600 

A-68
c find the total time from taxi till the application of limit load & beyond 

 tottime(1)=0.0 

do 60 i=1,nintervl 

 60 tottime(i+1)=tottime(i)+timeintv(i) 

 endif 

c if(iperiod.ne.3)return 

c 

c formulate transient cooling: 

c 

c initial cooling when the door is opened 

c 

 icool=1 

c sunfact is assumed to be always=1, during initial cooling 

 sunfact=1.0 

 if(itotitr.le.10*nperiods)then 

 do 15 i=1,2 

 write(13+i,*)'Total # of analysis=',itotitr 

 write(13+i,*)'TRANSIENT ANALYSIS-Temperatures' 

 15 write(13+i,*)'cooling when door is opened, iperiod=',iperiod 

c write(16,*)'Total # of analysis=',itotitr 

c write(16,*)'TRANSIENT ANALYSIS-Radiosity' 

c write(16,*)'cooling when door is opened, iperiod=',iperiod 

 write(10,*)'' 

 write(10,*)'' 

 write(10,*)'TRANSIENT TEMPERATURES FOR PERIOD:',iperiod 

 write(10,*)'' 

 write(10,*)'COOLING WHILE DOOR IS OPENED:' 

 write(10,*)'' 

 write(10,90)(i,i=1,53)

 write(14,90)(i,i=1,20)

 write(15,90)(i,i=21,53) 

c write(16,90)(i,i=54,67) 

 90 format(1x,'time(sec)',1x,500(1x,i5)) 

 endif 

 do 75 i=1,n 

 75 temptrn(i)=temp(i) 

 iprint=1 

 nprn=0 

 istep=0 

c find convective coefficient in the cabin; assume it to be a function of: 

 hhc=fhhc*(hhfo+hhwo+hhto)/3. 

 if(itotitr.eq.1)then 

c find weights for tend and tambcab 

 aa1=lfuselg*wfuselg 

 aa2=a1 

 aa3=lfuselg*hfuselg 

 aa4=aa3 

 aa5=a2 

 atot=aa1+aa2+aa3+aa4+aa5 

 cc1=aa1/atot

 cc2=aa2/atot

A-69
 cc3=aa3/atot

 cc4=aa4/atot

 cc5=aa5/atot

 aa6=hfuselg*wfuselg 

 aa7=aa6 

 atot1=aa1+aa2+aa3+aa4+aa5+aa6+aa7 

 dd1=aa1/atot1 

 dd2=aa2/atot1 

 dd3=aa3/atot1 

 dd4=aa4/atot1 

 dd5=aa5/atot1 

 dd6=aa6/atot1 

 dd7=aa7/atot1 

 endif 

 tempert2=1000.0 

 10 istep=istep+1 

 call lhswng1(itotitr,iperiod,istep,theta,ta,hhwo,rfw,absorbw, 

 & emisw,eemisw,sunfact) 

 call rhswng1(itotitr,iperiod,istep,theta,ta,hhwo,rfw,absorbw, 

 & emisw,eemisw,sunfact) 

 call fuslglr1(itotitr,icool,iperiod,istep,theta,ta,hhfo,hhc,rff, 

 & absorbf,eemisf,sunfact,sunfact2)

 call fuslgrf1(itotitr,icool,iperiod,istep,theta,ta,hhfo,hhc,qdir, 

 & qsky,qtarm,sunfact2) 

 call tarmac1(itotitr,icool,iperiod,istep,ta,ttarm,hhto) 

 call updradio(itotitr,iperiod,istep,qsky,qtarm,emisf, 

 & emisw,eemisf,eemisw,rff,rfw) 

 if(iprint.eq.istep)then 

 nprn=1+nprn 

 iprint=nprn*intvprnt+1 

 if(itotitr.le.10*nperiods)then 

 write(10,80)(istep-1)*deltime,(temptrn(i),i=1,53)

 write(14,80)(istep-1)*deltime,(temptrn(i),i=1,20)

 write(15,80)(istep-1)*deltime,(temptrn(i),i=21,53) 

c write(16,80)(istep-1)*deltime,(temptrn(i),i=54,67) 

 80 format(f10.4,2x,500(1x,f5.1)) 

 endif 

 endif 

 tend=cc1*t4(16)+cc2*tpg3+cc3*t3(12)+cc4*t3(16)+cc5*t4(12) 

 tempert1=tempert2 

 tambcab=dd1*t4(16)+dd2*tpg3+dd3*t3(12)+dd4*t3(16)+dd5*t4(12) 

 & +dd6*tend+dd7*tend 

 tempert2=tambcab 

 write(14,*)'istep,t1,t2, absdiff',istep,tempert1,tempert2, 

 & abs(tempert1-tempert2) 

 if(tambcab.le.amcabtmp)goto 100 

 goto 10 

c 

c cooling during taxi, stop, take-off, climb & cruise 

c 

 100 continue 

A-70
 if(itotitr.le.10*nperiods)then 

 do 16 i=1,3 

 16 write(13+i,*)'cooling during taxi, stop, take-off, climb & cruise, 

 &iperiod=',iperiod 

 write(10,*)'' 

 write(10,*)'COOLING DURING AIRCRAFT MANEUVERS:' 

 write(10,*)'' 

 write(10,90)(i,i=1,53)

 write(14,90)(i,i=1,20)

 write(15,90)(i,i=21,53) 

c write(16,90)(i,i=54,67) 

 endif 

 icool=2 

 sunfact=sunf 

 sunfact2=sunf2 

 vnatural=envtdata(iperiod,4) 

 v1=tvel(1) 

 time1=0.0 

 20 istep=istep+1 

c find the convective coefficients for the fuselage and the wing, hhf & hhw 

 time1=time1+deltime 

 if(time1.gt.tottime(nintervl+1))goto 120 

do 130 j=1,nintervl 

 if(time1.gt.tottime(j).and.time1.le.tottime(j+1))then 

 jj=j 

 goto 140 

 endif 

 130 continue 

 140 v2=tvel(jj)+(tvel(jj+1)-tvel(jj))/timeintv(jj)*(time1-tottime(jj)) 

 & + vnatural 

 v=(v1+v2)/2.

 v1=v2 

 call hcoolcon(itotitr,lplane,v,hhf) 

 call hcoolcon(itotitr,widthwng,v,hhw) 

c 

 call lhswng1(itotitr,iperiod,istep,theta,ta,hhw,rfw,absorbw, 

 & emisw,eemisw,sunfact) 

 call rhswng1(itotitr,iperiod,istep,theta,ta,hhw,rfw,absorbw, 

 & emisw,eemisw,sunfact) 

 call fuslglr1(itotitr,icool,iperiod,istep,theta,ta,hhf,hhc,rff, 

 & absorbf,eemisf,sunfact,sunfact2) 

 call fuslgrf1(itotitr,icool,iperiod,istep,theta,ta,hhf,hhc,qdir, 

 & qsky,qtarm,sunfact2) 

 call tarmac1(itotitr,icool,iperiod,istep,ta,ttarm,hhto) 

 call updradio(itotitr,iperiod,istep,qsky,qtarm,emisf, 

 & emisw,eemisf,eemisw,rff,rfw) 

 if(iprint.eq.istep)then 

 nprn=1+nprn 

 iprint=nprn*intvprnt+1 

 if(itotitr.le.10*nperiods)then 

 write(10,80)(istep-1)*deltime,(temptrn(i),i=1,53)

A-71
 write(14,80)(istep-1)*deltime,(temptrn(i),i=1,20)

 write(15,80)(istep-1)*deltime,(temptrn(i),i=21,53) 

c write(16,80)(istep-1)*deltime,(temptrn(i),i=54,67) 

 endif 

 endif 

 goto 20 

c replace n: note n was changed to 14 in the updradio subroutine 

 120 n=67 

 return 

 end 

c 

c 

 subroutine trnsien2(itotitr,iperiod,theta,ta,qsky, 

 & hhfo,hhwo,hhto,qdir,qtarm,ttarm,absorbf,emisf,absorbw, 

 & emisw,eemisf,eemisw,rff,rfw) 

c purpose: controls the transient analysis 

c called from: solution 

c calls: lhswng2, rhswng2, fuslglr2, fuslgrf2, tarmac2, updradio 

 common/c1/n,ntrial,tolx,tolf 

 common/c4/widthwng,lengtwng,hwng,locwng,lplane,lfuselg,hfuselg,

 & wfuselg 

 common/c7/nperiods,envtdata(100,20) 

 common/c12/a1,a2,a3,a4,a5,a6(100),a7(100) 

 common/c13/nintervl,vel(100),timeintv(100) 

 common/c16/ifindiff,deltime,amcabtmp,intvprnt 

 common/temp1/t1(1000) 

 common/temp2/t2(1000) 

 common/temp3/t3(1000) 

 common/temp4/t4(1000),tpg1,tpg2,tpg3 

 common/temp5/t5(1000) 

 common/radiosit/radiosj(14) 

 common/data1/ddeltime 

 common/result/temp(200),temptrn(200) 

 dimension tvel(100),tottime(100) 

 real lfuselg,lplane 

 parameter(fhhc=1.0) 

 parameter(sunf=0.5,sunf2=0.5) 

 parameter(temptoln=0.01) 

c parameter(temptoln=0.000001) 

c 

c Variables:

c t1()...t5()=temperatures 

c radiosj()=updated radiosity 

c deltime=time increment in transient analysis (sec) 

c temp=steady state temperature 

c temptrn=transient state temperature 

c amcabtmp=ambient cab temperature (Rankine) 

c fhhc=factor for convection in cabin

c temptoln=difference between the current ambient cabin temp and the 

c previous one: to prevent infinte loop

c intvprnt=intervals for printing purposes 

A-72
c sunf=assume there is a fifty percent chance that a spot will be under 

c direct sun throughout the cooling phase. 

c sunf2=only fifty percent of the sun gets into the cabin. or take it 

c the average between nothing geting through to all of the direct rays 

c getting through. 

c 

 if(itotitr.eq.1)then 

 ddeltime=deltime/3600 

c convert the velocity from mph to ft/sec 

do 50 i=1,nintervl+1 

 50 tvel(i)=vel(i)*5280./3600 

c find the total time from taxi till the application of limit load & beyond 

 tottime(1)=0.0 

do 60 i=1,nintervl 

 60 tottime(i+1)=tottime(i)+timeintv(i) 

 endif 

c if(iperiod.ne.3)return 

c 

c formulate transient cooling: 

c 

c initial cooling when the door is opened 

c 

 icool=1 

c sunfact is assumed to be always=1, during initial cooling 

 sunfact=1.0 

 if(itotitr.le.10*nperiods)then 

 do 15 i=1,2 

 write(13+i,*)'Total # of analysis=',itotitr 

 write(13+i,*)'TRANSIENT ANALYSIS-Temperatures' 

 15 write(13+i,*)'cooling when door is opened, iperiod=',iperiod 

 write(16,*)'Total # of analysis=',itotitr 

 write(16,*)'TRANSIENT ANALYSIS-Radiosity' 

 write(16,*)'cooling when door is opened, iperiod=',iperiod

 write(10,*)'' 

 write(10,*)'' 

 write(10,*)'TRANSIENT TEMPERATURES FOR PERIOD:',iperiod 

 write(10,*)'' 

 write(10,*)'COOLING WHILE DOOR IS OPENED:' 

 write(10,*)'' 

 write(10,90)(i,i=1,53)

 write(14,90)(i,i=1,20)

 write(15,90)(i,i=21,53) 

 write(16,90)(i,i=54,67) 

 90 format(1x,'time(sec)',1x,500(1x,i5)) 

 endif 

 do 75 i=1,n 

 75 temptrn(i)=temp(i) 

 iprint=1 

 nprn=0 

 istep=0 

c find convective coefficient in the cabin; assume it to be a function of: 

A-73
 hhc=fhhc*(hhfo+hhwo+hhto)/3. 

 if(itotitr.eq.1)then 

c find weights for tend and tambcab 

 aa1=lfuselg*wfuselg 

 aa2=a1 

 aa3=lfuselg*hfuselg 

 aa4=aa3 

 aa5=a2 

 atot=aa1+aa2+aa3+aa4+aa5 

 cc1=aa1/atot

 cc2=aa2/atot

 cc3=aa3/atot

 cc4=aa4/atot

 cc5=aa5/atot

 aa6=hfuselg*wfuselg 

 aa7=aa6 

 atot1=aa1+aa2+aa3+aa4+aa5+aa6+aa7 

 dd1=aa1/atot1 

 dd2=aa2/atot1 

 dd3=aa3/atot1 

 dd4=aa4/atot1 

 dd5=aa5/atot1 

 dd6=aa6/atot1 

 dd7=aa7/atot1 

 endif 

 tempert2=1000.0 

 10 istep=istep+1 

 call newraph2(itotitr,iperiod,istep,theta,ta,hhwo,rfw,absorbw, 

 & emisw,eemisw,sunfact,icool,hhfo,hhc,rff,absorbf, 

 & eemisf,sunfact2,qdir,qsky,qtarm,ttarm,hhto) 

 call updradio(itotitr,iperiod,istep,qsky,qtarm,emisf, 

 & emisw,eemisf,eemisw,rff,rfw) 

 if(iprint.eq.istep)then 

 nprn=1+nprn 

 iprint=nprn*intvprnt+1 

 if(itotitr.le.10*nperiods)then 

 write(10,80)(istep-1)*deltime,(temptrn(i),i=1,53)

 write(14,80)(istep-1)*deltime,(temptrn(i),i=1,20)

 write(15,80)(istep-1)*deltime,(temptrn(i),i=21,53) 

 write(16,80)(istep-1)*deltime,(temptrn(i),i=54,67) 

 80 format(f10.4,2x,500(1x,f5.1)) 

 endif 

 endif 

 tend=cc1*t4(16)+cc2*tpg3+cc3*t3(12)+cc4*t3(16)+cc5*t4(12) 

 tempert1=tempert2 

 tambcab=dd1*t4(16)+dd2*tpg3+dd3*t3(12)+dd4*t3(16)+dd5*t4(12) 

 & +dd6*tend+dd7*tend 

 tempert2=tambcab 

 write(14,*)'tempert1,tempert2, absdiff',tempert1,tempert2, 

 & abs(tempert1-tempert2) 

 if(tambcab.le.amcabtmp)goto 100 

A-74
 goto 10 

c 

c cooling during taxi, stop, take-off, climb & cruise 

c 

 100 continue 

 if(itotitr.le.10*nperiods)then 

 do 16 i=1,3 

 16 write(13+i,*)'cooling during taxi, stop, take-off, climb & cruise, 

 &iperiod=',iperiod 

 write(10,*)'' 

 write(10,*)'COOLING DURING AIRCRAFT MANEUVERS:' 

 write(10,*)'' 

 write(10,90)(i,i=1,53)

 write(14,90)(i,i=1,20)

 write(15,90)(i,i=21,53) 

 write(16,90)(i,i=54,67) 

 endif 

 icool=2 

 sunfact=sunf 

 sunfact2=sunf2 

 vnatural=envtdata(iperiod,4) 

 v1=tvel(1) 

 time1=0.0 

 20 istep=istep+1 

c find the convective coefficients for the fuselage and the wing, hhf & hhw 

 time1=time1+deltime 

 if(time1.gt.tottime(nintervl+1))goto 120 

do 130 j=1,nintervl 

 if(time1.gt.tottime(j).and.time1.le.tottime(j+1))then 

 jj=j 

 goto 140 

 endif 

 130 continue 

 140 v2=tvel(jj)+(tvel(jj+1)-tvel(jj))/timeintv(jj)*(time1-tottime(jj)) 

 & + vnatural 

 v=(v1+v2)/2.

 v1=v2 

 call hcoolcon(itotitr,lplane,v,hhf) 

 call hcoolcon(itotitr,widthwng,v,hhw) 

c 

 call newraph2(itotitr,iperiod,istep,theta,ta,hhw,rfw,absorbw, 

 & emisw,eemisw,sunfact,icool,hhf,hhc,rff,absorbf, 

 & eemisf,sunfact2,qdir,qsky,qtarm,ttarm,hhto) 

 call updradio(itotitr,iperiod,istep,qsky,qtarm,emisf, 

 & emisw,eemisf,eemisw,rff,rfw) 

 if(iprint.eq.istep)then 

 nprn=1+nprn 

 iprint=nprn*intvprnt+1 

 if(itotitr.le.10*nperiods)then 

 write(10,80)(istep-1)*deltime,(temptrn(i),i=1,53)

A-75
 write(14,80)(istep-1)*deltime,(temptrn(i),i=1,20)

 write(15,80)(istep-1)*deltime,(temptrn(i),i=21,53) 

 write(16,80)(istep-1)*deltime,(temptrn(i),i=54,67) 

 endif 

 endif 

 goto 20 

c restore n to 67: note n was changed to 120 in the newraph2 subroutine and 

c to 14 in the updradio subroutine 

 120 n=67 

 return 

 end 

c 

c 

 subroutine newraph2(itotitr,iperiod,istep,theta,ta,hhwo,rfw, 

 & absorbw,emisw,eemisw,sunfact,icool,hhfo,hhc,rff, 

 & absorbf,eemisf,sunfact2,qdir,qsky,qtarm,ttarm,hhto) 

c 

c purpose: this subroutine calls: linear solvers; ludcmp and lubksb. 

c given an initial guess x for a root in n dimensions, take 

c ntrial Newton-Raphson steps to improve the root. stop if the 

c root converges in either summed absolute variable increments tolx or summed 

c absolute function values tolf. 

c calls: lhswng2,rhswng2,fuslglr2,fuslgrf2,ludcmp,tarmac2,lubksb 

c called from: trnsien2 

 common/c1/n,ntrial,tolx,tolf 

 common/c2/x(200),fvec(200),jacob(200,200) 

 common/sc1/indx(200),p(200) 

 real jacob 

c 

c change n to 120 to reflect the 120 equations related to the 120 temperatures to be

c found for each time step, istep for the implicit finite difference method. 

c 

 n=120 

c 

c Variables:

c isolfoun=indicator of whether solution is found or not: =1 means found; 

c =0 means not found yet. 

c ipower=exponent of 10. to ensure solution converges in positive numbers 

c 

 isolfoun=0 

 ipower=1 

 50 do 14 k=1,ntrial 

c user subroutines supply function values at x in fvec and 

c Jacobian matrix in jacob. 

c 

c initialize jacobian jacob(120,120) 

c note: most jacobian values are 0's. also note that the local jacobians 

c tjacob in subroutines below are also initialized to 0's. this initialization 

c is redundant but simplifies programming. 

c 

 do 70 ijacob=1,n 

A-76
 do 70 jjacob=1,n 

 70 jacob(ijacob,jjacob)=0.0 

c 

 call lhswng2(k,itotitr,iperiod,istep,theta,ta,hhwo,rfw,absorbw, 

 & emisw,eemisw,sunfact,isolfoun) 

 call rhswng2(k,itotitr,iperiod,istep,theta,ta,hhwo,rfw,absorbw, 

 & emisw,eemisw,sunfact,isolfoun) 

 call fuslglr2(k,itotitr,icool,iperiod,istep,theta,ta,hhfo,hhc,rff, 

 & absorbf,eemisf,sunfact,sunfact2,isolfoun) 

 call fuslgrf2(k,itotitr,icool,iperiod,istep,theta,ta,hhfo,hhc,qdir 

 & ,qsky,qtarm,sunfact2,isolfoun) 

 call tarmac2(k,itotitr,icool,iperiod,istep,ta,ttarm,hhto,isolfoun)

c return to trnsien2 for the first step (istep=1) where only initialization has 

c taken place in the above called subroutines. note that the local temperature 

c variables have assumed the steady state temperature tt(i) (ie x(i)) in the 

c the above subroutines. after step 1, the steady state temperature values are 

c all destroyed, and instead filled up with the new temperature values, newt(i). 

c to retrieve the steady state temperatures, find them in /result/temp(i), 

c where temp(i) is in deg Fahrenheit, and not in absolute Rankine. 

 write(14,*)'istep,trial number=',istep,k 

c do 170 ijk=1,n 

c write(14,*)'row=',ijk 

c 170 write(14,*)(jacob(ijk,jjj),jjj=1,n)

 if(istep.eq.1)return 

c check function convergence 

 errf=0. 

do 11 i=1,n 

 errf=errf+abs(fvec(i)) 

 11 continue 

c if(errf.le.tolf) goto 20 

c right-hand side of linear equations 

do 12 i=1,n 

 p(i)=-fvec(i) 

 12 continue 

c solve linear eqns. using LU decomposition 

 call ludcmp 

c forward and back substitution 

 call lubksb 

c check root convergence; and update solution 

 errx=0. 

do 13 i=1,n 

 errx=errx+abs(p(i)) 

 x(i)=x(i)+p(i) 

 13 continue 

 write(14,*)'k(trial#),errx,errf',k,errx,errf 

if(errf.le.tolf.and.errx.le.tolx) goto 20 

 14 continue

 20 loopindc=0 

 do 30 i=1,n 

 if(x(i).lt.0.0)then 

 write(6,*)'i=',i,x(i) 

A-77
 x(i)=10**ipower 

 loopindc=1

 endif 

 30 continue

 if(loopindc.eq.1)then

 ipower=ipower+1 

 goto 50 

 endif 

c call the subroutines one more time, indicating a solution has been found 

c for the time step istep. this call also updates the previous temperatures 

c with the solved new (current) temperatures in these called subroutines.

 isolfoun=1 

 call lhswng2(k,itotitr,iperiod,istep,theta,ta,hhwo,rfw,absorbw, 

 & emisw,eemisw,sunfact,isolfoun) 

 call rhswng2(k,itotitr,iperiod,istep,theta,ta,hhwo,rfw,absorbw, 

 & emisw,eemisw,sunfact,isolfoun) 

 call fuslglr2(k,itotitr,icool,iperiod,istep,theta,ta,hhfo,hhc,rff, 

 & absorbf,eemisf,sunfact,sunfact2,isolfoun) 

 call fuslgrf2(k,itotitr,icool,iperiod,istep,theta,ta,hhfo,hhc,qdir 

 & ,qsky,qtarm,sunfact2,isolfoun) 

 call tarmac2(k,itotitr,icool,iperiod,istep,ta,ttarm,hhto,isolfoun)

 return 

 end 

c 

c 

c 

 subroutine lhswng1(itotitr,iperiod,istep,theta,ta,hhw,rfw, 

 & absorbw,emisw,eemisw,sunfact) 

c purpose: transient analysis of the left wing (the wing that is not under 

c shade) 

c called from: trnsien1 

c calls: none

 common/c2/tt(200),fn(200),jacob(200,200) 

 common/c3/tth(100),kk(100) 

 common/c6/absorbff,emisff,absorbww,emisww,absorbfl,emisfl,absorbfi 

 &,emisfi,absorbpg,emispg,absorftp,emisrftp,absorfbt,emisrfbt 

 &,absorbfb,emisfb 

 common/c10/aircondc,thdiffus,visckine,prandtl 

 common/c14/thicknes(100),density(100),conductv(100),sheatcap(100)

 common/reflecq/qreflin(6,100),qscabsin(6,100),qreflfwr(6,100), 

 &qreflfwl(6,100),qscabfwr(6,100),qscabfwl(6,100),qabstwa1(6,100), 

 &qabstwa2(6,100),qabstwb(6,100) 

 common/temp1/t(1000) 

 common/data1/deltime 

 common/radiosit/jj(14) 

 common/result/temp(200),temptrn(200) 

 dimension newt(100),k(100),th(100),dens(100),shc(100)

 real kk,k,jj,newt 

 parameter(sig=.1714e-8) 

 parameter(fpi=3.141593/180.) 

 parameter(nlayer=9) 

A-78
 parameter(lend=3*nlayer) 

 parameter(layair=5) 

 parameter(f=0.2) 

c 

c Variables:

c layair=the layer (or element) that is air space in the wing 

c f=fraction of an element thickness 

c 

 if(itotitr.eq.1.and.istep.eq.1)then 

c reference thickness (th): 

 th(1)=tth(1) 

 th(2)=tth(2) 

 th(3)=tth(3) 

 th(4)=tth(4) 

 th(5)=tth(29) 

 th(6)=tth(30) 

 th(7)=tth(5) 

 th(8)=tth(6) 

 th(9)=tth(7) 

c reference conductivity (k): 

 k(1)=kk(1) 

 k(2)=kk(2) 

 k(3)=kk(3) 

 k(4)=kk(4) 

 k(5)=kk(29) 

 k(6)=kk(30) 

 k(7)=kk(5) 

 k(8)=kk(6) 

 k(9)=kk(7) 

c reference density (dens): 

 dens(1)=density(1) 

 dens(2)=density(2) 

 dens(3)=density(3) 

 dens(4)=density(4) 

 dens(6)=density(6) 

 dens(7)=density(7) 

 dens(8)=density(8) 

 dens(9)=density(9) 

c reference specific heat capacity (shc): 

 shc(1)=sheatcap(1) 

 shc(2)=sheatcap(2) 

 shc(3)=sheatcap(3) 

 shc(4)=sheatcap(4) 

 shc(6)=sheatcap(6) 

 shc(7)=sheatcap(7) 

 shc(8)=sheatcap(8) 

 shc(9)=sheatcap(9) 

 endif 

 if(istep.eq.1)then 

c reference the initial temp. (t): 

 t(1)=tt(1) 

A-79
 t(2)=(tt(1)+tt(2))/2. 

 t(3)=tt(2) 

 t(4)=tt(2) 

 t(5)=(tt(2)+tt(3))/2. 

 t(6)=tt(3) 

 t(7)=tt(3) 

 t(8)=(tt(3)+tt(4))/2. 

 t(9)=tt(4) 

 t(10)=tt(4) 

 t(11)=(tt(4)+tt(62))/2. 

 t(12)=tt(62) 

 t(16)=tt(63) 

 t(17)=(tt(63)+tt(5))/2. 

 t(18)=tt(5) 

 t(19)=tt(5) 

 t(20)=(tt(5)+tt(6))/2. 

 t(21)=tt(6) 

 t(22)=tt(6) 

 t(23)=(tt(6)+tt(7))/2. 

 t(24)=tt(7) 

 t(25)=tt(7) 

 t(26)=(tt(7)+tt(8))/2. 

 t(27)=tt(8) 

c reference the initial radiosity (jj) 

 jj(1)=tt(40) 

 jj(2)=tt(41) 

 return 

 endif 

c 

c formulate transient cooling: 

c 

c upper wing surface temp (t1), as well as internal temp t2 and t3: 

 qnetsky=(sig*t(1)**4-rfw*jj(1))/((1.-rfw*eemisw)/eemisw) 

 newt(1)=t(1)+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((absorbw*qreflfwl(1,iperiod)*sin(fpi*theta)+ 

 & qscabfwl(1,iperiod))*sunfact+ 

 & (absorbw*qreflfwr(1,iperiod)*sin(fpi*theta)+ 

 & qscabfwr(1,iperiod))*(1-sunfact) 

 & -qnetsky-hhw*(t(1)-ta)- 

 & (t(1)-t(2))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 newt(2)=t(2)+deltime/(dens(1)*shc(1)*(1-f)*th(1))* 

 & ((t(1)-t(2))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (t(2)-t(3))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 newt(3)=t(3)+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((t(2)-t(3))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (t(3)-t(4))/(f/4*th(1)/k(1)+f/4*th(2)/k(2))) 

c internal nodes (all conduction): t4-t24: 

do 20 i=1,nlayer 

 if(i.eq.layair)goto 20 

 if(i.eq.1.or.i.eq.nlayer)goto 20 

 ii=(i-1)*3 

A-80
c beginning element within layer i: t4,t7,t10,t13,t16,t19,t22: 

 if(i.eq.layair+1)then 

 newt(ii+1)=t(ii+1)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((t(ii-3)-t(ii+1))/(f/4*th(i-2)/k(i-2)+th(i-1)/k(i-1)+ 

 & f/4*th(i)/k(i))-

 & (t(ii+1)-t(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 else 

 newt(ii+1)=t(ii+1)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((t(ii)-t(ii+1))/(f/4*th(i-1)/k(i-1)+f/4*th(i)/k(i))- 

 & (t(ii+1)-t(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 endif 

c middle element within layer i: t5,t8,t11,t14,t17,t20,t23: 

 newt(ii+2)=t(ii+2)+deltime/(dens(i)*shc(i)*(1-f)*th(i))* 

 & ((t(ii+1)-t(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (t(ii+2)-t(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

c ending element within layer i: t6,t9,t12,t15,t18,t21,t24: 

 if(i.eq.layair-1)then 

 newt(ii+3)=t(ii+3)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((t(ii+2)-t(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (t(ii+3)-t(ii+7))/(f/4*th(i)/k(i)+th(i+1)/k(i+1)+ 

 & f/4*th(i+2)/k(i+2))) 

 else 

 newt(ii+3)=t(ii+3)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((t(ii+2)-t(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (t(ii+3)-t(ii+4))/(f/4*th(i)/k(i)+f/4*th(i+1)/k(i+1))) 

 endif 

 20 continue 

c lower wing surface temp (t27), as well as internal temp t25 and t26: 

 newt(lend-2)=t(lend-2)+ 

 & deltime/(dens(nlayer)*shc(nlayer)*f/2*th(nlayer))* 

 & ((t(lend-3)-t(lend-2))/ 

 & (f/4*th(nlayer-1)/k(nlayer-1)+f/4*th(nlayer)/k(nlayer))- 

 & (t(lend-2)-t(lend-1))/ 

 & (f/4*th(nlayer)/k(nlayer)+(1-f)/2*th(nlayer)/k(nlayer))) 

 newt(lend-1)=t(lend-1)+ 

 & deltime/(dens(nlayer)*shc(nlayer)*(1-f)*th(nlayer))* 

 & ((t(lend-2)-t(lend-1))/ 

 & (f/4*th(nlayer)/k(nlayer)+(1-f)/2*th(nlayer)/k(nlayer))- 

 & (t(lend-1)-t(lend))/ 

 & (f/4*th(nlayer)/k(nlayer)+(1-f)/2*th(nlayer)/k(nlayer))) 

 qnetsky=(sig*t(lend)**4-jj(2))/((1.-emisw)/emisw) 

 newt(lend)=t(lend)+ 

 & deltime/(dens(nlayer)*shc(nlayer)*f/2*th(nlayer))* 

 & (qabstwa1(3,iperiod)-qnetsky-hhw*(t(lend)-ta)+ 

 & (t(lend-1)-t(lend))/ 

 & (f/4*th(nlayer)/k(nlayer)+(1-f)/2*th(nlayer)/k(nlayer))) 

 do 60 i=1,lend 

 60 t(i)=newt(i) 

 temptrn(1)=t(1)-459.67 

 temptrn(2)=(t(3)+t(4))/2.-459.67 

 temptrn(3)=(t(6)+t(7))/2.-459.67 

A-81
 temptrn(4)=(t(9)+t(10))/2.-459.67 

 temptrn(5)=t(12)-459.67 

 temptrn(6)=t(16)-459.67 

 temptrn(7)=(t(18)+t(19))/2.-459.67 

 temptrn(8)=(t(21)+t(22))/2.-459.67 

 temptrn(9)=(t(24)+t(25))/2.-459.67 

 temptrn(10)=t(27)-459.67 

 return 

 end 

c 

c 

 subroutine rhswng1(itotitr,iperiod,istep,theta,ta,hhw,rfw, 

 & absorbw,emisw,eemisw,sunfact) 

c purpose: transient analysis of the right wing (the wing that is under 

c shade) 

c called from: trnsien1 

c calls: none

 common/c2/tt(200),fn(200),jacob(200,200) 

 common/c3/tth(100),kk(100) 

 common/c6/absorbff,emisff,absorbww,emisww,absorbfl,emisfl,absorbfi 

 &,emisfi,absorbpg,emispg,absorftp,emisrftp,absorfbt,emisrfbt 

 &,absorbfb,emisfb 

 common/c10/aircondc,thdiffus,visckine,prandtl 

 common/c14/thicknes(100),density(100),conductv(100),sheatcap(100)

 common/reflecq/qreflin(6,100),qscabsin(6,100),qreflfwr(6,100), 

 &qreflfwl(6,100),qscabfwr(6,100),qscabfwl(6,100),qabstwa1(6,100), 

 &qabstwa2(6,100),qabstwb(6,100) 

 common/temp2/t(1000) 

 common/data1/deltime 

 common/radiosit/jj(14) 

 common/result/temp(200),temptrn(200) 

 dimension newt(100),k(100),th(100),dens(100),shc(100)

 real kk,k,jj,newt 

 parameter(sig=.1714e-8) 

 parameter(fpi=3.141593/180.) 

 parameter(nlayer=9) 

 parameter(lend=3*nlayer) 

 parameter(layair=5) 

 parameter(f=0.2) 

c 

c Variables:

c layair=the layer (or element) that is air space in the wing 

c f=fraction of an element thickness 

c 

 if(itotitr.eq.1.and.istep.eq.1)then 

c reference thickness (th): 

 th(1)=tth(1) 

 th(2)=tth(2) 

 th(3)=tth(3) 

 th(4)=tth(4) 

 th(5)=tth(29) 

A-82
 th(6)=tth(30) 

 th(7)=tth(5) 

 th(8)=tth(6) 

 th(9)=tth(7) 

c reference conductivity (k): 

 k(1)=kk(1) 

 k(2)=kk(2) 

 k(3)=kk(3) 

 k(4)=kk(4) 

 k(5)=kk(29) 

 k(6)=kk(30) 

 k(7)=kk(5) 

 k(8)=kk(6) 

 k(9)=kk(7) 

c reference density (dens): 

 dens(1)=density(1) 

 dens(2)=density(2) 

 dens(3)=density(3) 

 dens(4)=density(4) 

 dens(6)=density(6) 

 dens(7)=density(7) 

 dens(8)=density(8) 

 dens(9)=density(9) 

c reference specific heat capacity (shc): 

 shc(1)=sheatcap(1) 

 shc(2)=sheatcap(2) 

 shc(3)=sheatcap(3) 

 shc(4)=sheatcap(4) 

 shc(6)=sheatcap(6) 

 shc(7)=sheatcap(7) 

 shc(8)=sheatcap(8) 

 shc(9)=sheatcap(9) 

 endif 

 if(istep.eq.1)then 

c reference the initial temp. (t): 

 t(1)=tt(9) 

 t(2)=(tt(9)+tt(10))/2. 

 t(3)=tt(10) 

 t(4)=tt(10) 

 t(5)=(tt(10)+tt(11))/2. 

 t(6)=tt(11) 

 t(7)=tt(11) 

 t(8)=(tt(11)+tt(12))/2. 

 t(9)=tt(12) 

 t(10)=tt(12) 

 t(11)=(tt(12)+tt(64))/2. 

 t(12)=tt(64) 

 t(16)=tt(65) 

 t(17)=(tt(65)+tt(13))/2. 

 t(18)=tt(13) 

 t(19)=tt(13) 

A-83
 t(20)=(tt(13)+tt(14))/2. 

 t(21)=tt(14) 

 t(22)=tt(14) 

 t(23)=(tt(14)+tt(15))/2. 

 t(24)=tt(15) 

 t(25)=tt(15) 

 t(26)=(tt(15)+tt(16))/2. 

 t(27)=tt(16) 

c reference the initial radiosity (jj): 

 jj(3)=tt(42) 

 jj(4)=tt(43) 

 return 

 endif 

c 

c formulate transient cooling: 

c 

c upper wing surface temp (t1), as well as internal temp t2 and t3: 

 qnetsky=(sig*t(1)**4-rfw*jj(3))/((1.-rfw*eemisw)/eemisw) 

 newt(1)=t(1)+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((absorbw*qreflfwr(1,iperiod)*sin(fpi*theta)+ 

 & qscabfwr(1,iperiod))*sunfact+ 

 & (absorbw*qreflfwl(1,iperiod)*sin(fpi*theta)+ 

 & qscabfwl(1,iperiod))*(1-sunfact) 

 & -qnetsky-hhw*(t(1)-ta)- 

 & (t(1)-t(2))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 newt(2)=t(2)+deltime/(dens(1)*shc(1)*(1-f)*th(1))* 

 & ((t(1)-t(2))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (t(2)-t(3))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 newt(3)=t(3)+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((t(2)-t(3))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (t(3)-t(4))/(f/4*th(1)/k(1)+f/4*th(2)/k(2))) 

c internal nodes (all conduction): t4-t24: 

do 20 i=1,nlayer 

 if(i.eq.layair)goto 20 

 if(i.eq.1.or.i.eq.nlayer)goto 20 

 ii=(i-1)*3 

c beginning element within layer i: t4,t7,t10,t13,t16,t19,t22: 

 if(i.eq.layair+1)then 

 newt(ii+1)=t(ii+1)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((t(ii-3)-t(ii+1))/(f/4*th(i-2)/k(i-2)+th(i-1)/k(i-1)+ 

 & f/4*th(i)/k(i))-

 & (t(ii+1)-t(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 else 

 newt(ii+1)=t(ii+1)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((t(ii)-t(ii+1))/(f/4*th(i-1)/k(i-1)+f/4*th(i)/k(i))- 

 & (t(ii+1)-t(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 endif 

c middle element within layer i: t5,t8,t11,t14,t17,t20,t23: 

 newt(ii+2)=t(ii+2)+deltime/(dens(i)*shc(i)*(1-f)*th(i))* 

 & ((t(ii+1)-t(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (t(ii+2)-t(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

A-84
c ending element within layer i: t6,t9,t12,t15,t18,t21,t24: 

 if(i.eq.layair-1)then 

 newt(ii+3)=t(ii+3)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((t(ii+2)-t(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (t(ii+3)-t(ii+7))/(f/4*th(i)/k(i)+th(i+1)/k(i+1)+ 

 & f/4*th(i+2)/k(i+2))) 

 else 

 newt(ii+3)=t(ii+3)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((t(ii+2)-t(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (t(ii+3)-t(ii+4))/(f/4*th(i)/k(i)+f/4*th(i+1)/k(i+1))) 

 endif 

 20 continue 

c lower wing surface temp (t27), as well as internal temp t25 and t26: 

 newt(lend-2)=t(lend-2)+ 

 & deltime/(dens(nlayer)*shc(nlayer)*f/2*th(nlayer))* 

 & ((t(lend-3)-t(lend-2))/ 

 & (f/4*th(nlayer-1)/k(nlayer-1)+f/4*th(nlayer)/k(nlayer))- 

 & (t(lend-2)-t(lend-1))/ 

 & (f/4*th(nlayer)/k(nlayer)+(1-f)/2*th(nlayer)/k(nlayer))) 

 newt(lend-1)=t(lend-1)+ 

 & deltime/(dens(nlayer)*shc(nlayer)*(1-f)*th(nlayer))* 

 & ((t(lend-2)-t(lend-1))/ 

 & (f/4*th(nlayer)/k(nlayer)+(1-f)/2*th(nlayer)/k(nlayer))- 

 & (t(lend-1)-t(lend))/ 

 & (f/4*th(nlayer)/k(nlayer)+(1-f)/2*th(nlayer)/k(nlayer))) 

 qnetsky=(sig*t(lend)**4-jj(4))/((1.-emisw)/emisw) 

 newt(lend)=t(lend)+ 

 & deltime/(dens(nlayer)*shc(nlayer)*f/2*th(nlayer))* 

 & (qabstwa2(3,iperiod)-qnetsky-hhw*(t(lend)-ta)+ 

 & (t(lend-1)-t(lend))/ 

 & (f/4*th(nlayer)/k(nlayer)+(1-f)/2*th(nlayer)/k(nlayer))) 

 do 60 i=1,lend 

 60 t(i)=newt(i) 

 temptrn(11)=t(1)-459.67 

 temptrn(12)=(t(3)+t(4))/2.-459.67 

 temptrn(13)=(t(6)+t(7))/2.-459.67 

 temptrn(14)=(t(9)+t(10))/2-459.67 

 temptrn(15)=t(12)-459.67 

 temptrn(16)=t(16)-459.67 

 temptrn(17)=(t(18)+t(19))/2.-459.67 

 temptrn(18)=(t(21)+t(22))/2.-459.67 

 temptrn(19)=(t(24)+t(25))/2.-459.67 

 temptrn(20)=t(27)-459.67 

 return 

 end 

c 

c 

 subroutine fuslglr1(itotitr,icool,iperiod,istep,theta,ta,hhf,hhc, 

 & rff,absorbf,eemisf,sunfact,sunfact2) 

c purpose: transient analysis of the fuselage (left to right) 

c called from: trnsien1 

A-85
c calls: none

 common/c2/tt(200),fn(200),jacob(200,200) 

 common/c3/tth(100),kk(100) 

 common/c6/absorbff,emisff,absorbww,emisww,absorbfl,emisfl,absorbfi 

 &,emisfi,absorbpg,emispg,absorftp,emisrftp,absorfbt,emisrfbt 

 &,absorbfb,emisfb 

 common/c10/aircondc,thdiffus,visckine,prandtl 

 common/c14/thicknes(100),density(100),conductv(100),sheatcap(100)

 common/reflecq/qreflin(6,100),qscabsin(6,100),qreflfwr(6,100), 

 &qreflfwl(6,100),qscabfwr(6,100),qscabfwl(6,100),qabstwa1(6,100), 

 &qabstwa2(6,100),qabstwb(6,100) 

 common/temp3/t(1000) 

 common/data1/deltime 

 common/radiosit/jj(14) 

 common/result/temp(200),temptrn(200) 

 dimension newt(100),k(100),th(100),dens(100),shc(100)

 real kk,k,jj,newt,nusselt2 

 parameter(sig=.1714e-8,gravity=32.2) 

 parameter(fpi=3.141593/180.) 

 parameter(nlayer=9) 

 parameter(lend=3*nlayer) 

 parameter(layair=5) 

 parameter(f=0.2) 

c 

c Variables:

c layair=the layer (or element) that is air space (the fuselage width) 

c f=fraction of an element thickness 

c 

 if(itotitr.eq.1.and.istep.eq.1)then 

c reference thickness (th): 

 th(1)=tth(8) 

 th(2)=tth(9) 

 th(3)=tth(10) 

 th(4)=tth(23) 

 th(5)=tth(11) 

 th(6)=tth(23) 

 th(7)=tth(10) 

 th(8)=tth(9) 

 th(9)=tth(8) 

c reference conductivity (k): 

 k(1)=kk(8) 

 k(2)=kk(9) 

 k(3)=kk(10) 

 k(4)=kk(23) 

 k(5)=kk(11) 

 k(6)=kk(23) 

 k(7)=kk(10) 

 k(8)=kk(9) 

 k(9)=kk(8) 

c reference density (dens): 

 dens(1)=density(10) 

A-86
 dens(2)=density(11) 

 dens(3)=density(12) 

 dens(4)=density(13) 

 dens(6)=density(13) 

 dens(7)=density(12) 

 dens(8)=density(11) 

 dens(9)=density(10) 

c reference specific heat capacity (shc): 

 shc(1)=sheatcap(10)

 shc(2)=sheatcap(11)

 shc(3)=sheatcap(12)

 shc(4)=sheatcap(13)

 shc(6)=sheatcap(13)

 shc(7)=sheatcap(12)

 shc(8)=sheatcap(11)

 shc(9)=sheatcap(10)

 endif 

 if(istep.eq.1)then 

c reference the initial temp. (t): 

 t(1)=tt(17) 

 t(2)=(tt(17)+tt(18))/2. 

 t(3)=tt(18) 

 t(4)=tt(18) 

 t(5)=(tt(18)+tt(19))/2. 

 t(6)=tt(19) 

 t(7)=tt(19) 

 t(8)=(tt(19)+tt(20))/2. 

 t(9)=tt(20) 

 t(10)=tt(20) 

 t(11)=(tt(20)+tt(38))/2. 

 t(12)=tt(38) 

 t(16)=tt(39) 

 t(17)=(tt(39)+tt(21))/2. 

 t(18)=tt(21) 

 t(19)=tt(21) 

 t(20)=(tt(21)+tt(22))/2. 

 t(21)=tt(22) 

 t(22)=tt(22) 

 t(23)=(tt(22)+tt(23))/2. 

 t(24)=tt(23) 

 t(25)=tt(23) 

 t(26)=(tt(23)+tt(24))/2. 

 t(27)=tt(24) 

c reference the initial radiosity (jj) 

 jj(5)=tt(44) 

 jj(6)=tt(45) 

 jj(8)=tt(47) 

 jj(10)=tt(49) 

 return 

 endif 

 if(icool.eq.2)then 

A-87
 rayleig2=gravity*1./((t(12)+t(16))/2.)*abs(t(12)-t(16))* 

 & th(layair)**3/(thdiffus*visckine) 

 if(rayleig2.le.(1e3*(0.2+prandtl)/prandtl))then 

 nusselt2=1.0 

 elseif(rayleig2.gt.(1e3*(0.2+prandtl)/prandtl).and.rayleig2.lt.

 & 1.0e9)then 

 nusselt2=0.18*((prandtl/(0.2+prandtl))*rayleig2)**0.29 

 endif 

 endif 

c 

c formulate transient cooling: 

c 

c LHS surface temp (t1), as well as internal temp t2 and t3: 

 qnetsky=(sig*t(1)**4-rff*jj(5))/((1.-rff*eemisf)/eemisf) 

 newt(1)=t(1)+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((absorbf*qreflfwl(2,iperiod)*cos(fpi*theta)+ 

 & qscabfwl(2,iperiod))*sunfact+ 

 & qscabfwr(2,iperiod)*(1-sunfact) 

 & -qnetsky-hhf*(t(1)-ta)- 

 & (t(1)-t(2))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 newt(2)=t(2)+deltime/(dens(1)*shc(1)*(1-f)*th(1))* 

 & ((t(1)-t(2))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (t(2)-t(3))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 newt(3)=t(3)+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((t(2)-t(3))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (t(3)-t(4))/(f/4*th(1)/k(1)+f/4*th(2)/k(2))) 

c internal nodes (all conduction): t4-t24: 

do 20 i=1,nlayer 

 if(i.eq.layair)goto 20 

 if(i.eq.1.or.i.eq.nlayer)goto 20 

 ii=(i-1)*3 

c beginning element within layer i: t4,t7,t10,t16,t19,t22: 

 if(i.eq.layair+1)then 

 qnetcab=(sig*t(ii+1)**4-jj(10))/((1.-emisfi)/emisfi) 

 if(icool.eq.1)then 

 newt(ii+1)=t(ii+1)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & (-(t(ii+1)-t(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & qnetcab+absorbfi*qreflin(1,iperiod)*cos(fpi*theta)+ 

 & qscabsin(1,iperiod)-hhc*(t(ii+1)-ta)) 

 elseif(icool.eq.2)then 

 newt(ii+1)=t(ii+1)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((t(ii-3)-t(ii+1))/(f/4*th(i-2)/k(i-2)+ 

 & th(i-1)/(nusselt2*k(i-1))+f/4*th(i)/k(i))- 

 & (t(ii+1)-t(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & qnetcab+

 & (absorbfi*qreflin(1,iperiod)*cos(fpi*theta)+ 

 & qscabsin(1,iperiod))*sunfact2) 

c & (absorbfi*qreflin(3,iperiod)*cos(fpi*theta)+ 

c & qscabsin(3,iperiod))*(1-sunfact)) 

 endif 

 else 

A-88
 newt(ii+1)=t(ii+1)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((t(ii)-t(ii+1))/(f/4*th(i-1)/k(i-1)+f/4*th(i)/k(i))- 

 & (t(ii+1)-t(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 endif 

c middle element within layer i: t5,t8,t14,t17,t20,t23: 

 newt(ii+2)=t(ii+2)+deltime/(dens(i)*shc(i)*(1-f)*th(i))* 

 & ((t(ii+1)-t(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (t(ii+2)-t(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

c ending element within layer i: t6,t9,t12,t18,t21,t24: 

 if(i.eq.layair-1)then 

 qnetcab=(sig*t(ii+3)**4-jj(8))/((1.-emisfi)/emisfi) 

 if(icool.eq.1)then 

 newt(ii+3)=t(ii+3)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((t(ii+2)-t(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & qnetcab+absorbfi*qreflin(3,iperiod)*cos(fpi*theta)+ 

 & qscabsin(3,iperiod)-hhc*(t(ii+3)-ta)) 

 elseif(icool.eq.2)then 

 newt(ii+3)=t(ii+3)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((t(ii+2)-t(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (t(ii+3)-t(ii+7))/(f/4*th(i)/k(i)+ 

 & th(i+1)/(nusselt2*k(i+1))+f/4*th(i+2)/k(i+2))- 

 & qnetcab+ 

 & (absorbfi*qreflin(3,iperiod)*cos(fpi*theta)+ 

 & qscabsin(3,iperiod))*sunfact2) 

c & (absorbfi*qreflin(1,iperiod)*cos(fpi*theta)+ 

c & qscabsin(1,iperiod))*(1-sunfact)) 

 endif 

 else 

 newt(ii+3)=t(ii+3)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((t(ii+2)-t(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (t(ii+3)-t(ii+4))/(f/4*th(i)/k(i)+f/4*th(i+1)/k(i+1))) 

 endif 

 20 continue 

c RHS fuselage surface temp (t27), as well as internal temp t25 and t26: 

 newt(lend-2)=t(lend-2)+ 

 & deltime/(dens(nlayer)*shc(nlayer)*f/2*th(nlayer))* 

 & ((t(lend-3)-t(lend-2))/ 

 & (f/4*th(nlayer-1)/k(nlayer-1)+f/4*th(nlayer)/k(nlayer))- 

 & (t(lend-2)-t(lend-1))/ 

 & (f/4*th(nlayer)/k(nlayer)+(1-f)/2*th(nlayer)/k(nlayer))) 

 newt(lend-1)=t(lend-1)+ 

 & deltime/(dens(nlayer)*shc(nlayer)*(1-f)*th(nlayer))* 

 & ((t(lend-2)-t(lend-1))/ 

 & (f/4*th(nlayer)/k(nlayer)+(1-f)/2*th(nlayer)/k(nlayer))- 

 & (t(lend-1)-t(lend))/ 

 & (f/4*th(nlayer)/k(nlayer)+(1-f)/2*th(nlayer)/k(nlayer))) 

 qnetsky=(sig*t(lend)**4-rff*jj(6))/((1.-rff*eemisf)/eemisf) 

 newt(lend)=t(lend)+ 

 & deltime/(dens(nlayer)*shc(nlayer)*f/2*th(nlayer))* 

 & (qscabfwr(2,iperiod)*sunfact+ 

 & (absorbf*qreflfwl(2,iperiod)*cos(fpi*theta)+ 

A-89
 & qscabfwl(2,iperiod))*(1-sunfact) 

 & -qnetsky-hhf*(t(lend)-ta)+ 

 & (t(lend-1)-t(lend))/ 

 & (f/4*th(nlayer)/k(nlayer)+(1-f)/2*th(nlayer)/k(nlayer))) 

 do 60 i=1,lend 

 60 t(i)=newt(i) 

 temptrn(39)=t(1)-459.67 

 temptrn(40)=(t(3)+t(4))/2.-459.67 

 temptrn(41)=(t(6)+t(7))/2.-459.67 

 temptrn(42)=(t(9)+t(10))/2.-459.67 

 temptrn(43)=t(12)-459.67 

 temptrn(44)=t(16)-459.67 

 temptrn(45)=(t(18)+t(19))/2.-459.67 

 temptrn(46)=(t(21)+t(22))/2.-459.67 

 temptrn(47)=(t(24)+t(25))/2.-459.67 

 temptrn(48)=t(27)-459.67 

 return 

 end 

c 

c 

 subroutine fuslgrf1(itotitr,icool,iperiod,istep,theta,ta,hhf,hhc, 

 & qdir,qsky,qtarm,sunfact2) 

c purpose: transient analysis of the fuselage (top to bottom) 

c called from: trnsien1 

c calls: none

 common/c2/tt(200),fn(200),jacob(200,200) 

 common/c3/tth(100),kk(100) 

 common/c4/widthwng,lengtwng,hwng,locwng,lplane,lfuselg,hfuselg,

 & wfuselg 

 common/c6/absorbff,emisff,absorbww,emisww,absorbfl,emisfl,absorbfi 

 &,emisfi,absorbpg,emispg,absorftp,emisrftp,absorfbt,emisrfbt 

 &,absorbfb,emisfb 

 common/c5/trnsm,absorbta,emista,tarmd,densta,conducta,shcta 

 common/c10/aircondc,thdiffus,visckine,prandtl 

 common/c12/a1,a2,a3,a4,a5,a6(100),a7(100) 

 common/c14/thicknes(100),density(100),conductv(100),sheatcap(100)

 common/reflecq/qreflin(6,100),qscabsin(6,100),qreflfwr(6,100), 

 &qreflfwl(6,100),qscabfwr(6,100),qscabfwl(6,100),qabstwa1(6,100), 

 &qabstwa2(6,100),qabstwb(6,100) 

 common/temp4/t(1000),tpg1,tpg2,tpg3 

 common/data1/deltime 

 common/radiosit/jj(14) 

 common/result/temp(200),temptrn(200) 

 dimension newt(100),k(100),th(100),dens(100),shc(100)

 real kk,k,jj,newt,nusselt1,lfuselg,kpg,newtpg1,newtpg2,newtpg3 

 parameter(sig=.1714e-8,gravity=32.2) 

 parameter(fpi=3.141593/180.) 

 parameter(nlayer=13) 

 parameter(lend=3*nlayer) 

 parameter(layair1=5,layair2=10) 

 parameter(f=0.2) 

A-90
c 

c Variables:

c layair1=the layer (or element) that is air space (the fuselage height) 

c layair2=the layer (or element) that is air space (in between the floor 

c and belly-pan) 

c f=fraction of an element thickness 

c 

 if(itotitr.eq.1.and. istep.eq.1)then 

c reference thickness (th): 

 th(1)=tth(27) 

 th(2)=tth(26) 

 th(3)=tth(25) 

 th(4)=tth(28) 

 th(5)=tth(24) 

 th(6)=tth(21) 

 th(7)=tth(20) 

 th(8)=tth(19) 

 th(9)=tth(18) 

 th(10)=tth(31) 

 th(11)=tth(14) 

 th(12)=tth(13) 

 th(13)=tth(12) 

c reference conductivity (k): 

 k(1)=kk(27) 

 k(2)=kk(26) 

 k(3)=kk(25) 

 k(4)=kk(28) 

 k(5)=kk(24) 

 k(6)=kk(21) 

 k(7)=kk(20) 

 k(8)=kk(19) 

 k(9)=kk(18) 

 k(10)=kk(31) 

 k(11)=kk(14) 

 k(12)=kk(13) 

 k(13)=kk(12) 

c reference density (dens): 

 dens(1)=density(15) 

 dens(2)=density(16) 

 dens(3)=density(17) 

 dens(4)=density(18) 

 dens(6)=density(20) 

 dens(7)=density(21) 

 dens(8)=density(22) 

 dens(9)=density(23) 

 dens(11)=density(25) 

 dens(12)=density(26) 

 dens(13)=density(27) 

c reference specific heat capacity: 

 shc(1)=sheatcap(15)

 shc(2)=sheatcap(16)

A-91
 shc(3)=sheatcap(17)

 shc(4)=sheatcap(18)

 shc(6)=sheatcap(20)

 shc(7)=sheatcap(21)

 shc(8)=sheatcap(22)

 shc(9)=sheatcap(23)

 shc(11)=sheatcap(25) 

 shc(12)=sheatcap(26) 

 shc(13)=sheatcap(27) 

c also reference plexiglass thickness, conductivity, density 

c and specific heat capacity: 

 thpg=tth(22) 

 kpg=kk(22) 

 denspg=density(28) 

 shcpg=sheatcap(28)

c also find these values: 

 c1=a1/(a1+a2) 

 c2=a2/(a1+a2) 

 emispgrf=c1*emispg+c2*emisrfbt 

 endif 

 if(istep.eq.1)then 

c reference the initial temp. (t): 

 t(1)=tt(59) 

 t(2)=(tt(59)+tt(58))/2. 

 t(3)=tt(58) 

 t(4)=tt(58) 

 t(5)=(tt(58)+tt(57))/2. 

 t(6)=tt(57) 

 t(7)=tt(57) 

 t(8)=(tt(57)+tt(60))/2. 

 t(9)=tt(60) 

 t(10)=tt(60) 

 t(11)=(tt(60)+tt(61))/2. 

 t(12)=tt(61) 

 t(16)=tt(35) 

 t(17)=(tt(35)+tt(34))/2. 

 t(18)=tt(34) 

 t(19)=tt(34) 

 t(20)=(tt(34)+tt(33))/2. 

 t(21)=tt(33) 

 t(22)=tt(33) 

 t(23)=(tt(33)+tt(32))/2. 

 t(24)=tt(32) 

 t(25)=tt(32) 

 t(26)=(tt(32)+tt(31))/2. 

 t(27)=tt(31) 

 t(31)=tt(28) 

 t(32)=(tt(28)+tt(27))/2. 

 t(33)=tt(27) 

 t(34)=tt(27) 

 t(35)=(tt(27)+tt(26))/2. 

A-92
 t(36)=tt(26) 

 t(37)=tt(26) 

 t(38)=(tt(26)+tt(25))/2. 

 t(39)=tt(25) 

c also reference the intial plexiglass temp. 

 tpg1=tt(37) 

 tpg2=(tt(37)+tt(36))/2. 

 tpg3=tt(36) 

c reference the initial radiosity 

 jj(7)=tt(46) 

 jj(9)=tt(48) 

 jj(11)=tt(50) 

 return 

 endif 

 tpgrf=c1*tpg3+c2*t(12) 

 qnet=(sig*tpgrf**4-jj(9))/((1.-emispgrf)/emispgrf) 

c 

 if(icool.eq.2)then 

 rayleig1=gravity*1./((tpgrf+t(16))/2.)*(t(16)-tpgrf)* 

 & th(5)**3/(thdiffus*visckine) 

 if(rayleig1.le.3e5)then 

 nusselt1=1.0 

 elseif(rayleig1.gt.3e5.and.rayleig1.le.7e9)then 

 nusselt1=0.069*rayleig1**(1./3.)*prandtl**.074 

 endif 

 endif 

c 

c formulate transient cooling: 

c 

c first for the only three temperatures of plexiglass 

c 

c 

 newtpg1=tpg1+deltime/(a1*denspg*shcpg*f/2*thpg)* 

 & (emispg*qsky*a1+emispg*.5*(1.-emista)*qsky*a5 

 & +emispg*.5*qtarm*a5 

 & +f/2*absorbpg*qdir*cos(fpi*theta)*a6(iperiod) 

 & +f/2*absorbpg*qdir*sin(fpi*theta)*a7(iperiod) 

 & +f/2*absorbpg*.5*(1.-absorbta)*qdir*sin(fpi*theta)*a5 

 & +f/2*absorbpg*qreflin(4,iperiod)*sin(fpi*theta)*a3 

 & +f/2*absorbpg*qscabsin(4,iperiod)*lfuselg*wfuselg 

 & *a3/(absorbpg*a3+absorfbt*a4) 

 & -emispg*sig*tpg1**4*a1

 & -hhf*(tpg1-ta)*a1 

 & -(tpg1-tpg2)/(f/4*thpg/kpg+(1-f)/2*thpg/kpg)*a1) 

 newtpg2=tpg2+deltime/(a1*denspg*shcpg*(1-f)*thpg)* 

 & ((tpg1-tpg2)/(f/4*thpg/kpg+(1-f)/2*thpg/kpg)*a1 

 & -(tpg2-tpg3)/((1-f)/2*thpg/kpg+f/4*thpg/kpg)*a1 

 & +(1-f)*absorbpg*qdir*cos(fpi*theta)*a6(iperiod) 

 & +(1-f)*absorbpg*qdir*sin(fpi*theta)*a7(iperiod) 

 & +(1-f)*absorbpg*.5*(1.-absorbta)*qdir*sin(fpi*theta)*a5 

 & +(1-f)*absorbpg*qreflin(4,iperiod)*sin(fpi*theta)*a3 

A-93
 & +(1-f)*absorbpg*qscabsin(4,iperiod)*lfuselg*wfuselg 

 & *a3/(absorbpg*a3+absorfbt*a4)) 

 qpg=emispg*sig*tpg3**4-emispg/(a3*emispg+a4*emisrfbt)* 

 & (a3*emispg*sig*tpg3**4+a4*emisrfbt*sig*t(12)**4-qnet*(a3+a4)) 

 if(icool.eq.1)then 

 newtpg3=tpg3+deltime/(a1*denspg*shcpg*f/2*thpg)* 

 & ((tpg2-tpg3)/((1-f)/2*thpg/kpg+f/4*thpg/kpg)*a1 

 & -hhc*(tpg3-ta)*a1 

 & -qpg*a3 

 & +f/2*absorbpg*qdir*cos(fpi*theta)*a6(iperiod) 

 & +f/2*absorbpg*qdir*sin(fpi*theta)*a7(iperiod) 

 & +f/2*absorbpg*.5*(1.-absorbta)*qdir*sin(fpi*theta)*a5 

 & +f/2*absorbpg*qreflin(4,iperiod)*sin(fpi*theta)*a3 

 & +f/2*absorbpg*qscabsin(4,iperiod)*lfuselg*wfuselg 

 & *a3/(absorbpg*a3+absorfbt*a4)) 

 elseif(icool.eq.2)then

 newtpg3=tpg3+deltime/(a1*denspg*shcpg*f/2*thpg)* 

 & ((tpg2-tpg3)/((1-f)/2*thpg/kpg+f/4*thpg/kpg)*a1 

 & -(tpgrf-t(16))/(th(5)/(nusselt1*k(5)))*a3 

 & -qpg*a3 

 & +f/2*absorbpg*qdir*cos(fpi*theta)*a6(iperiod) 

 & +f/2*absorbpg*qdir*sin(fpi*theta)*a7(iperiod) 

 & +f/2*absorbpg*.5*(1.-absorbta)*qdir*sin(fpi*theta)*a5 

 & +f/2*absorbpg*qreflin(4,iperiod)*sin(fpi*theta)*a3 

 & +f/2*absorbpg*qscabsin(4,iperiod)*lfuselg*wfuselg 

 & *a3/(absorbpg*a3+absorfbt*a4)) 

 endif 

c 

c next for the roof structure and the rest 

c 

c first three temperatures of roof structure: 

 qnetsky=(emisrftp*sig*t(1)**4-emisrftp*qsky) 

 newt(1)=t(1)+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & (absorftp*qdir*sin(fpi*theta)-qnetsky-hhf*(t(1)-ta)- 

 & (t(1)-t(2))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 newt(2)=t(2)+deltime/(dens(1)*shc(1)*(1-f)*th(1))* 

 & ((t(1)-t(2))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (t(2)-t(3))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 newt(3)=t(3)+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((t(2)-t(3))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (t(3)-t(4))/(f/4*th(1)/k(1)+f/4*th(2)/k(2))) 

c internal nodes: t4-t36: 

do 20 i=1,nlayer 

 if(i.eq.layair1.or.i.eq.layair2)goto 20

 if(i.eq.1.or.i.eq.nlayer)goto 20 

 ii=(i-1)*3 

c beginning element within layer i: t4,t7,t10,t16,t19,t22,t25,t31,t34: 

 if(i.eq.layair1+1)then 

 qnetcab=(sig*t(ii+1)**4-jj(11))/((1.-emisfl)/emisfl) 

 if(icool.eq.1)then 

 newt(ii+1)=t(ii+1)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

A-94
 & (-(t(ii+1)-t(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & hhc*(t(ii+1)-ta)- 

 & qnetcab+absorbfl*qreflin(2,iperiod)*sin(fpi*theta)+ 

 & qscabsin(2,iperiod)) 

 elseif(icool.eq.2)then 

 newt(ii+1)=t(ii+1)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((tpgrf-t(ii+1))/(th(i-1)/(nusselt1*k(i-1)))- 

 & (t(ii+1)-t(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & qnetcab+

 & (absorbfl*qreflin(2,iperiod)*sin(fpi*theta)+ 

 & qscabsin(2,iperiod))*sunfact2) 

 endif 

 elseif(i.eq.layair2+1)then 

 newt(ii+1)=t(ii+1)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((t(ii-3)-t(ii+1))/(f/4*th(i-2)/k(i-2)+th(i-1)/k(i-1)+ 

 & f/4*th(i)/k(i))-

 & (t(ii+1)-t(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 else 

 newt(ii+1)=t(ii+1)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((t(ii)-t(ii+1))/(f/4*th(i-1)/k(i-1)+f/4*th(i)/k(i))- 

 & (t(ii+1)-t(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 endif 

c middle element within layer i: t5,t8,t11,t17,t20,t23,t26,t32,t35:

 newt(ii+2)=t(ii+2)+deltime/(dens(i)*shc(i)*(1-f)*th(i))* 

 & ((t(ii+1)-t(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (t(ii+2)-t(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

c ending element within layer i: t6,t9,t12,t18,t21,t24,t27,t33,t36:

 if(i.eq.layair1-1)then 

 qrfbt=emisrfbt*sig*t(ii+3)**4-emisrfbt/(a3*emispg+a4*emisrfbt)* 

 &(a3*emispg*sig*tpg3**4+a4*emisrfbt*sig*t(ii+3)**4-qnet*(a3+a4)) 

 if(icool.eq.1)then 

 newt(ii+3)=t(ii+3)+deltime/(a2*dens(i)*shc(i)*f/2*th(i))* 

 & ((t(ii+2)-t(ii+3))/((1-f)/2*th(i)/k(i)+f/4*th(i)/k(i))*a2- 

 & hhc*(t(ii+3)-ta)*a2 

 & -qrfbt*a4 

 & +absorfbt*qreflin(4,iperiod)*sin(fpi*theta)*a4 

 & +absorfbt*qscabsin(4,iperiod)*lfuselg*wfuselg*a4/ 

 & (absorbpg*a3+absorfbt*a4)) 

 elseif(icool.eq.2)then 

 newt(ii+3)=t(ii+3)+deltime/(a2*dens(i)*shc(i)*f/2*th(i))* 

 & ((t(ii+2)-t(ii+3))/((1-f)/2*th(i)/k(i)+f/4*th(i)/k(i))*a2- 

 & (tpgrf-t(ii+7))/(th(i+1)/(nusselt1*k(i+1)))*a4 

 & -qrfbt*a4 

 & +(absorfbt*qreflin(4,iperiod)*sin(fpi*theta)*a4 

 & +absorfbt*qscabsin(4,iperiod)*lfuselg*wfuselg*a4/ 

 & (absorbpg*a3+absorfbt*a4))*sunfact2) 

 endif 

 elseif(i.eq.layair2-1)then 

 newt(ii+3)=t(ii+3)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((t(ii+2)-t(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (t(ii+3)-t(ii+7))/(f/4*th(i)/k(i)+th(i+1)/k(i+1)+ 

A-95
 & f/4*th(i+2)/k(i+2))) 

 else 

 newt(ii+3)=t(ii+3)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((t(ii+2)-t(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (t(ii+3)-t(ii+4))/(f/4*th(i)/k(i)+f/4*th(i+1)/k(i+1))) 

 endif 

 20 continue 

c belly-pan temp (t39), as well as internal temp t37 and t38: 

 newt(lend-2)=t(lend-2)+ 

 & deltime/(dens(nlayer)*shc(nlayer)*f/2*th(nlayer))* 

 & ((t(lend-3)-t(lend-2))/ 

 & (f/4*th(nlayer-1)/k(nlayer-1)+f/4*th(nlayer)/k(nlayer))- 

 & (t(lend-2)-t(lend-1))/ 

 & (f/4*th(nlayer)/k(nlayer)+(1-f)/2*th(nlayer)/k(nlayer))) 

 newt(lend-1)=t(lend-1)+ 

 & deltime/(dens(nlayer)*shc(nlayer)*(1-f)*th(nlayer))* 

 & ((t(lend-2)-t(lend-1))/ 

 & (f/4*th(nlayer)/k(nlayer)+(1-f)/2*th(nlayer)/k(nlayer))- 

 & (t(lend-1)-t(lend))/ 

 & (f/4*th(nlayer)/k(nlayer)+(1-f)/2*th(nlayer)/k(nlayer))) 

 qnetsky=(sig*t(lend)**4-jj(7))/((1.-emisfb)/emisfb)

 newt(lend)=t(lend)+ 

 & deltime/(dens(nlayer)*shc(nlayer)*f/2*th(nlayer))* 

 & (qabstwb(3,iperiod)-qnetsky-hhf*(t(lend)-ta)+ 

 & (t(lend-1)-t(lend))/ 

 & (f/4*th(nlayer)/k(nlayer)+(1-f)/2*th(nlayer)/k(nlayer))) 

 tpg1=newtpg1 

 tpg2=newtpg2 

 tpg3=newtpg3 

 do 60 i=1,lend 

 60 t(i)=newt(i) 

 temptrn(21)=t(1)-459.67 

 temptrn(22)=(t(3)+t(4))/2.-459.67 

 temptrn(23)=(t(6)+t(7))/2.-459.67 

 temptrn(24)=(t(9)+t(10))/2.-459.67 

 temptrn(25)=t(12)-459.67 

 temptrn(26)=t(16)-459.67 

 temptrn(27)=(t(18)+t(19))/2.-459.67 

 temptrn(28)=(t(21)+t(22))/2.-459.67 

 temptrn(29)=(t(24)+t(25))/2.-459.67 

 temptrn(30)=t(27)-459.67 

 temptrn(31)=(t(27)+t(31))/2.-459.67 

 temptrn(32)=(t(27)+t(31))/2.-459.67 

 temptrn(33)=(t(27)+t(31))/2.-459.67 

 temptrn(34)=(t(27)+t(31))/2.-459.67 

 temptrn(35)=t(31)-459.67 

 temptrn(36)=(t(33)+t(34))/2.-459.67 

 temptrn(37)=(t(36)+t(37))/2.-459.67 

 temptrn(38)=t(39)-459.67 

 temptrn(49)=tpg1-459.67 

 temptrn(50)=tpg3-459.67 

A-96
 return 

 end 

c 

 subroutine tarmac1(itotitr,icool,iperiod,istep,ta,ttarm,hht) 

c purpose: transient analysis of the tarmac 

c called from: trnsien1 

c calls: none

 common/c4/widthwng,lengtwng,hwng,locwng,lplane,lfuselg,hfuselg,

 & wfuselg 

 common/c5/trnsm,absorbta,emista,tarmd,densta,conducta,shcta 

 common/c2/tt(200),fn(200),jacob(200,200) 

 common/c3/tth(100),kk(100) 

 common/c14/thicknes(100),density(100),conductv(100),sheatcap(100)

 common/reflecq/qreflin(6,100),qscabsin(6,100),qreflfwr(6,100), 

 &qreflfwl(6,100),qscabfwr(6,100),qscabfwl(6,100),qabstwa1(6,100), 

 &qabstwa2(6,100),qabstwb(6,100) 

 common/temp5/t(1000) 

 common/data1/deltime 

 common/radiosit/jj(14) 

 common/facttarm/fact 

 common/result/temp(200),temptrn(200) 

 dimension newt(100) 

 real jj,newt,lengtwng 

 parameter(sig=.1714e-8) 

 parameter(fpi=3.141593/180.) 

 parameter(afact=1.0,bfact=0.00001) 

c 

c Variables:

c afact=cooling when door is opened: no motion: therefore tarmac temperature 

c underneath the wing remains very constant 

c bfact=a small number to force the tarmac temperature underneath the wing 

c to be the same as the unsheltered tarmac temperature: in effect when 

c the aircraft is in motion 

c 

 if(itotitr.eq.1.and.istep.eq.1)then 

 rlwng=(lengtwng-wfuselg)/2. 

 endif 

 if(istep.eq.1)then 

c reference the initial temp.: 

 t(51)=tt(51) 

 t(52)=tt(52) 

 t(53)=tt(53) 

c reference the initial radiosity 

 jj(12)=tt(54) 

 jj(13)=tt(55) 

 jj(14)=tt(56) 

 return 

 endif 

 if(icool.eq.1)fact1=afact 

 if(icool.eq.2)fact1=bfact 

c 

A-97
c formulate transient cooling: 

c 

c LHS Wing/Tarmac t(51): 

 qnetsky=(sig*t(51)**4-jj(12))/((1.-emista)/emista) 

 newt(51)=t(51)+deltime/(densta*rlwng*widthwng*tarmd*shcta)* 

 & ((qabstwa1(1,iperiod)-qnetsky-hht*(t(51)-ta))*rlwng*widthwng+ 

 & 2.*(conducta/(fact1*fact*widthwng)*(ttarm-t(51)))*tarmd*rlwng+ 

 & conducta/(fact1*fact*widthwng)*(ttarm-t(51))*tarmd*widthwng- 

 & conducta/(.5*wfuselg+.5*rlwng)*(t(51)-t(52))*tarmd*widthwng) 

c RHS Wing/Tarmac t(53): 

 qnetsky=(sig*t(53)**4-jj(14))/((1.-emista)/emista) 

 newt(53)=t(53)+deltime/(densta*rlwng*widthwng*tarmd*shcta)* 

 & ((qabstwa2(1,iperiod)-qnetsky-hht*(t(53)-ta))*rlwng*widthwng+ 

 & 2.*(conducta/(fact1*fact*widthwng)*(ttarm-t(53)))*tarmd*rlwng+ 

 & conducta/(fact1*fact*widthwng)*(ttarm-t(53))*tarmd*widthwng- 

 & conducta/(.5*wfuselg+.5*rlwng)*(t(53)-t(52))*tarmd*widthwng) 

c MIDDLE Wing/Tarmac t(52): 

 qnetsky=(sig*t(52)**4-jj(13))/((1.-emista)/emista) 

 newt(52)=t(52)+deltime/(densta*wfuselg*widthwng*tarmd*shcta)* 

 & ((qabstwb(1,iperiod)-qnetsky-hht*(t(52)-ta))*wfuselg*widthwng+ 

 & 2.*(conducta/(fact1*fact*widthwng)*(ttarm-t(51)))*tarmd*wfuselg 

 & +conducta/(.5*wfuselg+.5*rlwng)*(t(51)-t(52))*tarmd*widthwng+ 

 & conducta/(.5*wfuselg+.5*rlwng)*(t(53)-t(52))*tarmd*widthwng) 

 t(51)=newt(51) 

 t(52)=newt(52) 

 t(53)=newt(53) 

 temptrn(51)=t(51)-459.67 

 temptrn(52)=t(52)-459.67 

 temptrn(53)=t(53)-459.67 

 return 

 end 

c 

 subroutine updradio(itotitr,iperiod,istep,qsky,qtarm,emisf, 

 & emisw,eemisf,eemisw,rff,rfw) 

c purpose: updates the radiosity (14 of them) during the cooling phase 

c ie:a*radiosity=tempfact 

c called from: trnsien1 and trnsien2

c calls: ludcmp,lubksb 

 common/c1/n,ntrial,tolx,tolf 

 common/c2/x(200),fvec(200),a(200,200) 

 common/sc1/indx(200),b(200) 

 common/c4/widthwng,lengtwng,hwng,locwng,lplane,lfuselg,hfuselg,

 & wfuselg 

 common/c5/trnsm,absorbta,emista,tarmd,densta,conducta,shcta 

 common/c6/absorbff,emisff,absorbww,emisww,absorbfl,emisfl,absorbfi 

 &,emisfi,absorbpg,emispg,absorftp,emisrftp,absorfbt,emisrfbt 

 &,absorbfb,emisfb 

 common/c7/nperiods,envtdata(100,20) 

 common/c12/a1,a2,a3,a4,a5,a6(100),a7(100) 

 common/temp1/t1(1000) 

 common/temp2/t2(1000) 

A-98
 common/temp3/t3(1000) 

 common/temp4/t4(1000),tpg1,tpg2,tpg3 

 common/temp5/t5(1000) 

 common/radiosit/radiosj(14) 

 common/vf1/fbwa(6,6),fbwb(6,6) 

 common/vf2/fa11,fa12,fa13,fa14,fa15,fa16,fa21,fa22,fa23,fa24, 

 & fa25,fa26,fa31,fa32,fa33,fa34,fa35,fa36,fa41,fa42, 

 & fa43,fa44,fa45,fa46,fa51,fa52,fa53,fa54,fa55,fa56, 

 & fa61,fa62,fa63,fa64,fa65,fa66 

 common/vf4/fa57,fa68 

 common/vf5/fc11,fc12,fc13,fc14,fc15,fc16,fc21,fc22,fc23,fc24, 

 & fc25,fc26,fc31,fc32,fc33,fc34,fc35,fc36,fc41,fc42, 

 & fc43,fc44,fc45,fc46,fc51,fc52,fc53,fc54,fc55,fc56, 

 & fc61,fc62,fc63,fc64,fc65,fc66 

 common/result/temp(200),temptrn(200) 

 dimension t(100),aa(14,14),iindx(14) 

 dimension tempfact(14) 

 parameter(sig=.1714e-8) 

 real lfuselg 

c 

c Variables:

c tempfact()=updated term that is a function of the previous temperature 

c aa(,)=matrix relating radiosity to tempfact; to be inverted 

c radiosity()=updated radiosity 

c 

 if(iperiod.eq.1.and.istep.eq.1)then 

c find the a matrix relating radiosity jj to the temperature term tempfact 

c because the a matrix is a constant, you need to find it and its 

c inverse only once for a given paint scheme 

c 

 n=14 

c initialize a(i,j) to zeroes initially because most of the coefficients are 

c zeroes 

 do 15 i=1,n 

do 15 j=1,n 

 a(i,j)=0.0 

 15 continue

c 

c LHS Wing/Fuselage :::Wing 

c J1:::a(1,1),a(1,5),tempfact(1); from fn(24): 

 a(1,1)=1/(1/fa12)+1/(1/fa13)+1/(1/fa14)+1/(1/fa15)+1/(1/fa16) 

 & +rfw/((1-rfw*eemisw)/eemisw) 

 a(1,5)=-1/(1/fa12) 

c LHS Wing/Fuselage :::Fuselage 

c J5:::a(5,1),a(5,5),tempfact(5); from fn(28): 

 a(5,1)=-1/(1/fa21) 

 a(5,5)=1/(1/fa21)+1/(1/fa23)+1/(1/fa24)+1/(1/fa25)+1/(1/fa26) 

 & +rff/((1-rff*eemisf)/eemisf) 

c RHS Wing/Fuselage :::Wing 

c J3:::a(3,3),a(3,6),tempfact(3); from fn(26): 

 a(3,3)=1/(1/fa12)+1/(1/fa13)+1/(1/fa14)+1/(1/fa15)+1/(1/fa16) 

A-99
 & +rfw/((1-rfw*eemisw)/eemisw) 

 a(3,6)=-1/(1/fa12) 

c RHS Wing/Fuselage :::Fuselage 

c J6:::a(6,3),a(6,6),tempfact(6); from fn(30): 

 a(6,3)=-1/(1/fa21) 

 a(6,6)=1/(1/fa21)+1/(1/fa23)+1/(1/fa24)+1/(1/fa25)+1/(1/fa26) 

 & +rff/((1-rff*eemisf)/eemisf) 

c LHS Wing/Tarmac :::Wing 

c J2:::a(2,2),a(2,12),a(2,13),a(2,14),tempfact(2); from fn(20): 

 a(2,2)=1/(1/fbwa(3,1))+1/(1/fbwa(3,2))+1/(1/fbwa(3,4))+ 

 & 1/(1/fbwa(3,5))+1/(1/fbwa(3,6))+1/((1-emisw)/emisw) 

 a(2,12)=-1/(1/fbwa(3,1)) 

 a(2,13)=-fbwb(4,1)/(1/fbwa(3,2)) 

 a(2,14)=-fbwa(4,1)*fbwb(4,2)/(1/fbwa(3,2)) 

c RHS Wing/Tarmac :::Wing 

c J4:::a(4,4),a(4,12),a(4,13),a(4,14),tempfact(4); from fn(22): 

 a(4,4)=1/(1/fbwa(3,1))+1/(1/fbwa(3,2))+1/(1/fbwa(3,4))+ 

 & 1/(1/fbwa(3,5))+1/(1/fbwa(3,6))+1/((1-emisw)/emisw) 

 a(4,12)=-fbwa(2,1)*fbwb(2,4)/(1/fbwa(3,4)) 

 a(4,13)=-fbwb(2,1)/(1/fbwa(3,4)) 

 a(4,14)=-1/(1/fbwa(3,1)) 

c MIDDLE Wing/Tarmac :::Wing 

c J7:::a(7,7),a(7,12),a(7,13),a(7,14),tempfact(7); from fn(46): 

 a(7,7)=1/(1/fbwb(3,1))+1/(1/fbwb(3,2))+1/(1/fbwb(3,4))+ 

 & 1/(1/fbwb(3,5))+1/(1/fbwb(3,6))+1/((1-emisfb)/emisfb) 

 a(7,12)=-fbwa(2,1)/(1/fbwb(3,4)) 

 a(7,13)=-1/(1/fbwb(3,1)) 

 a(7,14)=-fbwa(4,1)/(1/fbwb(3,2)) 

c LHS Wing/Tarmac :::Tarmac 

c J12:::a(12,2),a(12,4),a(12,7),a(12,12),tempfact(12); from fn(54): 

 a(12,2)=-1/(1/fbwa(1,3)) 

 a(12,4)=-fbwa(4,3)*fbwb(4,2)/(1/fbwa(1,2)) 

 a(12,7)=-fbwb(4,3)/(1/fbwa(1,2)) 

 a(12,12)=1/(1/fbwa(1,2))+1/(1/fbwa(1,3))+1/(1/fbwa(1,4))+ 

 & 1/(1/fbwa(1,5))+1/(1/fbwa(1,6))+1/((1-emista)/emista) 

c RHS Wing/Tarmac :::Tarmac 

c J14:::a(14,2),a(14,4),a(14,7),a(14,14),tempfact(14); from fn(56): 

 a(14,2)=-fbwa(2,3)*fbwb(2,4)/(1/fbwa(1,4)) 

 a(14,4)=-1/(1/fbwa(1,3)) 

 a(14,7)=-fbwb(2,3)/(1/fbwa(1,4)) 

 a(14,14)=1/(1/fbwa(1,2))+1/(1/fbwa(1,3))+1/(1/fbwa(1,4))+ 

 & 1/(1/fbwa(1,5))+1/(1/fbwa(1,6))+1/((1-emista)/emista) 

c MIDDLE Wing/Tarmac :::Tarmac 

c J13:::a(13,2),a(13,4),a(13,7),a(13,13),tempfact(13); from fn(55): 

 a(13,2)=-fbwa(2,3)/(1/fbwb(1,4)) 

 a(13,4)=-fbwa(4,3)/(1/fbwb(1,2)) 

 a(13,7)=-1/(1/fbwb(1,3)) 

 a(13,13)=1/(1/fbwb(1,2))+1/(1/fbwb(1,3))+1/(1/fbwb(1,4))+ 

 & 1/(1/fbwb(1,5))+1/(1/fbwb(1,6))+1/((1-emista)/emista) 

c FUSELAGE INTERIOR (CABIN) 

c LHS Fuselage Interior (Cabin) 

A-100
c J8:::a(8,8),a(8,9),a(8,10),a(8,11),tempfact(8); from fn(47): 

 a(8,8)=1/(1/fc41)+1/(1/fc42)+1/(1/fc43)+1/(1/fc45)+1/(1/fc46) 

 & +1/((1-emisfi)/emisfi) 

 a(8,9)=-1/(1/fc43) 

 a(8,10)=-1/(1/fc42) 

 a(8,11)=-1/(1/fc41) 

c RHS Fuselage Interior (Cabin) 

c J10:::a(10,8),a(10,9),a(10,10),a(10,11),tempfact(10); from fn(49): 

 a(10,8)=-1/(1/fc24) 

 a(10,9)=-1/(1/fc23) 

 a(10,10)=1/(1/fc21)+1/(1/fc23)+1/(1/fc24)+1/(1/fc25)+1/(1/fc26) 

 & +1/((1-emisfi)/emisfi) 

 a(10,11)=-1/(1/fc21) 

c BOTTOM (floor) Fuselage Interior (Cabin) 

c J11:::a(11,8),a(11,9),a(11,10),a(11,11),tempfact(11); from fn(50): 

 a(11,8)=-1/(1/fc14) 

 a(11,9)=-1/(1/fc13) 

 a(11,10)=-1/(1/fc12) 

 a(11,11)=1/(1/fc12)+1/(1/fc13)+1/(1/fc14)+1/(1/fc15)+1/(1/fc16) 

 & +1/((1-emisfl)/emisfl) 

c TOP (combined roof and plexiglass structure) Fuselage Interior (Cabin) 

c J9:::a(9,8),a(9,9),a(9,10),a(9,11),tempfact(9); from fn(48): 

 c1=a1/(a1+a2) 

 c2=a2/(a1+a2) 

 emispgrf=c1*emispg+c2*emisrfbt 

 a(9,8)=-1/(1/fc34) 

 a(9,9)=1/(1/fc31)+1/(1/fc32)+1/(1/fc34)+1/(1/fc35)+1/(1/fc36) 

 & +1/((1-emispgrf)/emispgrf) 

 a(9,10)=-1/(1/fc32) 

 a(9,11)=-1/(1/fc31) 

c if(itotitr.le.nperiods)write(17,*)'matrix a' 

c do 10 i=1,14 

c 10 if(itotitr.le.nperiods)write(17,80)(a(i,j),j=1,14) 

 80 format(2x,500(1x,e10.3)) 

 call ludcmp 

c if(itotitr.le.nperiods)write(17,*)'decomposed matrix a' 

c do 20 i=1,14 

c 20 if(itotitr.le.nperiods)write(17,80)(a(i,j),j=1,14) 

c replace a(i,j) with aa(i,j), bec. a(i,j) gets changed in the steady-state 

c analysis phase. same thing for index(i) 

 do 25 i=1,14 

 iindx(i)=indx(i) 

 do 25 j=1,14 

 25 aa(i,j)=a(i,j) 

 return 

 endif 

c 

c find the updated temperature term tempfact(i) 

c ie A*J=T, where T is the tempfact. then find updated J. A is constant 

c first relate t(i) to t1(i),t2(i),t3(i),tpg3,t4(i) and t5(i) obtained from 

c transient analysis 

A-101
 t(1)=t1(1) 

 t(8)=t1(27) 

 t(9)=t2(1) 

 t(16)=t2(27) 

 t(17)=t3(1) 

 t(38)=t3(12) 

 t(39)=t3(16) 

 t(24)=t3(27) 

 t(36)=tpg3 

 t(61)=t4(12) 

 t(35)=t4(16) 

 t(25)=t4(39) 

 t(51)=t5(51) 

 t(52)=t5(52) 

 t(53)=t5(53) 

c LHS Wing/Fuselage :::Wing 

c J1:::a(1,1),a(1,5),tempfact(1); from fn(24): 

 q1=qsky+fa57*(emisf*sig*t(17)**4 

 & +(1-emisf)*(qsky+.5*qtarm+.5*(1.-emista)*qsky)) 

 q2=qsky+fa68*(emisf*sig*t(17)**4 

 & +(1-emisf)*(qsky+.5*qtarm+.5*(1.-emista)*qsky)) 

 tempfact(1)=qsky/(1/fa13)+qsky/(1/fa14)+q1/(1/fa15)+q2/(1/fa16) 

 & +sig*t(1)**4/((1-rfw*eemisw)/eemisw) 

c LHS Wing/Fuselage :::Fuselage 

c J5:::a(5,1),a(5,5),tempfact(5); from fn(28): 

 q1=qsky+.5*qtarm+.5*(1-emista)*qsky 

 tempfact(5)=qsky/(1/fa23)+q1/(1/fa24)+q1/(1/fa25)+q1/(1/fa26) 

 & +sig*t(17)**4/((1-rff*eemisf)/eemisf) 

c RHS Wing/Fuselage :::Wing 

c J3:::a(3,3),a(3,6),tempfact(3); from fn(26): 

 q1=qsky+fa57*(emisf*sig*t(24)**4 

 & +(1-emisf)*(qsky+.5*qtarm+.5*(1.-emista)*qsky)) 

 q2=qsky+fa68*(emisf*sig*t(24)**4 

 & +(1-emisf)*(qsky+.5*qtarm+.5*(1.-emista)*qsky)) 

 tempfact(3)=qsky/(1/fa13)+qsky/(1/fa14)+q1/(1/fa15)+q2/(1/fa16) 

 & +sig*t(9)**4/((1-rfw*eemisw)/eemisw) 

c RHS Wing/Fuselage :::Fuselage 

c J6:::a(6,3),a(6,6),tempfact(6); from fn(30): 

 q1=qsky+.5*qtarm+.5*(1-emista)*qsky 

 tempfact(6)=qsky/(1/fa23)+q1/(1/fa24)+q1/(1/fa25)+q1/(1/fa26) 

 & +sig*t(24)**4/((1-rff*eemisf)/eemisf) 

c LHS Wing/Tarmac :::Wing 

c J2:::a(2,2),a(2,12),a(2,13),a(2,14),tempfact(2); from fn(20): 

c find qskytarm first 

 qskytarm=qsky+0.5*qtarm+0.5*(1.-emista)*qsky 

 tempfact(2)=1/(1/fbwa(3,2))* 

 & ((qskytarm*fbwa(4,2)+qskytarm*fbwa(4,5)+qskytarm*fbwa(4,6))* 

 & fbwb(4,2)+qskytarm*fbwb(4,5)+qskytarm*fbwb(4,6))+ 

 & qskytarm/(1/fbwa(3,4))+qskytarm/(1/fbwa(3,5))+ 

 & qskytarm/(1/fbwa(3,6))+sig*t(8)**4/((1-emisw)/emisw) 

c RHS Wing/Tarmac :::Wing 

A-102
c J4:::a(4,4),a(4,12),a(4,13),a(4,14),tempfact(4); from fn(22): 

 tempfact(4)=qskytarm/(1/fbwa(3,2))+1/(1/fbwa(3,4))* 

 & ((qskytarm*fbwa(2,4)+qskytarm*fbwa(2,5)+qskytarm*fbwa(2,6))* 

 & fbwb(2,4)+qskytarm*fbwb(2,5)+qskytarm*fbwb(2,6))+ 

 & qskytarm/(1/fbwa(3,5))+ 

 & qskytarm/(1/fbwa(3,6))+sig*t(16)**4/((1-emisw)/emisw) 

c MIDDLE Wing/Tarmac :::Wing 

c J7:::a(7,7),a(7,12),a(7,13),a(7,14),tempfact(7); from fn(46): 

 tempfact(7)=1/(1/fbwb(3,2))* 

 & (qskytarm*fbwa(4,2)+qskytarm*fbwa(4,5)+qskytarm*fbwa(4,6))+ 

 & 1/(1/fbwb(3,4))* 

 & (qskytarm*fbwa(2,4)+qskytarm*fbwa(2,5)+qskytarm*fbwa(2,6))+ 

 & qskytarm/(1/fbwb(3,5))+ 

 & qskytarm/(1/fbwb(3,6))+ 

 & sig*t(25)**4/((1-emisfb)/emisfb)

c LHS Wing/Tarmac :::Tarmac 

c J12:::a(12,2),a(12,4),a(12,7),a(12,12),tempfact(12); from fn(54): 

 tempfact(12)=1/(1/fbwa(1,2))* 

 & ((qsky*fbwa(4,2)+qsky*fbwa(4,5)+qsky*fbwa(4,6))* 

 & fbwb(4,2)+qsky*fbwb(4,5)+qsky*fbwb(4,6))+ 

 & qsky/(1/fbwa(1,4))+qsky/(1/fbwa(1,5))+ 

 & qsky/(1/fbwa(1,6))+sig*t(51)**4/((1-emista)/emista) 

c RHS Wing/Tarmac :::Tarmac 

c J14:::a(14,2),a(14,4),a(14,7),a(14,14),tempfact(14); from fn(56): 

 tempfact(14)=qsky/(1/fbwa(1,2))+1/(1/fbwa(1,4))* 

 & ((qsky*fbwa(2,4)+qsky*fbwa(2,5)+qsky*fbwa(2,6))* 

 & fbwb(2,4)+qsky*fbwb(2,5)+qsky*fbwb(2,6))+ 

 & qsky/(1/fbwa(1,5))+ 

 & qsky/(1/fbwa(1,6))+sig*t(53)**4/((1-emista)/emista) 

c MIDDLE Wing/Tarmac :::Tarmac 

c J13:::a(13,2),a(13,4),a(13,7),a(13,13),tempfact(13); from fn(55): 

 tempfact(13)=1/(1/fbwb(1,2))* 

 & (qsky*fbwa(4,2)+qsky*fbwa(4,5)+qsky*fbwa(4,6))+ 

 & 1/(1/fbwb(1,4))* 

 & (qsky*fbwa(2,4)+qsky*fbwa(2,5)+qsky*fbwa(2,6))+ 

 & qsky/(1/fbwb(1,5))+ 

 & qsky/(1/fbwb(1,6))+ 

 & sig*t(52)**4/((1-emista)/emista)

c FUSELAGE INTERIOR (CABIN) 

c find the weighted average temperatures tend and tpgrf first. 

 aa1=lfuselg*wfuselg 

 aa2=a1 

 aa3=lfuselg*hfuselg 

 aa4=aa3 

 aa5=a2 

 atot=aa1+aa2+aa3+aa4+aa5 

 cc1=aa1/atot 

 cc2=aa2/atot 

 cc3=aa3/atot 

 cc4=aa4/atot 

 cc5=aa5/atot 

A-103
 tend=cc1*t(35)+cc2*t(36)+cc3*t(38)+cc4*t(39)+cc5*t(61) 

 tpgrf=c1*t(36)+c2*t(61) 

c LHS Fuselage Interior (Cabin) 

c J8:::a(8,8),a(8,9),a(8,10),a(8,11),tempfact(8); from fn(47): 

 tempfact(8)=sig*tend**4/(1/fc45)+sig*tend**4/(1/fc46) 

 & +sig*t(38)**4/((1-emisfi)/emisfi) 

c RHS Fuselage Interior (Cabin) 

c J10:::a(10,8),a(10,9),a(10,10),a(10,11),tempfact(10); from fn(49): 

 tempfact(10)=sig*tend**4/(1/fc25)+sig*tend**4/(1/fc26) 

 & +sig*t(39)**4/((1-emisfi)/emisfi) 

c BOTTOM (floor) Fuselage Interior (Cabin) 

c J11:::a(11,8),a(11,9),a(11,10),a(11,11),tempfact(11); from fn(50): 

 tempfact(11)=sig*tend**4/(1/fc15)+sig*tend**4/(1/fc16) 

 & +sig*t(35)**4/((1-emisfl)/emisfl) 

c TOP (combined roof and plexiglass structure) Fuselage Interior (Cabin) 

c J9:::a(9,8),a(9,9),a(9,10),a(9,11),tempfact(9); from fn(48): 

 tempfact(9)=sig*tend**4/(1/fc35)+sig*tend**4/(1/fc36) 

 & +sig*tpgrf**4/((1-emispgrf)/emispgrf) 

c 

c now find the updated radiosity 

c first replace tempfact(i) with a column vector b(i) to be 

c input into lubksb 

 do 30 i=1,14 

 30 b(i)=tempfact(i) 

c if(itotitr.eq.1)write(17,80)(b(i),i=1,14) 

c put aa(i,j) into a(i,j); because a(i,j) has changed in the steady-state 

c phase. same thing for indx(i) 

 do 50 i=1,14 

 indx(i)=iindx(i) 

 do 50 j=1,14 

 50 a(i,j)=aa(i,j) 

c input n (n was 67) 

 n=14 

 call lubksb 

c lubksb now returns b(i) which is the updated radiosity 

c for clarity transfer b(i) to radiosj(i) 

 do 40 i=1,14 

 40 radiosj(i)=b(i) 

c if(itotitr.eq.1)write(17,80)(radiosj(i),i=1,14) 

 do 45 i=1,14 

 45 temptrn(53+i)=radiosj(i) 

 return 

 end 

c 

 subroutine hcoolcon(itotitr,l,v,h) 

c purpose: determines the convective coefficient according to the flight 

c speed 

c called from: trnsien1 and trnsien2

c calls: none

 common/c7/nperiods,envtdata(100,20) 

 common/c10/k,thdiffus,visckine,prandtl 

A-104
 real l,llamn,nusllamn,k 

 parameter(rellamn=5.e5) 

c 

c Variables:

c nusllamn=Nusselt number in the laminar zone 

c rellamn=Reynold's number for laminar flow 

c 

 rel=v*l/visckine 

c if(itotitr.le.nperiods)write(12,*)'coolingneg,l,v,h',l,v,h 

 if(rel.le.rellamn)then 

 nusllamn=0.664*rel**(1./2)*prandtl**(1./3) 

 h=k*nusllamn/l 

 elseif(rel.gt.rellamn)then 

 llamn=rellamn*visckine/v

 nusllamn=0.664*rellamn**(1./2)*prandtl**(1./3) 

 hlamn=k*nusllamn/llamn 

 hturb=0.0296*k/(l-llamn)*prandtl**(1./3)*(v/visckine)**(4./5) 

 & *(5./4)*(l**(4./5)-llamn**(4./5)) 

 h=1./l*(llamn*hlamn+(l-llamn)*hturb) 

 endif 

c if(itotitr.le.nperiods)write(12,*)'cooling,v,h',v,h 

 return 

 end 

c 

c 

c 

 subroutine lhswng2(ithtrial,itotitr,iperiod,istep,theta,ta,hhw, 

 & rfw,absorbw,emisw,eemisw,sunfact,isolfoun) 

c purpose: transient analysis of the left wing (the wing that is not under 

c shade) 

c called from: newraph2 

c calls: none

 common/c2/tt(200),ffn(200),jacob(200,200) 

 common/c3/tth(100),kk(100) 

 common/c6/absorbff,emisff,absorbww,emisww,absorbfl,emisfl,absorbfi 

 &,emisfi,absorbpg,emispg,absorftp,emisrftp,absorfbt,emisrfbt 

 &,absorbfb,emisfb 

 common/c10/aircondc,thdiffus,visckine,prandtl 

 common/c14/thicknes(100),density(100),conductv(100),sheatcap(100)

 common/reflecq/qreflin(6,100),qscabsin(6,100),qreflfwr(6,100), 

 &qreflfwl(6,100),qscabfwr(6,100),qscabfwl(6,100),qabstwa1(6,100), 

 &qabstwa2(6,100),qabstwb(6,100) 

 common/temp1/t(1000) 

 common/data1/deltime 

 common/radiosit/jj(14) 

 common/result/temp(200),temptrn(200) 

 dimension newt(100),k(100),th(100),dens(100),shc(100),fn(100) 

 dimension tjacob(27,27) 

 real kk,k,jj,newt,jacob

 parameter(sig=.1714e-8) 

 parameter(fpi=3.141593/180.) 

A-105
 parameter(f=0.2) 

c 

c Variables:

c f=fraction of an element thickness 

c 

c Note: 

c A total of 9 layers are used to model the wing. the fifth layer is usually either 

c air or fuel or a combination of both

c 

 if(itotitr.eq.1.and.istep.eq.1)then 

c reference thickness (th): 

 th(1)=tth(1) 

 th(2)=tth(2) 

 th(3)=tth(3) 

 th(4)=tth(4) 

 th(5)=tth(29) 

 th(6)=tth(30) 

 th(7)=tth(5) 

 th(8)=tth(6) 

 th(9)=tth(7) 

c reference conductivity (k): 

 k(1)=kk(1) 

 k(2)=kk(2) 

 k(3)=kk(3) 

 k(4)=kk(4) 

 k(5)=kk(29) 

 k(6)=kk(30) 

 k(7)=kk(5) 

 k(8)=kk(6) 

 k(9)=kk(7) 

c reference density (dens): 

 dens(1)=density(1) 

 dens(2)=density(2) 

 dens(3)=density(3) 

 dens(4)=density(4) 

 dens(5)=density(5) 

 dens(6)=density(6) 

 dens(7)=density(7) 

 dens(8)=density(8) 

 dens(9)=density(9) 

c reference specific heat capacity (shc): 

 shc(1)=sheatcap(1) 

 shc(2)=sheatcap(2) 

 shc(3)=sheatcap(3) 

 shc(4)=sheatcap(4) 

 shc(5)=sheatcap(5) 

 shc(6)=sheatcap(6) 

 shc(7)=sheatcap(7) 

 shc(8)=sheatcap(8) 

 shc(9)=sheatcap(9) 

 endif 

A-106
 if(istep.eq.1)then 

c reference the initial temp. (t): each layer is partitioned into 3 control volumes. 

c so the total number of temperatures = (nlayer=9)*3=27. note that after step 1, ie at 

c step 2 and beyond the steady state temperature tt(i) is replaced with newt(i). so 

c all steady state temperatures are lost. they can be retrieved, however from 

c the /result/temp(i) variable in the form of degrees Farenheit.

 t(1)=tt(1) 

 t(2)=(tt(1)+tt(2))/2. 

 t(3)=tt(2) 

 t(4)=tt(2) 

 t(5)=(tt(2)+tt(3))/2. 

 t(6)=tt(3) 

 t(7)=tt(3) 

 t(8)=(tt(3)+tt(4))/2. 

 t(9)=tt(4) 

 t(10)=tt(4) 

 t(11)=(tt(4)+tt(62))/2. 

 t(12)=tt(62) 

 t(13)=tt(62) 

 t(14)=(tt(62)+tt(63))/2. 

 t(15)=tt(63) 

 t(16)=tt(63) 

 t(17)=(tt(63)+tt(5))/2. 

 t(18)=tt(5) 

 t(19)=tt(5) 

 t(20)=(tt(5)+tt(6))/2. 

 t(21)=tt(6) 

 t(22)=tt(6) 

 t(23)=(tt(6)+tt(7))/2. 

 t(24)=tt(7) 

 t(25)=tt(7) 

 t(26)=(tt(7)+tt(8))/2. 

 t(27)=tt(8) 

c for the nonlinear solution of the new transient temperatures newt, one needs an 

c initial guess for the set of newt's. what better guess than the steady state 

c temperatures (as above) prior to the start of the transient analysis 

 do 10 i=1,27 

 10 newt(i)=t(i) 

c reference the initial radiosity (jj) 

 jj(1)=tt(40) 

 jj(2)=tt(41) 

 return 

 endif 

c 

c formulate transient cooling: develop the function ffn(i), which is the fvec 

c in the non-linear transient solution process. 

c 

c first need to replace newt(i) with the incoming tt(i), where tt(i) is 

c the solution from newraph2 subroutine for step=istep and at iteration=ithtrial. 

c note that the original tt(i), ie the steady state temperatures are all destroyed 

c after step 1, ie at step 2 and greater. 

A-107
c 

 if(ithtrial.eq.1)go to 65

do 50 i=1,27

 newt(i)=tt(i) 

 50 continue

c 

 65 if(isolfoun.eq.1)then 

 if(ithtrial.eq.1)then 

 do 55 i=1,27

 newt(i)=tt(i) 

 55 continue 

 endif 

 do 60 i=1,27 

 60 t(i)=newt(i) 

 temptrn(1)=t(1)-459.67 

 temptrn(2)=(t(3)+t(4))/2.-459.67 

 temptrn(3)=(t(6)+t(7))/2.-459.67 

 temptrn(4)=(t(9)+t(10))/2.-459.67 

 temptrn(5)=(t(12)+t(13))/2.-459.67 

 temptrn(6)=(t(15)+t(16))/2.-459.67 

 temptrn(7)=(t(18)+t(19))/2.-459.67 

 temptrn(8)=(t(21)+t(22))/2.-459.67 

 temptrn(9)=(t(24)+t(25))/2.-459.67 

 temptrn(10)=t(27)-459.67 

 return 

 endif 

c 

c upper wing surface temp (newt1), as well as internal temp newt2 and newt3: and fn(1), 

c fn(2) and fn(3). 

c 

 qnetsky=(sig*newt(1)**4-rfw*jj(1))/((1.-rfw*eemisw)/eemisw) 

 fn(1)=-newt(1)+t(1)+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((absorbw*qreflfwl(1,iperiod)*sin(fpi*theta)+ 

 & qscabfwl(1,iperiod))*sunfact+ 

 & (absorbw*qreflfwr(1,iperiod)*sin(fpi*theta)+ 

 & qscabfwr(1,iperiod))*(1-sunfact) 

 & -qnetsky-hhw*(newt(1)-ta)- 

 & (newt(1)-newt(2))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 fn(2)=-newt(2)+t(2)+deltime/(dens(1)*shc(1)*(1-f)*th(1))* 

 & ((newt(1)-newt(2))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (newt(2)-newt(3))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 fn(3)=-newt(3)+t(3)+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((newt(2)-newt(3))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (newt(3)-newt(4))/(f/4*th(1)/k(1)+f/4*th(2)/k(2))) 

c 

c internal nodes (all conduction): newt4-newt24: fn(4)-fn(24): layers 2 to 8. 

c 

do 20 i=2,8 

 ii=(i-1)*3 

c 

c beginning element within layer i: newt4,newt7,newt10,newt13,newt16,newt19,newt22: 

A-108
c 

 fn(ii+1)= 

 & -newt(ii+1)+t(ii+1)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((newt(ii)-newt(ii+1))/(f/4*th(i-1)/k(i-1)+f/4*th(i)/k(i))- 

 & (newt(ii+1)-newt(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

c 

c middle element within layer i: newt5,newt8,newt11,newt14,newt17,newt20,newt23: 

c 

 fn(ii+2)= 

 & -newt(ii+2)+t(ii+2)+deltime/(dens(i)*shc(i)*(1-f)*th(i))* 

 & ((newt(ii+1)-newt(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (newt(ii+2)-newt(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

c 

c ending element within layer i: newt6,newt9,newt12,newt15,newt18,newt21,newt24: 

c 

 fn(ii+3)= 

 & -newt(ii+3)+t(ii+3)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((newt(ii+2)-newt(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (newt(ii+3)-newt(ii+4))/(f/4*th(i)/k(i)+f/4*th(i+1)/k(i+1))) 

 20 continue 

c 

c lower wing surface temp newt27, as well as internal temp newt25 and newt26: fn(27), 

c fn(26) and fn(25). 

c 

 fn(25)=-newt(25)+t(25)+ 

 & deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & ((newt(24)-newt(25))/ 

 & (f/4*th(8)/k(8)+f/4*th(9)/k(9))- 

 & (newt(25)-newt(26))/ 

 & (f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

 fn(26)=-newt(26)+t(26)+ 

 & deltime/(dens(9)*shc(9)*(1-f)*th(9))* 

 & ((newt(25)-newt(26))/ 

 & (f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))- 

 & (newt(26)-newt(27))/ 

 & (f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

 qnetsky=(sig*newt(27)**4-jj(2))/((1.-emisw)/emisw) 

 fn(27)=-newt(27)+t(27)+ 

 & deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & (qabstwa1(3,iperiod)-qnetsky-hhw*(newt(27)-ta)+ 

 & (newt(26)-newt(27))/ 

 & (f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

c 

c transfer the fn(i) into the global ffn(i), which is used together with jacobian

c to find the p(i) and then the new tt(i) in newraph2 subroutine. 

c 

 do 30 i=1,27 

 30 ffn(i)=fn(i) 

c 

c now find the jacobian, global jacob(120,120) from the partial derivative of fn(i). 

c find tjacob(27,27) first. then put it into the global jacob(120,120). 

A-109
c 

c first initialize tjacob(27,27) to 0.0. 

c note: most of the tjacob(27,27) are 0.0 

c 

 do 70 itj=1,27 

 do 70 jtj=1,27 

 70 tjacob(itj,jtj)=0.0 

c 

c first find tjacob(1,),tjacob(2,),tjacob(3,) for the upper wing surface functions 

c fn(1),fn(2),fn(3) 

c 

c tjacob(1,1),tjacob(1,2):

c 

 tjacob(1,1)=-1+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & (-(4*sig*newt(1)**3)/((1.-rfw*eemisw)/eemisw)-hhw*(1)- 

 & (1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 tjacob(1,2)=deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & (-(-1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

c 

c tjacob(2,1),tjacob(2,2),tjacob(2,3): 

c 

 tjacob(2,1)=deltime/(dens(1)*shc(1)*(1-f)*th(1))* 

 & ((1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 tjacob(2,2)=-1+deltime/(dens(1)*shc(1)*(1-f)*th(1))* 

 & ((-1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 tjacob(2,3)=deltime/(dens(1)*shc(1)*(1-f)*th(1))* 

 & (-(-1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

c 

c tjacob(3,2),tjacob(3,3),tjacob(3,4): 

c 

 tjacob(3,2)=deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 tjacob(3,3)=-1+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((-1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (1)/(f/4*th(1)/k(1)+f/4*th(2)/k(2))) 

 tjacob(3,4)=deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & (-(-1)/(f/4*th(1)/k(1)+f/4*th(2)/k(2))) 

c 

c now find jacobians for the internal conduction layers: tjacob(4,) to tjacob(24,) 

c for functions fn(4) to fn (24): 

c 

do 120 i=2,8

 ii=(i-1)*3 

c 

c find tjacob(4,),tjacob(7,),tjacob(10,),tjacob(13,),tjacob(16,),tjacob(19,) 

c and tjacob(22,) for beginning element within 

c layer i: from fn(4),fn(7),fn(10),fn(13),fn(16),fn(19) and fn(22): 

c 

 tjacob(ii+1,ii)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((1)/(f/4*th(i-1)/k(i-1)+f/4*th(i)/k(i))) 

A-110
 tjacob(ii+1,ii+1)=-1+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((-1)/(f/4*th(i-1)/k(i-1)+f/4*th(i)/k(i))- 

 & (1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 tjacob(ii+1,ii+2)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & (-(-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i)))

c 

c find tjacob(5,),tjacob(8,),tjacob(11,),tjacob(14,),tjacob(17,),tjacob(20,) 

c and tjacob(23,) for middle element within 

c layer i: from fn(5),fn(8),fn(11),fn(14),fn(17),fn(20) and fn(23): 

c 

 tjacob(ii+2,ii+1)=deltime/(dens(i)*shc(i)*(1-f)*th(i))* 

 & ((1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 tjacob(ii+2,ii+2)=-1+deltime/(dens(i)*shc(i)*(1-f)*th(i))* 

 & ((-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 tjacob(ii+2,ii+3)=deltime/(dens(i)*shc(i)*(1-f)*th(i))* 

 & (-(-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

c 

c find tjacob(6,),tjacob(9,),tjacob(12,),tjacob(15,),tjacob(18,),tjacob(21,) 

c and tjacob(24,) for ending element within 

c layer i: from fn(6),fn(9),fn(12),fn(15),fn(18),fn(21) and fn(24): 

c 

 tjacob(ii+3,ii+2)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 tjacob(ii+3,ii+3)=-1+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (1)/(f/4*th(i)/k(i)+f/4*th(i+1)/k(i+1))) 

 tjacob(ii+3,ii+4)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & (-(-1)/(f/4*th(i)/k(i)+f/4*th(i+1)/k(i+1)))

 120 continue 

c 

c now find tjacob(25,),tjacob(26,),tjacob(27,) for the lower wing surface functions 

c fn(25),fn(26) and fn(27) 

c 

c tjacob(25,24),tjacob(25,25) and tjacob(25,26):

c 

 tjacob(25,24)=deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & ((1)/(f/4*th(8)/k(8)+f/4*th(9)/k(9))) 

 tjacob(25,25)=-1+deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & ((-1)/(f/4*th(8)/k(8)+f/4*th(9)/k(9))- 

 & (1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

 tjacob(25,26)=deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & (-(-1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

c 

c tjacob(26,25),tjacob(26,26) and tjacob(26,27):

c 

 tjacob(26,25)=deltime/(dens(9)*shc(9)*(1-f)*th(9))* 

 & ((1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

 tjacob(26,26)=-1+deltime/(dens(9)*shc(9)*(1-f)*th(9))* 

 & ((-1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))- 

 & (1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

A-111
 tjacob(26,27)=deltime/(dens(9)*shc(9)*(1-f)*th(9))* 

 & (-(-1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

c 

c tjacob(27,26) and tjacob(27,27): 

c 

 tjacob(27,26)=deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & ((1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

 tjacob(27,27)=-1+deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & (-(4*sig*newt(27)**3)/((1.-emisw)/emisw)-hhw*(1)+ 

 & (-1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

c 

c jacobian analysis is done. now put tjacob(i,j) into the global jacobian matrix 

c jacob(ijacob,jjacob), where ijacob starts and ends at 1 and 27, 

c and jjacob starts and ends at 1 and 27. note: most jacob values are 0's 

c 

 do 130 i=1,27 

 do 130 j=1,27 

 jacob(i,j)=tjacob(i,j) 

 130 continue 

 return 

 end 

c 

c 

 subroutine rhswng2(ithtrial,itotitr,iperiod,istep,theta,ta,hhw, 

 & rfw,absorbw,emisw,eemisw,sunfact,isolfoun) 

c purpose: transient analysis of the right wing (the wing that is under 

c shade) 

c called from: newraph2 

c calls: none

 common/c2/tt(200),ffn(200),jacob(200,200) 

 common/c3/tth(100),kk(100) 

 common/c6/absorbff,emisff,absorbww,emisww,absorbfl,emisfl,absorbfi 

 &,emisfi,absorbpg,emispg,absorftp,emisrftp,absorfbt,emisrfbt 

 &,absorbfb,emisfb 

 common/c10/aircondc,thdiffus,visckine,prandtl 

 common/c14/thicknes(100),density(100),conductv(100),sheatcap(100)

 common/reflecq/qreflin(6,100),qscabsin(6,100),qreflfwr(6,100), 

 &qreflfwl(6,100),qscabfwr(6,100),qscabfwl(6,100),qabstwa1(6,100), 

 &qabstwa2(6,100),qabstwb(6,100) 

 common/temp2/t(1000) 

 common/data1/deltime 

 common/radiosit/jj(14) 

 common/result/temp(200),temptrn(200) 

 dimension newt(100),k(100),th(100),dens(100),shc(100),fn(100) 

 dimension tjacob(27,27) 

 real kk,k,jj,newt,jacob

 parameter(sig=.1714e-8) 

 parameter(fpi=3.141593/180.) 

 parameter(f=0.2) 

c 

A-112
c Variables:

c f=fraction of an element thickness 

c 

c Note: 

c A total of 9 layers are used to model the wing. the fifth layer is usually either 

c air or fuel or a combination of both

c 

 if(itotitr.eq.1.and.istep.eq.1)then 

c reference thickness (th): 

 th(1)=tth(1) 

 th(2)=tth(2) 

 th(3)=tth(3) 

 th(4)=tth(4) 

 th(5)=tth(29) 

 th(6)=tth(30) 

 th(7)=tth(5) 

 th(8)=tth(6) 

 th(9)=tth(7) 

c reference conductivity (k): 

 k(1)=kk(1) 

 k(2)=kk(2) 

 k(3)=kk(3) 

 k(4)=kk(4) 

 k(5)=kk(29) 

 k(6)=kk(30) 

 k(7)=kk(5) 

 k(8)=kk(6) 

 k(9)=kk(7) 

c reference density (dens): 

 dens(1)=density(1) 

 dens(2)=density(2) 

 dens(3)=density(3) 

 dens(4)=density(4) 

 dens(5)=density(5) 

 dens(6)=density(6) 

 dens(7)=density(7) 

 dens(8)=density(8) 

 dens(9)=density(9) 

c reference specific heat capacity (shc): 

 shc(1)=sheatcap(1) 

 shc(2)=sheatcap(2) 

 shc(3)=sheatcap(3) 

 shc(4)=sheatcap(4) 

 shc(5)=sheatcap(5) 

 shc(6)=sheatcap(6) 

 shc(7)=sheatcap(7) 

 shc(8)=sheatcap(8) 

 shc(9)=sheatcap(9) 

 endif 

 if(istep.eq.1)then 

c reference the initial temp. (t): each layer is partitioned into 3 control volumes. 

A-113
c so the total number of temperatures = (nlayer=9)*3=27. note that after step 1, ie at 

c step 2 and beyond the steady state temperature tt(i) is replaced with newt(i). so 

c all steady state temperatures are lost. they can be retrieved, however from 

c the /result/temp(i) variable in the form of degrees Fahrenheit.

 t(1)=tt(9) 

 t(2)=(tt(9)+tt(10))/2. 

 t(3)=tt(10) 

 t(4)=tt(10) 

 t(5)=(tt(10)+tt(11))/2. 

 t(6)=tt(11) 

 t(7)=tt(11) 

 t(8)=(tt(11)+tt(12))/2. 

 t(9)=tt(12) 

 t(10)=tt(12) 

 t(11)=(tt(12)+tt(64))/2. 

 t(12)=tt(64) 

 t(13)=tt(64) 

 t(14)=(tt(64)+tt(65))/2. 

 t(15)=tt(65) 

 t(16)=tt(65) 

 t(17)=(tt(65)+tt(13))/2. 

 t(18)=tt(13) 

 t(19)=tt(13) 

 t(20)=(tt(13)+tt(14))/2. 

 t(21)=tt(14) 

 t(22)=tt(14) 

 t(23)=(tt(14)+tt(15))/2. 

 t(24)=tt(15) 

 t(25)=tt(15) 

 t(26)=(tt(15)+tt(16))/2. 

 t(27)=tt(16) 

c for the nonlinear solution of the new transient temperatures newt, one needs an 

c initial guess for the set of newt's. what better guess than the steady state 

c temperatures (as above) prior to the start of the transient analysis 

 do 10 i=1,27 

 10 newt(i)=t(i) 

c reference the initial radiosity (jj): 

 jj(3)=tt(42) 

 jj(4)=tt(43) 

 return 

 endif 

c 

c formulate transient cooling: develop the function fn(i), which is the fvec 

c in the non-linear transient solution process. 

c 

c first need to replace newt(i) with the incoming tt(i), where tt(i) is 

c the solution from newraph2 subroutine for step=istep and at iteration=ithtrial. 

c note that the original tt(i), ie the steady state temperatures are all destroyed 

c after step 1, ie at step 2 and greater. 

c 

 if(ithtrial.eq.1)go to 65

A-114
do 50 i=1,27

 newt(i)=tt(27+i) 

 50 continue

c 

 65 if(isolfoun.eq.1)then 

 if(ithtrial.eq.1)then 

 do 55 i=1,27

 newt(i)=tt(27+i) 

 55 continue 

 endif 

 do 60 i=1,27 

 60 t(i)=newt(i) 

 temptrn(11)=t(1)-459.67 

 temptrn(12)=(t(3)+t(4))/2.-459.67 

 temptrn(13)=(t(6)+t(7))/2.-459.67 

 temptrn(14)=(t(9)+t(10))/2.-459.67 

 temptrn(15)=(t(12)+t(13))/2.-459.67 

 temptrn(16)=(t(15)+t(16))/2.-459.67 

 temptrn(17)=(t(18)+t(19))/2.-459.67 

 temptrn(18)=(t(21)+t(22))/2.-459.67 

 temptrn(19)=(t(24)+t(25))/2.-459.67 

 temptrn(20)=t(27)-459.67 

 return 

 endif 

c 

c upper wing surface temp (newt1), as well as internal temp newt2 and newt3: and fn(1), 

c fn(2) and fn(3). 

c 

 qnetsky=(sig*newt(1)**4-rfw*jj(3))/((1.-rfw*eemisw)/eemisw) 

 fn(1)=-newt(1)+t(1)+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((absorbw*qreflfwr(1,iperiod)*sin(fpi*theta)+ 

 & qscabfwr(1,iperiod))*sunfact+ 

 & (absorbw*qreflfwl(1,iperiod)*sin(fpi*theta)+ 

 & qscabfwl(1,iperiod))*(1-sunfact) 

 & -qnetsky-hhw*(newt(1)-ta)- 

 & (newt(1)-newt(2))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 fn(2)=-newt(2)+t(2)+deltime/(dens(1)*shc(1)*(1-f)*th(1))* 

 & ((newt(1)-newt(2))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (newt(2)-newt(3))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 fn(3)=-newt(3)+t(3)+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((newt(2)-newt(3))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (newt(3)-newt(4))/(f/4*th(1)/k(1)+f/4*th(2)/k(2))) 

c 

c internal nodes (all conduction): newt4-newt24: fn(4)-fn(24): layers 2 to 8. 

c 

do 20 i=2,8 

 ii=(i-1)*3 

c 

c beginning element within layer i: newt4,newt7,newt10,newt13,newt16,newt19,newt22: 

c 

 fn(ii+1)= 

A-115
 & -newt(ii+1)+t(ii+1)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((newt(ii)-newt(ii+1))/(f/4*th(i-1)/k(i-1)+f/4*th(i)/k(i))- 

 & (newt(ii+1)-newt(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

c 

c middle element within layer i: newt5,newt8,newt11,newt14,newt17,newt20,newt23: 

c 

 fn(ii+2)= 

 & -newt(ii+2)+t(ii+2)+deltime/(dens(i)*shc(i)*(1-f)*th(i))* 

 & ((newt(ii+1)-newt(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (newt(ii+2)-newt(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

c 

c ending element within layer i: newt6,newt9,newt12,newt15,newt18,newt21,newt24: 

c 

 fn(ii+3)= 

 & -newt(ii+3)+t(ii+3)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((newt(ii+2)-newt(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (newt(ii+3)-newt(ii+4))/(f/4*th(i)/k(i)+f/4*th(i+1)/k(i+1))) 

 20 continue 

c 

c lower wing surface temp newt27, as well as internal temp newt25 and newt26: fn(27), 

c fn(26) and fn(25). 

c 

 fn(25)=-newt(25)+t(25)+ 

 & deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & ((newt(24)-newt(25))/ 

 & (f/4*th(8)/k(8)+f/4*th(9)/k(9))- 

 & (newt(25)-newt(26))/ 

 & (f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

 fn(26)=-newt(26)+t(26)+ 

 & deltime/(dens(9)*shc(9)*(1-f)*th(9))* 

 & ((newt(25)-newt(26))/ 

 & (f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))- 

 & (newt(26)-newt(27))/ 

 & (f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

 qnetsky=(sig*newt(27)**4-jj(4))/((1.-emisw)/emisw) 

 fn(27)=-newt(27)+t(27)+ 

 & deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & (qabstwa2(3,iperiod)-qnetsky-hhw*(newt(27)-ta)+ 

 & (newt(26)-newt(27))/ 

 & (f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

c 

c transfer the fn(i) into the global ffn(i), which is used together with jacobian

c to find the p(i) and then the new tt(i) in newraph2 subroutine. 

c 

 do 30 i=1,27 

 30 ffn(27+i)=fn(i) 

c 

c now find the jacobian, global jacob(120,120) from the partial derivative of fn(i). 

c find tjacob(27,27) first. then put it into the global jacob(120,120). 

c 

c first initialize tjacob(27,27) to 0.0. 

A-116
c note: most of the tjacob(27,27) are 0.0 

c 

 do 70 itj=1,27 

 do 70 jtj=1,27 

 70 tjacob(itj,jtj)=0.0 

c 

c first find tjacob(1,),tjacob(2,),tjacob(3,) for the upper wing surface functions 

c fn(1),fn(2),fn(3) 

c 

c tjacob(1,1),tjacob(1,2):

c 

 tjacob(1,1)=-1+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & (-(4*sig*newt(1)**3)/((1.-rfw*eemisw)/eemisw)-hhw*(1)- 

 & (1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 tjacob(1,2)=deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & (-(-1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

c 

c tjacob(2,1),tjacob(2,2),tjacob(2,3): 

c 

 tjacob(2,1)=deltime/(dens(1)*shc(1)*(1-f)*th(1))* 

 & ((1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 tjacob(2,2)=-1+deltime/(dens(1)*shc(1)*(1-f)*th(1))* 

 & ((-1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 tjacob(2,3)=deltime/(dens(1)*shc(1)*(1-f)*th(1))* 

 & (-(-1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

c 

c tjacob(3,2),tjacob(3,3),tjacob(3,4): 

c 

 tjacob(3,2)=deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 tjacob(3,3)=-1+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((-1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (1)/(f/4*th(1)/k(1)+f/4*th(2)/k(2))) 

 tjacob(3,4)=deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & (-(-1)/(f/4*th(1)/k(1)+f/4*th(2)/k(2))) 

c 

c now find jacobians for the internal conduction layers: tjacob(4,) to tjacob(24,) 

c for functions fn(4) to fn (24): 

c 

do 120 i=2,8

 ii=(i-1)*3 

c 

c find tjacob(4,),tjacob(7,),tjacob(10,),tjacob(13,),tjacob(16,),tjacob(19,) 

c and tjacob(22,) for beginning element within 

c layer i: from fn(4),fn(7),fn(10),fn(13),fn(16),fn(19) and fn(22): 

c 

 tjacob(ii+1,ii)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((1)/(f/4*th(i-1)/k(i-1)+f/4*th(i)/k(i))) 

 tjacob(ii+1,ii+1)=-1+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((-1)/(f/4*th(i-1)/k(i-1)+f/4*th(i)/k(i))- 

A-117
 & (1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 tjacob(ii+1,ii+2)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & (-(-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i)))

c 

c find tjacob(5,),tjacob(8,),tjacob(11,),tjacob(14,),tjacob(17,),tjacob(20,) 

c and tjacob(23,) for middle element within 

c layer i: from fn(5),fn(8),fn(11),fn(14),fn(17),fn(20) and fn(23): 

c 

 tjacob(ii+2,ii+1)=deltime/(dens(i)*shc(i)*(1-f)*th(i))* 

 & ((1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 tjacob(ii+2,ii+2)=-1+deltime/(dens(i)*shc(i)*(1-f)*th(i))* 

 & ((-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 tjacob(ii+2,ii+3)=deltime/(dens(i)*shc(i)*(1-f)*th(i))* 

 & (-(-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

c 

c find tjacob(6,),tjacob(9,),tjacob(12,),tjacob(15,),tjacob(18,),tjacob(21,) 

c and tjacob(24,) for ending element within 

c layer i: from fn(6),fn(9),fn(12),fn(15),fn(18),fn(21) and fn(24): 

c 

 tjacob(ii+3,ii+2)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 tjacob(ii+3,ii+3)=-1+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (1)/(f/4*th(i)/k(i)+f/4*th(i+1)/k(i+1))) 

 tjacob(ii+3,ii+4)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & (-(-1)/(f/4*th(i)/k(i)+f/4*th(i+1)/k(i+1)))

 120 continue 

c 

c now find tjacob(25,),tjacob(26,),tjacob(27,) for the lower wing surface functions 

c fn(25),fn(26) and fn(27) 

c 

c tjacob(25,24),tjacob(25,25) and tjacob(25,26):

c 

 tjacob(25,24)=deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & ((1)/(f/4*th(8)/k(8)+f/4*th(9)/k(9))) 

 tjacob(25,25)=-1+deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & ((-1)/(f/4*th(8)/k(8)+f/4*th(9)/k(9))- 

 & (1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

 tjacob(25,26)=deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & (-(-1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

c 

c tjacob(26,25),tjacob(26,26) and tjacob(26,27):

c 

 tjacob(26,25)=deltime/(dens(9)*shc(9)*(1-f)*th(9))* 

 & ((1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

 tjacob(26,26)=-1+deltime/(dens(9)*shc(9)*(1-f)*th(9))* 

 & ((-1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))- 

 & (1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

 tjacob(26,27)=deltime/(dens(9)*shc(9)*(1-f)*th(9))* 

A-118
 & (-(-1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

c 

c tjacob(27,26) and tjacob(27,27): 

c 

 tjacob(27,26)=deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & ((1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

 tjacob(27,27)=-1+deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & (-(4*sig*newt(27)**3)/((1.-emisw)/emisw)-hhw+ 

 & (-1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

c 

c jacobian analysis is done. now put tjacob(i,j) into the global jacobian matrix 

c jacob(ijacob,jjacob), where ijacob starts and ends at 28 and 54, 

c and jjacob starts and ends at 28 and 54. note: most jacob values are 0's 

c 

 do 130 i=1,27 

 do 130 j=1,27 

 jacob(27+i,27+j)=tjacob(i,j) 

 130 continue 

 return 

 end 

c 

c 

 subroutine fuslglr2(ithtrial,itotitr,icool,iperiod,istep,theta,ta, 

 & hhf,hhc,rff,absorbf,eemisf,sunfact,sunfact2,isolfoun) 

c purpose: transient analysis of the fuselage (left to right) 

c called from: newraph2 

c calls: none

 common/c2/tt(200),ffn(200),jacob(200,200) 

 common/c3/tth(100),kk(100) 

 common/c6/absorbff,emisff,absorbww,emisww,absorbfl,emisfl,absorbfi 

 &,emisfi,absorbpg,emispg,absorftp,emisrftp,absorfbt,emisrfbt 

 &,absorbfb,emisfb 

 common/c10/aircondc,thdiffus,visckine,prandtl 

 common/c14/thicknes(100),density(100),conductv(100),sheatcap(100)

 common/reflecq/qreflin(6,100),qscabsin(6,100),qreflfwr(6,100), 

 &qreflfwl(6,100),qscabfwr(6,100),qscabfwl(6,100),qabstwa1(6,100), 

 &qabstwa2(6,100),qabstwb(6,100) 

 common/temp3/t(1000) 

 common/data1/deltime 

 common/radiosit/jj(14) 

 common/result/temp(200),temptrn(200) 

 dimension newt(100),k(100),th(100),dens(100),shc(100),fn(100) 

 dimension tfn(100),tjacob(27,27),ttjacob(27,27) 

 real kk,k,jj,newt,nusselt2,jacob 

 parameter(sig=.1714e-8,gravity=32.2) 

 parameter(fpi=3.141593/180.) 

 parameter(f=0.2) 

c 

c Variables:

c f=fraction of an element thickness 

c 

A-119
c Note: 

c A total of 9 layers are used to model the fuselage cabin (widthwise). 

c the fifth layer is air space (maybe in turbulence). the fifth layer is 

c not modelled because of its complexity. any reference to the 13,14 and 15 

c control volume is only a dummy. 

c 

 if(itotitr.eq.1.and.istep.eq.1)then 

c reference thickness (th): 

 th(1)=tth(8) 

 th(2)=tth(9) 

 th(3)=tth(10) 

 th(4)=tth(23) 

 th(5)=tth(11) 

 th(6)=tth(23) 

 th(7)=tth(10) 

 th(8)=tth(9) 

 th(9)=tth(8) 

c reference conductivity (k): 

 k(1)=kk(8) 

 k(2)=kk(9) 

 k(3)=kk(10) 

 k(4)=kk(23) 

 k(5)=kk(11) 

 k(6)=kk(23) 

 k(7)=kk(10) 

 k(8)=kk(9) 

 k(9)=kk(8) 

c reference density (dens): 

 dens(1)=density(10) 

 dens(2)=density(11) 

 dens(3)=density(12) 

 dens(4)=density(13) 

 dens(5)=density(14) 

 dens(6)=density(13) 

 dens(7)=density(12) 

 dens(8)=density(11) 

 dens(9)=density(10) 

c reference specific heat capacity (shc): 

 shc(1)=sheatcap(10)

 shc(2)=sheatcap(11)

 shc(3)=sheatcap(12)

 shc(4)=sheatcap(13)

 shc(5)=sheatcap(14)

 shc(6)=sheatcap(13)

 shc(7)=sheatcap(12)

 shc(8)=sheatcap(11)

 shc(9)=sheatcap(10)

 endif 

 if(istep.eq.1)then 

c reference the initial temp. (t): each layer is partitioned into 3 control volumes. 

c so the total number of temperatures = (nlayer=9)*3=27. note that after step 1, ie at 

A-120
c step 2 and beyond the steady state temperature tt(i) is replaced with newt(i). so 

c all steady state temperatures are lost. they can be retrieved, however from 

c the /result/temp(i) variable in the form of degrees Fahrenheit.

 t(1)=tt(17) 

 t(2)=(tt(17)+tt(18))/2. 

 t(3)=tt(18) 

 t(4)=tt(18) 

 t(5)=(tt(18)+tt(19))/2. 

 t(6)=tt(19) 

 t(7)=tt(19) 

 t(8)=(tt(19)+tt(20))/2. 

 t(9)=tt(20) 

 t(10)=tt(20) 

 t(11)=(tt(20)+tt(38))/2. 

 t(12)=tt(38) 

 t(13)=tt(38) 

 t(14)=(tt(38)+tt(39))/2. 

 t(15)=tt(39) 

 t(16)=tt(39) 

 t(17)=(tt(39)+tt(21))/2. 

 t(18)=tt(21) 

 t(19)=tt(21) 

 t(20)=(tt(21)+tt(22))/2. 

 t(21)=tt(22) 

 t(22)=tt(22) 

 t(23)=(tt(22)+tt(23))/2. 

 t(24)=tt(23) 

 t(25)=tt(23) 

 t(26)=(tt(23)+tt(24))/2. 

 t(27)=tt(24) 

c for the nonlinear solution of the new transient temperatures newt, one needs an 

c initial guess for the set of newt's. what better guess than the steady state 

c temperatures (as above) prior to the start of the transient analysis 

 do 10 i=1,27 

 10 newt(i)=t(i) 

c reference the initial radiosity (jj) 

 jj(5)=tt(44) 

 jj(6)=tt(45) 

 jj(8)=tt(47) 

 jj(10)=tt(49) 

 return 

 endif 

c 

c formulate transient cooling: develop the function ffn(i), which is the fvec 

c in the non-linear transient solution process. 

c 

c first need to replace newt(i) with the incoming tt(i), where tt(i) is 

c the solution from newraph2 subroutine for step=istep and at iteration=ithtrial. 

c note that the original tt(i), ie the steady state temperatures are all destroyed 

c after step 1, ie at step 2 and greater. 

c note that the internal layer (cabin air) is not modelled due to its complexity. 

A-121
c result should be ok. so there are (27-3) control volumes = 24, rearranged as 

c shown here and at the end of this subroutine. 

c 

 if(ithtrial.eq.1)go to 65

 isum=0 

do 50 i=1,27

 if(i.eq.13.or.i.eq.14.or.i.eq.15)goto 50 

 isum=isum+1 

 newt(i)=tt(54+isum) 

 50 continue

c 

 65 if(isolfoun.eq.1)then 

 if(ithtrial.eq.1)then 

 isum=0 

 do 55 i=1,27

 if(i.eq.13.or.i.eq.14.or.i.eq.15)goto 55 

 isum=isum+1 

 newt(i)=tt(54+isum) 

 55 continue 

 endif 

 do 60 i=1,27 

 60 t(i)=newt(i) 

 temptrn(39)=t(1)-459.67 

 temptrn(40)=(t(3)+t(4))/2.-459.67 

 temptrn(41)=(t(6)+t(7))/2.-459.67 

 temptrn(42)=(t(9)+t(10))/2.-459.67 

 temptrn(43)=t(12)-459.67 

 temptrn(44)=t(16)-459.67 

 temptrn(45)=(t(18)+t(19))/2.-459.67 

 temptrn(46)=(t(21)+t(22))/2.-459.67 

 temptrn(47)=(t(24)+t(25))/2.-459.67 

 temptrn(48)=t(27)-459.67 

 return 

 endif 

c 

c find the rayleigh and nusselt number within the cabin when it is closed (icool=2) 

c and perhaps turbulence is present. these numbers are required to find the convective 

c coefficient within the cabin. note that when the cabin door is opened (icool=1) 

c the convective coefficient is hhc, coming from trnsien2 subroutine. 

c 

 if(icool.eq.2)then 

 rayleig2=gravity*1./((newt(12)+newt(16))/2.)*abs(newt(12)-newt(16))* 

 & th(5)**3/(thdiffus*visckine) 

 if(rayleig2.le.(1e3*(0.2+prandtl)/prandtl))then 

 nusselt2=1.0 

 elseif(rayleig2.gt.(1e3*(0.2+prandtl)/prandtl).and.rayleig2.lt.

 & 1.0e9)then 

 nusselt2=0.18*((prandtl/(0.2+prandtl))*rayleig2)**0.29 

 endif 

 endif 

c 

A-122
c formulate transient cooling: 

c 

c LHS surface temp of fuselage (newt1), as well as internal temp newt2 and newt3: as well 

c as functions fn(1), fn(2) and fn(3). 

c 

 qnetsky=(sig*newt(1)**4-rff*jj(5))/((1.-rff*eemisf)/eemisf) 

 fn(1)=-newt(1)+t(1)+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((absorbf*qreflfwl(2,iperiod)*cos(fpi*theta)+ 

 & qscabfwl(2,iperiod))*sunfact+ 

 & qscabfwr(2,iperiod)*(1-sunfact) 

 & -qnetsky-hhf*(newt(1)-ta)- 

 & (newt(1)-newt(2))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 fn(2)=-newt(2)+t(2)+deltime/(dens(1)*shc(1)*(1-f)*th(1))* 

 & ((newt(1)-newt(2))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (newt(2)-newt(3))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 fn(3)=-newt(3)+t(3)+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((newt(2)-newt(3))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (newt(3)-newt(4))/(f/4*th(1)/k(1)+f/4*th(2)/k(2))) 

c 

c internal nodes::: for layers 2 to 8 

c internal nodes::: beginning of control volume: newt(4,7,10,16,19,22): same as fn() 

c middle of control volume: newt(5,8,11,17,20,23): same as fn() 

c ending of control volume: newt(6,9,12,18,21,24): same as fn() 

c newt(13,14 and 16) for cabin air neglected: ie layer 5 

c 

do 20 i=2,8 

 if(i.eq.5)goto 20 

 ii=(i-1)*3 

c 

c beginning element within layer 2,3,4,6,7,8: 

c newt4,newt7,newt10,newt16,newt19,newt22: fn(4),fn(7),fn(10),fn(16),fn(19),fn(22) 

c 

 if(i.eq.6)then

 qnetcab=(sig*newt(ii+1)**4-jj(10))/((1.-emisfi)/emisfi) 

 if(icool.eq.1)then 

 fn(ii+1)= 

 & -newt(ii+1)+t(ii+1)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & (-(newt(ii+1)-newt(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & qnetcab+absorbfi*qreflin(1,iperiod)*cos(fpi*theta)+ 

 & qscabsin(1,iperiod)-hhc*(newt(ii+1)-ta)) 

 elseif(icool.eq.2)then 

 fn(ii+1)= 

 & -newt(ii+1)+t(ii+1)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((newt(ii-3)-newt(ii+1))/(f/4*th(i-2)/k(i-2)+ 

 & th(i-1)/(nusselt2*k(i-1))+f/4*th(i)/k(i))- 

 & (newt(ii+1)-newt(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & qnetcab+

 & (absorbfi*qreflin(1,iperiod)*cos(fpi*theta)+ 

 & qscabsin(1,iperiod))*sunfact2) 

c & (absorbfi*qreflin(3,iperiod)*cos(fpi*theta)+ 

c & qscabsin(3,iperiod))*(1-sunfact)) 

A-123
 endif 

 else 

 fn(ii+1)= 

 & -newt(ii+1)+t(ii+1)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((newt(ii)-newt(ii+1))/(f/4*th(i-1)/k(i-1)+f/4*th(i)/k(i))- 

 & (newt(ii+1)-newt(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 endif 

c 

c middle element within layer 2,3,4,6,7,8: 

c newt5,newt8,newt11,newt17,newt20,newt23: fn(5),fn(8),fn(11),fn(17),fn(20),fn(23) 

c 

 fn(ii+2)= 

 & -newt(ii+2)+t(ii+2)+deltime/(dens(i)*shc(i)*(1-f)*th(i))* 

 & ((newt(ii+1)-newt(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (newt(ii+2)-newt(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

c 

c ending element within layer 2,3,4,6,7,8: 

c newt6,newt9,newt12,newt18,newt21,newt24: fn(6),fn(9),fn(12),fn(18),fn(21),fn(24) 

c 

 if(i.eq.4)then

 qnetcab=(sig*newt(ii+3)**4-jj(8))/((1.-emisfi)/emisfi) 

 if(icool.eq.1)then 

 fn(ii+3)= 

 & -newt(ii+3)+t(ii+3)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((newt(ii+2)-newt(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & qnetcab+absorbfi*qreflin(3,iperiod)*cos(fpi*theta)+ 

 & qscabsin(3,iperiod)-hhc*(newt(ii+3)-ta)) 

 elseif(icool.eq.2)then 

 fn(ii+3)= 

 & -newt(ii+3)+t(ii+3)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((newt(ii+2)-newt(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (newt(ii+3)-newt(ii+7))/(f/4*th(i)/k(i)+

 & th(i+1)/(nusselt2*k(i+1))+f/4*th(i+2)/k(i+2))- 

 & qnetcab+ 

 & (absorbfi*qreflin(3,iperiod)*cos(fpi*theta)+ 

 & qscabsin(3,iperiod))*sunfact2) 

c & (absorbfi*qreflin(1,iperiod)*cos(fpi*theta)+ 

c & qscabsin(1,iperiod))*(1-sunfact)) 

 endif 

 else 

 fn(ii+3)= 

 & -newt(ii+3)+t(ii+3)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((newt(ii+2)-newt(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (newt(ii+3)-newt(ii+4))/(f/4*th(i)/k(i)+f/4*th(i+1)/k(i+1))) 

 endif 

 20 continue 

c 

c RHS surface temp of fuselage (newt27), as well as internal temp 

c newt25 and newt26: functions fn(27), fn(25) and fn(26). 

c 

 fn(25)=-newt(25)+t(25)+ 

A-124
 & deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & ((newt(24)-newt(25))/ 

 & (f/4*th(8)/k(8)+f/4*th(9)/k(9))- 

 & (newt(25)-newt(26))/ 

 & (f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

 fn(26)=-newt(26)+t(26)+ 

 & deltime/(dens(9)*shc(9)*(1-f)*th(9))* 

 & ((newt(25)-newt(26))/ 

 & (f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))- 

 & (newt(26)-newt(27))/ 

 & (f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

 qnetsky=(sig*newt(27)**4-rff*jj(6))/((1.-rff*eemisf)/eemisf) 

 fn(27)=-newt(27)+t(27)+ 

 & deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & (qscabfwr(2,iperiod)*sunfact+ 

 & (absorbf*qreflfwl(2,iperiod)*cos(fpi*theta)+ 

 & qscabfwl(2,iperiod))*(1-sunfact) 

 & -qnetsky-hhf*(newt(27)-ta)+ 

 & (newt(26)-newt(27))/ 

 & (f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

c 

c transfer fn(i) into tfn(i) so that fn(13),fn(14) and fn(15) are removed 

c and thus tfn(i) begins at tfn(1) and ends at tfn(24). fn(13),fn(14) and fn(15) 

c are dummies for the layer 5 (air in the cabin). 

c 

 isum=0 

 do 100 i=1,27 

 if(i.eq.13.or.i.eq.14.or.i.eq.15)goto 100 

 isum=isum+1 

 tfn(isum)=fn(i) 

 100 continue 

c 

c transfer the tfn(i) into the global ffn(i), which is used together with jacobian 

c to find the p(i) and then the new tt(i) in newraph2 subroutine. 

c 

 do 30 i=1,24 

 30 ffn(54+i)=tfn(i) 

c 

c now find the jacobian, global jacob(120,120) from the partial derivative of fn(i). 

c find tjacob(27,27) first. then put it into the global jacob(55..78,55..78). 

c note: tjacob(13,),tjacob(14,) and tjacob(15,) are just dummies to be removed later. 

c 

c first initialize tjacob(27,27) to 0.0. 

c note: most of the tjacob(27,27) are 0.0 

c 

 do 70 itj=1,27 

 do 70 jtj=1,27 

 70 tjacob(itj,jtj)=0.0 

c 

c first find tjacob(1,),tjacob(2,),tjacob(3,) for the LHS first layer functions 

c fn(1),fn(2),fn(3) 

A-125
c 

c tjacob(1,1) and tjacob(1,2): 

c 

 tjacob(1,1)=-1+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & (-(4*sig*newt(1)**3)/((1.-rff*eemisf)/eemisf)-hhf*(1)- 

 & (1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 tjacob(1,2)=deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & (-(-1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

c 

c tjacob(2,1),tjacob(2,2) and tjacob(2,3): 

c 

 tjacob(2,1)=deltime/(dens(1)*shc(1)*(1-f)*th(1))* 

 & ((1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 tjacob(2,2)=-1+deltime/(dens(1)*shc(1)*(1-f)*th(1))* 

 & ((-1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 tjacob(2,3)=deltime/(dens(1)*shc(1)*(1-f)*th(1))* 

 & (-(-1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

c 

c tjacob(3,2),tjacob(3,3) and tjacob(3,4): 

c 

 tjacob(3,2)=deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 tjacob(3,3)=-1+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((-1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (1)/(f/4*th(1)/k(1)+f/4*th(2)/k(2))) 

 tjacob(3,4)=deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & (-(-1)/(f/4*th(1)/k(1)+f/4*th(2)/k(2))) 

c 

c now find jacobians for the internal conduction layers: tjacob(4,) to tjacob(24,) 

c for functions fn(4) to fn (24): 

c 

do 120 i=2,8

 if(i.eq.5)goto 120 

 ii=(i-1)*3 

c 

c find tjacob(4,),tjacob(7,),tjacob(10,),tjacob(16,),tjacob(19,) 

c and tjacob(22,) for beginning element within 

c layer i: from fn(4),fn(7),fn(10),fn(16),fn(19) and fn(22): 

c note: tjacob(13) is skipped 

 if(i.eq.6)then

 if(icool.eq.1)then 

c tjacob(16,16),tjacob(16,17): 

 tjacob(ii+1,ii+1)=-1+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & (-(1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))-

 & (4*sig*newt(ii+1)**3)/((1.-emisfi)/emisfi)-hhc*(1)) 

 tjacob(ii+1,ii+2)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & (-(-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i)))

 elseif(icool.eq.2)then 

c tjacob(16,12),tjacob(16,16),tjacob(16,17): 

 if(nusselt2.eq.1.0)then 

A-126
 tjacob(ii+1,ii-3)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((1)/(f/4*th(i-2)/k(i-2)+

 & th(i-1)/(nusselt2*k(i-1))+f/4*th(i)/k(i))) 

 tjacob(ii+1,ii+1)=-1+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((-1)/(f/4*th(i-2)/k(i-2)+

 & th(i-1)/(nusselt2*k(i-1))+f/4*th(i)/k(i))- 

 & (1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))-

 & (4*sig*newt(ii+1)**3)/((1.-emisfi)/emisfi)) 

 tjacob(ii+1,ii+2)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & (-(-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 else

 const=0.18*( prandtl/(0.2+prandtl)*gravity*th(5)**3/ 

 & (0.5*(thdiffus*visckine)) )**0.29 

 dndt12=const*( ((newt(12)+newt(16))**(-0.29)*(0.29)* 

 & abs(newt(12)-newt(16))**(-0.71)*(1.0)) + 

 & (abs(newt(12)-newt(16))**0.29*(-0.29)* 

 & (newt(12)+newt(16))**(-1.29)*(1.0)) ) 

 dndt16=const*( ((newt(12)+newt(16))**(-0.29)*(0.29)* 

 & abs(newt(12)-newt(16))**(-0.71)*(-1.0)) + 

 & (abs(newt(12)-newt(16))**0.29*(-0.29)* 

 & (newt(12)+newt(16))**(-1.29)*(1.0)) ) 

 dcdt12=th(i-1)/k(i-1)*(-1.0)*nusselt2**(-2.0)*dndt12 

 dcdt16=th(i-1)/k(i-1)*(-1.0)*nusselt2**(-2.0)*dndt16 

 abcd=(f/4*th(i-2)/k(i-2)+ 

 & th(i-1)/(nusselt2*k(i-1))+f/4*th(i)/k(i)) 

 dydt12=abcd**(-1.0)*(1.0) + 

 & (newt(12)-newt(16))*( (-1.0)*abcd**(-2.0)*dcdt12 ) 

 dydt16=abcd**(-1.0)*(-1.0) + 

 & (newt(12)-newt(16))*( (-1.0)*abcd**(-2.0)*dcdt16 ) 

c 

 tjacob(ii+1,ii-3)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & (dydt12) 

 tjacob(ii+1,ii+1)=-1+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ( dydt16- 

 & (1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))-

 & (4*sig*newt(ii+1)**3)/((1.-emisfi)/emisfi) ) 

 tjacob(ii+1,ii+2)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ( -(-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i)) ) 

 endif

 endif 

 else 

 tjacob(ii+1,ii)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((1)/(f/4*th(i-1)/k(i-1)+f/4*th(i)/k(i))) 

 tjacob(ii+1,ii+1)=-1+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((-1)/(f/4*th(i-1)/k(i-1)+f/4*th(i)/k(i))- 

 & (1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 tjacob(ii+1,ii+2)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & (-(-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i)))

 endif 

c 

c find tjacob(5,),tjacob(8,),tjacob(11,),tjacob(17,),tjacob(20,) 

A-127
c and tjacob(23,) for middle element within 

c layer i: from fn(5),fn(8),fn(11),fn(17),fn(20) and fn(23): 

c note: tjacob(14) is skipped 

c 

 tjacob(ii+2,ii+1)=deltime/(dens(i)*shc(i)*(1-f)*th(i))* 

 & ((1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 tjacob(ii+2,ii+2)=-1+deltime/(dens(i)*shc(i)*(1-f)*th(i))* 

 & ((-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 tjacob(ii+2,ii+3)=deltime/(dens(i)*shc(i)*(1-f)*th(i))* 

 & (-(-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

c 

c find tjacob(6,),tjacob(9,),tjacob(12,),tjacob(18,),tjacob(21,) 

c and tjacob(24,) for ending element within 

c layer i: from fn(6),fn(9),fn(12),fn(18),fn(21) and fn(24): 

c note: tjacob(15) is skipped 

c 

 if(i.eq.4)then

 if(icool.eq.1)then 

c tjacob(12,11),tjacob(12,12): 

 tjacob(ii+3,ii+2)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 tjacob(ii+3,ii+3)=-1+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (4*sig*newt(ii+3)**3)/((1.-emisfi)/emisfi)-hhc*(1)) 

 elseif(icool.eq.2)then 

c tjacob(12,11),tjacob(12,12), tjacob(12,16): 

 if(nusselt2.eq.1.0)then 

 tjacob(ii+3,ii+2)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ( (1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i)) )

 tjacob(ii+3,ii+3)=-1+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ( (-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))-

 & (1)/(f/4*th(i)/k(i)+ 

 & th(i+1)/(nusselt2*k(i+1))+f/4*th(i+2)/k(i+2))- 

 & (4*sig*newt(ii+3)**3)/((1.-emisfi)/emisfi) ) 

 tjacob(ii+3,ii+7)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ( -(-1)/(f/4*th(i)/k(i)+ 

 & th(i+1)/(nusselt2*k(i+1))+f/4*th(i+2)/k(i+2))) 

 else 

 const=0.18*( prandtl/(0.2+prandtl)*gravity*th(5)**3/ 

 & (0.5*(thdiffus*visckine)) )**0.29 

 dndt12=const*( ((newt(12)+newt(16))**(-0.29)*(0.29)* 

 & abs(newt(12)-newt(16))**(-0.71)*(1.0)) + 

 & (abs(newt(12)-newt(16))**0.29*(-0.29)* 

 & (newt(12)+newt(16))**(-1.29)*(1.0)) ) 

 dndt16=const*( ((newt(12)+newt(16))**(-0.29)*(0.29)* 

 & abs(newt(12)-newt(16))**(-0.71)*(-1.0)) + 

 & (abs(newt(12)-newt(16))**0.29*(-0.29)* 

 & (newt(12)+newt(16))**(-1.29)*(1.0)) ) 

 dcdt12=th(i+1)/k(i+1)*(-1.0)*nusselt2**(-2.0)*dndt12 

 dcdt16=th(i+1)/k(i+1)*(-1.0)*nusselt2**(-2.0)*dndt16 

A-128
 abcd=(f/4*th(i)/k(i)+ 

 & th(i+1)/(nusselt2*k(i+1))+f/4*th(i+2)/k(i+2)) 

 dydt12=abcd**(-1.0)*(1.0) + 

 & (newt(12)-newt(16))*( (-1.0)*abcd**(-2.0)*dcdt12 ) 

 dydt16=abcd**(-1.0)*(-1.0) + 

 & (newt(12)-newt(16))*( (-1.0)*abcd**(-2.0)*dcdt16 ) 

c 

 tjacob(ii+3,ii+2)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ( (1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i)) ) 

 tjacob(ii+3,ii+3)=-1+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ( (-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (dydt12)- 

 & (4*sig*newt(ii+3)**3)/((1.-emisfi)/emisfi) ) 

 tjacob(ii+3,ii+7)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & (-(dydt16)) 

 endif 

 endif 

 else 

 tjacob(ii+3,ii+2)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 tjacob(ii+3,ii+3)=-1+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (1)/(f/4*th(i)/k(i)+f/4*th(i+1)/k(i+1))) 

 tjacob(ii+3,ii+4)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & (-(-1)/(f/4*th(i)/k(i)+f/4*th(i+1)/k(i+1)))

 endif 

 120 continue 

c 

c 

c now find tjacob(25,),tjacob(26,),tjacob(27,) for the RHS last layer surface functions 

c fn(25),fn(26) and fn(27) 

c 

c tjacob(25,24),tjacob(25,25) and tjacob(25,26):

c 

 tjacob(25,24)=deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & ((1)/(f/4*th(8)/k(8)+f/4*th(9)/k(9))) 

 tjacob(25,25)=-1+deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & ((-1)/(f/4*th(8)/k(8)+f/4*th(9)/k(9))- 

 & (1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

 tjacob(25,26)=deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & (-(-1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

c 

c tjacob(26,25),tjacob(26,26) and tjacob(26,27):

c 

 tjacob(26,25)=deltime/(dens(9)*shc(9)*(1-f)*th(9))* 

 & ((1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

 tjacob(26,26)=-1+deltime/(dens(9)*shc(9)*(1-f)*th(9))* 

 & ((-1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))- 

 & (1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

 tjacob(26,27)=deltime/(dens(9)*shc(9)*(1-f)*th(9))* 

 & (-(-1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

A-129
c 

c tjacob(27,26) and tjacob(27,27): outer most control volume

c 

 tjacob(27,26)=deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & ((1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

 tjacob(27,27)=-1+deltime/(dens(9)*shc(9)*f/2*th(9))* 

 & (-(4*sig*newt(27)**3)/((1.-rff*eemisf)/eemisf)-hhf*(1)+ 

 & (-1)/(f/4*th(9)/k(9)+(1-f)/2*th(9)/k(9))) 

c 

c jacobian analysis is done. now put tjacob(i,j) into the global jacobian matrix 

c jacob(ijacob,jjacob), where ijacob starts and ends at 55 and 78, 

c and jjacob starts and ends at 55 and 78. note: most jacob values are 0's 

c 

c 

c first transfer jacob(27,27) into ttjacob(24,24) so that rows and columns 

c 13, 14 and 15 are removed. 

c 

 isum1=0 

 do 140 i=1,27 

 if(i.eq.13.or.i.eq.14.or.i.eq.15)goto 140 

 isum1=isum1+1 

 isum2=0 

 do 125 j=1,27 

 if(j.eq.13.or.j.eq.14.or.j.eq.15)goto 125 

 isum2=isum2+1 

 ttjacob(isum1,isum2)=tjacob(i,j)

 125 continue 

 140 continue 

c 

c now put ttjacob into the global jacob 

c 

 do 130 i=1,24 

 do 130 j=1,24 

 jacob(54+i,54+j)=ttjacob(i,j) 

 130 continue 

 return 

 end 

c 

c 

 subroutine fuslgrf2(ithtrial,itotitr,icool,iperiod,istep,theta,ta, 

 & hhf,hhc,qdir,qsky,qtarm,sunfact2,isolfoun) 

c purpose: transient analysis of the fuselage (top to bottom) 

c called from: newraph2 

c calls: none

 common/c2/tt(200),ffn(200),jacob(200,200) 

 common/c3/tth(100),kk(100) 

 common/c4/widthwng,lengtwng,hwng,locwng,lplane,lfuselg,hfuselg,

 & wfuselg 

 common/c6/absorbff,emisff,absorbww,emisww,absorbfl,emisfl,absorbfi 

 &,emisfi,absorbpg,emispg,absorftp,emisrftp,absorfbt,emisrfbt 

 &,absorbfb,emisfb 

A-130
 common/c5/trnsm,absorbta,emista,tarmd,densta,conducta,shcta 

 common/c10/aircondc,thdiffus,visckine,prandtl 

 common/c12/a1,a2,a3,a4,a5,a6(100),a7(100) 

 common/c14/thicknes(100),density(100),conductv(100),sheatcap(100)

 common/reflecq/qreflin(6,100),qscabsin(6,100),qreflfwr(6,100), 

 &qreflfwl(6,100),qscabfwr(6,100),qscabfwl(6,100),qabstwa1(6,100), 

 &qabstwa2(6,100),qabstwb(6,100) 

 common/temp4/t(1000),tpg1,tpg2,tpg3 

 common/data1/deltime 

 common/radiosit/jj(14) 

 common/result/temp(200),temptrn(200) 

 dimension newt(100),k(100),th(100),dens(100),shc(100),fn(100) 

 dimension tfn(100),tjacob(42,42),ttjacob(42,42) 

 real kk,k,jj,newt,nusselt1,lfuselg,kpg,newtpg1,newtpg2,newtpg3 

 real jacob 

 parameter(sig=.1714e-8,gravity=32.2) 

 parameter(fpi=3.141593/180.) 

 parameter(f=0.2) 

c 

c Variables:

c f=fraction of an element thickness 

c 

c Note: 

c A total of 13 layers are used to model the fuselage cabin (heightwise), 

c the space between floor bottom and belly-pan top, and the belly itself. 

c the fifth layer is air space (maybe in turbulence). the fifth layer is 

c not modelled because of its complexity. any reference to the 13,14 and 15 

c control volume is only a dummy. 

c the tenth layer (or element) is air space (in between the floor bottom 

c and belly-pan top). it may be filled with fuel or just air 

c or both. this layer is modelled as conduction element using control volumes 

c 28, 29 and 30. 

 if(itotitr.eq.1.and. istep.eq.1)then 

c reference thickness (th): 

 th(1)=tth(27) 

 th(2)=tth(26) 

 th(3)=tth(25) 

 th(4)=tth(28) 

 th(5)=tth(24) 

 th(6)=tth(21) 

 th(7)=tth(20) 

 th(8)=tth(19) 

 th(9)=tth(18) 

 th(10)=tth(31) 

 th(11)=tth(14) 

 th(12)=tth(13) 

 th(13)=tth(12) 

c reference conductivity (k): 

 k(1)=kk(27) 

 k(2)=kk(26) 

 k(3)=kk(25) 

A-131
 k(4)=kk(28) 

 k(5)=kk(24) 

 k(6)=kk(21) 

 k(7)=kk(20) 

 k(8)=kk(19) 

 k(9)=kk(18) 

 k(10)=kk(31) 

 k(11)=kk(14) 

 k(12)=kk(13) 

 k(13)=kk(12) 

c reference density (dens): 

 dens(1)=density(15) 

 dens(2)=density(16) 

 dens(3)=density(17) 

 dens(4)=density(18) 

 dens(5)=density(19) 

 dens(6)=density(20) 

 dens(7)=density(21) 

 dens(8)=density(22) 

 dens(9)=density(23) 

 dens(10)=density(24) 

 dens(11)=density(25) 

 dens(12)=density(26) 

 dens(13)=density(27) 

c reference specific heat capacity: 

 shc(1)=sheatcap(15)

 shc(2)=sheatcap(16)

 shc(3)=sheatcap(17)

 shc(4)=sheatcap(18)

 shc(5)=sheatcap(19)

 shc(6)=sheatcap(20)

 shc(7)=sheatcap(21)

 shc(8)=sheatcap(22)

 shc(9)=sheatcap(23)

 shc(10)=sheatcap(24) 

 shc(11)=sheatcap(25) 

 shc(12)=sheatcap(26) 

 shc(13)=sheatcap(27) 

c also reference plexiglass thickness, conductivity, density 

c and specific heat capacity: 

 thpg=tth(22) 

 kpg=kk(22) 

 denspg=density(28) 

 shcpg=sheatcap(28)

c also find these values: 

 c1=a1/(a1+a2) 

 c2=a2/(a1+a2) 

 emispgrf=c1*emispg+c2*emisrfbt 

 endif 

 if(istep.eq.1)then 

c reference the initial temp. (t): 

A-132
 t(1)=tt(59) 

 t(2)=(tt(59)+tt(58))/2. 

 t(3)=tt(58) 

 t(4)=tt(58) 

 t(5)=(tt(58)+tt(57))/2. 

 t(6)=tt(57) 

 t(7)=tt(57) 

 t(8)=(tt(57)+tt(60))/2. 

 t(9)=tt(60) 

 t(10)=tt(60) 

 t(11)=(tt(60)+tt(61))/2. 

 t(12)=tt(61) 

 t(13)=tt(61) 

 t(14)=(tt(61)+tt(35))/2. 

 t(15)=tt(35) 

 t(16)=tt(35) 

 t(17)=(tt(35)+tt(34))/2. 

 t(18)=tt(34) 

 t(19)=tt(34) 

 t(20)=(tt(34)+tt(33))/2. 

 t(21)=tt(33) 

 t(22)=tt(33) 

 t(23)=(tt(33)+tt(32))/2. 

 t(24)=tt(32) 

 t(25)=tt(32) 

 t(26)=(tt(32)+tt(31))/2. 

 t(27)=tt(31) 

 t(28)=tt(31) 

 t(29)=(tt(31)+tt(28))/2. 

 t(30)=tt(28) 

 t(31)=tt(28) 

 t(32)=(tt(28)+tt(27))/2. 

 t(33)=tt(27) 

 t(34)=tt(27) 

 t(35)=(tt(27)+tt(26))/2. 

 t(36)=tt(26) 

 t(37)=tt(26) 

 t(38)=(tt(26)+tt(25))/2. 

 t(39)=tt(25) 

c also reference the intial plexiglass temp. 

 tpg1=tt(37) 

 tpg2=(tt(37)+tt(36))/2. 

 tpg3=tt(36) 

c for the nonlinear solution of the new transient temperatures newt, one needs an 

c initial guess for the set of newt's. what better guess than the steady state 

c temperatures (as above) prior to the start of the transient analysis. also

c assign initial guesses for newtpg's. 

 do 10 i=1,39 

 10 newt(i)=t(i) 

 newtpg1=tpg1 

 newtpg2=tpg2 

A-133
 newtpg3=tpg3 

c reference the initial radiosity 

 jj(7)=tt(46) 

 jj(9)=tt(48) 

 jj(11)=tt(50) 

 return 

 endif 

c 

c formulate transient cooling: develop the function ffn(i), which is the fvec 

c in the non-linear transient solution process. 

c 

c first need to replace newt(i) with the incoming tt(i), where tt(i) is 

c the solution from newraph2 subroutine for step=istep and at iteration=ithtrial. 

c note that the original tt(i), ie the steady state temperatures are all destroyed 

c after step 1, ie at step 2 and greater. 

c note that the internal layer (cabin air) is not modelled due to its complexity. 

c result should be ok. so there are (39-3) control volumes = 36, plus the

c three plexiglass control volumes with temperatures tpg1,tpg2 and tpg3.

c the plexiglass fvec are placed at the end of 

c fn ie: fn(40), fn(41) and fn(42). 

c 

 if(ithtrial.eq.1)go to 65

 isum=0 

do 50 i=1,39

 if(i.eq.13.or.i.eq.14.or.i.eq.15)goto 50 

 isum=isum+1 

 newt(i)=tt(78+isum) 

 50 continue

 newtpg1=tt(78+36+1) 

 newtpg2=tt(78+36+2) 

 newtpg3=tt(78+36+3) 

c 

 65 if(isolfoun.eq.1)then 

 if(ithtrial.eq.1)then 

 isum=0 

 do 55 i=1,39

 if(i.eq.13.or.i.eq.14.or.i.eq.15)goto 55 

 isum=isum+1 

 newt(i)=tt(78+isum) 

 55 continue 

 newtpg1=tt(78+36+1) 

 newtpg2=tt(78+36+2) 

 newtpg3=tt(78+36+3) 

 endif 

 do 60 i=1,39 

 60 t(i)=newt(i) 

 tpg1=newtpg1 

 tpg2=newtpg2 

 tpg3=newtpg3 

 temptrn(21)=t(1)-459.67 

 temptrn(22)=(t(3)+t(4))/2.-459.67 

A-134
 temptrn(23)=(t(6)+t(7))/2.-459.67 

 temptrn(24)=(t(9)+t(10))/2.-459.67 

 temptrn(25)=t(12)-459.67 

 temptrn(26)=t(16)-459.67 

 temptrn(27)=(t(18)+t(19))/2.-459.67 

 temptrn(28)=(t(21)+t(22))/2.-459.67 

 temptrn(29)=(t(24)+t(25))/2.-459.67 

 temptrn(30)=(t(27)+t(28))/2.-459.67 

 temptrn(31)=(t(27)+t(31))/2.-459.67 

 temptrn(32)=(t(27)+t(31))/2.-459.67 

 temptrn(33)=(t(27)+t(31))/2.-459.67 

 temptrn(34)=(t(27)+t(31))/2.-459.67 

 temptrn(35)=(t(30)+t(31))/2.-459.67 

 temptrn(36)=(t(33)+t(34))/2.-459.67 

 temptrn(37)=(t(36)+t(37))/2.-459.67 

 temptrn(38)=t(39)-459.67 

 temptrn(49)=tpg1-459.67 

 temptrn(50)=tpg3-459.67 

 return 

 endif 

c 

 tpgrf=c1*newtpg3+c2*newt(12) 

 qnet=(sig*tpgrf**4-jj(9))/((1.-emispgrf)/emispgrf) 

c 

 if(icool.eq.2)then 

 rayleig1=gravity*1./((tpgrf+newt(16))/2.)*(newt(16)-tpgrf)* 

 & th(5)**3/(thdiffus*visckine) 

 if(rayleig1.le.3e5)then 

 nusselt1=1.0 

 elseif(rayleig1.gt.3e5.and.rayleig1.le.7e9)then 

 nusselt1=0.069*rayleig1**(1./3.)*prandtl**.074 

 endif 

 endif 

c 

c formulate transient cooling: 

c 

c first for the only three temperatures of plexiglass: 

c fn(40),fn(41),fn(42): newtpg1,newtpg2,newtpg3. 

c note: newtpg1=outermost plexiglas temperature 

c note: newtpg3=innermost plexiglas temperature 

c note: newtpg2=middle plexiglas temperature 

c regard newtpg1,newtpg2 and newtpg3 as newt(40),newt(41) and newt(42) in the 

c jacobian tjacob matrix. 

c 

c 

 fn(40)=-newtpg1+tpg1+deltime/(a1*denspg*shcpg*f/2*thpg)* 

 & (emispg*qsky*a1+emispg*.5*(1.-emista)*qsky*a5 

 & +emispg*.5*qtarm*a5 

 & +f/2*absorbpg*qdir*cos(fpi*theta)*a6(iperiod) 

 & +f/2*absorbpg*qdir*sin(fpi*theta)*a7(iperiod) 

 & +f/2*absorbpg*.5*(1.-absorbta)*qdir*sin(fpi*theta)*a5 

A-135
 & +f/2*absorbpg*qreflin(4,iperiod)*sin(fpi*theta)*a3 

 & +f/2*absorbpg*qscabsin(4,iperiod)*lfuselg*wfuselg 

 & *a3/(absorbpg*a3+absorfbt*a4) 

 & -emispg*sig*newtpg1**4*a1 

 & -hhf*(newtpg1-ta)*a1 

 & -(newtpg1-newtpg2)/(f/4*thpg/kpg+(1-f)/2*thpg/kpg)*a1) 

 fn(41)=-newtpg2+tpg2+deltime/(a1*denspg*shcpg*(1-f)*thpg)* 

 & ((newtpg1-newtpg2)/(f/4*thpg/kpg+(1-f)/2*thpg/kpg)*a1 

 & -(newtpg2-newtpg3)/((1-f)/2*thpg/kpg+f/4*thpg/kpg)*a1 

 & +(1-f)*absorbpg*qdir*cos(fpi*theta)*a6(iperiod) 

 & +(1-f)*absorbpg*qdir*sin(fpi*theta)*a7(iperiod) 

 & +(1-f)*absorbpg*.5*(1.-absorbta)*qdir*sin(fpi*theta)*a5 

 & +(1-f)*absorbpg*qreflin(4,iperiod)*sin(fpi*theta)*a3 

 & +(1-f)*absorbpg*qscabsin(4,iperiod)*lfuselg*wfuselg 

 & *a3/(absorbpg*a3+absorfbt*a4)) 

 qpg=emispg*sig*newtpg3**4-emispg/(a3*emispg+a4*emisrfbt)* 

 & (a3*emispg*sig*newtpg3**4+a4*emisrfbt*sig*newt(12)**4-qnet* 

 & (a3+a4)) 

 if(icool.eq.1)then 

 fn(42)=-newtpg3+tpg3+deltime/(a1*denspg*shcpg*f/2*thpg)* 

 & ((newtpg2-newtpg3)/((1-f)/2*thpg/kpg+f/4*thpg/kpg)*a1 

 & -hhc*(newtpg3-ta)*a1 

 & -qpg*a3 

 & +f/2*absorbpg*qdir*cos(fpi*theta)*a6(iperiod) 

 & +f/2*absorbpg*qdir*sin(fpi*theta)*a7(iperiod) 

 & +f/2*absorbpg*.5*(1.-absorbta)*qdir*sin(fpi*theta)*a5 

 & +f/2*absorbpg*qreflin(4,iperiod)*sin(fpi*theta)*a3 

 & +f/2*absorbpg*qscabsin(4,iperiod)*lfuselg*wfuselg 

 & *a3/(absorbpg*a3+absorfbt*a4)) 

 elseif(icool.eq.2)then

c there is a problem here. correct the tpg2-tpg3 resistance 

 fn(42)=-newtpg3+tpg3+deltime/(a1*denspg*shcpg*f/2*thpg)* 

 & ((newtpg2-newtpg3)/((1-f)/2*thpg/kpg+f/4*thpg/kpg)*a1 

 & -(tpgrf-newt(16))/(th(5)/(nusselt1*k(5)))*a3 

 & -qpg*a3 

 & +f/2*absorbpg*qdir*cos(fpi*theta)*a6(iperiod) 

 & +f/2*absorbpg*qdir*sin(fpi*theta)*a7(iperiod) 

 & +f/2*absorbpg*.5*(1.-absorbta)*qdir*sin(fpi*theta)*a5 

 & +f/2*absorbpg*qreflin(4,iperiod)*sin(fpi*theta)*a3 

 & +f/2*absorbpg*qscabsin(4,iperiod)*lfuselg*wfuselg 

 & *a3/(absorbpg*a3+absorfbt*a4)) 

 endif 

c 

c next for the roof structure and the rest: 

c 

c first three temperatures of roof structure: fn(1),fn(2,fn(3): 

c newt(1),newt(2),newt(3). 

c 

 qnetsky=(emisrftp*sig*newt(1)**4-emisrftp*qsky) 

 fn(1)=-newt(1)+t(1)+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & (absorftp*qdir*sin(fpi*theta)-qnetsky-hhf*(newt(1)-ta)- 

A-136
 & (newt(1)-newt(2))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 fn(2)=-newt(2)+t(2)+deltime/(dens(1)*shc(1)*(1-f)*th(1))* 

 & ((newt(1)-newt(2))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (newt(2)-newt(3))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 fn(3)=-newt(3)+t(3)+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((newt(2)-newt(3))/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (newt(3)-newt(4))/(f/4*th(1)/k(1)+f/4*th(2)/k(2))) 

c 

c internal nodes::: for layers 2 to 12: newt4 to newt36: also fn(4) to fn(36) 

c where newt(13,14 and 16) and fn(13),fn(14) and fn(15) for cabin air (layer 5) neglected 

c internal nodes: beginning of control volume: newt(4,7,10,16,19,22,25,28,31,34): fn() 

c middle of control volume: newt(5,8,11,17,20,23,26,29,32,35): fn() 

c ending of control volume: newt(6,9,12,18,21,24,27,30,33,36): fn() 

c note: newt13,newt14 and newt15 are dummies: and so are fn(13),fn(14),fn(15). 

c 

do 20 i=2,12

 if(i.eq.5)goto 20 

 ii=(i-1)*3 

c 

c beginning element within layer 2,3,4,6,7,8,9,10,11 and 12: 

c newt4,newt7,newt10,newt16,newt19,newt22,newt25,newt28,newt31,newt34: 

c fn(4),fn(7),fn(10),fn(16),fn(19),fn(22),fn(25),fn(28),fn(31),fn(34): 

c 

 if(i.eq.6)then

 qnetcab=(sig*newt(ii+1)**4-jj(11))/((1.-emisfl)/emisfl) 

 if(icool.eq.1)then 

 fn(ii+1)=-newt(ii+1)+t(ii+1)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & (-(newt(ii+1)-newt(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))-

 & hhc*(newt(ii+1)-ta)- 

 & qnetcab+absorbfl*qreflin(2,iperiod)*sin(fpi*theta)+ 

 & qscabsin(2,iperiod)) 

 elseif(icool.eq.2)then 

 fn(ii+1)=-newt(ii+1)+t(ii+1)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((tpgrf-newt(ii+1))/(th(i-1)/(nusselt1*k(i-1)))- 

 & (newt(ii+1)-newt(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & qnetcab+

 & (absorbfl*qreflin(2,iperiod)*sin(fpi*theta)+ 

 & qscabsin(2,iperiod))*sunfact2) 

 endif 

 else 

 fn(ii+1)=-newt(ii+1)+t(ii+1)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((newt(ii)-newt(ii+1))/(f/4*th(i-1)/k(i-1)+f/4*th(i)/k(i))- 

 & (newt(ii+1)-newt(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 endif 

c 

c middle element within layer 2,3,4,6,7,8,9,10,11 and 12: 

c newt5,newt8,newt11,newt17,newt20,newt23,newt26,newt29,newt32,newt35: 

c fn(5),fn(8),fn(11),fn(17),fn(20),fn(23),fn(26),fn(29),fn(32),fn(35): 

c 

c234567812345678123456781234567812345678123456781234567812345678123456781234

5678 

A-137
 fn(ii+2)=-newt(ii+2)+t(ii+2)+deltime/(dens(i)*shc(i)*(1-f)*th(i))* 

 & ((newt(ii+1)-newt(ii+2))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (newt(ii+2)-newt(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

c 

c ending element within layer 2,3,4,6,7,8,9,10,11 and 12: 

c newt6,newt9,newt12,newt18,newt21,newt24,newt27,newt30,newt33,newt36: 

c fn(6),fn(9),fn(12),fn(18),fn(21),fn(24),fn(27),fn(30),fn(33),fn(36): 

c 

 if(i.eq.4)then 

 qrfbt=emisrfbt*sig*newt(ii+3)**4-emisrfbt/(a3*emispg+a4*emisrfbt)* 

 &(a3*emispg*sig*newtpg3**4+a4*emisrfbt*sig*newt(ii+3)**4-qnet* 

 &(a3+a4)) 

 if(icool.eq.1)then 

c234567812345678123456781234567812345678123456781234567812345678123456781234

5678 

 fn(ii+3)=-newt(ii+3)+t(ii+3)+ 

 & deltime/(a2*dens(i)*shc(i)*f/2*th(i))* 

 & ((newt(ii+2)-newt(ii+3))/((1-f)/2*th(i)/k(i)+f/4*th(i)/k(i))*a2-

 & hhc*(newt(ii+3)-ta)*a2 

 & -qrfbt*a4 

 & +absorfbt*qreflin(4,iperiod)*sin(fpi*theta)*a4 

 & +absorfbt*qscabsin(4,iperiod)*lfuselg*wfuselg*a4/ 

 & (absorbpg*a3+absorfbt*a4)) 

 elseif(icool.eq.2)then 

 fn(ii+3)=-newt(ii+3)+t(ii+3)+ 

 & deltime/(a2*dens(i)*shc(i)*f/2*th(i))* 

 & ((newt(ii+2)-newt(ii+3))/((1-f)/2*th(i)/k(i)+f/4*th(i)/k(i))*a2-

 & (tpgrf-newt(ii+7))/(th(i+1)/(nusselt1*k(i+1)))*a4 

 & -qrfbt*a4 

 & +(absorfbt*qreflin(4,iperiod)*sin(fpi*theta)*a4 

 & +absorfbt*qscabsin(4,iperiod)*lfuselg*wfuselg*a4/ 

 & (absorbpg*a3+absorfbt*a4))*sunfact2) 

 endif 

 else 

 fn(ii+3)=-newt(ii+3)+t(ii+3)+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((newt(ii+2)-newt(ii+3))/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (newt(ii+3)-newt(ii+4))/(f/4*th(i)/k(i)+f/4*th(i+1)/k(i+1))) 

 endif 

 20 continue 

c 

c belly-pan external temp (t39), as well as internal temp t37 and t38: 

c 

 fn(37)=-newt(37)+t(37)+ 

 & deltime/(dens(13)*shc(13)*f/2*th(13))* 

 & ((newt(36)-newt(37))/ 

 & (f/4*th(12)/k(12)+f/4*th(13)/k(13))- 

 & (newt(37)-newt(38))/ 

 & (f/4*th(13)/k(13)+(1-f)/2*th(13)/k(13))) 

 fn(38)=-newt(38)+t(38)+ 

 & deltime/(dens(13)*shc(13)*(1-f)*th(13))* 

 & ((newt(37)-newt(38))/ 

A-138
 & (f/4*th(13)/k(13)+(1-f)/2*th(13)/k(13))- 

 & (newt(38)-newt(39))/ 

 & (f/4*th(13)/k(13)+(1-f)/2*th(13)/k(13))) 

 qnetsky=(sig*newt(39)**4-jj(7))/((1.-emisfb)/emisfb) 

 fn(39)=-newt(39)+t(39)+ 

 & deltime/(dens(13)*shc(13)*f/2*th(13))* 

 & (qabstwb(3,iperiod)-qnetsky-hhf*(newt(39)-ta)+ 

 & (newt(38)-newt(39))/ 

 & (f/4*th(13)/k(13)+(1-f)/2*th(13)/k(13))) 

c 

c transfer fn(i) into tfn(i) so that fn(13),fn(14) and fn(15) are removed 

c and thus tfn(i) begins at tfn(1) and ends at tfn(39). fn(13),fn(14) and fn(15) 

c are dummies for the layer 5 (air in the cabin). and fn(40),fn(41),fn(42) 

c are equations for plexiglass 

c 

 isum=0 

 do 100 i=1,42 

 if(i.eq.13.or.i.eq.14.or.i.eq.15)goto 100 

 isum=isum+1 

 tfn(isum)=fn(i) 

 100 continue 

c 

c transfer the tfn(i) into the global ffn(i), which is used together with jacobian 

c to find the p(i) and then the new tt(i) in newraph2 subroutine. 

c 

 do 30 i=1,39 

 30 ffn(78+i)=tfn(i) 

c 

c now find the jacobian, global jacob(120,120) from the partial derivative of fn(i). 

c find tjacob(42,42) first. then put it into the global jacob(79..117,79..117). 

c note: tjacob(13,),tjacob(14,) and tjacob(15,) are just dummies to be removed later. 

c 

c first initialize tjacob(42,42) to 0.0. 

c note: most of the tjacob(42,42) are 0.0 

c 

 do 70 itj=1,42 

 do 70 jtj=1,42 

 70 tjacob(itj,jtj)=0.0 

c 

c first find tjacob for the only three temperatures of plexiglass: 

c tjacob(40,),tjacob(41,),tjacob(42,) 

c for fn(40),fn(41),fn(42): newtpg1,newtpg2,newtpg3. 

c note: newtpg1=outermost plexiglas temperature 

c note: newtpg3=innermost plexiglas temperature 

c note: newtpg2=middle plexiglas temperature 

c regard newtpg1,newtpg2 and newtpg3 as newt(40),newt(41) and newt(42) in the 

c jacobian tjacob matrix. 

c 

c jacobians tjacob(40,40) and tjacob(40,41): 

c 

 tjacob(40,40)=-1+deltime/(a1*denspg*shcpg*f/2*thpg)* 

A-139
 & (-4*emispg*sig*newtpg1**3*a1 

 & -hhf*(1)*a1 

 & -(1)/(f/4*thpg/kpg+(1-f)/2*thpg/kpg)*a1) 

 tjacob(40,41)=deltime/(a1*denspg*shcpg*f/2*thpg)* 

 & (-(-1)/(f/4*thpg/kpg+(1-f)/2*thpg/kpg)*a1) 

c 

c jacobians tjacob(41,40),tjacob(41,41), tjacob(41,42): 

c 

 tjacob(41,40)=deltime/(a1*denspg*shcpg*(1-f)*thpg)* 

 & ((1)/(f/4*thpg/kpg+(1-f)/2*thpg/kpg)*a1) 

 tjacob(41,41)=-1+deltime/(a1*denspg*shcpg*(1-f)*thpg)* 

 & ((-1)/(f/4*thpg/kpg+(1-f)/2*thpg/kpg)*a1 

 & -(1)/((1-f)/2*thpg/kpg+f/4*thpg/kpg)*a1) 

 tjacob(41,42)=deltime/(a1*denspg*shcpg*(1-f)*thpg)* 

 & (-(-1)/((1-f)/2*thpg/kpg+f/4*thpg/kpg)*a1) 

c 

 if(icool.eq.1)then 

c jacobians tjacob(42,12), tjacob(42,41) tjacob(42,42): 

 tjacob(42,12)=deltime/(a1*denspg*shcpg*f/2*thpg)* 

 & (-(-emispg/(a3*emispg+a4*emisrfbt)* 

 & (4*a4*emisrfbt*sig*newt(12)**3-(4*sig*(c1*newtpg3+ 

 & c2*newt(12))**3*(c2*(1)))/((1.-emispgrf)/emispgrf)*(a3+a4)))*a3) 

 tjacob(42,41)=deltime/(a1*denspg*shcpg*f/2*thpg)* 

 & ((1)/((1-f)/2*thpg/kpg+f/4*thpg/kpg)*a1) 

 tjacob(42,42)=-1+deltime/(a1*denspg*shcpg*f/2*thpg)* 

 & ((-1)/((1-f)/2*thpg/kpg+f/4*thpg/kpg)*a1 

 & -hhc*(1)*a1 

 & -(4*emispg*sig*newtpg3**3-emispg/(a3*emispg+a4*emisrfbt)* 

 & (4*a3*emispg*sig*newtpg3**3-(4*sig*(c1*newtpg3+ 

 & c2*newt(12))**3*(c1*(1)))/((1.-emispgrf)/emispgrf)*(a3+a4)))*a3) 

 elseif(icool.eq.2)then

c jacobians tjacob(42,12),tjacob(42,16),tjacob(42,41) tjacob(42,42): 

 if(nusselt1.eq.1.0)then 

 tjacob(42,12)=deltime/(a1*denspg*shcpg*f/2*thpg)* 

 & ( -(c2)/(th(5)/(nusselt1*k(5)))*a3 

 & -(-emispg/(a3*emispg+a4*emisrfbt)* 

 & (4*a4*emisrfbt*sig*newt(12)**3-(4*sig*tpgrf**3*c2)/ 

 & ((1.-emispgrf)/emispgrf)*(a3+a4)))*a3 ) 

 tjacob(42,16)=deltime/(a1*denspg*shcpg*f/2*thpg)* 

 & ( -(-1)/(th(5)/(nusselt1*k(5)))*a3 ) 

 tjacob(42,41)=deltime/(a1*denspg*shcpg*f/2*thpg)* 

 & ( (1)/((1-f)/2*thpg/kpg+f/4*thpg/kpg)*a1 ) 

 tjacob(42,42)=-1+deltime/(a1*denspg*shcpg*f/2*thpg)* 

 & ( (-1)/((1-f)/2*thpg/kpg+f/4*thpg/kpg)*a1 

 & -(c1)/(th(5)/(nusselt1*k(5)))*a3 

 & -(4*emispg*sig*newtpg3**3-emispg/(a3*emispg+a4*emisrfbt)* 

 & (4*a3*emispg*sig*newtpg3**3-(4*sig*tpgrf**3*c1)/ 

 & ((1.-emispgrf)/emispgrf)*(a3+a4)))*a3 ) 

 else 

 const=0.069*prandtl**.074*(gravity*th(5)**3/ 

 & (0.5*(thdiffus*visckine)))**(1./3.) 

A-140
 dndt12=const*( (tpgrf+newt(16))**(-1./3.)* 

 & ((1./3.)*(newt(16)-tpgrf)**(-2./3.)*(-c2)) + 

 & (newt(16)-tpgrf)**(1./3.)* 

 & ((-1./3.)*(tpgrf+newt(16))**(-4./3.)*(c2)) ) 

 dndt16=const*( (tpgrf+newt(16))**(-1./3.)* 

 & ((1./3.)*(newt(16)-tpgrf)**(-2./3.)*(1)) + 

 & (newt(16)-tpgrf)**(1./3.)* 

 & ((-1./3.)*(tpgrf+newt(16))**(-4./3.)*(1)) ) 

 dndt42=const*( (tpgrf+newt(16))**(-1./3.)* 

 & ((1./3.)*(newt(16)-tpgrf)**(-2./3.)*(-c1)) + 

 & (newt(16)-tpgrf)**(1./3.)* 

 & ((-1./3.)*(tpgrf+newt(16))**(-4./3.)*(c1)) ) 

 dcdt12=(-1.0)*(th(5)/k(5))*nusselt1**(-2.0)*dndt12 

 dcdt16=(-1.0)*(th(5)/k(5))*nusselt1**(-2.0)*dndt16 

 dcdt42=(-1.0)*(th(5)/k(5))*nusselt1**(-2.0)*dndt42 

 abcd=(th(5)/(nusselt1*k(5))) 

 dydt12=abcd**(-1.0)*(c2) + 

 & (tpgrf-newt(16))*((-1.0)*abcd**(-2.0)*dcdt12) 

 dydt16=abcd**(-1.0)*(-1.0) + 

 & (tpgrf-newt(16))*((-1.0)*abcd**(-2.0)*dcdt16) 

 dydt42=abcd**(-1.0)*(c1) + 

 & (tpgrf-newt(16))*((-1.0)*abcd**(-2.0)*dcdt42) 

c 

 tjacob(42,12)=deltime/(a1*denspg*shcpg*f/2*thpg)* 

 & ( -(dydt12)*a3 

 & -(-emispg/(a3*emispg+a4*emisrfbt)* 

 & (4*a4*emisrfbt*sig*newt(12)**3-(4*sig*tpgrf**3*c2)/ 

 & ((1.-emispgrf)/emispgrf)*(a3+a4)))*a3 ) 

 tjacob(42,16)=deltime/(a1*denspg*shcpg*f/2*thpg)* 

 & ( -(dydt16)*a3 ) 

 tjacob(42,41)=deltime/(a1*denspg*shcpg*f/2*thpg)* 

 & ( (1)/((1-f)/2*thpg/kpg+f/4*thpg/kpg)*a1 ) 

 tjacob(42,42)=-1+deltime/(a1*denspg*shcpg*f/2*thpg)* 

 & ( (-1)/((1-f)/2*thpg/kpg+f/4*thpg/kpg)*a1 

 & -(dydt42)*a3 

 & -(4*emispg*sig*newtpg3**3-emispg/(a3*emispg+a4*emisrfbt)* 

 & (4*a3*emispg*sig*newtpg3**3-(4*sig*tpgrf**3*c1)/ 

 & ((1.-emispgrf)/emispgrf)*(a3+a4)))*a3 ) 

 endif 

 endif 

c 

c jacobians for the roof structure tjacob(1,),tjacob(2,),tjacob(3,): 

c from the three temperatures and functions of roof structure: fn(1),fn(2,fn(3): 

c newt(1),newt(2),newt(3). 

c 

c 

c tjacob(1,1) and tjacob(1,2): 

c 

 tjacob(1,1)=-1+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & (-(4*emisrftp*sig*newt(1)**3)-hhf*(1)- 

 & (1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

A-141
 tjacob(1,2)=deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & (-(-1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

c 

c tjacob(2,1), tjacob(2,2) and tjacob(2,3): 

c 

 tjacob(2,1)=deltime/(dens(1)*shc(1)*(1-f)*th(1))* 

 & ((1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 tjacob(2,2)=-1+deltime/(dens(1)*shc(1)*(1-f)*th(1))* 

 & ((-1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 tjacob(2,3)=deltime/(dens(1)*shc(1)*(1-f)*th(1))* 

 & (-(-1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

c 

c tjacob(3,2), tjacob(3,3) and tjacob(3,4): 

c 

 tjacob(3,2)=deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))) 

 tjacob(3,3)=-1+deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & ((-1)/(f/4*th(1)/k(1)+(1-f)/2*th(1)/k(1))- 

 & (1)/(f/4*th(1)/k(1)+f/4*th(2)/k(2))) 

 tjacob(3,4)=deltime/(dens(1)*shc(1)*f/2*th(1))* 

 & (-(-1)/(f/4*th(1)/k(1)+f/4*th(2)/k(2))) 

c 

c now find jacobians tjacob(4) to tjacob(36) within layers 2 to 12. 

c note that tjacob(13,) tjacob(14,) and tjacob(15,) are skipped and are just dummies. 

c 

do 120 i=2,12

 if(i.eq.5)goto 120 

 ii=(i-1)*3 

c 

c find tjacob(4),tjacob(7),tjacob(10),tjacob(16),tjacob(19),tjacob(22) 

c and tjacob(25),tjacob(28),tjacob(31) and tjacob(34) for: 

c beginning element within layer 2,3,4,6,7,8,9,10,11 and 12: 

c newt4,newt7,newt10,newt16,newt19,newt22,newt25,newt28,newt31,newt34: 

c fn(4),fn(7),fn(10),fn(16),fn(19),fn(22),fn(25),fn(28),fn(31),fn(34): 

c 

 if(i.eq.6)then

 if(icool.eq.1)then 

c tjacob(16,16),tjacob(16,17): 

 tjacob(ii+1,ii+1)=-1+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & (-(1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))-

 & hhc*(1)- 

 & (4*sig*newt(ii+1)**3)/((1.-emisfl)/emisfl)) 

 tjacob(ii+1,ii+2)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & (-(-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i)))

 elseif(icool.eq.2)then 

c tjacob(16,12),tjacob(16,16),tjacob(16,17),tjacob(16,42) 

 if(nusselt1.eq.1.0)then 

 tjacob(ii+1,ii-3)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ( (c2)/(th(i-1)/(nusselt1*k(i-1))) )

 tjacob(ii+1,ii+1)=-1+deltime/(dens(i)*shc(i)*f/2*th(i))* 

A-142
 & ( (-1)/(th(i-1)/(nusselt1*k(i-1)))- 

 & (1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))-

 & (4*sig*newt(ii+1)**3)/((1.-emisfl)/emisfl) ) 

 tjacob(ii+1,ii+2)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ( -(-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i)) ) 

 tjacob(ii+1,42)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ( (c1)/(th(i-1)/(nusselt1*k(i-1))) ) 

 else

 const=0.069*prandtl**.074*(gravity*th(5)**3/ 

 & (0.5*(thdiffus*visckine)))**(1./3.) 

 dndt12=const*( (tpgrf+newt(16))**(-1./3.)* 

 & ((1./3.)*(newt(16)-tpgrf)**(-2./3.)*(-c2)) + 

 & (newt(16)-tpgrf)**(1./3.)* 

 & ((-1./3.)*(tpgrf+newt(16))**(-4./3.)*(c2)) ) 

 dndt16=const*( (tpgrf+newt(16))**(-1./3.)* 

 & ((1./3.)*(newt(16)-tpgrf)**(-2./3.)*(1)) + 

 & (newt(16)-tpgrf)**(1./3.)* 

 & ((-1./3.)*(tpgrf+newt(16))**(-4./3.)*(1)) ) 

 dndt42=const*( (tpgrf+newt(16))**(-1./3.)* 

 & ((1./3.)*(newt(16)-tpgrf)**(-2./3.)*(-c1)) + 

 & (newt(16)-tpgrf)**(1./3.)* 

 & ((-1./3.)*(tpgrf+newt(16))**(-4./3.)*(c1)) ) 

 dcdt12=(-1.0)*(th(5)/k(5))*nusselt1**(-2.0)*dndt12 

 dcdt16=(-1.0)*(th(5)/k(5))*nusselt1**(-2.0)*dndt16 

 dcdt42=(-1.0)*(th(5)/k(5))*nusselt1**(-2.0)*dndt42 

 abcd=(th(5)/(nusselt1*k(5)))

 dydt12=abcd**(-1.0)*(c2) + 

 & (tpgrf-newt(16))*((-1.0)*abcd**(-2.0)*dcdt12) 

 dydt16=abcd**(-1.0)*(-1.0) + 

 & (tpgrf-newt(16))*((-1.0)*abcd**(-2.0)*dcdt16) 

 dydt42=abcd**(-1.0)*(c1) + 

 & (tpgrf-newt(16))*((-1.0)*abcd**(-2.0)*dcdt42) 

c 

 tjacob(ii+1,ii-3)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ( (dydt12) ) 

 tjacob(ii+1,ii+1)=-1+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ( (dydt16)- 

 & (1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (4*sig*newt(ii+1)**3)/((1.-emisfl)/emisfl) ) 

 tjacob(ii+1,ii+2)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ( -(-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i)) ) 

 tjacob(ii+1,42)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ( (dydt42) ) 

 endif

 endif 

 else 

 tjacob(ii+1,ii)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((1)/(f/4*th(i-1)/k(i-1)+f/4*th(i)/k(i))) 

 tjacob(ii+1,ii+1)=-1+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((-1)/(f/4*th(i-1)/k(i-1)+f/4*th(i)/k(i))- 

 & (1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

A-143
 tjacob(ii+1,ii+2)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & (-(-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i)))

 endif 

c 

c find tjacob(5),tjacob(8),tjacob(11),tjacob(17),tjacob(20),tjacob(23) 

c and tjacob(26),tjacob(29),tjacob(32) and tjacob(35) for: 

c middle element within layer 2,3,4,6,7,8,9,10,11 and 12: 

c newt5,newt8,newt11,newt17,newt20,newt23,newt26,newt29,newt32,newt35: 

c fn(5),fn(8),fn(11),fn(17),fn(20),fn(23),fn(26),fn(29),fn(32),fn(35): 

c 

 tjacob(ii+2,ii+1)=deltime/(dens(i)*shc(i)*(1-f)*th(i))* 

 & ((1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 tjacob(ii+2,ii+2)=-1+deltime/(dens(i)*shc(i)*(1-f)*th(i))* 

 & ((-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 tjacob(ii+2,ii+3)=deltime/(dens(i)*shc(i)*(1-f)*th(i))* 

 & (-(-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

c 

c find tjacob(6),tjacob(9),tjacob(12),tjacob(18),tjacob(21),tjacob(24) 

c and tjacob(27),tjacob(30),tjacob(33) and tjacob(36) for: 

c ending element within layer 2,3,4,6,7,8,9,10,11 and 12: 

c newt6,newt9,newt12,newt18,newt21,newt24,newt27,newt30,newt33,newt36: 

c fn(6),fn(9),fn(12),fn(18),fn(21),fn(24),fn(27),fn(30),fn(33),fn(36): 

c 

 if(i.eq.4)then 

 if(icool.eq.1)then 

c tjacob(12,11),tjacob(12,12),tjacob(12,42) 

 tjacob(ii+3,ii+2)=deltime/(a2*dens(i)*shc(i)*f/2*th(i))* 

 & ((1)/((1-f)/2*th(i)/k(i)+f/4*th(i)/k(i))*a2) 

 tjacob(ii+3,ii+3)=-1+deltime/(a2*dens(i)*shc(i)*f/2*th(i))* 

 & ((-1)/((1-f)/2*th(i)/k(i)+f/4*th(i)/k(i))*a2- 

 & hhc*(1)*a2 

 & -(4*emisrfbt*sig*newt(ii+3)**3-emisrfbt/(a3*emispg+a4*emisrfbt)* 

 & (4*a4*emisrfbt*sig*newt(ii+3)**3-(4*sig*tpgrf**3*(c2*(1)))/ 

 & ((1.-emispgrf)/emispgrf)*(a3+a4)))*a4) 

 tjacob(ii+3,42)=deltime/(a2*dens(i)*shc(i)*f/2*th(i))* 

 & (-(-emisrfbt/(a3*emispg+a4*emisrfbt)* 

 & (4*a3*emispg*sig*newtpg3**3-(4*sig*tpgrf**3*(c1*(1)))/ 

 & ((1.-emispgrf)/emispgrf)*(a3+a4)))*a4) 

 elseif(icool.eq.2)then 

c tjacob(12,11),tjacob(12,12),tjacob(12,16),tjacob(12,42) 

 if(nusselt1.eq.1.0)then 

 tjacob(ii+3,ii+2)=deltime/(a2*dens(i)*shc(i)*f/2*th(i))* 

 & ( (1)/((1-f)/2*th(i)/k(i)+f/4*th(i)/k(i))*a2 ) 

 tjacob(ii+3,ii+3)=-1+deltime/(a2*dens(i)*shc(i)*f/2*th(i))* 

 & ( (-1)/((1-f)/2*th(i)/k(i)+f/4*th(i)/k(i))*a2- 

 & (c2)/(th(i+1)/(nusselt1*k(i+1)))*a4 

 & -(4*emisrfbt*sig*newt(ii+3)**3-emisrfbt/(a3*emispg+a4*emisrfbt)* 

 & (4*a4*emisrfbt*sig*newt(ii+3)**3-(4*sig*tpgrf**3*c2)/ 

 & ((1.-emispgrf)/emispgrf)*(a3+a4)))*a4 ) 

 tjacob(ii+3,ii+7)=deltime/(a2*dens(i)*shc(i)*f/2*th(i))* 

A-144
 & ( -(-1)/(th(i+1)/(nusselt1*k(i+1)))*a4 ) 

 tjacob(ii+3,42)=deltime/(a2*dens(i)*shc(i)*f/2*th(i))* 

 & ( -(c1)/(th(i+1)/(nusselt1*k(i+1)))*a4 

 & -(-emisrfbt/(a3*emispg+a4*emisrfbt)* 

 & (4*a3*emispg*sig*newtpg3**3-(4*sig*tpgrf**3*c1)/ 

 & ((1.-emispgrf)/emispgrf)*(a3+a4)))*a4 ) 

 else 

 const=0.069*prandtl**.074*(gravity*th(5)**3/ 

 & (0.5*(thdiffus*visckine)))**(1./3.) 

 dndt12=const*( (tpgrf+newt(16))**(-1./3.)* 

 & ((1./3.)*(newt(16)-tpgrf)**(-2./3.)*(-c2)) + 

 & (newt(16)-tpgrf)**(1./3.)* 

 & ((-1./3.)*(tpgrf+newt(16))**(-4./3.)*(c2)) ) 

 dndt16=const*( (tpgrf+newt(16))**(-1./3.)* 

 & ((1./3.)*(newt(16)-tpgrf)**(-2./3.)*(1)) + 

 & (newt(16)-tpgrf)**(1./3.)* 

 & ((-1./3.)*(tpgrf+newt(16))**(-4./3.)*(1)) ) 

 dndt42=const*( (tpgrf+newt(16))**(-1./3.)* 

 & ((1./3.)*(newt(16)-tpgrf)**(-2./3.)*(-c1)) + 

 & (newt(16)-tpgrf)**(1./3.)* 

 & ((-1./3.)*(tpgrf+newt(16))**(-4./3.)*(c1)) ) 

 dcdt12=(-1.0)*(th(5)/k(5))*nusselt1**(-2.0)*dndt12 

 dcdt16=(-1.0)*(th(5)/k(5))*nusselt1**(-2.0)*dndt16 

 dcdt42=(-1.0)*(th(5)/k(5))*nusselt1**(-2.0)*dndt42 

 abcd=(th(5)/(nusselt1*k(5)))

 dydt12=abcd**(-1.0)*(c2) + 

 & (tpgrf-newt(16))*((-1.0)*abcd**(-2.0)*dcdt12) 

 dydt16=abcd**(-1.0)*(-1.0) + 

 & (tpgrf-newt(16))*((-1.0)*abcd**(-2.0)*dcdt16) 

 dydt42=abcd**(-1.0)*(c1) + 

 & (tpgrf-newt(16))*((-1.0)*abcd**(-2.0)*dcdt42) 

c 

 tjacob(ii+3,ii+2)=deltime/(a2*dens(i)*shc(i)*f/2*th(i))* 

 & ( (1)/((1-f)/2*th(i)/k(i)+f/4*th(i)/k(i))*a2 ) 

 tjacob(ii+3,ii+3)=-1+deltime/(a2*dens(i)*shc(i)*f/2*th(i))* 

 & ( (-1)/((1-f)/2*th(i)/k(i)+f/4*th(i)/k(i))*a2- 

 & (dydt12)*a4 

 & -(4*emisrfbt*sig*newt(ii+3)**3-emisrfbt/(a3*emispg+a4*emisrfbt)* 

 & (4*a4*emisrfbt*sig*newt(ii+3)**3-(4*sig*tpgrf**3*c2)/ 

 & ((1.-emispgrf)/emispgrf)*(a3+a4)))*a4 ) 

 tjacob(ii+3,ii+7)=deltime/(a2*dens(i)*shc(i)*f/2*th(i))* 

 & ( -(dydt16)*a4 ) 

 tjacob(ii+3,42)=deltime/(a2*dens(i)*shc(i)*f/2*th(i))* 

 & ( -(dydt42)*a4 

 & -(-emisrfbt/(a3*emispg+a4*emisrfbt)* 

 & (4*a3*emispg*sig*newtpg3**3-(4*sig*tpgrf**3*c1)/ 

 & ((1.-emispgrf)/emispgrf)*(a3+a4)))*a4 ) 

 endif 

 endif 

 else 

 tjacob(ii+3,ii+2)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

A-145
 & ((1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))) 

 tjacob(ii+3,ii+3)=-1+deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & ((-1)/(f/4*th(i)/k(i)+(1-f)/2*th(i)/k(i))- 

 & (1)/(f/4*th(i)/k(i)+f/4*th(i+1)/k(i+1))) 

 tjacob(ii+3,ii+4)=deltime/(dens(i)*shc(i)*f/2*th(i))* 

 & (-(-1)/(f/4*th(i)/k(i)+f/4*th(i+1)/k(i+1))) 

 endif 

 120 continue 

c 

c belly-pan external temp (t39), as well as internal temp t37 and t38: 

c 

c find tjacob(37,),tjacob(38,) and tjacob(39,) for: 

c belly-pan elements within layer 13: 

c newt37,newt38 and newt39: 

c from fn(37),fn(38) and fn(39): 

c 

c 

c tjacob(37,36) and tjacob(37,37) and tjacob(37,38): 

c 

 tjacob(37,36)=deltime/(dens(13)*shc(13)*f/2*th(13))* 

 & ( (1)/(f/4*th(12)/k(12)+f/4*th(13)/k(13)) ) 

 tjacob(37,37)=-1+deltime/(dens(13)*shc(13)*f/2*th(13))* 

 & ( (-1)/(f/4*th(12)/k(12)+f/4*th(13)/k(13))- 

 & (1)/(f/4*th(13)/k(13)+(1-f)/2*th(13)/k(13)) ) 

 tjacob(37,38)=deltime/(dens(13)*shc(13)*f/2*th(13))* 

 & ( -(-1)/(f/4*th(13)/k(13)+(1-f)/2*th(13)/k(13)) ) 

c 

c tjacob(38,37) and tjacob(38,38) and tjacob(38,39): 

c 

 tjacob(38,37)=deltime/(dens(13)*shc(13)*(1-f)*th(13))* 

 & ( (1)/(f/4*th(13)/k(13)+(1-f)/2*th(13)/k(13)) ) 

 tjacob(38,38)=-1+deltime/(dens(13)*shc(13)*(1-f)*th(13))* 

 & ( (-1)/(f/4*th(13)/k(13)+(1-f)/2*th(13)/k(13))- 

 & (1)/(f/4*th(13)/k(13)+(1-f)/2*th(13)/k(13)) ) 

 tjacob(38,39)=deltime/(dens(13)*shc(13)*(1-f)*th(13))* 

 & ( -(-1)/(f/4*th(13)/k(13)+(1-f)/2*th(13)/k(13)) ) 

c 

c tjacob(39,38) and tjacob(39,39): 

c 

 tjacob(39,38)=deltime/(dens(13)*shc(13)*f/2*th(13))* 

 & ( (1)/(f/4*th(13)/k(13)+(1-f)/2*th(13)/k(13)) ) 

 tjacob(39,39)=-1+deltime/(dens(13)*shc(13)*f/2*th(13))* 

 & ( -(4*sig*newt(39)**3)/((1.-emisfb)/emisfb) 

 & -hhf*(1)+ 

 & (-1)/(f/4*th(13)/k(13)+(1-f)/2*th(13)/k(13)) ) 

c 

c 

c jacobian analysis is done. now put tjacob(i,j) into the global jacobian matrix 

c jacob(ijacob,jjacob), where ijacob starts and ends at 79 and 117, 

c and jjacob starts and ends at 79 and 117. note: most jacob values are 0's 

c 

A-146
c 

c first transfer tjacob(42,42) into ttjacob(39,39) so that rows and columns 

c 13, 14 and 15 are removed. 

c 

 isum1=0 

 do 140 i=1,42 

 if(i.eq.13.or.i.eq.14.or.i.eq.15)goto 140 

 isum1=isum1+1 

 isum2=0 

 do 125 j=1,42 

 if(j.eq.13.or.j.eq.14.or.j.eq.15)goto 125 

 isum2=isum2+1 

 ttjacob(isum1,isum2)=tjacob(i,j)

 125 continue 

 140 continue 

c 

c now put ttjacob into the global jacob 

c 

 do 130 i=1,39 

 do 130 j=1,39 

 jacob(78+i,78+j)=ttjacob(i,j) 

 130 continue 

 return 

 end 

c 

c 

 subroutine tarmac2(ithtrial,itotitr,icool,iperiod,istep,ta,ttarm, 

 & hht,isolfoun)

c purpose: transient analysis of the tarmac 

c called from: newraph2 

c calls: none

 common/c4/widthwng,lengtwng,hwng,locwng,lplane,lfuselg,hfuselg,

 & wfuselg 

 common/c5/trnsm,absorbta,emista,tarmd,densta,conducta,shcta 

 common/c2/tt(200),ffn(200),jacob(200,200) 

 common/c3/tth(100),kk(100) 

 common/c14/thicknes(100),density(100),conductv(100),sheatcap(100)

 common/reflecq/qreflin(6,100),qscabsin(6,100),qreflfwr(6,100), 

 &qreflfwl(6,100),qscabfwr(6,100),qscabfwl(6,100),qabstwa1(6,100), 

 &qabstwa2(6,100),qabstwb(6,100) 

 common/temp5/t(1000) 

 common/data1/deltime 

 common/radiosit/jj(14) 

 common/facttarm/fact 

 common/result/temp(200),temptrn(200) 

 dimension newt(100),fn(100) 

 real jj,newt,lengtwng,jacob 

 parameter(sig=.1714e-8) 

 parameter(fpi=3.141593/180.) 

 parameter(afact=1.0,bfact=0.00001) 

c 

A-147
c Variables:

c afact=cooling when door is opened: no motion: therefore tarmac temperature 

c underneath the wing remains very constant 

c bfact=a small number to force the tarmac temperature underneath the wing 

c to be the same as the unsheltered tarmac temperature: in effect when 

c the aircraft is in motion 

c 

 if(itotitr.eq.1.and.istep.eq.1)then 

 rlwng=(lengtwng-wfuselg)/2. 

 endif 

 if(istep.eq.1)then 

c reference the initial temp.: 

 t(51)=tt(51) 

 t(52)=tt(52) 

 t(53)=tt(53) 

c make the initial guess for the new temperature newt(i) 

 newt(51)=t(51) 

 newt(52)=t(52) 

 newt(53)=t(53) 

c reference the initial radiosity 

 jj(12)=tt(54) 

 jj(13)=tt(55) 

 jj(14)=tt(56) 

 return 

 endif 

 if(ithtrial.eq.1)goto 65 

 newt(51)=tt(117+1) 

 newt(52)=tt(117+2) 

 newt(53)=tt(117+3) 

 65 if(isolfoun.eq.1)then 

 if(ithtrial.eq.1)then 

 newt(51)=tt(117+1) 

 newt(52)=tt(117+2) 

 newt(53)=tt(117+3) 

 endif 

 t(51)=newt(51) 

 t(52)=newt(52) 

 t(53)=newt(53) 

 temptrn(51)=t(51)-459.67 

 temptrn(52)=t(52)-459.67 

 temptrn(53)=t(53)-459.67 

 return 

 endif 

c 

 if(icool.eq.1)fact1=afact 

 if(icool.eq.2)fact1=bfact 

c 

c formulate transient cooling: 

c 

c LHS Wing/Tarmac t(51): global temperature tt(118): global fvec ffn(118) 

c 

A-148
 qnetsky=(sig*newt(51)**4-jj(12))/((1.-emista)/emista) 

 fn(51)= 

 & -newt(51)+t(51)+deltime/(densta*rlwng*widthwng*tarmd*shcta)* 

 & ((qabstwa1(1,iperiod)-qnetsky-hht*(newt(51)-ta))*rlwng*widthwng+ 

 & 2.*(conducta/(fact1*fact*widthwng)*(ttarm-newt(51)))*tarmd*rlwng+ 

 & conducta/(fact1*fact*widthwng)*(ttarm-newt(51))*tarmd*widthwng- 

 &conducta/(.5*wfuselg+.5*rlwng)*(newt(51)-newt(52))*tarmd*widthwng) 

c 

c RHS Wing/Tarmac t(53): global temperature tt(120): global fvec ffn(120) 

c 

 qnetsky=(sig*newt(53)**4-jj(14))/((1.-emista)/emista) 

 fn(53)= 

 & -newt(53)+t(53)+deltime/(densta*rlwng*widthwng*tarmd*shcta)* 

 & ((qabstwa2(1,iperiod)-qnetsky-hht*(newt(53)-ta))*rlwng*widthwng+ 

 & 2.*(conducta/(fact1*fact*widthwng)*(ttarm-newt(53)))*tarmd*rlwng+ 

 & conducta/(fact1*fact*widthwng)*(ttarm-newt(53))*tarmd*widthwng- 

 &conducta/(.5*wfuselg+.5*rlwng)*(newt(53)-newt(52))*tarmd*widthwng) 

c 

c MIDDLE Wing/Tarmac t(52): global temperature tt(119): global fvec ffn(119) 

c 

 qnetsky=(sig*newt(52)**4-jj(13))/((1.-emista)/emista) 

 fn(52)= 

 & -newt(52)+t(52)+deltime/(densta*wfuselg*widthwng*tarmd*shcta)* 

 & ((qabstwb(1,iperiod)-qnetsky-hht*(newt(52)-ta))*wfuselg*widthwng+ 

 &2.*(conducta/(fact1*fact*widthwng)*(ttarm-newt(51)))*tarmd* 

 & wfuselg+

 &conducta/(.5*wfuselg+.5*rlwng)*(newt(51)-newt(52))*tarmd*widthwng+ 

 &conducta/(.5*wfuselg+.5*rlwng)*(newt(53)-newt(52))*tarmd*widthwng) 

c 

c transfer fn(51),fn(52),fn(53) into the global vector ffn(118),ffn(119) and ffn(120) 

c 

 ffn(118)=fn(51) 

 ffn(119)=fn(52) 

 ffn(120)=fn(53) 

c 

c find the jacobians, tjacb and put it into the global jacobian(120,120) 

c 

c now find tjacb151 and tjacb152 

c for LHS Wing/Tarmac t(51): global temperature tt(118): global fvec ffn(118) 

c 

 tjacb151=-1+deltime/(densta*rlwng*widthwng*tarmd*shcta)* 

 & ((-(4*sig*newt(51)**3)/((1.-emista)/emista)-hht*(1))*rlwng* 

 & widthwng+ 

 & 2.*(conducta/(fact1*fact*widthwng)*(-1))*tarmd*rlwng+ 

 & conducta/(fact1*fact*widthwng)*(-1)*tarmd*widthwng- 

 & conducta/(.5*wfuselg+.5*rlwng)*(1)*tarmd*widthwng) 

 tjacb152=deltime/(densta*rlwng*widthwng*tarmd*shcta)* 

 & (-conducta/(.5*wfuselg+.5*rlwng)*(-1)*tarmd*widthwng) 

c 

A-149
c now find tjacb251,tjacb252,tjacb253 

c MIDDLE Wing/Tarmac t(52): global temperature tt(119): global fvec ffn(119) 

c 

 tjacb251=deltime/(densta*wfuselg*widthwng*tarmd*shcta)* 

 & ((2.*(conducta/(fact1*fact*widthwng)*(-1)))*tarmd*wfuselg 

 & +conducta/(.5*wfuselg+.5*rlwng)*(1)*tarmd*widthwng) 

 tjacb252=-1+deltime/(densta*wfuselg*widthwng*tarmd*shcta)* 

 & ((-(4*sig*newt(52)**3)/((1.-emista)/emista)-hht*(1))* 

 & wfuselg*widthwng

 & +conducta/(.5*wfuselg+.5*rlwng)*(-1)*tarmd*widthwng+ 

 & conducta/(.5*wfuselg+.5*rlwng)*(-1)*tarmd*widthwng) 

 tjacb253=deltime/(densta*wfuselg*widthwng*tarmd*shcta)* 

 & (conducta/(.5*wfuselg+.5*rlwng)*(1)*tarmd*widthwng) 

c 

c now find tjacb352 and tjacb353 

c RHS Wing/Tarmac t(53): global temperature tt(120): global fvec ffn(120) 

c 

 tjacb352=deltime/(densta*rlwng*widthwng*tarmd*shcta)* 

 & (-conducta/(.5*wfuselg+.5*rlwng)*(-1)*tarmd*widthwng) 

 tjacb353=-1+deltime/(densta*rlwng*widthwng*tarmd*shcta)* 

 & ((-(4*sig*newt(53)**3)/((1.-emista)/emista)-hht*(1))* 

 & rlwng*widthwng+ 

 & 2.*(conducta/(fact1*fact*widthwng)*(-1))*tarmd*rlwng+ 

 & conducta/(fact1*fact*widthwng)*(-1)*tarmd*widthwng- 

 & conducta/(.5*wfuselg+.5*rlwng)*(1)*tarmd*widthwng) 

c 

c now put these jacobians into the global jacobian jacob(120,120) 

c the jacobians are: for ffn(118):jacob(118,118);jacob(118,119); 

c for ffn(119):jacob(119,118):jacob(119,119):jacob(119,120) 

c for ffn(120):jacob(120,119):jacob(120,120) 

c 

 jacob(118,118)=tjacb151 

 jacob(118,119)=tjacb152 

 jacob(119,118)=tjacb251 

 jacob(119,119)=tjacb252 

 jacob(119,120)=tjacb253 

 jacob(120,119)=tjacb352 

 jacob(120,120)=tjacb353 

 return 

 end 

c

A-150
A.3 REFERENCES. 

A-1. Govindarajoo, N., “THERMOD, An Enhanced Thermal Model for Determining Aircraft

Operational Temperatures User’s Manual,” FAA Report to be published. 

A-2. MSC.NASTRAN 70.5, “Quick Reference Guide,” MacNeal-Schwendler Corp., 815 

Colorado Boulevard, Los Angeles, CA, 1998. 

A-151/A-152