Kunio KATAOKACopyright© 2021 by the author. All rights reserved. No part of this book may be
reproduced in any form without the prior permission of the author.
The author sincerely gave a personal permission to the following three parties ororganizations only for education.
(1) Department of Chemical Science and Engineering, Kobe University, Japan(2) Kansai Chemical Engineering Co., Ltd., Japan(3) Department of Chemical Engineering, Louisiana State University, USA
This book will be placed as the PDF file in the respective web sites within their
homepages.
Within these parties, any teachers and students can freely access their individual
URL web site shown below to get any useful knowledge and/or information andmaterials for study and investigation.
Kansai Chemical Engineering: https://www.kce.co.jp/en/library
Kobe University: https://www.doi.org/10.24546/90008260
http://www.lib.kobe-u.ac.jp/handle_kernel/90008260
Permission for any other use must be obtained in writing from the author or fromone of the above three parties.

ⅰ

PREFACEThis book has been organized as a textbook of the three-year sequential course for students ofchemical engineering from the junior up to the first-year graduate level.
It is well known that in 1960, Professor R. B. Bird, Professor W. E. Stewart, and Professor E. N.
Lightfoot of the University of Wisconsin united fluid flow, heat transfer, and mass transfer into asingle concept with the publication of “Transport Phenomena.”
I believe that their concept remains unaltered and still alive.
I have 35 years’ teaching experience as a faculty member of the Chemical Engineering Department
of Kobe University, Kobe, Japan. Since 1968, in particular, I have had the privilege of lecturing forthis three-year course from the junior introductory to the first-year graduate advanced course.
Besides I have a fortunate experience of teaching as an invited professor the fundamentals ofchemical engineering by using their book “Transport Phenomena” at Louisiana State University forthe short term from 1978 to 1979. Some thoughtful comment and advice received from Professor
R. B. Bird during that term were invaluable and still remain as a concept of the author’s educational
standard for this book.
This is intended to serve as a textbook for an integrated understanding of fluid mechanics,
thermodynamics, heat transfer, and mass transfer from a viewpoint of process engineers. Radiation
is not included because the author thought that a brief summary of radiation transfer consistent withthe rest of the text was too difficult. I hope that this book will be useful not only as a reference but
also as a textbook to be used either by a class or for self-study.
The objective of the book is to present all the subjects of transport processes rearranged in thedeveloping sequence from the fundamentals to practical application.
Where this book is used for college instruction in chemical engineering, the whole PART I and
Chapter 20 of PART II are suitable for undergraduate courses; Chapters 11 through 16 of PART II
for graduate studies. The remaining part Chapters 17 through 19 are intended, in particular, forthose who expect to do a lot of academic research in those areas.
Chapter 2 of the book presenting basic definitions of physical transport properties is followed by
Chapter 3 simply and qualitatively explaining the fundamentals of fluid flow such as the generationof turbulent motion. Chapter 4 introduces a control volume approach for macroscopic balances ofmomentum, energy and mass, which give us very important fundamental equations practicallyapplicable to various chemical equipment designs. These three chapters are essential forundergraduate courses.
Chapter 5 introduces the microscopic shell balance of momentum, energy and mass, where
undergraduate and/or graduate students study how to construct the respective basic differential
equations such as the Navier-Stokes equation of motion. They must first learn a little vector andtensor analysis for mathematical generalization.
Chapter 6 provides several applications of those fundamental differential equations. The first two
examples deal with simple momentum equations for Newtonian fluids to be followed by two
non-Newtonian examples which may be appropriate for graduate course. The remaining parts
giving four examples on heat and mass transfer may also be useful for both course students. In
every examples, starting by the very difficult partial differential equations, we will study how tosimplify them into very simple ordinary differential equations.
Chapters 7 through 10 deal with interphase transfer of momentum, heat and mass, where they
will study learn “the film theory” and should learn how practically important it is to defineinterphase transfer coefficients. Therefore these chapters are the most importance as the primary
theme for the undergraduate course because most of chemical engineering processes can bemodeled by the control volume approach using the interphase transport concepts. Students will

ii

study how to design various chemical apparatuses from the concept of interphase transfer
coefficients. At the end of Chapter 9, the measuring principle of fluid velocities based on theconvective heat transfer around a circular cylinder is arranged for graduate-level researchers.
In Part II, Chapters 11 through 13 deal with distillation and humidification processes, both ofwhich proceed with simultaneous heat and mass transfer accompanied by phase transformation. In
Chapter 11, undergraduate students will study very simple and useful basic ideas of distillation froma viewpoint of equilibrium stage model. Chapter 12 provides graduate students with the advanced
concept of simultaneous heat and mass transfer, where the heat and mass transfer analysis of packed
column distillation can be challenged by means of a unique control volume approach. In Chapter 13,
undergraduate and/or graduate students will study how to practically design chemical equipmentusing air-water system such as air-conditioner, cooling tower, and evaporative cooler and condenser
from a viewpoint of simultaneous heat and mass transfer. Chapter 14 on ionic mass transfer isarranged especially for graduate students, which will be useful for experimental research on
liquid-phase mass transfer. Chapter 15 deals with condensation and boiling heat transfer. Chapter 16
provides the advanced analysis of diffusion processes accompanied with chemical reactions.
Because of the very difficult analytical analysis, a computer-aided numerical analysis may beutilized at present. Chapter 17 provides the fundamental aspect of turbulent transport phenomena
for graduate students as well as researchers. This will be able to provide various fundamental
knowledges as the background useful for the advanced academic research on transport processes.
The following Chapters 18 and 19 will give the advanced transport aspects of fluid flow to graduate
students. Finally Chapter 20 provides practical consideration of agitating equipment forundergraduate course. Because of the very complicated mixing phenomenon, the useful practicaldesign concept should be studied using a dimensional analysis from a viewpoint of transport
processes.
I would like to emphasize that it is getting more and more important to investigate various
chemical engineering process and equipment from a viewpoint of transport phenomena. I believethat it must effectively lead to process development, intensification, and technological innovation.
If the various mathematical approaches introduced in this book based on differential equations
are difficult to solve analytically, the computer-aided numerical analysis may be utilized as an
appropriate tool at the present time.
My friend, Emeritus Professor Dr. D. P. Harrison of Louisiana State University who was thechairman of Department of Chemical Engineering during my stay at LSU read the wholemanuscript from his stand of reaction engineering and gave me many helpful suggestions.
In closing, I would like to express my gratitude to all the faculty members and students of
Department of Chemical Science and Engineering, Kobe University and to The Society of Chemical
Engineers, Japan. In addition, Dr. H. Noda, the President of Kansai Chemical Engineering Co., Ltd.,gave me a lot of support and advice from a viewpoint of practical engineering. I would like toappreciate his kind encouragement and valuable comments
I would be very happy if many chemical engineering students will be encouraged by taking aninterest in this book.
KUNIO KATAOKA*Emeritus Professor
Dept. of Chemical Science & Engineering
Kobe University*Top Supervisor
Engineering Dept.
Kansai Chemical Engineering Co., Ltd.
Kobe
September 2020

iii

CONTENTSPART ⅠCHAPTER 1
INTRODUCTION
1.1
What do we study in Transport Process Chemical Engineering?
1.2
Concepts and Definitions(1) Two Viewpoints to Transport Science(2) Units(3) Three Categories of Physical Quantities(4) Transport Intensity(5) Fluid Forces
CHAPTER 2
2.1
2.2
2.3
2.4
2.5

FUNDAMENTAL LAWS OF
MOMENTUM, ENERGY AND MASS TRANSFER
Viscosity (Newton’s law of viscosity)
Thermal Conductivity (Fourier’s law of heat conduction)
Diffusivity (Fick’s law of diffusion)
Similarity among Molecular Transports of Momentum, Energy, and Mass
(Prandtl number and Schmidt number)
Non-Newtonian Fluids

1
1
3

7
7
9
11
13
14

CHAPTER 3
VISCOUS FLOW (LAMINAR AND TURBULENT FLOWS)
3.1
Laminar and Turbulent Flow in a Circular Pipe
3.2
Generation of Turbulent Motion (Transition to turbulent Flow)

16
16
17

CHAPTER 4
MACROSCOPIC BALANCES: CONTROL VOLUME APPROACH
4.1
Principles of Momentum, Energy, and Mass Conservation
4.2
Macroscopic Mass Balance
4.3
Macroscopic Momentum Balance
4.4
Macroscopic Energy Balance
4.5
Mechanical Energy Balance
4.6
Thermal Energy Balance for non-Isothermal System
4.7
Macroscopic Mass Balance of Individual Components

20
20
20
22
24
26
31
34

CHAPTER 5
MICROSCOPIC BALANCES: DIFFERENTIAL BALANCES
5.1
Differential Balances of Mass and Momentum
5.1-1 Differential mass balance (Equation of continuity)
5.1-2 Differential momentum balance (Navier-Stokes equation)
5.2
Differential Balance of Energy (Equation of energy)
5.3
Differential Balances of Mass (Equation of mass transport)

37
37
37
38
42
43

CHAPTER 6
APPLICATION OF DIFFERENTIAL TRANSPORT EQUATIONS
6.1
Application of the Equation of Motion (I)
6.2
Application of the Equation of Motion (II)
6.3
Application of the Equation of Motion (III) for non-Newtonian Fluid

46
46
49
52

iv

6.4
Application of the Equation of Motion (IV) for non-Newtonian Fluid
6.5
Application of the Equation of Energy (I)6.6
Application of the Equation of Energy (II)
6.6-1 Steady heat conduction
6.6-2 Unsteady heat conduction
6.7
Application of the Equation of Mass Transport

CHAPTER 7
INTERPHASE MOMENTUM TRANSPORT
7.1
Turbulent Flow Properties
7.2
Friction Factor and Pressure Drop for Channel Flows
7.3
Dimensional Analysis of Friction Factor for Channel Flows
7.4
Mechanical Energy Loss
7.4-1 Mechanical energy losses in pipelines
7.4-2 Mechanical energy losses due to pipe fittings
7.4-3 Non-circular channels (Equivalent diameter)
7.5
Drag Force on Submerged Objects

53
56
60
60
62
65

70
70
73
75
76
76
77
81
82

CHAPTER 8
INTERPHASE ENERGY TRANSPORT
84
8.1
Turbulent Heat Transfer and Definition of Heat Transfer Coefficient84
8.2
Application of the Equation of Energy Transport for Turbulent Heat Transfer 85
8.3
Overall Heat Transfer Coefficient and Heat Exchangers
87
8.3-1 Definition of overall heat transfer coefficient87
8.3-2 Logarithmic mean temperature difference
89
CHAPTER 9
HEAT TRANSFER EQUIPMENT
9.1
Shell-and-Tube Heat Exchanger
9.2
Tube-side Heat Transfer Coefficient
9.3
Heat Transfer Coefficient in Annular Space of Double-tube Exchangers
9.4
Shell-side Heat Transfer Coefficient
9.5
True Temperature Difference for 1-2 Exchangers
9.6
Engineering Design of a Shell-and-Tube Heat Exchanger
9.6-1 Thermal design procedure of double-tube exchangers
9.6-2 Thermal design procedure of shell-and-tube exchangers
9.7
Convective Heat Transfer around Submerged Objects
9.7-1 A circular cylinder in cross flow
9.7-2 Hot-wire anemometry
9.7-3 A circular sphere in cross flow

92
92
92
93
93
95
96
96
97
102
102
103
104

CHAPTER 10 INTERPHASE MASS TRANSPORT
10.1
Definition of Mass Transfer Coefficient
10.2
Analogy between Heat and Mass Transfer
10.3
Theory of Interphase Mass Transfer
10.3-1 Fundamentals --- Gas-liquid equilibrium for absorption
10.3-2 Interphase mass transfer for gas absorption
10.3-3 Mass transfer model for gas absorption
10.3-4 Mass transfer coefficients in a packed column absorber
10.4
Mass Transfer Correlations for Packed Columns
10.4.1 Height of a liquid-phase transfer unit
10.4.2 Height of a gas-phase transfer unit
10.5
Column Diameter and Pressure Drop of Packed Columns
10.6
Pressure Drop of Dry Packed Columns

106
106
107
108
108
109
110
114
114
114
116
116
118

v

PART ⅡCHAPTER 11 MASS TRAMSPORT EQUIPMENT
11.1
Distillation Fundamentals
11.1-1 Phase equilibria for distillation
11.1-2 Boiling-point diagram
11.2
Distillation Equipment
11.2.1 Continuous distillation plate column
11.2-2 Plate column fundamentals
11.2-2-1 Definition of ideal stage
11.2-2-2 Material balance
11.2-2-3 McCabe-Thiele method
11.2-2-4 McCabe-Thiele step-by-step calculation method
11.2-2-5 Reflux ratio
11.2-2-6 Enthalpy-composition method (Ponchon-Savarit step-by-step method)
11.3
Mass Transfer in Distillation Column (Plate column)
11.4
Tray Model (Plate efficiency)
11.4-1 Murphree plate efficiency
11.4-2 Mass transfer experiment
Design Calculation Procedure of Distillation Columns
11.5
11.6
Heat Balance of Distillation Column System

121
121
121
122
123
123
125
125
126
128
129
130
131
133
136
136
137
139
140

CHAPTER 12 SIMULTANEOUS HEAT AND MASS TRANSFER - I
12.1
Theory of Simultaneous Heat and Mass Transfer - I
12.2
Transport Phenomena in a Packed Column Distillation Process
12.2-1 Simultaneous mass and energy transfer model
12.2-2 Efficiency of packed distillation columns
12.3
Analogy between Mass and Enthalpy Transfer
in a Packed Column Distillation Process

143
143
145
145
147

CHAPTER 13
13.1

152

SIMULTANEOUS HEAT AND MASS TRANSFER - II
Theory of Simultaneous Heat and Mass Transfer II(Humidification and dehumidification)
13.2
Simultaneous Heat and Mass Transfer for Humidificationand Evaporative Cooling13.2-1 Theory of wet-bulb temperature
13.2-2 Humidity chart
13.3
Evaporative Cooling --- Water Cooling Tower ---CHAPTER 14
14.1
14.2
14.3
14.4

IONIC MASS TRANSPORTElectrolytic Cell
Ionic Mass Transport in an Electrochemical Reaction System
Mass Transfer Measurements by an Electrochemical Technique
Measurements of Velocity Gradient on a Wall

CHAPTER 15 HEAT TRANSFER WITH PHASE TRANSFORMATION
15.1
Condensation
15.1-1 Heat transfer for condensation
15.1-2 Film condensation of pure vapor on a vertical wall
15.1-3 Condensation heat transfer coefficient15.1-4 Engineering design of an overhead condenser
15.2
Evaporation
15.2-1 Heat transfer for boiling

148

152
154
154
157
159

163
163
164
165
167
169
169
169
170
171
172
174
174

vi

15.2-2 Pool boiling15.2-3 Heat transfer correlation for pool boiling15.2-4 Critical heat flux

174
176
177

CHAPTER 16 MASS TRANSFER WITH CHEMICAL REACTION
16.1
Diffusion with Homogeneous Chemical Reaction
16.1-1 Diffusion with a homogeneous reaction in a stagnant fluid----- Penetration theory ----16.1-2 Gas absorption with first-order reaction
16.2
Gas Absorption with Instantaneous Bimolecular Reaction
16.3
Design of Packed Absorption Towers

179
179

CHAPTER 17 TURBULENT TRANSPORT PHENOMENA
17.1
Fundamental Equations of Turbulent Transport
17.1-1 Fundamental properties of turbulent flows
17.1-2 Equation of motion for turbulent flows
17.1-3 Equations of energy and mass transport for turbulent flows
17.2
Phenomenological Understanding of Turbulent Transport
17.2-1 Effect of nonlinear interaction of turbulence
17.2-2 Mixing length theory and eddy diffusivity
17.2-3 Mixing length model based on turbulence correlation
Structure of Turbulence
17.3
17.3-1 Energy spectrum of kinetic energy
17.3-2 Spatial and temporal correlations ----- Definition of eddy sizes
17.4
Velocity Distribution of Turbulent Flow inside a Circular Pipe
17.5
Turbulent Structure and Role of Eddies
17.5-1 Turbulent structure in a circular pipe flow
17.5-2 Roles of turbulent eddies in transport processes

192
192
192
194
194
195
195
196
197
200
200
203
205
208
208
209

CHAPTER 18 BOUNDARY LAYER THEORY
18.1
Stream Function
18.2
Boundary Layer Solution of Laminar Flow along a Flat Plate
18.3
Integral Equation of Boundary Layer Flow
18.3-1 Momentum integral equation of boundary layer flow over a flat plater
18.3-2 Energy integral equation of boundary layer flow
18.3-3 Turbulent boundary layer flow
18.4
Application of Stream Function (Impinging flow)
18.5
Impinging Jet Heat Transfer
18.6
Boundary-layer Analysis for Velocity-gradient Measurement

211
211
212
217
217
218
218
220
222
223

CHAPTER 19 FREE CONVECTION
19.1
Boundary Layer Approach
19.2
Free Convection Heat Transfer

226
226
228

CHAPTER 20 AGITATION
20.1
Agitation and Mixing of Liquids
20.1-1 General structure of agitation equipment
20.1-2 Flow patterns in agitated vessels
20.2
Power Consumption in Agitated Vessels
20.2-1 Dimensional Analysis
20.2-2 Power correlations
20.3
Heat Transfer in an Agitated Tank
20.4
Scale-up of Agitated Tank Design

232
232
232
233
234
234
236
237
238

INDEX

179
181
182
185

PART ⅠCHAPTER 1
INTRODUCTION
1.1 What Do We Study in TRANSPORT PROCESSCHEMICAL ENGINEERING?
In order to understand what and how we will study, let us consider a process producing methanefrom heavy oil (higher hydrocarbons) as an example of chemical plants.
Figure 1 shows the flow sheet diagram. The heavy oil is usually withdrawn as one of the bottom
products from a distillation plant 1 of crude oil. The process consists of gasification, purification,and methanation.

Fig.1.1-1. Example of process flow diagram of a chemical plant

Heavy oil, oxygen, and steam are fed through heat exchangers 2 and 3 to the gasification reactor
4 where the higher hydrocarbons undergo complete reaction to produce CO and H2 without solid
catalyst. The outlet gas from the reactor 4 is cooled, passes through a soot separator 6, and enters asulfur removing equipment 7. The gas from the separator 7 consists of CO, H 2, and small amountof CO2. The gas is divided and fed into two sections: purification and methanation sections. In
the purification section, the gas is fed through a saturator 8 to a water-gas shift reactor 10. The
reactor consists of a fixed bed of solid catalyst particles through which the gas undergoes thereaction CO + H2O → CO2 + H2 in the presence of nickel based catalyst. The reaction is2Introduction

reversible and exothermic. The shifted gas is cooled by a contact cooler and heat exchanger 11,
and led to a CO2 absorber 12, where CO2 is removed from the shifted gas by water or some sort ofalkaline solution. The absorber is a column filled with some sort of device or packed with small
solids of varying shapes to increase the interfacial area between the trickling liquid and theflowing-up gas. The gas leaving the absorber is hydrogen rich gas. On the other hand, the rest of
the gas produced in the gasification process is mixed with the hydrogen rich gas from thepurification process, and fed to a series of methanation reactors 14. The reactors are a fixed bed ofnickel catalyst particles through which the reaction CO + 3 H2 → CH4 + H2O takes place. The
methane rich gas leaving the methanation reactors contains steam. The product of methane richgas is obtained after removing steam by condenser 15. A certain fraction of the product gas is fed
back to the methanation reactors 14 by a compressor 16.
As a prerequisite to the design of industrial equipment, an understanding of the fundamental
properties and characteristics of individual operations is required. This plant offers examples ofvarious momentum, energy, and mass transfer processes which will be studied in this course.
In the design of pipelines, we need to know1) mechanical energy losses of the flowing fluids (or pressure drops)2) power of pumps, compressors, and blowers and optimal pipe diameters required to attain thedesired flow rates
The mechanical energy balance (the modified Bernoulli equation) for calculation of the power
requirement and the concept of friction factor, equivalent length, and friction loss factor for
calculation of the pressure drops at the individual sections will be studied in this course.
In the design of heat exchangers, we need to know1) pressure drops of shell- and tube-side streams2) heat transfer area (i.e. the size of exchangers) required to attain the desired rate of heat transfer.
The energy balance for calculation of the discharge fluid temperatures and the concept of heattransfer coefficient will be studied to size the heat exchangers in this course.
In the design of gas absorbers, we need to know
1) operating conditions (liquid-gas flow ratio, temperature, pressure, etc.)2) height of absorption columns required to attain the desired separation efficiency
3) diameter of absorption columns to attain the desired gas-liquid contact condition
Phase equilibrium should be studied in thermodynamic course. The concept of mass transfer
coefficient will be studied for determination of mass transfer area i.e. the height of an absorption
column in this course.
Although the above example does not have a distillation process, distillation is one of the mostpopular unit operations. Many distillation columns are encountered in the oil refinery plants. In thedesign of distillation columns, we need to know
1) operating conditions (feed rate, reflux-ratio, temperature, pressure, liquid-gas flow ratio, etc.)
2) column structures (trayed column, packed column) suitable for the feed mixture properties
3) number of trays, height of packing section suitable for the desired separation efficiency
4) column diameter for attaining the desired gas-liquid contact condition
Phase equilibrium is also very important for distillation process engineering since the mostpopular design procedure is the equilibrium-stage model.
After learning the fundamentals of engineering calculation based on the equilibrium stage modelfor the practical design of distillation equipment, we try to understand the distillation processes
from a viewpoint of heat and mass transfer.

Concepts and Definitions

3

In the design of reactors, we need to know
1) operating conditions (temperature, pressure, feed composition, etc.)
2) size of reactors required to obtain the desired conversion and product yield3) selection of suitable catalyst
The reaction kinetics (reaction rate, chemical equilibrium, etc.) should be studied inanother course. The flow patterns and heat- and mass-transfer characteristics should also be
considered because actual rate of reaction depends on mass transfer of reactants as well as
temperature and pressure.
In the design of dust separators, we need to know1) minimum size of collectable particles
2) collection efficiency
The underlying theory on particle mechanics should be studied with the fluid mechanics.
The main problem in the design of such chemical engineering processes is to determine the sizeof equipment: to calculate how much contacting area is necessary for the required rate of heat or
mass transfer.
Most flows in chemical engineering equipment are turbulent. Of particular importance are themixing effects of turbulence on heat/mass transfer and chemical reaction. The phenomenologicalunderstanding of turbulent flows in this course will be helpful to the reader in the design of new
equipment and the improvement of equipment efficiency.
Unfortunately, no general approach to the solution of problems on the turbulent transportphenomena is available. Accurate quantitative predictions cannot be made without relying heavily
on empirical data. Recently many trials have been made for the computer-aided modeling of the
turbulent transport mechanism but those levels are beyond the concept of this course.
The Transport Process Chemical Engineering is one of the most basic and most importantscientific areas constituted for the understanding of individual operations from an engineering
viewpoint.

1.2 Concepts and Definitions(I)

Two Viewpoints for Transport Science
Lagrangian and Eulerian Viewpoints
There are two different viewpoints for representation of physical fields. In the Lagrangian
approach, the physical variation can be described for a particular element of interest as it moves.
The coordinates describe the time-dependent position of the moving element. This viewpoint ismainly used in particle and rigid-body dynamics. On the other hand, in the Eulerian viewpoint,
the physical variation is described on a given stationary line or plane of interest. In Transport
Science, therefore, fluid can be regarded as a continuum medium.
The Eulerian viewpoint is more useful for description of the physical variation in flowing
fluids. This approach gives the value of a fluid variable at a given position at a given time suchas the fluid velocity distribution on the cross section of a flow passage.
(II)
Units
The International Standard system of units (Systeme International), or the so-called “SI units”
will be used in this course. The basic units are mass in kilogram, (kg), length in meters, (m),and time in seconds, (s). The unit of force is newton, (N), which is easily defined using thesimplest form of Newton’s law of motion: 𝑭 = 𝑚 𝒂. One newton 1 N is the force required toaccelerate a mass of 1 kg at a rate of 1 m/s2. The unit of energy or work is joule, (J), which can

4Introduction

be easily understood as the product of displacement and force component in its direction.
Basically, degrees Kelvin, (K), will be used as the unit of absolute temperature, but degrees
Cerlsius, (℃), will be also used for convenience.
Another useful consideration in dealing with units is the concept of the molar units. In
processes including chemical reactions, kilogram moles, (kmol), may be much more convenient
for mass than kilograms.
(III) Three Categories of Physical Quantities
The physical quantities encountered in this course fall into one of three categories: (1) scalars
such as time, temperature, density, and energy; (2) vectors such as velocity, acceleration,
momentum, heat-flux and mass-flux; and (3) tensors such as shear stress or momentum-flux andrate of strain. Vectors have three components whereas tensors have nine components.
In rectangular coordinates, for example, a velocity vector is expressed as𝒗 = (𝑣𝑥 , 𝑣𝑦 , 𝑣𝑧 )(1.2-1)
And a stress tensor is given by𝜏𝑥𝑥 𝜏𝑥𝑦 𝜏𝑥𝑧𝜏
(1.2-2)
𝝉 = [ 𝑦𝑥 𝜏𝑦𝑦 𝜏𝑦𝑧 ]𝜏𝑧𝑥 𝜏𝑧𝑦 𝜏𝑧𝑧[Example 1.1-E1] Force Balance from the Lagranrian Viewpoint
As an example of Lagrangian methods, let us consider a problem for the trajectory ofa flying ball of mass 𝑚 shown below. The position (x, y) of the ball is a function of time𝑡.
The velocity 𝒗 at time t can be decomposed as𝑑𝑥
𝑑𝑦𝒗 = 𝑑𝑡 𝒊 + 𝑑𝑡 𝒋
(1.2-3)where 𝒊 and 𝒋 are unit vectors in the x- and y-directions, respectively.
The external forces acting on the ball are gravitational force 𝑚 𝑔 and the resistanceforce 𝑹 = 𝑅𝑥 𝒊 + 𝑅𝑦 𝒋 exerted by air.
Applying Newton’s second law of motion to the ball:𝑚 𝑑𝑡 2 = − 𝑅𝑥𝑑2 𝑥

(1.2-4)𝑑2 𝑦

𝑚 𝑑𝑡 2 = − 𝑅𝑦 − 𝑚 𝑔(1.2-5)

Fig.1.2-1. Equation of motion of a flying ball

If the resistance force is known as a function of velocity, the above set of equations
can be solved. The solution will be given by

Concepts and Definitions

5𝑥 = 𝑥(𝑡) and 𝑦 = 𝑦(𝑡)(1.2-6)
Eliminating the variable t, the trajectory is expressible as𝑓 (𝑥, 𝑦) = 0(1.2-7)This approach is applicable, for example, to calculation of the trajectories of liquid
droplets in spray dryers and solid particles in sedimenting centrifuges.
(IV) Transport Intensity
There are two solid plates in which heat is transferred at the same rate. The transport area of
one plate is double as much as that of the other plate. For this case, the heat transport intensity(called “heat-flux”) of the former plate is half as much as that of the latter plate.

flux =
rate =

a quantity of interest transferred(are)(time)
a quantity of interest transferredtime(1.2-8)(1.2-9)For example, there are two copper electrodes in an electrolytic solution of copper
sulfate, one of the electrodes (anode) has active surface much larger than the other one(cathode). In an electrolytic reaction occurring at some potential difference (emf)
between the electrodes, the cathode has a deposition of copper ions while the anodereleases copper ions at the same rate. In spite of the same ionic mass transfer rates at
the two electrodes, the ionic mass flux due to the copper deposition on the cathodebecomes much larger than the mass flux releasing copper ions from the anode. In thiscase, the resistance to mass transfer near the cathode becomes predominant.
(V)
Fluid Forces(5-1) Body Force and Surface Force
Body force such as gravitational force is proportional to the system’s mass. Surface force
such as pressure and friction force is proportional to the acting surface area and usually given interms of orthogonal components, tangent and normal to the surface. They are called tangential
and normal forces, respectively.
(5-2) Static Pressure
A fluid deforms continuously when it is subjected to shear stresses. In the case of anelement of stationary fluid, stress acts normal to the surface of the fluid element. This normal
stress, in the absence of motion, is called static pressure. In fluid flow, a normal stress generally
consists of static pressure and additional normal stress due to the fluid motion.
For example, one component of normal stress is expressible as𝜎𝑥𝑥 = 𝑝 + 𝜏𝑥𝑥(1.2-10)The pressure and stresses have units of N/m 2. Pascal (1 Pa = 1 N/m2) is often used for theunit of pressure.
(5-3) Hydrostatics
For both compressible and incompressible fluids, static pressure varies with vertical height as𝑑𝑝= 𝜌𝑔(1.2-11)𝑑𝑦

Fig.1.2-2. Hydrostatics of a water tank

6

Introduction

Let us consider a tank filled with water. What is the static pressure at a position ℎ below thefree surface of water? Water can be considered as an incompressible fluid: 𝜌 = const.
Integrating Eq.(1.2-11) with respect to 𝑦 gives𝑝ℎ𝑝 − 𝑝0 = ∫𝑝 𝑑𝑝 = ∫0 𝜌 𝑔 𝑑𝑦 = 𝜌 𝑔 ℎ
0(1.2-12)

where 𝑝0 is a pressure at the free surface and usually atmospheric pressure.

[Problem 1.2-P1]
Manometers are a device for measuring pressure difference between two points. A
fluid of density 𝜌 is flowing through a circular pipe. The U-tube manometer which
has a manometer fluid with density 𝜌𝑚 has two pressure taps at point A and B,
respectively. Determine the pressure difference 𝑝𝐴 − 𝑝𝐵 by the manometer reading ℎ.

Fig.1.2-P1. Pressure difference measured by a manometer[Problem 1.2-P2]
The cylindrical water tank (diameter 𝐷) shown below has a nozzle at the bottom
corner. Water is discharged due to the gravity in the following rule:𝑣 = 𝑐 √2 𝑔 ℎ
where 𝑣 is (cross-sectional are-averaged) velocity of water issuing the nozzle (nozzle
outlet diameter 𝑑) and ℎ the water surface height at time 𝑡 from the nozzle level. The
coefficient in front of the square-root is an assumed constant (usually 0.95). At the
beginning, the water height is ℎ0 .
How long does it take until the water level arrives at the nozzle height ℎ𝑁 ?

Fig.1.2-P2. Water discharge from a tank

CHAPTER 2
FUNDAMENTAL LAWS OF
MOMENTUM, ENERGY AND MASS TRANFER
All materials are generally composed of molecules. However any theory that would
predict the individual behaviors of these many molecules would be very difficult, far
beyond our abilities at the present level. Most engineering work is usually concernedwith the macroscopic behavior considering a fluid as a continuous distribution ofmatter or a continuum. Therefore we can use the continuum approach in this course.
We postulate that velocity, density, pressure, temperature, mass concentration,
momentum-flux, energy-flux, mass-flux etc. behave as continuum functions. An exact
derivation of these laws of molecular transport can be made by non-equilibrium
thermodynamics

2-1 Viscosity (Newton’s Law of Viscosity)
The viscosity can be understood with the aid of very simple flow problem. There is a
fluid --- either a gas or a liquid --- contained between two large parallel plates a small
distance ∆𝑌 apart, which extend indefinitely in the x- and z-directions.

Fig.2.1-1. Momentum transfer across fluid layer between two parallel plates

Initially, the fluid and two plates are at rest, but at time 𝑡 = 0 the upper plate is set
in motion in the positive x-direction at a constant velocity ∆𝑉 by a force 𝐹.
As time proceeds, the fluid distant from the upper plate gains velocity (momentum),and finally the steady-state linear velocity profile is attained. In Transport Science, weinterpret that the x-directed momentum 𝜌 𝑣𝑥 diffuses in the y-direction. If F is acting
on a definite area 𝐴 of the upper plate, we can get the following empirical relation
between 𝐹 and ∆𝑉:𝐹∆𝑉= − 𝜇 ∆𝑌
(2.1-1)𝐴
The constant of proportionality 𝜇 is known as the viscosity of the fluid. Even in steadystate the x-momentum 𝜌 𝑣𝑥 is steadily transferred in the y-direction to maintain the

8Phenomena

Fundamental Laws of Momentum, Energy, and Mass Transfer

linear velocity distribution:𝑣𝑥𝑦= 1 − ∆𝑌∆𝑉

(2.1-2)

In a differential form, Eq.(2.1-1) is expressible as𝑑𝑣
𝜏𝑦𝑥 = − 𝜇 𝑑𝑦𝑥(2.1-3)This is a form of Newton’s law of viscosity for one-dimensional flow. This law statesthat local shear stress 𝜏𝑦𝑥 at y is proportional to local velocity gradient 𝑑 𝑣𝑥 ⁄𝑑𝑦 at y.
The subscript yx indicates that the shear stress is in a plane perpendicular to y andparallel to x or that that the x-momentum diffuses in the y-direction. Since thedimensions of the shear stress is the same as those of momentum flux, it can beconsidered that local momentum-flux is proportional to local gradient of momentum
per unit volume:𝑑𝜚𝑣
𝜏𝑦𝑥 = − 𝜈 𝑑𝑦 𝑥
(2.1-4)
Here 𝜈 = 𝜇 ⁄𝜚 is called “kinematic viscosity,” which has the same units as diffusivity,
i.e. m2/s in SI units. The kinematic viscosity can be considered as molecular diffusivity
for momentum.
Fluids are classified as Newtonian or non-Newtonian, depending on the relation of
the shear stress with the rate of strain. The fluids which do not obey the Newton’s lawof viscosity are called “non-Newtonian fluid”. All kinds of gases and simple liquids
belong to Newtonian fluid. Polymer solutions and their melts, emulsions, slurries, andpastes often behave as non-Newtonian fluids. Some models for non-Newtonian fluids
will be described in Section 2.5.
The units of viscosity can be obtained as follows:
𝑀 𝐿 𝑇 −2

𝑀 𝐿 𝑇 −1

𝜏𝑦𝑥

𝑀 𝐿−1 𝑇 −2𝑁

(2.1-5)𝜏𝑦𝑥 (=) 𝐿2 (=) 𝐿2 𝑇 (=) 𝑚2
The units of shear stress become the same as that of pressure.
𝑀𝑘𝑔(2.1-6)𝜇 (=) (𝑑𝑣 ⁄𝑑𝑦) (=) (𝐿⁄𝑇) 𝐿−1 (=) 𝐿 𝑇 (=) 𝑚 𝑠
𝑥

𝐿2

𝑚2

𝜈 (=) 𝜇 ⁄𝜚 (=) 𝑇 (=) 𝑠(2.1-7)
where 𝑀, 𝐿, 𝑇 are characteristic dimensions of mass, length and time, respectively.
1 kg/m s (= Pa s) = 10 poise (= g/cm sec) = 1,000 centipoise (= cP).
General form of Newton’s law of viscosity is listed given in terms of rectangular, andcylindrical coordinates in Table 2.1-1.
Table 2.1-1 Newton’s law of viscosity[Rectangular coordinates(𝑥, 𝑦, 𝑧)]
𝜕𝑣
𝜕𝑣
2
2
𝜏𝑥𝑥 = − 𝜇 [2 𝑥 − (𝛻 ∙ 𝒗)] 𝜏𝑦𝑦 = − 𝜇 [2 𝑦 − (𝛻 ∙ 𝒗)]
𝜏𝑧𝑧 = − 𝜇 [2

𝜕𝑥
𝜕 𝑣𝑧
𝜕𝑧

−

3
2

3
𝜕 𝑣𝑦

𝜕𝑦

3
𝜕 𝑣𝑥

(𝛻 ∙ 𝒗)] 𝜏𝑥𝑦 = 𝜏𝑦𝑥 = − 𝜇 [

𝜕𝑦
𝜕 𝑣𝑧

𝜕 𝑣𝑧

+

𝜕 𝑣𝑦

𝜕𝑥
𝜕 𝑣𝑥

]

𝜏𝑦𝑧 = 𝜏𝑧𝑦 = − 𝜇 [+] 𝜏𝑧𝑥 = 𝜏𝑥𝑧 = − 𝜇 [ 𝜕𝑥 + 𝜕𝑧 ]𝜕𝑧𝜕𝑦𝜕𝑣𝑥 𝜕𝑣𝑦 𝜕𝑣𝑧(∇ ∙ 𝒗) =
+
+
𝜕𝑥
𝜕𝑦
𝜕𝑧
[Cylindrical coordinates (𝑟, 𝜃, 𝑧)]
𝜕𝑣
2
1 𝜕𝑣𝜃
𝑣
2
𝜏𝑟𝑟 = − 𝜇 [2 𝑟 − (𝛻 ∙ 𝒗)] 𝜏𝜃𝜃 = − 𝜇 [2 (+ 𝑟 ) − (𝛻 ∙ 𝒗)]
𝜏𝑧𝑧 = − 𝜇 [2

𝜕𝑟
𝜕𝑣𝑧
𝜕𝑧

−

3
𝜕𝑣𝜃

𝜏𝜃𝑧 = 𝜏𝑧𝜃 = − 𝜇 [(∇ ∙ 𝒗) =

1 𝜕
𝑟 𝜕𝑟

3
2

𝜕𝑧

𝑟 𝜕𝜃

(𝛻 ∙ 𝒗)]+(𝑟 𝑣𝑟 ) +

1 𝜕𝑣𝑧
𝑟 𝜕𝜃
1 𝜕𝑣𝜃
𝑟 𝜕𝜃

𝜏𝑟𝜃 = 𝜏𝜃𝑟 = − 𝜇 [𝑟
]
+

𝜕
𝜕𝑟

𝜕𝑣𝑧

𝑟

𝑣

𝑟
𝜕𝑣𝑟

𝜏𝑧𝑟 = 𝜏𝑟𝑧 = − 𝜇 [ +
]
𝜕𝑟
𝜕𝑧
𝜕𝑣𝑧
𝜕𝑧

3

( 𝜃) +

1 𝜕𝑣𝑟
𝑟 𝜕𝜃

]

Thermal Conductivity (Fourier’s Law of Heat Conduction)
7

9[Problem 2.1-P1] There are a pair of concentric cylinders, the outer one of which is rotated at a
speed of 20 RPS (rotation per second) and the inner one is at rest. The annular space is filled with
50% aqueous solution of glycerin which has viscosity of 5 cP or 0.0005 Pa s. Both cylinders are
500 mm long and the annular gap width is 2.5 mm.

Fig.2.1-P1. Velocity distribution in an annular gap between coaxial cylinders

The radii of the cylinders are 𝑅𝑖 = 25 mm and 𝑅𝑜 = 27.5 mm, respectively. The radial
distribution of circumferential velocity can be expressed as
𝑅𝑖 𝑅𝑜

𝑣𝜃 =

𝑟

𝑅𝑖

𝑅𝑖

𝑟

𝑅𝑜 𝜔𝑜 ( −𝑅𝑜 2 − 𝑅𝑖 2)

(2.1-P1)

What is the torque required to maintain the steady rotation?
[Problem 2.1-P2] There are two concentric cylinders, the outer one of which is rotated
at a rate of 10 RPS. The radii of the cylinders are 0.04 m and 0.05 m, respectively. The
lengths of the cylinders are about 0.75 m. The annulus between the cylinders is filledwith a liquid whose viscosity is unknown. The torque 𝑇𝑞 = 5.53 × 10−3 N-m isnecessary to keep the inner cylinder at rest. What is the viscosity of the fluid?
[Problem 2.1-P3]
expressed as
𝑣𝑧

2

𝑟

2

The velocity distribution in laminar flow through a circular annulus is𝑟= 𝐴 [1 − (𝑅 ) + 𝐵 ln 𝑅 ]𝑉𝑜𝑜(2.1-P2)

where 𝑉 = average velocity (constant)
𝐵 = (𝑟 ∗2 − 1)⁄ln 𝑟 ∗𝑟 ∗ = 𝑅𝑖 ⁄𝑅𝑜
Calculate the shear stresses on the inside surface of the outer tube and the outside surface of theinner tube.

2.2

Thermal Conductivity (Fourier’s Law of Heat Conduction)

Similarly to the definition of viscosity, we can define thermal conductivity. There are two verylarge parallel plates a small distance ∆𝑌 apart, which contain a slab of either solid or fluid between
them. Initially the slab has a uniform temperature 𝑇0. At time 𝑡 = 0, the upper surface has a stepchange in temperature from 𝑇0 to a higher constant temperature 𝑇1. As time proceeds, thermalenergy diffuses in the y-direction and ultimately a linear temperature profile is attained in steadystate. The slab material is stationary. The slab is so thin that we can neglect the effect of naturalconvection due to density change. We can also neglect the effect of radiative transport.

10
Phenomena

Fundamental Laws of Momentum, Energy, and Mass Transfer

Fig.2.2-1. Energy transfer in a slab put between two large parallel plates

We get the following relation between the rate of heat flow 𝑄 through the slab and thetemperature difference ∆𝑇 = 𝑇1 − 𝑇0 :𝑄∆𝑇= − 𝜅 ∆𝑌
(2.2-1)𝐴
The constant of proportionality 𝜅 is known as the thermal conductivity of the material. In adifferential form, the above equation can be written as𝑑𝑇
𝑞𝑦 = − 𝜅 𝑑𝑦
(2.2-2)This is Fourier’s law of heat conduction for one-dimensional heat flow. This law states that local
heat-flux 𝑞𝑦 is proportional to local gradient of temperature 𝑑𝑇⁄𝑑𝑦. The subscript y indicates the
direction of heat flow.
For the case of constant density and heat capacity, the above equation can be rewritten as

𝑞𝑦 = − 𝛼𝑑 (𝜌𝐶𝑝 𝑇)(2.2-3)𝑑𝑦This states that local heat-flux (or local enthalpy-flux) is proportional to local gradient of enthalpy
per unit volume. Here 𝛼 = 𝜅⁄𝜚𝐶𝑝 is called “thermal diffusivity”, which has the same units as
those of mass diffusivity.
The units of thermal conductivity and diffusivity can be obtained as follows:
𝑞𝑦 (=)

𝑀 𝐿 𝑇 −2 𝐿
𝐿2𝑇(=) 𝐽⁄𝑚2 𝑠 (=) 𝑊 ⁄𝑚2
𝑇−2 𝐿
𝑀𝐿 2
𝐿 𝑇

𝑀 𝐿2 𝑇 −2( =)(=) 𝐽⁄𝑚 𝑠 𝐾 (=) 𝑊 ⁄𝑚 𝐾
⁄
𝑡 𝐿
𝐿𝑇𝑡
𝑀 𝐿 𝑇 −2 𝐿 ⁄𝐿𝑇𝑡
𝐿2𝛼 (=) 𝜅⁄𝜌𝐶𝑝 (=) (𝑀⁄𝐿3 )(𝑀 𝐿2 𝑇−2⁄𝑀 𝑡) (=) 𝑇 (=) 𝑚2 ⁄𝑠𝑞𝑦𝜅 (=) (− 𝑑𝑇⁄𝑑𝑦) (=)(2. 2-4)(2.2-5)(2.2-6)

where 𝑡 is the characteristic dimension of temperature. General forms of Fourier’s law of heat
conduction are listed in Table 2.2-1.
Table 2.2-1 Fourier’s law of heat conduction[Rectangular coordinates (𝑥, 𝑦, 𝑧)]𝜕𝑇𝜕𝑇𝜕𝑇𝑞𝑥 = − 𝜅, 𝑞𝑦 = − 𝜅, 𝑞𝑧 = − 𝜅𝜕𝑥𝜕𝑦𝜕𝑧[Cylindrical coordinates (𝑟, 𝜃, 𝑧)]
𝜕𝑇
1 𝜕𝑇
𝜕𝑇
𝑞𝑟 = − 𝜅, 𝑞𝜃 = − 𝜅, 𝑞𝑧 = − 𝜅
𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑧

[Problem 2.2-P1] There are a pair of very large steel plates which are parallel placed 2 mmapart and contain mercury between them. The upper plate is kept at 303 K and the lower plate at
285 K. Mercury has a thermal conductivity of 8.4 W/m K. What is the steady-state heat-flux ?
Assume that mercury is stationary.
[Problem 2.2-P2] A window has a glass plate (area 𝐴 = 1 m2, thickness 𝛿 = 3 mm, thermal
conductivity 𝜅 = 0.814vW/m K). The inside surface is kept at 15 C and the outside surface at 5 C.
What is the heat loss through the glass plate ?

Diffusivity (Fick’s Law of Diffusion)

11[Problem 2.2-P3] Consider a hollow cylinder, then outside surface of which is kept at 𝑇𝑜 andthe inside surface of which at 𝑇𝑖 (𝑇𝑖 > 𝑇𝑜 ). The steady temperature distribution is given by
𝑇 −𝑇
𝑇 = 𝑇𝑜 + ln(𝑅𝑖 ⁄𝑅𝑜 ) ln(𝑟⁄𝑅𝑜 )𝑅𝑖 < 𝑟 < 𝑅𝑜
(2.2-P1)𝑖𝑜

The hollow cylinder has thermal conductivity 𝜅. Calculate the heat-flux on the outside surface.

2.3

Diffusivity (Fick’s Law of Diffusion)

Diffusion is similar to, but more complicated than the molecular transport of momentum andthermal energy because we have to deal with mixtures in which the velocities of the individual
species are different. The mass center may move as the diffusion process proceeds. The mixture
velocity must be evaluated by averaging the velocities of all of the species present.
Let us first consider two definitions of concentration for n-component mixture: the massconcentration of species i (the mass of species i per unit volume of mixture) 𝜌𝑖 ; the molecularconcentration of species i (the number of moles of species i per unit volume of mixture) 𝐶𝑖 =𝜌𝑖 /𝑀𝑖 .
Here 𝑀𝑖 is the molecular weight of species i.
The mass and molar densities of the mixture are respectively given by𝜌 = ∑𝑛𝑖=1 𝜌𝑖and𝑐 = ∑𝑛𝑖=1 𝐶𝑖(2.3-1)
Therefore the mass and mole fractions are given by
𝜔𝑖 = 𝜌𝑖 ⁄𝜌
and𝑥𝑖 = 𝐶𝑖 ⁄𝑐
Now consider the averaging of the velocities of species. The absolute velocity of species i isdenoted by 𝑣𝑖 with respect to stationary coordinate axes.
The mass-average velocity𝑣 = ∑𝑛𝑖=1 𝜌𝑖 𝑣𝑖 / ∑𝑛𝑖=1 𝜌𝑖(2.3-2)This implies the velocity of mass center which corresponds to 𝑣 for pure fluid.
The molar-average velocity 𝑣 ∗ is defined as𝑣 ∗ = ∑𝑛𝑖=1 𝐶𝑖 𝑣𝑖 / ∑𝑛𝑖=1 𝐶𝑖(2.3-3)This is the velocity of the mixture averaged with respect to the number of molecules.
Then two different diffusion velocities can be defined relative to these average velocities, that is,relative to the local motion of the fluid stream: 𝑣𝑖 − 𝑣 and 𝑣𝑖 − 𝑣 ∗ .
The diffusion flux of a given species is a vector quantity denoting the amount of the species thatpasses through a unit cross section normal to the average velocities.
For simplicity, we restrict our discussion to a binary system of species A and B.
The mass flux relative to the mass average velocity is defined as
𝑗𝐴 = 𝜌𝐴 (𝑣𝐴 − 𝑣)
(2.3-4)
Similarly the molar flux relative to the molar-average velocity is
𝐽𝐴 = 𝐶𝐴 (𝑣𝐴 − 𝑣 ∗)(2.3-5)Figure 2.3-1 shows time-dependent composition profiles spreading due to molecular diffusion froma line source.
From a viewpoint of thermodynamics, the driving force of the molecular diffusion is thechemical potential. For a homogeneous ideal solution at constant temperature and pressure, thechemical potential is defined by
𝜇𝑐 = 𝜇 0 + 𝑅 𝑇 ln 𝐶𝐴(2.3-6)

12

Fundamental Laws of Momentum, Energy, and Mass Transfer

Fig.2.3-1. Composition profiles expressed with light and shade, spreading with molecular
diffusion from a line source

The molar diffusion in y-direction occurs owing to the gradient of chemical potential:𝑑 𝜇𝑐
𝑣𝐴𝑦 − 𝑣 ∗ 𝑦 = − 𝑢𝐴𝑑𝑦

(2.3-7)

where 𝑢𝐴 is the “mobility” of component A, or the resulting velocity of the molecule A under the
influence of a unit driving force. From the above equation𝑑 𝜇𝑐
𝑅 𝑇 𝑑𝐶= 𝐶 𝑑𝑦𝐴(2.3-8)𝑑𝑦
𝐴

Then the y-component of molar flux is given by𝑑𝐶
𝑑𝐶
𝐽𝐴𝑦 = 𝐶𝐴 (𝑣𝐴𝑦 − 𝑣 ∗ 𝑦 ) = − 𝑢𝐴 𝑅 𝑇 𝑑𝑦𝐴 = − 𝐷𝐴𝐵 𝑑𝑦𝐴(2.3-9)This is Fick’s law in a binary mixture for one-dimensional diffusion relative to the molar-averagevelocity. Here the constant of proportionality 𝐷𝐴𝐵 is called the “diffusivity” of component A
diffusing through component B.
(In addition to concentration gradients, there are many other physical conditions which will produce
a chemical potential gradient: temperature gradients, pressure gradients, etc. Our discussion will
be restricted to the ordinary diffusion resulting from the concentration gradients.)
Under isothermal, isobaric conditions, the molar density 𝑐 is constant.
Then𝑑𝑥
𝐽𝐴𝑦 = − 𝑐 𝐷𝐴𝐵 𝑑𝑦𝐴(2.3-10)𝐶𝐴 (𝑣𝐴𝑦 − 𝑣 ∗𝑦 ) = − 𝑐 𝐷𝐴𝐵

𝑑 𝑥𝐴
𝑑𝑦

For the binary system,𝑣 ∗ 𝑦 = (𝐶𝐴 𝑣𝐴𝑦 + 𝐶𝐵 𝑣𝐵𝑦 )⁄𝑐
Therefore
𝐶𝐴 𝑣 ∗ 𝑦 = 𝑥𝐴 (𝐶𝐴 𝑣𝐴𝑦 + 𝐶𝐵 𝑣𝐵𝑦 )Substituting this relation
𝑑𝐶
𝑑𝐶
𝐶𝐴 𝑣𝐴𝑦 = 𝐶𝐴 𝑣 ∗ 𝑦 − − 𝐷𝐴𝐵 𝑑𝑦𝐴 = 𝑥𝐴 (𝐶𝐴 𝑣𝐴𝑦 + 𝐶𝐵 𝑣𝐵𝑦 ) − 𝐷𝐴𝐵 𝑑𝑦𝐴(2.3-11)(2.3-12)(2.3-13)(2.3-14)

The terms 𝐶𝐴 𝑣𝐴𝑦 and 𝐶𝐵 𝑣𝐵𝑦 imply the y-component molar fluxes of component A and B
relative to stationary coordinates. They are denoted by 𝑁𝐴𝑦 and 𝑁𝐵𝑦 .
Finally we obtain𝑑𝑥
𝑁𝐴𝑦 = 𝑥𝐴 (𝐶𝐴 𝑣𝐴𝑦 + 𝐶𝐵 𝑣𝐵𝑦 ) − 𝑐 𝐷𝐴𝐵 𝑑𝑦𝐴(2.3-15)

Similarity among Molecular Transports of Momentum, Energy, and Mass

13

This equation states that local molar flux at a plane fixed in the stationary coordinates consists ofthe bulk motion of the binary mixture (first term) and the molecular diffusion (second term) being
proportional to local gradient of concentration or mass of component A per unit volume.
In an analogous fashion, the Fick’s law can be expressed in terms of mass flux:𝑑𝜌𝑑𝜔
𝑗𝐴𝑦 = − 𝐷𝐴𝐵 𝑑𝑦𝐴 = − 𝜌𝐷𝐴𝐵 𝑑𝑦𝐴(2.3-16)𝑛𝐴𝑦 = 𝜔𝐴 (𝑛𝐴𝑦 + 𝑛𝐵𝑦 ) − 𝜌𝐷𝐴𝐵𝑑 𝜔𝐴

(2.3-17)𝑑𝑦

The subscript y indicates the direction of diffusion.
General forms of Fick’s law relative to the average velocity 𝑣 ∗ are listed in Table 2.3-1 .
Table 2.3-1 Fick’s law of molecular diffusion based on average velocity[Rectangular coordinates (𝑥, 𝑦, 𝑧)]
𝜕𝐶𝐴
𝜕𝐶𝐴
𝜕𝐶𝐴
𝑗𝑥 = − 𝐷𝐴𝐵, 𝑗𝑦 = −𝐷𝐴𝐵
, 𝑗𝑧 = − 𝐷𝐴𝐵𝜕𝑥𝜕𝑦𝜕𝑧[Cylindrical coordinates (𝑟, 𝜃, 𝑧)]𝜕𝐶𝐴
1 𝜕𝐶𝐴𝐵𝜕𝐶𝐴𝐵
𝑗𝑟 = − 𝐷𝐴𝐵, 𝑗𝜃 = − 𝐷𝐴𝐵, 𝑗𝑧 = − 𝐷𝐴𝐵
𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑧

It is noted that the Fick’s law has the same functional form as the Fourier’s law.
The units of mass flux, concentration, and diffusivity are𝑀
𝑘𝑚𝑜𝑙 𝑜𝑓 𝐴
𝐽𝐴𝑦 , 𝑁𝐴𝑦 (=) 2 (=)
2
𝐿 𝑇
𝑚 𝑠
𝑘𝑚𝑜𝑙 𝑜𝑓 𝐴

𝑀

𝐶𝐴 (=) 𝐿3 (=)
𝑗𝐴𝑦 , 𝑛𝐴𝑦 (=)

𝑚3
𝑘𝑔 𝑜𝑓 𝐴𝑚2 𝑠𝑘𝑚𝑜𝑙 𝑜𝑓 𝐴

𝑥𝐴 (=) 𝑘𝑚𝑜𝑙 𝑜𝑓 𝑚𝑖𝑥𝑡𝑢𝑟𝑒
𝑘𝑔 𝑜𝑓 𝐴𝜔𝐴 (=) 𝑘𝑔 𝑜𝑓 𝑚𝑖𝑥𝑡𝑢𝑟𝑒(2.3-18)

𝑘𝑔 𝑜𝑓 𝐴𝜌𝐴 (=) 𝑘𝑔 𝑜𝑓 𝑚𝑖𝑥𝑡𝑢𝑟𝑒
𝐽𝐴𝑦(𝑀⁄𝐿2 𝑇)𝐷𝐴𝐵 (=) (𝑑𝐶 ⁄𝑑𝑦) (=) (𝑀⁄𝐿3 /𝐿) (=)
𝐴

𝐿2
𝑇(=) 𝑚2 ⁄𝑠

Diffusion is confusion? The phenomenon of diffusion is so difficult that we will understand its
fundamental aspect with simplified models in this course.

2.4

Similarity among Molecular Transports of Momentum,
Energy, and Mass (Prandtl number and Schmidt number)

For constant physical property fluid (𝜌, 𝐶𝑝 : constant), one-dimensional transports of momentum,
energy, and mass can be rewritten as𝑑(𝜚 𝑣 )
𝜏𝑦𝑥 = − 𝜈 𝑑𝑦 𝑥
(2.4-1)𝑞𝑦 = − 𝛼𝑑(𝜚𝐶𝑝 𝑇)

𝐽𝐴𝑦 = − 𝐷𝐴𝐵

𝑑𝑦
𝑑 𝐶𝐴
𝑑𝑦

(2.4-2)(2.4-3)All these equations state the following common rule:(Flux of a quantity of interest) = (Diffusivity)(Gradient of the quantity per unit
volume)The ratio of the momentum to the energy diffusivity is called “Prandtl number”: 𝑃𝑟 = 𝜈 ⁄𝛼 .
If this dimensionless parameter is 1.0, heat and momentum diffuse through the fluid at the samerate.
The ratio of the momentum to the mass diffusivity is called “Schmidt number”: 𝑆𝑐 = 𝜈/𝐷𝐴𝐵 .
If 𝑆𝑐 = 1.0, mass and momentum also diffuse at the same rate. The ratio of the Prandtl to the
Schmidt number, called “Lewis number”, is also sometimes used: 𝐿𝑒 = 𝑃𝑟⁄𝑆𝑐 = 𝐷𝐴𝐵 ⁄𝛼.

14Fundamental Laws of Momentum, Energy, and Mass Transfer[Problem 2.4-P1] Compare Prandtl number of liquid mercury at 373 K with Prandtl number oflow pressure hydrogen gas at 10 mmHg and 10 K. Discuss heat transfer in fluid at very low Prandtl
numbers. The hydrogen gas has very low viscosity whereas the mercury has very large thermal
conductivity.
[Problem 2.4-P2] Compare Schmidt number of very viscous polymer fluid with Schmidt
number of an electrolytic solution. The polymer solution (density 𝜌 = 1,200 𝑘𝑔⁄𝑚3 ) has viscosityof 10 Pa s and diffusivity of 10-4 m2/s whereas the electrolytic solution (density 𝜌 =1,100 𝑘𝑔⁄𝑚3 ) has viscosity of 10-3 Pa s and diffusivity of 10-9 m2/s. Discuss mass diffusion at very
high Schmidt numbers taking into account the difference in flow condition.

2.5

Non-Newtonian Fluids

As studied in Section 2.1, Newton’s law of viscosity states that the momentum flux (shear stress)
in a usual fluid flow is proportional to velocity gradient (rate of strain). However Newton’s law ofviscosity does not always predict the shear stress in all fluids. In general, the relation between shear
stress and the rate of shearing strain can be expressed by𝑑𝑣
𝜏𝑦𝑥 = − 𝜂 𝑑𝑦𝑥(2.5-1)In the case of usual Newtonian fluids, 𝜂 can be regarded as a proportional constant, i.e.
viscosity 𝜂 = 𝜇. As shown in Fig.2.5-1, however, there are various industrial materials
having 𝜂 as a function of either 𝑑𝑣𝑥 ⁄𝑑𝑦 or 𝜏𝑦𝑥 and they are referred to as
non-Newtonian fluids.
There are numerous empirical models proposed but the subject of non-Newtonian flow is beyond
the objective of this course.

Fig.2.5-1. Schematic picture of momentum-flux against velocity gradient for non-Newtonian
fluidsAs slightly simpler representative example, the following two-parameter models are available:(1) The Bingham model𝑑𝑣
𝜏𝑦𝑥 = − 𝜇0 𝑑𝑦𝑥 + 𝜏0
if 𝜏𝑦𝑥 > 𝜏0
(2.5-2)𝑑𝑣𝑥
𝑑𝑦=0

if 𝜏𝑦𝑥 < 𝜏0

(2.5-3)This model states that when the shear stress 𝜏𝑦𝑥 is smaller than the yield stress 𝜏0 , the fluid
behaves like a rigid body but it flows like a Newtonian fluid when 𝜏𝑦𝑥 > 𝜏0 .
(2) The Power law modelNomenclature

15

𝑑𝑣 𝑛−1 𝑑𝑣𝑥

𝜏𝑦𝑥 = −𝑚 | 𝑑𝑦𝑥 |𝑑𝑦(2.5-4)This equation reduces to Newton’s law of viscosity with 𝑚 = 𝜇 when 𝑛 = 1. The behavior of
fluid is pseudo-plastic when 𝑛 < 1 and dilatant when 𝑛 > 1.
The laminar flow of a Bingham plastic (ideal plastic) fluid in a circular pipe will be studied in thelater section.
Nomenclature𝐴
surface area of two parallel plates, [m2]𝐶𝐴
concentration of component A, [kmol/m3]
𝐶𝑝
heat capacity, [J/kg K]
𝑐total molar density, [kmol/m3]
𝐷𝐴𝐵
diffusivity, [m2/s]
𝐹shear force, [N]𝑗𝐴 , 𝐽𝐴
molar flux relative to mass-average and molar-average velocity, [kg/m2s] or [kmol/m2s]𝑛𝐴 , 𝑁𝐴
mass flux, molar flux in stationary coordinates, [kg/m2s] or [kmol/m2s]
𝑃𝑟
Prandtl number, [ - ]𝑄heat transfer rate, [W]
𝑞𝑟 , 𝑞𝜃 , 𝑞𝑧 heat flux in cylindrical coordinates
𝑞𝑥 , 𝑞𝑦 , 𝑞𝑧 heat flux in rectangular coordinates𝑟, 𝜃, 𝑧cylindrical coordinates, [m, - , m]𝑆𝑐
Schmidt number, [ - ]𝑡time, [s]𝑣𝑖mass velocity, [m/s]𝑣∗
molar velocity, [m/s]𝑣𝑟 , 𝑣𝜃 , 𝑣𝑧 velocity component in cylindrical coordinates
𝑣𝑥 , 𝑣𝑦 , 𝑣𝑧 velocity component in rectangular coordinates, [m/s]𝑥, 𝑦, 𝑧,rectangular coordinates, [m]𝑥𝑖mole fraction, [ - ]∆𝑇constant temperature difference given to the upper plate, [K]∆𝑉
constant velocity given to the upper plate, [m/s]∆𝑌distance between two parallel plates, [m]𝛼thermal diffusivity, [m2/s]
𝜅thermal conductivity, [W/m K]
μviscosity, [kg/m s]
𝜇𝑐
chemical potential, [J/mol]𝜈kinematic viscosity or momentum diffusivity, [m2/s]𝜌density, [kg/m3]
𝜏𝑥𝑥 , 𝜏𝑦𝑥 , 𝜏𝑧𝑥 , − −momentum flux or shear stress, [N/m2] or [kg/s2m]
𝜏0yield stress, [N/m2]
𝜔mass fraction, [ - ]Subcripts
A
component Awwall

CHAPTER 3
VISCOUS FLOW(LAMINAR AND TURBULENT FLOWS)
3.1 Laminar and Turbulent Flows in a Circular Pipe
When a fluid moves through a system, either one of two different types of fluid flow may occur.
As shown in Fig.3.1-1, suppose that water flows steadily through a transparent pipe, and athreadlike stream of dye is injected into it. If the velocity of water is small enough, the dye will flowin parallel, straight lines. When the velocity exceeds a certain critical value, it is noted that the dye
introduced is instantly mixed across the entire cross section of the pipe and the entire mass of water
becomes colored.
The first type of flow is called laminar; the fluid particles move along streamlines in a laminatedform. In laminar flow, momentum, energy, and mass are transferred in the transverse direction due
to molecular effect only.
The second type is called turbulent; the fluid particles move irregularly with violent eddy motion.
Therefore momentum, energy, and mass can be radially transferred much more rapidly for turbulent
flow than for laminar flow.

Fig.3.1.1. Characteristics of laminar and turbulent flow (Reynolds’ experiment)

In 1883, this systematic experiment was first conducted by Osborne Reynolds1), who suggested
the dimensionless parameter 𝑅𝑒 = 𝑉 𝐷 𝜌⁄𝜇 as the criterion for predicting the type of flow. Here
𝑉 is the average velocity in the pipe, 𝐷 the pipe diameter, 𝜌 the density, and 𝜇 the viscosity of
the fluid. This parameter called the Reynolds number, is a basic parameter in the study of fluid
motion. In pipe flow, the transition from laminar to turbulent flow may occur at about 𝑅𝑒 = 2,100
for commercial pipes.
For laminar pipe flow the velocity profile is parabolic. The theoretical curve is expressible as𝑣𝑧 = 𝑣𝑚𝑎𝑥 [1 − (𝑟⁄𝑅)2 ](3.1-1)This analytical approach will be studied in Chapter 6.

Generation of Turbulent Motion

17[Problem 3.1-P1] A viscous liquid (𝜇 = 0.05 kg/m s, 𝜌 = 1050 kg/m3) is flowing in a 50 mm
ID circular tube. The distribution of axial velocity is described as𝑣𝑧 = 𝑣𝑧𝑚𝑎𝑥 [1 − (𝑟⁄𝑅)2 ] where the maximum velocity 𝑣𝑧𝑚𝑎𝑥 = 0.5 m/s and the pipe radius
𝑅 = 0.025 m. Calculate the friction force acting over 5 m of the inside surface of the tube, the flow
rate, and the Reynolds number.
1.

Reynolds, O., Trans. Roy. Soc. (London), 174A, 935 (1883)

3.2 Generation of Turbulent Motion(Transition to Turbulent Flow)
As shown in Fig.3.2-1, when the fluid flow rate is increased, the parabolic velocity profile forlaminar flow has an increase in maximum velocity accompanied with an increase in velocity
gradient near the pipe wall.
Let us consider a fluid particle moving along the streamlines near the pipe wall.
As shown in Fig. 3.2-2, the fluid velocity near the pipe wall becomes larger on the upper side thanon the lower side of the fluid particle. The friction force for acceleration is exerted on the upper side
whereas the friction force for deceleration is exerted on the lower side. As a result, these forces
cause angular acceleration for the rotating motion.

Fig.3.2-1. Instability of laminar flow in a circular pipe

Fig.3.2-2. Generation of turbulent motion in a viscous parallel flow

Therefore, when the flow rate or the velocity gradient near the wall exceeds some critical value,
the fluid particle undergoes a strong rotating angular acceleration due to the steep velocity gradient,so that eddy motion is generated by rotation of the fluid particle. This rotating condition is realized
when the velocity difference 𝑣1 − 𝑣2 between two vertical positions becomes large. This is aqualitative understanding of an origin of turbulence or a transition to turbulent flow.

18

Viscous Flow (Laminar and Turbulent Flows)

In turbulent flow, the velocity at a fixed position in the fluid fluctuates randomly about its
time-averaged value. This unsteady fluctuating motion comes from such a complicated eddy motion,
which gives a very large effect of mixing due to the transverse-directed interchange of fluid
particles.

Fig.3.2-3. Velocity fluctuations in turbulent flow

When the flow rate goes beyond a certain critical value, the laminar shear flow near the wall
becomes unstable due to the rotational moment causing eddy motion, and then the transition toturbulent flow occurs.

Fig.3.2-4. Characteristics of turbulent flow in a circular pipe

For turbulent pipe flow, therefore, the distribution of time-averaged velocity is much flatter in themain flow region and very steep near the pipe wall.
A semi-empirical equation of the time-averaged velocity distribution is given by(3.2-1)𝑣̅𝑧 = 𝑣𝑚𝑎𝑥 [1 − (𝑟⁄𝑅)]1⁄7thThis expression known as the “1/7 power law gives a good approximation for the Reynolds
number range 104 to 105 . It should be kept in mind that this law cannot describe precisely the
turbulent structure of the flow field.
The Reynolds number is a product of a characteristic velocity, a characteristic length, and thedensity of the fluid divided by its viscosity. The result is a dimensionless number which represents
the ratio of the inertial to the viscous forces in the fluid. The critical Reynolds number depends, ingeneral, upon the surface condition of the inside wall of the pipe.
𝑅𝑒 =

𝑉𝐷𝜌
𝜇

[− ] =

inertial force
viscous force(3.2-2)[Problem 3.2-P1] Obtain a relation between the mean (flow-area-average) velocity 𝑣𝑧 and the
maximum velocity 𝑣𝑧𝑚𝑎𝑥 for turbulent flow inside a circular pipe using the 1/7th power law.
[Problem 3.2-P2] An aqueous solution of sodium hydroxide is flowing at a volumetric flow rateof 50 m3/h in a circular pipe of inside diameter of 100 mm. The density and viscosity of the solution

Generation of Turbulent Motion (continued), Problems

19

at 50 C (= 323 K)b are 1,510 kg/m3 and 0.025 kg/m s, respectively. Calculate the Reynolds number..
Is the flow turbulent?
[Problem 3.2-P3] A viscous liquid (𝜇 = 50 cp. = 0.05 kg/m s, 𝜌 = 1,050 kg/m3) is flowing in a
50 mm ID circular tube. The velocity distribution is described by the following equation:𝑣𝑧 = 𝑣𝑧𝑚𝑎𝑥 [1 − (𝑟⁄𝑅)2 ]where 𝑣𝑧 is the axial velocity in m/s and 𝑟 the radial coordinate in m. The maximum velocity
𝑣𝑧𝑚𝑎𝑥 = 0.5 m/s and the pipe radius 𝑅 = 0.025 m. Calculate the friction force acting over 5 m ofthe inside surface of the tube, the flow rate, and the Reynolds number.
Answer: 25.12 N, 4.91 m3/s, and Re = 262.5.
[Problem 3.2-P4] In the above problem, the velocity gradient on the inner wall of the pipe isgiven for laminar flow in a term of average velocity by
4 𝑣𝑎𝑣
𝑑𝑣𝑧
|
=−
𝑅
𝑑𝑟 𝑟=𝑅
If the critical Reynolds number 𝑅𝑒𝑐𝑟 is given as 2,100, what is the critical velocity gradient?
Answer: - 320 (1/s)
Nomenclature𝐷𝑟, 𝜃, 𝑧
𝑅
𝑉
𝑣𝑧
𝑦
𝜏𝑞
𝑅𝑒

characteristic length or pipe diameter, [m]cylindrical coordinates, [m, - , m]pipe radius, [m]characteristic velocity, [m/s]axial velocity component in pipe flow, [m/s]
distance from pipe wall
torque, [Nm]Reynolds number, [ - ]CHAPTER 4
MACROSCOPIC BALANCES:
CONTROL VOLUME APPROACH
4.1 Principles of Momentum, Energy, and Mass Conservation
In this course, the laws of conservation of momentum, conservation of energy, and conservationof mass may be stated for a general control volume as follows:
𝑠𝑢𝑚 𝑜𝑓 𝑒𝑥𝑦𝑡𝑒𝑟𝑛𝑎𝑙𝑟𝑎𝑡𝑒 𝑜𝑓𝑟𝑎𝑡𝑒 𝑜𝑓𝑟𝑎𝑡𝑒 𝑜𝑓( 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 ) = (𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚) − (𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 ) + ( 𝑓𝑜𝑟𝑐𝑒𝑠 𝑎𝑐𝑡𝑖𝑛𝑔 )𝑜𝑛 𝑠𝑦𝑠𝑡𝑒𝑚
𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛
𝐼𝑁
𝑂𝑈𝑇
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑤𝑜𝑟𝑘
𝑟𝑎𝑡𝑒 𝑜𝑓𝑟𝑎𝑡𝑒 𝑜𝑓𝑟𝑎𝑡𝑒 𝑜𝑓( 𝑒𝑛𝑒𝑟𝑔𝑦 ) = (𝑒𝑛𝑒𝑟𝑔𝑦 ) − (𝑒𝑛𝑒𝑟𝑔𝑦 ) + ( 𝑑𝑜𝑛𝑒 𝑏𝑦 𝑡ℎ𝑒 )𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠
𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛
𝐼𝑁
𝑂𝑈𝑇
𝑟𝑎𝑡𝑒 𝑜𝑓𝑟𝑎𝑡𝑒 𝑜𝑓𝑟𝑎𝑡𝑒 𝑜𝑓() = ( 𝑚𝑎𝑠𝑠 ) − ( 𝑚𝑎𝑠𝑠 )𝑚𝑎𝑠𝑠𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛
𝐼𝑁
𝑂𝑈𝑇
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛
𝑟𝑎𝑡𝑒 𝑜𝑓𝑟𝑎𝑡𝑒 𝑜𝑓𝑟𝑎𝑡𝑒 𝑜𝑓𝑜𝑓 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐴 𝑏𝑦
( 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐴 ) = (𝑚𝑎𝑠𝑠 𝑜𝑓 𝐴) − (𝑚𝑎𝑠𝑠 𝑜𝑓 𝐴) + ()ℎ𝑜𝑚𝑜𝑔𝑒𝑛𝑒𝑜𝑢𝑠
𝑂𝑈𝑇
𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛𝐼𝑁𝑐ℎ𝑒𝑚𝑖𝑐𝑎𝑙 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛𝑠(4.1-1)(4.1-2)(4.1-3)(4.1-4)

Fig. 4.1-1. Shell balances of momentum, energy, and mass of component A over a control
volume4.2 Macroscopic Mass Balance
The mass balance should be set up over a control volume shown in Fig.4.2-1.
The mass present at 𝑡 in some arbitrary small volume within the control volume is 𝜌Δ𝑉.
Therefore the mass of the control volume at 𝑡 is𝑚 = ∫𝑉 𝜚 𝑑𝑉
where density 𝜌 = 𝜌(𝑥, 𝑦, 𝑧, 𝑡).

(4.2-1)

Macroscopic Mass Balance (continued)

21

Accumulation of mass in time Δ𝑡 is the difference between the mass in the control volume at 𝑡
and that at 𝑡 + Δ𝑡:∫𝑉 𝜚 𝑑𝑉 |𝑡+Δ𝑡𝑑− ∫𝑉 𝜚 𝑑𝑉 | = 𝑑𝑡 ∫𝑉 𝜚 𝑑𝑉 ∙ Δ𝑡𝑡

(4.2-2)

Fig.4.2-1. Macroscopic mass balance

The mass can be transferred by velocity component normal to the surface, so the velocity
component 𝑣𝑛 into the control volume can be expressed as⃗ ∙𝒏⃗⃗ ) = −𝑣 cos 𝛼
𝑣𝑛 = − (𝒗(4.2-3)⃗⃗ is the outward directed unit vector normal to the surface at that point and 𝛼 is the angle
where 𝒏⃗ and 𝒏⃗⃗ .
between 𝒗
The rate of mass transferred = (mass)/(unit time) = (mass/unit volume)(length/unit time)(area)
This is the net rate of mass in over the whole surface of the control volume = ∫𝑆 𝜌 𝑣𝑛 𝑑𝑆
Then application of the principle of mass conservation yields𝑑⃗ ∙𝒏⃗ ) 𝑑𝑆(4.2-4)∫ 𝜌 𝑑𝑉 = − ∫𝑆 𝜌 (𝒗𝑑𝑡 𝑉This is the macroscopic mass balance.
For a simpler flow system shown below= 𝜌1 〈𝒗𝟏 〉𝑆1 − 𝜌2 〈𝒗𝟐 〉𝑆2
𝑑𝑡
Introducing the symbol of mass flow rate
⃗ ∙𝒏⃗⃗ ) 𝑑𝑆𝑤 = 𝜌 〈𝑣 〉𝑆 = − ∫𝑆 𝜌 (𝒗to get the unsteady-state macroscopic mass balance:𝑑𝑚

Fig.4.2-2. Simpler pipe system with a single inlet and a single exit

(4.2-5)(4.2-6)

22

Macroscopic Balances: Control Volume Approach𝑑𝑚= − ∆𝑤
For steady state∆𝑤 = 0 or 𝑤2 = 𝑤1
𝑑𝑡

(4.2-7)(4.2-8)[Problem 4.2-P1] A pipe system shown is carrying water through section 1 at an averagevelocity of 0.3 m/s in steady state. The diameter at section l is 0.3 m. The same flow passes throughsection 2 where the diameter is 0.1 m. Find the average velocity at section 2.

Fig.4.2-P1. Mass balance in a horizontal pipe accompanied with contraction of cross section

4.3 Macroscopic Momentum Balance

Fig.4.3-1 Control volume for macroscopic momentum balance

Consider a small surface element of area ∆𝑆 of a control volume, through which momentum𝜌 ⃗𝒗 is entering or leaving. The rate of momentum entering or leaving through ∆𝑆 may be written
as⃗ )[− (𝒗⃗ ∙ ⃗𝒏)𝑑𝑆](𝜌 ⃗𝒗) 𝑣𝑛 𝑑𝑆 = (𝜌 𝒗
(4.3-1)⃗⃗ is outward directed. Therefore the net rate
Here the minus sign is necessary since the unit vector 𝒏of momentum in over the whole surface of the control volume is⃗ ∙ ⃗𝒏⃗ )𝑑𝑆− ∫𝑆 𝜌 ⃗𝒗 (𝒗(4.3-2)⃗ ∆𝑉 , the rate of momentum
Since the momentum in a small volume element ∆𝑉 is 𝜌 𝒗
accumulation in the control volume is𝑑𝑑𝑡(4.3-3)⃗ 𝑑𝑉∫𝑉 𝜌 𝒗⃗ .
The sum of external forces acting on the system is ∑ 𝑭
Applying the principle of momentum conservation,𝑑⃗ 𝑑𝑉 = − ∫ 𝜌 ⃗𝒗 (⃗𝒗 ∙ ⃗𝒏⃗ )𝑑𝑆 + ∑ ⃗𝑭
∫ 𝜌𝒗

(4.3-4)This is the macroscopic momentum balance corresponding to Newton’s second law of motion.
𝑑𝑡

𝑉

𝑆

Macroscopic Momentum Balance (continued)

23⃗ 𝑑𝑆, and
Note that the force term ∑ ⃗𝑭 comprises viscous forces ∑ ⃗⃗⃗⃗
𝑭𝒇 , pressure force − ∫𝑆 𝑝 𝒏⃗ acting on the fluid.
gravitational force 𝑚 ⃗𝒈
For steady state the left side of the above equation is zero⃗ (𝒗⃗ ∙ ⃗𝒏⃗ )𝑑𝑆 + ∑ ⃗𝑭 = 0− ∫𝑆 𝜌 𝒗(4.3-5)For turbulent pipe flow the following approximation may be employed for cross-sectional
averaging.
〈𝑣 〉 ≅ 〈𝑣 2〉/〈𝑣 〉(4.3-6)
For turbulent pipe flow, therefore, the force acting on the pipe wall is given by⃗⃗⃗⃗
𝑭′ = −⃗⃗⃗⃗⃗
𝑭𝒇 = 𝑤1 〈⃗⃗⃗⃗⃗⃗⃗⃗
𝒗𝟏 〉 − 𝑤2 〈⃗⃗⃗⃗⃗⃗⃗⃗
𝒗𝟐 〉 − 𝑝1 𝑆1 ⃗⃗⃗⃗
𝒏𝟏 − 𝑝2 𝑆2 ⃗⃗⃗⃗
𝒏𝟐 + 𝑚 ⃗𝒈(4.3-7)Note that the unit vectors are outward directed.
For an incompressible flow in a straight pipe (inside diameter 𝐷, length 𝐿) inclined by 𝜃, thefriction force acting on the inside surface of the pipe is given by the following familiar equation offorce balance:𝜋𝜋𝐹 ′ = (𝑝0 − 𝑝𝐿 ) 4 𝐷 2 + 4 𝐷 2 𝐿 𝜌𝑔 cos 𝜃 = 𝜏𝑤 𝜋𝐷𝐿(4.3-8)This result indicates that the net force acting downstream on the cylindrical fluid column by virtueof the pressure difference and gravitational acceleration is balanced by the friction force the pipe
wall exerts. The above equation is valid whether the flow is laminar or turbulent.
[Example 4.3-E1] Consider the problem of finding the force exerted on a horizontal 90 degreepipe bend. Water at 20℃ is flowing at a rate of 700 kg/s for cooling in an oil refinery. The pipeline
is 300 mm ID steel pipe. A diagram of the pipe bend and the quantities relating to the analysis areshown in Figure 4.3-E1.
[Solution] Actually there will be a small pressure drop caused by viscous force acting on thepiped bend wall. At this point, however, we assume negligibly small pressure drop between sections“1” and “2”. The first step is to choose as the control volume the region bounded by two planes “1”and “2” normal to the streamliners and the inner wall of the bend. Assuming turbulent flow (almost
flat velocity distribution), the overall momentum balance may be written, term by term, as

Fig.4.3-E1 Force acting on 90 degree bend⃗ (𝒗⃗ ∙ ⃗𝒏⃗ )𝑑𝑆 = − ∫𝑆 𝜌𝒗⃗ (𝒗⃗ ∙ ⃗𝒏⃗ )𝑑𝑆 − ∫𝑆 𝜌𝒗⃗ (𝒗⃗ ∙ ⃗𝒏⃗ )𝑑𝑆 = 𝑤1 𝒗⃗⃗⃗⃗𝟏 − 𝑤2 𝒗⃗⃗⃗⃗𝟐− ∫𝑆 𝜌𝒗(4.3-E1)
1
2⃗ and 𝒏⃗ ) is 180 degree at the inlet “1”There is a sign difference because 𝛼 (the angle between 𝒗
and 0 degree at the outlet 2”.
For the macroscopic mass balance
𝑤1 = 𝑤2 = 𝑤 = 𝑐𝑜𝑛𝑠𝑡

24

Macroscopic Balances: Control Volume Approach

And volumetric flow rate
𝑄1 = 𝑄2 = 𝑄 = 𝑐𝑜𝑛𝑠𝑡
Therefore for steady state, Eq.(4.3-4) becomes⃗𝑤 (⃗⃗⃗⃗
𝒗𝟐 − ⃗⃗⃗⃗𝒗𝟏 ) = ∑ 𝑭(4.3-E2)⃗⃗⃗⃗2 /𝑆
where ⃗⃗⃗⃗
𝒗𝟏 = 𝑄(− ⃗⃗⃗⃗
𝒏𝟏 )/𝑆 and ⃗⃗⃗⃗
𝒗𝟐 = 𝑄 𝒏
Then
𝑄∑ ⃗𝑭 = ( ) (⃗⃗⃗⃗⃗⃗⃗⃗𝟏 )𝑤
𝒏𝟐 + 𝒏(4.3-E3)
𝑆
Since the pipeline is horizontal, we do not have to consider the gravitational effect. The total force⃗⃗⃗⃗𝒃 .
on the fluid is made up of the pressure force ⃗⃗⃗⃗
𝑭𝒑 and the force exerted by the bend𝑭⃗⃗⃗⃗⃗⃗⃗⃗𝟏 + 𝒏⃗⃗⃗⃗𝟐 )
𝑭𝒑 = −𝑝 𝑆 (𝒏(4.3-E4)
Here we used the assumption of equal pressure at sections “1” and “2”. Recall that the force soughtis the reaction to 𝑭𝒃 and has components equal in magnitude and opposite in sense to 𝑭𝒃 .
𝑄
𝑤𝑄
⃗⃗⃗⃗
⃗⃗⃗⃗𝟏 + 𝒏⃗⃗⃗⃗𝟐 ) + 𝑝 𝑆 (⃗⃗⃗⃗⃗⃗⃗⃗𝟐 ) = ( + 𝑝 𝑆) (𝒏⃗⃗⃗⃗𝟏 + 𝒏⃗⃗⃗⃗𝟐 )
𝑭𝒃 = ( 𝑆 ) 𝑤(𝒏𝒏𝟏 + 𝒏(4.3-E5)𝑆
It remains to insert the numerical values:⃗⃗⃗⃗
𝑭𝒃 = [

(700
𝑘𝑔

𝑘𝑔 2
)
𝑠

105 𝑁

⃗⃗⃗⃗𝟏 + 𝒏⃗⃗⃗⃗𝟐 ) = 1.4 × 104 (𝒏⃗⃗⃗⃗𝟏 + ⃗⃗⃗⃗+ (1 × 𝑚2 )(0.0707 𝑚2 )] × (𝒏𝒏𝟐 ) N
2(1000 3)(0.0707 𝑚 )𝑚

The force acting on the bend ⃗⃗⃗⃗⃗⃗
𝑭′𝒃 = − ⃗⃗⃗⃗𝑭𝒃 .

[Problem 4.3-P1] Water 20 ℃ is flowing at a rate of 𝑄 = 5 × 10−2 𝑚3 ⁄𝑠 through ahorizontal convergent pipe shown below. The upstream pressure is 𝑝1 = 2 × 105 Pa. Calculate
the pressure 𝑝2 at the end of the converging section and the force acting on the convergent section.
Assume the friction loss negligibly small in the converging section.

Fig.4.3-P1 Fluid force acting on the wall of a horizontal convergent pipe

4.4 Macroscopic Energy Balance
In this section, we shall deduce a macroscopic balance of energy in general form from the
beginning to get the first law of thermodynamics. This deductive part may be skipped in case oftime limitation.

Macroscopic Energy Balance (continued)

25

Fig.4.4-1 Control volume for macroscopic energy balance

Energy associated with mass can be classified: potential energy 𝐸𝑝 , kinetic energy 𝐸𝑘 , andinternal energy 𝐸𝑢 .
Strictly speaking, the rate of the total energy 𝐸𝑇 = 𝐸𝑝 + 𝐸𝑘 + 𝐸𝑢 accumulation in the control
volume is given by𝑑(4.4-1)∫𝑉 𝜌 𝐸𝑇 𝑑𝑉
The net rate of energy in over the whole surface of the control volume is⃗ ∙ ⃗𝒏) 𝑑𝑆(4.4-2)∫𝑆 𝐸𝑇 𝜌 𝑣𝑛 𝑑𝑆 = − ∫𝑆 𝐸𝑇 𝜌(𝒗To get the mass of fluid inside the control volume, some of the fluid should be compressed into thecontrol volume. To do work on the surroundings, the boundary of the control volume should be
expanded. This is called “flow work”. The flow work required to add a unit mass to the control
volume can be written as (pressure) (specific volume) 𝑝 𝑉𝑠 like piston work. Therefore the flow
work is⃗ ∙𝒏⃗⃗ ) 𝑑𝑆
𝑊𝑓 = − ∫𝑆 𝑝 𝑉𝑠 𝜌(𝒗(4.4-3)
According to thermodynamics, enthalpy is expressed as 𝐻𝑒 = 𝐸𝑢 + 𝑝 𝑉𝑠 . Hence the net rate ofenergy transferred across the boundary with mass becomes⃗ ) 𝑑𝑆 + ∫𝑆 𝑝 𝑉𝑠 𝜌(𝒗⃗ ∙ ⃗𝒏⃗ ) 𝑑𝑆] = − ∫𝑆 (𝐻𝑒 + 𝐸𝑘 + 𝐸𝑝 )𝜌(𝒗⃗ ∙ ⃗𝒏⃗ ) 𝑑𝑆− [∫𝑆 𝐸𝑇 𝜌(⃗𝒗 ∙ 𝒏(4.4-4)𝑑𝑡

The thermal energy not associated with mass crossing the boundary per unit volume is 𝑄 (thermalenergy into the system is regarded positive). Finally applying the principle of energy conservation
we get𝑑⃗ ∙𝒏⃗ ) 𝑑𝑆 + 𝑄 − 𝑊(4.4-4)∫ 𝜌 𝐸𝑇 𝑑𝑉 = − ∫𝑆 (𝐻𝑒 + 𝐸𝑘 + 𝐸𝑝 )𝜌(𝒗𝑑𝑡 𝑉This is a general form of macroscopic energy balance.

<Simplification of the macroscopic energy balance>

For a flow system having a single fluid entrance (at plane “1” with cross-sectional area 𝑆1 ) and asingle exit (at plane “2” with cross-sectional area 𝑆2 ), the macroscopic energy balance reduces to𝑑⃗⃗⃗⃗𝟏 ∙ ⃗⃗⃗⃗
𝒏𝟏 )𝑑𝑆1∫𝑉 𝜌 𝐸𝑇 𝑑𝑉 = − ∫𝑆 (𝐻𝑒1 + 𝐸𝑘1 + 𝐸𝑝1 )𝜌1 (𝒗𝑑𝑡

1⃗⃗⃗⃗𝟐 ∙ ⃗⃗⃗⃗− ∫ (𝐻𝑒2 + 𝐸𝑘2 + 𝐸𝑝2 )𝜌2 (𝒗𝒏𝟐 )𝑑𝑆2 + 𝑄 − 𝑊
𝑆2= (𝐻𝑒1 + 𝐸𝑘1 + 𝐸𝑝1 )𝜌1 〈𝑣1 〉𝑆1 − (𝐻𝑒2 + 𝐸𝑘2 + 𝐸𝑝2 )𝜌2 〈𝑣2 〉𝑆2 + 𝑄 − 𝑊= − ∆[(𝐻𝑒 + 𝐸𝑘 + 𝐸𝑝 )𝑤] + Q – W(4.4-5)Here 〈 〉 means cross-sectional area-averaged value and 𝑤 = 𝜌〈𝑣 〉𝑆 is mass flow rate.
Strictly speaking, the above simplification is valid only when the density and other quantities

26

Macroscopic Balances: Control Volume Approach

𝐻𝑒 , 𝐸𝑘 , 𝐸𝑝 , 𝐸𝑇 do not vary very much across the cross section at planes “1” and “2”. Especially forthe kinetic energy term, the flow should be turbulent, which has nearly flat velocity profiles.
Usually the kinetic energy and potential energy in the constant gravitational field can be written as(1⁄2) 𝜌𝑣 2 + 𝜌 𝑔 ℎ.
Finally the above equation becomes𝑑𝐸𝑇𝑜𝑡
1= − ∆ [(𝐻𝑒 + 2 〈𝑣 〉2 + 𝑔 ℎ) 𝑤] + 𝑄 − 𝑊(4.4-6)𝑑𝑡
Here 𝐸𝑇𝑜𝑡 = ∫𝑉 𝜌 𝐸𝑇 𝑑𝑉 = ∫𝑉 𝜌(𝐸𝑢 + 𝐸𝑘 + 𝐸𝑝 )𝑑𝑉
This is a form of the first law of thermodynamics applied to a flow system, which may be regarded
as the starting equation for dealing with energy-related chemical engineering applications.
Dividing through by constant mass flow rate 𝑤 at steady state:1− ∆ (𝐻𝑒 + 2 〈𝑣 〉2 + 𝑔 ℎ) + 𝑄𝑚 − 𝑊𝑚 = 0(4.4-7)
Here 𝑄𝑚 is the heat added per unit mass of the fluid into the system and 𝑊𝑚 is the amount ofwork done per unit mass of the fluid on the surroundings.
This equation can be further simplified into a convenient form.
For ideal gases ( 𝑝 𝑉 = (1⁄𝑀) 𝑅 𝑇 )𝑇2
1(4.4-8)∫𝑇1 𝐶𝑝 𝑑𝑇 + ∆ 2 〈𝑣 〉2 + 𝑔 ∆ℎ = 𝑄𝑚 − 𝑊𝑚
For incompressible fluids (𝜌 = 𝑐𝑜𝑛𝑠𝑡, 𝐶𝑣 = 𝐶𝑝 )The enthalpy change is given as∆𝐻𝑒 = ∆𝐸𝑢 + ∆(𝑝 𝑉𝑠 ) = ∆𝐸𝑢 + 𝑉𝑠 ∆𝑝 = ∆𝐸𝑢 + (1⁄𝜌) ∆𝑝(4.4-9)
Then
𝑇2
𝑝 −𝑝
1
(4.4-10)∫𝑇 𝐶𝑝 𝑑𝑇 + 2 1 + ∆ 〈𝑣 〉2 + 𝑔 ∆ℎ = 𝑄𝑚 − 𝑊𝑚𝜌1

2

The above equation is usually applied for a flow system accompanied with heat input or output.

4.5 Mechanical Energy Balance
The equation of mechanical energy balance called “Bernoulli equation” is very important anduseful for practical applications to various piping designs.
For a flow system without heat exchange (𝑄𝑚 = 0)
∆𝐻𝑒 = ∆𝐸ℎ + ∆(𝑝𝑉𝑠 ) = 𝑇 ∆𝑆 + 𝑉𝑠 ∆𝑝(4.5-1)The presence of friction renders the process irreversible owing to the frictional heating. That is,
𝑇 ∆𝑆 = 𝑄𝑚 + 𝐹𝑟𝑚 Therefore ∆𝐻𝑒 = 𝑄𝑚 + 𝐹𝑟𝑚 + 𝑉𝑠 ∆𝑝.
(4.5-2)where 𝐹𝑟𝑚indicates the thermal energy generation transformed from mechanical energy.
Substituting this into the steady-state energy balance equation𝑝2
1(4.5-3)∫𝑝1 𝑉𝑠 𝑑𝑝 + ∆ 2 〈𝑣 〉2 + 𝑔 ∆ℎ = − 𝐹𝑟𝑚 − 𝑊𝑚
This is a generalized form of “Bernoulli equation” which is applicable to both ideal gases andincompressible fluids. Here 𝐹𝑟𝑚 is the rate of mechanical energy loss per unit mass due to thefrictional effect. This indicates that a part of mechanical energy is irreversibly converted to thermalenergy due to viscous dissipation. The mechanical energy loss 𝐹𝑟𝑚 is often called “friction loss”.
For steady-state isothermal case with no additional heat input, the rate of thermal energy producedby viscous dissipation becomes equal to the rate of heat removed from the control volume.
For ideal gases (𝑉𝑠 = (𝑅 𝑇⁄𝑀 )(1⁄𝑝)𝑅𝑇𝑝1
ln 𝑝2 + ∆ 2 〈𝑣 〉2 + 𝑔 ∆ℎ = − 𝐹𝑟𝑚 − 𝑊𝑚
(4.5-4)𝑀
1

For incompressible fluids (𝑉𝑠 = 1⁄𝜌 = 𝑐𝑜𝑛𝑠𝑡)
𝑝2 − 𝑝1
1+ ∆ 2 〈𝑣 〉2 + 𝑔 ∆ℎ = − 𝐹𝑟𝑚 − 𝑊𝑚𝜌

(4.5-5)

Problems, Example

27

For laminar flow the kinetic energy term ∆(1⁄2)〈𝑣〉2 should be replaced by ∆〈𝑣 〉2 because theparabolic velocity distribution gives the relation: 〈𝑣 3 〉⁄〈𝑣 〉 = 2 〈𝑣 〉2 .
[PROBLEM 4.5-P1] Ideal gas (molecular weight 𝑀 = 29) is in isothermal turbulent flow attemperature 𝑇 = 298 K (= 25℃) through a straight horizontal pipe with cross section 𝑆 = 3.14 ×
10−2 𝑚2 . The absolute pressures at reference planes “1” and “2” are 𝑝1 = 1.013 × 105 𝑃𝑎 (=1 𝑎𝑡𝑚) and 𝑝2 = 0.87 × 105 𝑃𝑎, respectively. The mass flow rate measured is 𝑤 = 0.15 𝑘𝑔⁄𝑠.
Evaluate 𝐹𝑟21 = 𝑤 𝐹𝑟𝑚21 in 𝑘𝑔 𝑚2 /𝑠 3 from the length 𝐿 = 20 𝑚 of pipe between “1” and “2”.
The gas constant is given by 𝑅 = 8.314 × 103 (𝑘𝑔 𝑚2 ⁄𝑠 2 )/𝑘𝑚𝑜𝑙 𝐾 .
[PROBLEM 4.5-P2] Water flows through a turbine at an average velocity of 10 m/s at theentrance of flow cross section 0.1 m2. The static pressures are 1 MPa at the entrance and 2 MPa at
the outlet of flow cross section 0.3 m2. The elevation difference between the entrance and outlet is
10 m. Calculate the theoretical power assuming negligibly small viscous dissipation.

Fig.4.4-P2. Calculation of power obtained by water turbine[EXAMPLE 4.5-E1] A fireboat has a circular convergent nozzle shown below. A jet of water( 𝜌 = 1,000 𝑘𝑔/𝑚3 ) issues from the nozzle at an average velocity of 100 𝑚/𝑠 . Assuming
negligible friction loss and gravitational effect, determine the static pressure 𝑝1 at plane “1” andthe horizontal component of the force required to keep the nozzle stationary. This nozzle is placedon an angle of 30 degree with the horizontal. The nozzle exit diameter is 80 mm and the diameter ofthe upstream pipe is 150 mm.

Fig.4.5-E2. Circular convergent nozzle

Solution: Assuming turbulent flow with incompressible fluid, the mechanical energy balance canbe applied to this system as
11𝑝 −𝑝〈𝑣2 〉2 − 〈𝑣1 〉2 + 2 1 = 0(4.5-E1)2
2𝜌where 𝑔 ∆ℎ = 0, 𝐹𝑟𝑚21 = 0 and 𝑊𝑚 = 0 are assumed.
Then for steady-state flow of incompressible fluid, the upstream pressure is given by28

Macroscopic Balances: Control Volume Approach1

𝑝1 = 2 𝜌 (〈𝑣2 〉2 − 〈𝑣1 〉2 ) + 𝑝2
𝑤 = 𝜌〈𝑣2 〉𝑆2 = 𝜌〈𝑣1 〉𝑆1 = 𝑐𝑜𝑛𝑠𝑡
At the nozzle exit the pressure 𝑝2 becomes atmospheric pressure.
Therefore〈𝑣1 〉 = (𝑆2 ⁄𝑆1 )〈𝑣2 〉 = (80⁄150)2 (100 𝑚⁄𝑠) = 28.4 𝑚/𝑠
Then
𝑝1 = (1⁄2)(1000 𝑘𝑔⁄𝑚3 )[(100 𝑚⁄𝑠 )2 − (28.4 𝑚⁄𝑠 )2 ] + 1.013 × 105 𝑃𝑎= 4.70 × 106 𝑃𝑎 = 4.70 𝑀𝑃𝑎
At the nozzle exit the pressure becomes atmospheric pressure.
The mass flow rate is calculated as
𝑤 = 𝜌〈𝑣 〉𝑆 = (1000 𝑘𝑔⁄𝑚3 )(100 𝑚⁄𝑠)(𝜋⁄4)(0.08 𝑚)2 = 502.7 𝑘𝑔⁄𝑠
Applying the momentum balance to this system with 𝑚𝑡𝑜𝑡 𝑔 = 0 assumed,⃗ = −(⃗⃗⃗⃗⃗⃗⃗⃗𝟏 𝑤) − (𝑝2 𝑆2 𝒏𝟐 + 𝑝1 𝑆1 𝒏𝟏 )𝑭
𝒗𝟐 𝑤 − 𝒗
The force acting on the nozzle in the direction of nozzle axis is given by
𝐹 = −(502.7 𝑘𝑔⁄𝑠)(100 𝑚⁄𝑠 − 28.4 𝑚⁄𝑠) − (1.013 × 105 𝑘𝑔⁄𝑚 𝑠 2 )× (𝜋⁄4)(0.08 𝑚)2 + (4.70 × 106 𝑘𝑔⁄𝑚 𝑠 2 )(𝜋⁄4)(0.15 𝑚)2= 4.66 × 104 𝑘𝑔 𝑚⁄𝑠 2 = 4.66 × 104 𝑁
The horizontal force required to keep the nozzle stationary is
𝐹𝑥 = 𝐹 cos 30° = (4.66 × 104 ) √3⁄2 = 4.04 × 104 𝑁(4.5-E2)(4.5-E3)(4.5-E4)(4.5-E5)(4.5-E6)(4.5-E7)(4.5-E8)(4.5-E9)[EXAMPLE 4.5-E2] To measure the friction loss of a glove valve, we use a horizontal pipeline
with a constant cross section shown. We use water (density 𝜌𝑤 ) as the flowing fluid and mercury(density 𝜌𝑚) as the manometer fluid. Determine the friction loss when the manometer reading is ℎ.
Assume that the friction losses in the sections 𝑙1 and 𝑙2 are small compared with that of the valve.
Solution: Use the macroscopic mechanical energy balance for incompressible, isothermal flow
between the pressure taps
𝑝2 − 𝑝1
1+ ∆ 2 〈𝑣 〉2 + 𝑔 ∆ℎ = − 𝐹𝑟𝑚 − 𝑊𝑚
(4.5-E10)𝜌𝑤

1

Since 〈𝑣1 〉 = 〈𝑣2 〉 , ∆ 2 〈𝑣 〉2 = 0 for a pipe with constant cross section.
∆ℎ = 0 for a horizontal pipeline. There is no power equipment in the system

𝑊𝑚 = 0

Fig.4.5-E2. Determination of friction loss for a globe valve

From the force balance of the stationary fluids in the manometer𝑝1 − 𝑝2 = (𝜌𝑚 − 𝜌𝑤 )𝑔 ℎ
Then we get the friction loss𝜌 −𝜌
𝐹𝑟𝑚 = 𝑚𝜌 𝑤 𝑔 ℎ
𝑤(4.5-E11)(4.5-E12)

Problems, Examples

29[EXAMPLE 4.5-E3] Pitot tube for velocity measurement
One of the most common, traditional devices for measuring velocity distributions in a flowing
stream is the pitot tube. As shown in Fig.4.5-E3, it consists of two concentric tubes, the inner one ofwhich is open at the end and the outer one is sealed at the same end of the inner tube. The side wallof the outer tube near the front tip has several small drilled holes.
Both tubes are filled with the flowing fluid and connected to the opposite ends of a manometer.
The pitot tube is placed to confront the opening of the inner tube (impact-tube opening) withthe velocity to be measured. As the fluid impinges on the impact-tube opening, its kinetic
energy is converted to pressure energy (sometimes called dynamic pressure).
Along a streamline the mechanical energy balance with an assumption of incompressible
fluid is applicable over the short distance of the approaching flow from 1 to 2:
1 2
𝑝 −𝑝
𝑣 − 0 + 𝑠 𝑡 = 0 or
𝜌

2

𝑣= √

2 (𝑝𝑡 − 𝑝𝑠)𝜌(4.5-E13)

where the impact pressure (1⁄2) 𝜌 𝑣 2 is the difference between the static pressure 𝑝𝑡 at the
impact-tube opening and the static pressure 𝑝𝑠 of the stream. The pressure acting on the side
holes of the outer tube is approximately equal to 𝑝𝑠 because the streamlines near the sidewall are almost parallel to the side wall.
The relation of manometer reading ℎ with the velocity 𝑣 is

𝑣= √2 (𝜌𝑚 − 𝜌)𝑔 ℎ𝜌

(4.5-E14)

where 𝜌𝑚 is the density of manometer fluid. The pitot tube responds slowly to changing
velocities owing to inertia effect, so it can measure only the time-averaged velocity in the
turbulent flow.

Fig.4.5-E3. Pitot tube

[PROBLEM 4.5-P3] The following table is some experimental data for a Pitot tube traverse
for the flow of water (temperature 20℃) inside a circular tube of 60 mm ID. The manometer
fluid is carbon tetra chloride (density 𝜌𝑚 = 1,600 kg/m3).
Calculate the local velocities and plot those data to obtain the power-law expression for thevelocity distribution. What is the Reynolds number?

30

Macroscopic Balances: Control Volume Approach

Position of Pitot tube
Radial distance fromthe pipe axis (mm)Manometer
readings (mm)

0.0
2.0
5.0
10.0
15.0
20.0
25.0
27.0
28.0
29.0
29.5
29.8

375
364
357
330
300
268
225
202
168
140
119
93

[EXAMPLE 4.5-E4] As shown in Fig.4.5-P1, let us consider an orifice-flowmeter installed in thestraight section of a horizontal circular tube (inside diameter 𝐷1 ).

Fig.4.5-E4. Orifice flowmeter

The orifice circular plate has a hole in the middle, the sharp edge of which has a very short straightsection of diameter 𝐷0 . The following mechanical energy balance of a constant density fluid flow is
set up over the region between planes “1” and “2” located at the two pressure taps under a conditionof no work effect and no potential energy change:〈𝑣𝑧2 〉22−〈𝑣𝑧1 〉22+

𝑝2 − 𝑝1
𝜌1+ 𝜍 2 〈𝑣𝑧2 〉2 = 0(4.5-E15)

where 𝜍 is called “friction loss factor” (dimensionless). The last term gives the friction loss causedby an abrupt change in the cross-sectional flow area A1 → A2. Usually for highly turbulent flow,
the flow cross-sectional area is changed from A0 to A2. Here the A2 indicates the flowcross-sectional area at the most-contracted position.
The minimum flow area is called “Vena-contracta.” Owing to a complicated flow pattern, only forsimplicity, the so-called contraction coefficients 𝛼1 are assumed to be unity in front of the firstterm. The contraction coefficient is defined as
𝛼 = 𝐴2 ⁄𝐴0, which is varied with 𝑚 = 𝐴0 ⁄𝐴1 . Therefore 〈𝑣𝑧2 〉 = 〈𝑣𝑧1 〉 𝐴1 ⁄𝐴2According to the continuity condition,
𝐴1 〈𝑣𝑧1〉 = 𝐴2 〈𝑣𝑧2 〉 . For engineering convenience, the discharge coefficient 𝐶𝑑 is introduced

Thermal Energy Balance for non-Isothermal System

31

instead of making 𝛼2 to be unity, and then 〈𝑣𝑧1〉 = 𝐶𝑑 𝑚 〈𝑣𝑧2〉. The discharge coefficient of agiven installation varies mainly with the Reynolds number. After all, the average velocity at themost-contracted position can be obtained as〈𝑣𝑧2 〉 =1√1− 𝐶𝑑 2 𝑚22(𝑝1− 𝑝2 )√

𝜌

(4.5-E16)

Then the velocity at the hole of the orifice is given by〈𝑣𝑧 〉0 = ( 𝜉𝐶𝑑 ⁄√1 − 𝐶𝑑 2 𝑚2 ) √2(𝑝1 − 𝑝2 )⁄𝜌

(4.5-E17)

As a result, the volumetric flow rate is given by the following equation:
𝑉 = 〈𝑣𝑧 〉0 𝐴0 = 𝐶𝑐 𝐴0 √2(𝑝1 − 𝑝2 )⁄𝜌

(4.5-E18)

Here 𝐶𝑐 is the correction coefficient taking into account the effect of 𝜉𝐶𝑑 ⁄√1 − 𝐶𝑑 2 𝑚2 .
Even for liquid flows, the calibration of flowmeters is necessary.
For compressible fluids, deviation from isentropic conditions is significant, and the calibration oforifice flowmeters to be used with gases must be carried out by means of the gasometer system.

4.6 Thermal Energy Balance for non-Isothermal System
For non-isothermal flow system, kinetic and potential energy effects are negligibly small
compared with thermal energy term. For many practical problems of heat transfer, therefore, theenergy balance reduces to∆𝐻𝑒 = 𝑄𝑚 − 𝑊𝑚
(4.6-1)Hence the friction loss term can be neglected except when the mechanical energy loss should bedetermined. If the system does not have a pump or other work-producing device (e.g. turbine andmixing stirrer), the enthalpy balance equation reduces to the following simple form:∆𝐻𝑚 = 𝑄𝑚
(4.6-2)[EXAMPLE 4.6-E1] Consider an incompressible fluid like water flowing through the pipeline
shown below, which comprises a pump and a heat exchanger connected by a circular pipe withconstant cross-sectional area.
(Ⅰ) pump: Consider the subsystem including the pump and the short horizontal pipe length
between planes “1” and “2”. The energy balance becomes1− ∆ (𝐻𝑚 + 2 〈𝑣 〉2 + 𝑔 ℎ) + 𝑄𝑚 − 𝑊𝑚 = 0(4.6-E1)
The mechanical energy balance is𝑝2
1(4.6-E2)∫𝑝 𝑉𝑠 𝑑𝑝 + ∆ 2 〈𝑣 〉2 = − 𝐹𝑟𝑚 − 𝑊𝑚
1

Since we do not have a heat exchanger in this subsystem, the term 𝑄𝑚 is generally small comparedwith the work done on the pump 𝑊𝑚 . The term ∆𝐻𝑒 in Eq.(4.4-9) is the enthalpy change of thefluid as it passes through the pump. Pumps for liquids and gases are usually designed so thatentering and leaving velocities are equal, i.e. ∆(1⁄2)〈𝑣〉2 = 032Macroscopic Balances: Control Volume Approach
27

Fig.4.6-E1. Macroscopic energy balance for a simple pipeline system

For incompressible fluid𝑝𝑝 −𝑝∆𝐻𝑚 = ∫𝑝 2 𝑉𝑠 𝑑𝑝 + 𝐹𝑟𝑚 = 2 𝜌 1 + 𝐹𝑟𝑚21
1

The power required for the pump is given by𝑝 −𝑝− 𝑊𝑚 = 2 𝜌 1 + 𝐹𝑟𝑚21(4.6-E3)(4.6-E4)

The pipe section is so short that the friction loss 𝐹𝑟𝑚21 can often be neglected.
(Ⅱ) heat exchanger: Typical heat exchangers are so designed that there is no shaft work andnegligibly small difference in elevation between the entrance and exit. Then
𝑊𝑚 = 0 and ∆ℎ = 0
For incompressible fluid, kinetic energy balance is also negligibly small. The friction loss term 𝐹𝑟𝑚
becomes important only when the power requirement is calculated by using the mechanical energybalance. Then𝑝2
𝑝 −𝑝(4.6-E5)∫𝑝 𝑉𝑠 𝑑𝑝 = − 𝐹𝑟𝑚 or 3 𝜌 2 = − 𝐹𝑟𝑚32
1

Regarding the total energy balance𝑇3
𝑝 −𝑝∫𝑇2 𝐶𝑝 𝑑𝑇 + 3 𝜌 2 = 𝑄𝑚

(4.6-E6)

Summing up those three equations, the overall mechanical energy balance becomes
𝑝 −𝑝
𝑝 −𝑝
+ 3 𝜌 2 + 4 𝜌 3 + 𝑔(ℎ4 − ℎ3 ) = −(𝐹𝑟𝑚21 + 𝐹𝑟𝑚32 + 𝐹𝑟𝑚43) − 𝑊𝑚𝜌

(4.6-E10)

The pressure change term is small compared with the enthalpy change due to the heat transfer. For
constant heat capacity
𝑄𝑚 = 𝐶𝑝 (𝑇3 − 𝑇2 ) i.e. 𝑄 = 𝑤 𝐶𝑝 (𝑇3 − 𝑇2 )(4.6-E7)
This equation indicates that the incompressible fluid gets an increase in enthalpy equal to the rate ofheat input. For example, in the case when another hot fluid is supplied into an annular space of
double-tube exchanger at flow rate 𝑤′
𝑄 = 𝑤 ′ 𝐶𝑝′ (𝑇 ′ 3 − 𝑇 ′ 2 ) = 𝑤 𝐶𝑝 (𝑇3 − 𝑇2 )(4.6-E8)
where 𝑇′3 and 𝑇′2 are the inlet and outlet temperatures of the hot fluid. This is the fundamental
heat balance equation for the design of heat exchangers.
(Ⅲ) pipe: Since this subsystem having a change in elevation is adiabatic, the term 𝑄𝑚 is zero.
There is no shaft work: 𝑊𝑚 = 0. The incompressible fluid does not have any velocity change
through the pipe with constant cross-sectional area, i.e. 〈𝑣 〉4 = 〈𝑣 〉3 .
The mechanical energy balance becomes
𝑝4 − 𝑝3+ 𝑔 (ℎ4 − ℎ3 ) = − 𝐹𝑟𝑚43(4.6-E9)𝜌
𝑝2 − 𝑝1

Finally we get+ 𝑔(ℎ4 − ℎ3 ) + ∑ 𝐹𝑟𝑚 = − 𝑊𝑚𝜌

𝑝4 − 𝑝1(4.6-E11)

Examples,

33

This can be considered as a mechanical energy balance equation applied between planes “1” and“4” and implies that the work done on the fluid − 𝑊𝑚 required to keep a given constant flow rateshould be equal to the sum of differences in flow work and potential energy and the total friction
loss (mechanical energy loss) due to viscous dissipation over the whole length of the pipeline.
This is the design equation of piping. The required power of pump is given by
𝑃𝑤 = −𝑤 𝑊𝑚(4.6-E12)[EXAMPLE 4.6-E2] We want to cool dry oxygen gas of 𝑊𝑜 kg/s from 𝑇𝑖 to 𝑇𝑜 𝐾 by meansof horizontal double-pipe heat exchanger shown. The cross sectional area of the inner pipe is 𝑆 𝑚2 .
If we supply coarse liquid nitrogen as the cooling medium into the annular space at its saturation
temperature, nucleate boiling occurs on the outer surface of the inner pipe. Its latent heat ofvaporization is 𝜆𝑁 𝐽⁄𝑘𝑔. The pressures at the inlet and outlet of the inner pipe are 𝑝𝑖 and 𝑝𝑜 𝑃𝑎,
respectively. Calculate the rate of energy removed by the coarse liquid nitrogen across the pipe wall.
Assume turbulent flow of ideal gas with constant heat capacity of oxygen 𝐶𝑝𝑂 𝐽⁄𝑘𝑚𝑜𝑙 𝐾 .

Fig.4.6-E1. Energy balance for cooling dry oxygen by liquid nitrogen

Solution: The macroscopic energy balance between the inlet and exit of the inner pipe is1− ∆ (𝐻𝑚 + 2 〈𝑣 〉2 + 𝑔 ℎ) + 𝑄𝑚 − 𝑊𝑚 = 0(4.6-E13)
For an ideal gas with constant capacity
𝑇𝑇
1
𝐶𝑝
Δ𝐻𝑚 = ∫𝑇 𝑜 𝐶𝑝𝑚 𝑑𝑇 = 𝑀 ∫𝑇 𝑜 𝐶𝑝𝑂 𝑑𝑇 = 𝑀𝑂 (𝑇𝑜 − 𝑇𝑖 )𝐽⁄𝑘𝑔
(4.6-E14)𝑖𝑖
From the mass balance 𝜌𝑖 〈𝑣𝑖 〉𝑆𝑖 = 𝜌𝑜 〈𝑣𝑜 〉𝑆𝑜 = 𝑊𝑜 .
The perfect gas law is
1
1 𝑅𝑇
= 𝑀 𝑝
(4.6-E15)
𝜌
Then
1
1
Δ 2 〈𝑣 〉2 = 2 (𝑊𝑜 ⁄𝑆)2 [(𝑅𝑇𝑜 ⁄𝑀 𝑝𝑜 )2 − (𝑅𝑇𝑖 ⁄𝑀 𝑝𝑖 )2 ] 𝐽⁄𝑘𝑔
(4.6-E16)
𝑊𝑚 = 0 and Δℎ = 0
Therefore
1
𝑄 𝑘𝐽⁄𝑠 = 𝑊𝑜 𝑄𝑚 = (𝑊𝑜 𝐶𝑝𝑂 ⁄𝑀 )(𝑇𝑜 − 𝑇𝑖 ) + 2 𝑊𝑜 (𝑊𝑜 ⁄𝑆)2 [(𝑅𝑇𝑜 ⁄𝑀 𝑝𝑜 )2 − (𝑅𝑇𝑖 ⁄𝑀 𝑝𝑖 )2 ](4.6-E17)
On the other hand, the coarse liquid nitrogen should be fed at a rate of 𝑄⁄𝜆𝑁 𝑘𝑔⁄𝑠.

[PROBLEM 4.6-P1] Mineral oil (𝜌 = 850 𝑘𝑔⁄𝑚3 , 𝐶𝑝 = 2.3 𝑘𝐽⁄𝑘𝑔 𝐾, 𝜇 = 0.03 𝑃𝑎 𝑠 and𝜅 = 0.143 𝑊 ⁄𝑚 𝐾 ) is flowing at an average velocity 〈𝑣 〉 = 0.4 𝑚⁄𝑠 in a 25 mm ID circular
steel pipe. The pipe wall is heated from outside at constant heat flux 𝑞𝑤 = 80 𝑊 ⁄𝑚2 (based on
the inside wall area). Calculate the Reynolds number. How many degrees does the averagetemperature rise per unit pipe length?

34

Macroscopic Balances: Control Volume Approach[PROBLEM 4.6-P2] A basic solution is to be fed to an absorption column operating at pressureof 0.4 MPa at a rate of 0.001 m3/s from a storage tank. The physical properties of the basic solution
can be assumed to be approximately equal to those of water. As a result of calculation, the total
friction loss of this system is given by 𝐹𝑟𝑚 = 18 𝐽⁄𝑘𝑔. Calculate the power of pump required forthe given feed rate when 30 mm ID pipe is used for the pipe line. Assume turbulent flow.
[PROBLEM 4.6-P3] Mineral oil (𝜌 = 800 𝑘𝑔⁄𝑚3 , 𝐶𝑝 = 2.1 𝑘𝐽⁄𝑘𝑔 𝐾 ) is transferred at a rateof 10 ton/h from a tank to a hydrogen plant. On the pipeline there is a heat exchanger to heat the oilfrom 308 K to 321 K. The pressure drop between the entrance and exit of the exchanger is 5 kPa.
Determine the friction loss in m 2/s2 and the heat transfer rate in J/h of the exchanger. The pressuresat the inlet (suction side) and outlet (discharge side) of the pump are 0.08 MPa and 1.8 MPa,
respectively. Determine the power of pump in kW.

4.7 Macroscopic Mass Balance of Individual Components
Even when the total mass is conserved, an individual species is not, in general, conserved underprocesses including chemical reaction. By applying the principle of mass conservation to the control
volume having a homogeneous chemical reaction, the mass balance of component A can be writtenas𝑑⃗ ∙ ⃗𝒏) 𝑑𝑆 + ∫𝑉 𝑟𝐴 𝑑𝑉(4.7-1)∫𝑉 𝐶𝐴 𝑑𝑉 = − ∫𝑆 𝐶𝐴 (𝒗𝑑𝑡where 𝐶𝐴 is the mass concentration of component A and 𝑟𝐴 is the reaction rate indicating the rateof mass generation per unit volume of component A.

[EXAMPLE 4.7-E1] A typical distillation plant shown below (Fig. 4.7-E1) comprises a
distillation column, a total overhead condenser, and a partial bottom reboiler. The feed of mixture 𝐹
of A and B, which is to be separated into fractions, is introduced to the column. Some of the lessvolatile component B is condensed from the vapor going up the column, and at the same time, someof the more volatile component A is vaporized from the liquid going down the column. These phasetransformations take place due to the giving and receiving of latent heat between A and B
components.

Fig.4.7-E1. Distillation plant

Macroscopic Mass Balance of Individual Components (continued)
35

35

The liquid reaching the bottom is partially vaporized in a reboiler heated by steam. The vapor issent back to the bottom of the column to provide the upflowing vapor in the stripping section(section below the feed stage). The remainder of the bottom liquid, which has high concentration of
the less volatile component B, is withdrawn as the bottom product. On the other hand, the vapor
reaching the top is completely condensed by an overhead condenser cooled by water. Some fractionof the condensate is returned by a reflux pump to provide the downflowing liquid (internal reflux
liquid) in the rectifying section (section above the feed stage). The distillation process cannot
proceed without the intimate contact of vapor with the internal reflux liquid.
The remainder of the condensate , which has high concentration of the more volatile component A,
is withdrawn as the overhead product. The column fed with 𝐹 kmol/h of mole fraction 𝑥𝐹 of
component A produces 𝐷 kmol/h of overhead product of mole fraction 𝑥𝐷 and 𝑊 kmol/h ofbottom product of mole fraction 𝑥𝑊 .
For steady-state operation, the overall mass balances can be written as
Total mass balance: 𝐹 = 𝐷 + 𝑊(4.7-E1)
Mass balance of component A: 𝐹 𝑥𝐹 = 𝐷 𝑥𝐷 + 𝑊 𝑥𝑊
The reflux liquid: 𝐿𝑅 = 𝑉𝑡 − 𝐷
where 𝑉𝑡 is the vapor rate from the column top to the overhead condenser.
(4.7-E2)[EXAMPLE 4.7-E2] Consider a flow system (gas absorption column) in which a gas stream is incontact with a liquid stream. The system, as shown in Fig.4.7-E2, has a single entrance and a single
exit. The cross-sectional area 𝑆 is constant. The gas stream contains component A soluble into theliquid.
The macroscopic mass balance of component A in gas phase is𝑑(4.7-E3)∫ 𝐶 𝑑𝑉 = 𝐶𝐴1 〈𝑣1 〉𝑆 − 𝐶𝐴2 〈𝑣2 〉𝑆 − 𝑊𝐴𝑖 + ∫𝑉 𝑟𝐴 𝑑𝑉
𝑑𝑡 𝑉 𝐴
Here 〈𝑣1〉 and 〈𝑣2 〉 are the “superficial velocities” at the entrance and exit (the average velocities
the gas would have in the column if the column were occupied with the gas only) and 𝑊𝐴𝑖 is therate of mass transfer of component A across the gas-liquid contacting interface to the liquid.
At steady state with no chemical reaction(𝐶𝐴1 〈𝑣1〉 − 𝐶𝐴2 〈𝑣2 〉)𝑆 = 𝑊𝐴𝑖(4.7-E4)

Fig.4.7-E2. Gas absorption column

Similarly the macroscopic mass balance in the liquid phase at steady state with no chemical reaction(𝐶′𝐴1 〈𝑣′1〉 − 𝐶′𝐴2 〈𝑣′2〉)𝑆 = 𝑊𝐴𝑖(4.7-E5)
Where the prime denotes the liquid phase. Here 〈𝑣′1 〉 and 〈𝑣′2 〉 are the superficial velocities

36

Macroscopic Balances: Control Volume Approach(the average velocities the liquid would have in the column if the column were occupied with theliquid only.).
[PROBLEM 4.7-P1] Nitrogen gas is flowing in a pipe of porous wall, through which hydrogengas is injected from outside into the pipe at a rate of 𝑚𝐻 kg/m2 s. At the entrance (𝐿 = 0) to theporous pipe, the flow rate of nitrogen is 𝑤𝑁 kg/s. he molecular weights of N2 and H2 are 𝑀𝑁 and
𝑀𝐻 , respectively. Obtain the expressions for the total mass flow rater as a function of pipe length 𝐿
and the gas composition in mole fraction at a position 𝐿 downstream of the porous pipe.

Fig.4.7-P1. Nirogen gas stream with hydrogen gas injected through porous pipe wall
Nomenclature𝐶𝐴molar concentration of component A, [kmol/m3]
𝐶𝑝
heat capacity, [J/kg K]
𝐷
pipe inside diameter, [m] or overhead product in distillation column, [kmol/s]
𝐸𝑘 , 𝐸𝑝 , 𝐸𝑢 kinetic, potential, internal energy, [J/kg]⃗
𝑭
external force vector,
𝐹
feed rate in distillation column, [kmol/s]𝐹𝑟friction loss or mechanical energy loss, [J/kg]
𝑔
gravitational acceleration, [m/s2]
𝐻𝑒enthalpy, [J/kg]ℎ
height, [m]𝐿pipe length, [m]𝐿𝑅
reflux liquid rate in distillation column, [kmol/s]𝑚mass of control volume, [kg]⃗𝒏unit vector𝑝pressure, [Pa]𝑄
volumetric flow rate, [m3/s] or heat input from surroundings, [J/s]𝑟𝐴
reaction rate, [kmol/m3s]𝑆surface area, [m2] or entropy, [J/kg K]
𝑡time, [s]⃗𝒗velocity vector
𝑉
volume, [m3]𝑉𝑠specific volume, [m3/kg]
𝑉𝑡vapor flow rate at top of distillation column, [kmol/s]𝑣velocity, [m/s]𝑊work done by fluid on surroundings, [W] or bottom product in distillation column, [kmol/s]𝑤
mass flow rate, [kg/s]
𝑥𝐹 , 𝑥𝐷 , 𝑥𝑊 mole fraction of more-volatile component of feed, overhead product, and bottom product, [ - ]𝜌density, [kg/m3]
𝜏𝑤wall shear stress, [N/m2]

CHAPTER 5
MICROSCOPIC DIFFERENTIAL BALANCES
5.1 Differential Balances of Mass and Momentum(Equations of Continuity and Motion)
In this section, we shall set up the equations of conservation of mass and momentum in generalform (Bird et al., 1960). Once we have developed these general equations, we can start with themand simplify them to fit the problem at hand.
1.

Bird, R.B., Stewart, W.E., and Lightfoot, E.N., Transport Phenomena, Wiley, New York, Chapt.3 (1960)

5.1-1 Differential mass balance (Equation of continuity)1)
Let us consider a differential mass balance over a stationary differential control volume ∆𝑥∆𝑦∆𝑧
shown in Fig.5.1-1.

Fig.5.1-1. Differential control volume in rectangular coordinate system

Generally the fluid is flowing through all six faces of the control volume. The fluid ishomogeneous, but the density changes with temperature and pressure.
The rate of mass transfer) can be written as(rate of mass in) = (mass transferred)/(unit time)= (mass/unit volume)(length/unit time)(area)
Therefore, as shown in Fig.5.1-2, mass is transferred across the pair of ∆𝑦∆𝑧 faces
perpendicular to the x-axis by 𝑣𝑥 only. Hence the rate of mass in through the face at x is
𝜌|𝑥 𝑣𝑥 |𝑥 ∆𝑦∆𝑧 and the rate of mass out through the face at 𝜌|𝑥+∆𝑥 𝑣𝑥 |𝑥+∆𝑥 ∆𝑦∆𝑧. Similar expressions
can be obtained for the remaining two pairs of faces.
The masses of the control volume at time 𝑡 and 𝑡 + ∆𝑡 are 𝜌|𝑡 ∆𝑥∆𝑦∆𝑧 and 𝜌|𝑡+∆𝑡 ∆𝑥∆𝑦∆𝑧,
repectively. Therefore the mass accumulation per unit time is given by𝜌|𝑡+∆𝑡 − 𝜌|𝑡() ∆𝑥∆𝑦∆𝑧∆𝑡

38

Microscopic Differential BalancesFig. 5.1-2 Mass flow rate transferred across planes 𝒙 and 𝒙 + ∆𝒙 of control volume

Then applying the principle of mass conservation, the following equation yields() ∆𝑥∆𝑦∆𝑧 = [𝜌|𝑥 𝑣𝑥 |𝑥 − 𝜌|𝑥+∆𝑥 𝑣𝑥 |𝑥+∆𝑥 ]∆𝑦∆𝑧
𝜌|𝑡+∆𝑡 − 𝜌|𝑡∆𝑡[𝜌|𝑦 𝑣𝑦 |𝑦 − 𝜌|𝑦+∆𝑦 𝑣𝑦 |𝑦+∆𝑦 ] ∆𝑧∆𝑥 − [𝜌|𝑧 𝑣𝑧 |𝑧 − 𝜌|𝑧+∆𝑧 𝑣𝑧 |𝑧+∆𝑧 ]∆𝑥∆𝑦(5.1-1)Dividing through by the volume ∆𝑥∆𝑦∆𝑧 and taking the limit as ∆𝑡, ∆𝑥, ∆𝑦, ∆𝑧 go to zero, we get
𝜕𝜌

𝜕 𝜌𝑣

= − ( 𝜕𝑥 𝑥 +
𝜕𝑡

𝜕 𝜌𝑣𝑦
𝜕𝑦

+

𝜕 𝜌𝑣𝑧
𝜕𝑧

(5.1-2))This is the equation of continuity in general form.
This equation can be written in vector notation:𝜕𝜌⃗= − ∇ ∙ 𝜌𝒗
(5.1-3)𝜕𝑡This equation describes the rate of change in density at a fixed position from the Eulerian view
point. That is, if the right side of the equation is expanded, we get
𝜕𝜌

𝜕𝜌

𝜕𝜌

𝜕𝜌

𝜕𝑣

𝜕 𝑣𝑦

𝜕𝑣

+ 𝑣𝑥 𝜕𝑥 + 𝑣𝑦 𝜕𝑦 + 𝑣𝑧 𝜕𝑧 = − 𝜌 ( 𝜕𝑥𝑥 + 𝜕𝑦 + 𝜕𝑧𝑧 )𝜕𝑡(5.1-4)The left side describes the rate of change in density observed by following the fluid motion from the
Lagrangian viewpoint. Therefore if the fluid is at rest, all terms vanish. The derivative of theleft-hand side is called the substantial derivative.
In engineering problems, we very frequently assume the fluid density to be constant asincompressible fluid. For constant density fluids, all the left-side terms vanish, and then the rightside becomes zero:
𝜕 𝑣𝑥
𝜕𝑥

𝜕 𝑣𝑦

𝜕𝑣

(5.1-5)+ 𝜕𝑦 + 𝜕𝑧𝑧 = 0

This is a very important form of the equation of continuity. In vector form it can be written as∇ ∙ ⃗⃗⃗
𝒗 =0(5.1-6)
General forms of the equation of continuity are listed in Table 5.1-1.
Table 5.1-1 Equation of continuity for fluid with constant density(1) rectangular coordinates
𝜕 𝑣𝑥

𝜕 𝑣𝑦

𝜕𝑣

+ 𝜕𝑦 + 𝜕𝑧𝑧 = 0𝜕𝑥(2) cylindrical coordinates
1 𝜕 𝑟𝑣𝑟
𝑟

𝜕𝑟

1 𝜕𝑣

𝜕𝑣

+ 𝑟 𝜕𝜃𝜃 + 𝜕𝑧𝑧 = 0

5.1-2 Differential momentum balance (Navier-Stokes equation)1)
Following the procedure established in the differential equation of continuity, let us apply theprinciple of momentum conservation to the same differential control volume. (see Fig.5.1-1).
1.

Bird, R.B., Stewart, W.E., and Lightfoot, E.N., Transport Phenomena, Wiley, New York, Chapt.3 (1960)

Differential Momentum Balance (Navier-Stokes Equation) (continued)
39

39

Since momentum is a vector, the conservation equation can be resolved into its three orthogonal
scalar momentum equations.
Consider the x momentum of each term. The x-momentums of the control volume at time 𝑡 and𝑡 + ∆𝑡 are (𝜌𝑣𝑥 )|𝑡 ∆𝑥∆𝑦∆𝑧 and (𝜌𝑣𝑥 )|𝑡+∆𝑡 ∆𝑥∆𝑦∆𝑧 , respectively. Then the accumulation of
x-momentum per unit time (i.e. the rate of x-momentum accumulation) becomes(𝜌𝑣 )|−(𝜌𝑣𝑥 )|𝑡
( 𝑥 𝑡+∆𝑡) ∆𝑥∆𝑦∆𝑧(5.1-7)∆𝑡
In general, momentum may be transferred into and out of the control volume by two mechanisms:
by convection (i.e. by virtue of fluid motion) and by diffusion (i.e. by virtue of molecular motion).

Fig.5.1-3 Control volume for differential momentum balanceFirst let us consider the convective transport of the x-momentum. Assuming the momentum flux tobe uniform on each face, the rate of the x-momentum entering the face at 𝑥 by 𝑣𝑥 |𝑥 is(𝜌𝑣𝑥 𝑣𝑥 )|𝑥 ∆𝑦∆𝑧, and similarly the rate of the x-momentum leaving the face at 𝑥 + ∆𝑥 by 𝑣𝑥 |𝑥+∆𝑥
is (𝜌𝑣𝑥 𝑣𝑥 )|𝑥+∆𝑥 ∆𝑦∆𝑧. The 𝜌𝑣𝑥 has dimensions of momentum per unit volume, the 𝑣𝑥 ∆𝑦∆𝑧 hasdimensions of volume per unit time, and then the 𝜌𝑣𝑥 𝑣𝑥 |𝑥 ∆𝑦∆𝑧 has dimensions of momentum
transferred per unit time (that is, the rate of momentum transferred).

Fig.5.1-4 Convective transport of x-momentum across faces at 𝒚 and 𝒚 + ∆𝒚Across the face of area ∆𝑧∆𝑥 at 𝑦, as shown in Fig.5.1-4, the x-momentum (𝜌𝑣𝑥 )|𝑦 can betransferred by the velocity component 𝑣𝑦 |𝑦 perpendicular to the face. The rate of the x-momentum
entering the face at 𝑦 by 𝑣𝑦 |𝑦 is (𝜌𝑣𝑥 𝑣𝑦 )|𝑦 ∆𝑧∆𝑥 and the rate of the x-momentum leaving theface 𝑦 + ∆𝑦 by 𝑣𝑦 |𝑦+∆𝑦 is (𝜌𝑣𝑥 𝑣𝑦 )|𝑦+∆𝑦 ∆𝑧∆𝑥 . Therefore the net rate of the x-momentum

40

Microscopic Differential Balancestransferred across all six faces by convection becomes[(𝜌𝑣𝑥 𝑣𝑥 )|𝑥 − (𝜌𝑣𝑥 𝑣𝑥 )|𝑥+∆𝑥 ]∆𝑦∆𝑧 + [(𝜌𝑣𝑥 𝑣𝑦) | − (ρ𝑣𝑥 𝑣𝑦 )|] ∆𝑧∆𝑥 + [(𝜌𝑣𝑥 𝑣𝑧 )|𝑧 − (𝜌𝑣𝑥 𝑣𝑧 )|𝑧+∆𝑧 ]∆𝑥∆𝑦𝑦+∆𝑦𝑦(5.1-8)
Next let us consider the diffusion (molecular transport) of x-momentum assuming the momentum
flux to be uniform on each face. The rate of x-momentum entering the face at 𝑥 by diffusion is𝜏𝑥𝑥 |𝑥 ∆y∆z, and the rate of x-momentum leaving the face at 𝑥 + ∆𝑥 is also 𝜏𝑥𝑥 |𝑥+∆𝑥 ∆y∆z. Notethat 𝜏𝑥𝑥 is the normal stress in the x direction on the face perpendicular to the x axis. Similarly therate of x-momentum entering the face at 𝑦 by diffusion is 𝜏𝑦𝑥 |𝑦 ∆z∆x , and the rate of
x-momentum leaving the face at y + ∆𝑦 is 𝜏𝑦𝑥 |𝑦+∆𝑦 ∆z∆x.

Recall that 𝜏𝑦𝑥 is the flux of x-momentum through a face perpendicular to the y axis. In other
word, 𝜏𝑦𝑥 is the shear stress on the face normal to the y axis and acting in the x direction.
Therefore the net rate of x-momentum transferred across all six faces by diffusion becomes(𝜏𝑥𝑥 |𝑥 − 𝜏𝑥𝑥 |𝑥+∆𝑥 )∆𝑦∆𝑧 + (𝜏𝑦𝑥 | − 𝜏𝑦𝑥 |) ∆𝑧∆𝑥 + (𝜏𝑧𝑥 |𝑧 − 𝜏𝑧𝑥 |𝑧+∆𝑧 )∆𝑥∆𝑦(5.1-9)𝑦𝑦+∆𝑦Fig.5.1-5 Molecular transport of x-momentum by diffusion in y direction and pressure force

Pressure is a scalar, but the pressure force is a vector perpendicular to the acting face. The net
effect of the x-directed pressure forces is(𝑝|𝑥 − 𝑝|𝑥+∆𝑥 )∆𝑦∆𝑧(5.1-10)The gravitational force acts on the mass center of the control volume as body force. The
x-component of the gravitational force is 𝜌𝑔𝑥 ∆𝑥∆𝑦∆𝑧.
Substituting all the foregoing contributions into the equation of momentum conservation law, anddividing the resulting equation by ∆𝑥∆𝑦∆𝑧, we get(𝜌𝑣𝑥 )|𝑡+∆𝑡 − (𝜌𝑣𝑥 )|𝑡=∆𝑡(𝜌𝑣𝑥 𝑣𝑥 )|𝑥+∆𝑥 − (𝜌𝑣𝑥 𝑣𝑥 )|𝑥

−[
𝜏𝑦𝑥 |

∆𝑥

𝑦+∆𝑦

− 𝜏𝑦𝑥 |

∆𝑦

𝑦

+

+

(𝜌𝑣𝑥 𝑣𝑦 )|

𝑦+∆𝑦

∆𝑦

𝜏𝑧𝑥 |𝑧+∆𝑧 − 𝜏𝑧𝑥 |𝑧∆𝑧− (𝜌𝑣𝑥 𝑣𝑦 )|]−𝑝|𝑥+∆𝑥 − 𝑝|𝑥
∆𝑥

𝑦

+

(𝜌𝑣𝑥 𝑣𝑧 )|𝑧+∆𝑧 − (𝜌𝑣𝑥 𝑣𝑧 )|𝑧
∆𝑧

𝜏

|

] − [ 𝑥𝑥 𝑥+∆𝑥

− 𝜏𝑥𝑥 |𝑥∆𝑥+(5.1-11)+ 𝜌𝑔𝑥

Taking the limit as ∆𝑡, ∆𝑥, ∆𝑦, ∆𝑧 go to zero, we obtain the x-component of the equation of
motion in general form:𝜕𝜌𝑣𝑥 𝑣𝑦
𝜕𝜏𝑦𝑥
𝜕
𝜕𝜌𝑣 𝑣
𝜕𝜌𝑣𝑥 𝑣𝑧𝜕𝜏𝜕𝜏𝜕𝑝(𝜌𝑣𝑥 ) = − ( 𝑥 𝑥 ++) − ( 𝑥𝑥 ++ 𝑧𝑥 ) −+ 𝜌𝑔𝑥(5.1-12)
𝜕𝑡

𝜕𝑥

𝜕𝑦

𝜕𝑧

The remaining two can be written by analogy.

𝜕𝑥

𝜕𝑦

𝜕𝑧

𝜕𝑥

Differential Momentum Balance

41

In vector symbolism, these equations can be combined into one vector equation:⃗
𝜕𝜌𝒗⃗𝒗⃗⃗⃗ − ∇ ∙ ⃗⃗𝝉 − ∇𝑝 + 𝜌𝒈⃗⃗
= − ∇ ∙ 𝜌𝒗

(5.1-13)
One of the most advantageous points of vector equation is that vector equation is valid for any
orthogonal coordinate system. The above vector equation is also valid for cylindrical and sphericalcoordinates. The tedious transformation from rectangular to curvilinear coordinates is not discussed
here. Keep in mind that these equations are valid for both Newtonian and non-Newtonian fluids.
In order to determine velocity distributions, the relation between stresses and velocity gradientsshould be plugged into the 𝜏- term of the equation of motion.
For simplicity, we assume Newtonian fluid with constant density and viscosity, and rewrite theequation of motion with the aid of the equation of continuity as follows:𝜕𝜌𝑣𝑥 𝑣𝑦𝜕𝜕𝜌𝑣 𝑣
𝜕𝜌𝑣𝑥 𝑣𝑧
𝜕𝑣
𝜕𝑣
𝜕𝑣
𝜕𝑣(𝜌𝑣𝑥 ) + ( 𝑥 𝑥 ++) = 𝜌 ( 𝑥 + 𝑣𝑥 𝑥 + 𝑣𝑦 𝑥 + 𝑣𝑧 𝑥 )
𝜕𝑡

𝜕𝑡

𝜕𝑥

𝜕𝑦

𝜕𝑧

𝜕𝑡

𝜕𝑥

𝜕𝑦

𝜕𝑧

𝜕𝜏𝑥𝑥 𝜕𝜏𝑦𝑥 𝜕𝜏𝑧𝑥
𝜕 2 𝑣𝑥 𝜕 2 𝑣𝑥 𝜕 2 𝑣𝑥++
= −𝜇 ( 2 +
+
)
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑥
𝜕𝑦 2
𝜕𝑧 2

Then the x-component of the equation of motion becomes
𝜕𝑣

𝜕𝑣

𝜕𝑣

𝜕𝑣

𝜕2 𝑣

𝜕𝑝

𝜕2 𝑣

𝜕2 𝑣

(5.1-14)𝜌 ( 𝜕𝑡𝑥 + 𝑣𝑥 𝜕𝑥𝑥 + 𝑣𝑦 𝜕𝑦𝑥 + 𝑣𝑧 𝜕𝑧𝑥 ) = − 𝜕𝑥 + 𝜇 ( 𝜕𝑥2𝑥 + 𝜕𝑦2𝑥 + 𝜕𝑧 2𝑥) + 𝜌𝑔𝑥

The remaining two can be written by analogy.
These equations for constant physical properties can be combined by the following vector
equation:⃗𝜕𝒗⃗ ∙ ∇ )𝒗
⃗ ] = − ∇𝑝 + 𝜇∇2 𝒗⃗ + 𝜌𝒈⃗⃗ρ [ + (𝒗(5.1-15)
𝜕𝑡This is well known as the Navier-Stokes equation. The differential operator ∇2 is called theLaplacian operator.
These equations derived in this section are tabulated in rectangular and cylindrical coordinates inTable 5.1-1.
Only for simplification and convenience, the equation of change for spherical coordinate system isomitted from the table. If we wish to rewrite these equations in spherical coordinates, we should
obtain the relations between (𝑥, 𝑦, 𝑧)and (𝑟, 𝜃, 𝜙) with the relations between(𝑣𝑥 , 𝑣𝑦 , 𝑣𝑧 ) and (𝑣𝑟 , 𝑣𝜃 , 𝑣𝜙 ). However wen will not have to wade through the details of this tedious
process.
Table 5.1-2 Equation of motion (Navier-Stokes equation) for a Newtonian fluid with constantdensity and viscosity(1) rectangular coordinates
x: 𝜌 (

𝜕𝑣𝑥

𝜕𝑣

𝜕𝑣

𝜕𝑣

𝜕𝑝

𝜕2 𝑣

𝜕2 𝑣

𝜕2 𝑣

y: 𝜌 (

𝜕𝑣𝑦

𝜕𝑣𝑦

𝜕𝑣𝑦

𝜕𝑣𝑦

𝜕𝑝

𝜕 2 𝑣𝑦𝜕 2 𝑣𝑦𝜕 2 𝑣𝑦

𝜕𝑣𝑧

𝜕𝑣

𝜕𝑣

𝜕𝑣

𝜕𝑝

𝜕2 𝑣

𝜕2 𝑣

𝜕2𝑣

z: 𝜌 (+ 𝑣𝑥 𝜕𝑥𝑥 + 𝑣𝑦 𝜕𝑦𝑥 + 𝑣𝑧 𝜕𝑧𝑥 ) = − 𝜕𝑥 + 𝜇 ( 𝜕𝑥2𝑥 + 𝜕𝑦2𝑥 + 𝜕𝑧 2𝑥 ) + 𝜌 𝑔𝑥
𝜕𝑡+ 𝑣𝑥 𝜕𝑥 + 𝑣𝑦 𝜕𝑦 + 𝑣𝑧 𝜕𝑧 ) = − 𝜕𝑦 + 𝜇 ( 𝜕𝑥2 + 𝜕𝑦2 + 𝜕𝑧 2 ) + 𝜌 𝑔𝑦
𝜕𝑡+ 𝑣𝑥 𝜕𝑥𝑧 + 𝑣𝑦 𝜕𝑦𝑧 + 𝑣𝑧 𝜕𝑧𝑧 ) = − 𝜕𝑧 + 𝜇 ( 𝜕𝑥2𝑧 + 𝜕𝑦2𝑧 + 𝜕𝑧2𝑧 ) + 𝜌 𝑔𝑧
𝜕𝑡(2) cylindrical coordinates
𝜕𝑣

𝜕𝑣

𝑣 𝜕𝑣

r: 𝜌 ( 𝜕𝑡𝑟 + 𝑣𝑟 𝜕𝑟𝑟 + 𝑟𝜃 𝜕𝜃𝑟 −
𝜕𝑣

𝜕𝑣

𝑣 𝜕𝑣

𝜌 ( 𝜕𝑡𝜃 + 𝑣𝑟 𝜕𝑟𝜃 + 𝑟𝜃 𝜕𝜃𝜃 +
𝜕𝑣

𝜕𝑣

𝑣 𝜕𝑣

𝑣𝜃 2
𝑟
𝑣𝑟 𝑣𝜃
𝑟

𝜕𝑣

𝜕𝑝

𝜕

𝜕𝑣

1 𝜕𝑝

1 𝜕2𝑣

1𝜕

2 𝜕𝑣

𝜕2𝑣

+ 𝑣𝑧 𝜕𝑧𝑟) = − 𝜕𝑟 + 𝜇 [𝜕𝑟 (𝑟 𝜕𝑟 (𝑟𝑣𝑟 )) + 𝑟 2 𝜕𝜃2𝑟 − 𝑟 2 𝜕𝜃𝜃 + 𝜕𝑧 2𝑟] + 𝜌 𝑔𝑟
𝜕

1 𝜕2𝑣

1𝜕

2 𝜕𝑣

𝜕2𝑣

+ 𝑣𝑧 𝜕𝑧𝜃) = − 𝑟 𝜕𝜃 + 𝜇 [𝜕𝑟 (𝑟 𝜕𝑟 (𝑟𝑣𝜃 )) + 𝑟 2 𝜕𝜃2𝜃 + 𝑟 2 𝜕𝜃𝑟 + 𝜕𝑧 2𝜃] + 𝜌 𝑔𝜃 
𝜕𝑣

𝜕𝑝

1𝜕

𝜕𝑣

1 𝜕2𝑣

𝜕2𝑣

z: 𝜌 ( 𝜕𝑡𝑧 + 𝑣𝑟 𝜕𝑟𝑧 + 𝑟𝜃 𝜕𝜃𝑧 + 𝑣𝑧 𝜕𝑧𝑧) = − 𝜕𝑧 + 𝜇 [𝑟 𝜕𝑟 (𝑟 𝜕𝑟𝑧 ) + 𝑟 2 𝜕𝜃2𝑧 + 𝜕𝑧2𝑧] + 𝜌 𝑔𝑧

42

Microscopic Differential Balances

5.2 Differential Balance of Energy (Equation of energy)1)
As aforementioned in macroscopic energy balance, the energy associated with mass consists ofpotential, kinetic and internal energies. In general, non-isothermal processes accompanied with fluid
motion take place undergoing various work effects such as viscous dissipation, pressure work andgravitational work. At this stage, it is too difficult to derive a general form of energy equation. In
this section, therefore, we shall develop the differential equation of thermal energy transfer written
for incompressible fluid with constant physical properties. This equation is widely used in
non-isothermal system with moderate change in temperature and velocity.
The following assumptions are made:(1) incompressible fluid with constant heat capacity and thermal conductivity(2) negligible kinetic- and potential-energy effects(3) negligible work done by static pressure force(4) negligible work done by viscous normal and shear forces(5) negligible work done by gravitational force(6) negligible radiative heat transfer

Fig.5.2-1 Enthalpy balance for convective and conductive transport
1.

Bird, R.B., Stewart, W.E., and Lightfoot, E.N., Transport Phenomena, Wiley, New York, Chapt.10 (1960)

Let us apply the principle of energy conservation to the differential control volume ∆𝑥∆𝑦∆𝑧
shown in Fig.5.2-1. The thermal energy can be written in a form of enthalpy 𝜌𝐶𝑝𝑇 for fluids
having constant properties.
The accumulation rate of thermal energy is(𝜌𝐶𝑝𝑇)|𝑡+∆𝑡 − (𝜌𝐶𝑝𝑇)|𝑡∆𝑥∆𝑦∆𝑧(5.2-1)∆𝑡
In general, thermal energy may be transferred into the control volume by convection andconduction (diffusion).
(1) Convective transport:
First let us consider the convective effect. The x-component of velocity 𝑣𝑥 brings in thermalenergy 𝜌𝐶𝑝𝑇 across the face ∆𝑦∆𝑧 at a rate of (𝜌𝐶𝑝𝑇)|𝑥 𝑣𝑥 |𝑥 ∆𝑦∆𝑧. Similarly the rate ofthermal energy brought out is (𝜌𝐶𝑝𝑇)|𝑥+∆𝑥 𝑣𝑥 |𝑥+∆𝑥 ∆𝑦∆𝑧. Therefore the net rate of thermalenergy transferred across all six faces by convection becomesDifferential Balances of Mass

43[(𝜌𝐶𝑝𝑇𝑣𝑥 )|𝑥 − (𝜌𝐶𝑝𝑇𝑣𝑥 )|𝑥+∆𝑥 ]∆𝑦∆𝑧 + [(𝜌𝐶𝑝𝑇𝑣𝑦 )| − (𝜌𝐶𝑝𝑇𝑣𝑦 )|] ∆𝑧∆𝑥 + [(𝜌𝐶𝑝𝑇𝑣𝑧 )|𝑧 −𝑦𝑦+∆𝑦(5.2-2)(𝜌𝐶𝑝𝑇𝑣𝑧 )|𝑧+∆𝑧 ]∆𝑥∆𝑦(2) Conductive transport:
Next let us consider the conductive effect. The rate of thermal energy entering the face ∆𝑦∆𝑧
at 𝑥 by conduction is 𝑞𝑥 |𝑥 ∆𝑦∆𝑧, and the rate of thermal energy leaving the face ∆𝑦∆𝑧 at
𝑥 + ∆𝑥 is 𝑞𝑥 |𝑥+∆𝑥 ∆𝑦∆𝑧. Note that 𝑞𝑥 is the heat flux through a face perpendicular to thex-axis.
Therefore the net rate of thermal energy transferred across all six faces by conduction becomes[𝑞𝑥 |𝑥 − 𝑞𝑥 |𝑥+∆𝑥 ]∆𝑦∆𝑧 + [𝑞𝑦 | − 𝑞𝑦 |] ∆𝑧∆𝑥 + [𝑞𝑧 |𝑧 − 𝑞𝑧 |𝑧+∆𝑧 ]∆𝑥∆𝑦(5.2-3)𝑦𝑦+∆𝑦After substituting all the foregoing contributions into the equation of energy conservation law,and dividing through the resulting equation by ∆𝑥∆𝑦∆𝑧 , if the limit is taken as∆𝑡, ∆𝑥, ∆𝑦, ∆𝑧 → 0, we get𝜕(𝜌𝐶𝑝𝑇)𝜕𝑡= −[𝜕(𝜌𝐶𝑝𝑇𝑣𝑥 )𝜕𝑥+𝜕(𝜌𝐶𝑝𝑇𝑣𝑦 )𝜕𝑦+𝜕(𝜌𝐶𝑝𝑇𝑣𝑧 )
𝜕𝑧

𝜕𝑞𝑥

]− (

𝜕𝑥

𝜕𝑞𝑦

𝜕𝑞

+ 𝜕𝑦 + 𝜕𝑧𝑧 )(5.2-4)In vector symbolism
𝜕(𝜌𝐶𝑝𝑇)⃗ ∙ 𝜌𝐶𝑝𝑇 ⃗𝒗) + (∇⃗ ∙ ⃗𝒒) = 𝟎+ (∇(5.2-5)𝜕𝑡
The vector equation is also valid for cylindrical coordinates.
The above equation is rewritten for incompressible fluid with the aid of the equation ofcontinuity and the Fourier’s law of conduction as

𝜌𝐶𝑝 (

𝜕𝑇
𝜕𝑡

+ 𝑣𝑥

𝜕𝑇
𝜕𝑥

+ 𝑣𝑦

𝜕𝑇
𝜕𝑦

+ 𝑣𝑧

In vector symbolism
𝜕𝑇⃗ )𝑻] = 𝜅 ∇2 𝑇⃗ ∙𝛁𝜌𝐶𝑝 [ 𝜕𝑡 + (𝒗𝜕𝑇𝜕𝑧)= 𝜅 (

𝜕2 𝑇

𝜕2 𝑇

𝜕2 𝑇

𝜕𝑥

𝜕𝑦

𝜕𝑧 2

+
2

+
2

(5.2-6))(5.2-7)The vector equation is valid for all kinds of orthogonal coordinate systems. These equations
are listed in Table 5.2-1, except for spherical coordinates. Note that since the rate of energy
transfer depends on the fluid velocity, it is necessary to solve the hydrodynamic problem
before the temperature distribution is calculated.
Table 5.2-1 Equation of energy for a Newtonian fluid of constant 𝜌 and 𝜅(1) rectangular coordinates
2
2
2
𝜕𝑇
𝜕𝑇
𝜕𝑇
𝜕𝑇
𝜕 𝑇 𝜕 𝑇 𝜕 𝑇𝜌𝐶𝑝 (+ 𝑣𝑥 + 𝑣 𝑦 + 𝑣 𝑧 ) = 𝜅 ( 2 +
+
)
2
2
𝜕𝑡
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑥
𝜕𝑦
𝜕𝑧
(2) cylindrical coordinates22
𝜕𝑇𝜕𝑇 𝑣 𝜕𝑇
𝜕𝑇1𝜕
𝜕𝑇1𝜕 𝑇 𝜕 𝑇𝜌𝐶𝑝 (+ 𝑣𝑟 + 𝜃 + 𝑣𝑧 ) = 𝜅 (
(𝑟
)+ 2
+
)
𝑟 𝜕𝜃
𝑟 𝜕𝑟
𝜕𝑡
𝜕𝑟
𝜕𝑟
𝜕𝑧
𝑟 𝜕𝜃2 𝜕𝑧2

5.3 Differential Balances of Mass (Equations of mass transport)1)
In this section, we shall develop the equation of mass A in a binary mixture A and B setting up themass balance over the same differential control volume. For convenience, we use the following formof Fick’s law written in terms of mass concentration in place of molar concentration:𝜕𝜔𝑛𝐴𝑥 = 𝜔𝐴(𝑛𝐴𝑥 + 𝑛𝐵𝑥 ) − 𝜌 𝐷𝐴𝐵 𝜕𝑥𝐴(5.3-1)
A/m3,

where 𝜔𝐴 = 𝜌𝐴 ⁄𝜌 = mass fraction, 𝜌𝐴 = mass density of component A in kg of
𝜌 = total
mass density in kg of (A+B)/m3, and 𝑛𝐴𝑥 = x-component of mass flux of species A in kg of A/m2/s.
This equation indicates that the mass flux in a stationary coordinate system consists of bothconvection (first term) and diffusion (second term).

44

Microscopic Differential Balances

The accumulation rate of species A is𝜌𝐴 |𝑡+∆𝑡 − 𝜌𝐴 |𝑡∆𝑥∆𝑦∆𝑧∆𝑡
The net rate of species A input due to both convection and diffusion is𝑛𝐴𝑦 |𝑦+∆𝑦 − 𝑛𝐴𝑦 |𝑦𝑛𝐴𝑥 |𝑥+∆𝑥 − 𝑛𝐴𝑥 |𝑥𝑛𝐴𝑧 |𝑧+∆𝑧 − 𝑛𝐴𝑧 |𝑧−[++] ∆𝑥∆𝑦∆𝑧∆𝑥∆𝑦∆𝑧
The rate of generation of mass A by chemical reaction is𝑟𝐴 ∆𝑥∆𝑦∆𝑧
Summing up all the contributions divided by ∆𝑥∆𝑦∆𝑧 and taking the limit as ∆𝑡, ∆𝑥, ∆𝑦, ∆𝑧 → 0,
we get𝜕𝜌𝐴𝜕𝑛= − ( 𝜕𝑥𝐴𝑥 +
𝜕𝑡

𝜕𝑛𝐴𝑦
𝜕𝑦

+

𝜕𝑛𝐴𝑧
𝜕𝑧

(5.3-2)) + 𝑟𝐴This is a general form of the equation of continuity for species A. In vector symbolism= − ∇ ∙ ⃗⃗⃗⃗⃗𝒏𝑨 + 𝑟𝐴(5.3-3)𝜕𝑡
The mass flux vector for a binary mixture can be related with fluid velocity:⃗𝑛⃗𝐴 + 𝑛⃗𝐵 = 𝜌 𝒗(5.3-4)
We shall get more convenient form of the equation using the following form of Fick’s law.
⃗⃗⃗⃗𝑛𝐴 = 𝜔𝐴 (𝑛⃗𝐴 + 𝑛⃗𝐵 ) − 𝜌 𝐷𝐴𝐵 ∇𝜔𝐴 = 𝜌𝐴 ⃗𝒗 − 𝜌𝐷𝐴𝐵 ∇𝜔𝐴
(5.3-5)
Substituting this expression into Eq.(5.3-3), we get𝜕𝜌𝐴⃗ ∙ ∇)𝜌𝐴 + 𝜌𝐷𝐴𝐵 ∇2𝜔𝐴 + 𝑟𝐴
= − 𝜌𝐴 (∇ ∙ ⃗𝒗) − (𝒗𝜕𝑡That is, in rectangular coordinates
𝜕𝜌 𝑣
𝜕𝜌𝐴
𝜕𝜌 𝑣
𝜕𝜌 𝑣
= − [ 𝐴 𝑥 + 𝐴 𝑦 + 𝐴 𝑧]
𝜕𝜌𝐴

𝜕𝑡

+[

𝜕

𝜕𝑥

𝜕𝑥
𝜕𝜔𝐴

(𝜌𝐷𝐴𝐵

𝜕𝑥

)+

𝜕𝑦
𝜕

𝜕𝑦

𝜕𝑧
𝜕𝜔𝐴

(𝜌𝐷𝐴𝐵

𝜕𝑦

)+

𝜕
𝜕𝑧

(𝜌𝐷𝐴𝐵

𝜕𝜔𝐴
𝜕𝑧

)] + 𝑟𝐴

For a fluid with constant 𝜌 and 𝐷𝐴𝐵 this equation can be simplified as follows.
𝜕𝜌𝐴⃗ ) + (𝒗⃗ ∙ ∇)𝜌𝐴 = 𝐷𝐴𝐵 ∇2𝜌𝐴 + 𝑟𝐴
+ 𝜌𝐴 (∇ ∙ 𝒗
𝜕𝑡
According to the equation of continuity for constant 𝜌, (∇ ∙ ⃗𝒗) = 0. Therefore𝜕𝜌𝐴⃗ ∙ ∇)𝜌𝐴 = 𝐷𝐴𝐵 ∇2 𝜌𝐴 + 𝑟𝐴
+ (𝒗(5.3-6)𝜕𝑡
Dividing through by molecular weight 𝑀𝐴 , we get𝜕𝐶𝐴⃗ ∙ ∇)𝐶𝐴 = 𝐷𝐴𝐵 ∇2 𝐶𝐴 + 𝑅𝐴
+ (𝒗(5.3-7)𝜕𝑡
where 𝐶𝐴 = 𝜌𝐴 ⁄𝑀𝐴 in kmol of A/m3 and 𝑅𝐴 = 𝑟𝐴 ⁄𝑀𝐴 in kmol of A/m3/s.
This is the diffusion equation for dilute solution at constant temperature and pressure where wecan assume constant density and diffusivity. The equation of continuity for species A written interms of molar fluxes can be obtained dividing Eq.(5.3-2) by molecular weight 𝑀𝐴 ,:
𝜕𝑁𝐴𝑦
𝜕𝐶𝐴
𝜕𝑁
𝜕𝑁
= − ( 𝐴𝑥 ++ 𝐴𝑧 ) + 𝑅𝐴
𝜕𝑡
𝜕𝑥
𝜕𝑦
𝜕𝑧

(5.3-8)

or

𝜕𝐶𝐴
𝜕𝑡

⃗ 𝐴 ) = 𝑅𝐴
+ (∇ ∙ 𝑁

The following tables give the diffusion equations in orthogonal coordinate systems.
Table 5.3 – 1 The equation of continuity of component A1)(1) rectangular coordinates (x, y, z)
𝜕𝑁
𝜕𝐶𝐴
𝜕𝑁
𝜕𝑁
+ ( 𝐴𝑥 + 𝐴𝑦 + 𝐴𝑧 ) = 𝑅𝐴
𝜕𝑡

𝜕𝑥

𝜕𝑦

𝜕𝑧

(2) cylindrical coordinates (𝑟, 𝜃, 𝑧)𝜕𝐶𝐴1 𝜕
1 𝜕𝑁𝐴𝜃
𝜕2 𝑁𝐴𝑧(𝑟 𝑁𝐴𝑟 ) ++ [+] = 𝑅𝐴
2
𝜕𝑡

𝑟 𝜕𝑟

𝑟

𝜕𝜃

𝜕𝑧

Nomenclature

45

Table 5.3 – 2 The equation of diffusion for constant 𝜌 and 𝐷𝐴𝐵 1)(1) rectangular coordinates (x, y, z)
𝜕𝐶𝐴
𝜕𝐶
𝜕𝐶
𝜕𝐶
+ 𝑣𝑥 𝐴 + 𝑣𝑦 𝐴 + 𝑣𝑧 𝐴
𝜕𝑡
𝜕𝑥
𝜕𝑦
𝜕𝑧

= 𝐷𝐴𝐵 (

𝜕 2 𝐶𝐴𝜕 2 𝐶𝐴𝜕 2 𝐶𝐴++) + 𝑅𝐴
𝜕𝑥 2
𝜕𝑦 2
𝜕𝑧 2

(2) cylindrical coordinates (𝑟, 𝜃, 𝑧)
𝜕𝐶𝐴
𝜕𝑇
𝑣 𝜕𝐶
𝜕𝐶
+ 𝑣𝑟 + 𝜃 𝐴 + 𝑣𝑧 𝐴
𝜕𝑡
𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑧

1.

= 𝐷𝐴𝐵 (1𝜕𝜕𝐶
1 𝜕2 𝐶𝜕 2 𝐶𝐴(𝑟 𝐴) + 2 2𝐴 +) + 𝑅𝐴
𝑟 𝜕𝑟
𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑧 2

Bird, R.B., Stewart, W.E., and Lightfoot, E.N., Transport Phenomena, Wiley, New York, Chapt.18 (1960)

Nomenclature
𝐶𝑝
heat capacity, [J/kg K]
𝑔gravitational acceleration, [m/s2]
𝑁𝐴𝑥 , 𝑁𝐴𝑦 , 𝑁𝑛𝐴𝑧
molar flux in rectangular coordinates, [kmol/m2s]
𝑛𝐴𝑥 , 𝑛𝐴𝑦 , 𝑛𝐴𝑧
mass flux in rectangular coordinates, [kg/m2s]𝑝pressure, [Pa]𝑞𝑟 , 𝑞𝜃 , 𝑞𝑧 heat flux in cylindrical coordinates, [J/m2s]
𝑞𝑥 , 𝑞𝑦 , 𝑞𝑧 heat flux in rectangular coordinates, [J/m2s]
𝑟𝐴 , 𝑅𝐴
reaction rate, [kg/m3s], [kmol/m3s]𝑟, 𝜃, 𝑧cylindrical coordinates, [m, - , m]𝑇
temperature, [K]𝑡time, [s]𝑣𝑟 , 𝑣𝜃 , 𝑣𝑧 velocity component in cylindrical coordinates
𝑣𝑥 , 𝑣𝑦 , 𝑣𝑧 velocity component in rectangular coordinates, [m/s]𝑥, 𝑦, 𝑧,rectangular coordinates, [m]𝜇viscosity, [kg/m s]𝜌density, [kg/m3]
𝜏𝑤wall shear stress, [N/m2]
𝜏𝑥𝑥 , 𝜏𝑦𝑥 , 𝜏𝑧𝑥 , − −momentum flux or shear stress, [N/m2] or [kg/s2m]

CHAPTER 6APPLICATION OF DIFFERENTIAL TRANSPORTEQUATIONS
6.1

Application of the Equation of Motion (Ⅰ)

In this section, we shall study how to use the differential equation of fluid motion.
One of the most useful velocity profiles is for flow in a circular pipe.

Fig.6.1-1 Flow development in the entrance region of a circular pipe

Generally speaking, it takes a certain length 𝐿𝑒𝑛𝑡 from the entrance until the velocity profile
becomes fully developed. At the entrance to the pipe as shown in Fig.6.1-1, the velocity profile isalmost uniform like a plug owing to the sudden contraction of flow area. As the fluid moves down
the pipe, a boundary layer of low-velocity fluid grows by deceleration due to the viscous force on
the wall surface. After the edge of the boundary layer coincides with the pipe axis, the downstream
velocity profile no longer varies with axial length 𝑧. We shall speak of the fully developed velocity
profile as the fixed velocity distribution in the fully developed region. When the Reynolds number
𝑅𝑒 is less than 2,100, the flow remains laminar and the fully-developed velocity profile is parabolic.
Let us consider the laminar incompressible flow inside an inclined pipe of length 𝐿 and radius 𝑅.
The pressure drop over the length 𝐿 is given by (𝑝0 − 𝑝𝐿 ).
What coordinate system is the most appropriate for this flow system? The cylindrical coordinates are the mostconvenient to describe positions in a pipe.

(1) What component of momentum (or the equation of motion) should be considered? We canassume only z-momentum to be important in laminar flow.
Therefore z-component of the equation of motion for constant 𝜌 and 𝜇 may be written
𝜕𝑣

𝜕𝑣

𝑣 𝜕𝑣

𝜕𝑣

𝜕𝑝

1𝜕

𝜕𝑣

1 𝜕2 𝑣

𝜕2 𝑣

𝜌 ( 𝜕𝑡𝑧 + 𝑣𝑟 𝜕𝑟𝑧 + 𝑟𝜃 𝜕𝜃𝑧 + 𝑣𝑧 𝜕𝑧𝑧 ) = − 𝜕𝑧 + 𝜇 [𝑟 𝜕𝑟 (𝑟 𝜕𝑟𝑧 ) + 𝑟 2 𝜕𝜃2𝑧 + 𝜕𝑧 2𝑧] + 𝜌 𝑔𝑧
(6.1-1)

Application of the Equation of Motion (I)(continued)

47

Fig.6.1-2 Laminar flow in a fully-developed region of circular pipe

(2) The next step is simplification of the above equation to fit the problem at hand. We
can easily imagine that in steady state laminar flow the fluid particles move along𝜕𝑣𝑧straight streamlines parallel to the pipe axis. Therefore
𝜕𝑣

The flow can be assumed axisymmetric, i.e. 𝜕𝜃𝑧 =𝜕𝑣

𝜕 2 𝑣𝑧𝜕𝜃2𝜕𝑡= 0 and 𝑣𝜃 = 𝑣𝑟 = 0.

=0

𝜕2 𝑣

The flow is fully developed: 𝜕𝑧𝑧 = 𝜕𝑧 2𝑧 = 0

𝜕

𝜕𝑣

𝑑

𝑑𝑣

Therefore we can expect 𝑣𝑧 to be a function of 𝑟 only: 𝜕𝑟 (𝑟 𝜕𝑟𝑧 ) = 𝑑𝑟 (𝑟 𝑑𝑟𝑧 )This type of internal flow is caused by the combined effect of pressure andgravitational forces. We assume the static pressure to be uniform over the pipe crosssection: 𝑝 = 𝑝(𝑧).
The z-component of the gravitational acceleration is given by 𝑔𝑧 = 𝑔 cos 𝛽.
Then the equation of motion becomes
𝑑𝑝
1 𝑑
𝑑𝑣
0 = − 𝑑𝑧 + 𝜇 𝑟 𝑑𝑟 (𝑟 𝑑𝑟𝑧 ) + 𝜌 𝑔 cos 𝛽(6.1-2)𝑑2 𝑝

Differentiate with respect to 𝑧, we get 𝑑𝑧 2 = 0.

𝑑𝑝

𝑑𝑝

If we integrate with respect to 𝑧, we can find 𝑑𝑧 = 𝑐𝑜𝑛𝑠𝑡. Thus 𝑑𝑧 =We get the second-order ordinary differential equation for 𝑣𝑧 :
1 𝑑
𝑑𝑣
𝑝 −𝑝
𝑷 −𝑷
𝜇 𝑟 𝑑𝑟 (𝑟 𝑑𝑟𝑧 ) = − 0 𝐿 𝐿 − 𝜌 𝑔 cos 𝛽 = − 𝟎 𝐿 𝑳

𝑝𝐿 − 𝑝0
𝐿(6.1-3)

Here the effective pressure 𝑷 is defined as 𝑷 = 𝑝 − 𝜌 𝑔 𝑧 𝑐𝑜𝑠 𝛽. If the reference plane at𝑧 = 0 is chosen at the position 1: 𝑷𝟎 = 𝑝0 and 𝑷𝐿 = 𝑝𝐿 − 𝜌 𝑔 𝐿 𝑐𝑜𝑠 𝛽.
(3) Integration of the above differential equation gives
𝑑𝑣𝑧
𝑑𝑟

= −

𝑷𝟎 − 𝑷𝑳
2𝜇𝐿

𝑟+

𝐶1
𝑟

(6.1-4)

We can use the boundary condition at the pipe axis to evaluate the integration constant 𝐶1 :𝑑𝑣𝑧
B.C.1 at 𝑟 = 0=0(6.1-5)𝑑𝑟(This boundary condition states that the velocity profile becomes flat at the center of the pipe.)

48

Application of Differential Transport Equations

Then 𝐶1 must be zero. Integration gives
𝑷 −𝑷
𝑣𝑧 = − 𝟎4𝜇𝐿 𝑳 𝑟 2 + 𝐶2

(6.1-6)The other boundary condition is𝑣𝑧 = 0
B.C.2 at 𝑟 = 𝑅(This boundary condition states that the fluid does not slip on the wall surface.)
Then 𝐶2 must be
𝑷 −𝑷
𝐶2 = 𝟎4𝜇𝐿 𝑳 𝑅2

(6.1-7)(6.1-8)Substituting this expression into E.(6.1-6), we get the following equation of parabolic velocity
distribution:𝑣𝑧 =𝑷 𝟎 − 𝑷𝑳4𝜇𝐿

𝑟 2

(6.1-9)𝑅2 [1 − ( ) ]𝑅This equation indicates that the maximum velocity occurs at the center of the pipe:
𝑷 −𝑷𝑣𝑧,𝑚𝑎𝑥 = 𝟎4𝜇𝐿 𝑳 𝑅2

Fig. 6.1-3.
pipe flow(6.1-10)

Parabolic distribution of velocity and linear distribution of shear stress in a laminar(4) In general, the average velocity taken over the flow cross section is defined as thearea-averaged value: 〈𝑣 〉 =∫𝐴 𝑣 𝑑𝐴∫𝐴 𝑑𝐴.

For axisymmetric flow in a circular pipe,

substituting the velocity distribution function into the integration, we get𝑅〈𝑣𝑧 〉 =

∫0 𝑣𝑧 2𝜋𝑟 𝑑𝑟𝑅∫0 2𝜋𝑟 𝑑𝑟=

𝑷𝟎 − 𝑷𝑳
8𝜇𝐿

𝑅2(6.1-11)

Note that the average velocity for laminar flow is just half of the maximum velocity.
(5) The volumetric flow rate can be calculated𝑅
𝜋(𝑷𝟎 − 𝑷𝑳 ) 4
𝑄 = ∫0 𝑣𝑧 2𝜋𝑟 𝑑𝑟 =𝑅
(6.1-12)8𝜇𝐿
This equation is known as the Hagen-Poiseuille equation which gives the relation
between the volumetric flow rate and the effective pressure drop. If we express theequation in terms of the average velocity or the Reynolds number

Application of the Equation of Motion (II)49
16 𝐿 1𝑷𝟎 − 𝑷𝑳 = 8𝜇𝐿 〈𝑣𝑧 〉⁄𝑅2 or 𝑷𝟎 − 𝑷𝑳 = 4 𝑅𝑒 𝐷 2 𝜌〈𝑣𝑧 〉2(6.1-13)This result gives a very important designing formula predicting the pressure drop
in a laminar pipe flow.
(6) The shear stress distribution 𝜏𝑟𝑧 may now be obtained with the aid of Newton’s lawof viscosity:𝜕𝑣𝑧 𝜕𝑣𝑟𝜏𝑟𝑧 = − 𝜇 (
+
)
𝜕𝑟
𝜕𝑧

In this case 𝑣𝑟 = 0 and 𝑣𝑧 = 𝑣𝑧 (𝑟). Therefore𝑑𝑣𝑧
𝜏𝑟𝑧 = − 𝜇𝑑𝑟Substitute the velocity distribution into the differentiation and get
𝑷 −𝑷𝜏𝑟𝑧 = 𝟎2𝐿 𝑳 𝑟(6.1-14)
Notice that the shear stress (the momentum flux) has the linear distribution andthe maximum value occurs at the wall of the pipe.
𝑷 −𝑷𝜏𝑟𝑧,𝑚𝑎𝑥 = 𝜏𝑤 = 𝟎2𝐿 𝑳 𝑅(6.1-15)(7) For circular pipe flow, we very frequently use the following definition of friction
factor called the Fanning friction factor:
𝐿 1
𝑃0 − 𝑃𝐿 = 4 𝑓 𝐷 2 𝜌〈𝑣𝑧 〉2(6.1-16)
The left side term is usually considered to be static pressure drop without thegravitational effect.
Eq.(6.1-13) indicates that for laminar flow inside a circular pipe the friction factor isgiven by16
𝑓 = 𝑅𝑒(6.1-17)[PROBLEM 6.1-P1] A viscous Newtonian fluid (density 𝜌 = 1,050 𝑘𝑔⁄𝑚3 , viscosity𝜇 =1.0 𝑝𝑜𝑖𝑠𝑒 = 0.10 𝑘𝑔⁄𝑚 𝑠) is transported through a horizontal 30-mm-ID circular pipe from astorage tank to an agitated tank reactor shown below. The volumetric flow rate is
2 × 10−4 𝑚3 ⁄𝑠. The effective length of the pipeline including the contribution from both ends
and a valve is 50 m. Determine the pressure drop between both ends of the pipeline. Calculate thepower of the pump.

Fig.6.1-P1 Design of feed-supplying pipeline for an agitated tank reactor

6.2 Application of the Equation of Motion (Ⅱ)
In this section, using the equation of motion in cylindrical coordinate system, we shall consider
shear flow of an incompressible Newtonian fluid in an annulus between two coaxial vertical long

50

Application of Differential Transport Equations

cylinders, the outer one of which is rotating with a constant angular velocity 𝜔. In steady state
there exists stable rotating flow.
Let us determine the velocity distribution neglecting end effects.
The 𝜃-component equation of motion is the most appropriate for this problem:
𝜕𝑣
𝜕𝑣
𝑣 𝜕𝑣
𝑣 𝑣
𝜕𝑣
1 𝜕𝑝
𝜌 ( 𝜕𝑡𝜃 + 𝑣𝑟 𝜕𝑟𝜃 + 𝑟𝜃 𝜕𝜃𝜃 + 𝑟𝑟 𝜃 + 𝑣𝑧 𝜕𝑧𝜃 ) = − 𝑟 𝜕𝜃𝜕

1 𝜕2 𝑣

1𝜕

2 𝜕𝑣

𝜕2 𝑣

+ 𝜇 [𝜕𝑟 (𝑟 𝜕𝑟 (𝑟𝑣𝜃 )) + 𝑟 2 𝜕𝜃2𝜃 + 𝑟 2 𝜕𝜃𝑟 + 𝜕𝑧2𝜃 ] + 𝜌 𝑔𝜃 

(6.2-1)All streamlines will form circles coaxial about the axis of rotation. Then the velocity
components 𝑣𝑟 and 𝑣𝑧 are zero and the peripheral velocity 𝑣𝜃 is a function of 𝑟 alone
because of axi-symmetry. Since these cylinders are placed vertically, 𝑔𝜃 = 0.

Fig.6.2-1 Laminar shear flow in an annular space between two concentric cylinders when theouter cylinder is rotating.

The 𝜃-component equation of motion then reduces to a second-order ordinary differential equationfor 𝑣𝜃 :𝑑
1 𝑑
0 = 𝑑𝑟 (𝑟 𝑑𝑟 (𝑟 𝑣𝜃 ))(6.2-2)Integration with respect to 𝑟 twice gives
𝐶𝐶𝑣𝜃 = 1 𝑟 + 2(6.2-3)2𝑟
We now have two integration constants to be evaluated by two boundary conditions:
B.C.1 at 𝑟 = 𝑅𝑖𝑣𝜃 = 0
B.C.2 at 𝑟 = 𝑅𝑜𝑣𝜃 = 𝑅𝑜 𝜔
We obtain
𝐶1

𝑅 2

𝑅𝜔

= 𝑅 −(𝑅𝑖 2⁄𝑅 ) = 𝑅 2 −𝑜 𝑅 2 𝜔
2𝑜

𝐶2 = −

𝑖

𝑜

𝑜

𝑖

𝑅𝑖 𝜔
(1⁄𝑅𝑖 )−(𝑅𝑖 ⁄𝑅𝑜 2)= −𝑅𝑖 2 𝑅𝑜 2
𝑅𝑜 2 − 𝑅𝑖 2

𝜔Then we get the hyperbolic velocity distribution:

𝑣𝜃 =

𝑅𝑖 𝑅𝑜
𝑅𝑜 2 − 𝑅𝑖 2

𝑟

𝑅𝑖

𝑅𝑖

𝑟

𝑅𝑜 𝜔 [ −](6.2-4)

According to Newton’s law of viscosity, the shear stress 𝜏𝑟𝜃 can be written as
𝜕 𝑣
1 𝜕𝑣
𝜏𝑟𝜃 = − μ [𝑟 𝜕𝑟 ( 𝑟𝜃 ) + 𝑟 𝜕𝜃𝑟 ](6.2-5)In this flow system𝑑 𝑣
𝜏𝑟𝜃 = − μ 𝑟 ( 𝜃 )𝑑𝑟(6.2-6)𝑟

Substituting the velocity distribution we get
𝑅 2𝑅 21

𝜏𝑟𝜃 = −2𝜇𝜔 𝑅 𝑖2 − 𝑜𝑅 2 𝑟 2𝑜𝑖(6.2-7)

Application of the Equation of Motion (II)(continued)

51

This equation indicates that 𝜏𝑟𝜃 is always negative, and that the 𝜃 -component of
momentum transfers in the negative r-direction (inwards).
We may need information about the torque required to turn the shaft of the outer cylinder. Recall⃗⃗ is a moment, that is, a vector product of force 𝐅 and lever arm 𝐫 from the center
that a torque 𝐓of rotation. In this system, the lever arm is the radius of the outer cylinder and the force is the shear
force acting on the inner surface of the outer cylinder:2⃗⃗ | = (2𝜋𝑅𝑜 𝐿)(−𝜏𝑟𝜃 )|𝑟=𝑅 ∙ 𝑅𝑜 = 4𝜋𝜇𝐿𝑅𝑜 2 𝜔2 2𝑅𝑖 2
𝑇𝑞 = |𝐓𝑜22= 8𝜋 𝜇𝐿𝑅𝑜 𝑁𝑟

𝑅𝑖 22
𝑅𝑜 − 𝑅𝑖 2

𝑅𝑜 − 𝑅𝑖(6.2-8)This result implies that the torque is proportional to the number of rotation 𝑁𝑟 (1/s) and theconstant of proportionality is the viscosity of the fluid. This is a principle of Couette-Hatschek
viscometers.
Laminar flow in this system is strongly stabilized by centrifugal forces. Thus a fluid particle near
the outer cylinder opposes being moved inwards since the centrifugal force acting on it is greaterthan on particle nearer the inner cylinder. As a result, the flow can remain laminar up to very highReynolds number.

Fig.6.2-2. Pairs of cellular vortices between two concentric cylinders when the inner cylinder isrotating.

On the other hand, it might be very interesting to consider the corresponding flow system in which
the inner cylinder is rotating and the outer one is at rest. The fluid velocity increases withdecreasing the radial distance. Hence the centrifugal forces tend to introduce instability.
We use a modified Reynolds number for the system:𝑇𝑎 =𝜌𝑅𝑖 𝜔𝑖 𝑑
𝜇

𝑑

√𝑅

𝑖

(6.2-9)This dimensionless parameter is sometimes called Taylor number in honor of the pioneer 1) in fluid
mechanics of this system. The curvature effect (centrifugal effect) can be taken into accountintroducing √𝑑 ⁄𝑅𝑖 . When the Taylor number exceeds a certain critical value (i.e. 𝑇𝑎𝑐𝑟 = 41.2 forvery small gap between two coaxial cylinders), there appear ring doughnut-shaped counter-rotating
cellular vortices arrayed axially in the annular space due to the centrifugal force. The critical Taylor
number for the onset of the cellular vortex flow depends slightly on the geometry 𝑑 ⁄𝑅𝑖 . For
example, 𝑇𝑎𝑐𝑟 = 51.4 for 𝑑 ⁄𝑅𝑖 = 1⁄3 . This vortex flow is not turbulent, but an elaborate
laminar flow with toroidal motion. After successive transitions, the purely turbulent flow can be

52

Application of Differential Transport Equations

attained at a high Taylor number. (e.g. 𝑇𝑎 > 𝑎𝑏𝑜𝑢𝑡 15,000 for relatively large gaps 𝑑 ⁄𝑅𝑖 = 0.62
to 1.14).
This flow system has a very interesting instability scenario 2) with increasing Taylor number,beginning from the Taylor instability (laminar Taylor vortex flow) via. two sequential wave
instabilities (wavy vortex flow, and amplitude-modulated wavy vortex flow) to the final transition
to chaotic turbulent flow. Local time-dependent mass transfer controlled by the axial array ofcellular vortices was observed on the inner wall of the stationary outer cylinder 2, 3) by using a
diffusion-controlled electrolytic reaction. (see Chapter 14)
1.
2.
3.

Taylor, G.I., Phil. Trans., A223, 289m (1923), Proc. Roy. Soc. (London), A151, 494 (1935)0
Kataoka, K., “Taylor vortices and instabilities in circular Couette flows,” Encyclopedia of Fluid Mechanics, Gulf Pub., Houston,ed. by N. P. Cheremisinoff, Chap. 9, pp.236 – 274 (1986)
Kataoka, K., Doi, H., and Komai, T., Int. J. Heat Mass Transfer, 20, 50 (1977)

6.3 Application of the Equation of Motion (Ⅲ)
for non-Newtonian Fluid
We read in section 2.5 that there are some particular fluids which cannot be
described by Newton’s law of viscosity. For reference, we shall study the Bingham
plastic fluid flow in a horizontal circular pipe.
The pressure drop over the length 𝐿 is given by (𝑝0 − 𝑝𝐿 )⁄𝐿.
In cylindrical coordinates, a simplified equation of the Bingham model is given by𝑑𝑣𝜏𝑟𝑧 = − 𝜇0 𝑑𝑟𝑧 + 𝜏0
if 𝜏𝑟𝑧 > 𝜏0
(6.3-1a)𝑑𝑣𝑧=0
if 𝜏𝑟𝑧 < 𝜏0
(6.3-1b)
In general, the z-component equation of motion (for a horizontal pipe flow 𝑔𝑧 = 0)
becomes in cylindrical coordinate system as follows:
𝑑𝑟

𝜌(

𝜕𝑣𝑧
𝜕𝑡

+ 𝑣𝑟

𝜕𝑣𝑧
𝜕𝑟

+

𝑣𝜃 𝜕𝑣𝑧
𝑟 𝜕𝜃

+ 𝑣𝑧

𝜕𝑣𝑧
𝜕𝑧

)= −

𝜕𝑝
𝜕𝑧

− (

1 𝜕
𝑟 𝜕𝑟

(𝑟 𝜏𝑟𝑧 ) +1 𝜕𝜏𝜃𝑧
𝑟

𝜕𝜃

+

𝜕𝜏𝑧𝑧
𝜕𝑧

)

(6.3-2)

As in section 6.1, assuming incompressible laminar flow with 𝑣𝜃 = 𝑣𝑟 = 0 , theequation of motion is simplified as
1 𝑑𝑝 −𝑝(𝑟 𝜏𝑟𝑧 ) = 0 𝐿(6.3-3)𝑟 𝑑𝑟
𝐿Integration with respect to 𝑟 gives𝑝 −𝑝𝐶𝜏𝑟𝑧 = 02𝐿 𝐿 𝑟 + 𝑟1
The shear stress can be expressed as𝜏𝑟𝑧 = − 𝜇0(6.3-4)𝑑𝑣𝑧
𝑝0 − 𝑝𝐿
𝐶1+ 𝜏0 =
𝑟+
𝑑𝑟2𝐿
𝑟

B.C.1 𝜏𝑟𝑧 = 𝜏0 at 𝑟 = 𝑟0
Therefore we can evaluate the integration constant as
𝑝 −𝑝
𝐶1 = 𝑟0 𝜏0 − 02𝐿 𝐿 𝑟0 2
In other words, this implies that the velocity gradient becomes zero at 𝑟 = 𝑟0 .
Then the following equation is obtained:𝑑𝑣𝑧
𝑝0 − 𝑝𝐿
𝜏0
𝐶1 1
= −
𝑟+
−
𝑑𝑟
2𝜇0 𝐿
𝜇0 𝜇0 𝑟Integration gives
𝑣𝑧 = −

𝑝0 − 𝑝𝐿 𝑟 2
2𝜇0 𝐿 2𝜏𝐶

+ 𝜇0 − 𝜇1 ln 𝑟 + 𝐶2
00

(6.3-5)

Application of the Equation of Motion (IV) for non-Newtonian Fluid

53

We have to solve the above ordinary differential equation by using the following
boundary conditions:
B.C.2 at 𝑟 = 𝑅
𝑣𝑧 = 0(no slip condition)(6.3-6)
Therefore𝑝 − 𝑝 𝑅2

𝜏𝐶

(6.3-7)0𝐿𝐶2 = 2𝜇+ 𝜇0 − 𝜇1 ln 𝑅
𝐿 2
0

0

0

Finally the velocity distribution can be expressed as𝑣𝑧 = −
𝑣𝑧 = −

𝑝0 − 𝑝𝐿 𝑟 2

2𝜇0 𝐿

𝜏

𝐶

+ 𝜇0 − 𝜇1 ln 𝑟 + 𝐶2

2𝜇0 𝐿 2𝑝0 − 𝑝𝐿 𝑟02
2

0

0

𝜏

𝐶

𝑟0 < 𝑟 < 𝑅+ 𝜇0 − 𝜇1 ln 𝑟0 + 𝐶2 = 𝑐𝑜𝑛𝑠𝑡
0

where 𝐶1 = 𝑟0 𝜏0 −0

𝑝0 − 𝑝𝐿
2𝐿

𝑟0 2

𝑝 −𝑝

𝑅2

0 < 𝑟 < 𝑟0𝜏

(6.3-8)(6.3-9)𝐶

0𝐿
𝐶2 = 2𝜇+ 𝜇0 − 𝜇1 ln 𝑅
𝐿 2
0

0

0

The radius 𝑟0 indicates the position where shear stress becomes the threshold value
𝜏0 .

Fig.6.3-1 Flow of a Bingham plastic fluid in a circular pipe

As can be seen in Fig.6.3-1, the fluid flows like a Newtonian fluid in the vicinity of thepipe wall 𝑟0 < 𝑟 < 𝑅 but like a plug with uniform velocity in the central region
0 < 𝑟 < 𝑟0 .

6.4 Application of the Equation of Motion (Ⅳ)
for non-Newtonian Fluid
As in Eq.(2.5-3) of section2.5, we should also take an interest in “power law model,
which is sometimes called the Ostwald-de Waele Model.
Some expressions of the power law are
𝑑𝑣 𝑛−1 𝑑𝑣𝑥(6.4-1)

𝜏𝑦𝑥 = −𝑚 | 𝑑𝑦𝑥 |𝑑𝑦𝑑𝑣𝑧 𝑛−1 𝑑𝑣𝑧

𝜏𝑟𝑧 = −𝑚 | 𝑑𝑟 |
𝑑

𝑑𝑟
𝑣

𝑛−1

𝜏𝑟𝜃 = −𝑚 |𝑟 𝑑𝑟 ( 𝑟𝜃 )|(6.4-2)𝑑𝑣

𝑟 𝑑𝑟 ( 𝑟𝜃 )(6.4-3)

The first equation can be used for flow inside rectangular channels, boundary layer
flow along a flat plate, film flow along a solid wall, and so on. The second equation canbe used for flow inside circular tubes and flow between concentric circular cylinders(the so-called annular flow). The third equation is for rotating flow such as tangential
flow in annular space between two concentric rotating cylinders.
For 𝑛 = 1, the power law model reduces to Newton’s law of viscosity with m = μ.

54Application of Differential Transport Equations[EXAMPLE 6.4-E1] Consider laminar flow of non-Newtonian fluid inside a horizontal circularpipe.
The power law model for this case should be of the form:𝑛𝑑𝑣(6.4-E1)

𝜏𝑟𝑧 = 𝑚 (− 𝑑𝑟𝑧 )The Navier-Stokes equation to be solved is Eq.(6.3-2):𝜕𝑣𝑧
𝜕𝑣𝑧 𝑣𝜃 𝜕𝑣𝑧
𝜕𝑣𝑧
𝜕𝑝
1 𝜕
1 𝜕𝜏𝜃𝑧 𝜕𝜏𝑧𝑧
(𝑟 𝜏𝑟𝑧 ) +
𝜌(
+ 𝑣𝑟
+
+ 𝑣𝑧
)= −
− (
+
)
𝜕𝑧
𝑟 𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑡
𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑧
𝜕𝑧

For steady state 𝜕𝑣𝑧 ⁄𝜕𝑡 = 0, For axisymmetric flow 𝜕𝑣𝑧 / ∂θ = 0 and 𝜕𝜏𝜃𝑧 ⁄∂θ = 0.
In the fully-developed flow region 𝜕𝑣𝑧 / ∂z = 0 and 𝜕𝜏𝑧𝑧 ⁄∂z = 0.
Since all streamlines are parallel to the tube wall, 𝑣𝑟 = 0
Then the equation becomes1 𝜕𝜕𝑝(𝑟 𝜏𝑟𝑧 ) = −(6.4-E2)𝑟 𝜕𝑟𝜕𝑧
For a horizontal tube flow
𝜕𝑝
𝑝 −𝑝
−
= 0 𝐿

(6.4-E3)

Then the equation becomes an ordinary differential equation:𝑝 −𝑝(𝑟 𝜏𝑟𝑧 ) = 0 𝐿
𝑟 𝑑𝑟
𝐿(6.4-E4)

𝜕𝑧

𝐿

1 𝑑

Integration with respect to 𝑧 gives𝑝 −𝑝𝐶𝜏𝑟𝑧 = 02𝐿 𝐿 𝑟 + 𝑟1
(6.4-E5)
Since 𝜏𝑟𝑧 does not become infinite at the tube axis, 𝐶1 must be zero. Then𝑝 −𝑝𝜏𝑟𝑧 = 02𝐿 𝐿 𝑟(6.4-E6)

Note that the linear distribution of shear stress is the same in form as Eq.(6.1-14)
for Newtonian fluid flow.
If the power law model can be used for 𝜏𝑟𝑧 , the above equation becomes𝑛𝑑𝑣𝑚 (− 𝑑𝑟𝑧 ) =𝑑𝑣

𝑝0 − 𝑝𝐿

𝑝 −𝑝

2𝐿
1⁄ 𝑛

− 𝑑𝑟𝑧 = ( 20 𝑚𝐿𝐿)𝑟

or
(6.4-E7)

𝑟 1⁄ 𝑛Integration gives

𝑝0 − 𝑝𝐿 1⁄𝑛 𝑛
𝑣𝑧 = − ()
𝑟 1+𝑛⁄𝑛 + 𝐶2
2 𝑚𝐿
1+ 𝑛
At 𝑟 = 𝑅, 𝑣𝑧 = 0. Therefore
𝑝0 − 𝑝𝐿 1⁄𝑛 𝑛
𝐶2 = ()
𝑅1+𝑛⁄𝑛2 𝑚𝐿
1+ 𝑛Then the laminar velocity distribution is
𝑝 −𝑝

1⁄ 𝑛

𝑣𝑧 = ( 20 𝑚𝐿𝐿 )𝑛⁄
𝑟 1+𝑛 𝑛𝑅1+𝑛⁄𝑛 [1 − (𝑅)1+ 𝑛](6.4-E8)

For 𝑛 = 1, it reduces to the parabolic velocity distribution for Newtonian fluid flow.
[PROBLEM 6.4-P1] An incompressible fluid of density 𝜌 and viscosity 𝜇 is in laminar flow
in a rectangular horizontal channel of width 𝑊 and height 2 𝐻, as shown. The static pressure
decreases at a rate of (𝑝0 − 𝑝𝐿 )⁄𝐿.
Obtain the expression for the velocity distribution in the fully developed region. What is the ratioof average to maximum velocity 〈𝑣𝑧 〉⁄𝑣𝑚𝑎𝑥 ? What is the shear stress acting on the upper wall?
Obtain the expression of volumetric flow rate.

Problems

55

Fig.6.4-P1 Laminar flow in a rectangular channel[PROBLEM 6.4-P2] A Newtonian fluid with constant 𝜌 and 𝜇 is in laminar liquidfilm flowing downward on the inside wall of a vertical, long circular tube shown below.
The radius of the tube is 𝑅 and the film thickness is 𝛿 in the fully-developed region.
Obtain the following steady velocity distribution in the fully-developed region:
𝑣𝑧 =

𝜌𝑔 𝑅2
4𝜇

𝑟 2

𝛿 2

𝑟

[(1 − ( ) + 2 (1 − ) ln ( )]𝑅
𝑅𝑅

(6.4-P1)

What is the Reynolds number?

Fig.6.4-P2. Liquid film stream falling along inside a tube wall

[PROBLEM 6.4-P3] We are now in a process to design an oil line shown below. Crude
oil at 15 ℃ is to be pumped through a horizontal 80 𝑚𝑚 ID smooth pipe at avolumetric flow rate 1 × 10−2 𝑚3 ⁄𝑠 . The viscosity and density of the crude oil at 15℃
are 8 × 10−2 𝑘𝑔⁄𝑚 𝑠 and 880 𝑘𝑔⁄𝑚3 , respectively. If pumps with 4 × 105 𝑃𝑎
pressure rise are used, how far apart can they be located?
Calculate the power of pumps, assuming that the pumps operate at an efficiency of
100 percent. Neglect the contribution from fittings and valves to the total pressure
drop.

56

Application of Differential Transport Equations

Fig.6.4-P3 Oil line to be designed and its pressure change

6.5 Application of the Equation of Energy (Ⅰ)
Let us consider the heat transfer problem in laminar flow inside a pipe. We assume
constant physical properties and negligibly small viscous dissipation. The equation ofenergy written in the cylindrical coordinate system can be employed:
𝜕𝑇

𝜕𝑇

𝑣 𝜕𝑇

𝜕𝑇

1𝜕

1 𝜕2 𝑇𝜕𝑇

𝜕2 𝑇

𝜌𝐶𝑝 ( 𝜕𝑡 + 𝑣𝑟 𝜕𝑟 + 𝑟𝜃 𝜕𝜃 + 𝑣𝑧 𝜕𝑧 ) = 𝜅 (𝑟 𝜕𝑟 (𝑟 𝜕𝑟 ) + 𝑟 2 𝜕𝜃2 + 𝜕𝑧 2 )(6.5-1)(1) In general, this equation cannot independently solved without information of thevelocity distribution. Let us restrict the problem to the fully-developed laminar flow
at steady state.
Therefore we start by noting that 𝑣𝑟 = 𝑣𝜃 = 0 and 𝜕𝑇⁄𝜕𝑡 = 0.
We assume axisymmetric heat transfer i.e. 𝜕𝑇⁄𝜕𝜃 = 𝜕 2 𝑇⁄𝜕𝜃 2 = 0.
Then the above equation becomes
𝜕𝑇

1 𝜕

𝜕𝑇

𝜕2 𝑇

𝜌𝐶𝑝 𝑣𝑧 𝜕𝑧 = 𝜅 [𝑟 𝜕𝑟 (𝑟 𝜕𝑟 ) + 𝜕𝑧2 ](6.5-2)

Usually the axial conduction (the term containing 𝜕 𝑇⁄𝜕𝑧 can be neglected relative
to radial conduction (the first term on the right side of Eq.(6.5-2)).
We finally obtain
1 𝜕
𝜕𝑇
𝑣 𝜕𝑇
(𝑟 𝜕𝑟 ) = 𝛼𝑧 𝜕𝑧(6.5-3)𝑟 𝜕𝑟where 𝛼 = 𝜅⁄𝜌𝐶𝑝.
(2) Let us further restrict the problem to fully-developed temperature field. First we
have to visualize that there exists, under certain heating conditions, adimensionless temperature profile invariant with pipe length.
The volumetric flow rate is given by𝑅

𝑄 = ∫0 𝑣𝑧 2𝜋𝑟 𝑑𝑟 =𝜋(𝒑𝟎 − 𝒑𝑳 )
8𝜇𝐿

2

2

𝑅4

For convenience the mean fluid temperature 𝑇𝑚 is defined as𝑅𝜌𝐶𝑝 𝑄 𝑇𝑚 = ∫0 𝜌𝐶𝑝 𝑇 𝑣𝑧 2𝜋𝑟 𝑑𝑟
For constant physical properties𝑅2
𝑇𝑚 = 〈𝑣 〉𝑅2 ∫0 𝑣𝑧 𝑇 𝑟𝑑𝑟𝑧(6.1-12)(6.5-4)(6.5-5)This temperature 𝑇𝑚 is sometimes referred to as the bulk fluid temperature or the
cup-mixing temperature.
The dimensionless temperature is also defined in terms of 𝑇𝑚 as(𝑇𝑤 − 𝑇)⁄(𝑇𝑤 − 𝑇𝑚 ).
If it is invariant in the axial direction

Application of the Equation of Energy (I)𝜕(continued)

57

𝑇 −𝑇

𝜕𝑧

(6.5-6)(𝑇 𝑤− 𝑇 ) = 0
𝑤

𝑚

𝑤

𝑚

Differentiate it with respect to 𝑟𝜕 𝜕𝑇𝑤 − 𝑇()=0𝜕𝑟 𝜕𝑧 𝑇𝑤 − 𝑇𝑚
Changing the order of differentiation, integrating with respect to 𝑧,𝜕𝑇 −𝑇( 𝑤 )|= 𝑐𝑜𝑛𝑠𝑡𝜕𝑟 𝑇 − 𝑇(6.5-7)𝑟=𝑅

By noting that 𝑇𝑚 and 𝑇𝑤 are not a function of 𝑟, we get−𝜕𝑇|𝜕𝑟 𝑟=𝑅𝑇𝑤 − 𝑇𝑚(6.5-8)= 𝑐𝑜𝑛𝑠𝑡Let us define a heat transfer coefficient in terms of the mean temperature:𝜕𝑇
𝑞𝑤 = ℎ (𝑇𝑤 − 𝑇𝑚 ) = − 𝜅 ( 𝜕𝑟 )|(6.5-9)𝑟=𝑅

Keep in mind that this is the case whether the flow is laminar or turbulent. Eqs.(6.5-4) to (6.5-9)
are valid even for turbulent flow. This definition implies that the heat flux from the pipe wall to
the fluid is proportional to the characteristic temperature difference. Since the temperature
gradient at the wall (𝜕𝑇⁄𝜕𝑟 )|𝑟=𝑅 is dependent on the flow condition, the heat transfer
coefficient is a function of flow condition (e.g. the Reynolds number) as well as physicalproperties (e.g. the Prandtl number).
Thus
ℎ
𝜅

=

−

𝜕𝑇
|
𝜕𝑟 𝑟=𝑅𝑇𝑤 − 𝑇𝑚= 𝑐𝑜𝑛𝑠𝑡or(6.5-10)ℎ = 𝑐𝑜𝑛𝑠𝑡

This indicates that the fully-developed temperature field can be observed far
downstream of the entrance for constant heat transfer coefficient conditions.
In fact, most heat exchangers are designed based on constant heat transfer
coefficient assumption.
Getting back to the definition of the fully-developed temperature profile𝜕𝑇𝑤 − 𝑇()=0𝜕𝑧 𝑇𝑤 − 𝑇𝑚
That is𝜕𝑇𝑑𝑇𝑤𝑇𝑤 − 𝑇 𝑑𝑇𝑤
𝑇𝑤 − 𝑇 𝑑𝑇𝑚
=
−
+
𝜕𝑧
𝑑𝑧
𝑇𝑤 − 𝑇𝑚 𝑑𝑧
𝑇𝑤 − 𝑇𝑚 𝑑𝑧
Substituting into the equation of energy
𝜕𝑇𝑣 𝑑𝑇
𝑇 −𝑇 𝑑𝑇𝑇 −𝑇 𝑑𝑇1 𝜕(𝑟 𝜕𝑟 ) = 𝛼𝑧 [ 𝑤 − 𝑇 𝑤− 𝑇 𝑑𝑧𝑤 + 𝑇 𝑤− 𝑇 𝑑𝑧𝑚]
(6.5-11)
𝜕𝑟
𝑟

𝑑𝑧

𝑤

𝑚

𝑤

𝑚

This equation can be solved by at least two boundary conditions:
𝜕𝑇
Case Ⅰ: constant heat flux 𝑞𝑤 = ℎ (𝑇𝑤 − 𝑇𝑚 ) = − 𝜅 (𝜕𝑟 )|= 𝑐𝑜𝑛𝑠𝑡(6.5-12)Case Ⅱ: constant wall temperature(6.5-13)𝑟=𝑅

𝑇𝑤 = 𝑐𝑜𝑛𝑠𝑡

The first heating condition is often encountered in counter-current heat exchangers
when the fluid capacity rates are the same for both fluids. 𝑤 𝑐𝑝 = 𝑊 𝐶𝑝 , where𝑤, 𝑊 are the mass flow rates of the hotter and the colder fluids and 𝑐𝑝, 𝐶𝑝 thecorresponding specific heats.
If ℎ = 𝑐𝑜𝑛𝑠𝑡, 𝑇𝑤 − 𝑇𝑚 = 𝑐𝑜𝑛𝑠𝑡. Therefore
𝑑𝑇𝑤
𝑑𝑧

−

𝑑𝑇𝑚
𝑑𝑧

= 0 i.e.

𝑑𝑇𝑚
𝑑𝑧

=

𝑑𝑇𝑤
𝑑𝑧

=

𝜕𝑇
𝜕𝑧

(6.5-14)

58

Application of Differential Transport Equations

The equation to be solved for constant heat flux is
1 𝜕
𝑟 𝜕𝑟

(𝑟

𝜕𝑇
𝜕𝑟

)=

𝑣𝑧 𝑑𝑇𝑚𝛼(6.5-15)𝑑𝑧

The second heating condition is also encountered in heat exchangers when one of
the two fluid streams has phase transformation such as condensation andevaporation. In this case𝑑𝑇𝑤
=0(6.5-16)𝑑𝑧
The equation of energy to be solved for constant wall temperature becomes
1 𝜕
𝑟 𝜕𝑟

(𝑟

𝜕𝑇
𝜕𝑟

)=

𝑣𝑧

𝑇𝑤 −𝑇

𝑑𝑇𝑚

(6.5-17)

𝛼 𝑇𝑤 − 𝑇𝑚 𝑑𝑧

We shall solve these differential equations Eq.(6.5-15) and Eq.(6.5-17) in thefollowing example.
The corresponding temperature variations with pipe length for these two cases areshown below.

Fig.6.5-1 Temperature variation along pipe axis

[EXAMPLE 6.5-E1]
Let us solve the first case:
𝜕𝑇
Case Ⅰ: constant heat flux 𝑞𝑤 = ℎ (𝑇𝑤 − 𝑇𝑚 ) = − 𝜅 (𝜕𝑟 )|𝑟=𝑅= 𝑐𝑜𝑛𝑠𝑡We assume the fully-developed parabolic velocity distribution and substitute it
into the equation to be solved:
1 𝜕𝜕𝑇(𝑟 𝜕𝑟 ) =𝑟 𝜕𝑟

2 〈𝑣𝑧 〉
𝛼𝑟 2 𝑑𝑇𝑚[1 − ( ) ]𝑅𝑑𝑧(6.5-E1)

Integrating with respect to 𝑟
𝜕𝑇
2 〈𝑣𝑧 〉 𝑟 2
𝑟 4 𝑑𝑇𝑚
[ −
]
𝑟
=
+ 𝐶1
𝜕𝑟
𝛼
2
4 𝑅2 𝑑𝑧
The following boundary condition is applicable to evaluate the integration constant
𝐶1 :𝜕𝑇
B.C.1 at 𝑟 = 0= 0.
Then 𝐶1 : must be zero. 𝐶1 = 0
𝜕𝑟
Integration gives𝑇=

2 〈𝑣𝑧 〉
𝛼

[

𝑟2

𝑟4

𝑑𝑇

− 16 𝑅2 ] 𝑑𝑧𝑚 + 𝐶2
4

Problem

59

The integration constant 𝐶2 can be evaluated from the following boundary
condition:
B.C.2 at 𝑟 = 𝑅
𝑇 = 𝑇𝑤 .
Thus
2 〈𝑣𝑧 〉 3 2 𝑑𝑇𝑚
𝐶2 = 𝑇𝑤 −𝑅
𝛼 16𝑑𝑧
Substituting this expression, we finally obtain the following temperature profile:
2 〈𝑣𝑧 〉

𝑇 = 𝑇𝑤 +

𝛼

𝑟2

𝑟4

4

16 𝑅 2

[ −

−

3
16

𝑅2 ]

𝑑𝑇𝑚
𝑑𝑧

According to the definition of the mean temperature
𝑅2
11 2 〈𝑣 〉 𝑑𝑇
𝑇𝑚 = 〈𝑣 〉𝑅2 ∫0 𝑣𝑧 𝑇 𝑟 𝑑𝑟 = 𝑇𝑤 − 96 𝛼 𝑧 𝑑𝑧𝑚 𝑅2(6.5-E2)(6.5-E3)𝑧

Then the driving force is given by
11 2 〈𝑣 〉 𝑑𝑇
Δ𝑇 = 𝑇𝑤 − 𝑇𝑚 = 96 𝛼 𝑧 𝑑𝑧𝑚 𝑅2(6.5-E4)
On the other hand, the temperature gradient at the wall is calculated as
〈𝑣 〉
𝜕𝑇
𝑑𝑇
= 2 𝑧𝛼 𝑅 𝑚(𝜕𝑟 )|(6.5-E5)𝑑𝑧
𝑟=𝑅

Using Eq.(6.5-E4), the heat flux becomes
11 2 〈𝑣𝑧 〉 2 𝑑𝑇𝑚
𝑅
96 𝛼
𝑑𝑧
Using Eq.(6.5-E5), the wall heat flux is also given by〈𝑣𝑧 〉 𝑑𝑇𝑚
𝜕𝑇
𝑞𝑤 = 𝜅 ( )|= 𝜅𝑅𝜕𝑟 𝑟=𝑅
2𝛼𝑑𝑧
Solving for ℎ from these two equations,
24 𝜅
ℎ=
11 𝑅
The dimensionless group called the Nusselt number becomes for laminar pipe flow
with constant heat flux :ℎ𝐷
48
𝑁𝑢 = 𝜅 = 11 = 𝑐𝑜𝑛𝑠𝑡(6.5-E6)
𝑞𝑤 = ℎ (𝑇𝑤 − 𝑇𝑚 ) = ℎ

The Nusselt number implies the ratio of convective to conductive heat transfer
rate:
ℎ Δ𝑇
𝑁𝑢 = 𝜅(Δ𝑇⁄𝐷)(6.5-E7)
The Stanton number is defined as𝑆𝑡 =ℎ𝜚𝐶𝑝〈𝑣𝑧

=
〉

𝑁𝑢
𝑅𝑒 𝑃𝑟

(6.5-E8)

Then the Stanton number becomes for heat transfer with constant heat flux inlaminar pipe flow:

𝑆𝑡 𝑃𝑟 =

(48⁄11)
𝑅𝑒(6.5-E9)

Note the similarity in form to the corresponding friction factor:
16
𝑓 = 𝑅𝑒(6.5-E10)
The above Nusselt number is a limiting value far downstream of the heating
section in a pipe.
Eq.(6.5-E6) is applicable to the thermal entry region:
𝐿𝑒𝑛𝑡> 0.05 𝑅𝑒 𝑃𝑟(6.5-E11)
𝐷

Experimental data are correlated by Sieder and Tate by the equation:
𝑁𝑢𝑚 = 1.86 [𝑅𝑒 Pr(𝐷⁄𝐿)]1⁄3 (𝜇 ⁄𝜇𝑤 )0.14(6.5-E12)

60

Application of Differential Transport Equations[PROBLEM6.5-P1] A Newtonian fluid of constant physical properties (𝜌, 𝜇, 𝜅) is inlaminar flow inside a rectangular horizontal channel.(see Problem 6.4-P1). Although
the heat transfer rate depends on the fluid flow rate, the upper wall is kept at 𝑇𝑤1 andthe lower wall is cooled at 𝑇𝑤2 . ( 𝑇𝑤1 > 𝑇𝑤2 ). Both the side walls are thermallyinsulated. Therefore it can be assumed that the fluid temperature does not change inthe transverse direction y. Obtain the temperature distribution in the vertical directionin the fully developed region. Using the velocity distribution obtained in Problem
6.4-P1, calculate the enthalpy flow rate across the cross-sectional area 2 𝑊 𝐻. What is
the heat flux required to keep the wall temperatures 𝑇𝑤1 and 𝑇𝑤2.?

Fig.6.5-P1 Temperature distribution of fully developed laminar flow inside a rectangularhorizontal channel.

6.6 Application of the Equation of Energy (Ⅱ)
6.6-1 Steady Heat Conduction
Let us consider the composite wall of blast furnace consisting of three different
materials as an example of heat conduction in hollow cylinders.

Fig.6.6-1 Temperature profile in insulating brick layers of furnace

For steady-state operation the inside surface of the fire-clay brick and the outside
surface of steel cover are held at constant temperature 𝑇1 and 𝑇4 , respectively. Since
the steel cover has very large thermal conductivity, that is, the heat conduction
resistance of the steel cover can be assumed negligibly small, so that the outer surface
temperature of the insulating brick can be assumed equal to 𝑇4 . (𝑇1 > 𝑇4 )

Application of the Equation of Energy (II)(continued)

61

The thermal conductivities of the fire-clay brick, Mullite (alumina brick), and theinsulating brick are 𝜅1 , 𝜅2 , 𝜅3 , respectively. Let us determine the heat loss over theheight 𝐿 from the outer surface of the furnace.
In the equation of energy in cylindrical coordinates, all velocity components vanish.
We can also assume the local temperature to be constant with respect to 𝑧 and 𝜃.
Therefore the temperature should be a function of 𝑟 only.
For this problem, the equation to be solved becomes𝑑𝑑𝑇(𝑟 𝑑𝑟 ) = 0(6.6-1)𝑑𝑟

Integrating twice with respect to 𝑟, the general solution is obtained:
𝑇 = 𝐶1 ln 𝑟 + 𝐶2
(6.6-2)
These two constants of integration can be evaluated in each brick layer from thefollowing boundary conditions:
B.C.1 at 𝑟 = 𝑟1 , 𝑇 = 𝑇1
B.C.2 at 𝑟 = 𝑟2 , 𝑇 = 𝑇2
B.C.3 at 𝑟 = 𝑟3 , 𝑇 = 𝑇3
B.C.4 at 𝑟 = 𝑟4 , 𝑇 = 𝑇4
Here the temperatures corresponding to radii
𝑟2 and 𝑟3 , are unknown andtemporarily denoted by 𝑇2 and 𝑇3, respectively.
Making use of a pair of terminal temperatures 𝑇1 and 𝑇2 , the temperature
distribution in the fire-clay brick layer can be written as
𝑇1 − 𝑇2
𝑇 = 𝑇2 + ln(𝑟ln(𝑟⁄𝑟2 )𝑟1 ≤ 𝑟 ≤ 𝑟2
(6.6-3)
⁄𝑟 )
1

2

2

3

Similarly
𝑇2 − 𝑇3
𝑇 = 𝑇3 + ln(𝑟ln(𝑟⁄𝑟3 )⁄𝑟 )𝑇3 − 𝑇4
𝑇 = 𝑇4 + ln(𝑟ln(𝑟⁄𝑟4 )3 ⁄𝑟4 )

𝑟2 ≤ 𝑟 ≤ 𝑟3

(6.6-4)𝑟3 ≤ 𝑟 ≤ 𝑟4

(6.6-5)For this problem, Eq.(6.6-1) implies𝑑(𝑟 𝑞𝑟 ) = 0(6.6-6)𝑑𝑟

By integrating, we find
(6.6-7)𝑟 𝑞𝑟 = 𝑟4 𝑞𝑟 |𝑟=𝑟4 = 𝑐𝑜𝑛𝑠𝑡
We now evaluate 𝑑𝑇⁄𝑑𝑟 at the surface 𝑟 = 𝑟4 with the aid of the temperature
distribution𝑑𝑇
𝑇3 − 𝑇4|=
𝑑𝑟 𝑟= 𝑟4
𝑟4 𝑙𝑛 (𝑟3 ⁄𝑟4 )Then we get
𝑇3 − 𝑇4
𝑟4 𝑞𝑟 |𝑟=𝑟4 = 𝜅3 ln(𝑟(6.6-8)
⁄𝑟 )
4

3

3

2

Similarly
𝑇2 − 𝑇3
𝑟3 𝑞𝑟 |𝑟=𝑟3 = 𝜅2 ln(𝑟⁄𝑟 )

(6.6-9)𝑇1 − 𝑇2
𝑟2 𝑞𝑟 |𝑟=𝑟2 = 𝜅1 ln(𝑟2 ⁄𝑟1 )

(6.6-10)

The above three equations are combined as

𝑟4 𝑞𝑟 |𝑟=𝑟4 =𝑇1 − 𝑇21𝑟ln( 2 )
𝜅1
𝑟1=

𝑇2 − 𝑇31𝑟ln( 3 )𝜅2
𝑟2=

𝑇3− 𝑇41𝑟ln( 4 )
𝜅3
𝑟3

= 1
𝜅1

𝑟
𝑟1

ln( 2 )+

𝑇1 − 𝑇41𝑟1𝑟ln( 3 )+ ln( 4 )
𝜅2
𝑟2
𝜅3
𝑟3

(6.6-11)Each denominator can be interpreted as the thermal resistance of a single layer.
The thermal resistance of the steel cover is so small that its term is omitted in the
denominator.
The total resistance of three layers in series is given by

62

Application of Differential Transport Equations
1

𝑟

1

𝑟

1

𝑟

1

1

2

2

3

3

𝑅𝑡𝑜𝑡𝑎𝑙 = 𝜅 ln (𝑟2 ) + 𝜅 ln (𝑟3 ) + 𝜅 ln (𝑟4 )(6.6-12)

Finally we obtain the heat loss 𝑄 from the outer surface of the furnace:
2𝜋𝐿 (𝑇1 − 𝑇4)
𝑟2
1
𝑟
1
𝑟
ln( )+ ln( 3 )+ ln( 4 )
𝜅1
𝑟1
𝜅2
𝑟2
𝜅3
𝑟3

𝑄 = 2𝜋𝑟4 𝐿𝑞𝑟 |𝑟=𝑟4 = 1(6.6-13)

The electrical analog for a three-layer cylinder with specified temperature is shown in
Fig.6.6-2.

Fig.6.6-2 Heat transfer resistances in series

As the thickness of insulating-brick increases (as 𝑟3 ⁄𝑟2 increases), the heat loss decreases. Froman economic viewpoint, the optimum thickness can be calculated by taking into account the energy
cost and the cost of insulating material. Obviously this approach can be extended to any number oflayers of material. We have tacitly assumed no contact resistance between the layers of different
materials. Even thin air gap formed in between the layers results in considerable contact resistanceto heat transfer.
[PROBLEM 6.6-P1] A cylindrical rod (radius 𝑅 m) of thermal conductivity 𝜅 W/m K is cooled
at steady state by air of temperature 𝑇𝑎𝑖𝑟 K in order to keep the rod surface at a safe
temperature 𝑇0 K decided by the safety regulation. This rod has a uniform volume productionof heat 𝑄𝐸 (W/m3) inside it.
(a) Derive the equation of heat conduction for obtaining the temperature profile.
(b) Obtain an expression for dimensionless temperature profile.
(c) Confirm that the heat released from the rod surface becomes equal to the heat generation.
6.6-2 Unsteady Heat Conduction1)
Let us consider a problem of unsteady cooling a finite slab having a thickness of 2𝑏 in y-direction.
The dimensions in x- and z-direction are infinitely large. The slab is initially kept at 𝑇0. At time
𝑡 = 0, both the surfaces at 𝑦 = ±𝑏 are instantaneously cooled to 𝑇𝑐 and kept at 𝑇𝑐 . We shall findthe time-dependent profile.

Unsteady Heat Conduction (continued)

63

Fig.6.6-3. Cooling a finite slab of solid

The differential equation of energy transport to be applied reduces to
𝜕2 𝑇

𝜕𝑇

(6.6-14)= 𝛼 𝜕𝑦2𝜕𝑡

We introduce the following dimensionless variables:
𝑇− 𝑇𝑐
Temperature 𝛩 =y-directional length𝛼𝑡𝑇0 − 𝑇𝑐𝑦𝜂= 𝑏(6.6-15)(6.6-16)

Time 𝜏 = 𝑏2
(6.6-17)
where 𝛼 = 𝜅⁄𝜌𝐶𝑝 is thermal diffusivity.
Therefore the dimensionless equation and the boundary conditions become simpler as follows:
𝜕2 𝛩

𝜕𝛩

= 𝛼 𝜕𝜂2
𝜕𝜏

(6.6-18)

I.C.
at 𝜏 = 0
𝛩=1(6.6-19)
B.C.1
at 𝜂 = +1
𝛩=0(6.6-20)
B.C.2
at 𝜂 = −1
𝛩=0(6.6-21)
Using the method of separation of variables, the following temperature function can be assumed:𝛩(𝜂, 𝜏) = 𝑓(𝜂)𝑔(𝜏)(6.6-22)
Substituting the function into Eq.(6.6-18)
1 𝑑𝑔
𝑔 𝑑𝜏

=

1 𝑑2 𝑓𝑓 𝑑𝜂2(6.6-23)

The left side is a function of 𝜏 only whereas the right side is a function of 𝜂 only.
This condition can be valid only if both sides equal a constant − 𝑐 2 . Therefore this problem
reduces to the following two ordinary equations:𝑑𝑔= − 𝑐2 𝑔(6.6-24)
𝑑𝜏
𝑑2 𝑓
𝑑𝜂2

= − 𝑐2 𝑓(6.6-25)

These equations can be integrated as
𝑓 = 𝐾1 sin 𝑐 𝜂 + 𝐾2 cos 𝑐 𝜂(6.6-26)
g = 𝐾3 exp(− 𝑐 2 𝜏)
(6.6-27)
It can be considered that the temperature profile should be symmetric about the centerline of theslab. Therefore 𝐾1 must be zero. From the boundary conditions, 𝑓𝜂=0 = 𝐾2 cos 𝑐 = 0 Thissuggests that𝑐 = (𝑛 + 1⁄2)𝜋𝑛 = 0, ±1, ±2, − − − − − −(6.6-28)
After all, the following form of the solution can be obtained:

64

Application of Differential Transport Equations

2 2
Θ = ∑∞(6.6-29)
𝑛=0 𝐶𝑛 exp(−(𝑛 + 1⁄2) 𝜋 𝜏 ) cos(𝑛 + 1⁄2) 𝜋 𝜂
The unknown constant 𝐶𝑛 should be determined by using the initial condition.
At 𝜏 = 0 Θ = 1 → 1 = ∑∞(6.6-30)
𝑛=0 𝐶𝑛 cos(𝑛 + 1⁄2) 𝜋 𝜂
Multiplying by cos(𝑚 + 1⁄2) 𝜋 𝜂 𝑑𝜂 and then integrating from 𝜂 = −1 to 𝜂 = +1
+1
+1
(6.6-31)∫−1 cos(𝑚 + 1⁄2) 𝜋 𝜂 𝑑𝜂 = ∑∞𝑛=0 𝐶𝑛 ∫−1 cos(𝑚 + 1⁄2) 𝜋 𝜂 cos(𝑛 + 1⁄2) 𝜋 𝜂 𝑑𝜂The integration on the right becomes zero except for the case 𝑛 = 𝑚. Eq. (6.6-32) becomes
𝜂= +1
1
1
( ) (𝑚 + 1/2)𝜋 𝜂 + ( )𝑠𝑖𝑛 (𝑚 + 1/2) 2𝜋 𝜂
4= 𝐶𝑚 ( 2)|(𝑚 + 1/2)𝜋
𝜂= −1
𝜂= +1

sin (𝑚 + 1⁄2) 𝜋 𝜂|(𝑚 + 1⁄2) 𝜋

𝜂= −1

Solving for 𝐶𝑚 , we get

2 (−1)𝑚= 𝐶𝑚(𝑚 + 1⁄2) 𝜋
Finally the temperature profile can be expressed as
𝑇− 𝑇𝑐
(−1)𝑛
1
𝜋𝑦
= 2 ∑∞
exp (−(𝑛 + )2 𝜋 2 𝛼𝑡⁄𝑏2 ) cos (𝑛 + 1⁄2)
𝑛=0
𝑇0 − 𝑇𝑐(𝑛+1⁄2) 𝜋2𝑏(6.6-32)

This solution can be used for many unsteady-state heat conduction problems. It can be kept inmind that for very short times the convergence becomes slow.
1.
2.

Bird, R.B., Stewart, W.E., and Lightfoot, E.N., Transport Phenomena, Wiley, New York, Chapt.3 (1960)
Carslaw, H. S., and Jaeger, J. C., Conduction of Heat in Solid, Oxford Univ. Press, p. 101 (1959)[PROBLEM 6.6-P2] A rectangular parallelepiped iron ingot with dimensions 0.3 m by 2 m by
10 m is fixed vertically and kept uniformly at a temperature 1,050℃. The hot ingot are rapidly anduniformly cooled by striking water jet (40℃) on the two side surfaces of 2 m by 10 m. Owing to thefilm boiling taking place on the ingot surface, the surface temperature is assumed to be kept at
180℃ for the cooling term. Find the time required for the center (𝑧 = 0) temperature of the ingot toreach 300℃. The density and thermal conductivity of the ingot can be assumed to be constant at 730
kg/m3 and 49 W/m K, respectively.

Fig. 6.6-P1. Cooling an iron ingot by striking water jets from two directions

Application of the Equation of Mass Transport

6.7

65

Application of the Equation of Mass Transport

Let us consider the mass transfer problem on the cylindrical surface of laminar liquid jet flowing
downward vertically through a vessel which contains the pure solute gas.
The carbon dioxide gas contained in a large vessel is absorbed by a water jet formed with asharp-edged convergent nozzle. The jet surface is smooth and free of ripples. The gas phaseresistance is negligible since the pure CO2 gas exists surrounding the liquid jet. A small amount ofcarbon dioxide is absorbed in the short contact time (order of 10 ms), so the CO 2 penetration depth
is very small relative to the jet diameter (order of 1 mm). Therefore the problem may be treated as
though the surface were flat. The water jet has uniform velocity distribution across the crosssection: 𝑣𝑧 = 𝑉 = 𝑐𝑜𝑛𝑠𝑡.. A well-designed convergent nozzle can discharge such a water jet withuniform velocity

Fig.6.7-1 Experimental apparatus for CO2 gas absorption

Fig.6.7-2 Situation of CO2 gas absorbed into a falling liquid jet

66

Application of Differential Transport Equations

The equation written in the rectangular coordinate system is applicable inside the liquid jet:
𝜕𝐶𝐴
𝜕𝑡

+ 𝑣𝑥

𝜕𝐶𝐴
𝜕𝑥

+ 𝑣𝑦

𝜕𝐶𝐴
𝜕𝑦

+ 𝑣𝑧

𝜕𝐶𝐴
𝜕𝑧

= 𝐷𝐴𝐵 (

𝜕2 𝐶𝐴

𝜕2 𝐶𝐴

𝜕2 𝐶𝐴

𝜕𝑥

𝜕𝑦

𝜕𝑧 2

+
2

+
2

) + 𝑅𝐴For steady state 𝜕𝐶𝐴 ⁄𝜕𝑡 = 0. We can ignore the chemical reaction between CO2 and H2O:
𝑅𝐴 = 0.
All streamlines are parallel to the jet axis. Then 𝑣𝑥 = 𝑣𝑦 = 0. We can expect that 𝐶𝐴 will bechanging both with 𝑥 and 𝑧. However 𝜕𝐶𝐴 ⁄𝜕𝑦 = 0.
Then the above equation reduces to
𝜕2 𝐶

𝜕𝐶

𝜕2 𝐶

𝑣𝑧 𝜕𝑧𝐴 = 𝐷𝐴𝐵 ( 𝜕𝑥2𝐴 + 𝜕𝑧 2𝐴 )(6.7-1)
The carbon dioxide is transferred in the z-direction primarily because of the convective flow of thejet, the diffusive contribution (the term containing 𝜕 2 𝐶𝐴 ⁄𝜕𝑧 2 ) can be neglected.
We get the partial differential equation for 𝐶𝐴
𝜕2 𝐶

𝜕𝐶

𝑉 𝜕𝑧𝐴 = 𝐷𝐴𝐵 𝜕𝑥2𝐴
(6.7-2)
The applicable boundary conditions are:
B.C.1 at 𝑧 = 0,
𝐶𝐴 = 0
B.C.2 at 𝑥 = 0,
𝐶𝐴 = 𝐶𝐴0(solubility of CO2 into H2O)
B.C.3 at 𝑥 = ∞,𝐶𝐴 = 0Since the jet length between the nozzle and receiver is short (order of 50 mm), CO 2
does not penetrate very far from the surface. The third boundary condition indicatesthat the problem can be treated as if the jet were of infinite thickness moving with theconstant velocity 𝑉.
Let us define dimensionless variables:𝐶𝑥
𝜙 = 𝐶𝐴𝜂=(6.7-3)⁄√4𝐷𝐴𝐵 𝑧 𝑉

𝐴𝑜

By the method of combination of variables the partial differential equation reduces tothe following ordinary differential equation:
𝜙 ′′ + 2𝜂 𝜙 ′ = 0(6.7-4)
where primes ′ indicates total differentiation with respect to 𝜂.
The boundary conditions become
B.C.2
at 𝜂 = 0,
𝜙=1(6.7-5)B.C.1 and 3 at 𝜂 = ∞,𝜙=0
(6.7-6)′
If 𝜙 is replaced by 𝜓, the above equation becomes𝜓′ + 2 𝜂 𝜓 = 0(6.7-7)′−𝜂2
Integration gives 𝜓 = 𝜙 = 𝐶1 𝑒
At this stage we do not have an appropriate boundary condition.
A second integration then gives2𝜂
𝜙 = 𝐶1 ∫0 𝑒 −𝜂 𝑑𝜂 + 𝐶2
(6.7-8)From the first boundary condition, 𝐶2 must be one.
Application of the second boundary condition gives
1
2
𝐶1 = − ∞ −𝜂2 = − 𝜋
∫0 𝑒

𝑑𝜂

√

The solution to the equation at hand is given by2
𝜂2
𝜙 = 1 − 𝜋 ∫0 𝑒 −𝜂 𝑑𝜂 = 1 − 𝑒𝑟𝑓 𝜂
or

𝐶𝐴
𝐶𝐴𝑜

√

= 1 − 𝑒𝑟𝑓𝑥
√4𝐷𝐴𝐵 𝑧⁄𝑉

Here erf 𝑋 is known as the “error function”, whose definition is2
𝜉2
𝑒𝑟𝑓 𝑋 = 𝜋 ∫0 𝑒 −𝜉 𝑑𝜉√(6.7-9)

Application of the Equation of Mass Transport (continued)

67

Once the concentration profile is known, the total absorption rate can be calculated.
Note that 𝑧⁄𝑉 is the time for fluid particle to move from the nozzle to axial distance 𝑧.
The molecular diffusion in the x-direction can be considered to be much larger thanconvective transport in the x-direction because of the low solubility of A in B.
Therefore the local mass flux at the surface (𝑥 = 0) can be expressed as
𝜕𝐶𝑁𝐴𝑥 |𝑥=0 = − 𝐷𝐴𝐵 𝜕𝑥𝐴 |

𝑥=0

𝐷

𝑉

(6.7-10)= 𝐶𝐴𝑜 √ 𝜋𝐴𝐵𝑧

Then the total absorption rate is𝐿

𝑊 = 𝜋𝐷 ∫0 𝑁𝐴𝑥 |𝑥=0 𝑑𝑧 = 𝜋𝐷𝐿 √4 𝐷𝐴𝐵 𝑉𝜋𝐿

𝐶𝐴𝑜

(6.7-11)

Let us define a mass transfer coefficient as the ratio of the mass flux to the driving
force (characteristic concentration difference):
𝑁𝐴𝑥 |𝑥=0 = 𝑘𝐿 (𝐶𝐴𝑜 − 𝐶𝐴∞ )(6.7-12)In this case, 𝑘𝐿 is the average mass transfer coefficient on the liquid side of theinterface. The CO2 concentration becomes approximately zero at a certain depth fromthe surface.
The depth 𝛿 is called a thickness of liquid film where the main resistance to masstransfer takes place. Then at 𝑥 = 𝛿
𝐶𝐴∞ = 0
Therefore Eq.(6.7-12) becomes
𝑁𝐴𝑥 |𝑥=0 = 𝑘𝐿 𝐶𝐴𝑜(6.7-13)Comparing this to the equation Eq,(6.7-11) of the total absorption rate𝑘𝐿 = √

4 𝐷𝐴𝐵 𝑉

(6.7-14)𝜋𝐿

Note that 𝐿⁄𝑉 corresponds to the contact time 𝑡𝑐 .
Much industrial gas-liquid contacting equipment operates with short contact time. As in thisexample, diffusion from interface to bulk fluid proceeds as a transport process. This is an example
of Higbie’s penetration theory 1) indicating that the mass transfer coefficient is proportional to thesquare root of diffusivity divided by contact time. This system is often used to measure molecular
diffusivity of CO2 into H2O.
1. Higbie, R., Trans. A.I.Ch.E., 31, 365 (1935)[EXAMPLE 6.7-E1]
Let us consider how to measure molecular diffusivity by using this system (Fig. 6.7-1).
The absorption rate 𝑊 can be determined from the feed rate of fresh CO 2 gas to the vessel(absorber) from a constant-pressure source. The jet surface area 𝜋𝐷𝐿 and contact time 𝑡𝑐 = 𝐿⁄𝑉
can be calculated directly from measurement of jet length, diameter, and water flow rate.
Then we determine the molecular diffusivity
𝜋𝐿

2

𝑤

𝐷𝐴𝐵 = 4𝑉 (𝜋𝐷𝐿 𝐶 )𝐴𝑜(6.7-E1)

In this case, the gas-phase resistance was negligible.

< Two film theory >If the ambient gas-phase includes inert gas (e.g. N2) insoluble to liquid-phase, there generates
another resistance to mass transfer on the gas side of the interface. The concentration gradient isformed in the gas-phase, too.
In general, a solute is transferred at constant rate through two resistances in series from gas- toliquid-phase.
We define two individual mass transfer coefficients:
𝑁𝐴 = 𝑘𝐺 (𝑦𝐴𝑏 − 𝑦𝐴𝑖 ) = 𝑘𝐿 (𝐶𝐴𝑖 − 𝐶𝐴𝑏 )(6.7-E2)

68

Application of Differential Transport Equations

Here 𝑦𝐴 and 𝐶𝐴 are mole fraction in gas-phase and molar concentration in liquid-phase ofcomponent A, respectively.
Assuming equilibrium at the interface
𝑦𝐴𝑖 = 𝐻 𝐶𝐴𝑖(6.7-E3)
This is known as the Henry’s law.
For practical purposes of process design, it is convenient to express transport rate ( mass-flux) interms of the bulk phase concentrations and an overall mass transfer coefficient. The definition is
𝑁𝐴 = 𝐾𝐺 (𝑦𝐴𝑏 − 𝐻𝐶𝐴𝑏 ) = 𝐾𝐿 (𝑦𝐴𝑏 ⁄𝐻 − 𝐶𝐴𝑏 )(6.7-E4)
The overall driving force should be used for definition of the overall mass transfer coefficients.
However, note that we cannot use the expression as 𝑁𝐴 = 𝐾𝐺 (𝑦𝐴𝑏 − 𝐶𝐴𝑏 ) because 𝐶𝐴 and 𝑦𝐴
use different bases for units of concentration. For unit conversion, Henry’s constant can be used.
∗
That is, 𝐻𝐶𝐴𝑏 is the imaginary concentration 𝑦𝐴𝑏
of gas-phase in equilibrium to 𝐶𝐴𝑏 whereas∗𝑦𝐴𝑏 ⁄𝐻 is the imaginary concentration 𝐶𝐴𝑏 of liquid-phase in equilibrium to 𝑦𝐴𝑏 .
Combining these equations, we get the flux equation analogous to Ohm’s law:∗𝑦𝐴𝑏 − 𝐻𝐶𝐴𝑏
𝑦𝐴𝑏 − 𝑦𝐴𝑖
𝐻𝐶𝐴𝑖 − 𝐻𝐶𝐴𝑏
𝑦𝐴𝑖 − 𝑦𝐴𝑏
𝑁𝐴 =
=
=
=
𝐻
𝐻
1
1
𝑘𝐿
𝑘𝐿
𝐾𝐺
𝑘𝐺
Therefore the following equation can be obtained by solving for 1⁄𝐾𝐺 :𝑁𝐴 =That is1

𝐾𝐺∗
𝑦𝐴𝑏 − 𝑦𝐴𝑏
1
𝐾𝐺

1

𝐻

𝐺

𝐿

=

𝑦𝐴𝑏 − 𝑦𝐴𝑖
1
𝑘𝐺

=

∗
𝑦𝐴𝑖 − 𝑦𝐴𝑏
𝐻
𝑘𝐿

=

∗
𝑦𝐴𝑏 − 𝑦𝐴𝑏
1
𝐻
+
𝑘𝐺 𝑘𝐿

(6.7-E5)(6.7-E6)=𝑘 + 𝑘This states that the overall mass transfer resistance is the sum of the resistances in series ofgas-phase and liquid-phase.
Similarly
1
1
1
=
+
𝐾
𝐻𝑘
𝑘
𝐿

𝐺

(6.7-E7)

𝐿

Since these two equations indicate the same meaning, either will do.

[PROBLEM6.7-P1] The diffusivity of CO2 into pure water is measured by using the falling
liquid jet method. The vessel containing deaerated water jet surrounded by pure CO 2 gas is kept at
𝑝 = 1 𝑎𝑡𝑚 (1.013 × 105 𝑃𝑎) and 20℃. The feed rate of pure water measured by a rotameter is
1.5 𝑚𝑙 ⁄𝑠. The diameter and length of the water jet are 1.2 𝑚𝑚 and 40 𝑚𝑚, respectively. The
CO2 absorption rate measured by gas burette is 0.038 𝑚𝑙 ⁄𝑠 .
Henry’s law for the CO2-H2O system is given by 𝑝 = 𝐻 𝐶𝐴 , where the constant𝐻 = 25.5 (𝑎𝑡𝑚 𝑚3 ⁄𝑘𝑚𝑜𝑙) at 20℃.
(1) Calculate the diffusivity of CO2 into H2O at 20℃.
(2) The penetration thickness 𝛿 is defined as that distance 𝑥 for which 𝐶𝐴 has dropped to avalue 0.01 𝐶𝐴𝑜 :𝛿 = 4 √𝐷𝐴𝐵 𝑧⁄𝑉
Estimate the penetration thickness just above the receiver, i.e. at the lowest point of
the water jet.
[Answer: 𝐷𝐴𝐵 = 1.7 × 10-9 m2/s,

𝛿 = 2.86 × 10-2 mm]

Nomenclature
Nomenclature
𝐶𝐴
𝐶𝑝
𝐷
𝐷𝐴𝐵
𝑓
𝑔
𝐻
ℎ
𝐾𝐺
𝑘𝐺
𝐾𝐿
𝑘𝐿
𝐿
𝑁𝐴
𝑁𝑢
𝑃𝑟
𝑝
𝑄
𝑞𝑤
𝑅
𝑅𝑖 , 𝑅𝑜𝑅𝑒𝑟, 𝜃, 𝑧
𝑆𝑡
𝑇
𝑇𝑎
𝑇𝑞
𝑉
𝑣𝑟 , 𝑣𝜃 , 𝑣𝑧
𝑊
𝛼
𝛿
𝜅
𝜇
𝜌
𝜏𝑟𝑧
𝜏0
𝜔

concentration of component A, [kg/m3]
heat capacity, [J/kg K]
pipe inside diameter, [m]diffusivity, [m2/s]
friction factor, [ - ]gravitational acceleration, [m/s2]
Henry’s constant, [m3/kmol]
heat transfer coefficient, [W/m2K]
overall mass transfer coefficient on vapor-phase concentration basis, [kmol/m2s]
gas-phase mass transfer coefficient, [kmol/m2s]
overall mass transfer coefficient on liquid-phase concentration basis, [m/s]liquid-phase mass transfer coefficient, [m/s]pipe length, [m] or liquid jet length, [m]mass flux of component A, [kg/m2s] or [kmol/m2s]
Nusselt number, [ - ]Prandtl number, [ - ]pressure, [Pa]volumetric flow rate, [m3/s] or heat loss, [J/s]wall heat flux, [W/m2]
pipe radius, [m]cylinder radius, inner and outer cylinders, [m]Reynolds number, [ - ]cylindrical coordinates, [m, - , m]
Stanton number, [ - ]temperature, [K]Taylor number, [ - ]torque, [N m]velocity of falling liquid jet, [m/s]velocity component in cylindrical coordinates
total absorption rate, [kg/s]
thermal diffusivity, [m2/s]
penetration thickness, [m]thermal conductivity, [W/m K]
viscosity, [kg/m s]density, [kg/m3]
shear stress at r in pipe flow, [N/m2]
yield stress, [N/m2]
angular velocity, [1/s]Brackets〈 〉averaged over flow cross section
Subscriptsmbulk or mixed mean
w
wall

69

CHAPTER 7
INTERPHASE MOMENTUM TRANSPORT
7.1 Turbulent Flow Properties
In laminar flow, all the streamliners have smooth curves or straight lines. Especially in steadylaminar flow, the streamlines are at rest. However in turbulent flow all the streamlines oscillate
irregularly in all directions. It is true that the equations of continuity and motion apply to turbulent
flow. Since the instantaneous velocity and pressure are irregularly oscillating function, any simplest
turbulent flow problems have not yet been solved exactly. For the purpose of application inengineering, other approaches have been necessary.
Let us consider a steady turbulent flow in a circular pipe as an example.
Suppose that we could stand at a fixed point in the flow field and observe the velocity and pressure
at the position. The instantaneous velocity 𝑣𝑧 and pressure 𝑝 are fluctuating at random about theirfinite mean values, whereas 𝑣𝑟 and 𝑣𝜃 are fluctuating around the zero mean values. The
oscillograms are shown in Fig.7.1-1.

Fig.7.1-1. Oscillograms of fluctuating velocity and pressure for a circular pipe flow

The time-averaged velocity and pressure are defined as
𝑡+𝑡
1𝑣̅𝑧 = 𝑡 ∫𝑡 0 𝑣𝑧 𝑑𝑡
0

𝑝̅ =

𝑡+𝑡0
𝑝 𝑑𝑡
∫
𝑡
𝑡0
1

(7.1-1)(7.1-2)

These are time-averages of instantaneous velocity and pressure taken over a time interval 𝑡0 froman arbitrary chosen reference time 𝑡. The time interval can be taken to be large compared to theperiodic time of fluctuations.
Then we can split up the instantaneous velocity and pressure into the time-averaged values and theirfluctuations:𝑣𝑧 (𝑟, 𝜃, 𝑧, 𝑡) = 𝑣̅𝑧 (𝑟, 𝜃, 𝑧) + 𝑣𝑧′ (𝑟, 𝜃, 𝑧, 𝑡)(7.1-3)𝑝(𝑟, 𝜃, 𝑧, 𝑡) = 𝑝̅ (𝑟, 𝜃, 𝑧) + 𝑝′(𝑟, 𝜃, 𝑧, 𝑡)(7.1-4)

Turbulent Flow Properties (continued)

71

Note that for steady turbulent flow in a pipe𝑣̅𝑧 = 𝑣̅𝑧 (𝑟, 𝑧) and 𝑝̅ = 𝑝̅ (r, z)
By definition,

1 𝑡+𝑡0 ′∫𝑣𝑧 𝑑𝑡 = 0𝑡0 𝑡
̅ = 0 𝑣̅𝑟 = ̅̅̅Similarly 𝑝′𝑣𝜃 = 0 and 𝑣̅𝑟′ = ̅̅̅𝑣𝜃′ = 0The instantaneous velocity fluctuations 𝑣𝑟′ , 𝑣𝜃′ , 𝑣𝑧′ can become negative very often, but
222
𝑣𝑟′ , 𝑣𝜃′ , 𝑣𝑧′ cannot be negative. Therefore ̅̅̅̅𝑣𝑧′ 2 will not be zero. As a measure of magnitude of𝑣̅𝑧′ =

the turbulence the level or intensity of turbulence is defined as

𝑇𝑢 =

̅̅̅̅̅
′2
√𝑣
𝑧
̅̅̅
𝑣𝑧

(7.1-5)This turbulence intensity may have values 0.01 to 0.10 in typical turbulent flow conditions.
Of the many methods for the measurement of turbulent velocity, the hot-wire anemometer is themost satisfactory. The detecting element consists of a very fine short metal wire (e.g. 5 𝜇𝑚 dia.
And 5 mm long platinum wire for air stream), which is heated by an electric current to a constanttemperature above the stream temperature.
The wire is placed perpendicular to the velocity component to be measured. The rate of heat loss tothe ambient stream from the wire is proportional to the square root of the stream velocity √𝑣 in theusual stream condition. The wire is of such low heat capacity that the temperature of the wire canfollow the rapid velocity fluctuations. The rate of heat loss is equal to the rate of heat generated by
the electric current through the wire 𝐼 2 𝑅 , where 𝐼 is the electric current and 𝑅 the electric
resistance of the wires. In the modern method, the electric resistance is kept constant as far as
possible by using an electronic feedback system. Instead, the feedback system changes the current
through the wire as soon as a variation in electric resistance occurs. The response time to the change
in approach velocity is of the order shorter than 0.1 ms.
Then we have the relation between 𝐼 2 and 𝑣:
𝐼 2 𝑅 = 𝛼 + 𝛽 √𝑣
(7.1-6)
Where the constants 𝛼 and 𝛽 are usually determined by experiment. If we substitute 𝐼 = 𝐼 ̅ + 𝐼 ′and 𝑣 = 𝑣̅ + 𝑣 ′ into the above equation, we get the approximate relation between 𝐼 ′ and 𝑣 ′:(𝐼 ̅ + 𝐼 ′ )2 𝑅 = 𝛼 + 𝛽 √𝑣̅ + 𝑣 ′The mean velocity is given by the equation
𝐼 2̅ 𝑅 = 𝛼 + 𝛽 √𝑣̅
(7,1-7)
The first approximation when 𝑣 ′ ≪ 𝑣̅ , i.e. 𝐼 ′ ≪ 𝐼 ̅ gives𝛽√𝑣̅

𝐼 ′ = 4 𝐼𝑅̅ 𝑣̅ 𝑣 ′(7.1-8)′
We can measure the fluctuating velocity 𝑣 by the use of the above relation.
A simple block diagram of hot-wire anemometer for constant temperature method is shown below.
[PROBLEM 7.1-1] A component of turbulent velocity at a position is assumed to be of the form𝑣𝑧 = 𝑎 + ∑∞(𝑎, 𝑏1 , − − −−, 𝑏𝑛 , − − −∶ 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)(7.1-P1)
𝑛=1 𝑏𝑛 cos 2𝜋𝑛 𝑡
Calculate the time-averaged velocity and the intensity of turbulence.
(Solution) 𝑣̅𝑧 = 𝑎,𝑇𝑢 =2√̅̅̅̅̅
𝑣𝑧′

̅̅̅
𝑣𝑧

=

1
√2 𝑎

2
√∑∞
𝑛=1 𝑏𝑛72

Interphase Momentum Transport

Fig.7.1-2 Block diagram of hot-wire anemometer for constant temperature method

Fig.7.1-3 Distribution of velocity fluctuations in a rectangular channel

As shown in Fig.7.1-3, the turbulence intensity in a rectangular channel becomes maximum near
the wall. This suggests that the turbulence is produced due to the instability of velocity gradient in
the vicinity of the pipe wall and diffuses both toward the center and the wall. The fluctuations in thestreamwise direction, or in the z direction, are greater than the fluctuations in the transverse
direction, or in the x direction. In the vicinity of the wall, molecular momentum transport becomes
important, but the fluctuations become very small. Turbulent flow in a pipe also has similartendency.
We usually consider turbulent pipe flow by dividing the flow region into three zones. (three-layer
concept)
The first layer very close to the wall is called the viscous sublayer, where Newton’s law of viscosity
can be applied to describe the flow. Since molecular transport is predominant, this sublayer causes
the main resistance to interphase transport. The second is called the buffer zone or transition zone,
where the laminar and turbulent effects are comparable and both important. And the third is called
the region of fully developed turbulence or the turbulent core, where viscous effects are negligibly
important.
[PROBLEM 7.1-2] Show how to measure the intensity of turbulence by use of hot-wireanemometer.
(Solution) 𝑇𝑢 =

̅̅̅̅̅
2
√𝑣′
𝑣̅

=

̅̅̅̅
2
4 𝐼 ̅ 𝑅 √𝐼′
𝐼 ̅ 𝑅− 𝛼Friction Factor and Pressure Drop for Channel Flows73

7.2 Friction Factor and Pressure Drop for Channel Flows
Most problems of engineering importance occur in the region of turbulent flow.
Let us consider the momentum transport in turbulent flow inside a circular pipe. Figure 7.2-1 shows
the time-averaged velocity distribution in the fully-developed region. The flow region can be
subdivided into three regions: the turbulent core, the buffer zone, and the viscous sublayer near thewall.

Fig.7.2-1 Time-averaged velocity profile and three subregions of turbulent pipe flow field

In the turbulent core, the velocity gradient is very small but momentum is transferred very rapidly
by virtue of eddy motion. However, close to the wall, eddy motion is suppressed by viscous effect.
Especially in the viscous sublayer, momentum is transferred by molecular diffusion only. Hence themain resistance to momentum transfer takes place in very thin film of the fluid near the wall, wheremost of the total velocity change occurs.
Then the momentum flux (i.e. the shear stress) at the pipe wall is given by̅̅̅𝜕𝑣
𝜏𝑤 = −𝜇 𝜕𝑟𝑧 |(7.2-1)
𝑟=𝑅

The velocity distribution near the wall can be approximated by
〈𝑣
̅̅̅
̅̅̅〉
𝜕𝑣
− 𝜕𝑟𝑧 |
= 𝛿𝑧
𝑟=𝑅

(7.2-2)

Here 𝛿 is the thickness of a fictitious viscous film.
The equation can be rewritten as〈𝑣
̅̅̅〉𝜏𝑤 = 𝜇 𝛿𝑧(7.2-3)
The film thickness is a complicated function of the flow condition, the fluid properties, and thegeometry of the flow system.
For flow in channels, pressure drop data are usually desired and correlations given in terms of afriction factor. This factor is defined by setting the magnitude of the acting force 𝐹 proportional tothe dynamic head or the characteristic kinetic energy 𝐾:
𝐹=𝑓𝐾𝐴(7.2-4)This is a useful definition because 𝑓 is only a function of 𝑅𝑒 for a given geometrical shape:𝑓 = 𝑓(𝑅𝑒)(7.2-5)For flow in a circular pipe, 𝐴 is taken to be the wetted surface 2𝜋𝑅 𝐿, 𝐾 is taken to be the kinetic
energy based on the average velocity, i.e. (1⁄2)𝜌〈𝑣̅𝑧 〉2. The force 𝐹′ acting on the inner wall is
𝐹 ′ = 2𝜋𝑅 𝐿 𝜏𝑤 . Then1𝜏𝑤 = 𝑓 2 𝜌〈𝑣̅𝑧 〉2(7.2-6)

74

Interphase Momentum TransportThis indicates that the wall shear stress for pipe flow is proportional to the kinetic energy.
Combining the equation with the foregoing equation for 𝛿, we get
〈𝑣
̅̅̅〉
1
𝜇 𝛿𝑧 = 𝑓 2 𝜌〈𝑣̅𝑧 〉2(7.2-7)
From the equation,
𝛿2
1=(7.2-8)𝐷
𝑅𝑒 𝑓(𝑅𝑒)

This suggests that the dimensionless film thickness is a function of the Reynolds number only.

Fig.7.2-2 Force balance of inclined circular pipe flow

According to the force balance on the fluid between 𝑧 = 0 and 𝐿 in the flow direction for fully
developed flow
2𝜋𝑅 𝐿 𝜏𝑤 = (𝑃0 − 𝑃𝐿 ) 𝜋 𝑅2
(7.2-9)Substituting the relation into the defining equation for 𝑓
𝐿 1
𝑃0 − 𝑃𝐿 = 4 𝑓
𝜌〈𝑣̅𝑧 〉2
(7.2-10)𝐷 2This is known as the Fanning friction factor. If the pipe is horizontal, the equation becomes𝐿 1
𝑝0 − 𝑝𝐿 = 4 𝑓 𝐷 2 𝜌〈𝑣̅𝑧 〉2(7.2-11)
Once we know as the relation between 𝑓 and 𝑅𝑒, we can calculate the pressure drop.
[EXAMPLE 7.2-1]

Fig.7.2-E1 Force balance on a cylindrical fluid element of radius r

As shown in Fig.7.2-E1, we can set up the force balance on the cylindrical fluid element of length
𝐿 and radius 𝑟 in the fully developed turbulent flow:(𝑃0 − 𝑃𝐿 ) 𝜋 𝑟 2 = 𝜏𝑟𝑧 2𝜋𝑟 𝐿(7.2-E1)
Here 𝜏𝑟𝑧 is the total apparent shear stress or the sum of molecular and turbulent contributions.
Then we divide by the foregoing equation for 𝜏𝑤
𝜏𝑟𝑧𝑟=(7.2-E2)
𝜏𝑅
𝑤

Note that in a fully-developed pipe flow, whether laminar or turbulent, the total apparent shear

Dimensional Analysis of Friction Factor for Channel Flows75

stress varies linearly from zero at the pipe axis to a maximum at the wall.
The friction factor can be measured experimentally by using the following equation:𝑓=

1 𝐷 𝑃0 − 𝑃𝐿
4 𝐿

(7.2-E3)

1
2
̅̅̅〉
𝜌〈𝑣
𝑧
2

7.3 Dimensional Analysis of Friction Factor for Channel Flows
Next let us consider the friction factor by applying a dimensional analysis to the equation of
motion. The turbulent effect on momentum transfer becomes of negligible importance in theviscous sublayer.
The following equation is valid in the viscous sublayer, whether the flow is laminar or turbulent:
𝜕𝑣

𝜕𝑣

𝑣 𝜕𝑣

𝜕𝑣

𝜕𝑝

1𝜕

𝜕𝑣

1 𝜕2 𝑣

𝜕2 𝑣

𝜌 ( 𝜕𝑡𝑧 + 𝑣𝑟 𝜕𝑟𝑧 + 𝑟𝜃 𝜕𝜃𝑧 + 𝑣𝑧 𝜕𝑧𝑧 ) = − 𝜕𝑧 + 𝜇 [𝑟 𝜕𝑟 (𝑟 𝜕𝑟𝑧 ) + 𝑟 2 𝜕𝜃2𝑧 + 𝜕𝑧 2𝑧 ] + 𝜌 𝑔𝑧 (7.3-1)
The equation could be solved with the following conditions:̅̅̅𝜕𝑣
𝑣𝑟 = 𝑣𝜃 = 𝑣𝑧 = 0 and 𝜏𝑤 = 𝜏𝑟𝑧 |𝑟=𝑅 = −𝜇 𝑧 |at 𝑟 = 𝑅𝜕𝑝𝜕𝑟 𝑟=𝑅

𝑃0 − 𝑃𝐿− 𝜕𝑧 + 𝜌𝑔𝑧 = 𝐿(7.3-2)
Next we introduce the dimensionless quantities𝑟 ∗ = 𝑟⁄𝐷, 𝑧 ∗ = 𝑧⁄𝐷, 𝑣𝑟∗ = 𝑣𝑟 ⁄〈𝑣̅𝑧 〉, 𝑣𝜃∗ = 𝑣𝜃 ⁄〈𝑣̅𝑧 〉, 𝑣𝑧∗ = 𝑣𝑧 ⁄〈𝑣̅𝑧 〉, 𝑝∗ = (𝑃0 − 𝑃𝐿 )⁄ 𝜌〈𝑣̅𝑧 〉2(7.3-3)

According to the definition of friction factor
1𝜏𝑤 = 𝑓 2 𝜌〈𝑣̅𝑧 〉2Then the friction factor has the following form:(7.3-4)̅̅̅̅𝜕𝑣𝑧|
𝜕𝑟 𝑟=𝑅
1
2
̅̅̅〉
𝜌〈𝑣
𝑧
2

(7.3-5)

𝑓= 1
2

𝜏𝑤

=
2

̅̅̅〉
𝜌〈𝑣
𝑧

−𝜇

=-

2

𝜕𝑣𝑧∗

|

𝑅𝑒 𝜕𝑟 ∗ 𝑟 ∗ =1⁄2

For steady state, the above momentum equation is made dimensionless𝜕𝑣 ∗𝑣 ∗ 𝜕𝑣 ∗

𝜕𝑣 ∗

𝜕𝑝∗

1

1 𝜕

𝜕𝑣 ∗

1 𝜕2 𝑣∗

𝜕2 𝑣∗

𝑣𝑟∗ 𝜕𝑟𝑧∗ + 𝜃∗ 𝜕𝜃𝑧 + 𝑣𝑧∗ 𝜕𝑧𝑧∗ = − 𝜕𝑧 ∗ + 𝑅𝑒 [𝑟 ∗ 𝜕𝑟 ∗ (𝑟 ∗ 𝜕𝑟𝑧 ) + 𝑟 ∗ 2 𝜕𝜃2𝑧 + 𝜕𝑧 ∗ 2𝑧](7.3-6)𝑟where 𝑅𝑒 = 𝐷〈𝑣̅𝑧 〉 𝜌/𝜇.
This equation is too difficult to solve analytically. But the solution can be expected to be of the form𝑣𝑟∗ = 𝑣𝑟∗(𝑟 ∗ , 𝜃, 𝑧 ∗ , 𝑅𝑒), 𝑣𝜃∗ = 𝑣𝜃∗ (𝑟 ∗ , 𝜃, 𝑧 ∗ , 𝑅𝑒), 𝑣𝑧∗ = 𝑣𝑧∗ (𝑟 ∗ , 𝜃, 𝑧 ∗ , 𝑅𝑒)(7.3-7)
We assume the fully-developed, axisymmetric velocity field:
𝜕𝑣𝑧∗
𝜕𝑣𝑧∗
=
0
and
=0
𝜕𝑧 ∗
𝜕𝜃
Then we get𝑣𝑧∗ = 𝑣𝑧∗(𝑟 ∗ , 𝑅𝑒)(7.3-8)𝜕𝑣𝑧∗
The velocity gradient at the wall 𝜕𝑟 ∗ | ∗is a function of 𝑅𝑒 only. Therefore the friction factor
𝑟 =1⁄2

is found to be a function of 𝑅𝑒 alone in the fully-developed velocity field:
𝑓 = 𝑓 (𝑅𝑒)(7.3-9)It is suggested that it is sufficient to plot only a single curve of 𝑓 against 𝑅𝑒 rather than determine
how 𝑓 varies for separate values of 𝐷, 〈𝑣̅𝑧 〉, 𝜌, 𝜇.
Fig.7.3-1 gives the friction factor chart or empirical correlation of 𝑓 and 𝑅𝑒 for circular tube flow.
For laminar flow, the friction factor can be derived theoretically from the Hagen-Poiseuille law:16
𝑓 = 𝑅𝑒(7.3-10)
The turbulent curve for hydraulically smooth tubes is curve-fitted from 𝑅𝑒 = 2.1 × 103 to
5 × 106 by the equation:1= 4.0 𝑙𝑜𝑔10 (𝑅𝑒 √𝑓 ) − 0.40
(7.3-11)√𝑓

76

A similar expression exists for 2.1 × 103 < 𝑅𝑒 < 5 × 105 :0.0791
𝑓 = 𝑅𝑒 1⁄4This is known as the Blasius formula.

Interphase Momentum Transport(7.3-12)Fig.7.3-1 Friction factor for smooth tubes

If pipes are rough, then in the turbulent flow higher pressure drop is required for a given flow rate
than that expected for smooth pipes. The friction factor for rough pipes is a function of not only the
Reynolds number but also the relative roughness 𝜖 ⁄𝐷, where 𝜖 is the height of a protrusion.
If the roughness-element height 𝜖 does not extend beyond the viscous sublayer, the roughness has
little effect on the turbulent-flow profiles. This condition is called “hydraulically smooth.” If theroughness extends partly into the buffer layer, there is an added resistance to flow due to the form
drag.
[PROBLEM 7.3-1] Gasoline (𝜇 = 0.4 𝑐𝑃, 𝜌 = 710 𝑘𝑔⁄𝑚3 ) is to be delivered at 10 m3 /minthrough a smooth straight pipe (400 mmID, length 𝐿 = 2 𝑘𝑚) from a refinery to an airport.
Calculate the pump horse power if the pump operates at an efficiency of 70%. For the case of
relative roughness 𝜖 ⁄𝐷 = 0.001, what is the power requirement of the pump?
[PROBLEM 7.3-2] Water (at 20℃) flows in a circular tube of 50 mmID at a flow rater of
0.15 𝑚3 /𝑚𝑖𝑛 . Calculate the thickness of the viscous film defined by Eq.(7.2-8).

7.4 Mechanical Energy Loss
7.4-1 Mechanical energy losses in pipelines
As shown in Fig.7.4-1, we set up the macroscopic momentum balance over the inclined pipe
section between planes 1 and 2.
− 𝐹 = 𝑤1 〈⃗⃗⃗⃗𝑣1 〉 − 𝑤2 〈⃗⃗⃗⃗𝑣2 〉 − 𝑝1 𝑛⃗⃗⃗⃗1 S− 𝑝2 ⃗⃗⃗⃗
𝑛2 S + ρV𝑔(7.4-1)(𝑃1 − 𝑃2 )𝑆 = (𝑝1 − 𝑝2 )𝑆 + 𝜌𝑉𝑔 𝑐𝑜𝑠 𝜃
𝐹 = 𝜋𝐷𝐿 𝜏𝑤 , 〈⃗⃗⃗⃗𝑣1 〉 = 〈𝑣⃗⃗⃗⃗2 〉,
Here 𝑃 = 𝑝 − 𝜌𝑔𝑧 𝑐𝑜𝑠 𝜃Mechanical Energy Loss

77

Then the momentum balance equation reduces to
𝐹 = 𝜋𝐷𝐿 𝜏𝑤 = (𝑃1 − 𝑃2 )𝑆(7.4-2)

Fig.7.4-1 Macroscopic momentum balance for turbulent flow in an inclined pipe section

Next we set up the macroscopic mechanical energy balance over the same pipe section:
1〈𝑣 〉2 = −𝐹𝑟𝑚 − 𝑊𝑚
+
∆
𝜌
2

∆𝑃

In this case 𝑊𝑚 = 0,= 𝐹𝑟𝑚𝜌𝑃1 − 𝑃2(7.4-3)〈𝑣1 〉 = 〈𝑣2 〉(7.4-4)Comparing Eq.(7.2-26) to Eq.(7.2-24), we get
𝐹 = 𝜌 𝐹𝑟𝑚 𝑆(7.4-5)
From the definition of friction factor
1
F = f 2 𝜌〈𝑣 〉2𝜋𝐷𝐿(7.4-6)
Finally we get
𝐿 1
𝐹𝑟𝑚 = 4𝑓 𝐷 2 〈𝑣 〉2(7.4-7)This is the defining equation of friction factor for turbulent pipe flow.
Once the friction factor is given in terms of the Reynolds number, we can calculate from thisequation the mechanical energy loss (friction loss) over the straight pipe section of length 𝐿.
In a general flow system the pipeline may not be of a uniform diameter and/or there may bevarious kinds of valves, elbows, and fittings which will cause mechanical energy losses. For most
calculations, losses caused by fittings and valves are treated by the equivalent length model. Thismodel is based on the assumption of high Reynolds number. Equivalent length 𝐿𝑒𝑞 for fittings andvalves is defined as a pipe length that would give the same flow rate. Equivalent lengths forstandard-size fittings and valves are given as
𝐿𝑒𝑞 1
𝐹𝑟𝑚 = 4𝑓 𝐷 2 〈𝑣 〉2(7.4-8)Then we obtain the useful equation of mechanical energy balance for isothermal system of
turbulent flows (This is called the modified Bernoulli equation).
𝐿𝑒𝑞 1
𝑝 𝑑𝑝
1
𝐿 1
∆ 2 〈𝑣 〉2 + 𝑔∆ℎ + ∫𝑝 2 𝜌 + 𝑊𝑚 + ∑𝑖 (4𝑓 𝐷 2 〈𝑣 〉2 ) + ∑𝑗 (4𝑓 𝐷 2 〈𝑣 〉2) = 0(7.4-9)1𝑖𝑗

7.4-2 Mechanical energy losses due to pipe fittings1,2)
We can also use the friction loss factor model for fittings, valves, and enlargements and contractionsof the flow cross section:
1
𝐹𝑟𝑚 = 𝜍 2 〈𝑣 〉2(7.4-10)where 𝜍 is the friction loss factor, dimensionless. This model is based on the fact that the friction
factor of rough pipes becomes almost constant in the range of turbulent flow.
(1) Sudden expansion of the flow cross section

78Interphase Momentum Transport
1

𝐹𝑟𝑚 = 2 〈𝑣

〉2

2
𝐷1 2

(7.4-11)(1 − ( ) )𝐷
2Here 〈𝑣 〉 is the upstream velocity, 𝐷1 the diameter of upstream pipe, and 𝐷2 the diameter ofdownstream pipe.
(2) Sudden expansion into very large regions, e.g. tanks
1
𝐹𝑟𝑚 = 2 〈𝑣 〉2(7.4-12)This implies 𝜍 = 1 for the case (𝐷2 ⁄𝐷1 → ∞).
Typical values for friction loss factor and/or equivalent length are listed below.
(1) Sudden contraction at a sharp-edged entrance

𝐴2⁄𝐴1
𝜍

0
0.5

0.2
0.45

0.4
0.36

0.6
0.21

0.8
0.07

1.0
0

(2) Contraction at a round entrance with a radius of rounding greater than 15%of the pipe diameter D
𝐹𝑟𝑚 = 𝜍

1
〈𝑣 〉2
2 2

𝜍 = 0.04(3) Sudden enlargement at a sharp-edged exit
𝐹𝑟𝑚 = 𝜍

1
〈𝑣 〉2
2 1

𝜍 = (1 −

𝐴1 2
)
𝐴2

1

(4) Valves
we use the equation for 𝜍: 𝐹𝑟𝑚 = 𝜍 2 〈𝑣2 〉2(i)Gate valve
Gate valve𝜍𝐿𝑒𝑞 /𝐷

(ii)

Fully
open
0.17
7

3/4
open
0.9
40

1/2
open
4.5
200

1/4 open
24.0
800

Diaphragm valve
Diaphragmvalve𝜍𝐿𝑒𝑞 /𝐷

Fully open

3/4 open

1/2 open

1/4 open

2.3
125

2.6
140

4.3
235

21.6
1140

Mechanical Energy Losses due to Pipe Fittings (continued)

(iii)

79

Globe valve
Bevel seat

Fully open
6.0

𝜍1/2 open
9.5

Composition seat𝜍𝐿𝑒𝑞 /𝐷

Plug disk
𝜍

(iv)

open
9.0

Fully open
1/2 open
6.0
8.5
330
470
3/4 open
1/2 open
13.0
36.0

1/4 open
112.0

Check valve (fully open only)
Check valve

swing
2.0
110

𝜍
𝐿𝑒𝑞 /𝐷

disk
10.0
500

ball
65
3500

(5) Pipe fittings1, 2)Standard tee

0.4
20

𝜍
𝐿𝑒𝑞 /𝐷

1.3
60

1.5
70

Standard elbow

𝜍𝐿𝑒𝑞 /𝐷

0.74
32

0.3
15

Bend

90˚ bend

180˚ bend

1.0
46

80

Interphase Momentum Transport

90˚ bend
0.5
36

R/D
𝐿𝑒𝑞 /𝐷

1.0
16.5

2.0
10

4.0
10

8.0
14.5

180˚ bend 𝑅 ≫ 𝐷
𝜍 = 1.2
1.
2.

𝐿𝑒𝑞
𝐷

= 50

Perry, R.H. and Chilton, C.H., Chemical Engineers’ Handbook, McGraw-Hill, New York, 5th ed. (1973)
Lydersen, A. L., Fluid Flow and Heat Transfer, John Wiley, New York, p.9(1979)[EXAMPLE 7.4-1] An example of local exhaust system is shown in Fig.7.4-E1.
A canopy type of hood draws in the air contaminated with lead fumes at a flow rate of
300 𝑚3 ⁄ℎ. The diameter of the pipeline is 100 𝑚𝑚. In gas flow we do not need toconsider the potential energy change. The air has a density of 0.89 𝑘𝑔⁄𝑚3 and a
viscosity of 0.022 cp or 2.2 × 10−5 kg⁄m s. The pressure drop at the hood is expressed as1∆𝑝ℎ𝑜𝑜𝑑 = 𝜍 2 𝜚〈𝑣 〉2(7.4-E1)
Here 〈𝑣 〉 is the downstream velocity. The friction loss factor for the hood is given by𝜍 = 0.4.
The pressure drop at the compact electrostatic precipitator is Δ𝑝𝐸𝑃 = 30 𝑚𝑚 𝐻2 𝑂 forthe given flow rate. The pipeline has five standard elbows and an exhaust stack. The
exhaust stack has pressure drop of 7 𝑚𝑚 𝐻2 𝑂. Calculate the power requirement of theturbo-blower assuming 60% efficiency.

Fig.7.4-E1 Pipeline of local exhaust system(Solution) The average velocity in the pipe is𝑄
300 𝑚34⁄〈𝑣 〉 = ⁄ 2 = ()
2 = 10.6 𝑚 𝑠(𝜋 4)𝐷3600 𝑠𝜋(0.10 𝑚)(7.4-E2)

The Reynolds number shown below indicates that the flow is turbulent.
𝐷 〈𝑣 〉𝜚
(0.10 𝑚)(10.6 𝑚⁄𝑠)(0.89 𝑘𝑔⁄𝑚3 )== 4.3 × 104 > 2,100
𝜇2.2 × 1 − 0−5 𝑘𝑔⁄𝑚 𝑠)
The reference planes are chosen outside the hood and exhaust stack. If we assume constant density
𝑅𝑒 =Non-circular Channels (Equivalent Diameter)81

and viscosity, the modified Bernoulli equation becomes1
𝐿 1(〈𝑣2 〉2 − 〈𝑣1〉2 ) + 𝑔(ℎ2 − ℎ1 ) + (𝑝2 − 𝑝1 )⁄𝜌 + 𝑊𝑚 + ∑𝑖 (4𝑓〈𝑣 〉2 )2𝐷 2

𝑖

+ ∑𝑗 (4𝑓

𝐿𝑒𝑞 1
𝐷 2

〈𝑣 〉2 ) + (∆𝑝𝐻 + ∆𝑝𝐸𝑃 + ∆𝑝𝐸𝑆 )⁄𝜌 = 0
𝑗(7.4-E3)

From the chart of f vs. Re, the friction factor 𝑓 = 0.0055 for
𝑅𝑒 = 4.3 × 104
The pipeline has a constant pipe diameter. The total length of straight pipe sections is given as
𝐿𝑡𝑜𝑡𝑎𝑙 = (2 + 20 + 2 + 5 + 10 + 1 + 5 + 5)𝑚 = 50 𝑚
The pressure drop at the hood is1∆𝑝𝐻 = 𝜁 2 𝜚 〈𝑣〉2 = (0.4)(1⁄2)(0.89)(10.6)2 = 20.0 𝑘𝑔⁄𝑚 𝑠 2(7.4-E4)
From the above table, the equivalent length for the standard elbow is given by 𝐿𝑒𝑞 = 3.20 𝑚 for= 0.10 𝑚 .
Δ𝑝𝐸𝑃 = 30 𝑚𝑚 𝐻2 𝑂 = (1000 𝑘𝑔⁄𝑚3 )(9.8 𝑚⁄𝑠 2 )(0.03 𝑚) = 294 𝑘𝑔⁄𝑚 𝑠 2(7.4-E5)
2⁄
Δ𝑝𝐸𝑆 = 7 𝑚𝑚 𝐻2 𝑂 = 68.6 𝑘𝑔 𝑚 𝑠(7.4-E6)
At planes 1 and 2 〈𝑣1 〉 = 〈𝑣2 〉 = 0 and 𝑝2 = 𝑝1
We neglect the potential energy effect: 𝑔(ℎ2 − ℎ1 ) ≅ 0
Then the Bernoulli equation reduces to
4𝑓 1∆𝑝 + ∆𝑝 + ∆𝑝𝐸𝑆
4(0.0055) 1− 𝑊𝑚 = 𝐷 2 〈𝑣 〉2 (∑𝑖 𝐿𝑖 + ∑𝑗 𝐿𝑒𝑞𝑗 ) + 𝐻 𝜌𝐸𝑃= 0.10 2 10.62 (50 + 5 × 3.20) +20+294+68.6= 816 + 430 = 1,246 𝑚2 ⁄𝑠 2
Therefore the power requirement can be calculated as𝑤(−𝑊𝑚 )
𝜌𝑄(−𝑊𝑚 )(0.89)(300⁄3600)(1246)−𝑊 ==== 154 𝑘𝑔 𝑚2 ⁄𝑠 3 = 0.154 𝑘𝑊
0.89

𝜂

𝜂

0.60

(7.4-E7)(7.4-E8)

7.4-3 Non-circular channels (Equivalent diameter)
The friction loss in long straight channels of noncircular cross section can be estimated by using
the equations for circular pipes if the equivalent diameter as the characteristic length and theaverage velocity as the characteristic velocity are used for the Reynolds number.
The equivalent diameter 𝐷𝑒𝑞 is defined as the ratio of four times the cross-sectional area of thechannel to the wetted perimeter of the channel:
4×𝑆
𝐷𝑒𝑞 = 𝐿 𝑐
(7.4-13)𝑤𝑝

Naturally for the special case of a circular pipe, the equivalent diameter becomes coincident withthe pipe diameter. The equivalent diameter is a useful parameter for turbulent flow, but the simple
equivalent-diameter rule does not apply to laminar flow without correction factor.
The equivalent diameter concept is based on the following information. For turbulent flow thevelocity distribution over most of the flow cross-section is flat and most of the velocity change is in
the viscous sublayer near the wall. Therefore, as long as the wall shear stress around the peripheryis almost uniform, the wall shear stress should be independent of the flow cross-sectional shape.
The Reynolds number is defined as
𝐷𝑒𝑞 𝐺
𝑅𝑒𝑒𝑞 = 𝜇(7.4-14)
Here 𝐺 = 𝜌 〈𝑣 〉.

[EXAMPLE 7.4-2]
Water is flowing at an average velocity 〈𝑣 〉 = 1.5 𝑚⁄𝑠 in a horizontal annular space formed
between two coaxial pipes shown below. The inner pipe has an outside diameter of 0.10 𝑚 and theouter pipe has an inside diameter of 0.20 𝑚. The temperature of water is 20 ˚C. Assuming the82Interphase Momentum Transport

fully-developed flow and the smooth wetted surfaces, calculate the pressure drop per unit pipe
length.
(Solution) The density and viscosity of water are𝜌 = 1,000 𝑘𝑔⁄𝑚3 and 𝜇 = 1 𝑐𝑃 = 1 × 10−3 𝑘𝑔⁄𝑚 𝑠
The equivalent diameter of the annulus is4(𝜋⁄4)(𝐷𝑜 2 − 𝐷𝑖 2 )𝐷𝑒𝑞 == 𝐷𝑜 − 𝐷𝑖 = 0.2 − 0.1 = 0.1 𝑚
𝜋(𝐷𝑜 + 𝐷𝑖 )The Reynolds number is
𝑅𝑒𝑒𝑞 =

𝐷𝑒𝑞 𝜌〈𝑣𝑧 〉𝜇=(0.10)(1000)(1.5)
1 × 10−3= 1.5 × 105(turbulent flow)

From the friction factor chart for circular pipes
𝑓 = 0.0041 at 𝑅𝑒𝑒𝑞 = 1.5 × 105
From the defining equation of Fanning friction factor for a horizontal pipe
𝑝0 − 𝑝𝐿
1 1
4(0.0041) 1(1000)(1.5)2 = 184.5 𝑃𝑎⁄𝑚
= 4𝑓
𝜌〈𝑣𝑧 〉2 =
𝐿
𝐷𝑒𝑞 2
0.1
2

7.5 Drag Force on Submerged Objects
Similarly to the defining equation Eq.(7.2-4) for a circular tube flow, the friction factor (called“drag coefficient”) for submerged objects can be defined as
𝐹𝐷 = 𝐶𝐷 𝐾 𝐴(7.5-1)
where 𝐹𝐷 is the drag force, 𝐾 the kinetic energy of the surrounding fluid flow as (1⁄2)𝜌𝑣∞ 2 ,and 𝐴 the characteristic area taken as the projected area of the submerged object.
Usually the coefficient 𝐶𝐷 is called the drag coefficient instead of the friction factor. As in pipe
flows, the drag coefficient is a function of the Reynolds number alone:
𝐶𝐷 = 𝐶𝐷 (𝑅𝑒)(7.5-2)
Let us consider a fluid flow past a sphere of diameter 𝐷𝑝 . In this case, the kinetic energy is
𝐾 = (1⁄2)𝜌𝑣∞ 2
and characteristic area is 𝐴 = (𝜋⁄4)𝐷𝑝 2. The Reynolds number is defined
as 𝑅𝑒 = 𝐷𝑝 𝑣∞ 𝜌⁄𝜇 .

Fig.7.5-1. Variation of drag coefficient with Reynolds number for a single sphereNomenclature

83

For a single sphere surrounded by the fluid flow of velocity 𝑣∞ , a chart of 𝐶𝐷 versus Re shownin Fig.7.5-1 is available. As distinct from a circular tube flow, there is no sharp transition fromlaminar to turbulent flow. However as the flow rate or 𝑣∞ increases, there is an increase in theamount of vortical motion behind the sphere. The sudden drop in the curve at about 𝑅𝑒 = 2 × 105
results from the shift of the separation position of the boundary layer from the front to the rear sideof the sphere. This sphere is subjected to both friction drag and form drag.
The following set of equations is the representative correlations obtained from many
experimental data:
24(1) 𝐶𝐷 =𝑅𝑒 < 0.1(7.5-3)𝑅𝑒
18.5(2) 𝐶𝐷 = 𝑅𝑒 3⁄5
2 < 𝑅𝑒 < 5 × 102(7.5-4)(3) 𝐶𝐷 ≅ 0.44
5 × 102 < Re < 2 × 105(7.5-5)
The first equation comes from Stokes’s law given by
𝐹𝐷 = 3𝜋𝜇𝐷𝑝 𝑣∞(7.5-6)The second equation is an approximate equation in the intermediate region.
The third equation known as the Newton’s law gives an approximately constant. This suggests thatthe drag force is proportional to the kinetic energy of the fluid flow.

Nomenclature
𝐷
𝐷𝑒𝑞
𝐹
𝐹𝑟
𝑓
𝐺
𝐻
𝐼
𝐿
𝐿𝑒𝑞
𝑝
𝑅
𝑅𝑒
𝑇𝑢
𝑡
𝑣𝑟 , 𝑣𝜃 , 𝑣𝑧
𝑊
𝑦
𝛿
𝜍
𝜇
𝜏𝑤

pipe inside diameter, [m]equivalent diameter, [m]external force, [N]friction loss, [m2/s2]
friction factor, [ - ]mass velocity, [kg/m2s]
height of rectangular channel, [m]electric current, [A]pipe length, [m]equivalent length, [m]pressure, [Pa]electric resistance of hot wire, [Ω] or Wheatstone bridge resistance, [Ω]Reynolds number, [ - ]turbulence intensity, [ - ]time, [s]velocity component in cylindrical coordinates
power requirement, [W]distance from wall, [m]thickness of fictitious viscous film, [m]friction loss factor, [ - ]viscosity, [kg/m s]wall shear stress, [N/m2]

Superscripts´
fluctuation̅time-averaged
Brackets〈 〉averaged over flow cross section

CHAPTER 8
INTERPHASE ENERGY TRANSPORT
8.1 Turbulent Heat Transfer and Definitionof Heat Transfer Coefficient
There are three distinct ways in which heat may transfer from a source to a receiver: conduction,
convection, and radiation. Conduction is the transfer of thermal energy due to molecular motion
independent of net bulk flow of the material. We already know that the Fourier’s law describes heattransfer by conduction. Radiation is the transfer of energy due to the emission and subsequentabsorption of electromagnetic radiation.
Convection is the transfer of energy due to fluid motion or bulk flow of the fluid. Let us consider
the heat flow from a flowing fluid to a pipe wall.
As shown in Fig.8.1-1, instantaneous fluid temperature is also fluctuating in a turbulent flow field
where convective heat transfer takes place. Therefore the following time-averaged temperature isused for analysis of convective heat transfer.
𝑡+𝑡1𝑇̅ = 𝑡 ∫𝑡 0 𝑇 𝑑𝑡
(8.1-1)0

Fig.8.1-1 Instantaneous fluid temperature in a turbulent flow field

Fig.8.1-2 Time-averaged temperature profile in the neighborhood of a heat transfer solid surface

Application of the Equation of Energy Transport to Turbulent Heat Transfer

85

For the case of forced convective heat transfer in a turbulent pipe flow, the time-averaged
temperature profile, as shown in Fig,8.1-2, is formed in the neighborhood of the solid surface forheat transfer. Convective and conductive heat transfers are intimately related because heattransferred by convection ultimately involves conduction.
In the turbulent core, the time-averaged temperature has almost uniform distribution due to mixing
by eddy motion. The temperature gradient is very small but energy is transferred very rapidly.
However in the viscous sublayer near the pipe wall, such an eddy motion is suppressed to havesteep linear temperature distribution by molecular transport effect only. The buffer zone has atransitional temperature profile, where the turbulent effect becomes of comparable order tomolecular transport effect.
Hence the main resistance to heat transfer takes place in very thin film of the fluid near the wall.
Then the heat-flux through the film toward the pipe wall can be expressed at the heat transfer
surface as
𝜕𝑇.𝑞𝑤 = 𝑞𝑟 |𝑟=𝑅 = − 𝜅 |(8.1-2)𝜕𝑟
𝑟=𝑅

Introducing a thermal film thickness 𝛿𝑇 , this temperature gradient can be approximated as
𝑇 −𝑇
𝑞𝑟 |𝑟=𝑅 = 𝜅 𝑚𝛿 𝑤
(8.1-3)𝑇

where 𝑇𝑚 is the mixed mean temperature (bulk temperature), 𝑇𝑤 the wall temperature, and 𝛿𝑇
the thickness of an fictitious conduction film. The definition of the mixed mean temperature is𝑅

𝑇𝑚 − 𝑇𝑤 =∫0 2𝜋𝑟 𝑣𝑧 𝜌𝐶𝑝(𝑇− 𝑇𝑤 )𝑑𝑟(8.1-4)𝑅∫0 2𝜋𝑟 𝑣𝑧 𝜌𝐶𝑝 𝑑𝑟

Then the equation becomes𝜅
𝑞𝑤 = 𝛿 (𝑇𝑚 − 𝑇𝑤 ) = ℎ (𝑇𝑚 − 𝑇𝑤 )(𝑇𝑚 > 𝑇𝑤 )𝑞𝑤 = 𝛿 (𝑇𝑤 − 𝑇𝑚 ) = ℎ (𝑇𝑤 − 𝑇𝑚 )(𝑇𝑚 < 𝑇𝑤 )

𝑇

𝜅

𝑇

(8.1-5)This is one of the defining equations for heat transfer coefficient:ℎ = 𝜅⁄𝛿𝑇(8.1-6)
The film thickness is a complicated function of the flow condition, the fluid properties, and thegeometry of the flow system, that is, a function of 𝑅𝑒 and 𝑃𝑟. Therefore the heat transfer
coefficient can be expressed to be a function of the same parameters. The heat transfer coefficient
has units of 𝐽⁄𝑚2 𝑠 𝐾 in SI unit system.

8.2 Application of the Equation of Energy
for Turbulent Heat Transfer
The turbulent effect on heat transfer becomes of negligible importance in the viscous sublayer.
The following equations are valid in the viscous sublayer, whether the flow is laminar or turbulent:
𝜕𝑣

𝜕𝑣

𝑣 𝜕𝑣

𝜕𝑣

𝜕𝑝

1𝜕

𝜕𝑣

1 𝜕2 𝑣𝜕𝑇1 𝜕2 𝑇

𝜕2 𝑇

𝜕2 𝑣

𝜌 ( 𝜕𝑡𝑧 + 𝑣𝑟 𝜕𝑟𝑧 + 𝑟𝜃 𝜕𝜃𝑧 + 𝑣𝑧 𝜕𝑧𝑧 ) = − 𝜕𝑧 + 𝜇 [𝑟 𝜕𝑟 (𝑟 𝜕𝑟𝑧 ) + 𝑟 2 𝜕𝜃2𝑧 + 𝜕𝑧 2𝑧 ]
𝜕𝑇

𝜕𝑇

𝑣 𝜕𝑇

𝜕𝑇

1𝜕

𝜌𝐶𝑝 ( 𝜕𝑡 + 𝑣𝑟 𝜕𝑟 + 𝑟𝜃 𝜕𝜃 + 𝑣𝑧 𝜕𝑧 ) = 𝜅 [𝑟 𝜕𝑟 (𝑟 𝜕𝑟 ) + 𝑟 2 𝜕𝜃2 + 𝜕𝑧 2 ](8.2-1)(8.2-2)In the last equation the viscous dissipation term and the pressure work term have been
neglected.
These equations could be solved with the following boundary conditions
𝜕𝑣
𝑣𝑧 = 𝑣𝑟 = 𝑣𝜃 = 0 and − 𝜇 𝜕𝑟𝑧 = 𝜏𝑤
at 𝑟 = 𝑅(8.2-3)
𝜕𝑇

𝑇 = 𝑇𝑤and− 𝜅 𝜕𝑟 = 𝑞𝑤
at 𝑟 = 𝑅
According to the definition of heat transfer coefficient
𝑞𝑟 |𝑟=𝑅 = 𝑞𝑤 = ℎ (𝑇𝑤 − 𝑇𝑚 )(8.2-4)(8.2-5)

86

Interphase Energy TransportThat is1
𝜕𝑇ℎ = 𝑇 − 𝑇 (− 𝜅 𝜕𝑟 )|(8.2-6)

𝑚

𝑟=𝑅

𝜕𝑇 ∗

𝑣 ∗ 𝜕𝑇 ∗

𝜕𝑇 ∗

1

𝜕𝑣 ∗

𝑣 ∗ 𝜕𝑣 ∗

𝜕𝑣 ∗

𝜕𝑝∗

𝑤

Then let us start with dimensional analysis for application of the equation of energy transport.
The following dimensionless quantities should be introduced:𝑟 ∗ = 𝑟⁄𝐷, 𝑧 ∗ = 𝑧⁄𝐷, 𝑇 ∗ = (𝑇𝑤 − 𝑇)⁄(𝑇𝑤 − 𝑇𝑚 ), 𝑣𝑟∗ = 𝑣𝑟 ⁄〈𝑣𝑧 〉,𝑣𝜃∗ = 𝑣𝜃 ⁄〈𝑣𝑧 〉,𝑣𝑧∗ = 𝑣𝑧 ⁄〈𝑣𝑧 〉, 𝑝∗ = (𝑝 − 𝑝0 )⁄(𝜌 〈𝑣𝑧 〉2 )〈𝑣𝑧 〉 is the representative velocity averaged over the cross section.
The above energy equation is made dimensionless
1

𝜕

𝜕𝑇 ∗

1

𝜕

𝜕𝑣 ∗

𝜕2 𝑇∗

𝜕2𝑇 ∗

𝑣𝑟∗ 𝜕𝑟∗ + 𝜃∗ 𝜕𝜃 + 𝑣𝑧∗ 𝜕𝑧 ∗ = 𝑅𝑒 𝑃𝑟 [𝑟 ∗ 𝜕𝑟 ∗ (𝑟 ∗ 𝜕𝑟 ∗ ) + 𝑟 ∗ 2 𝜕𝜃2 + 𝜕𝑧∗ 2 ](8.2-7)𝑟
where 𝑅𝑒 = 𝜌 𝐷 〈𝑣𝑧 〉⁄𝜇 and 𝑃𝑟 = 𝐶𝑝 𝜇⁄𝜅Note that these dimensionless parameters 𝑅𝑒 and 𝑃𝑟 appear automatically in the dimensional
analysis.
The equation of motion is also made dimensionless for steady state:
1

1

1

𝜕2 𝑣∗

𝜕2 𝑣∗

[
𝑣𝑟∗ 𝑧∗ + 𝜃∗ 𝑧 + 𝑣𝑧∗ 𝑧∗ = − ∗ +(𝑟 ∗ 𝑧∗ ) + ∗ 2 2𝑧 + ∗ 2𝑧](8.2-8)𝜕𝑟
𝜕𝑧
𝜕𝑧
𝑅𝑒 𝑟 ∗ 𝜕𝑟 ∗
𝜕𝑟
𝑟
𝜕𝜃
𝜕𝑧
𝑟 𝜕𝜃
In general, this equation is too difficult to solve analytically. However the velocity solution can be
expected to be of the form:𝑣𝑟∗ = 𝑣𝑟∗ (𝑟 ∗, 𝜃, 𝑧 ∗ , 𝑅𝑒)𝑣𝜃∗ = 𝑣𝜃∗ (𝑟 ∗, 𝜃, 𝑧 ∗ , 𝑅𝑒)(8.2-9)𝑣𝑧∗ = 𝑣𝑧∗ (𝑟 ∗, 𝜃, 𝑧 ∗ , 𝑅𝑒)
If the above dimensionless energy equation could be solved by substituting the dimensionlessvelocity solutions. Therefore the temperature solution can also be expected to be of the form:
𝑇 ∗ = 𝑇 ∗ (𝑟 ∗ , 𝜃, 𝑧 ∗ , 𝑅𝑒, 𝑃𝑟)(8.2-10)
We assume the fully-developed, axisymmetric temperature field
𝜕𝑇 ∗
𝜕𝑇 ∗
=
0
and
=0
𝜕𝑧 ∗
𝜕𝜃
∗
∗ ∗
Then we get 𝑇 = 𝑇 (𝑟 , 𝑅𝑒, 𝑃𝑟)𝜕𝑇 ∗The temperature gradient at the wall 𝜕𝑟 ∗ | ∗should be a function of 𝑅𝑒 and 𝑃𝑟 only.
𝑟 =1/2The boundary condition is made dimensionless:
𝑁𝑢 =

ℎ𝐷

𝜕(𝑇𝑤 −𝑇)
|
𝜕𝑟
𝑟=𝑅

𝜕𝑇 ∗

= (𝑇 − 𝑇 )⁄𝐷 = − 𝜕𝑟 ∗ | ∗
𝜅
𝑤

𝑚

𝑟 =

1
2

(8.2-11)

The new dimensionless parameter is known as the Nusselt number. The above equation states that
the Nusselt number is the dimensionless temperature gradient at the heat transfer surface. Therefore
the Nusselt number is found to be a function of dimensionless parameters 𝑅𝑒 and 𝑃𝑟 in the
fully-developed temperature field. Note that the Nusselt number gives a local heat transfer
coefficient.
𝑁𝑢 = 𝑁𝑢 (𝑅𝑒, 𝑃𝑟)(8.2-12)Similarly the Nusselt number averaged over the pipe length is also a function of 𝑅𝑒 and 𝑃𝑟.
It may be convenient to use a new dimensionless group defined asℎ𝐶𝑝 𝜇

𝑗𝐻 = 𝐶 𝐺 ( 𝜅 )
𝑝

𝑓

2⁄ 3

= 𝑁𝑢 𝑅𝑒 −1 𝑃𝑟 − 1⁄3(8.2-13)

where 𝐺 is mass velocity = (fluid density) x (average fluid velocity) 𝜌〈𝑣 〉.
This is called the j-factor for heat transfer. This parameter is often used when the convective
transport mechanism is discussed among momentum, energy, and mass transfer.
The subscript f denotes properties evaluated at the film temperature
𝑇𝑚 + 𝑇𝑤
𝑇𝑓 =2For large temperature changers, the variation in 𝜇 should be taken into account.

Overall Heat Transfer Coefficient and Heat Exchangers

87

Fig.8.2-1 j-factor correlation for heat transfer of circular pipe flows

Figure 8.2-1 shows the j-factor for heat transfer in circular pipes which includes the correction
factor 𝜙 for viscosity: 𝜙 = (𝜇𝑏 ⁄𝜇𝑤 )−0.14
In the highly turbulent flow range, the following empirical analogy for long smooth pipes is
proposed by Colburn:𝑓𝑗𝐻 =(8.2-14)2This analogy relation will be demonstrated for a laminar boundary layer flow over a flat plate in
Chapter 17.
The analogy permits one to predict heat transfer coefficients from friction data for the same flow
conditions. It should be noted that the laminar Nusselt number or j-factor depends on the ratio
𝐿⁄𝐷 . In the range of turbulent heat transfer, the following empirical correlation is available:
ℎ

𝐶𝑝 𝜇

( 𝜅 )
𝐶 𝐺
𝑝

𝑓

2⁄ 3

𝜇

−0.14

(𝜇 𝑏 )𝑤

𝐷𝐺= 0.026 ( 𝜇 )

− 0.2

𝑅𝑒 =

𝑏

𝐷𝐺
𝜇

> 104

0.6 < Pr < 100(8.2-15)

Colburn, A.P., Trans. A.I.Ch.E., 29, 174-210 (1933)
Sieder, E.N. and Tate, G.E., Ind. Eng. Chem., 28, 1429 (1936)

8.3 Overall Heat Transfer Coefficient and Heat Exchangers
8.3-1 Definition of overall heat transfer coefficientFig.8.3-1 Double-pipe heat exchanger

88Interphase Energy Transport

As a very simple and comprehensible example of heat exchangers, Figure 8.3-1 shows adouble-pipe heat exchanger consisting of two concentric pipes. For definiteness, we assume that the
hot fluid is flowing through the inner pipe and the cold fluid in the annular space; heat is transferred
rapidly across the wall of the inner pipe. If the two fluids flow in opposite directions, the type offlow is called “counter-current flow.” If the two fluids flow in the same direction, the flow is called“parallel flow.” Steady-state turbulent flow may be assumed here. Heat losses from the wall of theouter pipe to the surroundings may be neglected. The actual change in temperature with the radial
distance is shown schematically by the solid lines 𝑇ℎ , 𝑇𝑤ℎ , 𝑇𝑤𝑐 , 𝑇𝑐 . The dotted lines show thetemperature change modified by the film theory to define two heat transfer coefficients ℎ𝑖 , ℎ𝑜 .
The mixed mean temperature 𝑇ℎ𝑚 of the hot stream is somewhat lower than the maximum
temperature 𝑇ℎ , whereas the mixed mean temperature 𝑇𝑐𝑚 of the cold stream is somewhat higher
than the minimum temperature 𝑇𝑐 . The thickness and thermal conductivity of the inner pipe wall
are given by 𝛿𝑡 and 𝜅𝑡 , respectively.
It is convenient to utilize an overall resistance concept in calculation of the heat transfer rate
across many layers in series: the inside film, the tube wall, and the outside film resistances. The heat
flux can be expected to be proportional to the temperature difference 𝑇ℎ𝑚 − 𝑇𝑐𝑚 as the driving
force.
The rate of heat transfer over a very small pipe length 𝑑𝑧 can be expressed in terms of theindividual temperature drops in three layers:
𝑇 −𝑇
𝑑𝑄 = ℎ𝑖 𝜋𝐷𝑖 𝑑𝑧 (𝑇ℎ𝑚 − 𝑇𝑤ℎ ) = 𝜅𝑡 𝜋𝐷𝑎𝑣 𝑑𝑧 𝑤ℎ𝛿 𝑤𝑐 = ℎ𝑜 𝜋𝐷𝑜 𝑑𝑧 (𝑇𝑤𝑐 − 𝑇𝑐𝑚 )(8.3-1)This equation can be rewritten as𝑑𝑄 =𝑇ℎ𝑚− 𝑇𝑤ℎ1ℎ𝑖 𝜋𝐷𝑖𝑑𝑧=

𝑇𝑤ℎ − 𝑇𝑤𝑐
𝛿𝑡𝜅𝑡 𝜋𝐷𝑎𝑣 𝑑𝑧=𝑡

𝑇𝑤𝑐 − 𝑇𝑐𝑚
1ℎ𝑜 𝜋𝐷𝑜𝑑𝑧(8.3-2)where 𝐷𝑎𝑣 is the logarithmic mean of the outer and inner diameters of the inner pipe.
This equation indicates a heat transfer analog to Ohm’s law of electrical current. That is, the heattransfer rate 𝑑𝑄 corresponds to the electrical current. The individual temperature drops correspondto the voltage drops in the separate resistances operating in series. Each denominator indicates thecorresponding thermal resistance.
Adding these numerators and denominators separately, we get
𝑇ℎ𝑚 − 𝑇𝑐𝑚
𝑑𝑄 =(8.3-3)𝛿𝑡
1
1+ℎ𝑖 𝜋𝐷𝑖𝑑𝑧

𝜅𝑡𝜋𝐷𝑎𝑣 𝑑𝑧+ℎ𝑜𝜋𝐷𝑜𝑑𝑧

The denominator indicates the overall thermal resistance. The numerator indicates a local value ofthe overall temperature difference. Then we can define an overall heat transfer coefficient 𝑈𝑜based on the outside area of the inner pipe by the equation
𝑇 −𝑇
𝑑𝑄 = ℎ𝑚 1 𝑐𝑚 = 𝑈𝑜 𝜋𝐷𝑜 𝑑𝑧 (𝑇ℎ𝑚 − 𝑇𝑐𝑚 )(8.3-4)
𝑈𝑜 𝜋𝐷𝑜 𝑑𝑧

The denominator can be considered to be an overall resistance. Comparing this equation with
Eq.(8.3-3), we get the following relation:
1

𝑈𝑜 = 1 𝐷𝑜 𝛿𝑡 𝐷𝑜
ℎ𝑖 𝐷𝑖

+

𝜅𝑡 𝐷𝑎𝑣+1ℎ𝑜(8.3-5)The individual terms of the denominator on the right-hand side represent the individual resistancesof the two fluid films and of the tube wall. Note that 𝑈𝑜 means a local value of the overall
heat-transfer coefficient.
[Fouling]
Usually actual heat exchanger surfaces are subject to deposition of a film of foreign materials.
This phenomenon called “Fouling” very often occurs on the heat transfer surfaces on the coolingwater side when using natural water such as sea, river or well water. The film growing due to theaccumulation of unwanted materials on solid surfaces offers a thermal resistance.
Therefore the heat transfer resistance should be taken into account in the overall heat transfer

Logarithmic Mean Temperature Difference89

coefficient:
1
1 𝐷
1 𝐷
𝛿 𝐷
1
1
= ℎ 𝐷𝑜 + ℎ 𝐷𝑜 + 𝜅𝑡 𝐷 𝑜 + ℎ + ℎ𝑈(8.3-6)

𝑜

𝑖

𝑖

𝑠𝑖

𝑖

𝑡

𝑎𝑣

𝑜

𝑠𝑜

where ℎ𝑠𝑖 and ℎ𝑠𝑜 are the reciprocals of the fouling thermal resistances on the inner and outertube walls, respectively.

8.3-2 Logarithmic mean temperature difference1)
Since the overall temperature difference varies from point to point along the pipe, the heat flux
also varies with pipe length. The temperature variation for counter-current flow is shown as an
example in Figure 8.3-2.

Fig.8.3-2. Temperature variation with pipe length for countercurrent flow
in a double tube exchanger

To apply the defining equation for 𝑈𝑜 to the entire area of the exchanger, the equation must be
integrated with the simplifying assumptions: (1) the overall coefficient is constant and (2) the heat
capacities of the two fluids are kept constant.
Figure 8.3-3 shows a double-pipe counter-flow heat exchanger. The outside and inside diametersof the inner pipe are 𝐷𝑜 and 𝐷𝑖 and the inner pipe length for heat exchange is 𝐿.

Fig.8.3-3. Interphase heat transfer in a double-pipe countercurrent exchanger

The hot fluid is flowing through the inner pipe and the cold fluid through the annular space.
If we set up the differential energy balance over the volume element between 𝑧 and z + dz,
𝑑𝑄 = − 𝑤ℎ 𝐶𝑝ℎ 𝑑𝑇ℎ𝑚 = 𝑤𝑐 𝐶𝑝𝑐 𝑑𝑇𝑐𝑚
(8.3-7)
The flow rate of the cold fluid 𝑤𝑐 is negative since the cold fluid flows in opposite direction. The
temperature gradients are 𝑑𝑇ℎ𝑚⁄𝑑𝑧 < 0, 𝑑𝑇𝑐𝑚 ⁄𝑑𝑧 < 0 . The minus sign in front of 𝑤ℎ comesfrom the condition of positive 𝑑𝑄.
According to the equation for 𝑈𝑜 ,𝑑𝑄 = 𝑈𝑜 𝜋𝐷𝑜 𝑑𝑧 (𝑇ℎ𝑚 − 𝑇𝑐𝑚 )(8.3-8)From these equations we get the following equations for the hot and cold streams, respectively.
𝑑𝑇
𝜋𝐷𝑜 𝑑𝑧− 𝑇 −ℎ𝑚𝑇 = 𝑈𝑜 𝑤 𝐶𝑝
(8.3-9)ℎ𝑚

𝑑𝑇𝑐𝑚

𝑇ℎ𝑚 − 𝑇𝑐𝑚𝑐𝑚ℎ𝜋𝐷𝑜 𝑑𝑧= 𝑈𝑜 𝑤 𝐶𝑝𝑐𝑐ℎ(8.3-10)

90Interphase Energy Transport

By adding these equations, we obtain𝑑(𝑇 − 𝑇 )1
1
− 𝑇 ℎ𝑚− 𝑇 𝑐𝑚 = 𝑈𝑜 (𝑤 𝐶𝑝 + 𝑤 𝐶𝑝 ) 𝜋𝐷𝑜 𝑑𝑧
ℎ𝑚

𝑐𝑚

ℎ

ℎ

𝑐

𝑐

ℎ𝑚2

𝑐𝑚2

ℎ

ℎ

𝑐

𝑐

For constant coefficient 𝑈𝑜 , we can integrate between both ends of the exchanger:𝑇−𝑇11
ln (𝑇ℎ𝑚1 − 𝑇𝑐𝑚1 ) = 𝑈𝑜 (𝑤 𝐶𝑝 + 𝑤 𝐶𝑝 ) 𝜋𝐷𝑜 𝐿(8.3-11)(8.3-12)

According to the energy balance between 𝑧 = 0 and 𝐿, the total rate of heat exchange is
𝑄 = 𝑤ℎ 𝐶𝑝ℎ (𝑇ℎ𝑚1 − 𝑇ℎ𝑚2) = − 𝑤𝑐 𝐶𝑝𝑐 (𝑇𝑐𝑚1 − 𝑇𝑐𝑚2 )𝑇ℎ𝑚1 − 𝑇ℎ𝑚2=

1
𝑤ℎ 𝐶𝑝ℎ=-(𝑇𝑐𝑚1 − 𝑇𝑐𝑚2 )1
𝑤𝑐 𝐶𝑝𝑐Then we get
11+ 𝑤 𝐶𝑝 =𝑤ℎ 𝐶𝑝ℎ𝑐=(𝑇ℎ𝑚1 − 𝑇ℎ𝑚2 )− (𝑇𝑐𝑚1 − 𝑇𝑐𝑚2 )1
1+
𝑤ℎ 𝐶𝑝ℎ𝑤𝑐 𝐶𝑝𝑐[(𝑇ℎ𝑚1 − 𝑇𝑐𝑚1 )−(𝑇ℎ𝑚2 − 𝑇𝑐𝑚2 )]𝑄𝑐(8.3-13)(8.3-14)Substituting it into the foregoing equation, Eq,(8.3-12), we get

𝑄 = 𝑈𝑜 𝜋𝐷𝑜 𝐿 {(𝑇ℎ𝑚1 − 𝑇𝑐𝑚1 )−(𝑇ℎ𝑚2 − 𝑇𝑐𝑚2 )𝑇− 𝑇𝑐𝑚1)
𝑇ℎ𝑚2 − 𝑇𝑐𝑚2ln( ℎ𝑚1}

or
𝑄 = 𝑈𝑜 𝐴𝑜 (𝑇ℎ𝑚 − 𝑇𝑐𝑚 )𝑙𝑚 = 𝑈𝑜 𝐴𝑜 (∆𝑇𝑚 )𝑙𝑚
where∆𝑇 − ∆𝑇(∆𝑇𝑚 )𝑙𝑚 = 𝑚1∆𝑇𝑚1𝑚2
ln(8.3-15)(8.3-16)(8.3-17)∆𝑇𝑚2

This equation, valid for both parallel and counter-current flows, is the basic equation forengineering calculation of heat exchangers. The result indicates that we should use the logarithmic
mean of the mixed mean temperature differences at both ends (abbreviated LMTD) to determine thetotal area for the required heat transfer rate.
We can understand from this analysis for a double-pipe exchanger that the LMTD is appropriate as
the characteristic temperature difference for thermal engineering design of various exchangers
1.

Bird, R.B., Stewart, W.E. and Lightfoot, E.N., Transport Phenomena, Wiley, New York, Chapt.15, Example 15.4-2 (1960)

Nomenclature
𝐶𝑝
𝐷
𝑓
𝐺
ℎ
𝑗𝐻
𝑁𝑢
𝑃𝑟
𝑄
𝑞𝑟 , 𝑞𝑤𝑅𝑅𝑒𝑟, 𝜃, 𝑧𝑇𝑡𝑈𝑣𝑟 , 𝑣𝜃 , 𝑣𝑧
𝑤𝑐 , 𝑤ℎ
𝑧
𝛿𝑡
𝛿𝑇
𝜇
𝜅

heat capacity, [J/kg K]
pipe diameter, [m]friction factor, [ - ]mass velocity, [kg/m2s]
heat transfer coefficient, [W/m2K]
j-factor for heat transfer, [ - ]Nusselt number, [ - ]Prandtl number, [ - ]heat transfer rate, [W]heat flux at r, wall heat flux, [W/m2]
pipe radius, [m]Reynolds number, [ - ]cylindrical coordinates, [m, - , m]temperature, [K]time, [s]overall heat transfer coefficient, [W/m2K]
velocity component in cylindrical coordinatesmass flow rate of cold fluid, hot fluid, [kg/s]
axial coordinate, [m]tube wall thickness, [m]film thickness for heat transfer, [m]viscosity, [kg/m s]
thermal conductivity, [W/m K]

Nomenclature𝜌𝜙density, [kg/m3]
correction factor for viscosity, [ - ]Superscripts∗
dimensionless´
fluctuation̅time-averaged
Subscripts
c
cold
f
film
h
hot
i
inner tube
l.m.
logarithmic meanmbulk or mixed meanoouter tube or outside surface
t
tube
w
wall
Brackets〈 〉averaged over flow cross section91

CHAPTER 9HEAT TRANSFER EQUIPMENT
9.1 Shell-and-Tube Heat Exchanger
There are various kinds of heat exchangers in industrial processes: double-pipe exchanger,
shell-and-tube exchanger, cross-flow plate exchanger, etc.

1:tubes, 2:shell, 3:stationary tubesheet, 4:floating tubesheet, 5:floating head cover, 6:shell cover, 7:baffles, 8:shell
nozzle, 9:stationary head nozzle, 10:pass partition, 11:tie rods and spacers, 12:impingement baffle, 13:channel coverFig.9.1-1. Floating-head 1-2 heat exchanger

Figure 9.1-1 shows a floating-head one-shell-pass, two-tube-pass heat exchanger.
Here we deal with the thermal analysis of shell-and-tube exchangers, which are widely usedbecause they can be constructed with large heat-transfer surfaces in a relatively small volume. The
above figure is a sketch of a shell-and-tube heat exchanger with one shell pass and two tube passes.
Usually the hot fluid flows through the tubes and the cold fluid through the shell. The shell-side andtube-side heat transfer coefficients are of comparable importance, and both must be large to attain asatisfactorily large overall coefficient. Heat transfer tubes are laid out on either square or triangularpatterns.
The tube pitch 𝑃𝑡 is the shortest center-to-center distance between adjacent tubes. Usually we
use 1.25 times the outside diameter of the tubes for triangular pitch. Since the velocity on the shellside is usually low in comparison with that on the tube side, 25%-cut baffles are installed in theshell to decrease the flow cross-section and to cause the fluid to flow across the tube bank rather
than parallel with it. An increase in the shell-side coefficient is obtained by this type of cross flow
across the tube bank. Standard length of tubes is usually 5 m. Shells are fabricated from pipe or byrolling plate of steel or stainless steel. It is often difficult to obtain a high velocity when the
tube-side fluid flows through all the tubes in a single pass. Multipass construction by means of thepass partitions are used to increase the tube-side velocity with a corresponding increase in the
tube-side coefficient. The sizes of tubes which are frequently used are in the range 15 to 30 mm
OD.

9.2 Tube-side Heat Transfer Coefficient
In the last chapter, we concluded that the Nusselt number for fully-developed temperature field is

Heat Transfer Coefficient in Annular Space of Double-tube Exchangers

93

a function of Reynolds number and Prandtl number. The inside heat transfer coefficient ℎ𝑖 for the
tube-side fluid can also be correlated in the j-factor form:
𝑁𝑢

−0.14

𝜇

(9.2-1)

𝑗𝐻 = 𝑅𝑒 𝑃𝑟 −1/3 ( 𝜇𝑚)𝑤where 𝜇𝑚 , 𝜇𝑤 are the viscosities of the tube-side fluid at the mixed-mean temperature 𝑇𝑚 andwall temperature 𝑇𝑤 . Unless the temperature difference between 𝑇𝑚 and 𝑇𝑤 is very large, the−0.14
viscosity correction term 𝜙 = (𝜇𝑚 ⁄𝜇𝑤 )may be considered to be unity.
For highly turbulent pipe flow, we can use the empirical equation (Sieder and Tate correlation) 1)
𝐷 𝐺 0.8 𝐶𝑝 𝜇 1/3 𝜇𝑚 0.14
( 𝜅 ) (𝜇 )𝑤

ℎ𝑖 𝐷

(9.2-2)= 0.026 ( 𝜇 )𝜅or in the Colburn j-factor form
𝑗𝐻 = 0.026 𝑅𝑒 −0.2
(9.2-3)
which gives the heat transfer coefficient within ±20 % error in the range 𝑅𝑒 = 104 ~ 105 ,Pr = 0.6 ~ 100, and 𝐿⁄𝐷 > 10. In moderate heat transfer conditions 𝜇𝑚 ⁄𝜇𝑤 ≅ 1.
We can also use
𝐷 𝐺 0.8 𝐶𝑝 𝜇 1/3ℎ𝑖 𝐷= 0.023 ( 𝜇 )𝜅(9.2-4)( 𝜅 )For laminar flow we have to consider the boundary layer effect of entrance length developing fromthe pipe entrance. We can use the following equation
0.14

𝜇

ℎ𝑖 𝐷

(9.2-5)= 1.86 (𝑅𝑒 Pr 𝐷⁄𝐿 )1/3 ( 𝜇𝑚)𝜅𝑤

which is good within ±20 for 𝑅𝑒 Pr 𝐿⁄𝐷 > 10. Keep in mind that Eq.(9.2-5) gives an averageheat transfer coefficient in the developing temperature field.
Note that Figure 8.2-1 gives this laminar flow correlation in the form of parallel straight lines.
1.

Sieder, E.N. and Tate, G.E., Ind. Eng. Chem., 28, 1429 (1936)

9.3 Heat Transfer Coefficient in Annular Space of Double-tube
Exchangers1)
For evaluating the heat transfer coefficient of the annular side, we have to consider the hydraulicequivalent diameter for noncircular flow passage.
The heat transfer through the inner tube-side wall in turbulent annular flow can be correlated byequations similar to those used for circular tube flow accounting for the annular geometry.
Here the equivalent diameter 𝐷𝑒𝑞 is used.
One of the correlative equations isℎ𝑜 𝐷𝑒𝑞𝜅

𝐷𝑒𝑞 𝐺 0.8 𝐶𝑝 𝜇 0.4 𝐷2 0.45

= 0.023 (

𝜇

)

(

𝜅

)

(9.3-1)( )𝐷1

4𝐷𝑒𝑞 = 𝐷2 − 𝐷1 , 𝑅𝑒 > 10
where 𝐷2 and 𝐷1 are the inside diameter of the outer tube and the outside diameter of the innertube, respectively.
1.

Kern, D. Q., Process Heat Transfer, McGraw-Hill, New York, 137 (1950)

9.4 Shell-side Heat Transfer Coefficient
As shown in Fig.9.4-1, the triangular pitch gives high shell-side heat transfer coefficients because
the shell-side fluid flowing between adjacent tubes impinges directly on the succeeding tube row.

94

Heat Transfer Equipment

Fig.9.4-1. Two kinds of tube layouts for exchangers and equivalent diameter

Because the fluid flows partly parallel to the tube bank and partly across it, we would use acharacteristic mass velocity and an equivalent diameter appropriate for the flow pattern.
The mass velocity 𝐺𝑐 for cross flow is the mass flow rate divided by the total flow area 𝑆𝑐 fortransverse flow between the tubes in the row at or closest to the centerline of the exchanger.
𝑊
𝐺𝑐 = 𝑃 𝐷 − (𝐷 𝑠⁄𝑃 )𝑃 𝑑 for either square or triangular pitch(9.4-1)
𝑏 𝑠

𝑠

𝑡

𝑏 𝑜

where 𝑃𝑡 is the tube pitch and 𝑃𝑏 the baffle pitch.

Fig.9.4-2. Shell-side flow passage in tube bundles

We choose the cross-flow mass velocity 𝐺𝑐 for the characteristic mass velocity. The equivalentdiameter for the shell is taken as the equivalent diameter for the fictitious parallel flow without
baffles:

𝐷𝑒𝑞 =
𝐷𝑒𝑞 =

𝜋
1
√3
𝑃 2 − 𝑑𝑜 2 )
4 𝑡
4
2
1
𝜋 𝑑𝑜
2
𝜋
4(𝑃𝑡 2 − 𝑑𝑜2 )
4

4(

𝜋 𝑑𝑜

for triangular pitch(9.4-2)for square pitch(9.4-3)

If we use 𝐺𝑐 and 𝐷𝑒 as the mass velocity and equivalent diameter, the Nusselt number iscorrelated with the Reynolds number in the form
ℎ𝑜 𝐷𝑒𝑞𝜅= 0.36 (𝐷𝑒𝑞 𝐺𝑐 0.55𝜇)𝑅𝑒 = 2,000 ~ 1,000,000.

(

𝐶𝑝 𝜇 1/3 𝜇𝑚 0.14
𝜅

)

(

𝜇𝑤

)

(9.4-4)

True Temperature Difference for 1-2 Exchangers

95

9.5 True Temperature Difference for 1-2 Exchangers1)
The temperature-length curves for a 1-2 exchanger having the nozzle arrangement indicated areshown in Figure 9.5-1. The curve 𝑇ℎ𝑖 → 𝑇ℎ𝑡 and 𝑇𝑐𝑖 → 𝑇𝑐𝑜 can be considered as those of acounter-flow exchanger, and the curves 𝑇ℎ𝑡 → 𝑇ℎ𝑜 and 𝑇𝑐𝑖 → 𝑇𝑐𝑜 correspond to a parallel-flow
exchanger.
The LMTD for counter-flow can be formally written as

𝐿𝑀𝑇𝐷 = (∆𝑇)𝑙𝑚 =(𝑇ℎ𝑖 − 𝑇𝑐𝑜 )−(𝑇ℎ𝑜 − 𝑇𝑐𝑖 )𝑇 − 𝑇𝑐𝑜𝑇ℎ𝑜 − 𝑇𝑐𝑖

ln ℎ𝑖

(9.5-1)This cannot be used without correction to calculate the true mean temperature difference. A
correction factor 𝐹 is so defined that when it is multiplied by the counter-flow LMTD, the productis the correct mean temperature difference. Figure 9.5-2 shows the factor 𝐹 for 1-2 exchangers as afunction of the two dimensionless parameters. This correction factor shown in Fig.9.5-2 can also be
used for exchangers with one shell pass and even tube passes.
1.

Bowman, R.A., Mueller, A.C. and Nagle, W.M., Trans. A.S.M.E., 62, 283 (1940)

Fig.9.5-1. Temperature—pipe length curve for 1-2 exchanger

Fig.9.5-2 Correction factor plot for exchangers with one shell pass and two, four, or any multiple
passes96

Heat Transfer Equipment

Therefore the heat transfer surface area 𝐴𝑜 of a shell-and-tube exchanger can be calculated by
𝑄 = 𝑈𝑜 𝐴𝑜 (∆𝑇)𝑙𝑚 𝐹(9.5-2)where the LMTD to be used with this kind of correction factor is for a counter-flow double-pipe
heat exchanger with the same fluid inlet and exit temperatures. The correction factors 𝐹 forvarious and complex configurations are omitted here. If necessary, you should refer to thehandbooks.

9.6 Engineering Design of a Shell-and-Tube Heat Exchanger
Only for simplicity, we will study the fundamental calculation method.
(1) Pressure drop in heat exchanger(1-A) Tube-side pressure drop
The tube -side pressure drop can be calculated in a manner basically similar to the pressure
drop in a single straight pipe. That is, the tube-side pressure drop is the sum of the pressure
drop over the straight pipe section and the pressure drop due to the change of flow direction
caused in the front and rear heads when entering or leaving the tube-sheets.
The pressure drop over the straight pipe section is given by∆𝑝𝑡 =4 𝑓 𝐺𝑡 2 𝐿 𝑛
2 𝐷 𝜌 𝜙𝑡(9.6-1)

where 𝑛 is the number of tube passes. Therefore (1⁄2)𝐺𝑡 2 /𝜌 means (1⁄2)𝜌 𝑣𝑡 2 and−0.14
𝐿 𝑛 is the total length of path. The viscosity correction term 𝜙 = (𝜇𝑚 ⁄𝜇𝑤 )The pressure drop due to the abrupt change of direction in flowing from one pass into thenext pass can be considered as four velocity-heads per pass:1
Δ𝑝𝑟 = 4 𝑛 2 𝜚𝑣𝑡 2(9.6-2)(1-B) Shell-side pressure drop
It can be considered that the pressure drop is proportional to the number of times and the
distance the fluid crosses the tube bundle between baffles. Using a modification of Eq.(7.4-8),
an empirical correction available for calculation of the pressure drop through the shell side:𝐺𝑐 2 𝐷𝑠 (𝑁+1)(1⁄2) 𝜌 𝐷𝑒 𝜙𝑠⁄(1 2)𝐺𝑐 2 /𝜌 implies the kinetic energy∆𝑝𝑠 = 4 𝑓𝑠

(9.6-3)where(1⁄2)𝜌 𝑣𝑐 2 and 𝐷𝑠 (𝑁 + 1) is a kind of−0.14
equivalent length the fluid travels. The viscosity correction term is 𝜙 = (𝜇𝑚 ⁄𝜇𝑤 ). If 𝑁is the total number of baffles in the shell, the shell-side fluid crosses (N + 1) times the tube
bundle. That is, (N + 1) equals the ratio of tube length 𝐿 to baffle spacing 𝑃𝑏 . The
equivalent diameter 𝐷𝑒𝑞 is the same as for heat transfer. The above equation includes
entrance and exit losses.
The friction factor 𝑓𝑠 = 0.43 𝑅𝑒𝑠 −0.19
500 < 𝑅𝑒𝑠 < 105(9.6-4)
𝐷𝑒𝑞 𝐺𝑐where 𝑅𝑒𝑠 = 𝜇 .
9.6-1 Thermal design procedure of double-tube exchangers[EXAMPLE 9.6-1] 4,000 kg/h of 98 mol% methanol from an overhead condenser is to be cooled
from 62 C (=335 K) to 30 C (=3034 K) by a double-tube exchanger which consists of 50 mm OD,
2 mm thick copper tube and 80 mm ID copper tube. Methanol solution is placed on the tube side
and cooling water is supplied into the annulus in counter-current direction. The inlet watertemperature is 20 C (=293 K) and the outlet temperature is desired to be 25 C (=298 K). Obtain thenecessary total length of double tubes.
[Solution] The physical properties of methanol at mean temperature 46 C (=319 K) are:

Thermal Design Procedure of Double-tube Exchangers97

𝜌 = 780 𝑘𝑔⁄𝑚3 , 𝐶𝑝 = 2.59 × 103 𝐽⁄𝑘𝑔 𝐾 , 𝜇 = 0.0004 𝑘𝑔⁄𝑚 𝑠, and𝜅 = 0.196 𝑊 ⁄𝑚 𝐾.
Heat balance: 𝑄 = (4000)(2.59 × 103 )(335 − 303) = 𝑊 (4.183 × 103 )(298 − 293)
= 3.3165 × 109 𝐽⁄ℎ = 0.921 × 105 𝑊
From this equation, the required water rate is
𝑊 = 1.585 × 104 𝑘𝑔⁄ℎHeat transfer coefficient on methanol side:
Mass velocity(4000⁄3600)
𝐺 = (𝜋⁄4)(0.046)2 = 668.6 𝑘𝑔⁄𝑚2 𝑠
𝑅𝑒 = 𝐷 𝐺 ⁄𝜇 = (0.046) (668.6)⁄0.0004 = 76,900(turbulent flow)3
𝑃𝑟 = 𝐶𝑝 𝜇⁄𝜅 = (2.59 × 10 ) (0.0004⁄0.196) = 5.286
From Eq.(9.2-4)ℎ𝑖 = 0.023 (𝜅⁄𝐷) 𝑅𝑒 0.8 𝑃𝑟 1⁄3 = (0.023)(0.196⁄0.046)(76,900)0.8 (5.286)1⁄3= 1,380 𝑊 ⁄𝑚2 𝐾
Heat transfer coefficient on water side:
𝐷𝑒𝑞 = 𝐷2 − 𝐷10 = 0.08 − 0.05 = 0.03 𝑚(1.6585 × 104 ⁄3,600)

𝐺𝑜 = (𝜋⁄4)((0.08)2− (0.05)2) = 1,440 𝑘𝑔⁄𝑚2 𝑠

The physical properties of water at 22.5 C (=295.5 K) are: 𝜌 = 1,000 𝑘𝑔⁄𝑚3 ,𝐶𝑝 = 4.183 × 103 𝐽⁄𝑘𝑔 𝐾 , 𝜇 = 0.0009 𝑘𝑔⁄𝑚 𝑠, and 𝜅 = 0.616 𝑊 ⁄𝑚 𝐾 .
𝑅𝑒𝑒𝑞 = 𝐷𝑒𝑞 𝐺𝑜 ⁄𝜇 = (0.03)(1440⁄0.0009) = 48,000
𝑃𝑟 = 𝐶𝑝 𝜇⁄𝜅 = (4.183 × 103 ) (0.0009⁄0.616) = 6.11
From Eq.(876)0.45ℎ𝑜 = 0.023 (𝜅𝑏 ⁄𝐷𝑒𝑞 )(𝐷𝑒𝑞 𝐺𝑜 ⁄𝜇 )0.8 (𝐶𝑝 𝜇 ⁄𝜅)0.4 (𝐷2 ⁄𝐷1 )= 0.023 (0.616⁄0.03)(48000)0.8 (6.11)0.4 (80⁄50)0.45 = 6690 𝑊 ⁄𝑚2 𝐾
Neglecting the thermal resistance of copper tube wall and assuming the overall fouling
resistance of 1⁄ℎ𝑠 = 4.4 × 10−4 (𝑊 ⁄𝑚2 𝐾 )−1 , the overall coefficient is1
𝑈𝑜 == 726.1 𝑊 ⁄𝑚2 𝐾
1 0.0501−41380 0.046 + 4.4 × 10 + 6690
Temperature difference:(335−298)−(303−293)
𝐿𝑀𝑇𝐷 = (∆𝑇)𝑙𝑚 == 20.6 𝐾
335−298ln

303−293

Then the area required for the heat transfer rate is
𝐴𝑜 = 𝑄⁄𝑈𝑜 (∆𝑇)𝑙𝑚 = (0.921 × 105 )⁄(726.1)(20.6) = 6.16 𝑚2
The necessary total tube length is
𝐿 = 𝐴𝑜 ⁄𝜋𝐷 = 6.16⁄𝜋(0.05) = 39.2 𝑚
9.6-2 Thermal design procedure of shell-and-tube exchangers[EXAMPLE 9.6-2] 30,000 kg/h of liquid mixture of organic compounds is to be heated from
30℃(= 303 K) to 70℃ (= 343 K) by a shell-and-tube exchanger before feeding into a
distillation column.
The liquid mixture is placed on the shell side and as the heating medium, 42,000 kg/h oftoluene from the reboiler of another column is supplied at 105℃ (= 378 K) into the tube side.
Plant practice uses 25 mm OD, 22 mm ID smooth steel pipes 5 m long on 35 mm square
pitch and baffles 250 mm apart. A 1 × 105 Pa pressure drop is permissible on both streams.
Assume the overall fouling resistance factor 1⁄ℎ𝑠𝑡 to be 3.8 × 10−4 (W/m2K)-1 . Obtainthe necessary size of shell-and-tube exchanger for the requirement.

98

Heat Transfer Equipment

𝑇℃
30
50
70
90

𝜌

kg/m3
890
870
855
845

𝑇℃
50
70
90
110

𝜌

kg/m3
850
820
800
770

Table 9.6 Physical properties
Organic liquid mixture
𝜇 kg/m s
𝜅 W/m K
0.00170.150
0.00120.147
0.000940.145
0.00076
0.142
Toluene𝜇 kg/m s
𝜅 W/m K
0.000400.128
0.000340.124
0.000280.117
0.000210.114[Solution]
Average heat capacities are
Liquid mixture 𝐶𝑝𝑜 = 1.91 × 103

J/kg Kat

𝑇𝑐𝑜 =TolueneJ/kg Kat

𝑇𝑐𝑜 =𝐶𝑝𝑖 = 1.97 × 103

𝐶𝑝 J/kg K
1.85 x 103
1.91 x 103
1.98 x 103
2.04 x 103
𝐶𝑝 J/kg K
1.80 x 103
1.88 x 103
1.97 x 103
2.05 x 103

30+70= 50 ℃2
105+77.4
2= 91.2 ℃（Because the toluene side exit temperature is unknown, the above toluene heat capacity isdetermined by trial-and-error method applied to the following total heat balance）(i)
Heat Balance:𝐽
𝑄 = (30,000)(1.91 × 103 )(70 − 30) = 2.292 × 109 = 6.367 × 105 𝑊ℎ= (42,000) 𝐶𝑝𝑖 (105 − 𝑇𝑖2 )From the equation, the outlet temperature of the tube-side stream is 𝑇𝑖2 = 77.4 ℃.
(ii)
True temperature difference:(105 − 70) − (77.4 − 30)
𝐿𝑀𝑇𝐷 = (∆𝑇)𝑙𝑚 == 40.9 K
105 − 70
ln 77.4 − 30
105−77.4
70−30
For 1 – 4 exchanger
𝑍 = 70−30 = 0.69
𝑌 = 105−30 = 0.533
From Fig.9.5-2, F = 0.88
Therefore the true temperature difference ∆𝑇 = (∆𝑇)𝑙𝑚 𝐹 = (40.9)(0.88) = 36.0 𝐾(iii) Trial:
Assume 𝑈𝑜 = 300 𝑊 ⁄𝑚2 𝐾
Then the total heat-transfer area 𝐴𝑜 = 𝑄⁄𝑈𝑜 ∆𝑇 = 6.367 × 105 ⁄(300)(36.0) = 58.95 𝑚2
Outside surface area per tube 𝑎𝑜 = 𝜋𝐷𝑜 𝐿 = 𝜋(0.025)(5) = 0.393 𝑚2
Total number of tubes 𝑛𝑡 = 𝐴𝑜 ⁄𝑎𝑜 = 58.95⁄0.393 = 150 < 152
Each tube pass can accommodate 38 tubes.
Then a new overall heat transfer coefficient is re-assumed.
𝑈𝑜 = 𝑄⁄𝑛𝑡 𝑎𝑜 ∆𝑇 = 6.367 × 105 ⁄(152)(0.393)(36.0) = 296.1 𝑊 ⁄𝑚2 𝐾
Tube-side
Flow area per tube 𝑎𝑡 = (𝜋⁄4) 𝐷𝑖 2 = (𝜋⁄4)(0.022)2 = 3.80 × 10−4 𝑚2
Flow area per pass: 𝐴𝑡 = (152⁄4)(3.80 × 10−4 ) = 0.0144 𝑚2
Mass velocity 𝐺𝑖0 = 𝑊𝑖 ⁄𝐴𝑡 = (42,000⁄3600)⁄(0.0144) = 810.2 𝑘𝑔⁄𝑚2 𝑠
Toluene viscosity at 𝑇𝑐𝑖 = 91.2℃: 𝜇𝑖 = 0.000276 𝑘𝑔⁄𝑚 𝑠
Reynolds number on the tube side: 𝑅𝑒𝑖 = (0.022) (810.2)⁄0.000276 = 64,600
𝐿⁄𝐷𝑖 = 5⁄0.022 = 227,
Then from the turbulent flow curve of Fig.8.2-1, 𝑗𝐻𝑖 = 0.0028 at 𝑅𝑒𝑖 = 64,6000.
At 𝑇𝑐𝑖 = 91.2℃, 𝐶𝑝𝑖 = 1.97 × 103 J/kg K, 𝜅𝑖 = 0.117 𝑊 ⁄𝑚 𝐾

Thermal Design Procedure of Shell-and-Tube Exchangers (continued)

99

Prandtl number 𝑃𝑟𝑖 = 𝐶𝑝𝑖 𝜇𝑖 ⁄𝜅𝑖 = 4.65.
Then the inside heat-transfer coefficient isℎ𝑖 ⁄𝜙𝑖 = (𝑗𝐻𝑖 ⁄𝜙𝑖 )(𝜅𝑖 ⁄𝐷𝑖 )𝑅𝑒𝑖 𝑃𝑟𝑖 1⁄3 = (0.0028)(0.117⁄0.022)(64,600)(4.65)1⁄3= 1,606 𝑊 ⁄𝑚2 𝐾
Shell side
The inside diameter of shell accommodating 152 heat-transfer tubes of 25 mm OD can be
estimated from the tube layout on 35 mm square pitch. Since one square which has area of(0.035)2 = 0.00123 m2 can accommodate one tube. From the following equation,(𝜋⁄4) 𝐷𝑠 2 = (0.00123)(152) = 0.187 𝑚2
we get 𝐷𝑠 = 0.488 𝑚 as the first approximate value.
We need to avoid a collision between the inside wallof the shell and the outer-most tubes. A space for𝐷𝑜 ⁄2 is still necessary at both ends of the shell
diameter:
𝐷𝑠 = 0.488 + (0.0125 × 2) = 0.513 𝑚
Further about a tube diameter space for tie-rods
is necessary between the outer edge of the tube
bundle and the inside wall of the shell:
𝐷𝑠 = 0.513 + (0.025 × 2) = 0.563 𝑚
Equivalent diameter for square pitch is
4 (𝑃𝑡 2 − (𝜋⁄4) 𝐷𝑜 2 )4 ((0.035)2 − (𝜋⁄4) (0.025)2 )𝐷𝑒𝑞 === 0.0374 𝑚𝜋𝐷𝑜𝜋(0.025)
Mass velocity for crossflow position𝑊𝑜(30,000⁄3,600)
𝐺𝑐 ==(0.25)(0.563) − (0.563⁄0.035)(0.25)(0.25)
𝑃𝑏 𝐷𝑠 − (𝐷𝑠 ⁄𝑃𝑡 )𝑃𝑏 𝐷𝑜= 207.2 𝑘𝑔⁄𝑚2 𝑠
At 𝑇𝑐𝑜 = 50℃, 𝐶𝑝𝑜 = 1.91 × 103 𝐽⁄𝑘𝑔 𝐾, 𝜇𝑜 = 0.0012 𝑘𝑔⁄𝑚 𝑠,
𝜅𝑜 = 0.147 𝑊 ⁄𝑚 𝐾,
Prandtl number: 𝑃𝑟𝑜 = 15.6.
Reynolds number: 𝑅𝑒𝑜 = 𝐷𝑒𝑞 𝐺𝑐 ⁄𝜇𝑜 = (0.0374) (207.2)⁄0.0012 = 6,460
From Eq.9.4-4, we obtainℎ𝑜 ⁄𝜙𝑜 = 0.36 (𝜅𝑜 ⁄𝐷𝑒𝑞 ) 𝑅𝑒𝑜 0.55 𝑃𝑟𝑜 1⁄3= (0.36)(0.147⁄0.0374)(6460)0.55 (15.6)1⁄3 = 440.6 𝑊 ⁄𝑚2 𝑠 𝐾
Tube-wall temperature
From heat balance neglecting the fouling resistanceℎ𝑖1(𝑇𝑤 − 𝑇𝑐𝑖 ) = 𝜙𝑖 𝜙𝑜 (𝑇𝑐𝑜 − 𝑇𝑐𝑖 )𝜙𝑖That isℎ𝑖+ℎ𝑜ℎ𝑜 ⁄𝜙𝑜(𝑇 − 𝑇𝑐𝑖 )(ℎ𝑖 ⁄𝜙𝑖 ) + (ℎ𝑜 ⁄𝜙𝑜 ) 𝑐𝑜
= 91.2 + (440.6)(50 – 91.2)/(1606 + 440.6) = 82.3 C (= 355.3 K)
At 𝑇𝑤 = 82.3 𝐶 𝜇𝑤𝑜 = 0.000829 𝑘𝑔⁄𝑚 𝑠, 𝜇𝑤𝑖 = 0.000303 𝑘𝑔⁄𝑚 𝑠
0.14
0.14
𝜙𝑜 = (𝜇 ⁄𝜇𝑤 )= (0.0012⁄0.000829)= 1.050.14
0.14
𝜙𝑖 = (𝜇 ⁄𝜇𝑤 )= (0.000276⁄0.000 − 303)
= 0.99𝑇𝑤 = 𝑇𝑐𝑖 +Corrected heat –transfer coefficientsℎ𝑜 = (ℎ𝑜 ⁄𝜙𝑜 ) 𝜙𝑜 = (440.6)(1.05) = 462.6 W/m2K

100

Heat Transfer Equipmentℎ𝑖 = (ℎ𝑖 ⁄𝜙𝑖 ) 𝜙𝑖 = (1606)(0.99) = 1590 W/m2K
The thermal conductivity of steel tube: 𝜅𝑡 = 45.0 𝑊 ⁄𝑚 𝐾
Average tube-diameter (logarithmic mean)(𝐷𝑜 − 𝐷𝑖 )0.025 − 0.022
𝐷𝑎𝑣 === 0.0235 𝑚
𝐷𝑜
0.025
ln 𝐷
ln 0.022𝑖
Overall heat-transfer coefficient
1𝑈𝑜 =1 𝐷𝑜 𝛿𝑡 𝐷𝑜
1
1ℎ𝑖 𝐷𝑖 + 𝜅𝑡 𝐷𝑎𝑣 + ℎ𝑜 + ℎ𝑠𝑡
1= 1 0.025 0.0015 0.025 1= 303. 8 W/m2K (vs. 296 assumed)−4
1590 0.022+

45.0 0.0236

+

462.6

+3.8× 10

(If the calculated coefficient is greatly different from the assumed value, theforegoing calculation should be repeated with new values assumed.)
The overall coefficient calculated above is only 2.6% larger on the safe side than
the assumed value. The exact value of the shell diameter 𝐷𝑠 should bedetermined from drawing of tubes laid on 35 mm square pitch taking into
account the spaces for pass partition, tie rods, etc.
SHELL SIDE
1 passTUBE SIDE
4 tube passes

𝐷𝑠 = 0.563 m

𝐷𝑜 = 0.0-25 m, 𝐷𝑖 = 0.022 m, 𝐿 = 5 m

𝑇𝑜1 = 30 C

𝑇𝑖1 𝑇𝑖1 = 105 C

𝑇𝑜2 = 70 C

𝑇𝑖2 = 77.4 C

𝑇𝑐𝑜 = 50 C

𝑇𝑐𝑖 = 91.2 C

𝑅𝑒𝑜 = 6,460

𝑅𝑒𝑖 = 64,600ℎ𝑜 =462.6 W/m2Kℎ𝑖 = 1,590 W/m2K
𝑄 = 6.367 x 105 W∆𝑇 = 36.0 K, 𝑇𝑤 = 82.3 C
𝑛𝑡 = 152
𝑈𝑜 = 303.8 W/m2K(iv)

Pressure drop(a) Shell side:
From Eq.(9.6-4), 𝑓𝑠 = 0.43 𝑅𝑒𝑠 −0.19 = 0.0812 at 𝑅𝑒𝑠 = 6,460
In Eq.(9.6-3), 𝑁 = 20, 𝜙𝑠 = 1.05 , 𝐺𝑐 = 207.2 kg/m2 s, 𝐷𝑠 = 0.563 m
𝜚 = 870 𝑘𝑔⁄𝑚3 , 𝐷𝑒𝑞 = 0.0374 𝑚(207.2)2 (0.563)(21)∆𝑝𝑠 = 4 (0.0812) 2 (870)(0.0374)(1.05) = 2,400 < 105 𝑃𝑎6This value is smaller than the permissible limit.
(b) Tube side:
From Fig.7.3-1, 𝑓 = 0.0049
at 𝑅𝑒𝑖 = 64,600
In Eq.(9.6-1), 𝐺𝑡 = 810.2 𝑘𝑔⁄𝑚2 𝑠 , 𝐿 = 5 m, 𝑛 = 4, 𝜙𝑡 = 0.99 , 𝜚 = 798 𝑘𝑔⁄𝑚3 ,𝐷𝑖 = 0.022 𝑚
4 (0.0049)(810.2)2 (5)(4)∆𝑝𝑡 == 7,403 𝑃𝑎
2 (0.022)(798)(0.99)
In Eq.(9.6-4), 𝑣𝑡 = 𝐺𝑖 ⁄𝜌 = 810.2⁄798 = 1.02 𝑚⁄𝑠

Problems101∆𝑝𝑟 = 4 (4)(1⁄2)(798)(1.02)2 = 6,642 𝑃𝑎
Then the total pressure drop on the tube side is∆𝑝 = ∆𝑝𝑡 + ∆𝑝𝑟 = 7,403 + 6,642 = 1.04 × 104 < 105 𝑃𝑎
This value is also smaller than the permissible limit.
[PROBLEM 9.6-1]
A triangular pitched 1-2 shell-and-tube exchanger is operated to cool 4,000 kg/h of oil.
At the present time, the oil, which is the tube-side fluid, is being cooled from 90 C (=363
K) to 63 C (= 336 K), while water as the shell-side fluid is being heated from 30 C (= 303
K) to 38 C (= 311 K). The exchanger has the effective heat-transfer area of 10 m2 on thebasis of the outside diameter. The oil has average heat capacity 𝐶𝑝 = 2.2 × 103 𝐽⁄𝑘𝑔 𝐾
and density 𝜌 = 880 𝑘𝑔⁄𝑚3 . Determine the present value of the overall heat-transfer
coefficient.
(Answer: 162.2 W/m2K)[PROBLEM 9.6-2]
100,000 kg/h of crude oil is to be heated from 70 C (= 343 K) to 100 C (=373 K) by ashell-and-tube exchanger before feeding into the first distillation column of oil refinery.
The crude oil is placed on the tube side and as the heating medium, 65,000 kg/h ofkerosene is supplied at 200 C (= 473 K) into the shell side. Plant practice uses 25 mm
OD, 22 mm ID smooth steel pipes 5 m long on 35 mm triangular pitch and 25%-cut
bafflers 250 mm apart. Permissible pressure drops on the tube-side and the shell-sidestreams are 3 × 105 and 1 × 105 𝑃𝑎, respectively.
Calculate the heat transfer area required and obtain the size of the shell-and-tube
exchanger. Determine the appropriate number of the tube passes.
The following values of the fouling resistance factors can be used.
−1−11⁄ℎ𝑠𝑖 = 3.5 × 10−4 (𝑊 ⁄𝑚2 𝐾) ,1⁄ℎ𝑠𝑜 = 1.8 × 10−4 (𝑊 ⁄𝑚2 𝐾)
Table 9.6-2 Physical properties of kerosene and crude oil
𝑇 ℃
120
140
160
180
200

𝜌 𝑘𝑔⁄𝑚3

𝐶𝑝 𝑘𝐽⁄𝑘𝑔 𝐾

𝜇 𝑘𝑔⁄𝑚 𝑠κ 𝑊 ⁄𝑚 𝐾

740
720
710
690
680

2.39
2.47
2.55
2.63
2.71

0.000480.000390.000310.000260.00021

0.135
0.133
0.131
0.130
0.128

9.7

Convective Heat Transfer around Submerged Objects

𝑇 ℃
60
80
100
120
140
160

𝜌 𝑘𝑔⁄𝑚3

𝐶𝑝 𝑘𝐽⁄𝑘𝑔 𝐾

𝜇 𝑘𝑔⁄𝑚 𝑠κ 𝑊 ⁄𝑚 𝐾

820
810
790
780
770

2.08
2.16
2.24
2.34
2.43

0.00320.00230.00170.00130.000960.00071

0.132
0.131
0.130
0.129
0.128

9.7-1 A circular cylinder in cross flow1,2)
Let us consider a circular cylinder (diameter 𝐷) submerged in the free stream with uniform
velocity 𝑣∞ and temperature 𝑇∞. The cylinder surface is kept at a temperature 𝑇𝑆 .
The Nusselt number and Reynolds number are defined asℎ 𝐷
𝑣 𝐷𝜌
𝑁𝑢𝐷 = 𝑙𝑜𝑐
𝑅𝑒𝐷 = ∞𝜇 ∞𝜅𝑓𝑓where the viscosity 𝜇𝑓 and thermal conductivity 𝜅𝑓 should be evaluated at the film temperature.

102

Heat Transfer Equipment

Fig.9.7-1. A circular cylinder in a uniform flow normal to its axis

Fig.9.7-2 Local Nusselt numbers around a circular cylinder in a transverse air stream

When the Reynolds number is small enough, the boundary layer developing from the front
stagnation point is kept laminar and the flow in the separated region formed on the back side is stilllaminar.
Fig. 9.7-2 indicates a typical continuously decreasing variation in the Nusselt number at low
Reynolds numbers. We can see a slight rise in the separated wake region of the cylinder. When the
Reynolds number becomes sufficiently large, the boundary layer undergoes transition from laminar
to turbulent flow and the local distribution of heat transfer changes characteristically. The point of
separation shifts upstream to a value beyond 90 degree and the heat transfer coefficients become
very large in the separated region.
For engineering calculations, the main interest is in the total rate of heat transfer, or in the heattransfer coefficient ℎ𝑎𝑣 averaged over the circumstance.
The following equations are available as the empirical heat transfer correlations in two Reynolds
number ranges:
𝑁𝑢𝐷 = (0.43 + 0.50 𝑅𝑒𝐷0.5 )𝑃𝑟 0.38
1 < 𝑅𝑒𝐷 < 103(9.7-1)
0.6
0.38
𝑁𝑢𝐷 = 0.25 𝑅𝑒𝐷 𝑃𝑟
103 < 𝑅𝑒𝐷 < 2 × 105(9.7-2)
Here the ratio of Prandtl number 𝑃𝑟∞ ⁄𝑃𝑟𝑆 coming from the temperature difference between thewall and fluid bulk temperatures is assumed to be approximately unity. The Nusselt number isdefined asℎ 𝐷
𝑁𝑢𝐷 = 𝑎𝑣
(9.7-3)𝜅𝑓

Convective Heat Transfer around Submerged Objects

1.
2.

103

Eckert, E. R. G. and R. M. Drake, Jr.: “Analysis of Heat and Mass Transfer,” McGraw-Hill, Chap.9, 406 (1972)
Zhukauskas, A. A., V. Makarevicius and A. Shlanciauskas: “Heat Transfer in Banks of Tubes in Crossflow of Fluid,” Mintis,
Vilnius, Lithuania (12968)

9.7-2 Hot-wire anemometry
Eq. (9.7-1) is very important equation to be utilized for the principle of hot-wire anemometry.
Two basic forms; the constant-current type and the constant-temperature type are available. Bothutilize the same physical principle but in different ways.
The heat transfer from a fine, long wire (hot wire diameter 𝑑, length 𝑙, and electric resistance 𝑅𝑤 )of tungsten or platinum to a flowing air (velocity 𝑣∞ , temperature 𝑇∞) may be described by thefollow dimensionless equation:
𝑁𝑢𝑑 = 𝛼 + 𝛽√𝑅𝑒𝑑(9.7-4)This equation is one form of Eq. (9.7-1) applied in the range of very small Reynolds numbers,
where the effect of Prandtl number is also assumed to be constant. Only for simplicity, let us study
the constant-temperature mode.
The wire attains an equilibrium temperature when the heat 𝐼 2 𝑅𝑤 generated in it is just balanced by
the convective heat loss from its surface.
The wire temperature 𝑇𝑤 can be observed in terms of its electric resistance 𝑅𝑤 = 𝑅𝑤0 (1 + 𝑟0 (𝑇𝑤- 𝑇0 )) which can be measured by the bridge circuit shown in Fig.9.7-2. The constant 𝑟0 is givenby the temperature coefficient of the wire resistance.

Fig.9.7-3. Hot wire and bridge circuit for hot-wire anemometry

The heat transfer coefficient can be calculated by the following equation:
𝐼2 𝑅ℎ𝑎𝑣 = 𝜋𝑑𝑙(𝑇 𝑤
−𝑇 )
𝑤

∞

The Reynolds number is given by𝑣 𝑑𝜌𝑅𝑒𝑑 = ∞𝜇 ∞∞(9.7-5)(9.7-6)

Because of a very thin hot wire, the Reynolds number is kept very low.
Finally the approach velocity 𝑣∞ can be measured from the electric current 𝐼 measured by thebridge circuit with a balance-detecting galvanometer by using the following equation:
𝐼 2 = 𝛼 ′ + 𝛽′ √𝑣∞(9.7-7)
where the constant 𝛼 ′ and 𝛽′ must be determined by calibration. Once calibrated, the hot-wire104

Heat Transfer Equipment

probe can be used to measure an unknown local velocity by adjusting 𝑅𝐼 until the bridge balance isachieved.
9.7-3 A circular sphere in cross flow
Let us consider a circular sphere (diameter 𝐷) submerged in the free stream with uniform
velocity 𝑣∞ .
Similarly to the case of circular cylinders, the heat transfer distribution changes depending on the
Reynolds number. For engineering purposes, the main interest is in the heat transfer coefficient
averaged over the spherical circumference.
The famous heat transfer correlation applicable in a wide range of the Reynolds number is given by
Ranz and Marshall1):ℎ𝑚 𝐷
𝜅𝑓= 2.0 + 0.60 (𝐷 𝜌𝑓 𝑣∞
𝜇𝑓

1⁄2

)

(

𝐶𝑝 𝜇𝑓
𝜅𝑓

1⁄3

)

1 < 𝑅𝑒 < 7 × 104

0.6 < Pr < 400(9.7-8)where the physical properties are evaluated at the film temperature 𝑇𝑓 = (𝑇∞ + 𝑇𝑆 )⁄2.
This equation is practically very convenient for estimating the heat transfer coefficients of particles,
bubbles and droplets flying in an infinite fluid flow.
1. Ranz, W. E. and W.R. Marshall, Jr.: Chem. Eng. Prog., vol.48, 141, 173 (1962)

Let us consider again the convective heat transfer around a circular cylinder and a circular spherein a free stream with uniform velocity and temperature. Figure 9.7-4 shows a comparison between
those convective heat transfer correlation curves. It is very interesting that both heat transfer
correlations are correspondent exactly to each other.

Fig.9.7-4. Comparison of convective heat transfer correlations between two submerged objects.

[PROBLEM 9.7-1] A liquid droplet of toluene is suspended at the hot junction of a very thin wire
thermocouple in a superheated toluene vapor (pressure 1 atm, temperature 𝑇∞ = 125 ℃) flowing
upward at uniform velocity 1 m/s. The droplet can be assumed to be kept as a sphere in shape. The
droplet temperature is almost kept at the boiling point 𝑇𝑆 = 110℃. At the starting time 𝑡 = 0, thedroplet diameter D = 𝐷0 = 4 𝑚𝑚. How long does it take until the droplet disappears?

Nomenclature

Fig.9.7-P1. A single droplet suspended in a superheated vapor stream
Nomenclature
𝐴
𝐶𝑝
𝐷
𝐷𝑒𝑞
𝐷𝑠
𝑑
𝑑𝑜
𝐹
𝑓
𝐺
𝐺𝑐
ℎ
𝐼
𝑗𝐻
𝐿
𝑁
𝑁𝑢
𝑛
𝑛𝑡
𝑃𝑟
𝑃𝑏
𝑃𝑡
𝑄
𝑅𝑒
𝑇
𝑇𝑠
𝑈
𝜇
𝜙

heat transfer area, [m2]heat capacity, [J/kg K]
tube inside diameter, [m], cylinder diameter, [m] or sphere diameter, [m]equivalent diameter, [m]inside diameter of shell, [m]hot-wire diameter, [m]outside diameter of heat transfer tube, [m]correction factor for temperature difference, [ - ]friction factor, [ - ]mass velocity, [kg/m2s]
mass velocity in cross flow section, [kg/m2s]
heat transfer coefficient, [W/m2K]
electric current of hot wire, [A]j-factor for heat transfer, [ - ]tube length, [m]total number of baffles, [ - ]Nusselt number, [ - ]number of tube passes, [ - ]total number of heat transfer tubes, [ - ]Prandtl number, [ - ]baffle pitch of tube layout, [m]pitch of tube layout, [m]heat transfer rate, [W]Reynolds number, [ - ]temperature, [K]surface temperature, [K]overall heat transfer coefficient, [W/m2K]
viscosity, [kg/m s]correction factor for viscosity, [ - ]SubscriptsccoldDcylinder
f
film
h
hot
i
inner tube, inside surface or inletmbulk or mixed meanoouter tube, outside surface or outletsshell sidettube sidew
wall∞
free stream

105

CHAPTER 10
INTERPHASE MASS TRANSPORT
10.1 Definition of Mass Transfer Coefficient
Mass transfer mechanism also falls into one of the two large categories: molecular andconvective transports. The Fick’s law describes the ordinary diffusion as𝜕𝑥𝑁𝐴 − 𝑥𝐴 (𝑁𝐴 + 𝑁𝐵 ) = −𝑐 𝐷𝐴𝐵 𝐴(10.1-1)
𝜕𝑦where 𝑥𝐴 is mole fraction of component A and 𝑐 the total molar density. The second term of theleft side represents the contribution of the bulk motion.
Applying the Fick’s law to the interface
𝜕𝑥
𝑁𝐴𝑤 − 𝑥𝐴𝑤 (𝑁𝐴𝑤 + 𝑁𝐵𝑤 ) = −𝑐 𝐷𝐴𝐵 𝜕𝑦𝐴 |(10.1-2)𝑦=0

The mass transfer analog of the defining equation of heat transfer coefficient becomes(10.1-3)
𝑁𝐴𝑤 − 𝑥𝐴𝑤 (𝑁𝐴𝑤 + 𝑁𝐵𝑤 ) = 𝑘𝑥∗ (𝑥𝐴𝑚 − 𝑥𝐴𝑤 )Note that the mass transfer coefficient can be considered to be a function of the mass transfer rate.
Let us consider as an example the mass transfer of component A from a pipe wall to a flowing
fluid. Dissolution of benzoic acid (solid) from a pipe wall of cast benzoic acid to a turbulent flow ofwater is a good example. Fig.10.1-1 shows its concentration profile near the wall.

Fig.10.1-1. Time-averaged concentration profile near the wall of benzoic acid cast pipe

Then the mass flux at the pipe wall can be written by Fick’s law:𝜕𝑥
𝑁𝐴 |𝑟=𝑅 − 𝑥𝐴 (𝑁𝐴 + 𝑁𝐵 )|𝑟=𝑅 = −𝑐 𝐷𝐴𝐵 𝐴 |𝜕𝑟 𝑟=𝑅(10.1-4)

The main resistance to mass transfer takes place in very thin fluid film near the pipe wall, wheremost of the total concentration drop occurs. Therefore the following approximation can be madedefining 𝛿𝐷 the thickness of a fictitious diffusion film:
𝜕𝑥𝑥 −𝑥− 𝜕𝑟𝐴 |
= 𝐴𝑤𝛿 𝐴𝑚
(10.1-5)
𝑟=𝑅

𝐷

Analogy between Heat and Mass Transfer

107

Here 𝑥𝐴𝑚 is the mixed mean concentration (bulk concentration) of component A, 𝑥𝐴𝑤 the wallconcentration (solubility of benzoic acid in this case).
The definition of the mixed mean concentration in a circular pipe flow is𝑅

𝑥𝐴𝑤 − 𝑥𝐴𝑚 =∫0 2𝜋𝑟 𝑣𝑧 (𝑥𝐴𝑤 − 𝑥𝐴 )𝑑𝑟(10.1-6)𝑅∫0 2𝜋𝑟 𝑣𝑧 𝑑𝑟

The above equation, Eq.(10.1-4) becomes the defining equation of convective mass transfer
coefficient:[𝑁𝐴 − 𝑥𝐴 (𝑁𝐴 + 𝑁𝐵 )]𝑟=𝑅 = (𝑐 𝐷𝐴𝐵 ⁄𝛿𝐷 )(𝑥𝐴𝑤 − 𝑥𝐴𝑚 ) = 𝑘𝑥∗ (𝑥𝐴𝑤 − 𝑥𝐴𝑚 )(10.1-7)
In this case, 𝑘𝑥∗ has unit of kmol/m2s.
The mass transfer analog of the Nusselt number is defined by𝑘∗ 𝐷
𝑆ℎ = 𝑐 𝐷𝑥(10.1-8)
𝐴𝐵This dimensionless group is called the Sherwood number. As in the derivation of heat transfer
correlation, the mass transfer correlation for low mass transfer rate can be found to be of the form:
𝑆ℎ = 𝑆ℎ(𝑅𝑒, 𝑆𝑐, 𝐿⁄𝐷 )(10.1-9)Actually, most of the available mass transfer data have been correlated using the following
definition:
𝑁𝐴𝑤 = 𝑘𝑥 (𝑥𝐴𝑚 − 𝑥𝐴𝑤 )(10.1-10)
For the case of mass transfer in a gas phase, the partial pressure difference can be used as thedriving force
𝑁𝐴𝑤 = 𝑘𝐺 (𝑝𝐴𝑚 − 𝑝𝐴𝑤 )(10.1-11)
The 𝑘𝐺 has units of kmol/m2s atm.
If the molar concentration difference is used as the driving force for liquids
𝑁𝐴𝑤 = 𝑘𝐿 (𝐶𝐴𝑚 − 𝐶𝐴𝑤 )(10.1-12)
The 𝑘𝐿 has units of m/s.
It should be kept in mind that as distinct from 𝑘𝑥∗ , these coefficients include the contribution of thebulk fluid motion.
For the case in which one component is transferred and the other is stagnant i.e. 𝑁𝐵 = 0,𝑘∗𝑘∗𝑁𝐴𝑤 = (1− 𝑥𝑥 ) (𝑥𝐴𝑚 − 𝑥𝐴𝑤 ) = 𝑥 𝑥 (𝑥𝐴𝑚 − 𝑥𝐴𝑤 ) = 𝑘𝑥 (𝑥𝐴𝑚 − 𝑥𝐴𝑤 )(10.1-13)
𝐴𝑤

𝐵𝑤

For gases𝑘∗
𝑁𝐴𝑤 = 𝑥 (𝑥𝐴𝑚 − 𝑥𝐴𝑤 ) = 𝑘𝐺 (𝑝𝐴𝑚 − 𝑝𝐴𝑤 )𝑥𝐵𝑤

Here 𝑥𝐵𝑤 = 𝑝𝐵𝑤 ⁄𝑃 (𝑃 : total pressure).
𝑘𝑥∗ = 𝑘𝑥 𝑥𝐵𝑤 , 𝑘𝑥∗ = 𝑘𝐺 𝑝𝐵𝑤 , 𝑘𝑥∗ = 𝑘𝐿 𝐶𝐵𝑤
Gas absorption belongs to this case.
For liquids𝑘∗
𝑁𝐴𝑤 = 𝑥 (𝑥𝐴𝑚 − 𝑥𝐴𝑤 ) = 𝑘𝐿 (𝐶𝐴𝑚 − 𝐶𝐴𝑤 )𝑥𝐵𝑤

Here 𝑥𝐵𝑤 = 𝐶𝐵𝑤 ⁄𝑐 (𝑐 : total molar density).
For the case of equimolar counter-diffusion i.e. 𝑁𝐴 = − 𝑁𝐵
𝑘𝑥∗ = 𝑘𝑥 , 𝑘𝑥∗ = 𝑘𝐺 𝑃, 𝑘𝑥∗ = 𝑘𝐿 𝑐Distillation belongs to this case.

(10.1-14)(10.1-15)(10.1-16)(10.1-17)

10.2 Analogy between Heat and Mass Transfer
By dimensional analysis, we can also obtain the same functional similarity for 𝑁𝑢 and 𝑆ℎ:𝑁𝑢 = 𝑁𝑢(𝑅𝑒, 𝑃𝑟)
𝑆ℎ = 𝑆ℎ(𝑅𝑒, 𝑆𝑐)

A new dimensionless group for mass transfer is defined as

108Interphase Mass Transport
𝑘

𝜇

𝑀

𝐴𝐵

𝑗𝐷 = 𝐺 𝑥 (𝜌𝐷 )2⁄ 3(10.2-1)= 𝑆ℎ 𝑅𝑒 −1 𝑆𝑐 −1⁄3𝑓

The subscript f refers to properties evaluated at the film temperature 𝑇𝑓 = (𝑇𝑚 + 𝑇𝑤 )⁄2 and thefilm composition 𝑥𝐴𝑓 = (𝑥𝐴𝑚 + 𝑥𝐴𝑤 )⁄2.
The Chilton-Colburn analogy1) is one of the important empirical analogies:
𝑗𝐷 = 𝑗𝐻

or

𝑘𝑥
𝐺𝑀

𝜇

(𝜌𝐷 )
𝐴𝐵

𝑓

2⁄ 3

ℎ

𝐶𝑝 𝜇

= 𝐶𝑝 𝐺 ( 𝜅 )

2⁄ 3

𝑓

(10.2-2)

This analogy permits one to calculate mass-transfer coefficients from heat-transfer data for theequivalent boundary conditions. The analogy between 𝑗𝐷 and 𝑗𝐻 is also applicable tosimultaneous heat and mass transfer.
1.

Chilton, T. H. and A. P. Colburn, Ind. Eng. Chem., 26, 1183 (1934)

10.3 Theory of Interphase Mass Transfer
Most of mass transfer processes in chemical engineering involve mass transfer across theinterface between two phases, as in gas absorption and distillation. Interphase transfer occurs owing
to the concentration gradients resulting from a deviation from equilibrium.
10.3-1 Fundamentals --- Gas-liquid equilibrium relationship for absorption
In gas absorption, we usually neglect the solubility of the inert gas (e.g., air) and the presenceof vapor from the liquid (e.g., water) in the gas. Therefore there are four variables: pressure,temperature, and the concentrations of solute component A in liquid and gas. For simplicity, thetemperature and pressure are usually fixed: one concentration may be chosen as the remaining
independent variable. The equilibria of interest are those between a nonvolatile absorbing liquid anda solute gas. The equilibrium relationship in absorption can be expressed as the diagram of theconcentration of solute 𝑥𝑒 in the liquid against the concentration of solute 𝑦𝑒 in the gas. One of
the important fundamental expression is the Henry’s law:𝑦𝐴 = 𝐻 𝑥𝐴(10.3-1)or𝑝𝐴 = 𝐻′ 𝐶𝐴(10.3-2)
where 𝑥𝐴 the mole fraction of solute A, and 𝐻 the Henry’s constant.
In the second equation, 𝑝𝐴 is the partial pressure of the gas phase, and 𝐶𝐴 the molar concentrationof solute A, and the modified Henry’s constant.
Fig.10.3-1 is a schematic diagram of the solubility of gases into a solvent such as water.

Fig.10.3-1 Schematic picture of gas solubility

Interphase Mass Transfer for Gas Absorption

109

10.3-2 Interphase mass transfer for gas absorption
Consider a transferring component A in an inert gas B which is in contact with a nonvolatileliquid C.
A simple example is the removal of sulfur dioxide from stack gases by water using a wetted wallcolumn.
Water is flowing downward by gravity in the form of liquid film falling along the inside surface of acircular cylinder whereas the stack gas is flowing upward in the form of counter-current inside thecylinder.
In general, there exist three resistances in series to the interphase transfer of component A: thegas phase film, the interface, and the liquid phase film. The concentration gradients in the twophases are shown in Fig.10.3-2.
Note that components B and C do not diffuse in spite of the concentration gradients.
Two-film theory, initially suggested by Whitman, is appropriate at this stage to explain thisprocess.
First, three assumptions are made: (1) that there is no resistance to transport of component A acrossthe actual interface, (this is equivalent to assuming that the two phases are approximately inequilibrium just at the actual interface), (2) that the holdup of component A in the boundary layers
on both sides of the interface is negligibly small compared to the amount transferred, and (3) thatthe bulk of each phase is well mixed. For the case of packed columns for gas absorption, the gas
flows up and liquid trickles down.

Fig.10.3-2. Concentration profile in the neighborhood of gas-liquid contacting interface

The velocity distribution near the interface is also shown in Fig.10.3-2. Since there is no slip at theinterface, gas film is in enough contact with the liquid film to establish the phase equilibrium at theinterface.
Using mass transfer coefficients defined by the two-film theory, the steady-state mass flux is
expressible as
𝑁𝐴 = 𝑘𝑦𝐴 (𝑦𝐴𝑚 − 𝑦𝐴𝑖 ) = 𝑘𝑥𝐴 (𝑥𝐴𝑖 − 𝑥𝐴𝑚 )(10.3-3)Here 𝑦𝐴𝑚 and 𝑥𝐴𝑚 are the mixed mean concentrations (the bulk concentrations) and 𝑘𝑦𝐴 and𝑘𝑥𝐴 are the convective mass transfer coefficients having units of kmol of A/m 2 s in the gas andliquid phases, respectively.
According to the assumption made, the gas and liquid concentrations at the interface lie on theequilibrium curve given by
𝑦𝐴𝑖 = 𝑓 (𝑥𝐴𝑖 )(10.3-4)
where f is the equilibrium function. These equilibrium concentrations would be obtained if the two

110Interphase Mass Transport

phases had been in contact for an infinite period of time. Henry’s law is an example of theequilibrium function for dilute solution:
𝑦𝐴𝑖 = 𝑚 𝑥𝐴𝑖(10.3-5)
in which m is the Henry’s constant. For simplicity, the Henry’s law is assumed for the remaining
development.
It is convenient to define an overall mass transfer coefficient 𝐾𝑦𝐴 and 𝐾𝑥𝐴 based on an overall
driving force between the bulk concentration 𝑦𝐴𝑚 or 𝑥𝐴𝑚 because the interface concentrations arevery difficult to measure. This treatment is similar to that in defining the overall heat transfer
coefficient. However the bulk concentration 𝑦𝐴𝑚 and 𝑥𝐴𝑚 do not have the same units; 𝑦𝐴𝑚 hasunits of kmol of A/kmol of A+B whereas 𝑥𝐴𝑚 has units of kmol of A/kmol of A+C.
The above equation Eq.(10.3-3) can be rewritten as
𝑦 −𝑦
𝑥 −𝑥
𝑁𝐴 = 𝐴𝑚1 𝐴𝑖 = 𝐴𝑖 1 𝐴𝑚
(10.3-6)
𝑘𝑦𝐴

𝑘𝑥𝐴

Multiplying the numerator and denominator of the right-most term by m,a new fraction having the same value is obtained by the law of additivity as
𝑁𝐴 =𝑦𝐴𝑚 − 𝑦𝐴𝑖
1=𝑚(𝑥𝐴𝑖 − 𝑥𝐴𝑚 )

𝑘𝑦𝐴

𝑚
𝑘𝑥𝐴

=

∗
𝑦𝐴𝑚 − 𝑦𝐴𝑚
1

𝑘𝑦𝐴

+

𝑚
𝑘𝑥𝐴

∗
= 𝐾𝑦𝐴 (𝑦𝐴𝑚 − 𝑦𝐴𝑚)(10.3-7)∗
Here 𝑦𝐴𝑚= 𝑚 𝑥𝐴𝑚 is the bulk concentration of the liquid phase expressed with the units of thegas-phase concentration by using the Henry’s law. The overall mass transfer coefficient 𝐾𝑦𝐴 is
given by1
𝐾𝑦𝐴 = 1 𝑚
(10.3-8
𝑘𝑦𝐴

+

𝑘𝑥𝐴

Similarly the following formulae are available.
𝑁𝐴 =
𝐾𝑥𝐴 =

(𝑦𝐴𝑚 ⁄𝑚) −( 𝑦𝐴𝑖 ⁄𝑚)
1
𝑚𝑘𝑦𝐴

1
1
1
+
𝑚𝑘𝑦𝐴 𝑘𝑥𝐴=

𝑥𝐴𝑖 − 𝑥𝐴𝑚
1

𝑘𝑥𝐴

=

∗
𝑥𝐴𝑚
− 𝑥𝐴𝑚

1
1
+
𝑚𝑘𝑦𝐴 𝑘𝑥𝐴∗= 𝐾𝑥𝐴 ( 𝑥𝐴𝑚− 𝑥𝐴𝑚 )

(10.3-9)(10.3-10)

Notice that 𝐾𝑦𝐴 and 𝐾𝑥𝐴 represent the basic information in different units of concentration.
10.3-3 Mass transfer model for gas absorption
Gas absorption is the typical mass transfer operation in which one or more soluble gases areabsorbed from their mixture with an inert gas into a liquid. The absorption process may be purely
physical or involve chemical reactions in the liquid. A typical example is the washing sulfur dioxidefrom the boiler-exhaust gas by means of water or alkaline solution.
The gas absorption equipment is designed to provide intimate contact of a gas and a liquid. A
typical continuous gas-absorber is the packed column which consists of a vertical column filledwith a randomly arranged packing of small inert solid or equipped with structured packing. The
packing provides a large interfacial area of contact between the two phases. Figure 10.3-3 showssome example of random packings and structured packing.

Fig.10.3-3. Random packings and structured packing (a: Raschig ring, b: Berl saddle, c: Pall ring,and d: corrugated structured packing, e: partial structure of structured packing)Mass Transfer Model for Gas Absorption (continued)111

Fig.10.3-4. Packed column absorber equipped with Raschig rings

For simplicity, we will deal with the physical absorption of one soluble component A from a
mixture with an inert gas B by non-volatile liquid solvent C.
Fig.10.3-4 shows a conventional packed column equipped with Raschig rings. The liquid issprayed evenly over the top of the packed bed by the distributor and trickles down over the packing
surfaces by gravity. The solute-containing gas is fed into the column from below the packing andflows upward through the interstices due to the pressure drop through the bed of the packing. Thiscountercurrent direct contacting permits the transfer of component A from the gas to the liquid. The
liquid is enriched in component A as it flows down the packing, and the concentrated liquid leaves
the column through the liquid outlet. The bulk concentration of component A in the gas stream
decreases as the gas flows up, and the lean gas leaves the column through the gas outlet.
Let us develop an expression for the column height 𝑍𝑇 required to get the desired degree ofrecovery of component A. The column height is calculated from a viewpoint of mass-transfer area.
Assume that the mass transfer takes place isothermally. In usual absorption operations the liquidand gas flow rates 𝐿𝑀 and 𝐺𝑀 vary appreciably with the elevation from the bottom of the packed
section. It is convenient to use units of flow rate and concentration on a solute-free basis because
the molar flow rates of the inert gas B and nonvolatile liquid C are constant over the total height of′the packed column. The gas stream enters the column at molar velocity of 𝐺𝑀
on an A-free basis.
The mole fraction of A in the entering gas stream is 𝑦𝐴1 . In usual situation for the design of anabsorber, the designer is required to determine the molar liquid velocity 𝐿′𝑀 , the concentration 𝑥𝐴2
of which is usually known. The outlet gas concentration 𝑦𝐴2 is also given in terms of the specified
degree of recovery of component A.
The overall material balance of component A may be written as
𝑦𝐴1
𝑥
𝑦𝐴2
𝑥
′
′
𝐺𝑀
+ 𝐿′𝑀 1−𝐴2
= 𝐺𝑀
+ 𝐿′𝑀 1−𝐴1
(10.3-11)
1− 𝑦
𝑥
1− 𝑦
𝑥
𝐴1

𝐴2

𝐴2

𝐴1

Note that each term has units ofkmol of B
kmol of A
kmol of A=22(m of the bed)(s) kmol of B
(m of the bed)(s)′These molar velocities 𝐺𝑀
and 𝐿′𝑀 are the so-called superficial molar velocities the gas andliquid would have in the column on an A-free basis if no packing were present.
At the present stage, 𝐿′𝑀 and 𝑥𝐴1 are unknown. The above equation indicates that the outlet liquid
concentration 𝑥𝐴1 increases as the liquid flow rate is decreased. Choosing the portion of thecolumn above an arbitrary section as a control volume, as shown by the dotted line in Fig.10.3-5,
we get a material balance on component A:

112

Interphase Mass Transport

Fig.10.3-5. Control volume for mass balance and differential control volume of interphase masstransfer model for a packed column absorber
𝑦

𝑦

𝑥

𝑥

′
𝐺𝑀
(1− 𝐴𝑦 − 1− 𝐴2) = 𝐿′𝑀 (1−𝐴𝑥 − 1−𝐴2
)
𝑦
𝑥
𝐴

𝐴2

𝐴

𝐴2

(10.3-12)

This is the equation for the operating line which gives the relation between the bulk concentrationsof the gas and liquid at an arbitrary position. The liquid flow rate 𝐿′𝑀 can be determined from theviewpoint of the limiting liquid-gas ratio.

Fig.10.3-6. Concept of limiting liquid-gas ratio.

Figure 10.3-6 shows a graphical relation of the operating line with the equilibrium line for gas′
absorption. For a given gas flow rate 𝐺𝑀, a reduction in liquid flow rate 𝐿′𝑀 decreases the slope ofthe operating line. As the liquid flow rate is decreased fixing the terminal concentrations 𝑦𝐴2 , 𝑥𝐴2 ,and 𝑦𝐴1 , the upper end of the operating line approaches the equilibrium line. Just when theoperating line touches the equilibrium line, the minimum possible liquid flow rate (𝐿′𝑀 )𝑚𝑖𝑛 is
obtained.
At this condition, an infinitely long packed section is required owing to very small driving forces inthe vicinity of the bottom of the column.

Mass Transfer Model for Gas Absorption (continued)

113

In an actual column, the liquid rate must be greater than the minimum. Replacing 𝑦𝐴 and 𝑥𝐴 of∗Eq.(10.3-12) by 𝑦𝐴1 and 𝑥𝐴1, we obtain the limiting liquid-gas ratio:
𝐿′

𝑦

⁄(1−𝑦

)− 𝑦

⁄(1−𝑦

)

(10.3-13)

𝐴2
𝐴2
𝐴1
= 𝑥𝐴1
∗ ⁄(1−𝑥 ∗ )− 𝑥 ⁄(1−𝑥 )

( 𝑀
′ )
𝐺𝑀

𝐴2
𝐴2
𝐴1
𝐴1
𝑚𝑖𝑛
∗
Here 𝑥𝐴1= 𝑦𝐴1 ⁄𝑚′ is the liquid-phase concentration that would be in equilibrium with the inlet

gas-phase concentration. From the economical viewpoint, the optimum liquid rate is usually foundby making the operating line parallel to the equilibrium line:′ )(𝐿′𝑀⁄𝐺𝑀

=1(10.3-14)
The actual liquid rate is usually 25 to 100% greater than the minimum. Once the actual liquid rate isdecided in this way, we can calculate the concentration 𝑥𝐴1 in the exit liquid stream.
The equipment size required for the desired separation cannot be determined without considering
the relation between mass transfer rate and mass transfer area. The behavior of an actual packed
column is too complicated to be modeled in a straight-forward way. We simplify the actual column
to a system consisting of the two streams flowing side-by-side without back mixing and in contact
with one another.
The interfacial area “ 𝑎 ” per unit packed-bed volume and the mass transfer coefficients
𝑘𝑦 , 𝑘𝑥 , 𝐾𝑦 , 𝐾𝑥 are assumed to be constant over the packed section.
First we set up the mass balance within a differential height 𝑑𝑍 at an arbitrarily chosen distance𝑍 from the top of the packed bed:
𝑚′

𝑦

𝑦

′
𝑁𝐴 𝑎 𝑑𝑍 𝑆 = 𝐺𝑀
𝑆 (1−𝑦𝐴 |𝐴 𝑍+𝑑𝑍

That is′𝑁𝐴 𝑎 𝑑𝑍 = 𝐺𝑀
𝑑(

𝑦𝐴

1− 𝑦𝐴

𝑥

− 1−𝑦𝐴 | ) = 𝐿′𝑀 𝑆 (1−𝑥𝐴 |𝐴 𝑍) = 𝐿′𝑀 𝑑 (

𝑥𝐴
1− 𝑥𝐴

𝐴 𝑍+𝑑𝑍𝑥− 1−𝑥𝐴 | )
𝐴 𝑍

)

or
𝑑𝑦𝐴
𝑑𝑥
′
𝑁𝐴 𝑎 𝑑𝑍 = 𝐺𝑀
= 𝐿′𝑀 (1−𝑥𝐴 )2(1−𝑦 )2
𝐴

𝐴(10.3-15)(10.3-16)(10.3-17)

The cross-sectional area S of the packed section is usually determined by ban empirical relation
between the pressure drop per unit height of the packing and the gas velocity from a viewpoint of
the stable flow regime for intimate gas-liquid contact.
The interphase mass transfer is also expressed as
𝑁𝐴 𝑎 𝑑𝑍 = 𝑘𝑦 𝑎(𝑦𝐴 − 𝑦𝐴𝑖 )𝑑𝑍 = 𝑘𝑥 𝑎(𝑥𝐴𝑖 − 𝑥𝐴 )𝑑𝑍(10.3-18)∗∗
𝑁𝐴 𝑎 𝑑𝑍 = 𝐾𝑦 𝑎(𝑦𝐴 − 𝑦𝐴 )𝑑𝑍 = 𝐾𝑥 𝑎(𝑥𝐴 − 𝑥𝐴 )𝑑𝑍(10.3-19)
From one of these equations
𝑑𝑦𝐴∗
𝐺𝑀= 𝐾𝑦 𝑎(𝑦𝐴 − 𝑦𝐴∗ )𝑑𝑍(10.3-20)(1−𝑦 )2𝐴

The coefficients 𝑘𝑦 𝑎, 𝑘𝑥 𝑎, 𝐾𝑦 𝑎, 𝐾𝑥 𝑎 are called “volumetric mass transfer coefficients.”
Integrating the equation gives the column height of packing section required for the desired
separation:
̅̅̅̅
𝐺∗
𝑦
𝑑𝑦𝐴
𝑍𝑇 = 𝐾 𝑀𝑎 ∫𝑦 𝐴1 (1− 𝑦 )2(𝑦(10.3-21)− 𝑦∗ )
𝑦

𝐴2

𝐴

𝐴

𝐴

𝐴

𝐴

𝐴𝑖

Similar equations can also be obtained from the remaining three equations as follows.
̅̅̅̅
𝐺∗
𝑦
𝑑𝑦
𝑍𝑇 = 𝑘 𝑀𝑎 ∫𝑦 𝐴1 (1− 𝑦 )2 𝐴
(𝑦 − 𝑦 )
𝐴2
𝑦

̅̅̅̅
𝐿∗𝑀

𝑥
𝑑𝑥
= 𝐾 𝑎 ∫𝑥 𝐴1 (1− 𝑥 )2 (𝐴𝑥∗ −𝑥 )
𝐴2
𝑥
𝐴
𝐴
𝐴
̅̅̅̅
𝐿∗𝑀
𝑥𝐴1
𝑑𝑥𝐴
= 𝑘 𝑎 ∫𝑥 (1− 𝑥 )2 ( 𝑥 −𝑥 )
𝐴2
𝑥
𝐴
𝐴𝑖
𝐴
′ ⁄
′ ⁄
′
′
̅̅̅̅̅̅̅̅̅̅̅
̅̅̅̅̅̅̅̅̅̅̅
̅̅̅̅̅̅̅̅̅̅̅
̅̅̅̅̅̅̅̅̅̅
The quantities 𝐺
𝐾
𝑎
,
𝐺
𝑦
𝑀
𝑀 𝑘𝑦 𝑎 , 𝐿𝑀 ⁄𝐾𝑥 𝑎 , 𝐿𝑀 ⁄𝑘𝑥 𝑎(10.3-22)

are roughly approximated parameters called“Height of a Transfer Unit,” averaged between the top and bottom of the packing section. These
mass transfer capacity coefficients 𝐾𝑦 𝑎, 𝐾𝑥 𝑎, 𝑘𝑦 𝑎, 𝑘𝑥 𝑎 depend on the mass velocity of the gas

114Interphase Mass Transport′ ⁄𝐺𝑀 = 𝐺𝑀(1 − 𝑦𝐴 ) and decrease from bottom to top. From this viewpoint the precision may bebetter if the following expressions are used:
𝑦
𝐺
𝑑𝑦
̅̅̅̅
𝑍𝑇 = 𝐾 𝑀𝑎 ∫𝑦 𝐴1 (1− 𝑦 )(𝑦𝐴 − 𝑦∗ )(10.3-23)
𝐴2

𝑦

𝐴

𝐴

𝐴

[PROBLEM 10.3-1]
30 kmol/h of air contaminated by chlorine gas (mole fraction of Cl 2: 𝑦𝐴 = 0.06 ) is contactedwith water in a packed column to get the exit concentration 𝑦𝐴 = 0.005. The column operates at 1
atm and 20˚C(=293 K).
Determine the minimum liquid flow rate assuming physical absorption. The solubility of Cl 2 into
water (at 1 atm and 20˚C) is approximately expressed as 𝑝 = 1.92 × 109 𝑥 2.16
where 𝑥 = mole fraction of Cl2 in water and p = partial pressure (mmHg) of Cl2 in air.
10.3-4 Mass transfer coefficients in a packed column absorber
Usually the mass transfer data for packed-tower absorption are correlated in the form𝐺

𝐻𝑂𝐺 = 𝐾 𝑀𝑎𝑦,𝐺

𝐻𝐺 = 𝑘 𝑀𝑎𝑦,𝐿

𝐻𝑂𝐿 = 𝐾 𝑀𝑎𝑥,𝐿

𝐻𝐿 = 𝑘 𝑀𝑎
𝑥(10.3-24)

These parameters are called the height of a transfer unit (HTU). The concept is based on the idea ofdividing the packed section into a number of contact units called transfer units.
The following integrals are designated as the number of transfer units (NTU):𝑦𝑑𝑦
𝑁𝑂𝐺 = ∫𝑦 𝐴1 (1−𝑦 )(𝑦𝐴 −𝑦∗ )(10.3-25)
𝐴2

𝐴

𝐴

𝐴

𝑦
𝑑𝑦
𝑁𝐺 = ∫𝑦 𝐴1 (1−𝑦 )(𝑦𝐴 −𝑦 )
𝐴2
𝐴
𝐴
𝐴𝑖
𝑥
𝑑𝑥
𝑁𝑂𝐿 = ∫𝑥 𝐴1 (1−𝑥 )(𝑥𝐴∗ −𝑥 )
𝐴2
𝐴
𝐴
𝐴
𝑥𝐴1
𝑑𝑥𝐴
𝑁𝐿 = ∫𝑥 (1−𝑥 )(𝑥 −𝑥 )
𝐴2
𝐴
𝐴𝑖
𝐴

Therefore the tower height may be calculated as the product of the number of transfer units and theheight of a transfer unit (HTU):𝑍𝑇 = 𝐻𝑂𝐺 𝑁𝑂𝐺 = 𝐻𝐺 𝑁𝐺 = 𝐻𝑂𝐿 𝑁𝑂𝐿 = 𝐻𝐿 𝑁𝐿(10.3-26)
Using the two-film theory, the overall heights of transfer unit can be related to the single-phasetransfer units as
𝑚𝐺
𝐿
𝐻𝑂𝐺 = 𝐻𝐺 + 𝐿 𝑀 𝐻𝐿 and 𝐻𝑂𝐿 = 𝐻𝐿 + 𝑚 𝐺𝑀 𝐻𝐺(10.3-27)
𝑀

𝑀

These equations indicate that the overall resistance of interphase mass transfer consists of thegas-phase and liquid-phase resistances in series.

10.4 Mass Transfer Correlations for Packed Columns
As in Eq.(10.3-25), the overall resistance to the interphase mass transfer comprises the gas-phasefilm resistance 𝐻𝐺 and the liquid-phase film resistance (𝑚𝐺𝑀 ⁄𝐿𝑀 )𝐻𝐿 .
10.4-1 Height of a liquid-phase transfer unit
Sherwood and Holloway1) measured the desorption of oxygen, hydrogen, and carbon dioxidefrom water in the range of gas flow rate up to the loading point and obtained the following
correlation:
1

𝐿

𝑛

𝜇

0.5

𝐻𝐿 = 𝛼 (𝜇 ) (𝜌 𝐷𝐿 )𝐿

𝐿 𝐿

(10.4-1)

Temperature 278 < 𝑇 < 313𝐾 and liquid flow rate 0.556 < 𝐿 < 20.8 𝑘𝑔⁄𝑚2 𝑠, where 𝜇𝐿
is liquid viscosity (kg/m s), 𝐷𝐿 diffusivity for liquid (m2/s) and 𝜌𝐿 liquid density (kg/m 3).
1.

Sherwood, T.K. and Holloway, F.A.L.: Trans. Am. Inst. Chem. Engrs., vol.36, 39 (1940)

Mass Transfer Correlations for Packed Columns (continued)

Table 10.4-1 HTUL correlation parameters
Packingsize in.(I= 25.4 mm)
Raschig rings
3/8
1/2
1
1 1/2
2
Berl saddles
1/2
1
1 1/2

115

𝛼
3,100
1,400
430
380
340
600
780
730

n
0.46
0.35
0.22
0.22
0.22
0.28
0.28
0.28

Note that the transfer unit 𝐻𝐿 up to the loading point is independent of the gas flow rate.
Onda et al.1) obtained a dimensionless correlation:
𝜌1⁄ 3

𝑘𝐿 (𝜇 𝐿𝑔)
𝐿

= 0.0051 (𝑎

2⁄ 3

𝐿
𝑤

)
𝜇
𝐿

𝜇

𝐿
(𝜌 𝐷
)
𝐿

−0.5

𝐿

(10.4-2)(𝑎𝑡 𝐷𝑝 )0.4

where the wetted surface area 𝑎𝑤 is given taking into account the liquid surface tension by
0.75(𝐿⁄𝑎𝑡 𝜇𝐿 )0.1 (𝐿2 𝑎𝑡 ⁄𝜌𝐿 2 𝑔)−0.05 (𝐿2 ⁄𝜌𝐿 𝜎𝑎𝑡 )0.2 }𝑎𝑤 ⁄𝑎𝑡 = 1 − exp{− 1.45 (𝜎𝑐 ⁄𝜎)(10.4-3)
1.

Onda, K., Takeuchi, H., and Okumoto, Y., J. Chem. Eng. Japan, 1(1), 56 (1968)

The total surface area 𝑎𝑡 , the size 𝐷𝑝 of the packing are obtainable in relation with the void
fraction 𝜀 on the dry basis in Table 10.4-2.
Table 10.4-2 Geometrical data of Packings1)
Packing
Size in. (= 25.54 mm)
Steel Raschig rings
3/4
1
1 1/2
2
Steel Pall rings
5/8
1
1 1/2
2
Ceramic Raschig rings
1/4
1/2
3/4
1
1 1/2
2
Ceramic Berl saddles
1/4
1/2
3/4
1
1 1/2
2
1.

𝑎𝑡 m2 ⁄m3
364
184
128
95.1
341
207
128
102
712
367
243
190
121
91.9
899
466
285
249
151
105

0.73
0.86
0.90
0.92
0.93
0.94
0.95
0.96
0.62
0.64
0.72
0.74
0.73
0.74
0.60
0.62
0.66
0.68
0.71
0.72

𝜀 m3 ⁄m3

Perry, R. H., and Chilton, C. H., Chemical Engineers’ Handbook, McGraw-Hill, New York, 5th ed.,Table 18-6 (1973)

Hikita1) obtained the following Sherwood number correlation:
𝑘𝐿 𝐷𝑝
𝐷𝐿

4𝐿

0.45

= 𝑐 (𝑎 𝜇 )
𝑤 𝐿

0.5

𝜇𝐿

(𝜌 𝐷 )
𝐿 𝐿

Liquid Reynolds number

4𝐿

𝑎𝑤 𝜇𝐿

𝜌𝐿 2 𝑔 𝐷𝑝 3

(

𝜇𝐿 2

1⁄6

)

= 50 ~ 1,000

Here 𝑐 = 0.31 for Raschig rings, 𝑐 = 0.37 for Berl saddles.
𝐿 (kg/m2h): liquid flow rate, 𝐷𝑝 (cm): packing size, and 𝜎 (dyne/cm): surface tension.

(10.4-4)

116Interphase Mass Transport

The empirical relations between the wetted surface and the total surface area proposed by Hikita are
given as follows:−0.48 )
𝑎𝑤 ⁄𝑎𝑡 = 0.0406 𝐿0.455 𝜎 (−0.83 𝐷𝑝
for Raschig rings−0.98 )(−0.495𝐷
0.455
𝑝𝑎𝑤 ⁄𝑎𝑡 = 0.0078 𝐿𝜎
for Berl saddles(10.4-5)
1.

Hikita, H.: Kagakukogaku (Chem. Eng. Japan), vol.26, 725 (1962)

10.4-2 Height of a gas-phase transfer unit
Fellinger’s ammonia-air-water absorption data up to the loading point are correlated in theform1) :2⁄ 3

𝜇(10.4-6)

𝐻𝐺 = 𝑐 𝐺 𝑝 𝐿𝑞 (𝜌 𝐷𝐺 )𝐺 𝐺

where 𝐻𝐺 = height of a gas-phase transfer unit m, 𝐺 = gas flow rate 1,000 ~ 3,500 kg/m2 h, 𝐿 =liquid flow rate 2,500 ~ 7,500 kg/m2 h
1.

Fellinger, L. Sc.D. thesis in Chem. Eng. M.I.T. (1941) from Perry, J.H.: Chemical Engineers’ Handbook, 4th ed. Vol.18-42,
McGraw-Hill, New York (1963).

The unknown parameters 𝑝, q, c calculated by Hikita are given in Table 10.4-3.
Table 10.4-3 Parameters for correlation Eq.(10.4-6 )Packingsize in.(I= 25.4 mm)
Raschig rings
3/8
1/2
1
1 1/2
2
Berl saddles
1/2
1
1 1/2

𝑝
0.45
0.43
0.32
0.38
0.41
0.30
0.36
0.32

q
- 0.47- 0.60- 0.51- 0.66- 0.45- 0.24- 0.40- 0.45

c
0.85
4.2
3.07
9.59
1.44
0.262
0.745
2.20

Hikita, H.: Kagakukogaku (Chem. Eng. Japan), vol.26, 725 (1962)

The gas-phase mass transfer correlation is experimentally obtained by Onda et al.1):
𝑘𝐺

𝐺

𝑎𝑝 𝐷 𝐺

0.7= 5.23 (𝑎 𝜇 )
𝑝 𝐺

𝜇

1⁄ 3

(𝜌 𝐷𝐺 )𝐺 𝐺

(𝑎𝑝 𝐷𝑝 )−2(10.4-7)

The 𝐻𝐺 can be calculated by using the following defining equation with the superficial gas
velocity 𝑢𝑠𝑓
𝑢𝑠𝑓
𝐻𝐺 = 𝑘 𝑎(10.4-8)
𝐺 𝑒

Here 𝑎𝑒 is the effective interfacial area for gas-liquid contact per unit volume.
1.

Onda, K., Takeuchi, H., amnd Okamoto, Y., J. Chem. Eng. Japan, 1, No.1, 56 (1982)

10.5 Column Diameter and Pressure Drop of Packed Columns
The power of blower required to feed gas mixture at the specified rate is mainly related to thepressure drop in the packing section. The pressure drop per unit height of the packing section∆ 𝑝⁄𝑍𝑇 is characterized graphically in Figure 11.4-1. As can be seen, for the case of countercurrent
gas absorber, the pressure drop is increased due to reduction in free volume by the liquid.
When gas mass-velocity 𝐺 is increased at a constant liquid mass-velocity 𝐿 , three flow
conditions may occur successively. At very low gas velocities, the pressure drop is proportional

Column Diameter and Pressure Drop of Packed Columns (continued)

117

approximately to square of the gas velocity, as for dry packing. When 𝐺 is further raised, ∆ 𝑝⁄𝑍𝑇ceases to increase with 𝐺 2 . It is attributable to the fact that a portion of the kinetic energy of the gas
stream is used to support the liquid in the interstices of packing. This is called the “loading point.”
The transition from preloading to loading is usually gradual. When 𝐺 exceeds a certain criticalvalue, the liquid cannot flow downward over the packing, and then the column floods. This is called
the “flooding point.”

Fig.10.5-1. Pressure drop characteristics and flooding gas velocity of a packed column absorber

One of the earliest correlations of the flooding data, which has been modified later by Eckert, isin the form:𝜌𝐺 𝑢𝑠𝑓 2
𝜇𝐿 0.2
𝐿 𝜌𝐺
(
)
𝑣𝑠.
√
𝜌𝐿 𝑔(𝜀 3 ⁄𝑎𝑡 ) 𝜇𝑤
𝐺 𝜌𝐿
where 𝑢𝑠𝑓 = superficial gas-velocity at the flooding point (m/s), 𝜌𝐺 and 𝜌𝐿 = gas and liquiddensities (kg/m3), 𝜇𝐿 and 𝜇𝑤 = liquid and water viscosities (kg/m s), 𝑎𝑡 = total surface area
(m2/m3) and 𝜀 = void fraction (m3 interstices/m3 packed bed). Note that the left-side fraction
implies the ratio of the kinetic energy of the gas stream to the potential energy of the liquid stream.
For several kinds of packings, 𝑎𝑡 and 𝜀 are tabulated with respect to their sizes in the last section.
The pressure drop in packed columns can be expressed as the function of
𝐺 2 𝐹𝑝 𝜓 (𝜇𝐿 ⁄𝜇𝑤 )0.2
𝐿 𝜌𝐺
𝑣𝑠.
√
𝜌𝐺 𝜌𝐿 𝑔𝐺 𝜌𝐿
Here 𝜓 = 𝜌𝑤 ⁄𝜚𝐿 is the ratio of water density to liquid density and 𝐹𝑝 the packing factor in 1/m,
which should usually be given as the characteristics of various random packings. For example,
25-mm (1 in.) steel Pall rings give 𝐹𝑝 = 157 1/m.
It should be noted that as long as (𝜇𝐿 ⁄𝜇𝑤 )0.2 is regarded as a kind of correction factor, theremaining left-side term is a characteristic dimensionless kinetic energy 𝑢𝑠 2 group.
Fig.11.3-1 shows generalized correlations of pressure drop for various random packing towers
recalculated after Eckert correlation1).

118

Interphase Mass Transport

Figure 10.5-2. Generalized correlation for flooding and pressure drop in packed towers.
(After Eckert)
1.

Eckert, E. R. G., Chem. Eng. Progr., 66(3), 39 (1970)[PROBLEM 10.5-1] A gas absorber packed with 25-mm Pall rings (𝑭𝒑 = 𝟏𝟓𝟕 1/𝒎) deals with
500 m3 of entering gas per hour.
The SO2 content of the entering gas is 2 mol% and the remaining can be considered as air. Water isused as SO2-free absorbent. The temperature is 25˚C and pressure 1 atm. Ehen the ratio of gas flow
to liquid flow is 1 kg gas/1 kg water, the gas velocity is one-half of the flooding velocity.
(a) Calculate the tower diameter.
(b) What is the pressure drop when the packing section is 7 m high?

10.6 Pressure Drop of Dry Packed Columns
The pressure drop studied in the last section is for the gas-liquid countercurrent flow in a packed
column. Regarding the single phase fluid flow in a packed bed, the following empirical equation isavailable.1, 2)
𝑃0 − 𝑃𝐿𝐿=

150 𝜇𝑢𝑠 (1− 𝜖)2
𝐷𝑝

2

𝜖3

+

1.75 𝜌𝑢𝑠 2 (1− 𝜖)𝐷𝑝𝜖3(10.6-1)

This equation gives the pressure drop 𝑃0 − 𝑃𝐿 through the length 𝐿 of a packed bed. The firstterm is the Blake-Kozeny equation for laminar flow and the second term is the Burke-Plummer
equation for turbulent flow. As distinct from packed column gas absorbers, the packing material
may be spheres, cylinders, or various kinds of commercial packing. Here 𝑢𝑠 is the superficial
velocity, 𝜖 the void fraction, and 𝐷𝑝 is the mean packing particle diameter (e.g. sphere diameter).
The mean diameter is defined by(10.6-2)
𝐷𝑝 = 6⁄𝑎𝑣
where 𝑎𝑣 is the specific surface area of a nonspherical packing particle (particle surface/particle
volume).
We define the friction factor for the packed bed by the following equation:

Pressure Drop of Dry Packed Columns (continued)
𝑃0 − 𝑃𝐿
1
𝜌𝑢𝑠 2
2

=

𝐿
𝐷𝑝

119

(10.6-3)𝑓

Analogously to the friction factor for a circular tube flow, the first term can be rewritten as𝑓= ((1− 𝜖)2
𝜖3

75) 𝐷 𝜌𝑢 ⁄𝜇𝑝𝑠(10.6-4)

This equation corresponds to the following equation for the laminar flow in a circular tube:
16
𝑓 = 𝑅𝑒(6.1-17)
Similarly the second term may give1− 𝜖
𝑓 = 0.875 𝜖3(10.6-5)
Usually the friction factor becomes constant in the turbulent flow region, except for the turbulent
flow along smooth-walls.
For the case of smooth tubes, the Blasius equation for the turbulent tube flow is available as an
example.
0.0791
𝑓 = 𝑅𝑒 1⁄4(7.3-12)
The above correlation Eq.(10.6-5) can be understood from the following facts of various flows.
As can be seen in Fig.7.3-1, we note that for highly turbulent flow in rough tubes or pipe fittings
the friction factors become constant and a function of the roughness or the fittings structure only.
As will be studied in Chapter 20, the power number (a kind of friction factor) for baffled agitatedvessel also has the same tendency in the turbulent flow condition.
1.
2.

Bird, R. B., Stewart, W. E., and Lightfoot, E. N., “Transport Phenomena,” Wiley, New York (1960)
Ergun, S., Chem. Eng. Prog., 48, 89 (1952)

Nomenclature
𝑎, 𝑎𝑒
𝑎𝑡
𝑎𝑣
𝐶𝐴
𝐶𝑝
𝑐
𝐷
𝐷𝐴𝐵
𝐷𝑝
𝐹𝑝
𝑓
𝐺, 𝐿
𝐺𝑀 , 𝐿𝑀
𝐻, 𝐻′
𝐻𝐺 , 𝐻𝐿
𝐻𝑜𝐺 , 𝐻𝑜𝐿
𝑗𝐷 , 𝑗𝐻𝑘𝐺 , 𝑘𝐿
𝐾𝑦 , 𝐾𝑥
𝑘𝑦 , 𝑘𝑥𝐿𝑚𝑁𝐴
𝑁𝑢
𝑁𝐺 , 𝑁𝐿
𝑁𝑜𝐺 , 𝑁𝑜𝐿
𝑝
𝑝𝐴
𝑅
𝑅𝑒
𝑟
𝑆𝑐

effective interfacial area per unit volume of packed bed, [m2/m3]
total surface area of packing, [m2/m3]
specific surface area of non-spherical packing particles, [m2/m3]
molar concentration of solute A, kmol/m3 solution
heat capacity, [J/kg K]
total mole density, [kmol/m3]
pipe diameter, [m]diffusivity of component A, [m2/s]
size or diameter of packing particle, [m]packing factor, [1/m]friction factor, [ - ]mass velocity of gas and liquid, [kg/m2s] or [kg/m2h]
gas- and liquid-phase molar velocity, [kmol/m2s]
Henry’s constant, mole fraction gas-phase/mole fraction liquid-phase or Pa gas-phase/(kmol A/m3 liquid)gas- and liquid-phase HTU (Height of a Transfer Unit), [m]overall HTU, [m]j-factor for mass transfer, j-factor for heat transfer, [ - ]gas-phase, and liquid-phase mass transfer coefficient, [kmol/m2s Pa], [m/s]overall mass transfer coefficient using gas-phase, and liquid-phase concentration, [kmol/m2s]gas-, and liquid-phase mass transfer coefficient using mole fraction, [kmol/m2s]
pipe length, [m]Henry’s constant, [ - ]molar mass-flux of component A, [kmol/m2s]
Nusselt number, [ - ]gas- and liquid-phase NTU (Number of Transfer Units), [m]overall NTU, [m]pressure, [Pa]partial pressure of solute gas A, [Pa]pipe radius, [m]Reynolds number, [ - ]radial coordinate, [m]Schmidt number, [ - ]

120
𝑆ℎ
𝑥𝐴
𝑢𝑠
𝑢𝑠𝑓
𝑍
𝑦
𝛿𝐷
𝜀
𝜇
𝜅
𝜎

Interphase Mass Transport
Sherwood number, [ - ]mole fraction of component A, [ - ]superficial velocity of gas, [m/s]superficial gas velocity at flooding point, [m/s]height of packed section, [m]distance from pipe wall, [m]film thickness for mass transfer, [m]void fraction, [ - ]viscosity, [kg/m s]
thermal conductivity, [W/m K]
surface tension, [N/m]

Subscripts
f
film
G
gas
i
interfaceLliquidmbulk or mixed meanw
wall nor wetted

PART ⅡCHAPTER 11
MASS TRANSPORT EQUIPMENT
11.1 Distillation Fundamentals
11.1-1 Phase Equilibria for Distillation
The rate of inter-phase mass transport increases with the deviation from the vapor-liquid
equilibrium. We should have to some degree the knowledge of gas-liquid equilibrium. The
following phase rule is available to establish the number of independent variables or degree of
freedom in a specific situation:
𝑭 =𝑪−𝑷+𝟐(11.1-1)
Assume a binary solution having two components to be distilled, so 𝑪 = 2, 𝑷 = 2, and 𝑭 = 2.
Both components are found in both phases. It can be considered that there are four variables:pressure, temperature, and the concentrations of component A in the liquid and vapor phases. For
example, if the pressure is fixed, only one variable, e.g., liquid-phase concentration, can be changed
independently, and temperature and vapor-phase concentration follow. Therefore equilibrium data
are expressed in temperature-composition diagram under the condition of constant pressure. Suchan equilibrium curve is given by plotting 𝑦𝑒 , vapor-phase mole fraction , against 𝑥𝑒 , liquid-phasemole fraction. Here only a very simple equilibrium case is introduced.
At equilibrium the activities of component A in the vapor and liquid should be equal.
For ideal solutions, Raoult’s law can be applied:𝑝𝐴 = 𝑃𝐴 𝑥𝐴(11.1-2)
where 𝑝𝐴 is the partial pressure of component A in the vapor phase, 𝑃𝐴 the vapor pressure of purecomponent A, and 𝑥𝐴 the mole fraction of component A in the liquid. For the case of
two-component system (binary system), the total pressure is
𝑃 = 𝑝𝐴 + 𝑝𝐵 and 𝑥𝐴 + 𝑥𝐵 . Therefore the vapor concentration can be calculated as
𝑃 𝑥
𝑦𝐴 = 𝐴𝑃 𝐴(11.1-3)

Fig.11.1-1 Equilibrium diagram expressed mole fraction 𝒚𝑨 𝐚𝐠𝐚𝐢𝐧𝐬𝐭 𝒙𝑨

122

Mass Transport Equipment

The x-y diagram for binary equilibrium relationship is plotted for the more-volatile component A.
A more general equilibrium relation is usually expressed as
𝑦𝐴 = 𝐾𝐴 𝑥𝐴(11.1-4)
where 𝐾 is called the vapor-liquid equilibrium constant.
For arbitrarily-chosen two components A and B, the relative volatility is defined as𝑦 ⁄𝑥𝐾𝛼𝐴𝐵 = 𝐴 ⁄ 𝐴 = 𝐴(11.1-5)
𝑦𝐵 𝑥𝐵

𝐾𝐵

𝐴𝐵

𝐴𝐵

For a binary system, the vapor-liquid equilibrium relation can be written by using 𝛼𝐴𝐵 :𝛼 𝑥𝐴𝐵
𝑦𝐴 = 1+(𝛼𝐴𝐵 −1)(11.1-6)𝑥

Since the relative volatility does not vary so much in the small range of temperature, the equation
can be used with an assumption of constant 𝛼𝐴𝐵 . Eq.(11.1-6) is useful for analytical processsimulation.
11.1-2 Boiling-point diagram
As a simple example of equilibrium relation, a two-phase two-component system can beinterpreted in terms of the temperatures and mole fractions by reference to Fig.11.1-2 called“Boiling point diagram.”

Fig.11.1-2 Boiling-point diagram (constant pressure)

The boiling-point diagram at constant pressure for mixtures of component A (the more-volatilecomponent), boiling at temperature 𝑇𝐴, and component B, boiling at temperature 𝑇𝐵 . The diagram
consists of two curves, the ends of which coincide at the positions of pure A-component and pureB-component, respectively. The upper line (called the dew-point curve) represents the vapor 𝑦𝐴𝑒
that will just begin to condense at temperature 𝑇𝑑𝑒 (dew point). The lower line (called the
bubble-point curve) represents the liquid 𝑥𝐴𝑒 that will just begin to boil at temperature 𝑇𝑏𝑒 (bubble
point). Any two points d and b on the same horizontal line represent concentrations of liquid andvapor in equilibrium.at the temperature 𝑇𝑒 . For all points above the dew-point line, the mixture isentirely vapor whereas for all points below the bubble-point line, the mixture is liquid. For points
between the two lines, the mixture is partly liquid and partly vapor. As shown in Fig.11.1-2, thehorizontal line indicates the vapor-liquid equilibrium relation that the vapor 𝑦𝐴𝑒 is in equilibriumwith the liquid 𝑥𝐴𝑒 at temperature 𝑇𝑑𝑒 = 𝑇𝑏𝑒 .

Distillation Equipment

12311.2 Distillation Equipment
11.2-1 Continuous distillation plate column
Mass transfer equipment is designed to bring the two phases into intimate contact. Let us study
how to design a continuous fractionating plate column. A typical distillation column with the major
accessories is shown in Figure 11.2-1. The distillation system consists of the main fractionatingcolumn, an overhead condenser, and a bottom reboiler. Assume that the feed to be distilled is
supplied into a plate in the central portion of the column as the liquid saturated at its bubble point.
All plates above this feed plate constitute the rectifying section, and all plates below the feed plate
constitute the stripping section which usually includes the feed plate.
The liquid feed flows down the column by gravity to the bottom reboiler and is subjected torectification by the vapor rising from the reboiler. The vapor generated in the reboiler passes up theentire column, and is condensed in the overhead condenser. A portion of the condensate isreturned to the top plate to provide the downflowing liquid in contact with the upflowing vapor in
the rectifying section. This feedback liquid stream is called reflux. The bottom product liquid iswithdrawn from the liquid pool in the reboiler. The overhead product is also withdrawn from thecondensate collected in the accumulator.

Fig.11.2-1. A typical continuous distillation plate column system.

Various types of crossflow and dual-flow plates are available. As shown in Fig.11.2-2, vapor
flows upward through the passage of both types of plates. Liquid flows downward through the side
pipe called “downcomer” for crossflow plates. For dual-flow plates, liquid flows directly downward
from the perforations or openings of the plater itself.

124

Mass Transport Equipment

Fig.11.2-2. Vapor and liquid flows for crossflow and dual-flow trays

Bubble-plate and sieve-plate columns are commonly used in industrial distillation. A liquid
mixture to be distilled, for example, is fed continuously into the central portion of the column. All
plates above the feed plate constitute the rectifying section, and all plates below the feed plate(including the feed plate) constitute the stripping section. The liquid flows down the stripping
section by gravity to the bottom of the column. The liquid reaching the bottom is partially vaporized
by the reboiler and the vapor is sent back to the bottom of the column to provide the upflowing
vapor stream in the stripping section. A definite rate of liquid as the bottom product is withdrawn
from the pool of liquid in the reboiler. The vapor rises through the rectifying section after the
stripping section toward the top of the column. The vapor arriving at the top of the column is cooled
and completely condensed in the overhead condenser. Part of the condensate is returned to the top
plate of the column to provide the downflowing liquid stream in the rectifying section. This returnliquid is called “reflux.” Without the reflux there is no liquid stream in the rectifying section, and sono rectification would occur. A definite rate of the condensate is withdrawn as the overhead productfrom the condenser.

Fig.11.2-3. Typical traditional crossflow plates

Two typical traditional plates are shown in Fig.11.2-3. For cross-flow type plates, there are an
outlet weir and a downcomer for liquid overflowing and then downflowing to the downstair tray. At
each plate, bubbles of vapor are formed at the bottom of a liquid pool by forcing the vapor throughsmall holes drilled in the plate or under slotted caps immersed in the liquid. Therefore interphasemass transfer occurs across the bubble interface while the bubbles rise up through the liquid pool.
Some of the less-volatile component existing within the bubble is condensed while some of the
more-volatile component existing in the liquid pool is vaporized. The result is a vapor phase which

Plate Column Fundamentals

125

becomes richer in the more-volatile component as it passes up the column and a liquid phase whichbecomes richer in the less-volatile component as it cascades downward.
The mass transfer process in an actual distillation column is too complicated to model in any
direct way. For engineering design of actual distillation columns, therefore, the ideal plate concept
is introduced to overcome this difficulty, and then the plate efficiency concept is taken into account.
We shall firstly study the ideal plate model and engineering design method before considering
distillation from a viewpoint of mass transfer.
11.2-2 Plate column fundamentals
11.2-2-1 Definition of ideal stage
Let us consider a single plate (the nth plate from the top) in a column. As shown in Fig.11.2-4,
two fluid streams enter the nth plate, and two leave it. A stream of liquid, 𝐿𝑛−1 kmol/s, from theplate n-1, and a stream of vapor, 𝑉𝑛+1 kmol/s, from plate n+1, are brought into intimate contact on
the nth plate. A stream of vapor, 𝑉𝑛 and a stream of liquid, 𝐿𝑛 leave the nth plate. The definitionof ideal stage or plate states that the vapor and liquid leaving the nth stage or plate are in
equilibrium, so 𝑥𝐴𝑛 and 𝑦𝐴𝑛 represent equilibrium concentrations.

Fig.11.2-4 Definition of ideal plate or stage

Fig.11.2-5 Boiling-point diagram (constant pressure)Fig.11.2-5 shows the boiling-point diagram. Usually the concentrations of the volatile component inboth phases increase with the height of the column. Although the two streams 𝑦𝐴𝑛 and 𝑥𝐴𝑛 leaving
the nth plate are in equilibrium, those entering it 𝑦𝐴𝑛+1 and 𝑥𝐴𝑛−1 are not in general. The vapor
from plate n+1 and liquid from plate n-1 are brought into nth plate, and then some of the
more-volatile component A is vaporized from the liquid whereas some of the less-volatilecomponent B is condensed from the vapor. As a result, the liquid concentration decreases from𝑥𝐴𝑛−1 to 𝑥𝐴𝑛 and the vapor concentration increases from 𝑦𝐴𝑛+1 to 𝑦𝐴𝑛 . For an ideal solution
such as the benzene-toluene binary system, the latent heat necessary to vaporize component A can126

Mass Transport Equipment

be supplied by the latent heat released in the condensation of component B. This situation isexpressed in Fig.11.2-6.

Fig.11.2-6 Concentration change due to rectification on ideal plate in boiling-point diagram

11.2-2-2 Material balance
As shown in Fig.11.2-7, the column is fed with 𝐹 kmol/h of concentration 𝑥𝐹 (molefraction of more-volatile component A). The 𝐷 kmol/h of overhead product ofconcentration 𝑥𝐷 and the 𝐵 kmol/h of bottom product of concentration 𝑥𝐵 are
withdrawn. The vapor 𝑉𝑡 from the column top is condensed. Part of the condensate isreturned as the reflux liquid 𝑅 to the top plate.
The overall material balances can be written:
Total material balance
𝐹 =𝐷+𝐵(11.2-1)
Component A balance𝐹𝑥𝐹 = 𝐷𝑥𝐷 + 𝐵𝑥𝐵(11.2-2)
A material balance around the overhead condenser is
𝐷 = 𝑉𝑡 − 𝑅(11.2-3)
The (external) reflux ratio is defined as𝑅𝑉 −𝐷
𝑟 = 𝐷 = 𝑡𝐷
(11.2-4)

Fig.11-2-7. Distillation column for total material balances

Material Balance (continued)

127

Fig.11.2-8. Material balance in rectifying section

Let us consider the 𝑛th plate from the top, where the plates are numbered serially from the top
down.
A stream of liquid, 𝐿𝑛−1 kmol/h of concentration 𝑥𝑛−1 from the plate 𝑛 − 1, and a stream ofvapor, 𝑉𝑛+1 kmol/h of concentration 𝑦𝑛+1 from the plate 𝑛 + 1, are brought into intimate contact
on the 𝑛th plate. A stream of vapor, 𝑉𝑛 kmol/h of concentration 𝑦𝑛 , and a stream of liquid, 𝐿𝑛
kmol/h of concentration 𝑥𝑛 leave the 𝑛th plate. At each plate the liquid is at the bubbling point
and the vapor is at the dew point. Some of the less volatile component B is condensed from thevapor while some of the more volatile component A is vaporized by the latent heat released in thecondensation of component B.
Choosing the rectifying section above plate 𝑛 + 1 including the overhead condenser as a control
volume, we get material balances:
𝐷 = 𝑉𝑛+1 − 𝐿𝑛
(11.2-5)
𝐷 𝑥𝐷 = 𝑉𝑛+1 𝑦𝑛+1 − 𝐿𝑛 𝑥𝑛(11.2-6)This equation can be written as
𝐿𝐷
𝑦𝑛+1 = 𝑛 𝑥𝑛 +𝑥𝐷
(11.2-7)
𝑉𝑛+1

𝑉𝑛+1

This is the equation for the operating line in the rectifying section which gives the relation between
the concentrations of the entering vapor 𝑉𝑛+1 and the leaving liquid 𝐿𝑛 .

Fig.11-2-9 Material balance in stripping section

128

Mass Transport Equipment

Similarly choosing the stripping section below plate 𝑚 including the reboiler as a control volume,
we get material balances:
𝐵 = 𝐿′𝑚 − 𝑉′𝑚+1(11.2-8)
𝐵 𝑥𝐵 = 𝐿′𝑚 𝑥𝑚 − 𝑉′𝑚+1 𝑦𝑚+1
(11.2-9)The lower equation
𝐿′𝐵
𝑦𝑚+1 = 𝑉′ 𝑚 𝑥𝑚 − 𝑉′𝑥𝐵
(11.2-10)𝑚+1

𝑚+1This is the equation for the operating line in the stripping section.

11.2-2-3 McCabe-Thiele method
The number of ideal plates required to accomplish a specified separation can be computed by useof the McCabe-Thiele step-by-step construction of ideal plates. The ideal plate definition is that thevapor 𝑉𝑛 leaving the nth plate is in equilibrium with the liquid 𝐿𝑛 leaving the same plate.

Fig.11.2-10. Definition of ideal plate

Regarding the simplification of vapor-liquid equilibrium, the concept of volatility can beconsidered. The relative volatility for a binary solution A and B can be defined as𝑦 ⁄𝑥
𝛼 = 𝐴⁄ 𝐴(11.2-11)𝑦𝐵 𝑥𝐵

where 𝑦𝐵 = 1 − 𝑦𝐴 , 𝑥𝐵 = 1 − 𝑥𝐴 .
Assuming constant relative volatility 𝛼, the equilibrium can be approximated as
𝛼𝑥𝑦 = 1+(𝛼−1)𝑥(11.2-12)

The x-y diagram for binary equilibrium relationship is plotted for the more-volatile component.

The McCabe-Thiele method is based on the following two assumptions:(1) Equimolar heats of vaporization for the two components and(2) No heat leaks and no heat of mixing.
Then the concept of constant molar overflow is obtained:(Rectifying section)
𝑉𝑛 = 𝑉𝑛−1 = − − − − − = 𝑉1 = 𝑉𝑡 = 𝑉
𝐿𝑛 = 𝐿𝑛−1 = − − − − − = 𝐿1 = 𝑅 = 𝐿(Stripping section)
𝑉′𝑚 = 𝑉′𝑚+1 = − − − − − = 𝑉′
𝐿′𝑚 = 𝐿′𝑚+1 = − − − − − = 𝐿′
Therefore the equations for the operating lines become𝐿𝐷
𝑦𝑛+1 = 𝑉 𝑥𝑛 + 𝑉 𝑥𝐷
𝐿′

𝐵

𝑦𝑚+1 = 𝑉′ 𝑥𝑚 − 𝑉′ 𝑥𝐵(11.2-13)(11.2-14)(11.2-15)(11.2-16)(11.2-17)(11.2-18)

McCabe-Thiele Step-by-step Calculation Method

The slope 𝐿⁄𝑉 of the operating line is sometimes called the internal reflux ratio:𝐿𝑅𝑟= 𝑉 = 1+𝑟𝑉𝑡

129

(11.2-19)

A mass balance around the feed plate is𝐿′ = 𝐿 + 𝑞 𝐹(11.2-20)′()
𝑉 = 𝑉 + 1−𝑞 𝐹(11.2-21)
Here the 𝑞 is a measure of the thermal condition of the feed:
energy to convert 1 mole of feed to saturated vapor𝑞=molar heat of vaporization
If a saturated vapor mixture is fed, the 𝑞 becomes 0.
Then
𝐿′ = 𝐿𝑉 = 𝑉′ + 𝐹(11.2-22)
If a saturated liquid mixture is fed, the 𝑞 becomes 1.
Then
𝐿′ = 𝐿 + 𝐹
𝑉 = 𝑉′(11.2-23)
The q-line, as shown in Fig.11.2-11, passes through the intersection point of the two operating lines
at the feed plate.

Fig.11.2-11. Interrelation of q-line with two operating lines

The general equation of q-line is given by𝑞
1
𝑦 = 𝑞−1 𝑥 − 𝑞−1 𝑥𝐹(11.2-24)

When 𝑞 = 0, 𝑦 = 𝑥𝐹
When 𝑞 = 1, 𝑥 = 𝑥𝐹

(11.2-25)

11.2-2-4 McCabe-Thiele step-by-step calculation method1)
The operating line for the rectifying section intersects the diagonal (𝑦 = 𝑥) at the overhead
product concentration 𝑥𝐷 . Similarly, the operating line for the stripping section intersects the
diagonal at the bottom product concentration 𝑥𝐵 .

1.

McCabe, W. L., and Thiele, E. W., Ind. Eng. Chem., 17, 605 (1925)

130

Mass Transport Equipment

Fig.11.2-12. McCabe-Thiele step-by-step construction of ideal plates

Since, by definition of an ideal plate, 𝑦𝑛 and 𝑥𝑛 are the concentrations of the equilibrium vapor
and liquid leaving the nth plate, point A(𝑦𝑛 , 𝑥𝑛 ) on the equilibrium curve represents the equilibrium
plate n. The operating line represents the concentrations of all possible pairs of passing streams
within the section. A horizontal line AB at 𝑦𝑛 passing through the point B (𝑦𝑛 , 𝑥𝑛−1 ) on theoperating line gives the concentrations of vapor leaving and liquid entering plate n. A vertical line
AC at 𝑥𝑛 intersecting the point C((𝑦𝑛+1 , 𝑥𝑛 ) on the operating line gives the concentrations ofvapor entering and liquid leaving plate 𝑛 + 1. Likewise, we can count the ideal plates downward( or upward) from the starting point (𝑥𝐷 , 𝑥𝐷 ) through the rectifying section ( or stripping section)by alternating use of the equilibrium curve and the operating lines…
The number of ideal plates (hypothetical equilibrium stages) must be converted to the number ofactual plates by means of Murphree plate efficiency for practical column design, which will be
studied later..
11.2-2-5 Reflux ratio
As the reflux ratio is decreased, the intersection point of the two operating lines approaches theequilibrium curve. When the intersection point touches the equilibrium curve, an infinite number ofideal plates would be required to accomplish a specified separation. The reflux ratio corresponding
to this situation is the minimum reflux ratio. The economical (optimum) reflux ratio of actual
columns usually falls around 1.5 times the minimum reflux ratio.

Fig.11.2-13. McCabe-Thiele diagram for minimum reflux ratio

Enthalpy-composition method

131

11.2-2-6 Enthalpy-composition method (Ponchon-Savarit step-by-step method)1,2)
Analysis of fractionating columns can be done by using an enthalpy balance in conjunction withmaterial balances and phase equilibria. This method is based on the graphical analysis of anenthalpy-composition diagram. Fig.11.2.14 is its schematic picture. The abscissa indicates thecomposition of a binary solution and the ordinate gives the enthalpy of the mixture of component A
and B. Similarly to the boiling-point diagram, the curve KL is a dew point line and the curve MN
indicates a bubble point line. The enthalpy of the ordinate includes latent heat, heats of mixing, andsensible heats and all these effects are built into the diagram, so that none of them need beconsidered separately. The balances of mass and enthalpy are set up over a control volume shown in
Fig.11.2-8.

Fig.11.2-14. Enthalpy-composition diagram for Ponchon-Savarit method.

Regarding the rectifying section shown in Fig.11.2-14, the enthalpy of the vapor stream (flowrate 𝑉𝑛+1 ) rising from plate n+1 is given by 𝐻𝑉𝑛+1 and the enthalpy of the liquid stream (flow rate
𝐿𝑛 ) leaving plate n is given by 𝐻𝐿𝑛 . If the column is assumed adiabatic, points P, 𝑉𝑛+1 , 𝐿𝑛 arecollinear. This line is called “enthalpy operating line.” Point P represents a common operating point
for all values of 𝑉𝑛+1 and 𝐿𝑛 . According to the center-of gravity principle,
𝑉𝑛+1
𝐿𝑛

=

𝑃𝐿𝑛
𝑃𝑉𝑛+1

=

𝑥𝐷 − 𝑥𝑛
𝑥𝐷 − 𝑦𝑛+1

Similarly for the stripping section𝑥 −𝑥= 𝑦 𝑚 − 𝑥𝑤
𝐿′

𝑉′𝑚+1
𝑚

𝑚+1

𝑤

(11.2-26)(11.2-27)

In this rectifying section for adiabatic operation, the difference between the enthalpy carried
upward by the vapor stream and that carried downward by the liquid stream is constant. In thiscontrol volume, enthalpy is removed by the overhead condenser in an amount 𝑄𝑐 (usually fixed)
Setting up the enthalpy balance over the control volume
𝐿𝑛 𝐻𝐿𝑛 + 𝐷𝐻𝐷 + 𝑄𝑐 = 𝑉𝑛+1 𝐻𝑉𝑛+1(11.2-28)
𝐻𝑃 = 𝐻𝐷 + 𝑄𝑐 ⁄𝐷(11.2-29)
Then
𝐷𝐻𝑃 = 𝑉𝑛+1 𝐻𝑉𝑛+1 − 𝐿𝑛 𝐻𝐿𝑛(11.2-30)

132

Mass Transport Equipment

For adiabatic operation, 𝐷𝐻𝑃 (the difference between the enthalpy carried by the vapor and liquid
streams) becomes the same everywhere in the rectifying section.
Similarly in the stripping section we get
𝑊𝐻𝑊 = 𝐿′𝑚 𝐻𝐿𝑚 − 𝑉 ′ 𝑚+1 𝐻𝑉𝑚+1 + 𝑄𝑟
(11.2-31)
𝐻𝑄 = 𝐻𝑊 − 𝑄𝑟
(11.2-32)
𝑊𝐻𝑄 = 𝐿′𝑚 𝐻𝐿𝑚 − 𝑉 ′ 𝑚+1 𝐻𝑉𝑚+1(11.2-33)
Since 𝐷 and 𝑊 are constant, 𝐻𝑃 and 𝐻𝑄 are also constant.
𝑊𝐻𝑄 becomes the same everywhere in the stripping section.
In a manner similar to McCabe-Thiele method, the enthalpy operating lines are used alternately
with the tie-lines to give a step-by-step determination of the number of ideal plates necessary toaccomplish a specified separation. This is called the “Ponchon-Savarit method.”
Regarding the reflux ratio, the center-of-gravity principle gives
𝐻 −𝐻
𝑅 = 𝐻𝑃 − 𝐻𝑉1
𝑉1

𝐷(11.2-34)

This equation indicates that Point P is specified by the reflux ratio.
1)
2)

Randall, M. and Longtin, B., Ind. Eng. Chem., 30, 1063, 1188 (1938)
McCabe, W. L. and Smith, J. C., “Unit Operations of Chemical Engineering,” McGraw-Hill, 3rd ed., 571 – 585(1976)[EXAMPLE 12.2-1 ] A 2 kmol/h mixture of 0.4 mole fraction of methanol and 0.6 mole fraction ofwater is to be separated into overhead and bottom products of 0.85 and 0.2, respectively, by acolumn operated with a reflux ratio of 3 at 1 atm. The feed is supplied as a saturated liquid.
(Attention: The specified concentrations of the overhead and bottom products are madeintentionally moderate for easy explanation on the enthalpy-composition diagram, especially thebottom product concentration is specified larger than usual.)(1) Per kmol of overhead product, how much heat must be withdrawn at the overhead condenser
and how much heat per kmol of bottom product must be added at the reboiler?
(2) Explain the step-by-step method for obtaining the number of ideal plates on the diagram.
Solution:(1) From the overall material balance, 𝐷 = 0.615 and 𝑊 = 1.385 kmol/h, respectively.
Point F is placed on the bubble-point line at 𝑥𝐹 = 0.4. Points D and W are also located on the
bubble-point line at 𝑥𝐷 = 0.85 and 𝑥𝑊 = 0.2, respectively. The point P can be determinedfrom the specified reflux ratio 𝑅 = 3 as follows:
From Fig.11.2-E1, we obtain 𝐻𝑉1 = 42,610, and 𝐻𝐷 = 5,645 kJ/kmol . Then
𝐻𝑃 = 𝑅(𝐻𝑉1 − 𝐻𝐷 ) + 𝐻𝑉1 = 3 × (42,610 − 5,645) + 42,610 = 79,575 kJ/kmol
The point P is decided as 𝐻𝑃 = 79,575 kJ/kmol and 𝑥𝐷 = 0.85. Then
−𝑄𝑐 ⁄𝐷 = 79,575 − 5,645 = 73,930 kJ/kmol. The cooling duty becomes−𝑄𝑐 = 73,930 kJ⁄kmol × 0.615 kmol⁄h = 45,470 kJ/h
The point W is obtained at the intersection of the overall-enthalpy line PF with the line
𝑥𝑊 = 0.2, where 𝐻𝑊 = 6,245 kJ/kmol. Therefore
𝑄𝑟 ⁄𝑊 = 𝐻𝑊 − 𝐻𝑄 = 6,245 − (−27,000) = 33,240 kJ/kmol. The heat duty is
𝑄𝑟 = 33,240 kJ⁄kmol × 1.385 kmol⁄h = 46,040 kJ⁄h

Mass Transfer in Distillation Column (Plate Column)

133

Fig.11.2-E1. Ponchon-Savarit step-by-step method for a fractionating column(2) Since 𝑥𝐷 = 𝑦1 = 0.85, the point V1 is on the dew-point line at 𝑥𝐷 = 0.85. According to thevapor-liquid equilibrium relation, when 𝑦1 = 0.85 is in equilibrium with 𝑥1 = 0.656. This
indicates that point L1 should lie on the bubble-point line at 𝑥1 = 0.656. Therefore the arrow
V1→L1 shown in the figure is the tie-line. The next operating line through P and L1 intersects
the dew-point line at point V2 (𝑦2 = 0.75). Using the VL equilibrium relation, the next tie-line
gives point L2 on the bubble-point line at 𝑥2 = 0.442. Similarly point V3 is obtained at
𝑦3 = 0.655. The next tie-line V3→L3 crosses the overall-enthalpy line. It can be considered thatthe rectifying operating lines should be placed on the right of the overall-enthalpy line and thestripping lines to the left. The best location of the feed plate is on the first plate where the liquid
concentration is less than the abscissa of the intersection of the bubble-point line and theoverall-enthalpy line. Therefore the feed plate should be on the No.4 plate. Since L3 is at the leftof line PFQ, the next operating line should be drawn from point Q. This operating line intersects
the dew-point line at V4 (at 𝑦4 = 0.380). Since 𝑥4 = 0.0857 is less than 𝑥𝑤 , four steps can beconsidered to be sufficient. A reboiler plus three ideal plates should be specified with feed plate
located at No.3 plate. (Attention: Because of the concentration of the bottom product specified
larger than the usual value in this example, the feed plate is located at No.3 plate above thereboiler.)

11.3

Mass Transfer in Distillation Column (Plate Column)

In general, an actual column does not work well in the same manner as the ideal plate column
above mentioned. Let us consider the distillation efficiency of plate columns from a viewpoint ofmass transport phenomena. The plate efficiency calculation is based on the two-film theory.
Assuming equal molar heat of vaporization for a binary mixture of A and B, the basic equations canbe derived from the interphase mass transfer equations.
Three assumptions have been made:(1) the rate of mass transfer of a component within a phase is proportional to the difference in its
concentration in the bulk of the phase and at the vapor-liquid contacting interface,(2) the vapor and liquid at the interface are in equilibrium , and134

Mass Transport Equipment(3) the holdup of the transferring component in the boundary layers on both sides of the interface isnegligibly small compared to the amount transferred in the process.

Fig.11.3-1. Composition profile in the neighborhood of vapor-liquid interface

The rate of mass transfer per unit area or the mass flux may be expressed by
N A  K yA ( y*A  y A )  k yA ( y Ai  y A )  k xA ( x A  x Ai )(11.3-1)

where 𝑥𝐴 and 𝑦𝐴 are mole fractions of the more volatile component A.
The film coefficients of mass transfer 𝑘𝑦 , 𝑘𝑥 are defined as
𝑘𝑦𝐴 = 𝐷𝐺 ⁄𝛿𝐺𝑦 , 𝑘𝑥𝐴 = 𝐷𝐿 ⁄𝛿𝐿𝑥
(11.3-2)
According to the two-film theory, two resistances in series to interphase mass transfer can be
expressed as
1
1
𝑚
1
1
1
=
+
,
=
+
(11.3-3)
𝐾
𝑘
𝑘
𝐾
𝑚𝑘
𝑘
𝑦𝐴

𝑦𝐴

𝑥𝐴

𝑥𝐴

𝑦′

𝑑𝑦

𝐾𝑦𝐴 𝑎 𝑍𝑝

𝑛+1

𝐴

𝑦𝐴

𝑥𝐴

Consider the vapor bubbling up through a pool of liquid on plate n.
At the differential control volume 𝑑𝑧 × 1 × 1, the mass balance of more volatile component A is
𝐺𝑀 𝑑𝑦𝐴 = 𝐾𝑦𝐴 (𝑦𝐴∗ − 𝑦𝐴 )𝑎 𝑑𝑧(11.3-4)
23Here 𝑎 is the interfacial area per unit volume of gas and liquid holdup in m /m .
Integrating with assumption that 𝐾𝑦𝐴 𝑎⁄𝐺𝑀 is constant, we get
𝑁𝑂𝐺 = ∫𝑦 𝑛 𝑦∗ −𝑦 =𝐴𝐺𝑀

(11.3-5)This equation is the equation defining a local value of the number of transfer unit on plates of
distillation column. Here 𝑍𝑝 is the bubbling pool height on plate n.

Mass Transfer in Distillation Column (Plate Column)

135

𝐺𝑀 (1 × 1)𝑑𝑦 = 𝐾𝑦 (𝑦 ∗ − 𝑦)𝑎 (𝑑𝑧 × 1 × 1)
Fig.11.3-2. Material balance of component A over a differential control volume dz of bubbling
foam layer (height 𝒁𝒑) on a plate

The inlet vapor is assumed to be well mixed, so that 𝑦𝐴𝑛+1 can be regarded as a constant. The
concentration of the outlet vapor 𝑦′𝐴𝑛 is generally a function of the length from the outlet weir
𝐾𝑦𝐴 𝑎 𝑍𝑝
along plate n for this kind of crossflow plate. The right-hand side term 𝐻𝑇𝑈 = 𝐺
is called“Height of Transfer Unit.” Similar equations are obtained with the other transfer units:𝑘𝑦𝐴 𝑎 𝑍𝑝𝑦′𝑑𝑦
𝑁𝐺 = ∫𝑦 𝑛 𝑦 −𝑦 = 𝐺
𝑛+1

𝑥′′

𝑁𝑂𝐿 = ∫𝑥′ 𝑛
𝑁𝐿 =

𝐴𝑖

𝑑𝑥

𝐴

=

∗
𝑛 𝑥𝐴 −𝑥𝐴𝑥′′𝑛
𝑑𝑥∫𝑥′𝑛 𝑥 −𝑥 =𝐴
𝐴𝑖

𝑀
𝐾𝑥𝐴 𝑎 𝑍𝑝
𝐿∗ 𝑀
𝑘𝑥𝐴 𝑎 𝑍𝑝
𝐿∗ 𝑀

𝑀

(11.3-6)(11.3-7)(11.3-8)

Note that the above definition of local liquid-phase transfer units assuming the counter-current
flow does not describe actual mass transfer situations, since the actual liquid stream flows
horizontally on the plates.
We should interpret that 𝑥′𝑛 and 𝑥"𝑛 are hypothetical liquid concentrations at the top andbottom of the liquid pool and that 𝐿∗ 𝑀 is hypothetical molar flow rate divided by the effective platearea. In the later section, the average values of the transfer units will be defined.
The relation between hypothetical and actual molar flow rates is given by
𝐿𝑀 𝑍𝑝 = 𝐿∗ 𝑀 𝐿𝑡
(11.3-9)
where 𝐿𝑡 is the effective length of liquid travel on plate n.
The overall mass transfer resistance is expressed in terms of the individual gas-phase andliquid-phase resistances:
𝐺𝑀
𝐺𝑀
𝑚 𝐺𝑀
=
+
(11.3-10)
𝐾 𝑎
𝑘 𝑎
𝑘 𝑎
𝑦𝐴

𝑦𝐴

𝑥𝐴

where the vapor-liquid equilibrium curve is assumed to be locally a straight line.
𝑦∗ = 𝑚 𝑥 + 𝑏
𝐺
𝐺𝐿∗
Since 𝑍𝑝 = 𝐾 𝑀 𝑎 𝑁𝑂𝐺 = 𝑘 𝑀𝑎 𝑁𝐺 = 𝑘 𝑀𝑎 𝑁𝐿 ,
𝑦𝐴

𝑦𝑎

𝑥𝑎

𝐺𝑀

𝑍𝑝

𝑍𝑝

𝑚 𝐺𝑀

𝐾𝑦𝐴

= 𝑁 ,
𝑎

𝐺𝑀

𝑂𝐺

𝑘𝑦𝐴

𝐺

𝑘𝑥𝐴

𝑂𝐺

𝐺

𝐿

= 𝑁 ,
𝑎

𝐿∗

𝐺

(11.3-11)

𝑍𝑝

𝐺= (𝑚 𝐿∗𝑀 ) 𝑘 𝑀𝑎 = (𝑚 𝐿∗𝑀 ) 𝑁 ,𝑎

Thus Eq.(11.2-10) can be rewritten as
11
𝜆= 𝑁 + 𝑁
𝑁

𝑀

𝐴𝑥

𝑀

𝐿

(11.3-12)

where 𝜆 = 𝑚 𝐿 𝑀 ⁄𝐺𝑀 ) is the slope ratio of the equilibrium and operating liners called “thestripping factor.” The transfer units averaged over the effective plate area may be applied toEq.(11.2-10) for commercial-scale columns except for extremely large columns.
⁄( ∗

136

Mass Transport Equipment

11.4 Tray Model (Plate Efficiency)
11.4-1 Murphree Plate Efficiency
The following assumptions are made: (1) that the inlet and outlet vapors are well mixed withconcentrations of 𝑦𝑛+1 and 𝑦𝑛 respectively, and (2) that the inlet and outlet liquids are also well
mixed, but the liquid concentration will vary between 𝑥𝑛−1 and 𝑥𝑛 .
The Murphree plate efficiency is defined as𝑦 −𝑦
𝐸𝑀𝐺 = 𝑦 ∗𝑛 − 𝑦𝑛+1
(11.4-1)𝑛+1𝑛Fig. 11.3-1. Murphree plate efficiency and point efficiency

Here 𝑦 ∗ 𝑛 is the concentration of the vapor that would be in equilibrium with the liquid ofconcentration 𝑥𝑛 leaving the plate, i.e. 𝑦 ∗ 𝑛 = 𝑚 𝑥𝑛 + 𝑏. The 𝑦𝑛+1 can be considered as aconstant concentration since the inlet vapor is well mixed. As mentioned in the section of
McCabe-Thiele method, for an ideal plate the vapor rising from the plate n is in equilibriumwith the liquid of concentration 𝑥𝑛 leaving the plate. Therefore 𝐸𝑀𝐺 = 1.
Similarly the plate efficiency for the liquid phase is also defined as𝑥−𝑥
𝐸𝑀𝐿 = 𝑥 𝑛−1− 𝑥∗𝑛(11.4-2)𝑛−1𝑛

A point efficiency for the vapor phase can be defined as
𝐸𝑃𝐺 =𝑦 ′ 𝑛 − 𝑦𝑛+1

(11.4-3)𝑦 ′∗ 𝑛 − 𝑦𝑛+1

Similarly− 𝑥′𝑥

𝐸𝑃𝐿 = 𝑥 𝑛−1− 𝑥′∗𝑛𝑛−1′(11.4-4)𝑛′Here 𝑦 𝑛 and 𝑥 𝑛 are local concentrations of the leaving vapor and the crossing liquid at
distance 𝑙 from the outlet weir.
The point efficiency is related with the local value of the number of transfer unit on plates of
distillation column Eq. (11.3-5).1)
Assume that over the liquid-flow cross section the liquid is well mixed i.e. 𝑦 ′∗ 𝑛 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.
Combining the definition of 𝑁𝑂𝐺 with the definition of 𝐸𝑃𝐺𝑦′𝑦 ′∗ − 𝑦 ′𝑑𝑦

𝑁𝑂𝐺 = ∫𝑦 𝑛 𝑦∗−𝑦 = − ln 𝑦′∗ 𝑛− 𝑦 𝑛𝑛+1𝑦 ′∗ 𝑛− 𝑦 ′ 𝑛exp(− 𝑁𝑂𝐺 ) = 𝑦′∗ − 𝑦
𝑛

𝑛+1

𝑛

𝑛+1

= 1 − 𝐸𝑃𝐺

(11.4-5)This is the equation for calculating the point efficiency from the number of transfer units.
The transfer units averaged over the whole effective plate area can be expressed as𝑘𝑦 𝑎 𝑍𝑝𝑑𝑦
𝐾𝑦 𝑎 𝑍𝑝
𝑦
𝑦
𝑑𝑦
̅̅̅̅̅
𝑁 = ∫ 𝑛=, ̅̅̅̅
𝑁 = ∫ 𝑛 𝑦 −𝑦 = 𝐺
𝑂𝐺

𝑦𝑛+1 𝑦 ∗ −𝑦𝐺𝑀𝐺

𝑦𝑛+1 𝑖𝑀

Mass Transfer Experiment137

𝐾𝑥 𝑎 𝑍𝑝
𝑘𝑥 𝑎 𝑍𝑝𝑥𝑥𝑛−1 𝑑𝑥
𝑑𝑥
̅̅̅̅̅
̅̅̅̅̅
𝑁𝑂𝐿 = ∫𝑥 𝑛−1 𝑥− 𝑥∗ = 𝐿∗ , 𝑁= 𝐿∗
𝑂𝐿 = ∫𝑥
𝑥− 𝑥
𝑛

𝑀

𝑛

𝑖

𝑀

(11.4-6)

In the above definition of liquid-phase transfer units, the inlet and outlet concentrations
𝑥𝑛−1 , 𝑥𝑛 are adopted in place of the concentrations at the top and bottom of the liquid pool.
These averaged transfer units are more useful than the local transfer units because they aremeasurable by experiment.
In the case when the liquid on the plate is also completely mixed, the point and plate
efficiencies become identical :𝐸𝑃𝐺 = 𝐸𝑀𝐺 = 1 − exp(− ̅̅̅̅̅
𝑁𝑂𝐺 )(11.4-7)
1.

Kister, H. Z., Distillation Design, McGraw-Hill, Newb York, P.367 (1992)

11.4-2 Mass Transfer Experiment1)
In order to understand the mass transfer in plate column distillation processes, vapor-phase andliquid-phase mass transfer coefficients should be discussed separately. The numbers of transfer units
are a function of mass transfer coefficient and interfacial area. The transfer units can be consideredof the form̅̅̅̅𝑁𝐺 = ̅̅̅̅𝑁𝐺 (𝑆𝑐𝐺 , 𝑊𝑒, 𝑅𝑒𝐺 , ℎ𝑤 ⁄𝐿𝑡 , 𝑑 ⁄𝐿𝑡 , 𝐿𝑀 ⁄𝑉𝑀 , 𝑛 𝐿𝑡 2 )̅𝑁̅̅𝐿̅ = ̅𝑁
̅̅𝐿̅ (𝑆𝑐𝐿 , 𝑊𝑒, 𝑅𝑒𝐿 , ℎ𝑤 ⁄𝐿𝑡 , 𝑑 ⁄𝐿𝑡 , 𝐿𝑀 ⁄𝑉𝑀 , 𝑛 𝐿𝑡 2 )(11.4-8)
Here 𝑊𝑒 is known as the Weber number which represents the ratio of inertia to surface tension
force:𝑢2𝜌

𝑑(11.4-9)

𝑢𝐿0 𝜚𝐿 𝑑𝑒𝑞(11.4-10)

𝑊𝑒 = 𝐺0 𝜎 𝐺
The Reynolds numbers are𝑢 𝜚 𝑑
𝑅𝑒𝐺 = 𝐺0𝜇 𝐺
𝐺

𝑅𝑒𝐿 =

𝜇𝐿

The n , d and ℎ𝑤 are the number of holes per unit area of plate, the hole diameter, and the height ofthe outlet weir, respectively. The 𝑢𝐺0 is the vapor velocity in a hole and 𝑢𝐿 is the liquid velocity
along the plate deck. The 𝑑𝑒𝑞 is the equivalent hydraulic diameter on plate.
As an example, regarding the distillation column efficiency, the experimental data on bubble cap
tray reported very many years ago by AIChE Research Committee 1) is still available.
Fig.11.4-1 shows the geometry of a crossflow tray and bubbling foam layer.

Fig. 11.4-1 Geometry of a crossflow tray and bubbling foam or froth layer

The superficial vapor velocity 𝑢𝐺 (m/s) is based on the effective area of plate (bubbling area).
The correlation equation1,2) developed on the basis of the gas-phase efficiency was obtained
from the ammonia-air-water system and acetone-benzene system:
1
𝑁𝐺 = 𝑆𝑐 (0.776 + 4.57 ℎ𝑤 − 0.238 𝐹 + 105 𝐿)(11.4-11)√𝐺

where ℎ𝑤 is the outlet weir height, 𝐹 the F-factor given by 𝑢𝐺 √𝜌𝐺 in (m/s)(kg/m3)1/2, and138

Mass Transport Equipment

𝐿 the liquid volume flow-rate per unit width of plate in m3/m s.
The liquid holdup per unit bubbling area length ℎ𝐵 is also given byℎ𝐵 = 4.2 × 10−2 + 0.19 ℎ𝑤 − 1.36 × 10−2 𝐹 + 2.49 𝐿(11.4-12)
The number of transfer unit on the liquid-phase is given by the following correlation:
𝑁𝐿 = 2.08 × 104 √𝐷𝐿 (0.213 𝐹 + 0.15)𝑡𝐿(11.4-13)2Here 𝐷𝐿 is the diffusivity in the liquid in m /s. The residence time 𝑡𝐿 (s) is the time of liquidcrossing the liquid length 𝐿𝑡 on the bubbling area shown in Fig.11.4-1.
1.
2.

AIChE Research Committee, Bubble Tray Design Manual, New York (1958)
AIChE Research Committee, Tray Efficiencies in Distillation Columns, Final Report from the Univ. Delaware (1958)

Let us learn how to predict the plate efficiency from these mass transfer correlations expressed as
the number of transfer units.
[EXAMPLE 11.4-1] A liquid mixture of benzene and toluene is to be distilled by a bubble traycolumn which is 2 m in diameter and which contains eighteen trays on 0.50 m tray-to-tray spacing.
The feed containing 54 mol% benzene is supplied as the liquid at the bubbling point. The operating
conditions and tray geometry are listed in Table 11.3-1. Obtain the plate efficiency of the rectifying
section assuming that the liquid is completely mixed.
Table 11.4-1 Operating conditions
Feed rate
0.139 kmol/s
Feed composition
54 mol% benzene
Overhead rate
0.075 kmol/s
Overhead composition
99 mol% benzene
Reflux ratio
1.67
Rectifying section
Vapor flowrate
0.20 kmol/s (Rectifying section)
Liquid flowrate
0.125 kmol/s (Rectifying section)
Average temperature
98℃
Average pressure
0.15 MPa
Average liquid composition 90 mol% benzene
Average vapor composition 94 mol% benzene
Tray geometry2m
0.0762
triangular
1.7 m (average width)
1.2 m (average length)
Outlet weir height
0.08 mOutlet weir length
1.5 m

Tray diameter
Cap size
Cap layout
Bubbling area

Solution:(Step Ⅰ) Calculation of 𝑁𝐺 : The physical properties of the vapor at 98℃ and 0.15 MPa are𝜌𝐺 = 4.0 kg⁄m3 , 𝜇𝐺 = 8.5 × 10−6 kg⁄m s, and 𝐷𝐺 = 2.94 × 10−6 𝑚2 ⁄𝑠 , respectively. Thenthe Schmidt number is 𝑆𝑐𝐺 = 0.72. The average molecular weight of the vapor
𝑀𝐺 = 78.8 kg⁄kmol. Then the volumetric vapor flow rate is
𝐺 ′ = 𝐺𝑀 𝑀𝐺 ⁄𝜌𝐺 = (0.20)(78.8)⁄4 = 3.94 m3 ⁄s
The superficial velocity based on the bubbling area is
𝑢𝐺 = 𝐺′⁄𝐴 = 3.94⁄(1.2 × 1.7) = 1.93 m/s
Then the F-factor is obtained

Mass Transport Experiment (continued)

139

1⁄2

𝐹 = 𝑢𝐺 √𝜌𝐺 = (1.93)√4 = 3.86 (m⁄s)(kg⁄m3 )The average density of the liquid is 𝜌𝐿 = 800 kg⁄m3 . The average molecular weight is 𝑀𝐿 =79.4 kg⁄kmol
The volumetric liquid flow rate is𝐿′ = 𝐿𝑀 𝑀𝐿 ⁄𝜌𝐿 = (0.125)(79.4)⁄800 = 0.0124 m3 ⁄sThen the liquid rate per unit width of the outlet weir is obtained
𝐿 = 0.0124⁄1.5 = 0.00827 m3 ⁄s m
Then Eq.(11.4-11) gives
𝑁𝐺 = (0.72)−1⁄2 (0.776 + 4.57 (0.08) − 0.238 (3.86) + 105𝑏 (0.00827)) = 1.286(Step Ⅱ) Calculation of 𝑁𝐿 : From Eq.(11.4-12), the liquid holdup on the trays is calculated:ℎ𝐵 = 4.2 × 10−2 + 0.19 (0.08) − 1.36 × 10−2 (3.86) + 2.49 (0.00827) = 0.0253 m
The contact time of liquid should be calculated based on the bubbling area𝑡𝐿 = ℎ𝐵 𝐿𝑡 ⁄𝐿 = (0.0253)(1.2)⁄0.00827 = 3.67 𝑠
The liquid-phase diffusivity is given by
𝐷𝐿 = 6 × 10−9 m2 ⁄sThen Eq.(11.4-13) gives
𝑁𝐿 = (2.028 × 104 ) (6 × 10−9 )1⁄2 (0.213 × 3.86 + 0.15)(3.67) = 5.605(Step Ⅲ) Calculation of 𝑁𝑂𝐺 :The slope of the equilibrium curve gives 𝑚 = 0.5 at 𝑥 = 0.90. Then𝜆 = 𝑚 𝐺𝑀 ⁄𝐿𝑀 = (0.50) (0.20)⁄0.125 = 0.80
Eq.(11.3-12) gives
1
1
𝜆
1
0.80
= 𝑁 + 𝑁 = 1.286 + 5.605 = 0.920 or 𝑁𝑂𝐺 = 1.087
𝑁
𝑂𝐺

𝐺

𝐿

(Step Ⅳ) Calculation of point efficiency:
Eq.(11.4-7) gives
𝐸𝑃𝐺 = 1 − exp(− 𝑁𝑂𝐺 ) = 1𝑏 − exp(−1.087) = 0.663
For the case of completely mixed liquid flow the point and plate efficiencies become identical.
The plate efficiency is given by
𝐸𝑀𝐺 = 𝐸𝑃𝐺 = 0.663 or 66.3%

11.5 Design Calculation Procedure of Distillation Columns
The design calculation for the total number of actual plates required to accomplish a specified
separation should be done in the following order:(i)
Number of ideal plates
The step-by-step calculation needs the following information:
Vapor-liquid equilibrium relationship, material balances, optimum reflux ratio, operatinglines, and McCabe-Thiele stage-by-stage method based on stage equilibrium model for ideal
plate construction.
(ii)
Plate efficiency
The plate efficiency should be considered from mass transfer information:
Vapor-phase transfer unit, liquid-phase transfer unit, total number of transfer units, point
efficiency, and plate efficiency taking into account the mixing effect.
(iii) Overall efficiency
The overall efficiency is defined as the ratio of the number of ideal plates to the number ofactual plates. If the equilibrium and operating lines are straight, the overall efficiency
ln[1+ 𝐸𝑀𝐺 (𝜆−1)]
𝐸𝑂𝑉 =(11.5-1)ln 𝜆As an alternative method, the modified McCabe-Thiele method can be used. The number of140

Mass Transport Equipment

actual plates can be computed replacing the true equilibrium curve 𝑦𝑒 by an effective curve
𝑦𝑒′ with the aid of known Murphree plate efficiencies in the McCabe-Thiele step-by-stepmethod:𝑦𝑒′ = 𝑦 + 𝐸𝑀𝐺 (𝑦𝑒 − 𝑦)(11.5-2)
Figure 11.5-2 shows the modified McCabe-Thiele method.

11.6 Heat Balance of Distillation Column System
A continuous distillation column shown below is designed to separate 𝐹 (kmol/h) of a binary
mixture of 𝑥𝐹 (mole fraction) of the more volatile component X and (1 − 𝑥𝐹 ) of the less volatilecomponent Y into an overhead product containing 𝑥𝐷 of component X and a bottom productcontaining (1 − 𝑥𝐵 ) of component Y.

Fig.11.6-1. Flow sheet of distillation column system

(i)

(ii)

(iii)
(iv)

Calculate the overhead and bottom products.
A reflux ratio 𝑟 = 𝑅⁄𝐷 is to be used and the reflux is at its bubbling point. The
feed is to be heated from 𝑡𝐹0 to 𝑡𝐹1 (K) by exchanger I, from 𝑡𝐹1 to 𝑡𝐹2 (K) by
exchanger II, and from 𝑡𝐹2 to 𝑇𝐹 (K) by exchanger III, so that it can be introduced
into the column as saturated liquid at its bubbling point.
Obtain the equations of operating line and q-line.
The molar latent heat and heat capacity of the mixture can be approximated as
Δ𝐻𝑀 = 𝑥 Δ𝐻𝑋 + (1 − 𝑥 )Δ𝐻𝑌
(J/kmol)
𝐶𝑝𝑀 = 𝑥 𝐶𝑝𝑋 + (1 − 𝑥 )𝐶𝑝𝑌
(J/kmol K)
The vapor leaving the top of the column is to be completely condensed at its dew
point 𝑇𝐷 (K) by exchanger II and the overhead condenser,
If cooling water enters the overhead condenser at 𝑡𝑖 (K) and leaves at 𝑡𝑜 (K), howmuch water is required?
The bottom product leaves the reboiler at its bubbling point 𝑇𝐵 (K). If the heatreleased by steam at pressure 𝑃𝑆 (Pa) in the reboiler is assumed equal to the heat of

Heat Balance of Distillation column System (continued)

(v)
(vi)

141

vaporization 𝐻𝑆 (J/kg), how much steam is required?
What is the temperature of the bottom product at the outlet of exchanger III?
If a constant Murphree vapor-phase plate-efficiency 𝐸𝑀𝐺 (-) is given to each stage
excluding the reboiler, how many actual plates are needed? Where should the feed
be introduced?

Solution:(i)
Overall mass balances:
𝐹 =𝐷+𝐵(11.6-1)
𝐹 𝑥𝐹 = 𝐷 𝑥𝐷 + 𝐵 𝑥𝐵
(11.6-2)
From these equations, the overhead and bottom products 𝐷 and 𝐵 (kmol/h) arefound:
𝐷=(ii)

𝑥𝐹 − 𝑥𝐵𝐹,
𝑥𝐷 − 𝑥𝐵𝐵=

𝑥𝐷 − 𝑥𝐹𝐹
𝑥𝐷 − 𝑥𝐵

Since the feed is introduced as saturated liquid, 𝑞 = 1 and the q-line equation
becomes 𝑥 = 𝑥𝐹 . For 𝑞 = 1, the liquid and vapor flow rates are
𝐿′ = 𝐿 + 𝐹 and 𝑉 ′ = 𝑉
These quantities can be expressed in terms of the reflux ratio:
𝐿′ − 𝐹 = 𝐿 = 𝑅 = 𝑟 𝐷(11.6-3)′
𝑉 = 𝑉 = 𝑅 + 𝐷 = (𝑟 + 1)𝐷(11.6-4)
Then the equations for the operating lines are found:(Rectifying section)
𝐿𝐷𝑟1
𝑦 = 𝑉 𝑥 + 𝑉 𝑥𝐷 = 𝑟+1 𝑥 + 𝑟+1 𝑥𝐷(11.6-5)(Stripping section)
𝐿′
𝐵
𝑓+ 𝑟
𝑏
𝑦 = 𝑉′ 𝑥 − 𝑉 ′ 𝑥𝐵 = 𝑟+1 𝑥 − 𝑟+1 𝑥𝐵(11.6-6)
where 𝑓 = 𝐹 ⁄𝐷 and 𝑏 = 𝐵⁄𝐷 .
(iii) The vapor flow rate in the rectifying section is
𝑉 = (𝑟 + 1)𝐷
The total rate of heat released in condensation of the vapor is(𝑟 + 1)𝐷 [𝑥𝐷 ∆𝐻𝑋 + (1 − 𝑥𝐷 )∆𝐻𝑌 ]Assuming the average heat capacity not to be a function of temperature, therate of heat transfer at exchanger II is given by
𝐹 [𝑥𝐹 𝐶𝑝𝑋 + (1 − 𝑥𝐹 )𝐶𝑝𝑌 ](𝑡𝐹2 − 𝑡𝐹1 )By heat balance set up at the condenser, the water flow rate 𝑊 (kg/h) can becalculated as(𝑟 + 1)𝐷 [𝑥𝐷 ∆𝐻𝑋 + (1 − 𝑥𝐷 )∆𝐻𝑌 ] − 𝐹 [𝑥𝐹 𝐶𝑝𝑋 + (1 − 𝑥𝐹 )𝐶𝑝𝑌 ](𝑡𝐹2 − 𝑡𝐹1 )= 𝑊 𝐶𝑝𝑤 (𝑡𝑜 − 𝑡𝑖 )(11.6-7)
Here 𝐶𝑝𝑤 is heat capacity of water in (J/kg K).
(iv)
The heat required for vaporizing 𝑉′ (kmol/h) of liquid is𝑉 ′[𝑥𝐵 ∆𝐻𝑋 + (1 − 𝑥𝐵 )∆𝐻𝑌 ] = 𝑆 ∆𝐻′𝑆
(11.6-8)
From this equation, the steam needed, 𝑆 (kg/h) can be calculated.
(v)
By heat balance set up at exchanger III the outlet temperature 𝑡𝐵 (K) can becalculated.
𝐹 [𝑥𝐹 𝐶𝑝𝑋 + (1 − 𝑥𝐹 )𝐶𝑝𝑌 ](𝑇𝐹 − 𝑇𝐹2 ) = 𝐵[𝑥𝐵 𝐶𝑝𝑋 + (1 − 𝑥𝐵 )𝐶𝑝𝑌 ](𝑇𝐵 − 𝑡𝐵 )(11.6-9)(vi)
As shown in Fig.12.1-2, the effective equilibrium curve 𝑦′𝑒 vs. 𝑥 can beobtained from the true equilibrium curve 𝑦𝑒 vs. 𝑥 by using the following
equation:𝑦′𝑒 = 𝑦 + 𝐸𝑀𝐺 (𝑦𝑒 − 𝑦)(11.6-10)

142

Mass Transport Equipment

Fig.11.6-2. Modified McCabe-Thiele step-by-step method

As shown in Fig.11.6-2, the McCabe-Thiele step-by-step construction can be made between theeffective equilibrium curve and the operating lines to determine the number of actual plates. The
true equilibrium curve is used for the last step which corresponds to the reboiler. It is found that areboiler and N actual plates are needed and the feed should be introduced on the i th plate from thetop.
Nomenclature𝑎effective interfacial area per unit packed volume, [m2/m3]
𝐶𝑝
heat capacity, [J/kmol K]𝐷overhead product, [kmol/s]
𝐷𝐺 , 𝐷𝐿
diffusivity in gas- and liquid-phase, [m2/s]
𝐸𝑀𝐺
Murphree plate efficiency, [ - ]𝐸𝑃𝐺 , 𝐸𝑃𝐿 point efficiency for vapor- and liquid-phase, [ - ]𝐹
feed rate, [kmol/s] or F-factor [(m/s)(kg/m3)0.5]
𝐺𝑀
superficial molar gas-mass velocity, [kmol/m2s]
𝐻𝑋 , 𝐻𝑌
latent heat (enthalpy) of x- and y- component, [J/kmol]ℎ𝐸𝑇𝑃
HETP (Height Equivalent to a Theoretical Plate), [m]ℎ𝑤outlet weir height, [m]𝐾vapor-liquid equilibrium constant, [ - ]𝐾𝑥 , 𝐾𝑦
overall mass transfer coefficients defined by vapor-phase and liquid-phase concentrations [kmol/m2s]
𝑘𝑥 , 𝑘𝑦mass transfer coefficients of vapor-phase and liquid-phase film [kmol/m2s]𝑚slope of equilibrium curve, 𝑑𝑦⁄𝑑𝑥,
𝑁𝑂𝐺 , 𝑁𝐺 , 𝑁𝐿 number of transfer unit, OG:overall, G:gas phase, L:liquid phase, [ - ]𝑛theoretical plate/stage number, [ - ]𝑅𝑒Reynolds number, [ - ]𝑆𝑐Schmidt number, [ - ]𝑇, 𝑡temperature, [K]𝑥𝐴 , 𝑦𝐴mole fraction of component A, liquid- and vapor-phase, [ - ]𝑧height of packing section, [m]𝛼𝐴𝐵
relative volatility, [ - ]𝛿𝐺 , 𝛿𝐿
thicknesses of gas-phase and liquid-phase film, [m]𝜆stripping factor, [ - ]𝜇viscosity, [kg/m s]
Subscripts
A, B
component A, B
G, L
gas-phase, liquid-phase𝑖interfaceCHAPTER 12
SIMULTANEOUS HEAT AND MASS TRANSFER - I
12.1 Theory of Simultaneous Heat and Mass Transfer - I
As distinct from gas absorption, a distillation process takes place in the form of non-isothermal
counter-diffusion accompanied with the enthalpy releasing due to condensation of less-volatilecomponent B and the enthalpy receiving vaporization of more-volatile component A in theneighborhood of vapor-liquid contacting interface. Even at present, however, it is still very difficult
to confirm whether or not the enthalpy released by the component B is totally consumed for thevaporization of component A. This is an important theme for the investigators of non-equilibrium
thermodynamics.
In this chapter, we consider the distillation process by the simultaneous heat and mass transfer
based on the assumption that the total molar enthalpy 𝑁𝐵 ∆𝐻𝐵 released by condensation ofcomponent B is consumed for the vaporization of component A. The following enthalpy balanceshould exist:
𝑁𝐵 ∆𝐻𝐵 = 𝑁𝐴 ∆𝐻𝐴(12.1-1)
If ∆𝐻𝐵 = ∆𝐻𝐴 , 𝑁𝐵 = 𝑁𝐴 .
That is, this condition indicates the equimolar counter-diffusion.
For example, benzene and toluene have approximately equal molar enthalpy for condensationand vaporization.

Fig.12.1-1. Heat and mass transfer in a distillation process

As shown in Fig.12.1-1, in the counter-current condition of vapor and liquid streams, less-volatilecomponent B having enthalpy 𝐻𝐺𝐵 in the vapor bulk arrives by convective mass transfer at theinterface and then the latent heat 𝑁𝐵 ∆𝐻𝐵 released due to condensation of component B is absorbed
for vaporization by more-volatile component A which arrives at the interface by convective masstransfer in the opposite direction from the liquid bulk. At the same time, thermal energy istransported by convective heat transfer accompanied with the counter-diffusion from the vapor bulk
to the liquid bulk across the interface. Only for simplicity, it can be considered by the144

Simultaneous Heat and Mass Transfer - I

above-mentioned assumption that there is no temperature jump in the neighborhood of the interface.
We will pay attention to the mass transfer of component A from liquid to vapor phase in a packed
column (cross-section area 𝑆).
If their molar latent heats are equal: ∆𝐻𝐴 = ∆𝐻𝐵 (J/kmol), equimolar counter-diffusion takes place.
The interphase mass transfer of component A can be expressed by the following mass-flux equation
based on Fick’s law and the two film theory:𝑑𝑦𝑑𝑥
𝑁𝐴 = − 𝜌𝐺 𝐷𝐴𝐵𝐺 𝐴 |(12.1-2)= − 𝜌𝐿 𝐷𝐴𝐵𝐿 𝐴 |𝑑𝑧 𝑧=0𝑑𝑧 𝑧=0where 𝜌𝐺 and 𝜌𝐿 (kmol/m3) are molar densities of vapor and liquid, and 𝐷𝐴𝐵𝐺 and 𝐷𝐴𝐵𝐿 (m2/s)
are vapor-phase and liquid-phase diffusivities of component A in B, respectively.

Fig.12.1-2. Composition and temperature profiles in the neighborhood of vapor-liquid contacting
interface

Therefore the convective heat and mass transfer coefficients can be defined as𝑑𝑦
𝑁𝐴 = − 𝜌𝐺 𝐷𝐴𝐵𝐺 𝐴 |= 𝜌𝐺 (𝐷𝐴𝐵𝐺 ⁄𝛿𝐺 )(𝑦𝐴𝑖 − 𝑦𝐴 ) = 𝑘𝑦𝐴 (𝑦𝐴𝑖 − 𝑦𝐴 )𝑑𝑧 𝑧=0𝑑𝑥𝐴

𝑁𝐴 = − 𝜌𝐿 𝐷𝐴𝐵𝐿 𝑑𝑧 |

𝑧=0

𝑑𝑇

𝑄 = − 𝜅𝐺 𝑑𝑧𝐺|
𝑑𝑇

𝑧=0

𝑄 = − 𝜅𝐿 𝑑𝑧𝐿 |𝑧=0= 𝜌𝐿 (𝐷𝐴𝐵𝐿 ⁄𝛿𝐿 )(𝑥𝐴 − 𝑥𝐴𝑖 ) = 𝑘𝑥𝐴 (𝑥𝐴 − 𝑥𝐴𝑖 )(12.1-3)(12.1-4)= (𝜅𝐺 ⁄𝛿𝐺 )(𝑇𝐺 − 𝑇𝑖 ) = ℎ𝐺 (𝑇𝐺 − 𝑇𝑖 )(12.1-5)= (𝜅𝐿 ⁄𝛿𝐿 )(𝑇𝑖 − 𝑇𝐿 ) = ℎ𝐿 ((𝑇𝑖 − 𝑇𝐿 ))(12.1-6)

It should be noticed that as distinct from gas absorption, the mass transfer coefficients indistillation do not include the effect of bulk transfer owing to equimolar counter-diffusion.
As shown in Fig.12.1-2, the overall driving force can be considered to be the liquid bulk
concentration 𝑥𝐴𝑚 minus the vapor bulk concentration 𝑦𝐴𝑚 . However it is necessary to use aconcentration unit on the common basis.
If we use vapor-phase concentration unit in mole fraction, the overall driving force is given by𝑦𝐴∗ − 𝑦𝐴, where 𝑦𝐴∗ is fictitious concentration in equilibrium with the liquid bulk concentration𝑥𝐴 . Usually the liquid-bulk concentration 𝑥𝐴 is converted into 𝑦𝐴∗ by using a vapor-liquid
equilibrium relation such as 𝑦𝐴 = 𝑚 𝑥𝐴 .

Transport Phenomena in a Packed Column Distillation Process

145

12.2 Transport Phenomena in a Packed Column Distillation
Process
12.2-1 Simultaneous mass and energy transfer model
Unless a process fluid mixture to be distilled has any problem in thermal processes such as
fouling, polymerization, and/or thermal decomposition, a packed bed column is very often
employed for distillation operation.
Let us consider a packed column distillation. It is very instructive to reconsider a distillation
process from a standpoint of simultaneous heat and mass transfer. It should be kept in mind that as
distinct from gas absorption, a usual distillation process takes place under the condition ofequimolar counter-diffusion. That is, there is no stagnant incondensable component near thevapor-liquid interface.
Figure 12.2-1 shows a usual packed column, in which random packings or structured packings
are equipped in the three packing sections. The upper packing section above the feed stage is calledrectifying section and the lower one below the feed stage is called stripping section. In each empty
space above the respective packing beds, a liquid collector and a liquid distributor are installed inorder to avoid maldistribution of liquid stream in the packing section.

Fig.12.2-1. Packed distillation column

Fig.12.2-2. Shell balance of mass and enthalpy set up over a differential control volume of packed
distillation columns

146

Simultaneous Heat and Mass Transfer - I

Fig.12.2-2 shows a schematic picture of control volume having differential height 𝑑𝑧 for setting
up heat and mass balances.
For distillation processes, molar rates of vapor and liquid 𝐺𝑀 and 𝐿𝑀 can be assumed to be
kept constant due to equimolar counter-diffusion.
Therefore, referring to Eqs.(12.1-3~6), the differential heat and mass balance equations can be
written as
𝑁𝐴 𝑆 𝑑𝑧 = 𝐺𝑀 𝑑𝑦𝐴 𝑆 = 𝑘𝑦𝐴 𝑎 (𝑦𝐴𝑖 − 𝑦𝐴 )𝑆 𝑑𝑧 = 𝑘𝑥𝐴 𝑎 (𝑥𝐴 − 𝑥𝐴𝑖 )𝑆 𝑑𝑧(12.2-1)
𝑄 𝑆 𝑑𝑧 = 𝐺𝑀 𝑑𝐻𝐺 𝑆 = ℎ𝐺 𝑎 (𝑇𝐺 − 𝑇𝑖 )𝑆 𝑑𝑧 = ℎ𝐿 𝑎 (𝑇𝑖 − 𝑇𝐿 )𝑆 𝑑𝑧(12.2-2)
where 𝐺𝑀 and 𝐿𝑀 (𝑘𝑚𝑜𝑙 ⁄𝑚2 𝑠) are superficial molar velocities of vapor and liquid streams.
The molar mass-flux 𝑁𝐴 and enthalpy-flux 𝑄 have units of (𝑘𝑚𝑜𝑙 ⁄𝑚2 𝑠) and (𝑊 ⁄𝑚2 ) ,respectively.
Using the overall volumetric coefficients 𝐾𝑦𝐴 𝑎 and 𝑈𝑎
𝐺𝑀 𝑑𝑦𝐴 = 𝐾𝑦𝐴 𝑎 (𝑦𝐴∗ − 𝑦𝐴 )𝑑𝑧(12.2-3)()
𝐺𝑀 𝑑𝐻𝐺 = 𝑈 𝑎 𝑇𝐺 − 𝑇𝐿 𝑑𝑧(12.2-4)
In these equations, the effective interfacial area per unit packed-bed volume 𝑎 (m2/m3) isintroduced.
Usually it is not so easy to measure vapor-liquid-contacting interfacial area 𝑎 (m2/m3), so theoverall volumetric coefficients 𝐾𝑦𝐴 𝑎 and 𝐾𝑥𝐴 𝑎 are employed for the packed bed mass transfer.
Similarly it is possible to define the following overall coefficient 𝐾𝑥𝐴 based on liquid-phase
concentration:
𝑁𝐴 𝑎 𝑆𝑑𝑧 = 𝐾𝑥𝐴 𝑎(𝑥𝐴 − 𝑥𝐴∗ )𝑆𝑑𝑧
(12.2-5)∗
where 𝑥𝐴 is fictitious concentration in equilibrium with the vapor bulk concentration 𝑦𝐴 .
According to Eq.(11.2-3),𝐺𝑑𝑦𝑑𝑧 = 𝐾 𝑀𝑎 𝑦 ∗ − 𝐴𝑦
(12.2-6)
𝑦𝐴

𝐴

𝐴

𝐺𝑀

Assuming the coefficient𝑍𝑇 =to be constant, integration from column top to bottom gives

𝐾𝑦𝐴 𝑎
𝑍𝑇
𝑦𝐴2 𝑑𝑦𝐴
𝐺𝑀̅̅̅̅̅̅̅
∫0 𝑑𝑧 = 𝐾 𝑎 ∫𝑦𝐴1 𝑦∗ − 𝑦𝑦𝐴𝐴𝐴Here the quantity 𝐻𝑂𝐺 =𝐺𝑀̅̅̅̅̅̅

𝐾𝑦𝐴 𝑎(12.2-7)

averaged over the column height is called “Height of a Transfer

Unit.” Strictly speaking, HTU actually increases in the vertical direction from the bottom toward
the top of the packed bed. In order to calculate the total packed bed height 𝑍𝑇 required for aseparation specification, however, the overall volumetric coefficient 𝐾𝑦𝐴 𝑎 should be evaluated byconsidering the vapor-phase and the liquid-phase mass transfer, separately.
The interphase mass transfer resistance based on two film theory is expressed by
11𝑚= 𝑘 + 𝑘(12.2-8)
𝐾
𝑦𝐴

𝑦𝐴

𝑥𝐴

where the slope of equilibrium curve 𝑚 is used for conversion of liquid-phase to vapor-phase
concentration unit.
If the vapor-phase and liquid-phase volumetric coefficients 𝑘𝑦𝐴 and 𝑘𝑥𝐴 are given, the totalheight of the packed column can be calculated with the aid of Eq.(12.2-8).
Concerning the mass transfer resistance, the following parameters are defined: Height of Overall
Transfer Unit and Height of gas-phase and liquid-phase transfer units𝐺𝐿
𝐻𝑂𝐺 = 𝐾 𝑀𝑎 , 𝐻𝑂𝐿 = 𝐾 𝑀𝑎
(12.2-9)𝑦𝐴

𝐺𝑀

𝐻𝐺 = 𝑘

𝑦𝐴 𝑎

𝑥𝐴

,

𝐿𝑀

𝐻𝐿 = 𝑘𝑥𝐴 𝑎(12.2-10)

Therefore multiplying Eq.(11.2-8) by molar vapor velocity 𝐺𝑀 , the following equation is obtained
indicating the two mass transfer resistances in series in the neighborhood of the interface:
𝐻𝑂𝐺 = 𝐻𝐺 + 𝜆 𝐻𝐿
(12.2-11)

Efficiency of Packed Distillation Columns

147

where 𝜆 = 𝑚⁄(𝐿⁄𝑉 ) is the slope ratio of the equilibrium curve to the operating line, called“stripping factor.”.
This equation is very important to check which phase resistance is predominant over the interphasemass transfer, vapor-phase or liquid-phase.
Although these mass transfer parameters are available in some degree in the correlation databank,
we should consider it rather difficult to apply them for engineering column design without
experiment for checking.
12.2-2 Efficiency of packed distillation columns
The concept of plate efficiency cannot be applied to packed distillation columns.
In place of the Murphree plate efficiency, the distillation efficiency of a packed column can beevaluated by the HETP (Height Equivalent to Theoretical Plate). This is defined as the total heightof the packing section divided by the required number of theoretical stages. Therefore thischaracteristic value can be determined by comparing the result of distillation experiment conductedin a real packed column with the calculation result using the McCabe-Thiele stage-by-stage processsimulation calculation based on the equilibrium-stage model in an ideal column. However if we
consider local variation in distillation efficiency in a packed column, we should introduce a control
volume approach for definition of the local HETP. Fig. 12.2-3 indicates the significance of HETP
defined by using a control volume approach. In this case, the HETP can be defined as the control
volume height of a real column required for the condition that the vapor (concentration 𝑦𝐴𝑛 )leaving the control volume from the top should have an equilibrium relation with the liquid(concentration 𝑥𝐴𝑛 ) leaving from the bottom.
The mass balance of more volatile component A is set over the nth control volume as follows:∗
𝐺𝑀𝑛 𝑦𝐴𝑛 − 𝐺𝑀𝑛+1 𝑦𝐴𝑛+1 = 𝐾𝑦𝐴,𝑛 𝑎 (𝑦𝐴𝑛
− 𝑦𝐴𝑛 )𝑙.𝑚. ℎ𝐸𝑇𝑃,𝑛(12.2-12)
where logarithmic mean of the concentration difference as the mass transfer driving forces is used∗
between the top and bottom of each control volume. The concentration 𝑦𝐴𝑛
is the bulk
concentration of the liquid phase 𝑥𝐴𝑛 converted with the units of the gas-phase concentration.
This equation is similar to Eq. 12.2-3.

Fig. 12.2-3. Interrelation of HETP for a real packed column with equilibrium stage in an ideal
column148

Simultaneous Heat and Mass Transfer - I

If the required number of theoretical stages (i.e., control volumes) is N, the total height ofthe packed column 𝑍𝑇 for the given separation specification can be calculated as
𝐺

𝑦

−𝐺

𝑦

𝑀𝑛 𝐴𝑛
𝑀𝑛+1 𝐴𝑛+1
𝑁
𝑍𝑇 = ∑𝑁𝑛=1 ℎ𝐸𝑇𝑃,𝑛 = ∑𝑛=1∗

𝐾𝑦𝐴,𝑛 𝑎 (𝑦𝐴𝑛 − 𝑦𝐴𝑛)𝑙.𝑚.

(12.2-13)

Usually 𝐻𝑂𝐺 = 𝐺𝑀 ⁄𝐾𝑦𝐴 𝑎 varies with vertical height of the packing section. For ideal solutions,
𝐺𝑀𝑛 is kept almost constant. In addition, if we use an averaged value of 𝐻𝑂𝐺 , the above equation
can be simplified as
𝐺

𝑀
𝑍𝑇 = ∑𝑁𝑛=1 ℎ𝐸𝑇𝑃,𝑛 = (𝐾 𝑎)
𝑦𝐴

𝑎𝑣

∑𝑁
𝑛=1

𝑦𝐴𝑛 − 𝑦𝐴𝑛+1
∗ −𝑦
(𝑦𝐴𝑛
𝐴𝑛 )𝑙.𝑚.

(12.2-14)

This equation coincides with Eq.(12.2-7).
[PROBLEM 12.2-P1] Fig.12.2-P1 shows the relation of HETP with HOG. Derive the following
relation between HETP and HTUOG:ln 𝜆𝐻𝐸𝑇𝑃 = 𝜆−1 𝐻𝑂𝐺(12.2-P1)
where 𝜆 is the stripping factor defined by the slope ratio of equilibrium curve and operating line.

Fig. 12.2-P1. Relation of HETP with HTU.

12.3 Analogy between Mass and Enthalpy Transfer in a Packed
Column Distillation Process1,2)
A packed column distillation process takes place with non-isothermal and counter-diffusional
interphase mass transfer accompanied with phase transformation. Essentially the distillation
efficiency such as HETP should vary streamwise upward in the vertical direction. There is anexperimental work observing such a complicated transport phenomenon inside the packed
distillation column. This work1) deals with a semi-empirical model by making up the mutual defects
between the experiment in a real packed column and the computer-aided process simulation basedon the equilibrium model. It is instructive to introduce here the experimental results obtained in acommercial-scale column equipped with wire-mesh corrugated structured packing.
Let us consider the mass transfer of more-volatile component A from the liquid- to thevapor-phase through the interface. The total-reflux distillation experiment was performed withmethanol-ethanol binary system. Similarly to Eq.12.2-12, the interphase transfer of mass andenthalpy can be expressed by the following equations defining a cylindrical control volume shownin Fig. 12.3-1:

Analogy between Mass and Enthalpy Transfer in a Packed Column Distillation Process (continued)

149

Fig. 12.3-1. Control volume of j th ideal stage defined in a real column

The shell balances of mass and enthalpy are set up over the j-stage control volume:(Mass transfer)
𝐺𝑀𝑗 𝑦𝐴𝑗 − 𝐺𝑀𝑗+1 𝑦𝐴𝑗+1 = (𝑘𝑦𝐴 𝑎)𝑗 (𝑦𝐴𝑗,𝑖 − 𝑦𝐴𝑗 )ℎ𝐸𝑇𝑃𝑗(12.3-1)= (𝑘𝑥𝐴 𝑎)𝑗 (𝑥𝐴𝑗 − 𝑥𝐴𝑗,𝑖 )ℎ𝐸𝑇𝑃𝑗= (𝐾𝑦𝐴 𝑎)𝑗 (𝑦 ∗𝐴𝑗 − 𝑦𝐴𝑗 )

𝑙.𝑚.

ℎ𝐸𝑇𝑃𝑗

(Enthalpy transfer)(12.3-2)
𝐺𝑀𝑗+1 𝐻𝐺𝑗+1 − 𝐺𝑀𝑗 𝐻𝐺𝑗 = (ℎ𝐺 𝑎)𝑗 (𝑇𝐺𝑗 − 𝑇𝑖𝑗 )ℎ𝐸𝑇𝑃𝑗= (ℎ𝐿 𝑎)𝑗 (𝑇𝑖𝑗 − 𝑇𝐿𝑗 )ℎ𝐸𝑇𝑃𝑗= (𝑈𝑎)𝑗 (𝑇𝐺 − 𝑇𝐿 )𝑙.𝑚. ℎ𝐸𝑇𝑃𝑗
It is a key point of modeling to give the height ℎ𝐸𝑇𝑃𝑗 for defining the control volume.
Here 𝐺𝑀𝑗 , 𝐿𝑀𝑗 denote vapor and liquid molar velocity (kmol/m2s), 𝑘𝑦𝐴 𝑎, 𝑘𝑥𝐴 𝑎, 𝐾𝑦𝐴 𝑎 thevolumetric mass transfer coefficients (kmol/m 3s), ℎ𝐺 𝑎, ℎ𝐿 𝑎, 𝑈𝑎 the volumetric heat transfer
coefficients (W/m3K), 𝑦𝐴 , 𝑥𝐴 the vapor- and liquid-phase mole fractions of component A, and 𝑇𝐺 ,𝑇𝐿 the vapor- and liquid-phase temperatures (K). The subscript j indicates jth ideal stage and i
implies the vapor-liquid interface. The asterisk 𝑦 ∗𝐴𝑗 is the bulk concentration of the liquid phase
expressed with the units of the vapor-phase concentration.
In this work, local variations of not only the mass and enthalpy transfer coefficients but also the
HETPs are determined by the distillation experiment conducted in a real packed column with theaid of computer process simulation. An understanding of analogy between mass and enthalpy
transfer is possible because the enthalpy transfer is fulfilled primarily by the transferring volatilecomponents themselves.
It can be considered that the superficial vapor velocity is kept almost constant in the verticaldirection in the packing section, similarly to the free stream of the boundary layer flow over a flat
plate.
In order to consider the interrelation between mass and enthalpy transfer in the vapor-phase, thelocal j-factors for mass and enthalpy transfer are defined respectively as
𝑘𝑦𝐴 𝑎𝑗𝐷𝐺 = 𝑎 𝐺 𝑆𝑐𝐺 2⁄3(12.3-3)
𝑝 𝑀

ℎ 𝑎

𝐺
𝑗𝐻𝐺 = 𝑎 𝐶𝑝
𝑃𝑟𝐺 2⁄3𝐺𝑝

𝐺(12.3-4)

In a manner similar to the local analysis of boundary layer flow over a flat plate, the following local
length Reynolds number is defined on the basis of superficial relative velocity 𝑢𝑠 = 𝑢𝐺𝑠 − 𝑢𝐿𝑠 :𝑅𝑒𝑧𝐺 =𝑢𝑠 𝜌𝐺 (𝑍⁄𝑑𝑒𝑞 )𝑎𝑝 𝜇𝐺(12.3-5)

150

Simultaneous Heat and Mass Transfer - I

where 𝑍 is the vertically upward distance from the bottom of the packing section, 𝑑𝑒𝑞 theequivalent diameter of the structured packing, and 𝑎𝑝 the specific surface area per unit volume ofthe packing.
Fig.12.3-2 show one of the experimental results of the boundary-layer-like plot of j-factors
against local Reynolds number. Here F is the F-factor defined at the top of the packing section for
the experimental condition.
The F-factor defined by using the superficial vapor velocity 𝑢𝐺𝑠 measured at the top and its vapor
density 𝜌𝐺 is
𝐹 = 𝑢𝐺𝑠 √𝜌𝐺(12.3-6)
The position of the left-side ordinate shown in Fig.12.3-2 corresponds to the bottom of the packing
section. It has been found that the local coefficients of mass and enthalpy transfer decrease
downstream in the vertical direction like the boundary layer over a hot flat plate. If this control
volume approach based on the superficial vapor velocity is approved, as shown in Fig.12.3-3, local
variation of the j-factors in a packed distillation column resembles that of the single-phase boundary
layer flow over a flat plate. Only the difference between the two cases is that the vapor stream withconstant superficial velocity has streamwise variation of vapor composition in the packed column
whereas the free stream has uniform constant concentration in the boundary layer flow. It can beconjectured that the same boundary-layer-like tendency can also be observed in the wetted-wall
distillation column.

Fig.12.3-2. Similarity plot of j-factors against length Reynolds number

Fig.12.3-3. Comparison of packed column distillation with boundary layer flow over a flat plate

Nomenclature151

It is very interesting that the j-factors 𝑗𝐷𝐺 , 𝑗𝐻𝐺 indicate the same dependency of local length
Reynolds numbers. This suggests that local similarity exists between simultaneous mass andenthalpy transfer in the vapor-phase film. Therefore 𝑗𝐻𝐺 ⁄𝑗𝐷𝐺 = C (constant) leads to :
𝑗𝐻𝐺
𝑗𝐷𝐺

=

(ℎ𝐺 ⁄𝐶𝑝𝐺 )𝑘𝑦𝐴(

𝑃𝑟𝐺 2⁄3
𝑆𝑐𝐺)(12.3-7)This implies that the distillation process takes place with the similar transfer processes expressedbyℎ𝐺 ⁄𝐶𝑝𝐺= 𝐶 𝐿𝑒 −2⁄3
(12.3-8)𝑘𝑦𝐴

The liquid-phase j-factors also indicate similar dependency but the results are omitted owing to
the complicated mechanism beyond the scope of the book.
Although the bottom edge of the packing section should correspond to the sharp leading edge ofthe flat plate, it can be taken into account that the fictitious origin 𝑧 = 0 slightly shifts dependingon the top F-factor.
1.
2.

Nishimura, G., Kataoka, K., Noda, H., and Ohmura, N., Proc. 30 th European Symposium on Computer Aided Process
Engineering (ESCAPE30), 1-6, May 24-27, Milano, Italy (2020)
Nishimura, G., Kataoka, K., Noda, H., and Ohmura, N., Journal of Advanced Chemical Engineering, Vol.11, Issue 2, (2021)

Nomenclature𝑎effective interfacial area per unit packed volume, [m2/m3]
𝑎𝑝
specific surface area per unit packed volume, [m2/m3]
𝐶𝑝
heat capacity, [J/kmol K]
𝐷𝐴𝐵𝐺 , 𝐷𝐴𝐵𝐿 diffusivity in gas- and liquid-phase, [m2/s]
𝑑𝑒𝑞
equivalent diameter, [m]𝐹F-factor [(m/s)(kg/m3)0.5]
𝐺
superficial vapor mass velocity, [kg/m2s]𝐺𝑀superficial molar vapor velocity, [kmol/m2s]
𝐻𝐴 , 𝐻𝐵
latent heat (enthalpy) of component A, B, [J/kmol]
𝐻𝑂𝐺 , 𝐻𝐺 , 𝐻𝐿 height of transfer unit, OG:overall, G:vapor phase, L:liquid phase, [ - ]𝐻𝑉 , 𝐻𝐿
enthalpy of vapor and liquid, [J/kmol]ℎ𝐸𝑇𝑃
HETP (Height Equivalent to a Theoretical Plate), [m]ℎ𝐺 , ℎ𝐿vapor-phase and liquid-phase heat transfer coefficient, [W/m2K]

𝑗𝐷𝐺 , 𝑗𝐻𝐺 j-factor for mass and enthalpy transfer in vapor phase, [ - ]𝐾𝑥 , 𝐾𝑦
𝑘𝑥 , 𝑘𝑦𝐿𝑀𝑚
𝑁𝐴 , 𝑁𝐵𝑛
𝑄
𝑅𝑒𝑥 , 𝑅𝑒𝑧𝑆
𝑆𝑐
𝑇𝐺 , 𝑇𝐿𝑈
𝑥𝐴 , 𝑦𝐴𝑧
𝑍𝑇𝛿𝐺 , 𝛿𝐿
𝜅𝐺 , 𝜅𝐿
𝜚𝐺 , 𝜌𝐿

overall mass transfer coefficients defined by vapor-phase and liquid-phase concentrations [kmol/m2s]
mass transfer coefficients of vapor-phase and liquid-phase film [kmol/m2s]
superficial molar liquid velocity, [kmol/m2s]
slope of equilibrium curve, 𝑑𝑦⁄𝑑𝑥,
molar mass flux of component A, B, [kmol/m2s]
control volume/stage number, [ - ]enthalpy flux, [J/m2s]
length Reynolds number, [ - ]cross-sectional area of packed columns, [m2]Schmidt number, [ - ]temperature of vapor- and liquid-phase, [K]overall enthalpy transfer coefficient, [W/m2K]
mole fraction of component A in liquid- and vapor-phase, [ - ]vertical distance from the bottom of packing section, [m]total height of packing section, [m]thickness of gas-phase and liquid-phase film, [m]thermal conductivity of vapor- and liquid-phase, [J/m s K]
molar density of vapor and liquid, [kmol/m3]

Subscripts
A, B
component A, B
G, L
gas-phase, liquid-phase𝑖interfaceTtemperature

CHAPTER 13
SIMULTANEOUS HEAT AND MASS TRANSFER - II13.1 Theory of Simultaneous Heat and Mass Transfer II(Humidification and Dehumidification)
When a condensable vapor A is condensed from a gas mixture, enthalpy transfer occurs by two
effects: one is the conductive heat transfer due to the temperature gradient; the other is the enthalpy
transfer due to the mass transfer.

Fig.13.1-1. Schematic temperature and concentration profiles in the neighborhood of a coolingpipe wall

Fig-13.1 shows schematic picture of temperature and concentration profiles in the neighborhoodof a cooling pipe wall, on which condensate liquid is falling filmwise.
Let us consider the enthalpy transfer through a gas-phase film of thickness 𝛿 on the gas-liquidinterface. The total flux of enthalpy into the gas-phase film is made up of the conductive heat flux
−𝜅(𝜕𝑇⁄𝜕𝑧 ) and the enthalpy flux due to diffusion 𝑁𝐴 𝐶𝑝𝐴 (𝑇 − 𝑇0 ) + 𝑁𝐵 𝐶𝑝𝐵 (𝑇 − 𝑇0 ).
Here 𝑇0 is a standard-state temperature to be selected later, and 𝐶𝑝𝐴 , 𝐶𝑝𝐵 are the molar heat
capacities of components A and B. The vertical variation in the thickness and composition of thegas-phase film is assumed to be small compared to the variation in the transverse z-direction.
From the mass balance over a differential control volume of thickness 𝑑𝑧 × 1 × 1
𝑁𝐴 |𝑧+𝑑𝑧 − 𝑁𝐴 |𝑧 = 0 →

𝜕𝑁𝐴
𝜕𝑧
𝜕𝑁𝐵

=0

𝑁𝐵 |𝑧+𝑑𝑧 − 𝑁𝐵 |𝑧 = 0 → 𝜕𝑧 = 0
Therefore
𝑁𝐴 = 𝑁𝐴0 = 𝑐𝑜𝑛𝑠𝑡
𝑁𝐵 = 𝑁𝐵0 = 𝑐𝑜𝑛𝑠𝑡
According to the Fick’s law(13.1-1)(13.1-2)

Theory of Simultaneous Heat and Mass Transfer- II (continued)

153

𝜕𝑦

𝑁𝐴 − 𝑦𝐴 (𝑁𝐴 + 𝑁𝐵 ) = −𝑐 𝐷𝐴𝐵 𝜕𝑧𝐴(13.1-3)
where 𝑦𝐴 is mole fraction of component A and 𝑐 the molar density of the gas-phase film.
Since 𝑁𝐴 = 𝑁𝐴0 and 𝑁𝐵 = 𝑁𝐵0 , the equation becomes𝜕𝑦𝑁𝐴0 − 𝑦𝐴 (𝑁𝐴0 + 𝑁𝐵0 ) = −𝑐 𝐷𝐴𝐵 𝜕𝑧𝐴
Then the equation to be solved for the concentration distribution is
𝑑𝑦𝐴
𝑁 +𝑁
𝑁
− 𝐴0 𝐵0 𝑦𝐴 = − 𝐴0
(13.1-4)
𝑑𝑧

𝑐𝐷𝐴𝐵

𝑐𝐷𝐴𝐵

Integration gives𝑁
𝑁 + 𝑁𝐵0𝑦𝐴 = 𝑁 +𝐴0𝑁 + 𝐶1 𝑒𝑥𝑝 ( 𝐴0
𝑧)
𝑐𝐷
𝐴0

𝐵0

(13.1-5)𝐴𝐵

The integration constant 𝐶1 can be evaluated using the boundary condition 𝑦𝐴 = 𝑦𝐴0 at 𝑧 = 0:𝑁𝐴0𝐶1 = 𝑦𝐴0 −𝑁𝐴0 + 𝑁𝐵0
Then the solution for concentration distribution in the gas film is
𝑁𝐴0
𝑁𝐴0
𝑁 +𝑁
𝑦𝐴 =
+ (𝑦𝐴0 −) 𝑒𝑥𝑝 ( 𝐴0 𝐵0 𝑧)(13.1-6)
𝑁𝐴0 + 𝑁𝐵0

𝑁𝐴0 + 𝑁𝐵0

That is𝑁𝐴0
𝑁𝐴0 + 𝑁𝐵0𝑁𝐴0
𝑦𝐴0 −𝑁𝐴0 + 𝑁𝐵0

𝑦𝐴 −

𝑁

+ 𝑁𝐵0

= 𝑒𝑥𝑝 ( 𝐴0
𝑐𝐷

𝐴𝐵

𝑐𝐷𝐴𝐵

(13.1-7)𝑧)

On the other hand, setting up the enthalpy balance over the same control volume,𝜕𝑇 𝜕 2 𝑇
𝜕𝑇−𝜅 ( +𝑑𝑧)+𝜅+ 𝑁𝐴 𝐶𝑝𝐴(𝑇 − 𝑇0 )|𝑧+𝑑𝑧 + 𝑁𝐵 𝐶𝑝𝐵 (𝑇 − 𝑇0 )|𝑧+𝑑𝑧
𝜕𝑧
𝜕𝑧 2
𝜕𝑧
− 𝑁𝐴 𝐶𝑝𝐴(𝑇 − 𝑇0 )|𝑧 − 𝑁𝐵 𝐶𝑝𝐵 (𝑇 − 𝑇0 )|𝑧 = 0

Then

𝑑2 𝑇

𝑁 𝐶𝑝 + 𝑁 𝐶𝑝

𝑑𝑇(13.1-8)− 𝑑𝑧 2 + 𝐴 𝐴 𝜅 𝐵 𝐵 𝑑𝑧 = 0
The equation to be solved for the temperature distribution is
𝑑2 𝑇
𝑑𝑧 2

𝑑𝑇

(13.1-9)− 𝐾 𝑑𝑧 = 0

𝑁𝐴 𝐶𝑝𝐴 + 𝑁𝐵 𝐶𝑝𝐵

where 𝐾 =
𝜅The boundary conditions are
𝑇 = 𝑇𝑏at z = δ
𝑇 = 𝑇0at z = 0
The solution is
1−exp(𝐶0 𝑧⁄𝛿)𝑇 = 𝑇0 + (𝑇𝑏 − 𝑇0 )(13.1-10)(13.1-11)

1−exp(𝐶0 )

where

𝐶0 =

𝑁𝐴 𝐶𝑝𝐴 + 𝑁𝐵 𝐶𝑝𝐵𝜅⁄𝛿The temperature gradient can be calculated as
𝜕𝑇− (𝐶0⁄𝛿 ) exp(𝐶0 𝑧⁄𝛿)= (𝑇𝑏 − 𝑇0 )𝜕𝑧(13.1-12)

1− exp(𝐶0 )Using the heat transfer coefficient of the gas-phase film defined by the film theory, the heat flux atthe interface is𝐶0𝜕𝑇𝐶= − (𝜅⁄𝛿 )(𝑇𝑏 − 𝑇0 ) 1− exp(𝐶𝑞𝑤 = 𝜅 𝜕𝑧 |= ℎ (𝑇𝑏 − 𝑇0 ) exp(𝐶0 )−1(13.1-13)
)
𝐶0

𝑧=0

𝜓 = exp(𝐶 )−10

00(13.1-14)

If we take the interface temperature 𝑇0 as the standard-state temperature, 𝑞𝑤 becomes equal to thetotal enthalpy flux at the interface. The function 𝜓 is called the Ackermann correction factor forthe mass transfer. If the Ackermann correction factor is plotted against the mass transfer rate factor
𝐶0 , we get Figure 13.1-2 .

154

Simultaneous Herat and Mass Transfer- II

Fig. 13.1-2. Variation of the Ackermann correction factor with the mass transfer rate factor

In the case when the mass transfer of A and B is toward the interface, as in condensation, 𝐶0 is
negative. Then 𝜓 > 1. This suggests that the apparent heat transfer coefficient ℎ 𝜓 is increased by
the mass transfer. In the case when the mass transfer of A and B is in opposite directions, as inevaporation, 𝐶0 is positive. Then 𝜓 < 1. This suggests that the apparent heat transfer coefficientℎ 𝜓 is decreased by the mass transfer.
For processes in which a phase change occurs at the interface, as in evaporation or condensation,an additional latent enthalpy effect should be taken into account. The total flux of enthalpy into thesurface of condensate film which receives mass fluxes 𝑁𝐴0 and 𝑁𝐵0 is the sum of 𝑞𝑤 and thelatent enthalpy change
𝑄 = 𝑞𝑤 − (𝑁𝐴 ∆𝐻𝐴 + 𝑁𝐵 ∆𝐻𝐵 )(13.1-15)
Note that 𝑁𝐴 and 𝑁𝐵 are negative for condensation, but that 𝑄 and 𝑞𝑤 are positive.
If only one component A condenses, 𝑁𝐵 = 0. Then1
𝜕𝑦𝐴
𝑁𝐴 = −𝑐 𝐷𝐴𝐵
1 − 𝑦𝐴 𝜕𝑧
Since 𝑁𝐴 = 𝑁𝐴0 = 𝑐𝑜𝑛𝑠𝑡
𝛿

𝑦𝐴𝑏∫ 𝑁𝐴0 𝑑𝑧 = 𝑐 𝐷𝐴𝐵 ∫0

𝑦𝐴0

That is
𝑐𝐷
1− 𝑦
1− 𝑦
𝑁𝐴0 = 𝛿𝐴𝐵 ln 1− 𝑦𝐴𝑏 = 𝑘𝑦 ln 1− 𝑦𝐴𝑏
𝐴0

−

𝑑𝑦𝐴
1 − 𝑦𝐴(13.1-16)𝐴02

Here 𝑘𝑦 is the mass transfer coefficient in the gas-phase film in kmol A/m s.
The total enthalpy flux at the interface where only one component A condenses is

𝑄 = ℎ (𝑇𝑏 − 𝑇0 )𝐶0exp(𝐶0 )−1

Here 𝐶0 = 𝑁𝐴 𝐶𝑝𝐴 ⁄ℎ .

− ∆𝐻𝐴 𝑘𝑦 𝑙𝑛
1− 𝑦

1− 𝑦𝐴𝑏1− 𝑦𝐴0(13.1-17)

Note that 𝑙𝑛 1− 𝑦𝐴𝑏 is negative for condensation.
𝐴0

13.2 Simultaneous Heat and Mass Transfer for Humidificationand Evaporative Cooling13.2-1 Theory of wet-bulb temperature
Let us consider a wet- and dry-bulb thermos meter shown in Fig.13.2-1.

Simultaneous Heat and Mass Transfer for Humidification and Evaporative Cooling (continued)

155

Fig. 13.2-1. Wet- and dry-bulb thermometers

One thermometer bulb (dry bulb) is left bare, and the other (wet bulb) is wrapped with cotton
wick that is kept wet with a pure liquid A (Usually water). A gas mixture of condensable gas A(Usually water vapor) and non-condensable gas B (Usually air) flows over the thermometer bulbs.
The dry-bulb temperature is the same as the temperature of the approaching gas mixture 𝑇∞. The
wet-bulb temperature 𝑇𝑤𝑏 , which is approximately equal to the surface temperature of the wet wick,is kept lower than 𝑇∞ owing to the steady evaporation of liquid A; heat is transferred to thegas-liquid interface of the wet wick and fresh liquid A is continuously supplied to the interface fromthe reservoir by capillary effect. At steady state the heat flux to the interface becomes equal to theenthalpy requirement for the evaporation of liquid A.
The mass flux due to evaporation at the interface is given by Eq. (13.1-16)
1− 𝑦
𝑁𝐴0 = 𝑘𝑦 ln 1− 𝑦 𝐴∞
𝐴𝑤𝑏

where 𝑘𝑦 is the mass transfer coefficient in the gas-phase film and 𝑦𝐴𝑤𝑏 is the gas concentration
at the interface. The heat flux to the interface for evaporation is given by (13.1-13):
𝐶0
𝑞𝑤 = ℎ𝐺 (𝑇∞ − 𝑇𝑤𝑏 )exp( 𝐶0 ) − 1
where 𝑇𝑤𝑏 is the temperature at the interface. Thus𝐶
1− 𝑦ℎ𝐺 (𝑇∞ − 𝑇𝑤𝑏 ) exp(𝐶0 )−1 = ∆𝐻𝑒𝑣 𝑘𝑦 ln 1− 𝑦 𝐴∞
(13.2-1)
0

𝐴𝑤𝑏

where ∆𝐻𝑒𝑣 is the molar heat of evaporation. In this case 𝑁𝐵 = 0 and 𝐶0 = 𝑁𝐴 𝐶𝑝𝐴 ⁄ℎ𝐺 is
positive. According to the Colburn’s analogy between heat and mass transfer
𝑗𝐻 = 𝑗𝐷

or

𝑘𝑦

ℎ𝐺

(13.2-2)

𝑃𝑟 2⁄3 = 𝐺 𝑆𝑐 2⁄3
𝐶𝑝 𝐺
𝑀

2

Here 𝐺𝑀 is the molar velocity having units of kmol/m s.
Substituting the analogy into the above equation(𝑇∞ − 𝑇𝑤𝑏 )𝐶0∆𝐻𝑃𝑟 2/3= 𝐶𝑝𝑒𝑣 ( 𝑆𝑐 )exp(𝐶 )−1
0

𝑚

1− 𝑦

ln 1− 𝑦 𝐴∞

𝐴𝑤𝑏

(13.2-3)

where 𝐶𝑝𝑚 = 𝑀𝑎𝑣 𝐶𝑝 is the molar specific heat of the approaching gas mixture. The ratio Pr/Sc
is known as Lewis number 𝐿𝑒. The mole fraction 𝑦𝐴𝑤𝑏 at the interface is the equilibrium
concentration, which can be determined by the vapor pressure 𝑝𝐴𝑣𝑎𝑝 of liquid a at 𝑇𝑤𝑏 and thetotal pressure 𝑃:
𝑦𝐴𝑤𝑏 = 𝑝𝐴𝑣𝑎𝑝 ⁄𝑃(13.2-4)
The approaching gas concentration 𝑦𝐴∞ can be calculated using the above equation with thetemperatures 𝑇∞ and 𝑇𝑤𝑏 measured by the dry- and wet-bulb thermometer.

156

Simultaneous Heat and Mass Transfer - II

As a simpler example, consider the same wet- and dry-bulb thermometer for the air-water
system. Since the Ackermann correction factor 𝜓 is very close to unity in usual mass transfer
situations, the heat input to the interface is
𝑞𝐺 = ℎ𝐺 (𝑇𝐺 − 𝑇𝑤𝑏 )(J⁄m2 s)(13.2-5)Introducing the humidity instead of the mole fraction of water vapor 𝑦𝐴 ,𝑀𝐴 𝑦𝐴
kg−water vapor
𝐻 = 𝑀 (1−()
(13.2-6)
𝑦 )kg−dry air
𝐵𝐴

where 𝑀𝐴 and 𝑀𝐵 are the molecular weights of water and air, respectively.
The mass transfer of water vapor from the interface is
𝑁𝐴 = 𝑘𝐻 (𝐻𝑤𝑏 − 𝐻𝐺 )(13.2-7)
where 𝐻𝑤𝑏 is the equilibrium humidity at the wet bulb temperature and 𝑘𝐻 the mass transfer
coefficient in kg-dry air/m2s. At steady state all of the heat input is consumed for the evaporation ofwater and the interface temperature is kept at 𝑇𝑤𝑏 . Then the steady-state enthalpy balance isℎ𝐺 (𝑇𝐺 − 𝑇𝑤𝑏 ) = Δ𝐻𝑒𝑣 𝑘𝐻 (𝐻𝑤𝑏 − 𝐻𝐺 )(13.2-8)
where ∆𝐻𝑒𝑣 is the heat of evaporation of water at 𝑇𝑤𝑏 (J/kg-water).
Introducing the mass humid heat capacity
J
𝐶𝐻 = 𝐶𝑝𝑎 + 𝐻 𝐶𝑝𝑤((kg−dry air)K)(13.2-9)
The Colburn’s analogy (𝑗𝐻 = 𝑗𝐷 ) can be written asℎ𝐺𝑘
𝑃𝑟 2/3 = 𝐺𝐻 𝑆𝑐 2/3
𝐶 𝐺
𝐻

ℎ𝐺

𝑘𝐻 𝐶𝐻

or
(13.2-10)= 𝐿𝑒 −2/3

Here 𝐺 is the mass velocity of the approaching gas on dry basis.
For the air-water system, the Lewis number is very close to unity. Thenℎ𝐺
≅1(air-water system)𝑘 𝐶(13.2-11)

𝐻 𝐻

This is known as the Lewis relation.
Using the Lewis relation,∆𝐻
𝑇𝐺 − 𝑇𝑤𝑏 = 𝐶 𝑒𝑣 (𝐻𝑤𝑏 − 𝐻𝐺 )𝐻or

𝐶𝐻 𝑇𝐺 + ∆𝐻𝑒𝑣 𝐻𝐺 = 𝐶𝐻 𝑇𝑤𝑏 + ∆𝐻𝑒𝑣 𝐻𝑤𝑏(13.2-12)
This is an important equation for calculating the humidity 𝐻𝐺 of the approaching air at 𝑇𝐺 fromthe wet-bulb temperature 𝑇𝑤𝑏 .
The total enthalpy of humid air is defined as𝑖𝐻 = 𝐶𝐻 (𝑇 − 𝑇0 ) + ∆𝐻𝑒𝑣 𝐻(13.2-13)
where 𝑇0 is the standard-state temperature. Then the above equation becomes
𝑖𝐻𝐺 = 𝑖𝐻𝑤𝑏
(13.2-14)
The equation indicates that the enthalpy of the bulk humid air is equal to that of air saturated withwater. The equation can be rewritten as𝐶
𝐻𝐺 = − ∆𝐻𝐻 (𝑇𝐺 − 𝑇𝑤𝑏 ) + 𝐻𝑤𝑏(13.2-15)
𝑒𝑣

If unsaturated air is brought into contact with the surface covered with liquid water, the air-waterinterface will be kept at the constant wet-bulb temperature 𝑇𝑤𝑏 while 𝑇𝐺 and 𝐻𝐺 change.

Next consider a spray chamber shown in Fig.13.2-2.
Unsaturated air entering at 𝐻 and 𝑇 is cooled and humidified by sprayed water. Water iscirculated by a pump and small amount of water is continuously supplied to make up the evaporated
water. At steady state the water temperature is kept at a definite saturation temperature 𝑇𝑎𝑠 , andthe air leaving the chamber is in equilibrium with the water. This is called “adiabatic-saturation
temperature.”
For the air-water system, the adiabatic-saturation temperature becomes essentially equal to thewet-bulb temperature.

Humidity Chart

157

Fig.13.2-2. Spray chamber for air humidification

13.2-2 Humidity chart
A humidity chart for the air-water system at 1 atm is shown in Fig.13.2-3.
The line marked 100% gives the humidity of saturated air as a function of air temperature. Any
point (𝐻, 𝑇) below the saturation line represents an undersaturated mixture of air and water vapor.
The slanting lines are called adiabatic-cooling lines, each of which is drawn with the slope of− 𝐶𝐻 ⁄∆𝐻𝑒𝑣 for a given constant adiabatic-saturation temperature. The relative humidity 𝐻𝑅 is
defined as the ratio of the partial pressure of the water vapor to the vapor pressure of water at thegas temperature.

Fig.13.2-3. Humidity Chart (Air-water system at 1 atm)

158

Simultaneous Heat and Mass Transfer – II

[EXAMPLE 13.2-1] A water droplet initially having a diameter 𝐷𝑝 = 3 mm is suspended in thehumid hot air stream (temperature 𝑇𝑏 = 98 𝐶 = 371 𝐾, humidity 𝐻𝑏 = 0.03 kg-water/kg-dry air,pressure 𝑝 = 1 atm = 1.0133 × 105 Pa, average velocity 𝑣∞ = 0.3 m/s). What are thetemperature and humidity of the droplet surface? How long does it take for the droplet to becompletely evaporated? The average mass transfer coefficient around a sphere is given by the Ranz
and Marshall equation:
𝑣𝑓 𝑘𝐻 𝐷𝑝
𝐷𝐴𝐵

= 2 + 0.6 (𝜌𝑓 𝑣∞ 𝐷𝑝
𝜇𝑓

1/2

)

𝑆𝑐 1/3The physical properties for the humid air are given by
𝜇 = 10.53 × 10−6 √𝑇
𝐷𝐴𝐵 𝑣 = 5.01 × 10−9 √𝑇 3
𝑆𝑐 = 0.60

Fig.13.2-E1. Water droplet surrounded by hot humid air stream

Solution:
The interface temperature should be the wet-bulb temperature 𝑇𝑤𝑏 and the interface humidity is
the equilibrium humidity 𝐻𝑤𝑏 corresponding to 𝑇𝑤𝑏 . From the humidity chart, 𝑇𝑤𝑏 = 42 C = 315
K, and 𝐻𝑤𝑏 = 0.054 kg-water/kg-dry air. The film temperature and humidity are
𝑇𝑓 = (𝑇𝑤𝑏 + 𝑇𝑏 )⁄2 = 70 C = 343 K, 𝐻𝑓 = (𝐻𝑤𝑏 + 𝐻𝑏 )⁄2 = 0.041 kg-water/kg-dry air. At thisfilm condition, the specific volume 𝑣𝑓 = 1.03 m3/kg-dry air. The density is𝜌𝑓 = (1 kg-dry air/kg-dry air + 0.041 kg-water/kg-dry air )/(1.03 m3 humid air/kg-dryair) = 1.01 kg/m3.
The viscosity is
𝜇𝑓 = 10.53 × 10−6 √343 = 1.95 × 10−5 kg/m s
𝐷𝐴𝐵 = 5.01 × 10−9 (343)3/2 = 0.318 × 10−4 m2/s
From the Ranz and Marshall equation
1𝑣∞ 𝐷𝑝 𝜌𝑓 2 1
𝐷𝐴𝐵
𝑘𝐻 =(2 + 0.6 () 𝑆𝑐 3 )
𝑣𝑓 𝐷𝑝
𝜇𝑓
1

=

1
0.318 × 10−4
0.3×𝐷𝑝×1.01 2
1(2 + 0.6 () (0.6)3 ) =(0.6175 × 10−4 + 0.003245 √𝐷𝑝 )−5
1.03 𝐷𝑝
1.95×10𝐷𝑝

The mass transfer of water vapor can be expressed for the droplet:𝑑𝜋( 𝐷𝑝3 𝜌𝐿 ) = 𝑘𝐻 (𝜋 𝐷𝑝2 )(𝐻𝑤𝑏 − 𝐻𝑏 )𝑑𝑡 6
where 𝜌𝐿 is the water density. From the equation,

(1)
(2)

Evaporative Cooling --- Water Cooling Tower
𝑑 𝐷𝑝
𝑑𝑡

=

2(𝐻𝑤𝑏 − 𝐻𝑏 )
𝜌𝐿

𝑘𝐻 =

2(0.054−0.03)
1000

159

𝑘𝐻 = 4.8 × 10−5 𝑘𝐻

Substituting Eq. (1) into Eq. (3)𝑑 𝐷𝑝
1= 𝐷𝑝 (2.964 × 10−9 + 1.5576 × 10−7 √𝐷𝑝 )
𝑑𝑡

(3)
(4)

This equation can be numerically integrated to get the required time. The result is given in thefollowing table.
Table 13.2-E1 Numerical calculation of time required for evaporation of a water droplet
t (min)T (s)Dp (m)Δt (s)ΔDp (m)
0
0
60
0.003
1
60
60
0.002770.000230
2
120
60
0.002530.0002423
180
60
0.002270.0002564240
60
0.00200
0.0002745300
60
0.001700.0002986360
600.001370.000331742060
0.000988
0.000382848060
0.000511
0.000477
8.66
519.4
39.4
0
0.000511

The time required for the evaporation of the droplet is 8.66 min.

13.3 Evaporative Cooling-----

Water Cooling Tower-----

Water can be cooled by exposing its surface to air. Packed towers are widely used in order to getcool water to be recirculated for heat exchangers and other process equipment.
Fig. 13.3-1 shows a water cooling tower. Water to be cooled is brought into contact with air
whose adiabatic-saturation temperature is lower than the water temperature. The air entering at theside of the tower flows up across the water stream as it falls through the packing.

Fig. 13.3-1. Water cooling tower

160

Simultaneous Heat and Mass Transfer - II

The material balances of water are
𝐿2 − 𝐿1 = 𝐺 (𝐻2 − 𝐻1 )(13.3-1)
𝐿2 − 𝐿 = 𝐺 (𝐻2 − 𝐻)
(13.3-2)
where 𝐺 is the mass velocity of the air on dry basis.
Similarly the enthalpy balances are
𝐺 (𝑖𝐺2 − 𝑖𝐺1 ) = 𝐿2 𝐶𝑝𝐿 𝑇𝐿2 − 𝐿1 𝐶𝑝𝐿 𝑇𝐿1(13.3-3)
𝐺 (𝑖𝐺2 − 𝑖𝐺 ) = 𝐿2 𝐶𝑝𝐿 𝑇𝐿2 − 𝐿𝐶𝑝𝐿 𝑇𝐿(13.3-4)
The rate of water evaporation (𝐿2 − 𝐿1 ) is small compared with the liquid flow rate 𝐿2 :𝐿2 ≅ 𝐿1 ≅ 𝐿(13.3-5)
Then the above enthalpy balance equation can be approximated as
𝐺 (𝑖𝐺2 − 𝑖𝐺1 ) ≅ 𝐿2 𝐶𝑝𝐿 (𝑇𝐿2 − 𝑇𝐿1 )(13.3-6)and
𝐺 (𝑖𝐺2 − 𝑖𝐺 ) ≅ 𝐿2 𝐶𝑝𝐿 (𝑇𝐿2 − 𝑇𝐿 )(13.3-7)The last equation can be rewritten as
𝐿 𝐶𝑝
𝑖𝐺 = 2 𝐿 (𝑇𝐿 − 𝑇𝐿2 ) + 𝑖𝐺2(13.3-8)
𝐺This is called the operating line equation.
The mass balance taken around a differential section of height 𝑑𝑧 is
𝑑𝐿 = 𝐺 𝑑𝐻 = 𝑘𝐻 𝑎 (𝐻 ∗ − 𝐻 ) 𝑑𝑧
(13.3-9)
The transfer rate of sensible heat in the gas-phase film is
𝑑𝑞𝐺 = ℎ𝐺 𝑎 (𝑇 ∗ − 𝑇𝐺 ) 𝑑𝑧 = 𝐺 𝐶𝐻 𝑑𝑇𝐺(13.3-10)

Fig. 13.3-2. Shell balance of mass and enthalpy

The rate of heat transfer in the liquid-phase film is
𝑑𝑞𝐿 = ℎ𝐿 𝑎 (𝑇𝐿 − 𝑇 ∗ )𝑑𝑧 = 𝐿 𝐶𝑝𝐿 𝑑𝑇𝐿
Introducing the total enthalpy of humid air
𝑖𝐺 = 𝐶𝐻 (𝑇𝐺 − 𝑇0) + ∆𝐻𝑒𝑣 𝐻(13.3-11)(13.3-12)

The enthalpy transfer due to the evaporated water is
𝐺 𝑑𝐻 [∆𝐻𝑒𝑣 + 𝐶𝑝𝑊 (𝑇 ∗ − 𝑇𝐺 )](13.3-13)
Here 𝑇 ∗ is the interface temperature, which is kept at the wet-bulb temperature. The 𝐶𝑝𝑊 is theEvaporative Cooling (continued)

161

heat capacity of water vapor. The enthalpy balance around the same differential section of height𝑑𝑧 is
𝐺 𝑑𝑖𝐺 = ℎ𝐺 𝑎(𝑇 ∗ − 𝑇𝐺 )𝑑𝑧 + 𝑘𝐻 𝑎 (𝐻 ∗ − 𝐻 )𝑑𝑧 [∆𝐻𝑒𝑣 + 𝐶𝑝𝑊 (𝑇 ∗ − 𝑇𝐺 )]ℎ= 𝑘𝐻 𝑎 𝑑𝑧 [(𝑖𝐺∗ − 𝑖𝐺 ) + ( 𝐺 𝐶𝐻 − 𝐶𝑝𝑎 )(𝑇 ∗ − 𝑇𝐺 ))](13.3-14)
𝑘𝐻 𝐶𝐻

For the air-water system
ℎ𝐺
ℎ𝐺
𝐶
−
𝐶𝑝
≅
𝐶𝑝
(
− 1) ≅ 0
𝐻
𝑎
𝑎
𝑘 𝐶
𝑘 𝐶0
𝐻 𝐻

𝐻

𝐻

Then we get
𝐺 𝑑𝑖𝐺 = 𝑘𝐻 𝑎(𝑖𝐺∗ − 𝑖𝐺 ) 𝑑𝑧
The required height of packing is𝑍𝑖𝐺𝑑𝑖
𝑍𝑇 = ∫0 𝑇 𝑑𝑧 = 𝑘 𝑎 ∫𝑖 𝐺2 𝑖 ∗ −𝐺𝑖 = 𝐻𝑇𝐺 𝑁𝐺
where
𝑖
𝑑𝑖
𝑁𝐺 = ∫𝑖 𝐺2 𝑖 ∗ −𝐺𝑖
𝐺1

𝐺1

𝐻

and

𝐺

𝐺

𝐺

𝐺 𝑙.𝑚.

𝐺𝐺

𝑑𝑇𝐺

𝑖 `∗ − 𝑖𝑘 𝐶

𝐻𝑖 `∗ − 𝑖= ( 𝐻 𝐻 ) 𝑇𝐺∗ − 𝑇𝐺 ≅ 𝑇𝐺∗ − 𝑇𝐺
ℎ𝐺

𝐺

𝐺

Similarly
𝐺 𝑑𝑖𝐺 = 𝑘𝐻 𝑎(𝑖𝐺∗ − 𝑖𝐺 ) 𝑑𝑧
𝐿 𝐶𝐿 𝑑𝑇𝐿 = ℎ𝐿 𝑎 (𝑇𝐿 − 𝑇 ∗ ) 𝑑𝑧
Dividing the former equation by the latter equation
𝑑𝑖𝐺

𝐿 𝐶𝐿 𝑘𝐻

𝐿

𝐿

=
𝑑𝑇

(13.3-17)𝐺

Fig. 13.3-3 shows the relation between the equilibrium line and operating line.
𝐺 𝑑𝑖𝐺 = 𝑘𝐻 𝑎(𝑖𝐺∗ − 𝑖𝐺 ) 𝑑𝑧
𝐺 𝐶𝐻 𝑑𝑇𝐺 = ℎ𝐺 𝑎 (𝑇 ∗ − 𝑇𝐺 )𝑑𝑧
Dividing the former equation by the latter equation
𝑑𝑖𝐺(13.3-16)

𝐻𝑇𝐺 = 𝑘 𝑎

The above equation can be approximated as𝐺𝑖 −𝑖
𝑍𝑇 = 𝑘 𝑎 (𝑖 `∗𝐺2− 𝑖 )𝐺1
𝐻(13.3-15)

𝑖 `∗ − 𝑖( ℎ ) 𝑇𝐺 − 𝑇𝐺∗
𝐺
𝐿(13.3-18)(13.3-19)(13-.3-20)(13.3-21)(13.3-22)

This is called the tie line equation which gives the relation between (𝑖𝐺 , 𝑇𝐿 ) and (𝑖𝐺∗ , 𝑇 ∗ ).

Fig.13.3-3. Relation of operating line with equilibrium line

162

Simultaneous Heat and Mass Transfer - II

Nomenclature
𝑎
𝐶𝐻
𝐶𝑝
𝐷𝐴𝐵
𝐷𝑝
𝐺
𝐻
𝐻𝑇𝐺
ℎ
ℎ𝐺
𝑖𝐺 , 𝑖𝐻
𝑘𝐻
𝑘𝑦
𝐿𝑒
𝐿
𝑁𝐺
𝑁𝐴 , 𝑁𝐵
𝑝𝐴𝑣𝑎𝑝
𝑃𝑟
𝑞𝐺
𝑞𝑤
𝑆𝑐
𝑇𝐺 , 𝑇𝐿
𝑇∞ , 𝑇𝑤
𝑡
𝑣
𝑦
𝑧

∆𝐻𝑒𝑣
𝛿
𝜅

𝜓

effective interfacial area, [m2/m3 packed volume]
mass humid heat capacity, [J/(kg-dry air)K]
heat capacity, [J/kg K]
diffusivity, [m2/s]
diameter of droplet, [m]flow rate of dry air, [kg/s]
humidity, [kg-water vapor/kg-dry air]
HTU (Height of Transfer Unit) of packed bed, [m]heat transfer coefficient, [W/m2K]
gas-film heat transfer coefficient, [W/m2K]
total enthalpy of humid air, [J/kg-dry air]mass transfer coefficient, [kg-dry air/m2s]
gas-film mass transfer coefficient, [kg/m2s/mass fraction]
Lewis number, [ - ]liquid (water) flow rate, [kg/s]
NTU (Number of Transfer Unit) of packed bed, [ - ]mass flux of component A, B, [kg/m2s]
vapor pressure of water
Prandtl number. [ - ]enthalpy flux, [J/m2s]
heat flux at wall, [J/m2s]
Schmidt number, [ - ]gas and liquid temperature, [K]temperature at gas bulk and wall, [K]time, [s]velocity
mass fraction, [ - ]distance from condensate film surface, [m]heat of evaporation of water, [J/kg]

thickness of gas-phase film, [m]thermal conductivity of gas phase, [J/m s K]
Ackermann correction factor, [ - ]Subcripts
𝑓
H
L
W
wb
0
1, 2

film
humid
liquidwater
wet-bulb
surface of condensate film
inlet, outlet

CHAPTER 14
IONIC MASS TRANSPORT
14.1 Electrolytic Cell
It should be kept in mind that the transport phenomena occurring during electrolysis are socomplicated that we had better study a very simple case of ionic mass transfer in an aqueous
electrolytic cell.
Let us consider an electrolytic cell in which two electrodes are separated by an ionicallyconducting liquid such as an aqueous solution of electrolyte. If an electric current generated by an
external emf (electromotive force such as potential difference or voltage) flows through theelectrolytic cell, an electrode reaction takes place by the following three steps:(1) Transfer of ions from the bulk of the solution to the surface of either of the two electrodes(2) Electrochemical reactions at both electrodes(3) Formation of reaction products and their deposition on the surface of the electrodes or theirremoval from that surface
This phenomenon can be considered as a special case of a heterogeneous chemical reaction.
Generally the diffusivity 𝐷𝐴𝐵 m2/s for a single electrolyte at infinite dilution is given by the
Nernst-Haskell equation:
𝐷𝐴𝐵 =𝑅𝑇 (1⁄𝑛+ + 1⁄𝑛− )
𝐹𝑎2 (1⁄𝜆0+ + 1⁄𝜆0− )where 𝑇 is temperature (K), 𝑅 is gas constant (J/K kmol), 𝜆0+ , 𝜆0− are the limiting ionic
conductances of cation and anion, respectively (A/m 2)(V/m)(kg-equiv/m3), 𝑛+ , 𝑛− are thevalences of cation and anion, respectively ( - ), and 𝐹𝑎 is Faraday constant (A s/kg-equiv.).
In a system of mixed electrolytes, the unidirectional diffusion of each ion species results from a
combination of electrical potential and concentration gradients.
As shown in Fig. 14.1-1, an anode of metal M undergoes the dissolution by the anode reaction to
generate M+ ions. The M+ ions arriving at the cathode receive electrons due to the cathode reaction(discharge of metal ions) and the deposition of the reduced M occurs on the cathode surface.
The molar flux of M+ ions transferring in the electrolytic solution at rest is given by
𝑀 𝑥 𝜖
𝑁𝐴 = − 𝑐 𝐷𝐴𝐵 [∇𝑥𝐴 + 𝑅𝐴𝑇𝐴 𝑚𝐴 ∇Φ](14.1-1)
𝐴

where 𝑥𝐴 is mole fraction of M+ ions, 𝑀𝐴 the molecular weight, 𝜖𝐴 the ionic charge, and 𝑚𝐴
the ionic mass.

164

Ionic Mass Transport

Fig. 14.1-1. Electrolytic cell (aqueous solution of electrolytic ions M + and X-)

The first term implies the ordinary diffusion due to the concentration gradient ∇𝑥𝐴 of ions and thesecond term the electric diffusion due to the electrostatic potential ∇Φ. The ions are moved by the
Coulomb force generated due to the gradient of the electrostatic potential.
The rate of the electrochemical reaction, i.e., the electric current flowing through the cell increases
rapidly with the applied emf.

14.2 Ionic Mass Transport in an Electrochemical Reaction
SystemLet us consider an aqueous solution of salts in an electrolytic cell. In general, the individual
species diffuse irreversibly when they are subjected to several unequal external forces. For
simplicity, we neglect the effects of pressure, gravity, and thermal diffusions. Then we shallconsider the ordinary diffusion due to the concentration gradients and the forced diffusion due to theelectrostatic force. We shall furthermore confine our attention to a dilute solution of copper sulfate
with sulfuric acid serving as the unreactive supporting electrolyte.

Fig. 14.2-1. Electrolytic cell filled with aqueous solution of electrolytes: 𝐂𝐮++ , 𝐇 +, 𝐒𝐎𝟒−−

Ionic Mass Transport in an Electrochemical Reaction System

165

Let us consider a simple cell (Fig. 14.2-1) filled with a ternary electrolytic solution of three ionicspecies: Cu++, H+ , SO4 −−. A voltage is imposed upon the cell that is sufficient to cause thecopper ion concentration at the cathode to drop essentially to zero. The only electrode reactions
are dissolution of the anode (copper plate) and deposition of Cu++ on the cathode. The molar fluxof the Cu++, that is, the electrode reaction rate is proportional to the current density in the cell.
When the voltage imposed between the anode and cathode is increased, the reaction rate isincreased. The distribution of electrostatic potential shown in Fig.14.2-1 is formed in the condition
when the electrical conductivity is not very high owing to the dilute solution. However theconcentration of the unreactive supporting electrolyte becomes so high that the gradient ofelectrostatic potential reduces to almost zero owing to very high electrical conductivity of thesolution. In this condition, the copper ions diffuse only due to the concentration gradient of thereactive ions Cu++.

14.3 Mass Transfer Measurements by an Electrochemical
Technique1,2,3,4)
If the concentration of sulfuric acid is sufficiently high, the electric conductivity becomes sohigh that the gradient of the electrostatic potential becomes negligibly small over the entire electricfield in the cell. In this state the copper ions diffuse only due to the ordinary diffusion. Most of thevoltage drop between the anode and cathode occurs in the electric double layers formed on thesurfaces of these electrodes. The thickness of the electric double layers is negligibly small ascompared to the diffusion layer. When the electrode voltage 𝐸 is increased, the current density 𝐼
shows the variation shown in Fig.14.3-1. When the electrode potential 𝐸 exceeds the characteristic
discharge potential 𝑉0 , the deposition of Cu starts at the cathode. When 𝐸 becomes sufficiently
large (larger than 𝐸𝑑 ), the current density no longer increases owing to the insufficient supply ofCu++ by mass transfer. This is called the diffusion-controlling condition or the limiting current
condition. That is, the electrode reaction is controlled by the mass transfer of Cu++ ions.
In this condition, the concentration of Cu++ becomes essentially zero on the cathode surface.

Fig. 14.3-1. Variation of current density with electrode voltage

If the applied emf becomes further greater than the next discharge potential 𝑉1, the next electrodereaction will occur, as shown by the dotted line in Fig.14.3-1. For example, in this case, hydrogen
bubbles due to the discharge of hydrogen ions may be released from the cathode surface.
It is possible to observe time-dependent, local coefficients of convective mass transfer if we
utilize the electrochemical technique above mentioned. 1,4)
For example, the working fluid is an aqueous solution having equimolar concentrations (0.01 M)

166

Ionic Mass Transportof potassium ferri- and ferro-cyanide and a 1 M concentration of potassium hydroxide KOH as an
unreactive supporting electrolyte.2,3)

Fig. 14.3-2.

Position of main and isolated test cathodes.

Dimensions given are in mm.

Its Schmidt number is kept at 𝑆𝑐 = 1,800 by controlling the fluid temperature.
As shown in Fig. 14.3-2, a nickel plate is stuck on the surface of the test section serving as themain cathode. Another large plate is placed as the anode, usually somewhere on the downstreamside.
The following electrode reaction takes place under the diffusion-controlling condition:
4−
3−
−
−
[cathode] Fe(CN)3−[anode] Fe(CN)4−
6 + e → Fe(CN)6
6 → Fe(CN)6 + e
In this case, an electrode reaction does not involve deposition of material, but merely a change inthe valence of an ion.
A small circular cathode made from nickel wire is embedded into a circular drilled hole drilled on
the wall of the main cathode. The surface area of the main cathode is much smaller than that of theanode. Therefore the current density 𝐼 A/m2 at the anode becomes so small that the convective
ionic mass transfer on the cathode only can be observed. In addition, if the small circular cathode(test cathode) is held at the same potential as the main cathode, local time-dependent mass transfer
can be observed at the position of the test cathode.
The ionic mass flux 𝑁𝐴 is given by
𝑁𝐴 = 𝐼𝑑 ⁄𝐹𝑎 = 𝑘(𝐶𝐴 − 𝐶𝐴𝑤 )(14.3-1)
3−
Under the diffusion-controlling condition, the concentration of Fe(CN)6 on the cathode surface
𝐶𝐴𝑤 becomes zero. Then the local mass transfer coefficient 𝑘 m/s is calculated by the equation:𝑘 = 𝐼𝑑 ⁄𝐹𝑎 𝐶𝐴(14.3-2)2
where 𝐼𝑑 (A/m ) is the limiting current density on the test cathode, 𝐹𝑎 the Faraday constant (=96,500 C/kg-equiv.) and 𝐶𝐴 the bulk concentration of Fe(CN)3−
in kg-equiv./m3. Since the6concentration becomes zero on the cathode surfaces in the limiting current condition, the term 𝐶𝐴
in Eq. (14.3-1) implies the concentration difference as the driving force.

Fig. 14.3-3 Simple circuit for observing time-dependent ionic mass transfer

Measurements of Velocity Gradient on a Wall167

The fluctuating electric current which the Fe(CN)3−
6 ions discharge on the test cathode (surface
area 𝑆) can be observed as the time-dependent local mass flux of Fe(CN)3−
6 ions. A simple circuit
is shown in Fig. 14.3-3. The time-dependent current density 𝑖𝑑 = (𝑉𝑠 ⁄𝑅𝑠 )/𝑆 at the point cathodecan be observed from the voltage drop 𝑉𝑆 at the standard resistance 𝑅𝑆 .
1.
2.
3.
4.

Kataoka, K., H. Doi and T. Komai: Int. J. Heat Mass Transfer, vol.20, pp.57-63 (1977).
Kataoka, K., Y. Kamiyama, S. Hashimoto and T. Komai: J. Fluid Mech., vol.119, pp.91-105 (1982).
Hanratty, T. J.: Phys. Fluids Suppl., vol.10, S126 (1967).
Mizushina, T.: Advances in Heat Transfer (1971), Vol.7, p.87. Academic.

14.4 Measurements of Velocity Gradient on a Wall2,3,4)
In addition, this electrochemical method can also measure local, time-dependent values of thevelocity gradient on a solid surface for liquid flow observation. As shown in Fig.14.4-1, rectangulartest cathodes are fabricated by inserting L mm thick nickel sheets into 𝐿𝐵 mm long slits. The
velocity-gradient measurement is carried out by measuring the limiting current between the test
cathode and the anode.

Fig.14.4-1 Isolated cathode for measuring velocity gradient on the wall

Owing to the high Schmidt number (𝑆𝑐 = 1,800 for an electrolytic solution of Fe(CN)3−
6 andFe(CN)4−),
theconcentration
boundarylayerdevelopingfromtheleading
edge
of
a
test
cathodeis6much smaller in the thickness than the velocity boundary layer. Therefore the velocity can beassumed to be linear near the wall.
By measuring the current density 𝐼𝑑 on each test cathode under the limiting current condition,
local time-dependent velocity gradient on the wall s can be determined as𝑠 = 1.90 (𝐼𝑑 ⁄𝐹𝑎 𝐶𝐴 )3 (𝐿⁄𝐷𝐴𝐵 2 )(14.4-1)
3−
where 𝐷𝐴𝐵 is the diffusivity of Fe(CN)6 ion.
The above equation can be obtained by the boundary layer analysis, which will be described in
Chapter 18.
This electrochemical method can observe fluctuating velocity-gradients within the viscous
sublayer of turbulent flows. It should be kept in mind that there usually appear definite velocity
fluctuations even in the viscous sublayer due to the influence of the turbulence generated in thebuffer zone. For an example,1) by using this velocity-gradient measurement, local variation of
the near-wall velocity gradient is observed in the impingement region struck by a free jet of an
4−electrolytic solution of Fe(CN)3−
6 and Fe(CN)6 . If you refer to the journal shown below, youwill find the lateral distribution of velocity-gradient on the wall of the impingement region and its
turbulence intensity.
1.

Kataoka, K., Y. Kamiyama, S. Hashimoto and T. Komai: J. Fluid Mech., vol.119, pp.91-105 (1982).

168

Ionic Mass Transport[PROBLEM 14-1] An aqueous solution of electrolytes consisting of 0.02 kg-equiv./m3 Fe(CN)3−
6
3
and Fe(CN)4−withthesupporting
electrolyteof1
kmol/mKOH
isflowing
at
a
flow
rate
of
0.001
6
m3/s in a rectangular channel shown in Fig.14-P.1. The density, viscosity, and diffusivity of thesolution are 1,010 kg/m3, 0.0015 Pa s, and 8.3 × 10−10 m2/s, respectively. Owing to the
supporting electrolyte the ionic diffusion of Fe(CN)3−
due to the electrical potential can be
6neglected. The bulk concentration of the reactant ion Fe(CN)3−
can be assumed to be 0.02
6kg-equiv./m3. The active surface area of each test cathode is 0.5 mm x 3 mm = 1.5 mm 2.
(a) Examine whether this flow is turbulent or not.
(b) When the main cathode is made active at the same as the EMF (V) of the test cathodes, the twotest cathodes indicate the (time-averaged) limiting currents of
8 × 10−2 and 5.3 × 10−2 𝜇𝐴 , respectively.
Calculate the local mass transfer coefficient 𝑘 (m⁄s) at the position of each test cathode.
(c) When the main cathode is made inactive, those two test cathodes indicate the (time-averaged)
limiting currents of 5.2 and 4.7 𝜇𝐴, respectively.
Calculate the local velocity-gradient 𝑠 (1⁄s) at the position of each test cathode on the inside wall of thechannel.

Fig.14-P1. Measurement of mass transfer and velocity-gradient at wall by an electrochemical
methodNomenclature
𝑐
𝐶𝐴
𝐷𝐴𝐵
𝐸
𝐹𝑎
𝐼
𝐼𝑑
𝑘
𝐿
𝑀𝐴
𝑁𝐴
𝑅
𝑆
𝑠
𝑇
𝑥𝐴
𝜖
𝛷

total molar density, [kmol/m3]
concentration of reactant ion, [kmol/m3]
diffusivity of ion A, [m2/s]
electrode potential, [V]Faraday constant = 96,500 A s/kg-equiv.
current density, [A/m2]
limiting current density, [A/m2]
ionic mass transfer coefficient, [kmol/m2s]
length of test cathode in stream direction, [m]molecular weight, [kg/kmol]
molar flux of reactant ion A, [kmol ion/m2s]
gas constant, [J/kmol K]
electrode surface area, [m2]velocity gradient at electrode surface, [1/s]temperature, [K]molar concentration (fraction) of reactant ion, [kmol/kmol]ionic charge
electrostatic potential, [EMF]

Subcripts
Dw

limiting current
electrode surface

CHAPTER 15
HEAT TRANSFER WITH PHASE
TRANSFORMATION
15.1 Condensation
15.1-1 Heat transfer for condensation
Let us study the condensation of pure vapor.
When a pure vapor saturated at a given pressure comes in contact with a cooled surface, the change
from vapor to liquid occurs isothermally at the saturation or equilibrium temperature.
The pressure in the bulk of the vapor phase is very slightly greater than the saturation pressure of
the condensate surface. As shown in Fig. 15.1-1, the mass transfer toward the vapor-liquid interface
occurs due to the self-diffusion. The latent heat released must flow through the condensate to thecooled surface. The main resistance to heat transfer lies in the condensate film and the vapor phase
is assumed at the uniform equilibrium temperature.

Fig.15.1-1. Pure vapor condensation

If the condensate does not wet the cooled surface, individual droplets grow by coalescence and run
down the surface under the influence of gravity. This is called “dropwise condensation.” If thecondensate wets the cooled surface, a continuous liquid film is formed over the cooled surface. Thisis called “filmwise condensation.” Dropwise condensation gives much higher rate of condensation
than filmwise condensation because bare cooled surface of the condenser is directly exposed to thevapor stream. Steam is the only one pure vapor known to condense in a dropwise manner in theparticular condition of the cooled surface. The dropwise condensation is very difficult to achieve or
maintain in commercial equipment; it is customary in condenser design to assume filmwise
condensation.

170

Heat Transfer with Phase Transformation

15.1-2 Film condensation of pure vapor on a vertical wall
Let us consider the problem of filmwise condensation of a pure saturated vapor on a vertical wallshown in Fig.15.1-2.

Fig. 15.1-2. Laminar falling film of condensate in filmwise condensation

To illustrate the classical Nusselt approach the following assumptions are made:(1) The pure vapor is at the saturation temperature 𝑇𝑠𝑡(2) The heat delivered by the condensing vapor is latent heat only and the latent heat released
at the interface is transported solely by the heat conduction through the condensate film.
(3) The condensate film drains in non-rippling laminar flow.
(4) The condensate film is so thin that the temperature distribution in it is linear.
(5) The cooled surface of the solid is at a constant temperature 𝑇𝑤 .
(6) The shear at the liquid-vapor interface is negligible.
(7) The curvature of the condensate film is negligible, so that the heat flux is very nearly normal to
the wall.
The film thickness increases from top to bottom cumulatively, so the condensing coefficient(heat transfer coefficient) for a vapor condensing on a vertical surface decreases from top to bottom.
The equation of motion to be applied is
𝜕𝑣𝑧
𝜕𝑣𝑧
𝜕𝑣𝑧
𝜕𝑝
𝜕 2 𝑣𝑧 𝜕 2 𝑣𝑧 𝜕 2 𝑣𝑧
𝜕𝑣𝑧
+ 𝑣𝑥
+ 𝑣𝑦
+ 𝑣𝑧
)= −
+ 𝜇𝐿 ( 2 ++) + ( 𝜌𝐿 − 𝜌𝑉 ) 𝑔𝑧
𝜌𝐿 (
𝜕𝑡
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑧
𝜕𝑥
𝜕𝑦 2
𝜕𝑧 2

(1) The z- component velocity 𝑣𝑧 does not change in the transverse direction 𝑥,
𝜕𝑝(2) The static pressure does not change greatly in the flow direction; 𝜕𝑧 ≅ 0, and𝜕𝑣(3) The velocity distribution does not change in the streamwise direction; 𝜕𝑧𝑧 ≅ 0,
Under steady-state conditions the equation of motion on the laminar falling film of the condensate
is, neglecting inertia (convective) terms
𝜕2 𝑣

𝜇𝐿 𝜕𝑦2𝑧 = ( 𝜌𝐿 − 𝜌𝑉 ) 𝑔𝑧

(15.1-1)

At high operating pressures, buoyancy forces occur in the liquid layer and the right side term has
the density difference ( 𝜌𝐿 − 𝜌𝑉 ) in place of 𝜌𝐿 . The equation of motion reduces to
𝑑2 𝑣𝑧
𝑑𝑦 2

𝜌 − 𝜌𝑉= − ( 𝐿𝜇𝐿)𝑔

Assuming the film thickness 𝛿 at 𝑧, the boundary conditions are(15.1-2)

Condensation Heat Transfer Coefficient171

at 𝑦 = 0
at 𝑦 = 𝛿(15.1-3)𝑣𝑧 = 0𝑑 𝑣𝑧=0𝑑𝑦

Integration gives the velocity distribution in the condensate film:𝜌 − 𝜌𝑉𝑣𝑧 = ( 𝐿𝜇𝐿1(15.1-4)) 𝑔 (𝛿𝑦 − 2 𝑦 2 )Using the velocity profile, the mass flow rate per unit width perpendicular to yz plane is𝛿𝛤 = ∫0 𝜌𝐿 𝑣𝑧 𝑑𝑦 =𝜌𝐿 ( 𝜌𝐿 − 𝜌𝑉 )𝑔 𝛿 3(15.1-5)

3 𝜇𝐿

Differentiating with respect to 𝛿
𝑑𝛤

=
𝑑𝛿

𝜌𝐿 ( 𝜌𝐿 − 𝜌𝑉 )𝑔 𝛿 2(15.1-6)𝜇𝐿

The following boundary conditions for applying the equation of energy to the condensate film areused:
𝑇 = 𝑇𝑤
at 𝑦 = 0
𝑇 = 𝑇𝑠𝑡at 𝑦 = 𝛿(15.1-7)
As a result, the following linear temperature distribution can be obtained:𝑦𝑇 − 𝑇𝑤 = (𝑇𝑠𝑡 − 𝑇𝑤 ) 𝛿(15.1-8)
To consider the temperature drop within the condensate film, the average enthalpy change(modified latent heat) Δ𝐻′𝑓𝑔 of the vapor in condensing to liquid and subcooling to the averageliquid temperature of the condensate film can be calculated as𝛿∫0 𝜌𝐿 𝑣𝑧 𝐶𝑝𝐿 (𝑇𝑠𝑡 − 𝑇)𝑑𝑦
Δ𝐻′𝑓𝑔 = Δ𝐻𝑓𝑔 + 𝐶𝑝𝐿 (𝑇𝑠𝑡 − 𝑇𝑚 ) = Δ𝐻𝑓𝑔 +𝛿∫0 𝜌𝐿 𝑣𝑧 𝑑𝑦
1 𝛿= Δ𝐻𝑓𝑔 + ∫ 𝜌𝐿 𝑣𝑧 𝐶𝑝𝐿 (𝑇𝑠𝑡 − 𝑇)𝑑𝑦
𝛤 0
3
= Δ𝐻𝑓𝑔 + 8 𝐶𝑝𝐿 (𝑇𝑠𝑡 − 𝑇𝑤 )(15.1-9)
where Δ𝐻𝑓𝑔 is the latent heat at the saturation temperature.
15.1-3 Condensation heat transfer coefficient
In a segment of the falling film the vapor condenses at a rate of d𝛤 and the heat liberated∆𝐻′𝑓𝑔 d𝛤 must flow through the film 𝛿 from 𝑇𝑠𝑡 to 𝑇𝑤 by conduction.
𝑇𝑠𝑡 − 𝑇𝑤(𝑑𝑧 × 1) = Δ𝐻′𝑓𝑔 d𝛤 × 1
𝑞𝑦 (𝑑𝑧 × 1) = 𝜅𝐿𝛿
Substituting Eq.(15.1-6) into the above equation𝜅 𝜇 (𝑇 − 𝑇𝑤 )
𝛿 3 𝑑𝛿 = 𝜌 ( 𝜌𝐿 −𝐿 𝜌𝑠𝑡)𝑔 Δ𝐻′
𝑑𝑧(15.1-10)
𝐿

𝐿

𝑉

𝑓𝑔

Integration from 𝛿 = 0 (at 𝑧 = 0) to 𝛿 = 𝛿 (at 𝑧 = 𝑧) gives
1/4

4𝜅 𝜇 (𝑇 − 𝑇 )𝑤
𝛿 = [𝜌 ( 𝜌𝐿 −𝐿 𝜌 𝑠𝑡)𝑔 Δ𝐻′
𝐿

𝐿

𝑉

𝑓𝑔

(15.1-11)𝑧]

The local condensing coefficient can be defined as ℎ𝑐 = 𝜅𝐿 ⁄𝛿 . Thenℎ𝑐 = [𝜌𝐿 ( 𝜌𝐿 − 𝜌𝑉 )𝑔 Δ𝐻′𝑓𝑔 𝜅𝐿 3
4𝜇𝐿 (𝑇𝑠𝑡 − 𝑇𝑤 )𝑧

1/4(15.1-12)]

The average condensing coefficient ̅̅̅ℎ𝑐 for a vapor condensing on a vertical plate of height 𝐿 is
obtained integrating the local value ℎ𝑐 over the height 𝐿:
3 1/4

1 𝐿
2 √2 𝜌𝐿 ( 𝜌𝐿 − 𝜌𝑉 )𝑔 Δ𝐻′𝑓𝑔 𝜅𝐿̅̅̅]ℎ𝑐 = 𝐿 ∫0 ℎ𝑐 𝑑𝑧 = 3 [𝜇 (𝑇 − 𝑇 )𝐿
𝐿

𝑠𝑡

𝑤

All the physical properties are evaluated at the arithmetic mean temperature
𝑇𝑓 = (𝑇𝑠𝑡 + 𝑇𝑤 )⁄2(15.1-13)

172

Heat Transfer with Phase Transformation

This is the analytical result Nusselt 1) first achieved.
Although the foregoing analysis was made for a vertical flat plate, the development is also valid
for the inside and outside surfaces of vertical tubes if the tubes are large in diameter, compared withthe film thickness.
1.

Nusselt, W.; Z. Ver. Dtsch. Ing., 60, 541-546, 569-575 (1916)

15.1-4 Engineering design of an overhead condenser
By using a vertical 1-1 condenser shown below, 3,000 kg/h of a saturated vapor mixture of 98
mol% of n-pentane which comes from the top of a distillation column at 55℃ (= 327 K) and 1.7
atm (= 0.172 MPa) is to be condensed completely at 51.7 ℃ (= 324.7 K).

Fig. 15.1-3. Vertical 1-1 condenser

As the cooling medium, 70,000 kg/h of water at 32℃ (= 305 K) will be used. Only for simplicity
of calculation, shell baffles are omitted in the equipment. There will exist a short zone of subcoolingat the bottom, but we neglect the contribution of heat transfer from this zone to heat balance.
Determine the number of heat transfer tubes if 3 m long steel tubes of 25 mm OD and 22 mm ID areused.
From enthalpy chart, the enthalpy difference between n-pentane vapor at 55 ℃ and liquid at
51.7 ℃ is given by
J/kg∆𝐻𝑓𝑔 = 3.37 × 105
This is a little higher than the latent heat of propane vapor at the saturation temperature owing to theeffect of subcooling.
Solution:
The heat to be absorbed by cooling water is
𝑄 = (3000 kg⁄h)(h⁄3,600 s)(3.37 × 105 J⁄kg) = 2.81 × 105 W
The outlet water temperature is
𝑡2 = (𝑄⁄𝑊 𝐶𝑝𝑤 ) + 𝑡1
2.81 × 105= (70,000 kg⁄h)(h⁄3,600 s)(1 kcal⁄kg K)(4.184 × 103 J⁄kcal) + 32 = 35.5℃

The average temperature of condensing vapor is𝑇𝑠𝑡 = (55 + 51.7)⁄2 = 53.4 ℃
The average temperature of water is
𝑡𝑐 = (35.5 + 32)⁄2 = 33.8 ℃
Assume the overall heat transfer coefficient
𝑈𝑜 = 525 J⁄m2 s K

Engineering Design of an Overhead Condenser (continued)

173

The logarithmic mean of temperature differences is(𝑇 − 𝑡1 ) − (𝑇𝑠𝑡 − 𝑡2 )(∆𝑇)𝑙.𝑚. = 𝑠𝑡𝑇 − 𝑡ln 𝑇𝑠𝑡 − 𝑡1
𝑠𝑡2
(53.4 − 32) − (53.4 − 35.5)== 19.6 𝐾
53.4 − 32
ln 53.4 − 35.5
The heat transfer surface for condensation is𝑄
2.81 × 105 J⁄s
𝐴𝑜 === 27.3 𝑚2
𝑈𝑜 (∆𝑇)𝑙.𝑚.
(525 J⁄m2 s K)(19.6 K)
The outside surface of a tube for heat transfer is
𝑎𝑜 = 𝜋𝑑𝑜 𝐿 = 𝜋(0.025m)(3 m) = 0.236 𝑚2
The total number of tubes is
𝑁 = 𝐴𝑜 ⁄𝑎𝑜 = (27.3 m2 )⁄(0.236 m2 ) = 115.6 < 116
The following empirical equation is used to calculate the inside heat transfer coefficient:ℎ𝑖 𝑑𝑖
𝐺𝑖 𝑑𝑖 0.8 1⁄3) 𝑃𝑟= 0.023 (𝜅𝑤
𝜇𝑤
The mass velocity is𝑊(70,000 kg⁄h)(h⁄3,600 s)
𝐺𝑖 == 441 kg⁄m2 s
2 =(𝜋⁄4)(0.022 m)2 (116)(𝜋⁄4)𝑑𝑖 𝑁
The physical properties of water at 𝑡𝑐 = 33.8 ℃ are
𝜇𝑤 = 8 × 10−4 kg⁄m s, 𝜅𝑤 = 0.622 J⁄m s K, Pr = 4.5
𝑅𝑒𝑖 = 𝐺𝑖 𝑑𝑖 ⁄𝜇𝑤 =(441 kg⁄m2 s)(0.022 m)= 1.21 × 104
8 × 10−4 kg⁄m s(turbulent flow)

The inside heat transfer coefficient is
0.622 J⁄m s Kℎ𝑖 = (0.023) () (12,100)0.8 (4.5)1⁄3 = 1,980 J⁄m2 s K
0.022 mAssume the condensing coefficient̅̅̅ℎ𝑐 = 737 J⁄m2 s K
The tube wall temperature can be calculated with omission of the resistance of the tube metal as
𝑇𝑠𝑡 − 𝑡𝑐
𝑇𝑤 − 𝑡𝑐
𝑞=
=
1
1 𝑑𝑜
1 𝑑𝑜
̅̅̅𝑐 + ℎ𝑖 𝑑𝑖ℎ𝑖 𝑑𝑖ℎFrom the equation
1 0.0251980
0.022(53.4 − 33.8) = 39.6℃
𝑇𝑤 = 33.8 +1
1 0.025
737 + 1980 0.022
The physical properties of liquid n-pentane at 𝑇𝑓 = (53.4 + 39.6)⁄2 = 46.6 ℃ are
𝜅𝐿 = 0.133 J⁄m s K,
𝜇𝐿 = 1.9 × 10−4 kg⁄m s,
𝜌𝐿 = 600 J⁄m3
𝐶𝑝𝐿 = 2.39 × 103 J⁄kg K
At the operating pressure𝜌𝐿 ≅ 𝜌𝐿 − 𝜌𝑉
The average enthalpy change ∆𝐻"𝑓𝑔 is3∆𝐻′𝑓𝑔 = ∆𝐻𝑓𝑔 + 𝐶𝑝𝐿 (𝑇𝑠𝑡 − 𝑇𝑤 )8
= 3.37 × 105 + (3⁄8)(2.39 × 103 )(53.4 − 39.6) = 3.49 × 105 J⁄kg
Using Eq. (15.1-13)

,

174

Heat Transfer with Phase Transformation
3 1/4

)𝑔

(

2√2 𝜌𝐿 𝜌𝐿 − 𝜌𝑉 Δ𝐻′𝑓𝑔 𝜅𝐿̅̅̅̅]ℎ𝑐 = 3 [𝜇 (𝑇 − 𝑇 )𝐿
𝐿

𝑠𝑡

𝑤

2

1⁄ 4

(600 kg⁄m3 ) (9.8 m⁄s 2 )(3.49 × 105 J⁄kg)3()= 0.943 [× (0.133 J⁄m s K) ](1.9 × 10−4 kg⁄m s)(3 m)(53.4 − 39.6)= 734.6 𝐽⁄𝑚2 𝑠 𝐾
The calculated value is almost equal to the assumed value 737 J⁄m2 s K .
Then the overall heat transfer coefficient can be obtained with the calculated ℎ𝑖 and ℎ𝑐 :1
𝑈𝑜 == 519 J⁄m2 s K
1
1 0.025734.6 + 1980 0.022
2
The assumed value 525 J⁄m s K is within 2% of the calculated value.
The total number of heat transfer tubes is
𝑄
𝑁=
= 117
𝑈𝑜 (∆𝑇)𝑙.𝑚. 𝑎𝑜
For simplicity we neglected the heat transfer resistance due to the fouling and tube metal effects,
but the omission of these resistances should sometimes be checked in design calculation.

15.2 Evaporation
15.2-1 Heat transfer for boiling
Boiling is another example of heat transfer processes with phase change.
There are two general boiling operations: pool boiling and forced boiling. Boiling water in a kettle,
which is at or near saturation temperature without forced convective flow or agitation, is the firstexample, where many water vapor bubbles generating are agitating the liquid pool very strongly.
Water, which flows in heat exchangers with vapor formation, is the second example. The heat flux
is much larger than that obtainable without phase change.

Fig. 15.2-1. Pool boiling and forced boiling15.2-2 Pool boiling
The pool-boiling heat transfer can be understood in the form of the so-called boiling curve, i.e.,the curve of heat flux versus the temperature difference ∆𝑇𝑒𝑥 between the heated wall and thesurrounding saturated fluid (simply called the excess temperature).

Pool Boiling (continued)

175

Fig. 15.2-2. Nukiyama’s experimental system for observing various boiling regimes

To acquire a physical understanding of pool boiling, we shall consider the famous Nikiyama’s
experiment1).
His experimental setup is shown in Fig. 15.2-2. A thin platinum wire is immersed in a pool of
distilled water controlled to be kept at a given temperature 𝑇𝑓 < 𝑇𝑠𝑡 at atmospheric pressure,which is not much lower than the saturation temperature. Let us consider the electrically heatedplatinum resistance wire. We can easily obtain the boiling curve by measuring the surface area of
the wire, the electric power input, and the temperature of the wire with the aid of thetemperature-resistance relationship.

Fig. 15.2-3. A typical boiling curve for a wire immersed in a pool of water
1.

Nukiyama, S., J. Soc. Mech. Eng., Japan, 37(206), 367 (1934)

Fig. 15.2-3 is a schematically drawn typical boiling curve for a wire, where the heat flux isplotted as a function of the excess temperature. As long as the temperature of the wire does notexceed the boiling point by more than a few degrees (AB range), heat is transferred to the water
bulk by free convection without phase change. As the temperature of the wire is increased, a point
B is reached where the energy level of water adjacent to the wire surface becomes so high thatwater vapor bubbles are generated at a small number of discrete sites. This is the beginning of thenucleate boiling regime (BC Range). As vapor bubbles form and grow on the heating surface, they
push hot water from the vicinity of the wire into the colder water bulk. As the heat flux is raised andthe number of bubbles is increased, the bubbles begin to coalesce and the heat transfer by

176

Heat Transfer with Phase Transformation

evaporation becomes predominant. As the excess temperature further increases, the number of sites
at which bubbles form increases rapidly, and then the bubbles coalesce to form continuous vapor
columns which causes a reduction of effectiveness of each site. The heat flux no longer increases,
i.e., when the heat flux reaches a maximum at point C, a further increase of the temperature causes a
decrease in the rate of heat flow (dotted line). This maximum heat flux occurs at the critical excess
temperature (referred to as the burnout point). When the excess temperature goes beyond Point C,
provided that the melting point of the wire is sufficiently high, a transition from nucleate to film
boiling will take place. This case corresponds to a transition from point C to E in Fig. 15.2-3.
Provided that the wire heater has a low melting point, the heater will melt, i.e., burnout occurs. The
remaining regime of the curve (beyond D) is the fully developed film boiling. The heat transport
through the vapor film is by conduction and radiation to the vapor-liquid interface where
vaporization takes place.
The forced convective boiling curve is similar to that for pool boiling. But the regimes are madecomplicated by the effects of velocity, subcooling, and different vapor-liquid flow patterns.
15.2-3 Heat transfer correlation for pool boiling
In boiling heat transfer, except for the physical properties of the vapor and liquid, the latent
heat of vaporization, the surface tension, the surface characteristics, and the pressure should beconsidered. Owing to the great number of variables involved, there is no single correlation equation
applicable to the entire range of pool boiling. Owing to the complicatedness, only the following
widely accepted semiempirical equation proposed by Rohsenow1) is introduced:

𝑞 = 𝜇𝐿 ∆𝐻𝑓𝑔 √(𝜌𝐿− 𝜌𝑉)𝑔𝜎[

𝐶𝑝𝐿 (𝑇𝑤 − 𝑇𝑠𝑡)
∆𝐻𝑓𝑔𝑃𝑟𝐿 𝑠𝐶𝑠𝑓

3

(15.2-1)]

where 𝐶𝑝𝐿 : specific heat of saturated liquid in J/kg K
𝐶𝑠𝑓 : experimental constant, dimensionless𝑔 ∶ gravitational acceleration in m/s2
∆𝐻𝑓𝑔 : enthalpy of vaporization in J/kg
𝑃𝑟𝐿 ∶ Prandtl number of saturated liquid, dimensionless𝑞 ∶heat flux per unit area of heated surface in J/m2s
𝑇𝑤 − 𝑇𝑠𝑡 : excess temperature in K
𝜇𝐿 ∶ liquid viscosity in kg/m s
𝜎 ∶ surface tension in N/m𝜌𝐿 , 𝜌𝑉 ∶ densities of saturated liquid and vapor in kg/m3
1.

Rohsenow, W. M., Trans. ASME, 74, 969 (1952)

Table 15.2-1 Experimental constants for nucleate boiling1)
Fluid- Heated Surface Combination
Constant 𝐶𝑠𝑓
Water on Ground and Polished Stainless Steel
0.0080
Water on Mechanically Polished Stainless Steel
0.0132
Water – brass, Water – nickel
0.0060
Water – platinum
0.0130
Water – copper
0.0130Ethanol – chromium
0.0027Isopropanol – copper
0.00225
35% K2CO3 – copper
0.0054
Benzene – chromium
0.0100n-Pentane – chromium
0.015
These constants are picked up from the table of Kreith’s book.
1.

Kreith, F.; “Principles of Heat Transfer,” 3trd ed., Intext Press, Inc., 1973

Exponent 𝑠
1.0
1.0
1.0
1.0
1.0
1.7
1.7
1.7
1.7
1.7

Critical Heat Flux177

Eq.(15.2-1) indicates that the heat flux in nucleate boiling is proportional to the cube of the excess
temperature. The experimental constant 𝐶𝑠𝑓 depends on the particular fluid- heated surface
combination and is affected by the surface roughness of the heater.
Table 15.2-1 gives the experimental values. The exponent 𝑠 of the Prandtl number is 1.0 forwater but 1.7 for all other fluids.
15.2-4 Critical Heat Flux1,2,3)
The critical heat flux is a very important characteristic giving the heat flux at which a transition
occurs from nucleate to film boiling. The prediction of the critical heat flux is very important fromtwo engineering viewpoints because this value indicates the maximum performance for manysystems and the great drop in heat flux beyond this point sometimes results in the destruction(burnout) of the heating surface owing to the accompanied rise in surface temperature.
One of many proposed models is Zuber’s one1):𝜋𝑞𝑚𝑎𝑥 = 24 𝜌𝑉 ∆𝐻𝑓𝑔 [𝜎(𝜌𝐿 − 𝜌𝑉 )𝑔 1/4
𝜌𝑉

2

]

−1/2

𝜌

(15.2-2)(𝜌 +𝐿𝜌 )𝐿

𝑉This equation was obtained based on hydrodynamic instability theory postulating that the volume
flow rate of vapor from the heated surface should be equal to that of liquid flow toward the surface.
Any other correlations do not differ greatly in the form from Zuber’s equation.
1.
2.
3.

Zuber, N. and M. Tribus; “Further Remarks on the Stability of Boiling Heat Transfer,” Rep.58-5, Dept of Eng., Univ. of
Calif., Los Angeles (1958)
Rohsenow, W. M.; “A Method of Correlating Heat-Transfer Data for Surface Boiling Liquids,” Trans. ASME, 74, pp. 969(1952)
Rohsenow, W. M.; “Boiling Heat Transfer,” Dev. In Heat Transfer, W. M. Rohsenow, ed. (Cambridge, Massachusetts: MIT
Press, 1964), pp.169-260[EXAMPLE 15.3-1] Estimate the heat flux for water boiling on a mechanically polished clean
stainless steel at 1 atm and 110℃ (= 383 K) surface temperature and compare it with the criticalheat flux to confirm the nucleate boiling.
Solution:
From Table 15.3-1, 𝐶𝑠𝑓 = 0.0132 for the fluid-surface combination.
The appropriate physical properties at 100℃ and 1 atm are∆𝐻𝑓𝑔 = 539 kcal/kg = 2.26 × 106 J/kg, 𝜌𝐿 = 961.8 kg/m3, 𝜌𝑉 = 0.598 kg/m3 (fromsteam table)
𝑃𝑟𝐿 = 1.78, 𝜇𝐿 = 0.130 × 10−3 kg/m s, 𝐶𝑝𝐿 = 4.18 × 103 J/kg K, 𝜎 = 0.0588
The excess temperature ∆𝑇𝑒𝑥 = 110 − 100 = 10 𝐾.
From Eq. (15.2-1)
1

3(961.8 − 0.598)(9.8) 2(4.18 × 103 )(10)𝑞 = (0.130 × 10−3 )(2.26 × 106 ) [] [](2.26 × 106 )(1.78)(0.0132)
0.0588= 5.736 × 104 J/m2s
From Zuber’s equation

1

1

2(0.0588)(961.8 − 0.598)(9.8) 4𝜋
961.8(0.598)(2.26 × 106 ) × [𝑞𝑚𝑎𝑥 =] () = 1.11 × 106
2
0.598
24
961.8 + 0.6

It has been confirmed that the assumption of nucleate boiling is valid because 𝑞 < 𝑞𝑚𝑎𝑥 .
Nomenclature
𝐴𝑜
𝑎𝑜
𝐶𝑝
𝑑𝑖

total heat transfer surface area, [m2]outside surface area of a heat transfer tube, [m2]heat capacity, [J/kg K]
inside diameter of heat transfer tube, [m]

178
𝑔
ℎ𝑐
ℎ𝑖
𝐿
𝑁
𝑃𝑟
𝑝
𝑞
𝑡
𝑣𝑥 , 𝑣𝑦 , 𝑣𝑧
𝑈𝑜𝑊𝑥, 𝑦, 𝑧
Γ
𝛿

∆𝐻𝑓𝑔
𝛥𝑇
𝜅
𝐺
𝜇
𝜌
𝜎

Nomenclaturegravitational acceleration, [m/s2]
condensing coefficient, [J/m2sK]
inside heat transfer coefficient, [J/m2sK]
height of vertical plate, [m] or tube length, [m]number of tubes, [ - ]Prandtl number, [ - ]pressure, [Pa]heat flux, [J/m2s]time, [s] or temperature, [K]velocity components in rectangular coordinates, [m/s]overall heat transfer coefficient on outside surface basis, [W/m2K]
mass flow rate, [kg/s]
rectangular coordinates, [m]mass flow rate per unit width of falling condensate film, [kg/s m]thickness of condensate film, [m]latent heat of evaporation, [J/kg]

temperature difference, [K]thermal conductivity of condensate liquid, [J/m s K]
mass velocity, [kg/m2s]
viscosity, [kg/m s]density, [kg/m3]
surface tension, [N/m]Subcripts
f
L
l.m.
st
V
W

film
liquidlogarithmic meansaturated
vapor
wall

CHAPTER 16
MASS TRANSFER WITH CHEMICAL REACTION
16.1 Diffusion with Homogeneous Chemical Reaction
16.1-1

Diffusion with a homogeneous reaction in a stagnant fluid------- Penetration theory -------

In gas absorption process, chemical reaction can be used to provide greater capacity of solvent
for solute gas and greater rate of absorption than could be obtained by pure physical absorption
only.
Much industrial gas-liquid contacting equipment usually operates with repeated short contacts ofthe two phases. Therefore let us consider the unsteady diffusion of dissolved gas from interface intoa semi-infinite stagnant liquid mixture by use of the Higbie’s penetration theory. The irreversible
first-order chemical reaction occurs between the dissolved gas A and one component B in the liquid
phase with the reaction - rate constant 𝑘1 .

Fig. 16.1-1. Gas absorption accompanied with a first-order chemical reaction in liquid phase

Setting up mass balance over a differential control volume 𝑆 𝑑𝑦 apart 𝑦 from the interface
𝜕𝐶𝐴
𝜕𝐶𝐴
𝜕𝐶𝐴
| + 𝐷𝐴𝑚|)𝑆 =𝑆 𝑑𝑦 + 𝑘1 𝐶𝐴 𝑆 𝑑𝑦
𝜕𝑦 𝑦
𝜕𝑦 𝑦+𝑑𝑦𝜕𝑡(− 𝐷𝐴𝑚or𝜕2 𝐶

𝐷𝐴𝑚 𝜕𝑦2𝐴 =𝜕𝐶𝐴𝜕𝑡

Introducing a new variable𝜕 2 𝐶𝐴 ′𝐷𝐴𝑚 𝜕𝑦2 =

𝜕𝐶𝐴 ′
𝜕𝑡

(16.1-1)+ 𝑘1 𝐶𝐴𝐶𝐴′ =+ 𝑘1 𝐶𝐴 ′The boundary conditions are

𝐶𝐴 − 𝐶𝐴∞(16.1-2)

180

Mass Transfer with Chemical Reaction

𝐶𝐴′ = 𝐶𝐴𝑖 − 𝐶𝐴∞at 𝑦 = 0𝐶𝐴′ = 0at 𝑦 = ∞Taking the s-multiplied Laplace transform of Eq. (16.1-2)
2𝐶 ′
2𝐶 ′
̅̅̅̅̅̅
̅̅̅̅̅̅𝑑𝑑
𝑘 +𝑠𝐴𝐷𝐴𝑚 𝑑𝑦2𝐴 = 𝑠 𝐶𝐴 ′ + 𝑘1 𝐶𝐴 ′ or
− ( 𝐷1 ) 𝐶𝐴 ′ = 0
𝑑𝑦 2

𝑘 +𝑠
𝐴𝑚

(16.1-5)𝑘 +𝑠𝑦) + 𝑏 𝑒𝑥𝑝 (− √ 𝐷1

𝐴𝑚

Using the boundary conditions
𝑘 +𝑠

𝐶𝐴 ′ = (𝐶𝐴𝑖 − 𝐶𝐴∞ ) 𝑒𝑥𝑝 (− √ 𝐷1

𝐴𝑚

(16.1-4)

𝐴𝑚

The boundary conditions are
𝐶𝐴 ′ = 𝐶𝐴𝑖 − 𝐶𝐴∞at 𝑦 = 0
𝐶𝐴 ′ = 0
at 𝑦 = ∞The general solution to Eq. (16.1-4) is
𝐶𝐴 ′ = 𝑎 𝑒𝑥𝑝 (√ 𝐷1(16.1-3)(16.1-6)𝑦)(16.1-7)𝑦)

The mass flux at the interface is expressible as𝑑𝐶 ′𝑁𝐴 = − 𝐷𝐴𝑚 𝑑𝑦𝐴 |(16.1-8)= (𝐶𝐴𝑖 − 𝐶𝐴∞ ) √(𝑘1 + 𝑠)𝐷𝐴𝑚

𝑦=0

Taking the inverse Laplace transform, the mass flux becomes
𝑁𝐴 = (𝐶𝐴𝑖 − 𝐶𝐴∞ )√𝑘1𝐷𝐴𝑚 [erf(𝑘1 𝑡)1/2 +exp (− 𝑘1 𝑡)√𝜋 𝑘1 𝑡(16.1-9)]

For pure physical diffusion, i.e., 𝑘1 = 0
𝑁𝐴0 = (𝐶𝐴𝑖 − 𝐶𝐴∞ ) √𝑠 𝐷𝐴𝑚
or𝐷(16.1-10)𝑁𝐴0 = (𝐶𝐴𝑖 − 𝐶𝐴∞ ) √ 𝜋𝐴𝑚𝑡

Note that Eq. (16.1-10) is equivalent with Eq. (6.7-10) if the exposure time 𝑡 is replaced by 𝑧⁄𝑉 .
The average rate of absorption over the total exposure time 𝑡 can be calculated as𝑡𝑡1

exp (− 𝑘1 𝑡)1√𝜋 𝑘1 𝑡𝑊𝐴 = ∫0 𝑁𝐴 𝑑𝑡⁄∫0 𝑑𝑡 = (𝐶𝐴𝑖 − 𝐶𝐴∞ ) √𝐷𝐴𝑚 𝑘1 [(1 + 2 𝑘 𝑡) erf (𝑘1 𝑡)1⁄2 +The corresponding time-average mass transfer coefficient 𝑘𝐿 is
1

exp (− 𝑘1 𝑡)1√𝜋 𝑘1 𝑡𝑘𝐿 = √𝐷𝐴𝑚 𝑘1 [(1 + 2 𝑘 𝑡) erf (𝑘1 𝑡)1⁄2 +Taking the time-average of Eq. (16.1-10)]

4𝐷](16.1-11)(16.1-12)(16.1-13)

𝑘𝐿0 = √ 𝜋 𝐴𝑚𝑡Here 𝑘𝐿0 is the mass transfer coefficient without chemical reaction.
Introducing a ratio 𝛽 of the mass transfer rate with and without reaction
𝜋𝑘 𝑡

1

exp (− 𝑘1 𝑡)1√𝜋 𝑘1 𝑡𝛽 = √ 41 [(1 + 2 𝑘 𝑡) erf (𝑘1 𝑡)1⁄2 +For small values of 𝑘1 𝑡
𝑘 𝑡]

(16.1-14)(𝑘 𝑡)21
𝛽 = 1 + 31 − 30
+ −−−
(16.1-15)
This reaction factor indicates the deviation from the purely physical mass transfer (𝛽 = 1).
For large values of 𝑘1 𝑡𝜋

𝛽 = √ 4 [√𝑘1𝑡 + 2

1
√𝑘1 𝑡

− − − −]

(16.1-16)

If the exposure time and reaction-rate constant are given, the reaction factor can be calculated. The
reaction factor calculated by the penetration theory is in good agreement with that calculated byother theories (Film theory and surface renewal theory).

Gas Absorption with First-order Chemical Reaction

181

16.1-2 Gas absorption with first-order chemical reaction
Let us consider a packed column gas absorber shown in Fig. 16.1-2, in which the first-orderchemical reaction of a dissolved gas A is carried out with a continuous stream of liquid mixture.
Setting up mass balance over a differential control volume 𝑆 𝑑𝑧 in the packing section
𝑘𝐿 𝑎 (𝐶𝐴𝑖 − 𝐶𝐴𝐿 )𝑆 𝑑𝑧 + (𝐶𝐴𝐿 |𝑧 − 𝐶𝐴𝐿 |𝑧+𝑑𝑧 )𝐿𝑀 𝑆⁄𝜌𝑀 − 𝑘1 𝐶𝐴𝐿 𝑆 𝑑𝑧 𝜀 = 0(16.1-17)
or𝐿 𝑑𝐶
𝑘𝐿 𝑎 (𝐶𝐴𝑖 − 𝐶𝐴𝐿 ) = 𝜌𝑀 𝑑𝑧𝐴𝐿 + 𝜀 𝑘1 𝐶𝐴𝐿(16.1-18)
𝑀

where 𝑘𝐿 is the liquid-phase mass transfer coefficient with reaction, 𝜌𝑀 the total molar density,and 𝜀 the volume fraction of liquid in the packing section.
The total number of transfer units 𝑁𝐿 = 𝑍𝑇 𝜌𝑀 𝑘𝐿° 𝑎⁄𝐿𝑀 and the number of transfer units
𝑛𝐿 = 𝑧 𝜌𝑀 𝑘𝐿° 𝑎⁄𝐿𝑀 are introduced using the liquid-phase mass transfer coefficient 𝑘𝐿° for purely
physical absorption.

Fig. 16.1-2. Packed column gas absorber accompanied by a first-order homogeneous chemical
reaction

The total holdup time (i.e., the so-called residence time) is calculated as𝜃 = 𝜀 𝑆 𝑍𝑇 𝜌𝑀 ⁄𝐿𝑀 𝑆 = 𝜀 𝑍𝑇 𝜌𝑀 ⁄𝐿𝑀
The above equation becomes(16.1-19)

𝑑 𝐶𝐴𝐿(𝑎 + 𝑘1 𝜀) 𝐶𝐴𝐿 = 𝑘𝐿 𝑎 𝐶𝐴𝑖 𝐿𝑀(16.1-20)

𝑘(16.1-21)

𝑑𝑧

+

𝑘𝐿 𝜌𝑀
𝐿𝑀

𝑘

𝜌

𝐿

𝑀

Using the number of transfer units 𝑛𝐿 and the time of exposure 𝜃
𝑑 𝐶𝐴𝐿
𝑑𝑛𝐿

+ (𝛽 + 𝑁1 𝜃) 𝐶𝐴𝐿 = 𝛽 𝐶𝐴𝑖
𝐿

The boundary condition is
𝐶𝐴𝐿 = 0 at 𝑧 = 0
The following solution indicates the vertical variation in the bulk concentration of component A:
𝛽𝑁
𝐶𝐴𝐿 = 𝐶𝐴𝑖 (𝑘 𝜃+ 𝛽𝐿 𝑁 ) {1 − exp[− (𝑘1 𝜃 + 𝛽 𝑁𝐿 )(𝑛𝐿 ⁄𝑁𝐿 )]}(16.1-22)
1

𝐿

Then the total rate of absorption per unit cross-sectional area can be calculated as
𝑍
𝑁𝐿
𝑁𝐴𝑇 = ∫0 𝑇 𝑘𝐿 𝑎(𝐶𝐴𝑖 − 𝐶𝐴𝐿 ) 𝑑𝑧 = 𝛽 𝑀 ∫0 𝐿 (𝐶𝐴𝑖 − 𝐶𝐴𝐿 ) 𝑑𝑛𝐿
𝐿𝑀

𝛽 𝑁𝐿

𝑀

𝐿

𝛽 𝑁𝐿

𝜌𝑀

= 𝐶𝐴𝑖 𝜌 (𝛽 𝑁 + 𝑘 𝜃) {𝑘1 𝜃 + 𝛽 𝑁 + 𝑘 𝜃 [1 − exp(− (𝛽 𝑁𝐿 + 𝑘1 𝜃))]}1

𝐿

1(16.1-23)

If the reaction factor 𝛽 and the total holdup time 𝜃 are given, the total rate of absorption can be

182

Mass Transfer with Chemical Reaction

calculated using the physical absorption data. However the exposure time is generally difficult toestimate as for the film thickness in the film theory.
For practical design problem, the following equations can be used:
𝑁𝐴 = 𝐾𝐺 (𝑝𝐴 − 𝑝𝐴∗`) = 𝑘𝐺 (𝑝𝐴 − 𝑝𝐴𝑖 ) = 𝛽 𝑘𝐿° (𝐶𝐴𝑖 − 𝐶𝐴𝐿 )(16.1-24)
and
1
1
𝐻
=
+
(16.1-25)
𝐾
𝑘
𝛽 𝑘°
𝐺

𝐺

𝐿

Here 𝐻 is the Henry’s constant for purely physical equilibrium and 𝑝𝐴∗` is the concentration of
unreacted gas A expressed in units of gas-phase concentration, i.e., 𝑝𝐴∗` = 𝐻 𝐶𝐴𝐿 .

16.2 Gas Absorption with Instantaneous Bimolecular Reaction
Many industrially important absorption processes are accompanied by bimolecular reaction
A + ν B → products
The irreversible chemical reaction such as SO2 + 2 NaOH → Na2 SO3 + H2 O occurs in theliquid phase between a dissolved gas reactant A and a nonvolatile reactant B which is initiallypresent in the liquid phase.
The problem of liquid-phase diffusion with bimolecular reactions has not yet been solved
analytically, except for some simple cases. One of the extreme cases occurs when the reaction rateis so large that A and B react immediately and completely on a plane (or a very thin zone) whichmoves with time. The reaction plane which was initially located at the interface is now at a distance𝑧𝑟 (𝑡) from the interface.

Fig. 16.2-1. Schematic composition profile of gas absorption with bimolecular reaction

As shown in Fig.16.2-1, the reaction rate 𝑟 = 𝑘2 𝐶𝐴 𝐶𝐵 is zero everywhere, except at the reaction
plane. In the region between the interface and the moving reaction plane, the concentration of A is
obtained by solving the transient diffusion equation for A without reaction; the same is true for B in
the semi-infinite liquid body on the right side of the moving reaction plane. Therefore thepenetration theory will also be used.
The set of differential equations to be solved are
𝜕 𝐶𝐴

𝜕 2 𝐶𝐴𝜕 𝐶𝐵

𝜕𝑧 2
𝜕 2 𝐶𝐵

= 𝐷𝐴𝑤
𝜕𝑡
𝜕𝑡

= 𝐷𝐵𝑤

𝜕𝑧 2

𝐶𝐵 = 00 ≤ 𝑧 ≤ 𝑧𝑟 (𝑡)(16.2-1)

𝐶𝐴 = 0𝑧𝑟 (𝑡) ≤ 𝑧 ≤ ∞(16.2-2)

Gas Absorption with Instantaneous Bimolecular Reaction (continued)

183

where 𝐷𝐴𝑤 and 𝐷𝐵𝑤 are diffusivities of A and B in the solution, respectively. Note that any
reaction does not appear in the above equations.
The general solutions to the above two equations are of the form𝐶𝐴𝑧
= 𝐾1 + 𝐾2 erf 2 𝐷 𝑡
0 ≤ 𝑧 ≤ 𝑧𝑟 (𝑡)(16.2-3)
𝐶
√ 𝐴𝑤
𝑧

𝐴𝑖

𝐶𝐵
𝐶𝐵∞

= 𝐾3 + 𝐾4 erf 2 𝐷(16.2-4)

𝑧𝑟 (𝑡) ≤ 𝑧 ≤ ∞√ 𝐵𝑤 𝑡At the reaction plane, A and B react immediately.
𝐶𝐴 = 𝐶𝐵 = 0
at 𝑧 = 𝑧𝑟 (𝑡)Therefore from perfect differentia
𝜕𝐶𝜕𝐶
𝑑𝐶𝐴 = ( 𝜕𝑧𝐴 ) 𝑑𝑧𝑟 + ( 𝜕𝑡𝐴 ) 𝑑𝑡 = 0
𝑟(16.2-6)𝑧𝑟𝑡

From this equation𝑑𝑧𝑟(16.2-5)

𝜕𝐶

𝜕𝐶(16.2-7)= − ( 𝜕𝑡𝐴 ) ⁄( 𝜕𝑧𝐴 )
𝑑𝑡
𝑧𝑟

𝑟

From Eq. (16.2-3)

(

𝑡

𝜕𝐶𝐴
2
1
𝑧𝑟 2
𝑧𝑟
) = 𝐾2 𝐶𝐴𝑖(−) exp (−)
2𝑡
4 𝐷𝐴𝑤 𝑡 2 √𝐷𝐴𝑤 𝑡
𝜕𝑡 𝑧𝑟
√𝜋
𝜕𝐶𝐴
2 1
𝑧𝑟 2
𝑧𝑟
(
) = 𝐾2 𝐶𝐴𝑖( ) exp (−)
𝜕𝑧𝑟 𝑡
4 𝐷𝐴𝑤 𝑡 2 √𝐷𝐴𝑤 𝑡√𝜋 𝑧𝑟

Substitute these relations into Eq. (16.2-7)
𝑑𝑧𝑟𝑧= 2 𝑟𝑡(16.2-8)𝑑𝑡
Integration gives𝑧𝑟 (𝑡) = √4 𝛼 𝑡(16.2-9)
where √4 𝛼 is an arbitrarily chosen integration constant. These five integration constants
𝐾1 , 𝐾2 , 𝐾3 , 𝐾4 , and 𝛼 can be evaluated by the following five initial and boundary conditions:
I.C. or B.C.1 at 𝑡 = 0 𝐶𝐵 = 𝐶𝐵∞
0 ≤𝑧 ≤ ∞orat 𝑧 = ∞ 𝐶𝐵 = 𝐶𝐵∞
𝑡 >0
B.C.2
at 𝑧 = 0 𝐶𝐴 = 𝐶𝐴𝑖
B.C.3
at 𝑧 = 𝑧𝑟
𝐶𝐴 = 0(16.2-10)
B.C.4
at 𝑧 = 𝑧𝑟
𝐶𝐵 = 0𝜕𝐶
𝜕𝐶
B.C.5
at 𝑧 = 𝑧𝑟− 𝜈 𝐷𝐴𝑤 𝜕𝑧𝐴 = 𝐷𝐵𝑤 𝜕𝑧𝐵
The last boundary condition comes from the stoichiometric requirement that one mole of A reacts
with 𝜈 moles of B.
From B.C.2
𝐾1 = 1
1
1
From B.C.3
𝐾2 = −
= −
𝑧𝑟
𝛼
erf

2 √𝐷𝐴𝑤𝑡

erf √From B.C.1

𝐾3 = 1 − 𝐾4

From B.C.4

0 = (1 − 𝐾4 ) + 𝐾4 erf √𝐷

𝛼
𝐵𝑤

1

Therefore 𝐾4 =Then 𝐾3 = 1 −
𝜕𝐶𝐴

|

𝜕𝑧 𝑧𝑟

= −

𝐷𝐴𝑤

𝛼
𝐷𝐵𝑤

1−erf √1𝛼𝐷𝐵𝑤1−erf √

1
erf √

𝛼
𝐷𝐴𝑤

𝐶𝐴𝑖

2
√𝜋

exp (−

𝑧𝑟 2
4 𝐷𝐴𝑤 𝑡)1
2 √𝐷𝐴𝑤 𝑡(16.2-11)

184
𝜕𝐶𝐵

Mass Transfer with Chemical Reaction

|

1

=

𝜕𝑧 𝑧𝑟

1−erf √

𝐶𝐵∞

𝛼
𝐷𝐵𝑤

2
√𝜋

𝑧𝑟 2

exp (−

4 𝐷𝐵𝑤 𝑡)1

(16.2-12)

2 √𝐷𝐵𝑤 𝑡

Substitute these relations with Eq. (16.2-9) into B.C.5𝛼𝛼

exp (𝐷 ) [1 − erf √𝐷
𝐵𝑤

𝐵𝑤] = exp (

𝛼

𝛼

𝐷𝐴𝑤

) erf √𝐷

𝐷

𝐴𝑤

𝐶

√𝐷𝐵𝑤 𝜈 𝐶𝐵
𝐴𝑤

𝐴𝑖

This equation should be solved by trial and error method to determine 𝛼.
The concentration profiles of A and B are given by
𝐶𝐴
1
𝑧
=
1
−
erf
0 ≤ 𝑧 ≤ 𝑧𝑟 (𝑡)
𝛼
𝐶
2 𝐷
𝑡
𝐴𝑖

𝐶𝐵
𝐶𝐵∞

√ 𝐴𝑤

erf √

𝐷𝐴𝑤

= 1−

erf

1

+

𝛼
𝐷𝐵𝑤

1−erf √𝑧
2 √𝐷𝐵𝑤 𝑡

𝑧𝑟 (𝑡) ≤ 𝑧 ≤ ∞𝛼𝐷𝐵𝑤

1− erf √(16.2-13)(16.2-14)(16.2-15)

Then the rate of absorption can be calculated as
𝑁𝐴 |𝑧=0 = − 𝐷𝐴𝑤𝜕𝐶𝐴|𝜕𝑧 𝑧=0

=

1

𝐷

erf √

(16.2-16)√ 𝐴𝑤 𝐶𝐴𝑖

𝛼

𝜋𝑡

𝐷𝐴𝑤

The average rate of absorption over the total exposure time 𝑡 is𝑡𝑡

𝑊𝐴 = ∫0 𝑁𝐴 |𝑧=0 𝑑𝑡⁄∫0 𝑑𝑡 =

1
erf √

4𝐷

𝛼

(16.2-17)√ 𝜋𝑡𝐴𝑤 𝐶𝐴𝑖

𝐷𝐴𝑤

The coefficient in front of 𝐶𝐴𝑖 is the mass transfer coefficient with bimolecular chemical reaction:
4𝐷𝛼𝑘𝐿 = √ 𝜋𝑡𝐴𝑤⁄erf √𝐷(16.2-18)

𝐴𝑤

Comparing with the result of purely physical diffusion, Eq. (16.1-13), the reaction factor is given by1
𝛽=(16.2-19)
𝛼
erf √

𝐷𝐴𝑤

From here, the case when 𝛼 is sufficiently small in comparison with 𝐷𝐴𝑤 and 𝐷𝐵𝑤 will beconsidered to obtain the reaction factor as a function of known variables. It will be shown later that
this situation can be obtained when 𝐶𝐵∞ is sufficiently large with respect to 𝐶𝐴𝑖 .
For small 𝛼
𝛼
𝛼
1 𝛼 2
exp (
+ (
)= 1+
) + −−−−−𝐷𝐴𝑤
𝐷𝐴𝑤 2 𝐷𝐴𝑤
erf √

𝛼
𝐷𝐴𝑤

=

2
√𝜋

[(

𝛼
𝐷𝐴𝑤

)

1⁄2

1 − erf √

−

1
3

(

𝛼
𝐷𝐴𝑤

)

3⁄2

+

1
10

(

𝛼

𝐷𝐴𝑤

)

5⁄2

− − − −]𝛼2
𝛼 1⁄2 1 𝛼 3⁄2
1
𝛼 5⁄2
=1−
[(
) − (
) +
(
) − − − −]
𝐷𝐵𝑤
3 𝐷𝐵𝑤
10 𝐷𝐵𝑤√𝜋 𝐷𝐵𝑤

Substitute these relations into Eq. (16.2-13)[1 −𝛼 1⁄2 1 𝛼 3⁄2
1
𝛼 5⁄2
[(
) − (
) +
(
) − − − −]]3 𝐷𝐵𝑤
10 𝐷𝐵𝑤√𝜋 𝐷𝐵𝑤
𝛼1 𝛼 2
2
𝛼 1⁄2 1 𝛼 3⁄21
𝛼 5⁄2= [1 ++ ([() + − − −]
) − (
) +
(
) − −
𝐷𝐴𝑤 2 𝐷𝐴𝑤
3 𝐷𝐴𝑤
10 𝐷𝐴𝑤√𝜋 𝐷𝐴𝑤2− −] √

𝐷𝐵𝑤 𝐶𝐵∞
𝛼
1 𝛼 2
[1 +
+ (
) + − − −]𝐷𝐴𝑤 𝜈 𝐶𝐴𝑖
𝐷𝐵𝑤 2 𝐷𝐵𝑤

For 𝛼 ≪ 𝐷𝐴𝑤 and 𝐷𝐵𝑤 , the above equation can be approximated and rearranged as
1−

The equation can be solved

2

𝛼
2
𝛼
𝐷𝐵𝑤 𝐶𝐵∞√≅√√√𝜋 𝐷𝐵𝑤√𝜋 𝐷𝐴𝑤 𝐷𝐴𝑤 𝜈 𝐶𝐴𝑖

Design of Packed Absorption Towers𝜋1

𝛼= 4 [ 𝐷

√ 𝐵𝑤

+

1

𝐷

√𝐷𝐴𝑤

−2

𝐶

]
√𝐷𝐵𝑤 𝜈 𝐵∞
𝐶
𝐴𝑤

185

𝐴𝑖

for small 𝛼(16.2-20)

This equation suggests that 𝛼 becomes small when 𝐶𝐵∞ is sufficiently large compared with 𝐶𝐴𝑖 .
The larger 𝐶𝐵∞ is, the closer the reaction plane remains to the interface.
The reaction factor can also be approximated for small 𝛼:
1
1
𝛽=
≅ 2 𝛼
(16.2-21)
𝛼
erf √

𝐷𝐴𝑤

√𝜋

√𝐷

𝐴𝑤

Substitute Eq. (16.2-20) into the above equation
𝐷

𝐷

𝐷𝐵𝑤

𝐷𝐴𝑤 𝜈 𝐶𝐴𝑖

β = √ 𝐴𝑤 + √ 𝐵𝑤

𝐶𝐵∞

for large 𝐶𝐵∞

In the case when 𝐷𝐴𝑤 ≅ 𝐷𝐵𝑤
𝐶
β = 1 + 𝜈 𝐵∞
for the case of large 𝐶𝐵∞ and 𝐷𝐴𝑤 ≅ 𝐷𝐵𝑤
𝐶
𝐴𝑖(16.2-22)(16.2-23)

Unlike the results obtained for purely physical absorption or absorption with first-order reaction, therate of absorption with second-order reaction is not proportional to the concentration of thesubstance being absorbed. It should be noticed that even when the interfacial concentration 𝐶𝐴𝑖 isvery small owing to small concentration of A in the gas phase, the absorption rate may becomelarge owing to the effect of 𝐶𝐵 . According to Eq. (16.2-22),
The reaction factor is a function of diffusivities 𝐷𝐴𝑤 , 𝐷𝐵𝑤 , bulk concentration 𝐶𝐵 , and interfacialconcentration 𝐶𝐴𝑖 only. This result can be applied to any type of absorption equipment since 𝛼
is not a function of equipment geometry and flow conditions.
Regarding the global climate change by CO2 emission due to fossil fuel use, many chemical andelectric power companies were challenging the technical development of reactive gas absorption
using aqueous amine solvents for CO2 capture. One possible reaction between CO2 andmonoethanol amine in the aqueous solution can be considered as
2 RNH2 + CO2 ↔ RNHCOO− + RNH3+
This is an endothermic bimolecular reaction. This topic is not dealt with here owing to its difficult
mechanism.

16.3 Design of Packed Absorption Towers
Let us study a practical design of gas absorbers by using the following example.
[EXAMPLE 16.2-1] A packed absorption tower is to be designed for the removal of SO2 fromair by chemical absorption in an aqueous solution of sodium hydroxide. The tower has aninside diameter of 0.5 m and is packed with 1-in. ceramic Raschig rings. An absorption may be
assumed isothermal at 25℃(= 298 K) and 1 atm (= 1.013 × 105 Pa). The air stream
containing 10 mol% of SO2 enters the tower from the bottom at a rate of 40 kmol/m 2h. The
aqueous solution containing 2 M-NaOH (=7.4 wt% = 3.47 mol%) enters the tower from the top
at a rate of 400 kmol/m2h. Estimate the height of packing required to remove 95% of SO 2 fromthe airstream.
Solution:(1) First step: Calculation of mass transfer coefficient 𝑘𝐺 , 𝑘𝐿° for physical absorption(a) Gas-phase mass transfer coefficient 𝑘𝐺 :Due to the low concentration of SO2 in air, the viscosity and density of air can be used forthe airstream:
𝜇𝐺 = 1.85 × 10−5 kg/m s, 𝜌𝐺 = 1.19 kg/m3. The diffusivity of SO2 in air can be calculated
by Hirschfelder et al. equation: 𝐷𝐴𝑔 = 1.27 × 10−5 m2/s.

186

Mass Transfer with Chemical Reaction

The Schmidt number is𝜇
𝑆𝑐𝐺 = 𝜌 𝐷𝐺 = 1.22
𝐺 𝐴𝑔

The mass velocities are
𝐺 = (40 kmol⁄m2 h)[(0.90)(29) + (1 − 0.90)(64)] kg⁄kmol = 1,300 kg⁄m2 h
L = (400 kmol⁄m2 h)[(0.965)(18) + (0.0347)(40)] kg⁄kmol = 7,500 kg⁄m2 h
Gas-phase mass transfer coefficient
For these values of mass velocities, Eq. (10.4-6) can be used to estimate 𝑘𝐺 :𝜇2⁄ 3

(16.3-1)

𝐻𝐺 = 𝑐 𝐺 𝑝 𝐿𝑞 (𝜌 𝐷𝐺 )𝐺 𝐺

For 1-in. Raschig ring, 𝑝 = 0.32, 𝑞 = −0.51, 𝑐 = 3.07. Then
𝐻𝐺 = (3.07)(1300)0.32 (7500)−0.51(1.22)2⁄3 = 0.365 m
From Table 10.4-2, 𝐷𝑝 = 0.0254 m, 𝑎𝑡 = 190 m2 ⁄m3 .
The surface tension of water at 25℃ is
𝜎 = 72.0 dyne⁄cm = 0.072 N⁄m.
The interfacial area 𝑎 is calculated by using Eq. (10.4-5) as follows:𝑎 = 0.0406 (7500)0.455(72.0)−0.83(2.54)−0.48 = 46.2 m2 ⁄m3
From the definition of 𝐻𝐺 : 𝐻𝐺 = 𝐺𝑀 ⁄𝑘𝑦 𝑎
Assuming 𝐺𝑀 = 𝐺𝑀1
𝐺(16.3-2)(40 kmol⁄m2h)

𝑘𝑦 = 𝐻 𝑀𝑎 = (0.365 m)(46.2 m2⁄m3 ) = 2.37 kmol⁄m2 h
𝐺Therefore
𝑘𝐺 = 𝑘𝑦∗ ⁄𝑝𝐵 = 𝑦𝐵 𝑘𝑦 ⁄𝑝𝐵 = 𝑘𝑦 ⁄𝑃 = 2.37 kmol⁄m2 h atm(16.3-3)°(b) Liquid-phase mass transfer coefficient 𝑘𝐿 :The viscosity and density of water at 25℃ can be used:
𝜇𝐿 = 0.895 cP = 0.895 × 10−3 kg⁄m s, 𝜌𝐿 = 997 kg⁄m3 .
The diffusivity of SO2 in water 𝐷𝐴𝑤 = 1.85 × 10−9 m2 ⁄s.
The Schmidt number is0.895 × 10−3
𝑆𝑐𝐿 =
= 485
(997)(1.85 × 10−9 )Liquid-phase mass transfer coefficient
𝐿 = 7500⁄3600 = 2.083 kg⁄m2 s. Eq. (10.4-1) can be used to estimate 𝑘𝐿° for 1-in. Raschigrings (𝛼 = 430, 𝑛 = 0.22):
1

2.083

0.22

𝐻𝐿 = 430 (0.895 × 10−3 )(485)0.5 = 0.282 m

From the definition of 𝐻𝐿 : 𝐻𝐿 = 𝐿𝑀 ⁄𝑘𝑥 𝑎. Assume 𝐿𝑀 = 𝐿𝑀2𝐿𝑀
400 kmol⁄m2 h
𝑘𝑥 === 30.7 kmol⁄m2 h
𝐻𝐿 𝑎(0.282 𝑚)(46.2 m2 ⁄m3 )Therefore(16.3-4)(2) Second step: Equilibrium relationship of SO2 – electrolytic solution system:
Strictly speaking, the SO2 – water system does not follow Henry’s law. For simplicity, over arange of very low partial pressure (0_ ≤ 𝑝𝐴 ≤ 0.1) the equilibrium curve can be roughly
approximated by a linear relationship, i.e., Henry’s law:𝑝𝐴 = 0.54 𝐶𝐴(16.3-5)3Here the Henry’s constant 𝐻𝑤 = 0.54 atm m /kmol.
(a) Effect of ions in an electrolytic solution
In general, the solubility of gas is decreased by the presence of ions in an electrolytic solution.
According to van Krevelen and Hoftijzer1,2), the Henry’s constant for the solution can be relatedby the following empirical equation:
𝑘𝐿 = 𝑘𝑥∗ ⁄𝐶𝐵 = 𝑥𝐵 𝑘𝑥 ⁄𝐶𝐵 = 𝑘𝑥 ⁄𝑐 = (18 m3⁄1000 kmol)(30.7 kmol⁄m2 h) = 0.553 m⁄hDesign of Packed Absorption Towers (continued)

187

Table 16.3-1 Equilibrium relation of SO2 – water system
𝑝𝐴 atm
𝐶𝐴 kmol⁄m3
0.793
1.17
0.519
0.783
0.248
0.391
0.143
0.243
0.09080.156
0.05990.109
0.04080.07790.02230.04690.01340.03120.009170.02350.005200.01560.00191
0.007830.000724
0.00312Fig. 16.3-1. Approximation of equilibrium curve𝐻

log 𝐻𝑤 = − ∑𝑗 ℎ𝑗 𝐼𝑗(16.3-6)
Here 𝐼𝑗 is the ionic strength and ℎ𝑗 = ℎ+ + ℎ− + ℎ𝐺 is the empirical constant.
The ionic strength can be calculated by1
𝐼 = 2 ∑𝑖 𝑧𝑖 2 𝐶𝑖
(16.3-6)
where 𝑧𝑖 is the number of positive or negative charges on an ion having molarity 𝐶𝑖 .
1.
2.
3.

Van Krevelen, D. W. and P.J. Hoftijzer: Rec. Trav. Chim., vol.67, 563 (1943)
Van Krevelen, D. W. and P.J. Hoftijzer: Chim. Ind. XXI Congr. Int., Chim. Ind., p.168 (1948)
Sherwood, T. K., R. L. Pigford, and C.R. Wilke: “Mass Transfer,” p. McGraw-Hill, New York (1975)

The ℎ values are listed in Table 16.3-2.
Table 16.3-2 Constants in Eq. (16.3-6)
cationsℎ+ m3 ⁄kg − ion anions
+
0.0
𝐻
𝑂𝐻−
+
0.091
𝑁𝑎
𝐶𝑙−
+
0.074
𝐾
𝑆𝑂4 −−+
0.028
𝐶𝑂3 −−
𝑁𝐻4
0.051
𝑀𝑔++
𝐵𝑟 −
++
0.053
𝐶𝑎
𝑆𝑂3 −−ℎ− m3 ⁄kg − ion
0.066
0.021
0.022
0.021
0.012
0.001

gas
𝐻2
𝑂2
𝐶𝑂2
𝑁𝐻3
𝑆𝑂2
𝐻2 𝑆

ℎ𝐺 m3 ⁄kmol-0.0020.022-0.019-0.054-0.103-0.033

188

Mass Transfer with Chemical Reaction

The above ℎ𝐺 values were picked up from the data taken at 25℃.
1.

Danckwerts, P.V.: “Gas‒Liquid Reaction,” p.19, McGraw‒Hill, Nerw York (1970)(b) At the top of the tower
𝑧1 = 1For NaOH 𝐶1 = 2 kg − ion Na+ /m3−
3
𝑧2 = 1𝐶2 = 2 kg − ion OH /m
1
𝐼 = ((2)(1)2 + (2)(1)2 ) = 2
2
From Table 16.2-2,ℎ = ℎ𝐺 + ℎ+ + ℎ− = −0.103 + 0.091 + 0.066 = 0.054
𝐻𝑤
log 𝐻 = − ∑𝑗 ℎ𝑗 𝐼𝑗 = −(0.054)(2) = −0.108
Then the Henry’s constant for the electrolytic solution is calculated as𝐻𝑤= 0.78 → 𝐻 = 𝐻𝑊 /0.78 = 0.54/0.78 = 0.69 atm m3/kmol
𝐻
Molar velocity of NaOH at the top = (400)(0.0347) = 13.88 kmol/m 2hRate of SO2 removed from air = (40)(0.10)(0.95) = 3.8 kmol/m2h
Molar velocity of SO2 at the top = (40)(0.10)(1 – 0.95) = 0.2 kmol/m2h
Rate of NaOH consumed for the reaction = (2)(3.8) = 7.6 kmol/m2h(c) At the bottom of the tower
Molar velocity of NaOH = 13.88 – 7.6 = 6.28 kmol/m2h
Concentration of NaOH ≅ 6.28⁄400 = 0.0157 (mole fraction)
1000 0.0157
𝐶𝑁𝑎𝑂𝐻 ≅ 18 1−0.0157 = 0.89 kmol/m3
Molar velocity of Na2SO3 = 3.8 kmol/m2h
Concentration of Na2SO3 ≅ 3.8⁄400 = 0.0095 (mole fraction)
1000 0.0095
𝐶𝑁𝑎2 𝑆𝑂3 ≅ 18 1−0.0095 = 0.53 kmol/m3
For NaOH ℎ = −0.103 + 0.091 + 0.066 = 0.054
𝐼 = (1⁄2)((0.89)(1)2 + (0.89)(1)2 ) = 0.89
For Na2SO3 ℎ = −0.103 + 0.091 + 0.001 = − 0.011
𝐼 = (1⁄2)((0.53)(2)2 + (0.53 × 2)(1)2 ) = 1.59
Then
𝐻
log 𝐻𝑤 = − ((0.054)(0.89) + (−0.011)(1.59)) = −0.0306
Averaging between the top and bottom𝐻𝑤
0.108 + 0.0306
log= −= −0.0693
𝐻
2
The Henry’s constant averaged over the tower is calculated as𝐻𝑤= 0.852 → 𝐻 = 𝐻𝑊 /0.78 = 0.54/0.852 = 0.634 atm m3/kmol
𝐻(16.3-7)(16.3-8)(16.3-9)(3) Third step: Absorption rate
𝐷𝐷𝐶

] 𝑘𝐿° 𝐶𝐴𝑖
𝑁𝐴 = 𝑘𝐺 (𝑝𝐴 − 𝑝𝐴𝑖 ) = 𝛽 𝑘𝐿° 𝐶𝐴𝑖 = [ √𝐷𝐴𝑤 + √𝐷𝐵𝑤 𝜈 𝐵∞
𝐶
𝐵𝑤

𝐷𝐵𝑤 𝐶𝐵∞

√𝐷

𝐴𝑤

𝜈

𝐴𝑤

𝐴𝑖

𝐷

= 𝑘𝐿° [ √𝐷𝐴𝑤 𝐶𝐴𝑖 +𝐵𝑤] (16.3-10)

Diffusivity of NaOH in water is given from literature: 𝐷𝐵𝑤 = 1.25 10−9 m2 ⁄s (at 25℃).
The effect of Na2 SO3 on the liquid-phase diffusivities 𝐷𝐴𝑤 , 𝐷𝐵𝑤 is assumed negligibly
small due to the low concentration.
For the present case 𝐷𝐴𝑤 ≅ 𝐷𝐵𝑤 .
Therefore

Design of Packed Absorption Towers (continued)

𝑁𝐴 = 𝑘𝐺 (𝑝𝐴 − 𝑝𝐴𝑖 ) = 𝑘𝐿° (𝐶𝐴𝑖 +=𝑝𝐴 − 𝑝𝐴𝑖=

1𝑘𝐺

𝐻 𝐶𝐴𝑖 +𝐻 𝐶𝐵∞ /𝜈
𝐻
𝑘°𝐿

=

𝐶𝐵∞

189

)

𝜈
𝑝𝐴 +𝐻 𝐶𝐵∞ /𝜈(16.3-11)

𝐻
1
+
𝑘𝐺 𝑘°
𝐿

(4) Fourth step: Overall mass transfer coefficient 𝐾𝐺
1
1
𝐻
1
0.66
= 𝑘 + 𝑘 ° = 2.37 + 0.553 = 1.62
𝐾
𝐺

𝐺

𝐿

𝐾𝐺 = 0.62 kmol⁄m2 h atm(5) Fifth step: Mass balance
Mass balance can be set up over the packing section between 𝑧 = 0 and 𝑧 to get theoperating-line equation:
2 × (molar rate of SO2 absorbed) = ( molar rate of NaOH consumed)𝑝𝐴𝑝′𝜈 𝐺𝑀 (1− 𝑝 − 1− 𝐴2) = 𝐿𝑀2 𝑥𝐵2 − 𝐿𝑀 (𝐶𝐵∞ ⁄𝑐 )(16.3-12)𝑝𝐴

𝐴2

where 𝑐 is the molar density of the solution in kg/m3.
Due to the very low concentration of ions, the molar velocity 𝐿𝑀 and molar density 𝑐 can beconsidered constant over the whole height of the packing section.
𝑐 ≅

1000
+ 2 = 57.6 kmol⁄m3 at the top
181000+ 0.89 + 0.53 + 0.53 = 57.5 kmol⁄m3 at the bottom
18
𝐿𝑀 ≅ 𝐿𝑀2 = 400 kmol⁄m2 h
Mass balance over the differential volume element 𝑑𝑧 is
𝐻𝐶
𝑝
𝑑𝑝𝐴
′
′
𝜈 𝐺𝑀
= 𝐾𝐺 𝑎 𝑑𝑧 (𝑝𝐴 + 𝜈𝐵∞ )(16.3-13)
𝑑 ( 𝐴 ) = 𝜈 𝐺𝑀
(1− )2
𝑐 ≅

1− 𝑝𝐴

𝑝𝐴

The height of the packing section is given by
𝜈 𝐺′

𝑝

𝑑𝑝𝐴
𝐻 𝐶𝐵∞
𝐴2 (𝑝𝐴 +)(1− 𝑝𝐴 )2
𝜈
′
𝜈 𝐺𝑀

(16.3-14)

𝑍𝑇 = 𝐾 𝑎𝑀 ∫𝑝 𝐴1
𝐺Using the HTU = 𝐾 𝑎 averaged over the packing section from the top through the bottom
𝐺

𝑍𝑇 =

′
𝜈 𝐺𝑀

𝑝𝐴1
𝑑𝑝𝐴
∫
𝐾𝐺 𝑎 𝑝𝐴2 (𝑝𝐴 + 𝐻 𝐶𝐵∞ )(1− 𝑝𝐴 )
𝜈

(16.3-15)

The reaction has the following form of irreversible bimolecular reaction
SO2 + 2 NaOH → Na2 SO3 + H2 O
Therefore 𝜈 = 2
The modified Henry’s constant 𝐻 = 0.66 atm m3 /kmol
and the overall coefficient 𝐾𝐺 = 0.62 kmol⁄m2 h atm
𝐺𝑀 = (𝐺𝑀1 + 𝐺𝑀2 )⁄2 = (40 + 36.2)⁄2 = 38.1 kmol⁄m2 hFirst the bulk concentration of NaOH corresponding to the partial pressure of SO 2 iscalculated at each section by using the operating-line equation:
𝑐
𝑝
𝑝
𝑝
′
[ 𝐴 − 𝐴2 ]} = 2.06 − 10.4 𝐴
𝐶𝐵∞ = 𝐿 {𝐿𝑀2 𝑥𝐵2 − 𝜈 𝐺𝑀
(16.3-16)
1− 𝑝
1− 𝑝
1− 𝑝
𝑀

𝐴

𝐴2

𝐴

Next the equation shown below can be integrated numerically or graphically to get the height ofpacking:
𝑍𝑇 =0.10(2)(38.1 kmol⁄m2 h)
𝑑𝑝𝐴
∫
2
2
3
0.66
𝐶𝐵∞
⁄
⁄
(0.62 kmol m h atm)(46.2 m m ) 0.00552 (𝑝 +) (1 − 𝑝𝐴 )𝐴2′
+ 0.2) = 0.2⁄(36 + 0.20) = 0.00552 atm
Here 𝑝𝐴2 = 0.2⁄( 𝐺𝑀

190

Mass Transfer with Chemical Reaction

𝑝𝐴
0.00552
0.010
0.015
0.020
0.025
0.030
0.035
0.040
0.045
0.050
0.055
0.060
0.065
0.070
0.075
0.080
0.,085
0.090
0.095
0.100
∑

𝐶𝐵∞
2.00
1.955
1.902
1.848
1.793
1.738
1.683
1.627
1.570
1.513
1.455
1.396
1.337
1.277
1.217
1.156
1.094
1.031
0.968
0.904

1⁄(𝑝𝐴 + 0.33 𝐶𝐵∞ )(1 − 𝑝𝐴 )
1.511
1.542
1.580
1.620
1.663
1.708
1.755
1.806
1.860
1.916
1.977
2.043
2.113
2.188
2.268
2.355
2.450
2.554
2.666
2.789

∆𝑝𝐴
(𝑝𝐴 + 0.33 𝐶𝐵∞)(1 − 𝑝𝐴 )11
= ( ) (1.511 + 1.542)(0.01 − 0.00552) + ( ) (1.542 + 1.580)(0.015 − 0.01)
22
1+ ⋯ ⋯ ⋯ + ( ) (2.666 + 2.789)(0.1 − 0.095) = 0.190 atm−1
2

Finally the required height of the packing section is
𝑍𝑇 = (2.66 m atm)(0.190 atm−1 ) = 0.505 m
From a viewpoint of practical engineering design, this result seems to be smaller than
expected. It is usually necessary to take into account the safety factor.

Nomenclature𝑎𝐶𝐴
𝐻
𝐷𝐴𝑚 , 𝐷𝐴𝑤
𝐷𝐵𝑤
𝐺
𝐻𝐺
𝐼𝑗
𝑘1
𝑘𝐿
𝐾𝐺 , 𝑘𝐺
𝐿
𝐿𝑀
𝑁𝐴
𝑁𝐿
𝑝
𝑟𝐴
𝑆
𝑆𝑐
𝑊
𝑡
𝑦
𝑧

effective interfacial area in packing section, [m2/m3]
molar concentration of component A, [kmol/m3]
Henry’s constant, [ - ]diffusivity of component A, [m2/s]
diffusivity of component B, [m2/s]
mass velocity of gas, [kg/m2s]
HTU (Height of Transfer Unit) of gas phase, [m]ionic strength,
reaction-rate constant of first-order reaction, [1/s]liquid-phase mass transfer coefficient, [m/s]overall and gas-phase mass transfer coefficient, [m/s] or [kmol/m2s Pa]
mass velocity of liquid, [kg/m2s]
superficial molar velocity of liquid, [kmol/m2s]
mass flux, [kmol/m2s]
liquid-phase number of transfer units, [ - ]partial pressure, [Pa]rate of first-order reaction, [kmol/m3s]
area of surface perpendicular to mass transfer, [m2]Schmidt number, [ - ]mass flow rate, [kg/s]
time, [s]liquid-side coordinate perpendicular to interface, [m]vertical coordinate of packed column, [m]Nomenclature
𝑍𝑇
𝑧𝑖
𝑧𝑟

height of packed column, [m]number of positive or negative charges, [ - ]position of reaction plane, [m]

𝛼
𝛽
𝜀
𝜃
∞

integration constant, [m2/s]
reaction factor, [ - ]volume fraction of liquid in packing section, [ - ]total holdup time, [s]bulkSubcripts
A
B
f
G
i
L
M

component A
component B
film
gas phaseinterfaceliquid phase
molar

191

CHAPTER 17
TURBULENT TRANSPORT PHENOMENA
17.1 Fundamental Equations of Turbulent Transport
17.1-1 Fundamental properties of turbulent flows
Most flows occurring in practical engineering processes are turbulent. Turbulent flow always
occurs at high Reynolds number. We know that laminar motion in a circular tube tends to become
unstable at Reynolds numbers above about 2,100 in the presence of small disturbances, and atransition to turbulent flow will occur.
One characteristic of turbulent flows is the irregularity: the velocity at any point varies with timein both magnitude and direction. Turbulence is rotational and three dimensional. Fig.17.1-1 is afluctuating curve imitating the real oscillogram of one component of the fluctuating velocities
taken by a hot-wire anemometer placed at a point in the turbulent flowfield. Regarding the principleof hot-wire anemometry, you can refer to Section 9.7-2 of Chapter 9. The static pressure has alsosimilar fluctuations, which can be observed by using a piezo-electric pressure transducer.

Fig.17.1-1. Oscillogram of velocity and pressure of turbulent flow

We can notice from these fluctuating oscillograms that they are not without a certain degree of
regularity from a statistical point of view. The instantaneous velocity 𝑣 can be decomposed into atime-averaged velocity 𝑣̅ and velocity fluctuations 𝑣′, such that𝑣(𝑥, 𝑦, 𝑧, 𝑡) = 𝑣̅ (𝑥, 𝑦, 𝑧) + 𝑣 ′(𝑥, 𝑦, 𝑧, 𝑡)(17.1-1)
The time average is defined by
1 𝑡 +𝑇𝑣̅ = lim 𝑇→∞ 𝑇 ∫𝑡 0 𝑣 𝑑𝑡
(17.1-2)0𝜕𝑣̅

At steady state=0𝜕𝑡
The velocity measured by a pitot tube is approximately equal to the time-averaged velocity, except
for turbulent flows which have extremely high turbulence intensity. Turbulence consists of theFundamental Properties for Turbulent Flows (continued)

193

superposition of various, small periodic motions on large-scale intermittent motions. It is like theorchestral sound consisting of the superposition of various, high-frequency sounds on
low-frequency sounds.
The instantaneous velocity fluctuations can become negative very often.
Therefore the time-averaged value of a fluctuating quantity is zero by definition, for example,
̅ = lim 𝑇→∞ 1 ∫𝑡0 +𝑇(𝑣 − 𝑣̅ )𝑑𝑡 = lim 𝑇→∞ 1 ∫𝑡0 +𝑇 𝑣 𝑑𝑡 − 𝑣̅ = 0𝑣′(17.1-3)
𝑇 𝑡0
𝑇 𝑡0
However the squared values cannot be negative. As a measure of the magnitude of the turbulence,
the intensity of turbulence is defined as√𝑣′2

𝑇𝑢 = 𝑣(17.1-4), which may have values 0.01 to 0.10 in typical turbulent pipe flows.
Of many methods for the measurement of turbulent velocity, the hot-wire anemometer is the mostuseful. As described in Section 9.7-2, the principle of hot-wire anemometry is based on theconvective heat transfer from a very fine heated wire to the approaching stream. The detecting
element consists of a very fine short metal wire (e.g. 5 μm dia. And 5 mm long platinum wire for
air stream), which is heated by an electric current to a constant temperature above the stream
temperature. The wire is placed perpendicular to the velo0city component to be measured. The rateof heat loss to the ambient stream from the wire is proportional to the square root of the stream
velocity √𝑣 in the usual stream condition. The wire is of such low heat capacity that thetemperature of the wire can follow the rapid velocity fluctuations. The rate of heat loss is equal to
the rate of heat generated by the electric current through the wire, 𝐼 2 𝑅, where 𝐼 is the electric
current and 𝑅 the electric resistance of the wire. In the usual measuring method, the electric
resistance is controlled to be kept constant as far as possible by using an electronic feedback system.
Instead, the feedback system changes the current through the wire as soon as a variation in electric
resistance occurs. The response time to the change in approach velocity is of the order shorter than
0.1 ms.
Then we have the relation between 𝐼 2 and 𝑣:
𝐼 2 𝑅 = 𝛼 + 𝛽 √𝑣
(17.1-5)
where 𝛼 and 𝛽 are usually determined by experiment.
If we substitute 𝐼 = 𝐼 + 𝐼′ and 𝑣 = 𝑣 + 𝑣′ into the Eq.(17.1-5), we get the approximate relation
between 𝐼′ and 𝑣′:
2

𝐼 𝑅 = 𝛼 + 𝛽 √𝑣
𝛽√𝑣

𝐼 ′ = 4 𝐼 𝑅 𝑣 𝑣′

(mean velocity)(velocity fluctuation)

The last equation is obtained as the first approximation when 𝑣 ′ ≪ 𝑣, i.e. 𝐼 ′ ≪ 𝐼.
The usual block diagram of hot-wire anemometry is shown in Fig.17.1-2.

Fig. 17.1-2. Block diagram of hot-wire anemometer for constant temperature method(17.1-6)

194

Turbulent Transport Phenomena

Fig. 17.1-3. Distribution of velocity fluctuations and Reynolds stress in a rectangular channel

As shown in Fig.17.1-3, the turbulence intensity in a rectangular channel becomes maximum near
the wall.
The fluctuations √𝑣𝑧 ′2 in the streamwise direction, or in the z direction, are greater than thefluctuations √𝑣𝑥 ′2 in the transverse direction, or in the x direction. In the vicinity of the wall,
molecular momentum transport becomes important but the fluctuations become very small.
Turbulent flow in a pipe also has similar tendency.
Such properties and structure of turbulent flows are described in detail in Sec. 17.2~ 17.5.
17.1-2 Equation of motion for turbulent flows
Turbulence is a continuum phenomenon taking place in a continuum fluid. It is true that theequation of motion applies to turbulent flow, for example,
𝜕(𝜌𝑣𝑥 )
𝜕𝑡+𝜕(𝜌𝑣𝑥 𝑣𝑥 )𝜕𝑥+𝜕(𝜌𝑣𝑥 𝑣𝑦 )𝜕𝑦+𝜕(𝜌𝑣𝑥 𝑣𝑧 )
𝜕𝑧

= −

𝜕𝑝
𝜕𝑥

+ 𝜇(

𝜕2 𝑣𝑥
𝜕𝑥 2

+

𝜕2 𝑣𝑥
𝜕𝑦 2

+

𝜕2 𝑣𝑥
𝜕𝑧 2

)

(17.1-7)

The instantaneous velocity and pressure can be decomposed into a time-averaged value and its
fluctuations, such that𝑣𝑥 = 𝑣
̅̅̅𝑥 + 𝑣𝑥 ′𝑣𝑦 = 𝑣
̅̅̅𝑦 + 𝑣𝑦 ′𝑣𝑧 = 𝑣̅𝑧 + 𝑣𝑧 ′𝑝 = 𝑝̅ + 𝑝′

(17.1-8)

Substituting these instantaneous quantities into Eq.(17.1-7) and averaging the resultant equation
with respect to time, we get the equation of motion governing the mean (time-averaged) flow:
̅̅̅̅
𝜕(𝜌𝑣
̅̅̅𝑥 𝑣
̅̅̅)
𝜕(𝜌𝑣
𝜕(𝜌𝑣
̅̅̅𝑥 𝑣̅̅̅)𝜕(𝜌𝑣̅̅̅𝑥 ̅̅̅̅
𝑣𝑧 )
𝜕𝑝̅
𝜕2 𝑣
̅̅̅𝑥 𝜕2 𝑣
̅̅̅𝑥 𝜕2 𝑣
̅̅̅𝑥
𝑦
𝑥)
𝑥
+
+
+
=−
+ 𝜇( 2 +
+
)
𝜕𝑡
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑥
𝜕𝑥
𝜕𝑦 2
𝜕𝑧 2
−(

̅̅̅̅̅̅̅̅̅)𝜕(𝜌𝑣𝑥 ′𝑣𝑥 ′
𝜕𝑥+̅̅̅̅̅̅̅̅̅)𝜕(𝜌𝑣
𝑥 ′𝑣𝑦 ′𝜕𝑦+̅̅̅̅̅̅̅̅̅̅
𝜕(𝜌𝑣𝑥 ′𝑣𝑧 ′)
𝜕𝑧)

(17.1-9)̅̅̅̅̅)
𝜕(𝜌𝑣

In steady turbulent flow field, 𝜕𝑡 𝑥
In the averaging process, the terms consisting of a product of a mean value and a fluctuation vanishin the following way:𝑣𝑥 𝑣𝑦 = (𝑣𝑥 + 𝑣𝑥′ )(𝑣𝑦 + 𝑣𝑦′ ) = 𝑣𝑥 𝑣𝑦 + 𝑣𝑥 𝑣𝑦′ + 𝑣𝑥′ 𝑣𝑦 + 𝑣𝑥 ′ 𝑣𝑦′ = 𝑣𝑥 𝑣𝑦 + 𝑣𝑥 ′ 𝑣𝑦′

Here the mean value is regarded as a mere constant coefficient as far as the averaging is concerned.
17.1-3 Equations of energy and mass transport for turbulent flows
The equation of energy can also be used for turbulent heat transport with constant properties:

Equation of Motion for Turbulent Flows𝜕(𝜌𝐶𝑝𝑇)𝜕𝑡+𝜕(𝜌𝐶𝑝𝑣𝑥 𝑇)𝜕𝑥+𝜕(𝜌𝐶𝑝𝑣𝑦 𝑇)
𝜕𝑦

+

195
𝜕(𝜌𝐶𝑝𝑣𝑧 𝑇)
𝜕𝑧

𝜕2 𝑇

𝜕2 𝑇

𝜕2 𝑇

= 𝜅 ( 𝜕𝑥2 + 𝜕𝑦 2 + 𝜕𝑧 2 )(17.1-10)

The instantaneous temperature can be decomposed as(17.1-11)

𝑇 = 𝑇 + T′

In similar manner of averaging, we get̅̅̅ 𝜕(𝜌𝐶𝑝𝑣̅̅̅̅
𝜕(𝜌𝐶𝑝𝑇)̅̅̅𝑥 𝑇̅) 𝜕(𝜌𝐶𝑝𝑣
𝜕(𝜌𝐶𝑝𝑣̅𝑧 ̅̅̅𝑇)𝜕2 𝑇̅ 𝜕2 𝑇̅ 𝜕2 𝑇̅
𝑦 𝑇)
+
+
+
= 𝜅 ( 2 +
+
)
𝜕𝑡
𝜕𝑥
𝜕𝑧
𝜕𝑥
𝜕𝑦 2
𝜕𝑧 2
𝜕𝑦
−(

̅̅̅̅̅̅̅
𝜕(𝜌𝐶𝑝𝑣𝑥 ′𝑇′)
𝜕𝑥+̅̅̅̅̅̅̅̅
𝜕(𝜌𝐶𝑝𝑣𝑦 ′𝑇′)𝜕𝑦+̅̅̅̅̅̅̅̅̅
𝜕(𝜌𝐶𝑝𝑣𝑧 ′𝑇′)
𝜕𝑧(17.1-12))̅̅̅𝜕(𝜌𝐶𝑝𝑇)

In steady turbulent temperature field, 𝜕𝑡 = 0
Similarly for turbulent mass transport without chemical reaction

̅̅̅
̅̅̅̅
̅̅̅
̅̅̅
̅̅̅
̅̅̅
𝜕(𝑣
̅̅̅
𝜕 ̅̅̅
𝐶𝐴 𝜕(𝑣
̅̅̅𝑥 𝐶𝜕(𝑣̅𝑧 𝐶
𝜕2 𝐶
𝜕2 𝐶
𝜕2 𝐶
𝑦 𝐶𝐴 )
𝐴)
𝐴)
𝐴
𝐴
𝐴
+
+
+
= 𝐷 ( 2 +
+
)
2
𝜕𝑡
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑥
𝜕𝑦
𝜕𝑧 2
−(

̅̅̅̅̅̅̅̅̅
𝜕(𝑣𝑥 ′𝐶𝐴 ′)
𝜕𝑥+̅̅̅̅̅̅̅̅̅
𝜕(𝑣𝑦 ′𝐶𝐴 ′)𝜕𝑦+̅̅̅̅̅̅̅̅̅̅
𝜕(𝑣𝑧 ′𝐶𝐴 ′)

(17.1-13))𝜕𝑧

In steady turbulent concentration field,𝜕 ̅̅̅̅
𝐶𝐴
𝜕𝑡=0

17.2 Phenomenological Understanding of Turbulent Transport
17.2-1 Effect of Nonlinear interaction of turbulence
In those three equations Eqs.(17.1-6), (17.1-9), and (17.1-10), new terms indicated by underlines
arise.
They are concerned with the turbulent transport by eddy motion coming from nonlinear
interaction of fluctuations. If these nonlinear effects of turbulence were not existent, thetime-averaged distributions of velocity, temperature, and concentration would become the same as
those for laminar transport.
The following group forms nine components of the turbulent momentum-flux tensor:(𝑡)̅̅̅̅̅̅̅̅
𝜏𝑥𝑥 = − 𝜌𝑣
𝑥 ′𝑣𝑥 ′(𝑡)(𝑡)̅̅̅̅̅̅̅̅
𝜏𝑥𝑦 = 𝜏𝑦𝑥 = − 𝜌𝑣
𝑦 ′𝑣𝑥 ′(𝑡)(𝑡)̅̅̅̅̅̅̅
𝜏𝑥𝑧 = 𝜏𝑧𝑥 = − 𝜌𝑣𝑧 ′𝑣𝑥 ′(𝑡)̅̅̅̅̅̅̅̅
𝜏𝑦𝑦 = − 𝜌𝑣𝑦 ′𝑣𝑦 ′(𝑡)(𝑡)̅̅̅̅̅̅̅̅
𝜏𝑦𝑧 = 𝜏𝑧𝑦 = − 𝜌𝑣𝑦 ′𝑣𝑧 ′(𝑡)̅̅̅̅̅̅̅
𝜏𝑧𝑧 = − 𝜌𝑣𝑧 ′𝑣𝑧 ′

(𝑡)
(𝑡)
(𝑡)
These are usually referred to as the Reynolds stresses. The diagonal components 𝜏𝑥𝑥, 𝜏𝑦𝑦 , 𝜏𝑧𝑧
are turbulent normal stresses. In many turbulent flows they contribute little to the momentum
(𝑡)
(𝑡)
(𝑡)
(𝑡)
(𝑡)
(𝑡)
transfer. The off-diagonal components 𝜏𝑥𝑦= 𝜏𝑦𝑥 , 𝜏𝑦𝑧 = 𝜏𝑧𝑦 , 𝜏𝑥𝑧 = 𝜏𝑧𝑥 are turbulent shear stresses.
They play a dominant role in momentum transfer, except in the viscous sublayer near the wall
where molecular transport is predominant.
We know that the distribution of time-averaged velocity for turbulent flow in a circular pipe ismuch flatter in the main flow region than for laminar flow. That fact is due to the Reynolds stresses.
We can interpret the effect of mixing due to eddy motion large in the transverse direction in main
flow region.

The following group forms three components of turbulent heat-flux vector:(𝑡)𝑞𝑥̅̅̅̅̅̅
= − 𝜌𝐶𝑝𝑣
𝑥 ′𝑇′

(𝑡)

𝑞𝑦

̅̅̅̅̅̅̅
= − 𝜌𝐶𝑝𝑣
𝑦 ′𝑇′

(𝑡)

𝑞𝑧

̅̅̅̅̅̅
= − 𝜌𝐶𝑝𝑣
𝑧 ′𝑇′

They play an important role in heat transfer.
Similarly the following group forms three components of turbulent mass-flux vector:(𝑡)𝑗𝑥= − ̅̅̅̅̅̅̅̅
𝑣𝑥 ′𝐶𝐴 ′(𝑡)𝑗𝑦= − ̅̅̅̅̅̅̅̅
𝑣𝑦 ′𝐶𝐴 ′(𝑡)𝑗𝑧= − ̅̅̅̅̅̅̅̅
𝑣𝑧 ′𝐶𝐴 ′196

Turbulent Transport Phenomena

17.2-2 Mixing length theory and eddy diffusivity
A complete understanding of turbulent transport phenomena needs a quantitative description of
turbulence, including the turbulence intensity and the size of eddies. Many direct numericalanalyses based on physical models have been developed. Unfortunately, however, these approaches
are beyond the level of this course.
Historically speaking, the Prandtl’s mixing length model continues to serve as a backgrounduseful for the understanding of turbulent transport. One approach to the solution of turbulent(𝑡)momentum transport problems is to postulate a relation between the turbulent momentum-flux 𝜏𝑥𝑦
and the mean rate of strain (velocity gradient) 𝜕𝑣𝑥 ⁄𝜕𝑦. It is clear that the greater the velocity(𝑡)̅̅̅̅̅̅̅̅
= − 𝜌𝑣
gradient, the larger will be the 𝑣𝑥 ′ induced by a 𝑣𝑦 ′, and hence the larger will be 𝜏𝑦𝑥
𝑦 ′𝑣𝑥 ′.
The random motion of fluid lumps (eddies) in turbulent flow results in turbulent transfer ofmomentum, energy, and mass. In analogy with the mean free path in the gas kinetic theory, theso-called Prandtl mixing length 𝑙 is a measure of the distance a large scale eddy travels before it
breaks up and loses its identity.
Fig. 17.2-1 indicates a model of eddy interchange between two parallel layers set apart by amixing length.

Fig.17.2-1 Mixing length theory

The transverse transport of momentum, energy, and mass is caused as a result of eddy
interchange between two parallel layers as follows:(𝑡)𝜏𝑦𝑥 = 𝜌𝑣𝑥 |𝑦 𝑣𝑦′ − 𝜌𝑣𝑥 |𝑦+∆𝑦 𝑣𝑦′ = − 𝜌𝑣𝑦′ 𝑙𝑀
(𝑡)

𝑑 𝑣𝑥

𝑞𝑦

= 𝜌𝐶𝑝𝑇|𝑦 𝑣𝑦′ − 𝜌𝐶𝑝𝑇|𝑦+∆𝑦 𝑣𝑦′ = − 𝜌𝐶𝑝𝑣𝑦′ 𝑙𝐻(𝑡)= 𝐶𝐴 |𝑦 𝑣𝑦′ − 𝐶𝐴 |𝑦+∆𝑦 𝑣𝑦′ = − 𝑣𝑦′ 𝑙𝐷

𝑗𝑦

𝑑 𝐶𝐴
𝑑𝑦

(17.2-1)𝑑𝑦𝑑𝑇
𝑑𝑦(17.2-2)(17.2-3)

in which the mixing length 𝑙 is defined as the differential distance between the two layers ∆𝑦.
If we assume that 𝑣𝑥 |𝑦+∆𝑦 − 𝑣𝑥 |𝑦 ≅ 𝑣𝑥′ , 𝑇|𝑦+∆𝑦 − 𝑇|𝑦 ≅ 𝑇′, and 𝐶𝐴 |𝑦+∆𝑦 − 𝐶𝐴 |𝑦 ≅ 𝐶𝐴 ′,the turbulent transport fluxes can be obtained as(𝑡)𝜏𝑦𝑥 = − 𝜌𝑣𝑦 ′𝑣𝑥 ′(17.2-4)(𝑡)𝑞𝑦 = − 𝜌𝐶𝑝𝑣𝑦 ′𝑇′(𝑡)𝑗𝑦 = − 𝑣𝑦 ′𝐶𝐴 ′(17.2-5)(17.2-6)

In Eqs. (17.2-1), (17.2-2), and (17.2-3), the products of velocity fluctuation and mixing length𝑣𝑦′ 𝑙𝑀 , 𝑣𝑦′ 𝑙𝐻 , 𝑣𝑦′ 𝑙𝐷 have the unit of diffusivity 𝑚2 ⁄𝑠. Therefore the eddy diffusivities are defined as(𝑡)𝜏𝑦𝑥 = − 𝜌𝜀𝑀

𝑑 𝑣𝑥

𝑑𝑦
𝑑𝑇
(𝑡)
𝑞𝑦 = − 𝜌𝐶𝑝𝜀𝐻
𝑑𝑦
(𝑡)

𝑗𝑦

= − 𝜀𝐷𝑑 𝐶𝐴
𝑑𝑦(17.2-7)(17.2-8)(17.2-9)

They are not a property of the fluid but vary with the flow conditions. They are much greater thanmolecular diffusivities ν, 𝛼, 𝐷𝐴𝐵 in the turbulent flow region. These eddy diffusivities 𝜀𝑀 , 𝜀𝐻 , 𝜀𝐷
are not only a property of fluids but a flow parameter as well. When molecular motion still prevails,

Mixing Length Theory and Eddy Diffusivity (continued)

197

the total momentum-flux can be expressed as
(𝑙)

(𝑡)

𝜏𝑦𝑥 = 𝜏𝑦𝑥 + 𝜏𝑦𝑥 = − 𝜌(𝜈 + 𝜀𝑀 )Similarly(𝑙)(𝑡)𝑞𝑦 = 𝑞𝑦 + 𝑞𝑦
(𝑙)

(𝑡)

𝑗𝑦 = 𝑗𝑦 + 𝑗𝑦

𝑑 𝑣𝑥
𝑑𝑦

= − 𝜌𝐶𝑝(𝛼 + 𝜀𝐻 )= −(𝐷𝐴𝐵 + 𝜀𝐷 )

𝑑𝑇

𝑑𝑦
𝑑 𝐶𝐴
𝑑𝑦

(17.2-10)(17.2-11)(17.2-12)

In a fully-developed pipe flow, the shear stress distribution is linear whether the flow islaminar or turbulent:
𝜏𝑟𝑧 =

𝑟

𝑅

(17.2-13)

𝜏𝑤

Therefore the eddy diffusivity 𝜀𝑀 can be evaluated from the measured velocity distribution byusing the following equation:
𝜀𝑀

=
𝜈

𝑟
𝑅
𝑑𝑣
−𝜇 𝑧
𝑑𝑟

𝜏𝑤

−1

(17.2-14)

Fig. 17.2-2 shows the radial distribution of eddy diffusivity in the fully developed flow region ina circular pipe. These curves are schematically drawn taking into consideration many empirical data.
The maximum of 𝜀𝑀 ⁄𝜈 occurs at the intermediate position between the axis and wall. This may be
due to the fact that the turbulent eddy motion is produced there. As the axis is approached, 𝜀𝑀 ⁄𝜈decreases from the maximum value. In the neighborhood of the pipe axis, it is too difficult todetermine the eddy diffusivity owing to the fact that both the denominator and numerator of Eq.
(17.2-14) become almost zero. Still 𝜀𝑀 ⁄𝜈 is much larger than unity, except very near the pipe wall.
We should notice that these 𝜀𝑀 ⁄𝜈 distribution curves resemble the turbulence intensity curves inFig. 7.1-3. However it should be kept in mind that the turbulent kinetic energy is generated muchcloser to the pipe wall.

Fig. 17.2-2. Radial distribution of eddy diffusivity for momentum in a circular tube flow

17.2-3 Mixing length model based on turbulence correlation
We have already understood a qualitative explanation of the generation of turbulent motion in
Chapter 3. At this section, let us consider the concept of “mixing length” again by using the more
rigorous theory of turbulence.
Consider a traveling fluid element which starts from a level 𝑦 = 0 at time 𝑡 = 0 and passes anarbitrary level 𝑦 at time 𝑡. Figure 18.2-3 is the turbulent velocity field for explanation of themixing length theory.

198

Turbulent Transport Phenomena

Fig. 17.2-3. Mixing length model based on turbulence correlation

If the fluid element does not lose its original momentum 𝜌 𝑣𝑥 (0, 0) as it moves in the transverse
direction, a momentum deficit at 𝑦 and 𝑡 is
Δ𝑀 = 𝜌 𝑣𝑥 (𝑦, 𝑡) − 𝜌 𝑣𝑥 (0, 0) = 𝜚(𝑣𝑥 (𝑦) − 𝑣𝑥 (0)) + 𝜚(𝑣𝑥′ (𝑦, 𝑡) − 𝑣𝑥 ′(0, 0))(17.2-15)

Since the intensity of turbulence in many turbulent flows has values in the neighborhood of 1 to
10%, the second term is small compared to the first one:
Δ𝑀 ≅ 𝜚(𝑣𝑥 (𝑦) − 𝑣𝑥 (0))

For small values of 𝑦, the velocity difference can be approximated by the first term of the Taylor
series:Δ𝑀 ≅ 𝜚[𝑣𝑥 (𝑦) − 𝑣𝑥 (0)] = 𝜌

𝑑 𝑣𝑥

|

𝑑𝑦 𝑦=0𝑦(17.2-16)

The volume of the fluid transported per unit area and unit time is the transverse component 𝑣𝑦 ′ of
the instantaneous velocity of the traveling fluid element.
From Lagrangian viewpoint
𝑣𝑦′ =

𝑑𝑦

(17.2-17)𝑑𝑡

where 𝑦 is the position of the fluid element. Since the momentum flux has dimensions of(momentum/volume)(volume/area/time), the average turbulent momentum-flux at 𝑦 = 0 can be
expressed as
(𝑡)

𝜏𝑦𝑥 |

𝑦=0

= ∆𝑀 𝑣𝑦 ′ = 𝜌

𝑑𝑣
1
𝑑𝑣
𝑑 𝑦2
𝑑𝑦
= 𝜌 𝑥|
𝑣 ′𝑦 = 𝜌 𝑥 |𝑦 |||𝑑𝑦 𝑦=0 𝑦
𝑑𝑦 𝑦=0𝑑𝑡 𝑦=02
𝑑𝑦 𝑦=0 𝑑𝑡 𝑦=0

𝑑 𝑣𝑥(17.2-18)

The overbar denotes an average taken over a large number of moving fluid elements starting from𝑦 = 0.

Fig. 17.2-4 Significance of length scale and time scale

Mixing Length Theory Model Based on Turbulence Correlation (continued)

199

Therefore we can assume that the correlation 𝑣𝑦 ′𝑦 becomes essentially zero at a distance 𝑦comparable to a transverse length scale 𝑙 . If the RMS velocity √𝑣𝑦 ′2 is denoted by 𝑢𝑦 ′, thecorrelation 𝑣𝑦 ′𝑦 is of order 𝑢𝑦 ′𝑙:𝑣𝑦 ′𝑦 = 𝐾1 𝑢𝑦 ′𝑙This transverse length scale is called “mixing length.”
can be written as(17.2-19)
Therefore the turbulent momentum-flux

𝑑 𝑣𝑥
𝑑𝑦

(17.2-20)
Usually the mixing length 𝑙 absorbs the unknown numerical coefficient because it is of order one:
𝐾1 𝑙 = 𝑙𝑀 .
If the mixing length 𝑙 and the RMS velocity turbulence 𝑢𝑦 ′ were known over the entire flow field,
the equation of motion could be solved. The mixing length can be considered to be proportional to
the size of the larger eddies which play a predominant role in the turbulent momentum transport.
The distance 𝑦 traveled for time 𝑡 by the fluid element is𝑡
𝑦 = ∫0 𝑣𝑦′ (𝑡 ′)𝑑𝑡′(17.2-21)
The correlation 𝑣𝑦 ′𝑦 becomes(𝑡)𝜏𝑦𝑥 = 𝐾1 𝜌𝑢𝑦′ 𝑙𝑡𝑡(17.2-22)
The averaging is performed with respect to the number of the moving fluid elements starting from𝑣𝑦 ′(𝑡)𝑦 = 𝑣𝑦′ (𝑡) ∫0 𝑣𝑦′ (𝑡 ′ )𝑑𝑡 ′ = ∫0 𝑣𝑦′ (𝑡)𝑣𝑦′ (𝑡 ′ )𝑑𝑡 ′𝑦 = 0.

The correlation 𝑣𝑦′ (𝑡)𝑣𝑦′ (𝑡 ′ ) does not depend on the reference time but on the time difference𝑡 − 𝑡 ′ = 𝜏.
Then the correlation becomes a Lagrangian correlation coefficient:
𝑅𝐿2 (𝜏) =𝑣𝑦′ (𝑡)𝑣𝑦′ (𝑡− 𝜏)
𝑢𝑦 ′2(17.2-23)

The Lagrangian integral time scale ℑ2 can be regarded as a measure of the time interval which islong enough for fluid elements to travel the length scale (mixing length) 𝑙:∞(17.2-24)ℑ2 = ∫0 𝑅𝐿2 (𝜏) 𝑑𝜏
The Lagrangian correlation implies that as the time interval 𝜏 becomes large, the velocity
fluctuation 𝑣𝑦 ′(𝑡) does not resemble 𝑣𝑦 ′(𝑡 − 𝜏) in wave form, that is, the degree of correlation
between 𝑣𝑦 ′(𝑡) and 𝑣𝑦 ′(𝑡 − 𝜏) falls.
Then the turbulent momentum-flux is expressed as
∞
∞
𝑑𝑣
𝑑𝑣
𝑑𝑣
(𝑡)
𝜏𝑦𝑥 = 𝜌 𝑥 ∫0 𝑣𝑦′ (𝑡)𝑣𝑦′ (𝑡 − 𝜏)𝑑𝜏 = 𝜌 𝑥 ∫0 𝑢𝑦 ′2 𝑅𝐿2 (𝜏)𝑑𝜏 = 𝜌 𝑥 𝑢𝑦 ′2 ℑ2(17.2-25)𝑑𝑦𝑑𝑦𝑑𝑦
A Lagrangian integral length scale 𝑙𝐿 is used as𝑙𝐿 = 𝑢𝑦′ ℑ2(17.2-26)
On the other hand, the Eulerian integral scale 𝑙 is defined by∞𝑢𝑦 ′2 𝑙 = ∫0 𝑣𝑦 ′(𝑦)𝑣𝑦 ′(0) 𝑑𝑦(17.2-27)
The overbar denotes an average taken over a long time interval with zero time delay between thetwo velocity fluctuations. If the 𝑙𝐿 and 𝑙 are assumed of the same order of magnitude,𝑑𝑣(𝑡)𝜏𝑦𝑥 = 𝐾𝑀 𝜌 𝑢𝑦′ 𝑙 𝑥(17.2-28)𝑑𝑦This suggests that the turbulent momentum-flux can be estimated using the Eulerian integral scale 𝑙.
From the definition of Reynolds stresses(𝑡)𝜏𝑦𝑥 = 𝜌 𝑣𝑦 ′𝑣𝑥 ′(17.2-29)
Comparing these equations, it is found that𝑑𝑣𝑢𝑥′ = 𝐾 𝑙 𝑥(17.2-30)𝑑𝑦

We can obtain similar expressions of turbulent heat- and mass-fluxes:

200

Turbulent Transport Phenomena(𝑡)𝑞𝑦 = 𝐾𝐻 𝜌𝐶𝑝 𝑢𝑦′ 𝑙(𝑡)𝑗𝑦 = 𝐾𝐷 𝑢𝑦′ 𝑙𝑑𝑇𝑑𝑦(17.2-31)

𝑑𝐶
𝑑𝑦(17.2-32)

The numerical coefficients 𝐾𝑀 , 𝐾𝐻 , 𝐾𝐷 are of order one.
Finally it has been found that the eddy diffusivities are of order 𝑢𝑦′ 𝑙:𝜀𝑀 = 𝑢𝑦′ 𝑙𝑀(17.2-33)′
𝜀𝐻 = 𝑢𝑦 𝑙𝐻(17.2-34)′
𝜀𝐷 = 𝑢𝑦 𝑙𝐷(17.2-35)
The ratio of the Reynolds stress to the laminar viscous stress is usually very large in most part of theflow field except near the wall.
(𝑡)

𝜏𝑦𝑥
(𝑙)
𝜏𝑦𝑥

=

𝑑 𝑣𝑥
𝑑𝑦
𝑑 𝑣𝑥
𝜌𝜈
𝑑𝑦

𝜌𝜀𝑀

=

𝜀𝑀
𝜈

= 𝐾1

′ 𝑙
𝑢𝑦

(17.2-36)

𝜈

The term 𝑢𝑦 ′𝑙⁄𝜈 is sometimes called “eddy Reynolds number.”

17.3 Structure of Turbulence
17.3-1 Energy spectrum of kinetic energy
It will be instructive to find the regularity in seemingly-chaotic turbulent flows by a statistical
approach. If the statistical properties such as 𝑣𝑥 ′2, 𝑣𝑥 ′𝑣𝑦 ′, etc. are time-independent, the turbulent
flow is characterized as being stationary. If the statistical properties do not vary in space, it is said
that the turbulent flow is homogeneous. If the statistical properties are invariant with direction, it issaid that the turbulent flow is isotropic: 𝑣𝑥 ′2 = 𝑣𝑦 ′2 = 𝑣𝑧 ′2.
It is well known that the turbulent flow behind a uniform grid has a turbulent structure close to
locally homogeneous isotropic turbulence. Turbulent flow in a circular pipe, especially that near to
the wall, is neither homogeneous nor isotropic.
A constant average value of 𝑣′2 exists in the flowfield statistically homogeneous with respect totime.
The kinetic energy of turbulence is expressible as1
𝜌 (𝑣𝑥 ′2 + 𝑣𝑦 ′2 + 𝑣𝑦 ′2 )2To describe turbulent motion quantitatively, it is necessary to introduce a concept of the intensity
and scale of turbulence. The intensity of the turbulent velocity fluctuations is defined as the
root-mean-square value divided by the mean flow velocity:
𝑇𝑢𝑣 =√𝑣𝑥 ′2𝑣𝑥(17.3-1)

We usually use the velocity fluctuation component in the same direction as the main flow.
Turbulence consists of the superposition of various, small periodic motions on large-scale motions.
The small periodic motions come from a certain vortex motion, the extent of which is called “eddy.”
In other words, turbulence consists of the superposition of eddies of various sizes. The eddy size
corresponds to the frequency of the periodic motion. The smaller an eddy, the higher the frequency.
As shown in Fig. 17.3-1, a large scale eddy consists of a large number of small eddies. All eddies
are moving approximately at the average velocity 𝑣𝑥 . They can be considered to be transported by
the mean flow. When a hot-wire is placed perpendicular to the velocity 𝑣𝑥 , the time required for asmall eddy to pass by the wire is shorter than that for a large eddy. Each amplitude of theoscillations corresponds to the turbulent kinetic energy each eddy has.

Energy Spectrum of Kinetic Energy (continued)

201

Fig. 17.3-1. Structure of a large-scale eddy containing various-sized small eddy motions

Fig. 17.3-2 explains the relation between eddy size and fluctuation frequency. An energy spectrum
describes the distribution of the kinetic energy 𝑣𝑥 ′2 of the turbulence in the frequency domain. This
analysis of turbulence is similar to the frequency analysis applied to light waves and sound waves.

Fig. 17.3-2. Relation between eddy size and fluctuation frequency

Fig. 17.3-3. Schematic curve of energy spectrum

202

Turbulent Transport Phenomena

Fig. 17.3-3 is a one-dimensional energy spectrum, where the distribution function 𝐸1 (𝑛) is similar
to a probability density function.
The contribution to 𝑣𝑥 ′2 of the frequencies between 𝑛 and 𝑛 + 𝑑𝑛 is given by 𝑣𝑥 ′2 𝐸1 (𝑛)𝑑𝑛.
∞
By definition, ∫0 𝐸1 (𝑛)𝑑𝑛 = 1
The one-dimensional energy spectrum can be measured by a hot-wire anemometer. Signals from thewire are time-averaged to get 𝑣𝑥 and the original signals 𝑣𝑥 are subtracted 𝑣𝑥 to get 𝑣𝑥 ′. Then thefluctuations 𝑣𝑥 ′ are passed through a filter circuit which can filtrate the oscillations between 𝑛 and
𝑛 + 𝑑𝑛. The filtrated oscillations are squared and time-averaged to get 𝑣𝑥 ′2 𝐸1 (𝑛).
At the present time, a digital computer is often used for the statistical analyses of this kind. Signalsof fluctuating velocities are first digitalized by an AD converter and stored as a time history into thememory of the computer. The energy spectrum can be obtained from the Fourier transforms of thefluctuating velocity vs. time record.
Turbulence is dissipative. The smaller an eddy, the greater the velocity gradient within the eddy.
Therefore the viscous shear stress counteracts the very high frequency eddying motion. There is astatistical lower limit to the size of the smallest eddy. The highest frequency range in the energy
spectra corresponds to the smallest eddies. Even the smallest eddies are far larger than the mean
free path of the molecules of the fluid. That is, turbulence is a continuum phenomenon. The kinetic
energy of turbulence is always converted into the internal energy (heat) of the fluid owing to thedeformation work by viscous shear stresses in the smallest eddies.
Therefore turbulence needs a continuous supply of energy to make up for the kinetic energy loss.
The kinetic energy is extracted from the mean flow by large eddies which correspond to low
frequency range in energy spectra. The generation of small-scale fluctuations is due to thenonlinear interaction of large-scale fluctuations.
Suppose two components of velocity fluctuations to be of the form:′
𝑣𝑥 = ∑𝑚 𝐴𝑚 cos 2𝜋𝑚𝑡
(17.3-2)
𝑣𝑦′ = ∑𝑛 𝐵𝑛 cos 2𝜋𝑛𝑡
(17.3-3)
The product of these two functions implies the nonlinear interaction as follows:𝑣𝑥′ 𝑣𝑦 ′ = ∑𝑚 ∑𝑛 𝐴𝑚 𝐵𝑛 cos 2𝜋𝑚𝑡 cos 2𝜋𝑛𝑡
(17.3-4)
1= ∑𝑚 ∑𝑛 2 𝐴𝑚 𝐵𝑛 (cos 2𝜋 (𝑚 + 𝑛)𝑡 + cos 2𝜋(𝑚 − 𝑛)𝑡)
Here the higher frequency fluctuations cos 2𝜋(𝑚 + 𝑛)𝑡 and the lower frequency fluctuationscos 2𝜋(𝑚 − 𝑛)𝑡 are produced due to the nonlinear interactions cos 2𝜋𝑚𝑡 and cos 2𝜋𝑛𝑡. In the casewhen 𝑚 = 𝑛, the time-average lower frequency component cos 2𝜋(𝑚 − 𝑛)𝑡 implies the effect ofvelocity fluctuations on the mean motion:(𝑚 ≠ 𝑛 )𝑣𝑥′ 𝑣𝑦 ′ = 01= ∑𝑛 2 𝐴𝑛 𝐵𝑛(𝑚 = 𝑛)(17.3-5)This suggests that the mean velocity distribution is distorted owing to the nonlinear interaction. As
shown in Fig. 17.3-4, in this manner, a spectral energy transfer from the energy-containing eddies(large eddies) to smaller eddies occurs. Finally the viscous dissipation occurs by the smallest eddies
approximately at the same rate as the production of the kinetic energy. This also suggests that thestructure of turbulence is in dynamical equilibrium in which local inputs (production) of kinetic
energy balance local losses of kinetic energy.
Therefore we need a measure of eddy size and often determine the extent of the flowfield where thefluid motion (e.g. velocities) have a certain correlation with each other.

Spatial and Temporal Correlations---

Definition of Eddy Sizes

203

Fig. 17.3-4. Spectral energy transfer by nonlinear interaction of velocity fluctuations

17.3-2 Spatial and temporal correlations ----- Definition of eddy sizes
The size and orientation of eddies can be defined with the aid of the spatial correlation between
fluctuations measured simultaneously at two positions.
From the Eulerian point of view, a longitudinal velocity correlation coefficient 𝑓(𝑟) is defined
between the longitudinal velocity fluctuation components measured at two points, the distance of
which is 𝑟:
𝑓 (𝑟 ) =𝑣𝑥 ′(𝑥)𝑣𝑥 ′(𝑥+𝑟)(17.3-6)√𝑣𝑥 ′(𝑥)2 √𝑣𝑥 ′(𝑥+𝑟)2

In homogeneous turbulent flowfield√𝑣𝑥 ′(𝑥)2 = √𝑣𝑥 ′(𝑥 + 𝑟)2Then𝑓 (𝑟 ) =𝑣𝑥 ′(𝑥)𝑣𝑥 ′(𝑥+𝑟)𝑣𝑥 ′2

Fig.17.3-5 is a qualitative sketch of longitudinal velocity correlation.

Fig. 17.3-5. Longitudinal velocity correlation(17.3-7)

204

Turbulent Transport Phenomena

Similarly a lateral velocity correlation coefficient 𝑔(𝑟) is defined between the lateral velocity
fluctuation components measured at two points, the distance of which is 𝑟:
𝑔 (𝑟 ) =𝑣𝑦 ′(𝑥)𝑣𝑦 ′(𝑥+𝑟)𝑣𝑦 ′2(17.3-8)

Fig.17.3-6 is a sketch of lateral velocity correlation.

Fig.17.3-6. Lateral velocity correlation

These space correlations can be measured by using two sets of hot-wire anemometers. One
hot-wire is fixed at a position 𝑥 and the other is placed at any arbitrary position 𝑥 + 𝑟. Signals𝑣𝑥 (𝑥) and 𝑣𝑥 (𝑥 + 𝑟) from the two hot-wires are subtracted their time-averaged values𝑣𝑥 (𝑥) and 𝑣𝑥 (𝑥 + 𝑟) to get their fluctuations 𝑣𝑥 ′(𝑥) and 𝑣𝑥 ′(𝑥 + 𝑟). Their fluctuations are multipliedeach other with zero time delay and averaged with respect to time to get the correlation𝑣𝑥 ′(𝑥)𝑣𝑥 ′(𝑥 + 𝑟). If the two fluid elements are very close together, they must be moving at nearly thesame velocities. That is, there is a high degree of correlation between the two fluid elements.
From definition, 𝑓(0) = 1 and g(0) = 1
As the distance 𝑟 increases, the correlation becomes low. We often need a measure of the size oflarger eddies which contribute very much to convective transfer of momentum, heat, and mass. In
particular, the surface renewal motion due to those large-scale eddies in the neighborhood of theinterface is very effective in interphase heat and mass transport.
As the Eulerian macroscale of turbulence, two integral length scales are defined:∞
Λ𝑓 = ∫0 𝑓(𝑟)𝑑𝑟(17.3-9)∞Λ𝑔 = ∫0 𝑔(𝑟)𝑑𝑟(17.3-10)
They are considered to be a measure of the largest space interval in which two velocity fluctuations
are correlated with each other, i.e., the fluid elements at two points move substantially in the same
direction.
From Lagrangian point of view, another correlation coefficient can be defined between velocity
fluctuations of the same fluid particle at different times:

𝑅𝐿 (𝜏) =𝑣𝑥 ′(𝑥,𝑦,𝑧,𝑡)𝑣𝑥 ′(𝑥,𝑦,𝑧,𝑡+𝜏)𝑣𝑥 ′(𝑥,𝑦,𝑧)2This is known as the Lagrangian autocorrelation coefficient.

(17.3-11)

Spatial and Temporal Correlations---

Definition of Eddy Sizes

205

Fig. 17.3-7. Lagrangian autocorrelation

Fig. 17.3-7 is a sketch of Lagrangian autocorrelation. This autocorrelation can be calculated withthe aid of time-delay circuit from the fluctuations measured by a hot-wire anemometer.
A Lagrangian integral time scale is defined by∞ℑ = ∫0 𝑅𝐿 (𝜏) 𝑑𝜏
(17.3-12)
This can be considered to be a measure of the longest time interval during which 𝑣𝑥 ′(𝑡) iscorrelated with itself, i.e., a fluid element persists in a motion in a given direction.
If a homogeneous turbulent flowfield has a constant mean velocity 𝑣𝑥 ,
Λ𝑓 = ℑ 𝑣𝑥(17.3-13)
These macroscales of turbulence are of great importance in connection with turbulent transport ofmomentum, heat, and mass.

17.4 Velocity Distribution of Turbulent Flow inside a CircularPipe
Let us derive the turbulent velocity distribution in the fully-developed region.
The total shear stress has a linear distribution whether the flow is laminar or turbulent:𝑟(17.4-1)
We may use three layer concept although the velocity change in the radial direction is continuous:
the viscous sublayer, where viscous effect is predominant; the buffer layer, where the viscous andturbulent effects are of comparable importance; and the turbulent core, where purely viscous effectis of negligible importance. It can be considered that the time-smoothed velocity at a point is afunction of local conditions from the wall of the pipe. Therefore the distance from the wall𝑦 = 𝑅 − 𝑟 is used instead of 𝑟 as the transverse coordinate.

𝜏𝑟𝑧 =

𝜏
𝑅 𝑤

[EXAMPLE 17.4-1] Dimensional Analysis
Show that the time-smoothed velocity distribution in turbulent shear layer should be a functionof two dimensionless groups by application of dimensional analysis:𝑣𝑥𝑦 √𝜏 ⁄𝜌=𝑓( 𝑤 )
(17.4-E1)
⁄
𝜈
√𝜏 𝜌
𝑤

Solution: If we denote characteristic velocity and length by 𝑈 and 𝑌, respectively, we canformally express the velocity distribution as
𝑣𝑥
𝑦
=𝑓( )
(17.4-E2)
𝑈𝑌
The time-smoothed velocity is a function of local conditions from the nearest wall. Thus𝑦
𝑀 𝑎 𝑀 𝑏𝑀 𝑐
= 𝑦 𝜏𝑤 𝑎 𝜚𝑏 𝜇𝑐 = 𝐿 ( 2 ) ( 3 ) ( ) = 𝐿1−𝑎−3𝑏−𝑐 𝑀𝑎+𝑏+𝑐 𝑇 −2𝑎−𝑐
𝑌
𝐿𝑇
𝐿
𝐿𝑇

(17.4-E3)

206

Turbulent Transport Phenomena

In order that the right side of Eq.(17.3-E3) should be dimensionless, 𝑎 = 1⁄2 , 𝑏 = 1⁄2 ,and 𝑐 = −1.
This implies𝑦=𝑌𝑦 √𝜏𝑤 ⁄𝜚(17.4-E4)= 𝑦+𝜈

Similarly𝑣𝑥
𝑈(17.4-E5)= 𝑣𝑥 𝜏𝑤 𝐴 𝜚 𝐵 𝜇𝐶 = 𝐿1−𝐴−3𝐵−𝐶 𝑀𝐴+𝐵+𝐶 𝑇 −1−2𝐴−𝐶

The right side should be dimensionless. Then A = − 1⁄2, B= 1⁄2 , and 𝐶 = 0.
This implies
𝑣𝑥
𝑈

=

𝑣𝑥
√𝜏𝑤 ⁄𝜚

(17.4-E6)= 𝑢+

As a result
𝑢+ = 𝑓(𝑦 + )(17.4-E7)
We have found that the velocity distribution should be expressed using the above functional
form.
Viscous sublayer
In the region very close to the pipe wall, the shear stress will vary only slightly from 𝜏𝑤 andthe turbulent effect is negligibly small. Thus𝑑𝑣𝜏𝑤 = 𝜚 𝜈 𝑧(17.4-2)𝑑𝑦
Rearranging the terms d𝑣𝑧 = 𝜚𝑤𝜈 𝑑𝑦 and integrating in dimensionless form,𝜏𝑣𝑧√𝜏𝑤 ⁄𝜚

=𝑦 √𝜏𝑤 ⁄𝜚𝜈

+ 𝐶1Using the boundary condition that 𝑣𝑧 = 0 at 𝑦 = 0,
We get the velocity distribution in the viscous sublayer:

𝐶1 must be zero.

𝑢+ = 𝑦 +(17.4-3)

The term √𝜏𝑤 ⁄𝜚, which has the dimension of velocity, is called the friction velocity.
The friction velocity is denoted by 𝑢∗ .
Buffer layer
In this layer, the laminar and turbulent effects are both important. We cannot derive analyticallythe velocity distribution. One of empirical equations is
𝑢+ = −3.05 + 5.00 ln 𝑦 +(17.4-4)
Turbulent core
In this region, the turbulent effect is predominant. The total shear stress can be written as𝑑𝑣(𝑡)𝜏𝑟𝑧 = 𝜏𝑟𝑧 = 𝜚 𝑢𝑦′ 𝑙𝑀 𝑧
(17.4-5)𝑑𝑦′Assuming that the RMS velocity 𝑢𝑦 in the transverse direction is proportional to the RMS velocity
𝑢𝑧′ in the axial direction:𝑢𝑦′ = 𝐶2 𝑢𝑧′(17.4-6)
For small 𝑦,𝑑 𝑣𝑧
𝑢𝑧′ ≅ 𝐾 𝑙𝑀
(17.4-7)𝑑𝑦Then we obtain(𝑡)𝜏𝑟𝑧 = 𝜏𝑟𝑧 = 𝜚 𝑙𝑀 2 (𝑑 𝑣𝑧 2)𝑑𝑦(17.4-8)

The numerical coefficients 𝐶2 and 𝐾 are absorbed in the unknown mixing length 𝑙𝑀 .
From the shear stress distribution𝜏𝑟𝑧 = 𝜏𝑤 (

𝑅−𝑦
)=
𝑅

𝜚 𝑙𝑀 2 (𝑑 𝑣𝑧 2)𝑑𝑦

The equation of motion for turbulent pipe flows can also be written as(17.4-9)

Velocity Distribution of Turbulent Flow inside a Circular Pipe (continued)

0=𝑃0 − 𝑃𝐿
𝐿

1 𝑑

207

(17.4-10)− 𝑟 𝑑𝑟 (𝑟 𝜏𝑟𝑧 )(𝑙)(𝑡)where 𝜏𝑟𝑧 = 𝜏𝑟𝑧+ 𝜏𝑟𝑧
Integrating with the boundary condition𝜏𝑟𝑧 =𝜏𝑟𝑧 = 0 at 𝑟 = 0,(𝑃0 − 𝑃𝐿 )𝑅 𝑟
𝑟
= 𝜏𝑤
2𝐿
𝑅
𝑅

That is(𝑃 − 𝑃 )𝑅𝜏𝑤 = 0 𝐿(17.4-11)
2𝐿This implies that the frictional force at the wall balances the pressure force acting on the flowcross-section whether the flow is laminar or turbulent. It is known that the turbulent velocity profileis almost flat in most part of the flow cross-section and that the steep velocity change takes placenear the wall.
Then Prandtl assumed that 𝜏𝑟𝑧 do not vary markedly from its wall value 𝜏𝑤𝜏𝑟𝑧 ≅ 𝜏𝑤 = 𝜚 𝑙𝑀 2 (

𝑑 𝑣𝑧
)
𝑑𝑦

2

(17.4-12)

This assumption made in the turbulent core should be considered as a very rough estimation.
Further, he assumed that the mixing length 𝑙𝑀 is simply proportional to the distance from the wall,that is𝑙𝑀 = 𝜅 𝑦(17.4-13)
Thus the equation to be solved becomes
𝜏𝑤

𝑑𝑣

2

= 𝜅 2 𝑦 2 ( 𝑑𝑦𝑧 )𝜌(17.4-14)𝑢∗(17.4-15)

Taking the square root
𝑑 𝑣𝑧
𝑑𝑦

=

𝜅𝑦

Integrating from the outer edge of the buffer layer 𝑦𝑏 ,1
𝑦𝑣𝑧 − 𝑣𝑧,𝑏 = 𝜅 𝑢∗ ln 𝑦𝑏(17.4-16)

In dimensionless form

𝑢+ − 𝑢𝑏+ =

1
𝜅

𝑦+

ln 𝑦+
𝑏

(17.4-17)

We may expect that experimental turbulent velocity profiles will give a universal curve in 𝑢+ , 𝑦 +coordinates that is logarithmic over the turbulent core and linear immediately adjacent to the pipewall.
The universal velocity profile empirically obtained is expressed by
Viscous sublayer𝑦+ ≤ 5𝑢+ = 𝑦 +(17.4-18)+++
Buffer layer
5 ≤ 𝑦 ≤ 30
𝑢 = −3.05 + 5.0 ln 𝑦(17.4-19)
Turbulent core𝑦 + ≥ 30
𝑢+ = 5.5 + 2.5 ln 𝑦 +(17.4-20)
From many experimental data, it is known that the universal constant 𝜅 is given by 0.4. The
above semi-empirical universal velocity distribution law is valid over a wide range 𝑅𝑒 > 20,000
in the magnitude of time-smoothed velocity. However the three-layer model involves discontinuities
in eddy diffusivity 𝜀𝑀 and non-zero velocity gradient at the pipe axis because the model does nottake into consideration the continuity in velocity gradients at the boundaries of these three layers.
In particular, Eq.(17.4-20) cannot meet the real boundary condition 𝜕𝑣𝑧 ⁄𝜕𝑟 = 0 at 𝑟 = 0.
[PROBLEM 17.4-P1] As a set of universal velocity distribution for a circular tube flow, Eqs.
(17.4-18~20) were obtained. Plot the dimensionless velocity against the dimensionless distancefrom the wall 𝑢+ vs. log 𝑦 + in a semi-logarithmic coordinates. A Newtonian fluid( 𝜌 = 1,000 kg⁄m3 , 𝜇 = 0.0012 Pa s) is flowing in at an average velocity 〈𝑣𝑧 〉 = 1.5 m⁄s in acircular tube (inside diameter 𝐷𝑖 = 1.0 m). What is the friction factor? What is the wall shear
stress 𝜏𝑤 ? In this fully-developed turbulent flow region, obtain a radial distribution of thetime-averaged axial velocity 𝑣𝑧 ⁄〈𝑣𝑧 〉 vs. 𝑟⁄𝑅.

208

Turbulent Transport Phenomena

17.5 Turbulent Structure and Role of Eddies
17.5-1 Turbulent structure in a circular pipe flow
Let us study again a turbulent flow in a circular pipe from a viewpoint of chemical engineering
processes. Fig. 17.5-1 shows the radial distributions of the turbulence intensity and the Reynolds
stress.

Fig. 17.5-1. Schematic picture of radial distributions of turbulence intensity and Reynolds stress

The turbulence intensity of the axial component velocity fluctuations 𝑣𝑧 ′2 is much larger than thoseof other components 𝑣𝜃 ′2 and 𝑣𝑟 ′2 in the main part of the flowfield and becomes a maximum inthe vicinity of the pipe wall. However, as the pipe axis is approached, the flow becomes nearly
isotropic: 𝑣𝑧 ′2 = 𝑣𝜃 ′2 = 𝑣𝑟 ′2.
The Reynolds stress also shows a maximum in the vicinity of the wall. The buffer layer can beconsidered as a constant source for the turbulent kinetic energy. The turbulent kinetic energy 𝑣𝑧 ′2
is produced from the gradient of the mean velocity 𝜕𝑣𝑧 ⁄𝜕𝑟 through the Reynolds stress − 𝑣𝑟 ′𝑣𝑧 ′mainly in the buffer layer (𝑦 + ≅ 15), where both 𝜕𝑣𝑧 ⁄𝜕𝑟 and − 𝑣𝑟 ′𝑣𝑧 ′ are very large. Large eddies
generated in this region are called “the energy-containing eddies” since they have the largest energy
density 𝐸1 (𝑛)𝑚𝑎𝑥 in the energy spectrum.
The integral scale Λ𝑓 is determined mainly by the size of the larger energy-containing eddies.
Even in the fully-developed flow, there exist more permanent largest eddies, which contain muchless energy than the energy-containing eddies. In the vicinity of the wall the spectral energy-transfer
occurs from the energy-containing eddies (low frequency) to smaller and smaller eddies (higher andhigher frequencies).
The rate of energy production is nearly in dynamical equilibrium with the rate of energy
dissipation. We should know that this kinetic energy balance is utilized for the principle of scale-up
design of an agitated vessel. (see Section 20.3)
Still there is a turbulent diffusion flux of the residual kinetic energy from the energy-producing zone
toward the pipe axis. This is the only supply of kinetic energy in the central part where there is nosource of turbulence energy. Thus the energy transferred by turbulent diffusion is consumed bydissipation. This implies that the energy-containing eddies in the energy-producing zone are muchlarger than those in the central part and that the dissipating eddies in the energy-producing zone areTurbulent Structure in a Circular Pipe Flow

209

much smaller than those in the central part. In the turbulent core large eddies are elongated in thez-direction: the integral scale Λ𝑓 in the z-direction is larger than the integral scales Λ𝑔 in the r- andθ-direction. It is known that at high Reynolds numbers, for example, Λ𝑓 becomes of the order of
0.5 R whereas Λ𝑔 is of the order of 0.2 R.
17.5-2 Roles of turbulent eddies in transport processes
When we study various processes involving heat and mass transports, it is necessary to
distinguish fluid particles (small eddies) from fluid lumps (large eddies).
Fluid particle is a very small volume of the fluid (of the order of microscale), but far larger thanthe mean free path of the fluid molecules. Within such a fluid particle, the velocity, temperature, andconcentration can be assumed uniform. Fluid lump is a much larger volume of the fluid which
consists of a large number of fluid particles. The size of such a fluid lump is of the order of theintegral scale and as big as the width of turbulent flowfield (e.g. the radius for pipe flows, thehalf-radius for free jet flows, and the thickness for boundary layer flows).
In usual turbulent flows, the turbulent transport of momentum, heat, and mass is controlled
mainly by fluid lumps. However the size of the eddies which play a predominant role in heat andmass transfer may be slightly different from that of the eddies controlling momentum transfer.
Turbulent diffusion is one of the most important factors in chemical engineering. Chemicalengineers make use of turbulence to mix and homogenize fluid mixtures as well as to accelerate therates of heat and mass transfer. Chemical reaction rates are often accelerated by turbulent mixing.
For heterogeneous chemical reactions such as the liquid-liquid reaction in an agitated vessel, therelative size between the discrete liquid particles and the surrounding eddies is an important factor
in relation with the rate of reaction. If the Reynolds number is raised in an apparatus, the integral
scale remains almost constant but the microscale tends to become small: a turbulent flow at a
relatively high Reynolds number has a relatively “fine” small-scale structure. Combustion processes
also involve turbulence and often depend on the turbulent mixing and diffusion of fuel gases andoxidizer. For mist flow in pipes of boiler, the relative size between small water droplets andturbulent eddies of steam stream may be an important factor in consideration of the turbulent
diffusion of water droplets toward the heat transfer wall.
In this text, the film theory was mainly adopted for the study of heat and mass transfer across theinterfaces separating phases.
However it should be kept in mind that the recent hydrodynamic studies pay attention to theturbulent eddying motion near the interface. There are many attempts to relate the turbulent energy
spectrum to mass transport behavior near the gas-liquid interface from a viewpoint of large
energy-containing eddies which control the mass transport rate.
Very recently various sophisticated statistical studies have demonstrated with the aid ofvisualization techniques the existence of “coherent” (or well-ordered) structure in the wall layer ofwall-bounded turbulent flows: there are intermittent large-scale interactions between wall layer andouter layer. The low-speed streaks of fluid are lifted up (ejected) from the wall by the stretchedvortex-like structure, and then the high-speed outer layer appears by the sweep motion called“bursting.” It plays an important role in the production of turbulence and the turbulent transport
processes. There are some attempts to relate the interphase transport rates to the intermittent
renewal of fluid lumps near the interface by bursting motion.
Nomenclature𝐶𝐴𝐶𝑝

𝐸1 (𝑛)𝑓(𝑟)𝑔(𝑟)𝐼

molar concentration of component A, [kmol/m3]
heat capacity, [J/kg K]

one-dimensional energy spectrum, [ - ]longitudinal velocity correlation coefficient, [ - ]lateral velocity correlation coefficient, [ - ]electric current, [A]210

Turbulent Transport Phenomena𝑗𝑙𝑀 , 𝑙𝐻 , 𝑙𝐷
𝑙𝐿
𝑝
𝑞
𝑅
𝑅𝑒
𝑅𝐿2
𝑟
𝑇
𝑇𝑢
𝑡
𝑢∗

mass flux, [kmol/m2s]
mixing length of momentum, heat, and mass, [m]Lagrangian integral length scale, [m]pressure, [Pa]heat flux, [J/m2s]
electric resistance, [Ω] or pipe radius, [m]Reynolds number, [ - ]Lagrangian correlation, [ - ]radial coordinate, [m], or distance between two points of lateral correlation, [m]temperature, [K]turbulence intensity, [ - ]time, [s]friction velocity, [m/s]𝑢+dimensionless velocity =𝑣𝑥 , 𝑣𝑦 , 𝑣𝑧
x, y, zvelocity component in rectangular coordinates, [m/s]rectangular coordinates, [m]𝑦+Δ𝑀𝜀𝑀 , 𝜀𝐻 , 𝜀𝐷𝜅𝑤
dimensionless distance from wall =𝜈2momentum deficit, [kg/m s]
eddy diffusivity of momentum, heat, and mass, [m2/s]
thermal conductivity, [W/m K]

Λ𝑓
Λ𝑔
𝜇
𝜈
𝜌
𝜏

ℑ2

𝑦 √𝜏 ⁄𝜚

Eulerian integral longitudinal length scale, [m]Eulerian integral lateral length scale, [m]viscosity, [kg/m s]
kinematic viscosity or diffusivity of momentum, [m2/s]:
density, [kg/m3]
momentum flux or shear stress, [kg/s 2m] or [N/m2] or time difference, [s]time scale, [s]Superscripts‘
fluctuation
time-averaged(l)laminar(t)turbulent
Subcripts
Aw𝑣𝑧√𝜏𝑤 ⁄𝜚

component A

wall

CHAPTER 18
BOUNDARY LAYER THEORY
18.1 Stream Function1)
For a two-dimensional, incompressible flow, i.e., ρ = const , 𝑎𝑛𝑑 𝜕 ⁄𝜕𝑧 = 0, the equations ofcontinuity and motion are given by
𝜕𝑣𝑦

𝜕𝑣𝑥

(18.1-1)+ 𝜕𝑦 = 0
𝜕𝑥
𝜕𝑣

𝜕𝑣

1 𝜕𝑝

𝜕2 𝑣

𝜕2 𝑣

(18.1-2)

𝜕𝑣𝑦

𝜕𝑣𝑦

1 𝜕𝑝

𝜕 2 𝑣𝑦𝜕 2 𝑣𝑦(18.1-3)𝑣𝑥 𝜕𝑥𝑥 + 𝑣𝑦 𝜕𝑦𝑥 = − 𝜌 𝜕𝑥 + 𝜈 ( 𝜕𝑥2𝑥 + 𝜕𝑦2𝑥)
𝑣𝑥 𝜕𝑥 + 𝑣𝑦 𝜕𝑦 = − 𝜌 𝜕𝑦 + 𝜈 ( 𝜕𝑥2 + 𝜕𝑦2 )The set of equations can be simplified into one equation with one unknown variable byintroducing a new function 𝜓(𝑥, 𝑦) called “stream function.”
The function is defined so that the equation of continuity is satisfied automatically:𝜕𝜓𝜕𝜓𝑣𝑥 = 𝜕𝑦 and 𝑣𝑦 = − 𝜕𝑥(18.1-4)
Since 𝜓 is a function of 𝑥 and 𝑦, the total derivative is
𝑑𝜓 =

𝜕𝜓
𝜕𝜓
dx +
dy = − 𝑣𝑦 dx + 𝑣𝑥 dy𝜕𝑥𝜕𝑦

If 𝜓 = const (i.e., 𝑑𝜓 = 0)
𝑣𝑦
𝑑𝑦
|
= 𝑣
𝑑𝑥
𝜓=𝑐𝑜𝑛𝑠𝑡𝑥(18.1-5)

The path 𝜓 = 𝑐𝑜𝑛𝑠𝑡 is seen to represent a streamline since the streamline is tangent to the velocity
vector𝑣⃗ = 𝑣𝑥 𝑖⃗ + 𝑣𝑦 𝑗⃗(18.1-6)
Fig. 18.1-1 indicates definition of stream function.

Fig. 18.1-1. Definition of stream function 𝜓

212

Boundary Layer Theory

Differentiating Eq. (18.1-2) with respect to 𝑦 and Eq. (18.1-3) with respect to 𝑥 and combining
these two equations to eliminate the pressure terms, we get𝜕𝑣𝑥 𝜕𝑣𝑦 𝜕𝑣𝑥 𝜕𝑣𝑦𝜕𝜕𝑣𝑥 𝜕𝑣𝑦𝜕+−+ 𝑣𝑦 ) (−()() + (𝑣𝑥
)
𝜕𝑥
𝜕𝑦
𝜕𝑥
𝜕𝑦
𝜕𝑥
𝜕𝑦 𝜕𝑦
𝜕𝑥

= 𝜈 (

𝜕𝑣𝑦
𝜕2
𝜕2
𝜕𝑣
+ 2) ( 𝑥 −
)
2
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑥

(18.1-7)

Substituting 𝜓 into Eq. (18.1-6)

𝜕𝜓 𝜕
𝜕𝜓 𝜕
𝜕 2𝜓 𝜕 2𝜓
𝜕2
𝜕2
𝜕 2𝜓 𝜕 2𝜓
(−)( 2 +
)
=
𝜈
(
+
)
(
+
)
𝜕𝑦 𝜕𝑥
𝜕𝑥 𝜕𝑦 𝜕𝑥
𝜕𝑦 2
𝜕𝑥 2 𝜕𝑦 2 𝜕𝑥 2
𝜕𝑦 2

Then we get a fourth-order differential equation with one unknown, 𝜓:−𝜕(𝜓,∇2𝜓)𝜕(𝑥,𝑦)(18.1-8)= 𝜈 ∇4 𝜓Still its general solution is very difficult owing to its nonlinearity.
Here
𝜕2

𝜕2

𝜕2

𝜕2

𝜕2

𝜕2

(18.1-9)∇2 = 𝜕𝑥2 + 𝜕𝑦2 and ∇4 = (𝜕𝑥2 + 𝜕𝑦2 ) (𝜕𝑥2 + 𝜕𝑦2)𝜕𝑓⁄𝜕𝑥
𝜕(𝑓,𝑔)= |𝜕(𝑥,𝑦)
𝜕𝑔⁄𝜕𝑥

𝜕𝑓⁄𝜕𝑦
𝜕𝑓 𝜕𝑔
𝜕𝑓 𝜕𝑔
| =
−
𝜕𝑥 𝜕𝑦
𝜕𝑦 𝜕𝑥
𝜕𝑔⁄𝜕𝑦

The equation for two-dimensional flow (i.e., 𝑣𝑧 = 0 and 𝜕 𝑣𝑟 ⁄𝜕𝑧 = 𝜕 𝑣𝜃 ⁄𝜕𝑧 = 0) can be
expressed in cylindrical coordinates:
1 𝜕(𝜓,∇2𝜓)−𝑟

𝜕(𝑟,𝜃)(18.1-10)= 𝜈 ∇4 𝜓
∇2 =

𝜕2
𝜕𝑟

Here the stream function 𝜓 is defined as
1 𝜕𝜓𝜕𝜓𝑣𝑟 = 𝑟 𝜕𝜃 and 𝑣𝜃 = − 𝜕𝑟

2+

1 𝜕2

1 𝜕
+
𝑟 𝜕𝑟 𝑟 2 𝜕𝜃 2

[PROBLEM 18.1-1]
For an inviscid uniform fluid flow over a stationary cylinder (radius 𝑅), the stream function canbe expressed as

𝑅2
𝜓 = 𝑣∞ (𝑟 −) sin 𝜃𝑟
The circle 𝑟 = 𝑅 itself must be a streamline.
Draw the streamlines around the cylinder and obtain the velocity components 𝑣𝑟 and 𝑣𝜃 .

18.2 Boundary Layer Solution of Laminar Flow along a Flat Plate2)
This example is instructive for the understanding of local variations of friction factor and Nusselt
number in developing flows.
As shown in Fig. 18.1-2, the flat plate is heated at constant wall temperature 𝑇𝑤 . The

Boundary Layer Solution of Laminar Flow along a Flat Plate (continued)

213

approaching fluid has uniform velocity 𝑣∞ and uniform temperature 𝑇∞.
The thickness of the velocity boundary layer is defined as follows.
Usually the thickness 𝛿 is taken as the distance away from the surface of the flat plate where thevelocity reaches 99% of the free-stream velocity. Similarly the thickness of the temperatureboundary layer 𝛿𝑇 is defined as the distance away from the surface (the outer edge of the boundary
layer) where the fluid temperature reaches 99% of the free-stream temperature.
Obtain the expressions of local friction factor and local Nusselt number in terms of local
Reynolds number.
1.
2.

Bird, R. B., Stewart, W. E., and Lightfoot, E. N., “Transport Phenomena,” Wiley, New York (1960)
Schlichting, H., “Boundary Layer Theory,” 4th ed., McGraw-Hill, New York (1960)

Fig. 18.2-1. Boundary layer flow along a heated flat plate

The local Reynolds number called “length Reynolds number” is defined using the distance 𝑥
from the leading edge:𝑣 𝑥
𝑅𝑒𝑥 = ∞𝜈
(18.2-1)
The following two-dimensional flow equations can be applied:
𝜕𝑣

𝜕𝑣

1 𝜕𝑝

𝜕𝑇

𝜕𝑇

𝜕2 𝑇

𝜕2 𝑣

𝜕2 𝑣

𝑣𝑥 𝜕𝑥𝑥 + 𝑣𝑦 𝜕𝑦𝑥 = − 𝜌 𝜕𝑥 + 𝜈 ( 𝜕𝑥2𝑥 + 𝜕𝑦2𝑥 )(18.2-2)𝑣𝑥 𝜕𝑥 + 𝑣𝑦 𝜕𝑦 = 𝛼 (𝜕𝑥2 + 𝜕𝑦2)

(18.2-3)

The boundary layer equations to be solved become𝜕𝑣+ 𝜕𝑦𝑥 = 0𝜕𝑥(18.2-4)

𝜕2 𝑇

where 𝜈 = 𝜇 ⁄𝜌 𝑎𝑛𝑑 𝛼 = 𝜅⁄𝜌 𝐶𝑝.
The free stream velocity outside the boundary layer and pressure are kept at
𝑑𝑃
𝑣∞ = 𝑐𝑜𝑛𝑠𝑡 and=0𝑑𝑥
For convenience, the temperature 𝑇 is replaced by the temperature difference 𝑇𝑤 − 𝑇.
When the Reynolds number is sufficiently large, the thickness of the boundary layer is very small
as compared to the x-directed (streamwise) distance. Therefore the variation of the x-velocity andtemperature is very small in the x direction than in the y direction.
𝜕𝑣𝑥
𝜕𝑣
𝜕𝑇
𝜕𝑇
≪ 𝜕𝑦𝑥 and 𝜕𝑥 ≪ 𝜕𝑦
𝜕𝑥
𝜕𝑣𝑥

𝜕𝑣

𝜕𝑣

𝑣𝑥 𝜕𝑥𝑥 + 𝑣𝑦 𝜕𝑦𝑥 = 𝜈

𝜕 2 𝑣𝑥𝜕𝑦 2(18.2-5)

214

𝑣𝑥

Boundary Layer Theory
𝜕(𝑇𝑤 − 𝑇)𝜕𝑥+ 𝑣𝑦𝜕(𝑇𝑤 − 𝑇)𝜕𝑦= 𝛼𝜕 2 (𝑇𝑤 − 𝑇)𝜕𝑦 2

(18.2-6)

If the Prandtl number is unity, i.e., Pr = 1, Eqs. (18.2-5) and (18.2-6) are the same in form. Thissuggests that the dimensionless velocity profiles are coincident with the dimensionless temperature
profiles.
From the equation of continuity Eq. (18.2-4),𝑦 𝜕𝑣𝑣𝑦 = − ∫0 𝜕𝑥𝑥 𝑑𝑦(18.2-7)
Substituting this equation into Eqs. (18.2-5) and (18.2-6),
𝜕𝑣

𝜕𝑣

𝜕2 𝑣

𝑦 𝜕𝑣

(18.2-8)𝑥
𝑣𝑥 𝜕𝑥𝑥 − 𝜕𝑦𝑥 ∫0 𝜕𝑥𝑥 𝑑𝑦 = 𝜈𝜕𝑦2𝜕(𝑇𝑤 − 𝑇)𝜕 2 (𝑇 − 𝑇)𝑦 𝜕𝑣

𝜕(𝑇𝑤 − 𝑇)(18.2-9)∫0 𝜕𝑥𝑥 𝑑𝑦 = 𝛼 𝜕𝑦𝑤2𝜕𝑥𝜕𝑦The boundary conditions are
B.C.1 𝑦 = 0: 𝑣𝑥 = 0
B.C.2 𝑦 = 𝛿: 𝑣𝑥 = 𝑣∞
(18.2-10)
B.C.3 𝑦 = 0: 𝑇𝑤 − 𝑇 = 0
B.C.4 𝑦 = 𝛿𝑇 : 𝑇𝑤 − 𝑇 = 𝑇𝑤 − 𝑇∞(18.2-11)
It is reasonable to assume that the velocity profiles and the temperature profiles arerespectively similar at various distances from the leading edge. This implies that the velocity
distribution curves and the temperature distribution curves for various distances can be madecoincident with one another by selecting the free-stream quantities 𝑣∞ and 𝑇𝑤 − 𝑇∞ , and theboundary layer thickness 𝛿 and 𝛿𝑇 , respectively. The principle of similarity in the velocity andtemperature profiles can be analytically expressed as𝑣𝑦𝜙 (𝜂 ) = 𝑥
𝜂=(18.2-12)
𝑣𝑥

−

𝑣∞
𝑇𝑤 −𝑇

Θ(𝜂𝑇 ) = 𝑇 − 𝑇
𝑤

∞

𝛿(𝑥)

𝑦

(18.2-13)

𝜂𝑇 = 𝛿 (𝑥)𝑇

Fig. 18.2-2 Velocity and temperature profiles based on similarity transformation

The boundary layer thickness 𝛿 and 𝛿𝑇 are a function of x. Eqs. (18.2-12) and (18.2-13)
constitute a very important set of variables for a variable transformation, known as the similarity
transformation. It is further assumed that the ratio of 𝛿𝑇 to 𝛿 is a constant independent of thedeveloping distance 𝑥:
Δ = 𝛿𝑇 ⁄𝛿 = 𝑐𝑜𝑛𝑠𝑡(18.2-14)
Using the new variables, the boundary layer equations can be transformed as follows.
𝜕𝑣𝑣𝑥 𝜕𝑥𝑥 = − 𝜂 𝜙 ′ 𝜙𝑣∞ 2 𝑑 𝛿
𝛿

𝑑𝑥

−

𝜂
𝜕𝑣𝑥 𝑦 𝜕𝑣𝑥
𝑣∞ 2 𝑑 𝛿∫
𝑑𝑦 = (𝜙 ′ ∫ 𝜂 𝜙 ′ 𝑑𝜂 )𝜕𝑦 0 𝜕𝑥
𝛿 𝑑𝑥
0

𝜕 2 𝑣𝑥
𝑣∞
𝜈
=
𝜈
𝜙"
𝜕𝑦 2
𝛿2

Boundary Layer Solution of Laminar Flow along a Flat Plate (continued)

Then Eq. (18.2-8) becomes𝜂𝑑𝛿(𝜙 ′ ∫0 𝜂 𝜙 ′ 𝑑𝜂 − 𝜂 𝜙 ′ 𝜙)𝛿
=
𝑑𝑥

𝜈
𝑣∞

(18.2-15)

𝜙"

where 𝜙 ′ = 𝑑𝜙⁄𝑑 𝜂.
Using 𝛿𝑇 = 𝛿 ∆,𝜕(𝑇 − 𝑇)𝑣 (𝑇 −𝑇 ) 𝑑 𝛿𝑣𝑥 𝑤= − 𝜂𝑇 Θ′ 𝜙 ∞ 𝑤𝛿 ∞ 𝑑𝑥
𝜕𝑥−𝜕(𝑇𝑤 − 𝑇)
𝜕𝑦

𝑦 𝜕𝑣

𝜂

∫0 𝜕𝑥𝑥 𝑑𝑦 = Θ′ ∫0 𝜂 𝜙 ′ 𝑑𝜂

Then Eq. (18.2-9) becomes215

𝑣∞ (𝑇𝑤 −𝑇∞ )

𝑑𝛿

∆𝛿

𝑑𝑥

𝜕 2 (𝑇𝑤 − 𝑇)𝛼(𝑇𝑤 − 𝑇∞ ) "
𝛼
=
Θ
2
𝜕𝑦
∆2 𝛿 2Θ′ 𝜂𝑑𝛿𝛼∫ 𝜂 𝜙 ′ 𝜙 𝑑𝜂 − 𝜂 𝑇 Θ′ 𝜙] 𝛿 𝑑𝑥 = 𝑣 Θ"
∆ 0∞
where 𝛩′ = 𝑑𝛩⁄𝑑𝜂 𝑇 .

(18.2-16)∆2 [If Eq. (18.2-15) is integrated over the boundary layer thickness (from 𝜂 = 0 to 1), a
first-order ordinary differential equation for 𝛿 can be obtained:1𝜂𝜈𝑑𝛿= − 𝑣 𝜙′(0)(18.2-17)∫0 (𝜙 ′ ∫0 𝜂 𝜙 ′ 𝑑𝜂 − 𝜂 𝜙 ′ 𝜙) 𝑑𝜂 𝛿𝑑𝑥∞

in which the boundary condition 𝜕𝑣𝑥 ⁄𝑦 = 0 at 𝑦 = 𝛿 has been applied. If the velocity
distribution function 𝜙(𝜂) is known, the above equation can be solved. Therefore a polynomial of
the fourth degree is assumed:𝜙 (𝜂 ) = 𝑎 + 𝑏 𝜂 + 𝑐 𝜂 2 + 𝑑 𝜂 3 + 𝑒 𝜂 4(18.2-18)
Five boundary conditions necessary to evaluate the five unknown constants, 𝑎, 𝑏, 𝑐, 𝑑, 𝑒 areconsidered for an approximate distribution function, shown in Fig. 18.2-3.

B.C.1
B.C.2
B.C.3
B.C.4
B.C.5

Fig. 18.2-3.

𝜂=0 ∶ 𝜙=0
𝜂 = 0 ∶ 𝜙" = 0
𝜂=1 ∶ 𝜙=1
𝜂 = 1 ∶ 𝜙′ = 0
𝜂 = 1 ∶ 𝜙" = 0

Approximate velocity distribution function

The second boundary condition can be obtained by taking the limit of the boundary layer equation
as 𝑦 goes to zero. The condition states that the second derivative of the velocity 𝑣𝑥 at the wall isequal to the pressure gradient. The fifth condition implies that the velocity curve has a point of
inflexion at the outer edge of the boundary layer.
Thus the approximate velocity distribution is obtained:𝜙 (𝜂 ) = 2 𝜂 − 2 𝜂 3 + 𝜂 4(18.2-19)
Substituting the above function into Eq. (18.2-17)𝑑𝛿630 𝜈𝛿 𝑑𝑥 = 37 𝑣
(18.2-20)∞

Integrating with respect to 𝑥
1260 𝜈

𝛿= √

37 𝑣∞

𝑥

(18.2-21)

in which the initial condition (at 𝑥 = 0, 𝛿 = 0) has been used. The local friction factor 𝑓𝑥 can be
defined as

216

Boundary Layer Theory𝜏 (𝑥)(18.2-22)

𝑓𝑥 = 1 𝑤
2𝜌 𝑣∞ 2

Applying the polynomial velocity function to the Newton’s law of viscosity,
𝜕𝑣

𝜇𝑣

𝜏𝑤 (𝑥 ) = 𝜇 𝜕𝑦𝑥 |𝑦=0

37 𝜌𝜇 𝑣∞ 3𝑥

(18.2-23)∞= 𝛿(𝑥)
𝜙 ′ (0) = √315

Substituting into Eq. (18.2-22), the local friction factor can be obtained in a function of length
Reynolds number:𝑓𝑥 =√148⁄315√𝑣∞ 𝑥⁄𝜈≅

0.685(18.2-24)

𝑅𝑒𝑥 1/2

Regarding the temperature boundary layer, in the same manner as in the approximation of velocity
function, we get the following polynomial function for the temperature distribution:
𝛩(𝜂𝑇 ) = 2 𝜂𝑇 − 2 𝜂𝑇 3 + 𝜂𝑇 4
(18.2-25)
The boundary layer thickness ratio ∆ has two possibilities : ∆ ≤ 1 and ∆ ≥ 1. If Pr = 1, ∆becomes unity. We consider the case when the temperature boundary layer is thinner than thevelocity boundary layer: ∆ ≤ 1.
The relation between 𝜂 and 𝜂𝑇 is𝜂 = 𝜂𝑇 ∆(18.2-26)
Integrating Eq.(18.2-16) over the temperature boundary layer thickness (from 𝜂𝑇 = 0 to 1),
1 Θ′

𝜂 ∆

𝑑𝛿

𝛼

∆2 ∫0 [ ∆ ∫0 𝑇 𝜂 𝜙 ′ 𝜙 𝑑𝜂 − 𝜂𝑇 Θ′ 𝜙(𝜂𝑇 ∆)] 𝑑𝜂𝑇 𝛿 𝑑𝑥 = − 𝑣 Θ′ (0)∞(18.2-27)

in which the boundary condition (𝜕𝑇⁄𝜕𝑦 = 0 at 𝑦 = 𝛿𝑇 ) has been applied.
Dividing Eq.(18.2-27) by Eq.(18.2-17)
1 Θ′ 𝜂𝑇 ∆𝜂 𝜙′ 𝜙 𝑑𝜂− 𝜂𝑇 Θ′ 𝜙(𝜂𝑇 ∆)]𝑑𝜂𝑇
∫
∆ 0
1
𝜂
∫0 (𝜙′ ∫0 𝜂 𝜙′ 𝑑𝜂 − 𝜂 𝜙′ 𝜙)𝑑𝜂

∆2 ∫0 [=

1 Θ′ (0)𝑃𝑟 𝜙′(0)(18.2-28)

Substituting Eqs.(18.2-19) and (18.2-25) into Eq.(18.2-28) and calculating the integrations, we get
1
3
1
37 1
∆3 − 280 ∆5 + 360 ∆6 = 630 𝑃𝑟(∆ ≤ 1)(18.2-29)
15
The six-order algebraic equation can be approximated within 5% error by∆ = 𝑃𝑟 −1⁄3
(18.2-30)
According to Fourier’s law
𝜕𝑇2𝜅
𝑞𝑥 (𝑥 ) = 𝜅 𝜕𝑦|= ∆ 𝛿 (𝑇𝑤 − 𝑇∞ )(18.2-31)
𝑦=0

The local heat transfer coefficient and the local Nusselt number are defined as
𝑞𝑤 = ℎ𝑥 (𝑇𝑤 − 𝑇∞ )ℎ 𝑥
𝑞 𝑥
𝑁𝑢𝑥 = 𝜅𝑥 = 𝜅 (𝑇 𝑤− 𝑇 )𝑤∞(18.2-32)(18.2-33)

By using Eqs. (18.2-21), Eq.(18.2-30), and (18.2-31), Eq.18.2-33) becomes2

37

𝑁𝑢𝑥 = ∆ 𝛿 = 2 √1260 𝑅𝑒𝑥 1⁄2 𝑃𝑟 1⁄3 = 0.343 𝑅𝑒𝑥 1⁄2 𝑃𝑟 1⁄3(18.2-34)

It has been found that the local friction factor is inversely proportional to the square root of thelocal Reynolds number (length Reynolds number) and that the local Nusselt number is proportional
to the square root of the local Reynolds number and the 1/3 power of the Prandtl number.
The j-factor for heat transfer is defined as𝑁𝑢
𝑗𝐻𝑥 = 𝑅𝑒 𝑃𝑟𝑥1⁄3(18.2-35)𝑥

If the j-factor is calculated from Eq.(18.2-34)
0.343
𝑗𝐻𝑥 =
1⁄2

(18.2-36)

From Eq.(17-2-24)
𝑓𝑥
0.343
=
1⁄2
2

(18.2-37)

𝑅𝑒𝑥

𝑅𝑒𝑥

It has been demonstrated that the analogy 𝑗𝐻 = 𝑓𝑥 ⁄2 between momentum and heat transfer holds

Integral Equation of Boundary Layer Flow217

exactly between momentum and heat transfer for laminar boundary layer flow over a flat plate.
If we continue this approach of the boundary layer over a flat plate similarly for mass transfer, thefollowing result can be obtained for laminar flow:
𝑆ℎ𝑥 = 0.343 𝑅𝑒𝑥 1⁄2 𝑆𝑐 1⁄3(18.2-38)
where the Sherwood number is defined as
𝑐𝑘 𝑥
𝑆ℎ𝑥 = 𝐷 𝑥
(18.2-39)
𝐴𝐵

It has been confirmed that the Chilton-Colburn analogy relation exists between heat and masstransfer
𝑗𝐷 = 𝑗𝐻(18.2-40)
where the j-factor for mass transfer is defined as𝑆ℎ𝑗𝐷𝑥 = 𝑅𝑒 𝑆𝑐𝑥1⁄3(18.2-41)𝑥

Similar method is available for turbulent boundary layer flow.
This boundary layer theory gives a valuable concept for the analogy between momentum and heattransfer. The section of convective heat transfer in a circular pipe flow in Chapter 8 also gives thesame analogy relation Eq.(8.2-14).

18.3 Integral Equation of Boundary Layer Flow
18.3-1 Momentum integral equation of boundary layer flow over a flat plate
It is usually difficult to obtain an exact solution of the boundary layer equations.
An approximate method is available to overcome this difficulty.
The boundary layer equation for steady two-dimensional flow in the x-direction is
𝜕𝑣

𝜕𝑣

𝜕2 𝑣

𝑑𝑝

(18.3-1)

𝜌 (𝑣𝑥 𝜕𝑥𝑥 + 𝑣𝑦 𝜕𝑦𝑥 ) = − 𝑑𝑥 + 𝜇 𝜕𝑦2𝑥

The equation of motion for external flow (potential flow) is
𝑑𝑣
𝑑𝑝𝜌 𝑣∞ 𝑑𝑥∞ = − 𝑑𝑥
Substituting Eq. (18.3-2) into Eq. (18.3-1)
𝑣𝑥

𝜕𝑣𝑥
𝜕𝑥

+ 𝑣𝑦

𝜕𝑣𝑥
𝜕𝑦

= 𝑣∞

𝑑𝑣∞

+ 𝜈

𝑑𝑥

𝜕 2 𝑣𝑥(18.3-2)(18.3-3)

𝜕𝑦 2Integrating this equation with respect to y from y = 0 (wall) to 𝑦 = 𝛿 (the outer edge of theboundary layer)
𝛿

𝜕𝑣

𝜕𝑣

𝜕𝑣 𝛿

𝑑𝑣

𝜏

∫0 (𝑣𝑥 𝜕𝑥𝑥 + 𝑣𝑦 𝜕𝑦𝑥 − 𝑣∞ 𝑑𝑥∞ ) 𝑑𝑦 = 𝜇 𝜕𝑦𝑥 | = − 𝜌𝑤
0(18.3-4)𝑦 𝜕𝑣

Substituting the equation of continuity: 𝑣𝑦 = − ∫0 𝜕𝑥𝑥 𝑑𝑦
𝛿

𝜕𝑣

𝜕𝑣

𝑦 𝜕𝑣

𝑑𝑣

𝜏

∫0 (𝑣𝑥 𝜕𝑥𝑥 − 𝜕𝑦𝑥 ∫0 𝜕𝑥𝑥 𝑑𝑦 − 𝑣∞ 𝑑𝑥∞ ) 𝑑𝑦 = − 𝜌𝑤

(18.3-5)

The second term is integrated by parts𝛿𝛿𝛿
𝜕𝑣𝑥 𝑦 𝜕𝑣𝑥
𝜕𝑣𝑥
𝜕𝑣𝑥
∫ (
∫
𝑑𝑦) 𝑑𝑦 = 𝑣∞ ∫𝑑𝑦 − ∫ 𝑣𝑥
𝑑𝑦𝜕𝑦 0 𝜕𝑥
𝜕𝑥
0
0 𝜕𝑥
0

Then Eq. (18.3-5) becomes𝛿∫ (2𝑣𝑥
0

𝜕𝑣𝑥
𝜕𝑣𝑥
𝑑𝑣∞
𝜏𝑤
− 𝑣∞
− 𝑣∞
) 𝑑𝑦 = −
𝜕𝑥
𝜕𝑥
𝑑𝑥
𝜌

This equation can be contracted to
𝛿 𝜕
𝛿
𝜏
𝑑𝑣
∫0 𝜕𝑥 (𝑣𝑥 (𝑣∞ − 𝑣𝑥 )) 𝑑𝑦 + 𝑑𝑥∞ ∫0 (𝑣∞ − 𝑣𝑥 )𝑑𝑦 = 𝜌𝑤

The following two meaningful thicknesses can be introduced(18.3-6)

218

Boundary Layer Theory𝛿𝑣∞ 𝛿𝐷 = ∫ (𝑣∞ − 𝑣𝑥 )𝑑𝑦0𝛿𝜌𝑣∞ 𝑣∞ 𝛿𝑚 = 𝜌 ∫ 𝑣𝑥 (𝑣∞ − 𝑣𝑥 )𝑑𝑦0

where 𝛿𝐷 and 𝛿𝑚 are the displacement thickness and momentum thickness.
The integral equation becomes𝜏𝑤
𝑑𝑑𝑣= 𝑑𝑥 𝑣∞ 2 𝛿𝑚 + 𝑣∞ 𝛿𝐷 𝑑𝑥∞𝜌(18.3-7)This is the boundary-layer equation in integral form, which is valid both for laminar and turbulent
flows.
Therefore this equation is useful for a macroscopic analysis of turbulent flows.
18.3-2 Energy integral equation of boundary layer flow
The thermal boundary layer equation is given by
𝜕𝑇

𝜕𝑇

𝜕2 𝑇

(18.3-8)𝑣𝑥 𝜕𝑥 + 𝑣𝑦 𝜕𝑦 = 𝛼 𝜕𝑦2Substituting the equation of continuity
𝜕𝑇

𝜕𝑇

𝜕2 𝑇

𝑦𝜕𝑣

(18.3-9)𝑣𝑥 𝜕𝑥 − 𝜕𝑦 ∫0 𝜕𝑥𝑥 𝑑𝑦 = 𝛼 𝜕𝑦2Integrating from 𝑦 = 0 to 𝑦 = 𝛿𝑇 (the outer edge of the thermal boundary layer)
𝛿𝑇

∫
0

(𝑣𝑥

𝜕 𝑇 𝜕 𝑇 𝑦 𝜕 𝑣𝑥
𝑞𝑤
𝜕 𝑇 𝛿𝑇−∫𝑑𝑦) 𝑑𝑦 = 𝛼| = −𝜕𝑥𝜕𝑦 0 𝜕𝑥𝜌𝐶𝑝
𝜕𝑦 0

Using a temperature difference 𝛩 = 𝑇∞ − 𝑇 in place of 𝑇𝛿𝑇𝜕𝛩𝜕 𝛩 𝑦 𝜕 𝑣𝑥𝑞𝑑𝑦) 𝑑𝑦 = − 𝑤∫0 (𝑣𝑥 𝜕𝑥 −
∫0
𝜕𝑦

𝜕𝑥

The second term is integrated by parts
𝛿𝑇

∫
0

𝜌𝐶𝑝

(18.3-10)

𝛿𝑇𝛿𝑇𝜕 𝛩 𝑦 𝜕 𝑣𝑥
𝜕 𝑣𝑥
𝜕 𝑣𝑥
∫
𝑑𝑦) 𝑑𝑦 = 𝛩∞ ∫𝑑𝑦 − ∫ 𝛩𝑑𝑦(
𝜕𝑦 0 𝜕𝑥𝜕𝑥
𝜕𝑥
00

Since the first term becomes zero, the equation reduces to𝛿𝑇∫0That is
𝛿

𝑑

(𝑣𝑥

𝜕𝛩
𝜕 𝑣𝑥
𝑞𝑤
+ 𝛩
) 𝑑𝑦 = −
𝜕𝑥
𝜕𝑥
𝜌𝐶𝑝

𝑞

(18.3-11)𝑇𝑤∫ 𝛩 𝑣𝑥 𝑑𝑦 = − 𝜌𝐶𝑝
𝑑𝑥 0This is the boundary-layer energy equation in integral form, which is also valid both for laminar andturbulent flows.
18.3-3 Turbulent boundary layer flow 1)
As shown in Fig. 18.3-1, in the initial section of laminar boundary layer, the boundary layer
thickness increases with the length Reynolds number 𝑅𝑒𝑥 . Eventually its instabilities cause theboundary layer to become turbulent. For the level of free-stream turbulence 0.5%, the critical length
Reynolds number for transition on a smooth flat plate with a sharp leading edge can be assumed as
3 × 105 ≤ 𝑅𝑒𝑥 ≤ 5 × 105

1.

Schlichting, H., “Boundary Layer Theory,” 4th ed., McGraw-Hill, New York (1960)

Turbulent Boundary Layer Flow (continued)

219

Fig. 18.3-1. Transition to turbulent boundary layer on a flat plate

The transition region, which is unstable and oscillatory in nature, has a finite length. Owing to thetransverse exchange of fluid elements, the velocity distribution in the turbulent boundary layer ismuch flatter than that in the laminar boundary layer.
If a thin wire (called “tripping wire”) is mounted near the leading edge of the flat plate, aturbulent boundary layer starts to grow from the beginning.
The integral form of the boundary layer equation can be used as a simple tool for analyzing
difficult turbulent boundary layer flows. In addition, a simple solution can be obtained if a
one-seventh power law is assumed for the velocity profile in the turbulent boundary layer.
For moderately high Reynolds numbers, the one-seventh power law is of the form:
1/7
𝑢+ = 8.74 𝑦 +From the definition of 𝑢+ and 𝑦 +⁄𝑦 𝑣∗ 1 7(18.3-12)𝑣𝑥 = 8.74 𝑣 ∗ ( 𝜈 )where 𝑣 ∗ = √𝜏𝑤 ⁄𝜌 is called “friction velocity.”At the outer edge of the boundary layer
𝑢+ = 𝑣∞ + at 𝑦 = 𝛿Applying Eq. (18.3-12) at the outer edge (𝑦 = 𝛿), the wall shear stress is calculated as
𝛿𝑣

−1⁄4

𝜏𝑤 = 0.0225 𝜌 𝑣∞ 2 ( 𝜈∞ )(18.3-13)
Substituting the power law into the defining equations of displacement and momentum thicknesses,
𝛿

1
𝑣 𝛿
8 ∞
0
𝛿
𝜌 𝑣∞ 2 𝛿𝑚 = ρ ∫0 𝑣𝑥 (𝑣∞ − 𝑣𝑥 )𝑑𝑦 = 0.0972 𝜌 𝑣∞ 2 𝛿𝑣∞ 𝛿𝑑 = ∫ (𝑣∞ − 𝑣𝑥 ) 𝑑𝑦 =For this case (uniform constant velocity free stream)
𝑑 𝑣∞
=0
𝑑𝑥
Therefore the integral momentum equation becomes𝜏𝑤𝑑= 𝑑𝑥 𝑣∞ 2 𝛿𝑚
𝜌

(18.3-14)(18.3-15)

Substituting Eq.(18.3-13) into Eq.(18.3-15)
𝑑𝛿

𝜈

1⁄ 4

𝛿 1⁄4 𝑑𝑥 = 0.231 (𝑣 )∞Integrating this with the boundary condition 𝛿 = 0 at 𝑥 = 0,
4 5⁄ 4
𝜈 1⁄4
𝛿
= 0.231 ( ) 𝑥
5𝑣∞
From this equation, the following equation is obtained as a result:(18.3-16)

220

Boundary Layer Theory𝛿= 0.37 𝑅𝑒𝑥 − 1⁄5(18.3-17)
It has been found that the dimensionless thickness of the turbulent boundary layer 𝛿 ⁄𝑥 decreases
with 𝑅𝑒𝑥 − 1⁄5.
Combining the defining equation of local friction factor with Eq.(18.3-13),𝑥

𝑓𝑥 = 1
2

𝜏𝑤
𝜌 𝑣∞

⁄
𝛿 𝑥 𝑣∞ −1 4

= 2 × 0.0225 (𝑥2𝜈)= 0.058 𝑅𝑒𝑥 − 1⁄5(18.3-18)

This is a very important result. As distinct from the laminar boundary layer, the local friction factor
is correlated well with 𝑅𝑒𝑥 − 1⁄5.
This equation is in good agreement with experiments for 𝑅𝑒𝑥 up to several millions.
For the case of turbulent heat transfer from a heated flat plate,
𝑁𝑢𝑥
𝑗𝐻 =~𝐶 𝑅𝑒𝑥 − 1⁄5⁄
1
3
𝑅𝑒𝑥 𝑃𝑟It should be kept in mind that the analogy between momentum and heat transfers still holds forturbulent boundary layer flow. The empirical coefficient C is difficult to determine because of theuncertainty in the location of the transition region.

18.4 Application of Stream Function (Impinging Flow)
Plane stagnation flow shown in Fig.18.4-1 can be solved using the stream function. Fluid
approaching from the positive y direction is turned aside by the plane 𝑦 = 0. The free stream
outside the boundary layer has the following velocity parallel to the plane:
𝑣𝑥𝑝 → 𝑐 𝑥 as 𝑦 → ∞(18.4-1)
This velocity can be obtained by the potential flow theory assuming non-viscous flow.
However the no-slip” condition should be satisfied at the surface of the plane.

Fig.18.4-1 Stagnation flow impinging on a flat plane

The stream function is defined by Eq. (18.1-4):𝜕𝜓𝜕𝜓
𝑣𝑥 = 𝜕𝑦 and 𝑣𝑦 = − 𝜕𝑥

The equation to be solved is−𝜕 (𝜓,∇2 𝜓)
𝜕 (𝑥,𝑦)= 𝜈 ∇4 𝜓(18.4-2)

Application of Stream Function (Impinging Flow) (continued)

The boundary conditions are𝜕𝜓→ 𝑐 𝑥 as 𝑦 → ∞
𝜕𝑦
𝜕𝜓

221

(18.4-3)𝜕𝜓= 𝜕𝑥 = 0 at 𝑦 = 0(18.4-4)
The first boundary condition suggests that the solution should be of the form
𝜓 = 𝑥 𝑓(𝑦)(18.4-5)
With this assumption, the main differential equation reduces to− 𝑓 𝑓 ′′′ + 𝑓 ′ 𝑓 ′′ = 𝜈 𝑓 𝑖𝑣
(18.4-6)
where the primes denote differentiation with respect to 𝑦. This can be integrated once with respectto 𝑦(𝑓′)2 − 𝑓 𝑓 ′′ − 𝜈 𝑓 ′′′ = 𝐾 2(18.4-7)2
where 𝐾 is an integration constant. This equation can be simplified by a variable transformation.
Let us define
𝜕𝑥

𝜑=

𝑓
√𝐾𝜈

𝐾

and 𝜂 = √ 𝜈 𝑦Then Eq.(18.3-7) becomesφ′′′ + 𝜑 𝜑 ′′ − (𝜑 ′ )2 + 1 = 0
Here the primes denote differentiation with respect to 𝜂.
The corresponding boundary conditions are
𝜑 ′ → 1 as 𝜂 → ∞𝜑 = 𝜑 ′ = 0 at 𝜂 = 0
The solution numerically calculated from Eq. (18.4-9) is shown in Fig.18.4-2.

(18.4-8)(18.4-9)(18.4-10)(18.4-11)

Fig.18.4-2. Numerically calculated solution of Eq.(18.4-9)

The x-component velocity 𝜑 ′ = 𝑣𝑥 ⁄𝑣𝑥𝑝 becomes 0.99 when 𝜂 = 2.4. This indicates that the
thickness of the boundary layer is given by𝛿 = 𝜂𝛿 √𝜈⁄𝐾 = 2.4 √𝜈⁄𝐾
(18.4-12)
It should be noted that the boundary layer thickness remains constant in the impingement region.
This velocity boundary layer thickness 𝛿 also has a relation with the thermal boundary layer
thickness 𝛿𝑇 . The integration constant 𝐾 depends on the flow condition such as the approach
velocity 𝑣𝑦∞ .

222

Boundary Layer Theory

18.5 Impinging Jet Heat Transfer1,2)This kind of impinging flows are very often practically applied for heating/cooling and drying
technologies such as the drying of paper or film. For example, a free air jet issuing from a
convergent nozzle is made striking normally onto a large flat plate, as shown in Fig.18.5-1.

Fig.18.5-1. Schematic picture of jet impingement on a flat plate with heat transfer

When the nozzle-to-plate spacing is small, the free jet impinges on the flat plate without loss of
its initial velocity before it becomes developed. The central jet impingement region around thestagnation point resembles the central region of impinging flow above-mentioned. An importantquestion arises as to where the flat plane as a heat transfer surface should be placed to maximize thestagnation-point heat flux. When the spacing is large, the free jet becomes fully developed before it
impinges. In the fully developed free jet region, the jet velocity decays in inverse proportional to theaxial distance from the nozzle exit. The following impingement jet heat transfer correlation at thestagnation point is available as an example 1,2):
𝑁𝑢𝑠
𝑁𝑢= 𝐶 (1 + 𝜖 ) ( 1⁄2 1⁄2 )(18.4-13)
1⁄2
1⁄2
𝑅𝑒𝑠

𝑃𝑟

𝑅𝑒𝑠

𝑃𝑟

𝑇𝐹

where the constant 𝐶 was empirically obtained to be 0.88 for gas flow Pr ≤ 1 in our research.
The Reynolds number 𝑅𝑒𝑠 = 𝜌𝐷 𝑈⁄𝜇 and the turbulence intensity 𝑇𝑢 = √𝑢′2 ⁄𝑈 are defined as
the quantities of free jet at the position of the heat transfer plate.
The coefficient 𝜖 gives a factor of heat transfer enhancement due to the turbulence effect, which
can be considered as a function of 𝑇𝑢 √𝑅𝑒𝑠 defined on the centerline of the approaching jet. The
subscript TF implies the turbulence-free condition (laminar flow solution).
We should know that it is one of the appropriate methods effective for the augmentation of
convective heat transfer to strike a turbulent stream on a heat transfer surface like the above
example.
For example, a shell-and-tube heat exchanger usually has the shell-side structure striking theshell-side stream on the tube bundles perpendicularly by the installed baffle plates.
1.
2.

Kataoka, K., Sahara, R., Ase, H. and Harada, T., J. Chem. Eng. Japan, 20(1), 71 (1987)f
Kataoka, K., Suguro, M., Degawa, H., Maruo, K. and Mihata, I., Int. J. Heat Mass Transfer, 30(3), 559 (1987)

Boundary-layer Analysis for Velocity-gradient Measurement223

18.6 Boundary-layer Analysis for Velocity-gradient Measurement
In Chapter 14.4 it was introduced that an electrochemical method can measure local
velocity-gradient on the wall of liquid flow. A rectangular test cathode has ionic mass transfer on its
surface under the diffusion-controlling condition, where the bulk concentration of reacting ion is 𝐶𝐴
but the concentration at the wall (cathode surface) 𝐶𝐴𝑤 becomes zero. Therefore the ionic massflux on the cathode is given by
𝑁𝐴 = 𝐼𝑑 ⁄𝐹𝑎 = 𝑘𝐶𝐴
The velocity gradient s on the wall is given by𝑠 = 1.90 (𝐼𝑑 ⁄𝐹𝑎 𝐶𝐴 )3 (𝐿⁄𝐷𝐴𝐵 2 ) = 1.90 𝑘 3 (𝐿⁄𝐷𝐴𝐵 2 )(14.4-1)
Let us study again how this equation can be obtained by using the boundary layer theory in thissection.
The two-dimensional boundary equation of laminar mass transfer can be written as𝑣𝑥

Using the equation of continuity𝜕𝐶𝑣𝑥 𝜕𝑥𝐴 −

𝜕𝐶𝐴

𝑦 𝜕𝑣

𝜕𝐶𝐴
𝜕𝐶𝐴
𝜕 2 𝐶𝐴
+ 𝑣𝑦
= 𝐷𝐴𝐵
𝜕𝑥
𝜕𝑦
𝜕𝑦 2

𝜕 2𝐶

∫ 𝑥 𝑑𝑦 = 𝐷𝐴𝐵 𝜕𝑦2𝐴
𝜕𝑦 0 𝜕𝑥

(18.6-1)

Fig.18.6-1 Velocity profile and concentration profile on an isolated cathodeAs shown in Fig.18.6-1, the concentration boundary layer begins from the leading edge of thecathode while the velocity boundary layer has been developing from far ahead. Therefore within theconcentration boundary layer on the test cathode the velocity profile can be assumed linear:𝑣𝑥 = 𝑠 𝑦, where s is the velocity gradient on the wall. In addition, the velocity change over the
streamwise length of the test cathode is very small: 𝜕𝑣𝑥 ⁄𝜕𝑥 ≅ 0.
The boundary layer equation reduces to
𝜕𝐶

𝜕2 𝐶

𝑠𝑦 𝜕𝑥𝐴 = 𝐷𝐴𝐵 𝜕𝑦2𝐴

(18.6-2)

Dividing the equation by y to define a new (mass flux) function 𝑓 = 𝐷𝐴𝐵 (𝜕𝐶𝐴 ⁄𝜕𝑦) and thendifferentiating with respect to y:
𝜕𝑓
𝜕
1 𝜕𝑓
𝑠 𝜕𝑥 = 𝐷𝐴𝐵 𝜕𝑦 (𝑦 𝜕𝑦)(18.6-3)
By the following variable transformation

224

Boundary Layer Theory𝜓 = 𝑓 ⁄𝑓0 , 𝜂 = 𝑦⁄𝛿𝑐 , 𝜆 = 𝐷𝐴𝐵 𝑥⁄𝑠 𝛿𝑐 3
where 𝛿𝐶 is the concentration boundary layer thickness
The above equation is made dimensionless:
𝜕𝜓
𝜕
1 𝜕𝜓
= 𝜕𝜂 (𝜂 𝜕𝜂 )𝜕𝜆With boundary conditions
B.C. 1:at 𝜆 = 0, 𝜓 = 0
B.C. 2:at 𝜂 = 0, 𝜓 = 1
B.C. 3:at 𝜂 = ∞, 𝜓 = 0(18.6-4)(18.6-5)(18.6-6)This equation can be solved analytically by the combination of variables by using then following
independent variable:𝜂χ= 3(18.6-7)√9 𝜆Finally the mass flux equation becomes𝜒 𝜓" + (3 𝜒 3 − 1) 𝜓′ = 0
The boundary conditions become:
B.C. 1:at 𝜒 = 0, 𝜓 = 1
B.C. 2:at 𝜒 = ∞, 𝜓 = 0
The primes mean differentiation with respect to : 𝜓′ = 𝑑𝜓⁄𝑑𝜒The solution is
𝜓=

∞

3

∞

3

∫𝜒 𝜒 𝑒 −𝜒 𝑑𝜒

3

∞

3

(18.6-9)(18.6-10)= Γ(2⁄3) ∫𝜒 𝜒 𝑒 −𝜒 𝑑𝜒∫0 𝜒 𝑒 −𝜒 𝑑𝜒(18.6-8)

Here the calculation procedure for solving Eq.(18.6-8) by the method of combination of variables isomitted owing to the limitation of page allocation.
The concentration profile can be obtained by integration:𝐶𝐴∞∞1(18.6-11)∫𝐶 𝑑𝐶𝐴 = − 𝐷 ∫𝑦 𝑓 𝑑𝑦 = (𝐶𝐴∞ − 𝐶𝐴 (𝑦))𝐴𝐴𝐵

The dimensionless concentration is given by∞𝐶 −𝐶3
𝛬 = 𝛿 𝐴𝑓 ⁄𝐷𝐴∞ = √9𝜆 ∫𝜒 𝜓 𝑑𝜒
𝑐 0

(18.6-12)

𝐴𝐵

where
𝜕𝐶
𝑓0 = 𝐷𝐴𝐵 𝐴 |(18.6-13)

𝜕𝑦 𝑦=0Substituting Eq.(18.6-10) into the integration of Eq.(18.6-12), the following equation is obtained:323

𝛬 = √9𝜆 [3Γ( ,𝜒 )
𝑒 −𝜒
3
{1
−
𝜒
−
2 }]
Γ(2⁄3)
Γ( )(18.6-14)3

where Γ(𝑝, 𝑥) is called “incomplete gamma function,” which is defined as∞Γ(𝑝, 𝑥 ) = ∫𝑥 𝑡 𝑝−1 𝑒 −𝑡 𝑑𝑡(18.6-15)

2

Since the mass transfer wall lies at 𝜒 = 0, Γ(3 , 𝜒 3) becomes Γ(2⁄3) and the second term
disappears.
The gamma function is also defined as∞
Γ(𝑝) = ∫0 𝑥 𝑝−1 𝑒 −𝑥 𝑑𝑥(18.6-16), which gives Γ(2⁄3) ≅ 1.35411.
Then Eq.(18.6-13) can be written at the wall 𝜒 = 0 as3𝛬𝜒=0 =

𝐶𝐴∞ − 𝐶𝐴𝑤
√9𝜆= Γ(2⁄3) =⁄𝛿𝑐𝑓0 𝐷𝐴𝐵

3

𝐷

𝑥

1⁄3

√9( 𝐴𝐵3 )𝑠 𝛿𝑐

Γ(2⁄3)=1because according to the film theory the following relation is valid at any position x
𝜕𝐶𝐴
𝐶𝐴∞ − 𝐶𝐴𝑤|
𝑓0 = 𝐷𝐴𝐵= 𝐷𝐴𝐵
𝜕𝑦 𝑦=0𝛿𝑐(18.6-17)

Boundary-layer Analysis for Velocity-gradient Measurement (continued)

Then from Eq.(18.6-14)
3

√9

3

3

√9

𝐷

3

𝐷

3

𝑠 = [Γ(2⁄3)] ( 𝐴𝐵3 ) 𝑥 = [Γ(2⁄3)] ( 𝐴𝐵) 𝐷
𝛿𝑐

𝛿𝑐

3

√9

𝑥
𝐴𝐵

3

3
2 = [Γ(2⁄3)] 𝑘

225

𝑥(18.6-18)

𝐷𝐴𝐵 2

The velocity-gradient at the wall of the test cathode averaged over the electrode length from x = 0 to
x = L is,3√93

𝑠 = [Γ(2⁄3)] 𝑘 3 𝐷

1
𝐴𝐵

1

𝐿

1

3

√9

3

3
2 𝐿 ∫0 𝑥 𝑑𝑥 = 2 [Γ(2⁄3)] 𝑘
𝐷

𝐿
𝐴𝐵

2

≅ 1.8124 𝑘 3 𝐷

𝐿
𝐴𝐵

2

(18.4-19)

This result coincides to Eq. (14.4-1) but the coefficient 1.8124 is slightly different from thecoefficient 1.90 of Eq. (14.4-1).
It is interesting that this electrochemical method can observe fluctuating velocity gradients in theviscous sublayer of various turbulent flows.
Nomenclature
𝐶𝐴
𝐶𝑝
𝐷𝐴𝐵
𝐹𝑎
𝑓
𝑓𝑥
𝐼𝑑
𝑗𝐷𝑥
𝑗𝐻𝑥
𝐿
𝑁𝑢𝑥
𝑃𝑟
𝑝
𝑞𝑥
𝑅𝑒𝑥
𝑟, 𝜃, 𝑧Sc𝑆ℎ𝑥𝑠𝑇

molar concentration of reactant ion A, [kmol/m3]
heat capacity, [J/kg K]
diffusivity of component A, [m2/s]
Faraday constant (96,500 C/kg-equiv.),
mass-flux function, [kmol/m2s]
local friction factor, [ - ]limiting current density, [A/m2]
local j-factor for mass transfer, [ - ]local j-factor for heat transfer, [ - ]length of rectangular electrode in flow direction, [m]local Nusselt number, [ - ]Prandtl number, [ - ]pressure, [Pa]local heat flux, [J/m2s]
local length Reynolds number, [ - ]cylindrical coordinates, [m, - , m]
Schmidt number, [ - ]local Sherwood number, [ - ]velocity gradient at wall, [1/s]temperature, [K]𝑢+dimensionless velocity =𝑢∗𝑣𝑥 , 𝑣𝑦 , 𝑣𝑧𝑣∞x, y, zfriction velocity, [m/s]velocity component in rectangular coordinates, [m/s]free-stream velocity, [m/s]rectangular coordinates, [m]

𝑦+
𝛼
∆
𝛿, 𝛿𝑇
𝛿𝑐
𝛿𝐷
𝛿𝑚
ϵ
𝛩
𝛬
𝜈
𝜌
𝜓

𝑤
dimensionless distance from wall =𝜈
2thermal diffusivity, [m /s]
thickness ratio of thermal to velocity boundary layer, [ - ]thickness of velocity and temperature boundary layer, [m]concentration boundary layer thickness, [m]displacement thickness of boundary layer, [m]momentum thickness of boundary layer, [m]heat transfer enhancement factor, [ - ]temperature difference, [K]dimensionless concentration difference, [ - ]kinematic viscosity or diffusivity of momentum, [m2/s]:
density, [kg/m3]
stream function, [m2/s]𝑣𝑧√𝜏𝑤 ⁄𝜚

Subcriptsppotential flowsstagnation point of impinging jetw
wall∞
free stream𝑦 √𝜏 ⁄𝜚

CHAPTER 19
FREE COVECTION
19.1 Boundary Layer Approach
Fluid motion caused by the gradients of the fluid density is called “natural or free convection.”
As a simple example we shall consider a natural laminar flow along a vertical flat plate which isheated at a constant temperature 𝑇𝑤 and surrounded by a large volume of the gas fluid at a constanttemperature 𝑇∞. When 𝑇𝑤 > 𝑇∞ , in the neighborhood of the plate the fluid flows upward due tothe buoyancy force and has a boundary layer structure as shown in Fig.19.1-1. For simplicity thephysical properties are assumed constant except for the fluid density in the buoyancy term.

Fig. 19.1-1. Natural flow along a vertical hot plate(The following theoretical approach is not so easy to follow for this course. This section can beskipped, except for the heat transfer correlations given in the latter part, which are useful forengineering calculations.)
The following two-dimensional flow equations can be applied1,2):
𝜌∞ ( 𝑣𝑥

𝜕𝑣𝑥
𝜕𝑥

+ 𝑣𝑦

𝜕𝑣𝑥
𝜕𝑦

)= −

𝜕𝑝
𝜕𝑥

𝜕 2 𝑣𝑥

𝜕 2 𝑣𝑥

𝜕𝑥

𝜕𝑦 2

+ 𝜇 (

2 +

) − 𝜌𝑔

The pressure in each horizontal plane is equal to the gravitational pressure𝑝 = − 𝜌∞𝑔 𝑥(19.1-1)

Boundary Layer Approach (continued)

227

where 𝜌∞ is the density at 𝑇∞. The density 𝜌 at 𝑇 can be expanded in a Taylor series about
the reference temperature 𝑇∞:𝜕𝜌𝜌 = 𝜌∞ + | (𝑇 − 𝑇∞ ) + − − − ≅ 𝜌∞ − 𝜌∞ 𝛽(𝑇 − 𝑇∞ )(19.1-2)
𝜕𝑇
𝑇∞

1

𝜕𝜌

where 𝛽 = − 𝜌is the coefficient of volume expansion.

|∞ 𝜕𝑇 𝑇∞

Therefore the buoyancy force is given by
𝜕𝑝

𝑇− 𝑇∞

− 𝜕𝑥 − 𝜌𝑔 = 𝜌∞ 𝛽(𝑇 − 𝑇∞ )𝑔 = 𝜌∞ 𝑔

For gases 𝛽 =1
𝑇∞(19.1-3)𝑇∞

Thus we get the following set of equations:

𝜕𝑣𝑥
𝜕𝑥

𝜕𝑣𝑦

+ 𝜕𝑦 = 0

𝜕𝑣

𝜕𝑣

𝜕𝑇

𝜕𝑇

𝜕2 𝑣

𝜕2 𝑣

𝑇− 𝑇∞

𝑣𝑥 𝜕𝑥𝑥 + 𝑣𝑦 𝜕𝑦𝑥 = 𝜈 ( 𝜕𝑥2𝑥 + 𝜕𝑦2𝑥 ) + 𝑔
𝜕2 𝑇

𝑇∞

𝜕2 𝑇

(19.1-4)𝑣𝑥 𝜕𝑥 + 𝑣𝑦 𝜕𝑦 = 𝛼 (𝜕𝑥2 + 𝜕𝑦2)where 𝜈 = 𝜇 ⁄𝜌 𝑎𝑛𝑑 𝛼 = 𝜅⁄𝜌 𝐶𝑝.
By the boundary layer approximation

𝜕 2 𝑣𝑥

𝜕2 𝑇≅ 0 and≅0𝜕𝑥 2
Here in place of 𝑇 , the dimensionless local temperature 𝛩 = (𝑇 − 𝑇∞ )⁄(𝑇𝑤 − 𝑇∞ ) is
introduced.
Thus the equations to be solved are
𝜕𝑥 2

𝜕𝑣𝑦

𝜕𝑣𝑥

(19.1-5)+ 𝜕𝑦 = 0
𝜕𝑥
𝜕𝑣

𝜕2 𝑣

𝜕𝑣

𝑣𝑥 𝜕𝑥𝑥 + 𝑣𝑦 𝜕𝑦𝑥 = 𝜈 𝜕𝑦2𝑥 + 𝑔𝜕𝛩(𝑇𝑤 − 𝑇∞ )
𝑇∞

𝜕2 𝛩

𝜕𝛩

(19.1-6)

𝛩(19.1-7)𝑣𝑥 𝜕𝑥 + 𝑣𝑦 𝜕𝑦 = 𝛼 𝜕𝑦2

The boundary conditions are𝑣𝑥 = 𝑣𝑦 = 0,
𝛩=1
at 𝑦 = 0𝑣𝑥 = 0 𝛩 = 0 at 𝑦 = ∞(19.1-8)
Firstly we should consider what similarity transformation is appropriate for this problem.
The stream function is introduced
𝜓(𝑥,𝑦)𝑦= 𝜁(𝜂)
𝜂 = 𝐿(𝑥)(19.1-9)𝑓(𝑥)This is a kind of the similarity transformation. Here 𝜁(𝜂) is a dimensionless stream function
which is a function of 𝜂 alone, 𝜂 the dimensionless y-coordinate, 𝐿 an unknown characteristic
length relating to the boundary layer thickness, and 𝑓(𝑥) the unknown function to be determined.
The velocity components are𝜕𝜓1
𝑣𝑥 = 𝜕𝑦 = 𝑓 (𝑥 )𝜁′ 𝐿(19.1-10)𝜕𝜓

𝑓(𝑥) 𝑑𝐿(19.1-11)

𝑣𝑦 = − 𝜕𝑥 = −𝑓 ′(𝑥 ) 𝜁 (𝜂) + 𝜂𝜁′ 𝐿 𝑑𝑥
where 𝜁′ = 𝑑𝜁 ⁄𝑑𝜂 and 𝑓 ′(𝑥 ) = 𝑑𝑓 ⁄𝑑𝑥.
Substituting into the equation of motion,
𝑓 𝑓′

𝑓2 𝑑𝐿

𝐿

𝐿3 𝑑𝑥

[(𝜁′ )2 − 𝜁′′ 𝜁] −2

(𝜁 ′ )2 =Dividing through by 𝑓 𝑓′ 𝐿

𝜈𝑓
𝐿3

𝜁′′′ + 𝑔⁄ 2[(𝜁′ )2 − 𝜁′′ 𝜁] −𝑓𝑑𝐿

𝑓′ 𝐿 𝑑𝑥(𝜁 ′ )2 =𝜈𝐿 𝑓′

𝜁′′′ + 𝑔

Each term should be dimensionless. Thus
𝑓𝑑𝐿

𝑓′ 𝐿 𝑑𝑥= 𝐾1 ,𝜈
𝐿 𝑓′= 𝐾2 , 𝑔

𝑇𝑤 − 𝑇∞ 𝐿2
𝑇∞

𝑓 𝑓′

= 𝐾3

From the three equations of Eq. (19.1-14)

𝑇𝑤 − 𝑇∞𝑇∞𝛩(19.1-12)𝛩(19.1-13)

𝑇𝑤 − 𝑇∞ 𝐿2
𝑇∞

𝑓 𝑓′

(19.1-14)

228

Free Convection𝑑𝐿

𝐿3 𝑑𝑥 =

𝐾1 𝐾3

𝜈2

𝐾2 2

𝑇 −𝑇
𝑔 𝑤 ∞
𝑇∞

(19.1-15)

The arbitrary constant 𝐾1 𝐾3 ⁄𝐾2 2 can be taken as unity without loss of generality. Integration
with the initial condition 𝐿 = 0 at 𝑥 = 0 gives
4 𝜈2 𝑇44

𝐿 = √𝑔 (𝑇 −∞𝑇 ) √𝑥 =
𝑤

∞

1 4
𝛾

(19.1-16)√𝑥This suggests that the velocity and temperature boundary layer thicknesses are proportional to𝑥 . For convenience 𝐾2 is taken as 1/3 in the second equation of Eq.(19.1-14). Then𝑓 (𝑥 ) = 4 𝜈 𝛾 𝑥 3⁄ 4
(19.1-17)
It has been found that the similarity transformation is of the form:𝑦𝜂=𝛾 4(19.1-18)1⁄ 4√𝑥

𝜓(𝑥, 𝑦) = 4 𝜈 𝛾 𝑥 3⁄4 𝜁(𝜂)(19.1-19)
The velocity components become
𝑣𝑥 = 4 𝜈 𝛾 2 𝑥 1⁄2 𝜁′(19.1-20)
𝑣𝑦 = 𝜈 𝛾 𝑥 −1⁄4 (𝜂𝜁′ − 3 𝜁)
(19.1-21)
The boundary layer equations to be solved become a set of ordinary differential equation:
𝜁′′′ + 3 𝜁′ 𝜁 − 2 (𝜁′ )2 + 𝛩 = 0(19.1-22)
𝛩′′ + 3 𝑃𝑟 𝜁 𝛩′ = 0(19.1-23)
where primes denote differentiation with respect to 𝜂.
The boundary conditions are𝜁 = 𝜁′ = 0 𝛩 = 1
at 𝜂 = 0(19.1-24)
𝜁′ = 0 𝛩 = 0
at 𝜂 = ∞(19.1-25)
This set of equations is very difficult to solve analytically. Owing to the limitation of this course,
the numerical solutions are not shown here. It is known that the numerical solutions are in good
agreement with the experimental distributions obtained by Schmidt and Beckmann 1).
1.
2.

Schmidt, E. and Beckmann, W., Tech. Mech. U. Thermodynamik, 1, 341 and 391 (1930)
Schuh, H., Boundary Layers of Temperature, in W. Tollmien (ed.), “Boundary Layers,” British Ministry of Supply, German
Document Center, Ref. 3220T (1948)

19.2 Free Convection Heat Transfer
By using the numerical solutions, the wall heat-flux can be calculated as𝜕𝑇𝜕𝑇 𝑑𝛩 𝜕𝜂𝛾 𝑑𝛩𝑞𝑤 = − 𝜅 𝜕𝑦|= − 𝜅 𝜕𝛩 𝑑𝜂 𝜕𝑦|= − 𝜅(𝑇𝑤 − 𝑇∞) 4 𝑑𝜂 |
𝑦=0

√𝑥

𝑦=0

𝜂=0

(19.2-1)

For Pr = 0.73 (air) the dimensionless temperature gradient at the wall is given by the numericalsolution:𝑑𝛩|= −0.508(19.2-2)𝑑𝜂 𝜂=0The heat transfer coefficient can be defined as𝑞ℎ𝑥 = 𝑇 −𝑤𝑇
𝑤(19.2-3)∞

Then the local Nusselt number becomes

𝑁𝑢𝑥 =

ℎ𝑥 𝑥
𝜅

0.508

= 0.508𝛾 𝑥 3⁄4 = 41⁄4 [𝑔 𝑥 3 (𝑇𝑤 − 𝑇∞ )where 𝐺𝑟𝑥 is the local Grashof number.

𝜈2 𝑇∞

]

1⁄4

= 0.359 𝐺𝑟𝑥 1⁄4

The average Nusselt number can be determined by the following integration:𝐻ℎ 𝐻
𝑁𝑢,𝑚 = 𝑚𝜅 = 0.508 𝛾 ∫0 𝑥 − 1⁄4 𝑑𝑥 = 0.677 𝐺𝑟𝐻 1⁄4(19.2-4)(19.2-5)

Free Convection Heat Transfer (continued)

229

Here the Grashof number is the ratio of the buoyant to the viscous force:𝐺𝑟𝐻 =𝑔 𝐻 3 (𝑇𝑤 − 𝑇∞ )
𝜈2 𝑇∞

(19.2-6)This result, Eq.(19.2-5) is very important to understand the empirical heat transfer correlations
for the laminar free convection. This suggests that the average Nusselt number is proportional to
one-fourth power of the Grashof number for laminar free convection.
The transition from laminar to turbulent flow occurs in the range 108 < 𝐺𝑟𝐻 Pr < 1010 .
The following empirical equations are recommended by Eckert and Jackson1):
𝑁𝑢𝑚 = 0.555 (𝐺𝑟𝐻 𝑃𝑟)1⁄4
𝐺𝑟𝐻 Pr < 109(Laminar flow)(19.2-7)⁄5
2(Turbulent flow)(19.2-8)
𝑁𝑢𝑚 = 0.021 (𝐺𝑟𝐻 𝑃𝑟)
𝐺𝑟𝐻 Pr > 109
1.

Eckert, E. R. G., and Jackson, T. W., NACA RFM 50 D25, July (1950)

These correlation equations Eqs.(19.2-7) and (19.2-8) are also applicable to a vertical hot
cylinder.
For a long horizontal hot cylinder, experimental data have been correlated by McAdams1):ℎ 𝐷
𝑁𝑢𝑚 = 𝑚𝜅 = 0.525 (𝐺𝑟𝐷 𝑃𝑟)1⁄4
𝐺𝑟𝐷 Pr > 104(19.2-9)
where 𝐺𝑟𝐷 =1.

𝑔 𝐷3 (𝑇𝑤 − 𝑇∞ )
𝜈2 𝑇∞

McAdams, W. A., Heat Transmission, Second ed., McGraw-Hill, New York (1942)[EXAMPLE 19.2-1]
Calculate the heat loss from the outside surface of a big cylindrical furnace, which consists of thecomposite wall of three materials: fire-clay brick, insulating brick, and iron. As shown in Fig.
19.2-E1, the melted iron well mixed at a constant temperature 𝑇𝑖 keeps the inside surface of thefire-clay brick at 𝑇𝑖 . The surrounding air is assumed to be stationary at a constant temperature 𝑇∞.
In the neighborhood of the furnace, the air rises owing to the buoyancy force. The furnace is 𝐻 inheight and 𝐷𝑜 in outside diameter. The heat loss from the top of the furnace is neglected only forsimplicity. For this case there exist four heat-transfer resistances in series.

Fig.19.2-E1. Heat loss from the outside vertical wall of a furnace

230Free Convection

The overall heat-transfer resistance is expressed as1(𝐷 −𝐷 )⁄2 𝐷𝑜(𝐷 −𝐷 )⁄2 𝐷𝑜(𝐷 −𝐷 )⁄2 𝐷= ℎ + 𝑜𝜅3
+ 3𝜅2+ 2𝜅 𝑖 𝐷𝑜
𝑈
𝐷
𝐷
1

𝑜

where

𝐷3𝑎𝑣 =

𝑜

3

3𝑎𝑣

2

𝐷𝑜 − 𝐷3
𝐷3 − 𝐷2, 𝐷2𝑎𝑣 =,ln(𝐷𝑜 ⁄𝐷3 )ln(𝐷3 ⁄𝐷2 )2𝑎𝑣

𝐷1𝑎𝑣 =

1

1𝑎𝑣(19.2-E1)

𝐷2 − 𝐷𝑖ln(𝐷2 ⁄𝐷𝑖 )Here the outside heat-transfer coefficient ℎ𝑜 can be obtained by calculating from Eq. (19.2-7) orEq.(19.2-8).
The temperature 𝑇𝑜 at the outside surface of the furnace is necessary to determine the Grashof
number.
The following heat-balance equation can be used:
𝑈𝑜 𝜋𝐷𝑜 𝐻 (𝑇𝑖 − 𝑇∞ )𝜋𝐷𝑜 𝐻(𝑇𝑖 − 𝑇∞)(𝐷 −𝐷 )⁄2 𝐷𝑜(𝐷3 −𝐷2 )⁄2 𝐷𝑜
+
+ 2 𝑖
𝐷3𝑎𝑣
𝜅2
𝐷2𝑎𝑣
𝜅1
𝐷1𝑎𝑣

= (𝐷𝑜−𝐷3)⁄2 𝐷𝑜
𝜅3(19.2-E2)

The outside surface temperature 𝑇𝑜 can be calculated by the use of trial and error method in theabove equation.
Then the total heat-loss is given by
𝑄 = 𝑈𝑜 𝜋𝐷𝑜 𝐻 (𝑇𝑖 − 𝑇∞ )(19.2-E3)[PROBLEM]
An LD converter is a furnace used for reduction of the carbon concentration in melted iron. A
supersonic oxygen gas jet impinges on the melted iron to remove carbon as CO from it. The wallconsists of three layers: refractory brick layer (𝜅1 = 6.1 W/m K), insulating brick layer (𝜅2 = 3.1
W/m K), and a steel plate (𝜅3 = 44 W/m K). Their thicknesses are 450 mm, 200 mm, and 50 mm,
respectively. The melted iron is kept at uniform temperature 1,800 K owing to the strong jetagitation. The surrounding air is at 𝑇∞ = 313 𝐾.
Calculate the heat loss from the vertical wall of this LD converter. What is the surface
temperature of the steel plate?

Fig. 19.2-P1.

Heat Loss from an LD converter. Dimensions given are in mm

Nomenclature
Nomenclature
𝐷
𝑔
𝐺𝑟𝐻
𝐺𝑟𝑥
ℎ𝑥
𝐿
𝑁𝑢𝑥
𝑃𝑟
𝑝
𝑄
𝑞𝑤
𝑅𝑒𝑥
𝑇
𝑈
𝑣𝑥 , 𝑣𝑦 , 𝑣𝑧
x, y, z
𝛼
𝛽
𝛩
𝜅
μ
𝜈
𝜌
𝜓

cylinder diameter, [m]gravitational acceleration, [m/s2]
Grashof number, [ - ]local Grashof number, [ - ]local heat transfer coefficient, [W/m2K]
unknown characteristic length, [m]local Nusselt number, [ - ]Prandtl number, [ - ]pressure, [Pa]heat loss, [J/s]wall heat flux, [J/m2s]
length Reynolds number, [ - ]temperature, [K]overall heat transfer coefficient, [W/m2K]
velocity component in rectangular coordinates, [m/s]rectangular coordinates, [m]thermal diffusivity, [m2/s]
coefficient of volume expansion, [1/K]
dimensionless local temperature, [ - ]thermal conductivity, [W/m K]
viscosity, [kg/m s]
kinematic viscosity, [m2/s]:
density, [kg/m3]
stream function, [m2/s]

Subcripts
Dcylinderhhot platemaveragedwwall∞bulk fluid

231

CHAPTER 20
AGITATION
20.1 Agitation and Mixing of Liquids
Agitation technology is very often utilized for the following purposes:(1) Suspending solid particles or dissolving solid into liquid, (2) blending miscible liquids, (3)dispersing a second liquid, immiscible with the first one to form an emulsion, (4) dispersing a
gas as the bubbles through the liquid, and (5) promoting heat transfer or chemical reaction.
20.1-1 General structure of agitation equipment
A typical agitation vessel is shown in Fig.20.1-1. An impeller is mounted on an overhung shaft,which is driven by a motor. The proportions of the tank vary widely, depending on various purposes.
For the case of a standardized practical design, accessories such as inlet and outlet nozzles, coils,
jackets, measuring-device wells are necessary.

Fig.20.1-1. Standard structure of agitation vessel

The impeller agitators are classified into two types: the axial-flow impeller generates currents
parallel with the impeller shaft axis and the radial-flow impeller generates currents in a tangential orradial direction.
Several propeller designs efficient in liquids of low viscosity are illustrated in Fig.20.1-2: (a)standard screw-type, three blades, (b) open straight blade-type, four blades, (c) turbine bladed
disk-type, eight blades, (d) vertical curved blade-type, eight blades.

Flow Patterns in Agitated Vessels

233

Two agitators employed for high viscosity or non-Newtonian liquids are shown in Fig.20.1-3: (a)anchor-type, (b) helical ribbon-type.
Flows in these kinds of agitating vessels are three-dimensional and very complicated. Therefore
to obtain theoretical solutions by using the equation of motion is very difficult, especially for
turbulent flows. In addition, the structure of agitated vessels is so complicated that it is also very
difficult to specify the boundary conditions. The commercial simulator packages based on the
CFD models (computational fluid dynamics) are available for very viscous or laminar Newtonian
fluid flows. Although the main flow caused by the rotation of an impeller is tangential, thetangential velocity component is not so effective for mixing. The radial and longitudinalcomponents are important for the mixing action. In addition, the rotational fluid flow following acircular path around the shaft creates an undesirable vortex at the surface of the liquid.
In addition, a ring doughnut-shaped isolated mixing region having independent circular motion
is formed in each central region above and below the impellers for unbaffled tanks. The effect of
mixing is not so good in the isolated mixing region. Generally the preferable method of suppressing
the vortex formation is to install baffles.
Usually several vertical baffle plates are fixed to the inner wall of the tank.

(a)

(b)

(c)

(d)

(e)

Fig.20.1-2. Impellers for low viscosity fluids: (a) marine-type propeller, three blades, (b) open
straight blade turbine, four blades, (c) bladed disk turbine, eight blades, (d) vertical curved blade
turbine, eight blades, (e) pitched-blade turbine, four blades

(a)

(b)

Fig.20.1-3. Impellors for high viscosity or non-Newtonian fluids: (a) anchor, (b) helical ribbon

20.1-2 Flow patterns in agitated vessels
Let us consider the flow pattern of a turbine-type agitated vessel as an example. It is necessary to234Agitationobserve the radial and longitudinal velocity components. Figure 20.1-4 shows the flow currents
formed in a cylindrical, baffled agitated vessel equipped with a turbine agitator. The plane ofobservation passes through the impeller shaft and in front of baffle plates. Fluid leaves the impeller
outward in a radial direction, separates into longitudinal streams flowing upward or downward overthe baffle, flows inward toward the rotation axis, and ultimately returns to the impeller intake. In thebottom region below the agitator, similar circulating flow is formed.

Fig.20.1-4. Schematic picture of circulation flow pattern in a turbine agitator tank

20.2 Power Consumption in Agitated Vessels
20.2-1 Dimensional analysis
In the design of an agitated vessel, it is very important to consider the power required to drive theimpeller.
The power consumption 𝑃 in agitation for a certain type of impeller can be obtained by thefollowing dimensional analysis. When the impeller of the lever length 𝑅 is rotated at an angularvelocity 𝜔, the form drag acting on the blade is proportional to the kinetic energy per unit volume(1⁄2)𝜌𝑣𝜃 2 (i.e., a kind of pressure) and the acting surface area 𝐴, where 𝑣𝜃 = 𝑅𝜔 and 𝐴 are thecharacteristic (peripheral) velocity and area, respectively. The torque acting on the impeller can beconsidered as (torque) = (force)·(lever length), i.e., (1⁄2)𝜌𝑣𝜃 2 𝐴 × 𝑅. Therefore the power input(work/time) should be a function of torque times angular velocity, i.e., (1⁄2)𝜌𝑣𝜃 2 𝐴 × 𝑅𝜔.

Fig.20.2-1. Relation of torque with rotating impellerPower Consumption in Agitated Vessels (continued)

235

Fig.20.2-2. Dimensions of rotating turbine impeller in a tank

Since the peripheral speed of the tip of the impeller is given by 2𝜋 (𝑑 ⁄2)𝑁, the Reynolds
number of agitators is defined as𝑅𝑒 =𝑑2 𝑁 𝜌
𝜇(20.2-1)

where 𝑑 is the impeller diameter and 𝑁 the rotation number. Thus the power consumption
should have the following functional form:
𝑃 = 𝐶 (1⁄2)𝜌 𝑣𝜃 3 𝐴
In the dimensional analysis, the common variables are: the length (blade diameter), 𝑑; thevelocity (peripheral velocity), 𝑑 𝑁; the fluid viscosity, 𝜇; the density, 𝜌; the acceleration of gravity,𝑔; the area, 𝐴 ~ 𝑑 2 ;Here the power consumption is made dimensionless as
𝑃
𝑁𝑝 = 𝜌𝑁3𝑑5(20.2-2)
This is called “the power number,” which can be considered to be analogous in physical meaning
to the friction factor in a circular pipe flow.

(Dimensional analysis)
Many chemical engineering problems cannot be solved theoretically unless enough is known
about the physical situations. The physics of mixing in an agitated tank is also so difficult that wecannot help relying on the following dimensional analysis. It may be expected that the power
consumption depends on the following variables:
𝑁𝑝 = 𝑎′ 𝑑 𝑏′𝑁 𝑐′ 𝜌 𝑑′𝜇 𝑒′𝑔 𝑓′= 𝑎′ 𝐿𝑏′𝑡 −𝑐′(𝑀 ⁄𝐿3 )𝑑′ (𝑀 ⁄𝐿𝑡)𝑒′(𝐿⁄𝑡 2 ) 𝑓′
(20.2-3)
where 𝑀, 𝐿, 𝑡 are the dimensions of mass, length, and time, respectively.
Buckingham’s theorem states: if an equation is dimensionally homogeneous, it can be reduced toa relationship among a complete set of dimensionless products.
In order for the above equation to be dimensionally homogeneous, the following conditions
should hold:
M: 0 = 𝑑 ′ + 𝑒′
L: 0 = 𝑏 ′ − 3𝑑 ′ − 𝑒 ′ + 𝑓′
t: 0 = −𝑐 ′ − 𝑒 ′ − 2𝑓′
From these equations, the three of the five unknowns can be expressed in terms of the remaining
two unknowns:
𝑐 ′ = 2𝑏′ − 3𝑑′𝑒 ′ = −𝑑′236Agitation𝑓 ′ = −𝑏′ + 2𝑑′
Finally it has been found that the power number equation should have the following functional
form:
′
′
𝑁𝑝 = 𝑎′ 𝑑 𝑏′𝑁 2𝑏 −3𝑑′𝜌 𝑑′𝜇 −𝑑′𝑔−𝑏′+2𝑑
𝜌𝑑2 𝑁

𝑑′

= 𝑎′ ( 𝜇 )𝑑 𝑁2(𝑏 ′ −2𝑑′ )(20.2-4)( 𝑔 )The first term is the Reynolds number and the second term is called “Froude number.” These
dimensionless groups have the physical meanings: the Reynolds number 𝑅𝑒 =inertial force

𝑉2

and the Froude number 𝐹𝑟 = gravity force = 𝐿𝑔 =

𝑑 𝑁2
𝑔

.

inertial force=
viscous force

𝑑2 𝑁𝜌
𝜇

(20.2-5)

The effect of the Froude number is usually very small if baffles are installed.
The unknowns 𝑎′ , 𝑑 ′, 𝑏′ − 2𝑑′ should be determined by experiment.
Except for the Reynolds number and the power number, the pumping number sometimes called
the flow number used as one more dimensionless key variable is defined as𝑞
𝑁𝑞 =(20.2-6)
𝑁 𝑑3
The volumetric flow rate 𝑞 through the impeller can be assumed proportional to the peripheral
velocity 𝜋𝑑 𝑁 times the area 𝜋𝑑 𝑏 swept out by the tips of the impeller blades: 𝑞 ~ 𝑁 𝑑 3
because of 𝑏~𝑑. As the pumping number is increased, the time required for one round of thecirculating stream is decreased. Therefore the pumping number is also very important for promoting
the mixing effect.
The pumping number tends to increase with the Reynolds number, and becomes almost constantin the turbulent flow, where it depends on the ratio of impeller diameter to tank diameter 𝑑 ⁄𝐷.
20.2-2 Power correlations
In the last section (20.2-1), we have learnt that the power number is a function of the Reynolds
number and the Froude number. However if baffle plates are installed in a tank, almost flathorizontal liquid level is kept. In this condition, the power number does not depend on the Froudenumber. Only for simplicity, Fig.20.2-3 shows the empirical correlations of the power number forsome different impellers given for the case of baffled tanks.

Fig. 20.2-3. Correlations of Impeller power number with Reynolds number for agitator tanks.
(Np for baffled, Ф for unbaffled.)

Heat Transfer in an Agitated Tank

237

In the range of high Reynolds numbers 𝑅𝑒 ≥ 10,000, the power number becomes almost
constant and independent of the Reynolds number. This effect due to baffles may be similar to theconstant friction factor in the range of high Reynolds numbers for circular rough pipes. For theradial-flow type impellers, at low Reynolds numbers 𝑅𝑒 ≤ 500, the lines of 𝑁𝑝 vs. 𝑅𝑒 for bothbaffled and unbaffled tanks coincide, and the slope of the line on logarithmic coordinates is – 1.
This implies the laminar flow range. When the power requirement is calculated, we should consider
the effect of system geometry in the terms of shape factors such as
𝑆1 = 𝐷⁄𝑑 , 𝑆2 = 𝐻𝑝⁄𝑑, 𝑆3 = 𝑙 ⁄𝑑, 𝑆4 = 𝑏 ⁄𝑑 , 𝑆5 = 𝐻⁄𝑑 . These dimensions are given in
Fig.20.2-2.
Usually typical proportions of agitated tanks are given to dimensions of the agitated tank andlocation and dimensions of the impeller for standardization.
For vigorous agitation in a baffled tank, the liquid surface is slightly disturbed but usually kept
flat, so that the effect of the Froude number on the power number is not necessary to consider.
For the case of low-viscosity liquid in an unbaffled agitated tank, there is vortex formation on theliquid surface owing to centrifugal force. The effect of radially-varying liquid depth is related with
the Froude number. Therefore the effect of the Froude number on the power number should betaken into account by the following modified power number:
𝛷 = 𝑁𝑝 𝐹𝑟 𝑚
(20.2-7)
Here 𝑚 is a function of the Reynolds number.
𝑎−log 𝑅𝑒𝑚=
(20.2-8)𝑏
For example, for propellers ( 𝑑 ⁄𝐷 = 1⁄3) , 𝑎 = 2.1 and 𝑏 = 18.0 and for disk turbine(𝑑 ⁄𝐷 = 1⁄3), 𝑎 = 1.0 and 𝑏 = 40.0.

20.3 Heat Transfer in an Agitated Tank
Generally a chemical tank reactor has complicated structures of not only the heating jacket but
also the tube coils shown in Fig.20.3-1. The tube coils are often used for the heat input into or
removal from the central region of reacting fluids. In the engineering book specialized for agitationand mixing, various heat transfer correlations are available. Owing to the limitation of this bookscope, many heat transfer correlations cannot be treated.

Fig.20.3-1 Ordinary chemical batch reactor

238

Agitation

In this section, let us consider a simple flow reactor of jacketed agitated vessel without coils andbaffle plates shown in Fig.20.3-2. No chemical reaction is assumed to take place in the vessel only
for simplicity and convenience. There is neither endothermic nor exothermic reaction within thevessel.

Fig.20.3-2 Heat transfer in a continuous flow reactor of jacketed tank unbaffled

A liquid is being continuously heated by the steam jacket of an agitated tank. Heat is supplied bycondensation of steam from the outside jacket (i.e. from the inside wall of the vessel.). Usually thethermal resistance of the condensate film and tank wall can be assumed small compared to that of
the liquid in the tank. Therefore the main resistance lies in the convective heat transfer on the insidewall of the reactor. The unjacketed portion of the tank is also assumed to be well insulated.
One example of heat correlation for constant physical properties is given by the following
equation1, 2):
𝑁𝑢 =

ℎ𝑖 𝐷

𝜇

0.14

= 𝛼 𝑅𝑒 2⁄3𝑃𝑟 1⁄3 (𝜇 )𝜅𝑤

(20.3-1)

Depending on various agitators and arrangements of the vessels, the coefficient 𝛼 varies
considerably. For the case of pitched-blade turbine agitator tank unbaffled , 𝛼 = 0.44 can be usedin the above equation.
1.
2.

Uhl, V. W., and Gray, J. B., “Mixing,” vol.1, p.284, Academic, New York (1966)
Chilton, T. H., Drew, T. B., and Jebens, R.H., Ind. Eng. Chem., 36, 510 (1944)

20.4 Scale-up of Agitated Tank Design
The performance characteristics are generally determined empirically by correlating measured
data as a function of dimensionless numbers. Those correlation equations consisting ofdimensionless groups are employed for scale-up design. Under the condition of geometric similarity,
appropriate correlations or rules should be selected depending on the physical purpose of thechemical equipment to be scaled-up. We know that the power given by an impeller is
continuously dissipated in the form of irreversible degradation of mechanical energy to thermalenergy of the liquid in a tank. Therefore the amount of power consumed by the specified impeller
per unit volume of liquid can be considered as a measure of mixing effectiveness. Usually many
power correlations are available for scale-up of various agitators and vessels.
Let us consider the scale-up design of low-viscosity liquid mixing. The subscript of variables 1

Scale-up of Agitated Tank Design (continued)

239

denotes the smaller-scale system such as laboratory and/or pilot-plant tanks and 2 denotes thelarger-scale system to be scaled-up.
The following three conditions should be considered as the fundamental rule of scale-up from thesmaller-scale to the larger-scale system, i.e. (1) to (2):The power numbers and flow numbers of the system 1 and 2 should be equal to each other
𝑁𝑝1 = 𝑁𝑝2
(20.4-1)
𝑁𝑞1 = 𝑁𝑞2
(20.4-2)(1) Low-viscosity homogeneous liquid solution:
The power consumption per unit volume 𝑃𝑣 (W⁄m3 ) should be equal to each other:
𝑃𝑣1 = 𝑃𝑣2
(20.4-3)This scale-up condition implies that the rate of the mechanical energy dissipated per unit
volume of liquid should be equal to each other between the smaller-scale and larger-scale systems.
In other words, the flow condition in a unit volume should be the same in the both systems.
However strictly speaking, even in this condition, it is too difficult to make the flow structure same
between the two systems because the rotation number obtained by the scale-up rules is usuallymuch smaller in the larger-scale system than in the smaller-scale system.
The liquid level is assumed equal to the tank diameter: 𝐻⁄𝐷 = 1. If the ratio of the impeller
diameter to the tank diameter is given by 𝑘 = 𝑑 ⁄𝐷, the power consumption per unit volume can be
expressed as
𝑃𝑣 = 𝑃 ⁄[(𝜋⁄4)𝐷 2𝐻 ] = 𝜌𝑁 3 𝑑 5 𝑁𝑝⁄[(𝜋⁄4)(𝑑 ⁄𝑘 )2 (𝑑 ⁄𝑘)] ~ 𝑁 3 𝑑 2 𝑁𝑝
(20.4-4)
Therefore the rotation number of the larger-scale system is given by
𝑁2 = (𝑑1 ⁄𝑑2 )2⁄3 (𝑁𝑝2 ⁄𝑁𝑝1)1⁄3 𝑁1(20.4-5)
As can be seen in Fig.20.2-3, the power number becomes almost constant in the wide range of the
Reynolds number in the turbulent flow condition. Therefore even if the Reynolds numbers of thetwo systems are different, the power numbers can be assumed to be equal to each other: 𝑁𝑝2 =𝑁𝑝1
By using this relation, the Reynolds number and power consumption become
𝑅𝑒2 = 𝑑2 2 𝑁2 𝜌⁄𝜇 = (𝑑2 ⁄𝑑1 )2 (𝑁2 ⁄𝑁1 ) 𝑅𝑒1 = (𝑑2 ⁄𝑑1 )4⁄3𝑅𝑒1(20.4-6)533
53
𝑃2 = 𝑁𝑝2 𝜌𝑁2 𝑑2 = (𝑁2 ⁄𝑁1 ) (𝑑2 ⁄𝑑1 ) 𝑃1 = (𝑑2 ⁄𝑑1 ) 𝑃1
(20.4-7)
According to these results, when the impeller diameter is scale-upped by eight times, i.e.,𝑑2 = 8 × 𝑑1with the tank diameter 𝐷2 = 8 × 𝐷1 , the rotation number reduces to 𝑁2 = (1⁄4) × 𝑁1.
However the Reynolds number and power consumption should become very large as 𝑅𝑒2 = 16 ×𝑅𝑒1 𝑎𝑛𝑑 𝑃2 = 512 × 𝑃1 .
In scale-up of tanks assuming equal power consumption per unit volume (sometimes calledagitation intensity), there is a possibility of forming the weak mixing regions here and there in thelarger-scale system owing to small rotation number of the agitator of the larger-scale system.
(2) Mixing of liquid-liquid dispersion:
The gradient of the peripheral velocity near the tips of the impeller is important to break liquid
lumps into small droplets. Therefore the scale-up rule for this purpose is given by
𝑁2 = (𝑑1 ⁄𝑑2 )𝑁1(20.4-8)
In the turbulent flow region, we can assume 𝑁𝑝2 = 𝑁𝑝1
Therefore the power consumption is given by
𝑃2 = (𝑁2 ⁄𝑁1 )3 (𝑑2 ⁄𝑑1 )5 𝑃1 = (𝑑2 ⁄𝑑1 )2 𝑃1(20.4-9)
The Reynolds number also becomes
𝑅𝑒2 = (𝑑2 ⁄𝑑1 )2 (𝑁2 ⁄𝑁1 ) 𝑅𝑒1 = (𝑑2 ⁄𝑑1 )𝑅𝑒1
(20.4-10)
Emulsified polymerization reactors belong to this case. However the reactor tank should be
unbaffled because a baffled reactor has a problem of undesirable polymer deposition in the concave
stagnant region between the baffle plates.

240

Agitation(3) Heat transfer of jacketed vessels
Firstly we should consider the power consumption per unit volume to be equal between thesystems 1 and 2.
The flow condition will be fully turbulent. Therefore 𝑁𝑝2 = 𝑁𝑝1 .
The rotation number for the larger-scale system is given by using Eq. (20.4-5):
𝑁2 ⁄𝑁1 = (𝑑1 ⁄𝑑2 )2⁄3
(20.4-5)
In usual scale-up of jacketed vessels with heating or cooling jackets, the heat exchange raterequirement can be expressed as
𝑄2 ⁄𝑄1 = ℎ2 𝐷2 𝐻2 (𝑇𝑤 − 𝑇𝐿 )2 ⁄ℎ1 𝐷1 𝐻1 (𝑇𝑤 − 𝑇𝐿 )1(20.4-11)
where ℎ1 , ℎ2 are the heat transfer coefficients of the two systems, respectively.
From the purpose of scale-up, the temperature difference ∆𝑇 = 𝑇𝑤 − 𝑇𝐿 should be equal toeach other between the smaller-scale and the larger-scale systems.
The geometric similarity is adopted for standardization of system design: 𝑑 ~ 𝐷 and 𝐻 ~ 𝑑.
The new length ratio of scale-up is given by
𝜆 = 𝐷2 ⁄𝐷1 = 𝑑2 ⁄𝑑1(20.4-12)
Therefore
𝑄2 ⁄𝑄1 = ℎ2 𝐷2 2 ⁄ℎ1 𝐷1 2
(20.4-13)
According to the heat transfer correlation equation Eq. (20.3-1),ℎ𝑖 𝐷
𝜇 0.14= 𝛼 𝑅𝑒 2⁄3 𝑃𝑟 1⁄3 ( )
𝜅
𝜇𝑤

𝑁𝑢 =

In these two systems, 𝑃𝑟2 = 𝑃𝑟1 and (𝜇 ⁄𝜇𝑤 )2 = (𝜇 ⁄𝜇𝑤 )1 . Then from the above equation, theheat transfer coefficient is
2⁄ 3

ℎ2

𝐷

𝑅𝑒

1

2

1

𝑅𝑒2

𝑑

1

1

(20.4-14)= (𝐷1 ) (𝑅𝑒2 )ℎ

From Eq. (20.4-5), 𝑁2 ⁄𝑁1 = (𝑑1 ⁄𝑑2 )2⁄3 under the condition of 𝑁𝑝2 = 𝑁𝑝1 . Then Eq.(20.4-6)
becomes
2

4⁄ 3

𝑁

𝑑

1

1

= (𝑑2 ) (𝑁2 ) = (𝑑2 )𝑅𝑒(20.4-15)= 𝜆4⁄3The Reynolds number becomes large but the rotation number of the agitator is made small.
Finally
𝑄2

ℎ

2

𝐷

𝐷

𝑅𝑒

2

1

2 ⁄3

= ℎ2 (𝐷2 ) = [(𝐷1 ) (𝑅𝑒2 )
𝑄
1

1

1

𝐷

2

𝑑

8⁄ 9 𝐷
2

1

1] ( 2) = ( 2)𝐷𝑑1

(𝐷 ) = 𝜆17⁄9 ~ 𝜆1.89(20.4-16)

From Eq.(20.4-14), the heat transfer coefficient is given by
ℎ2
ℎ1

𝐷1

⁄
𝑅𝑒2 2 3= (𝐷 ) (𝑅𝑒 )
2

1

𝐷1

⁄ 2⁄3
𝑑2 4 3= (𝐷 ) [(𝑑 )21]= 𝜆−1⁄9(20.4-17)

It has been found that the heat transfer coefficient for the larger-scale system becomes a little bit
smaller than that for the smaller-scale system.

Nomenclature
𝐷
𝐹𝑟
𝑔
𝑏
𝑑
𝐻
𝐻𝑝
ℎ𝑖
𝐽
𝑙

agitated tank/vessel diameter, [m]Froude number, [ - ]gravitational acceleration, [m/s2]
impeller width, [m]diameter of impeller, [m]depth of liquid in vessel, [m]height of impeller above vessel floor, [m]heat transfer coefficient on the inside surface of vessel, [W/m2K]
width of baffles, [m]length of impeller blades, [m]Nomenclature
𝑁
𝑁𝑝
𝑁𝑞
𝑁𝑢
𝑃
𝑃𝑟
𝑃𝑣
𝑄
𝑅
𝑅𝑒
𝑇
𝑣𝜃
𝜅
𝜆
μ
𝜌
𝛷
𝜔

rotation number of agitator, [1/s]
power number, [ - ]flow number, [ - ]Nusselt number, [ - ]power consumption, [W]Prandtl number, [ - ]power consumption per unit volume, [W/m3]
heat exchange rate, [W]length of rotating lever (= d/2), [m]Reynolds number, [ - ]temperature, [K]peripheral velocity, [m/s]thermal conductivity, [W/m K]
scale-up length ratio, [ - ]viscosity, [kg/m s]density, [kg/m3]
modified power number, [ - ]angular velocity, [1/s]

Subscripts
w
wall
1
small-scale system
2large-scale system scaled up

241

242

243

INDEX

Absorption, 109, 110
Chemical absorption, 179~190,
Design (chemical absorber), 161,
Gas solubility in electrolytic solution, 187,
Higbie’s penetration theory, 67,
HTU, 114, 115,
Limiting liquid-gas ratio, 113,
in liquid jet, 65,
Mass transfer model, 110,
Mass transfer coefficient, 106, 109, 114,
Mass transfer correlations, 114~116,
NTU, 114,
Packed columns, 111, 114, Column diameter, 116,
Pressure drop, 116,
Packing, 110, (random, structured)
Reaction factor, 185,with chemical reaction, 179~190,
Ackermann correction factor, 153,
(see Humidification)
Agitation/agitators, 232,
Agitators, 233,
Dimensional analysis, 235,
Flow pattern, 233, Froude number, 236,
Heat transfer correlation, 238,
Power consumption, 235, Power correlation, 236,,
Power number, 235, modified, 237,
Reynolds number, 235,
Scale-up, 238,
Analogy
between momentum and heat transfer, 216,
between heat and mass transfer, 150, 216,
in packed column distillation, 145, 150,
Chilton-Colburn, 217, Colburn, 156,
Length Reynolds number, 149, 213, 216,
Annulus/Annular flow passage
Double tube exchanger, 89, 93,
Heat transfer, 93,
Rotating cylinders, 9, 50,
Bend, 23, 79, (see Pipe fittings)
Bernoulli equation, 26, 77, (see Mechanical energy)
Bingham model, 14, (see non-Newtonian)Bingham plastic, 14, 52,
Blasius formula, 76, (see Friction factor)
Boiling, 174,
Boiling curve, 175, Boiling heat transfer, 176,
Critical heat flux, 177,
Heat transfer correlation, 176,
Nucleate boiling, 175, Pool boiling, 175,
Boundary layer, 211,
Analogy theory, 216, Energy boundary layer, 213,
Energy integral equation, 218,
Free convection boundary layer, 226,
Friction factor, 216,
Integral boundary layer equation,217,
of Ionic mass transfer, 223, j-factor, 216,
Laminar flow over a flat plate, 212,
Length Reynolds number, 213,
Stream function, 211,

Thermal boundary layer, 213,
Turbulent boundary layer, 218,
Buffer zone/layer, 206, (see Universal velocity profile)
Buoyancy forces, 226, (see Free convection)in flow near heated plate, 226, 228,
Chemical reaction, 181~185,
Electrochemical reaction, 164~165,
Gas absorption, 179, Reaction factor, 181, 184, 185,
Condensation, 169,
Dropwise, 169, Filmwise, 169, 170,
Heat transfer, 169, 171,
Overhead condenser, 171,
Conduction
Conductivity, 9,
Steady, 60, Unsteady, 62,Conservation law, 20, (see Macroscopic balance)
Control volume/control volume approach, 20,
Energy balance, 24, Mass balance, 20, Momentum
balance, 22,
Packed column model, 147,
Convective heat transfer
Circular pipe, 87, 93, Submerged objects, 102,
Cylinder, 102, sphere, 104,
Convergent nozzle, 24, 222,
Critical heat flux, 177, (see Boiling),
CylinderCoaxial rotating cylinder, 50,
Heat transfer in cross flow, 102,
Submerged cylinder, 102,
Differential balance, 37, (see Microscopic balance)
Diffusion
Diffusion-controlling, 165,
Equimolar counter, 145, Fick’s law, 12,
Limiting current, 165, Unsteady diffusion, 182,
with chemical reaction, 182,
Diffusivity (Fick’s law), 12,
Definition, 12, Ionic diffusion, 163, Mass flux, 13,
Molar flux, 13,
Dimensional analysis,of Agitated tanks, 235,
of Channel/pipe flows, 75,
Dissipation, 203, (see Turbulence)
Distillation,
Analogy, mass and enthalpy, 150,
Boiling-point diagram, 125,
Column design 139, Condenser, 171,
Continuous distillation, 123,
Control volume approach, 147, 149, (packed column)
Design calculation, 139,
Efficiency
HETP, 147, 149, (packed column)
Murphree efficiency, 136, modified 141,
Enthalpy balance, 131, 143, 146,
F-factor, 137, 138,
Heat/enthalpy balance, 140, (plate), 149, (packed)
Ideal plate or stage, 125,
j-factor, 149,
Mass transfer model,
Plate column, 133, 136, Packed column, 145~148,

244
HTU, 146, NTU, 137,
McCabe-Thiele method, 128, 129,
Operating line, 127, 128,
Overall transfer coefficient, 146,
Overhead condenser, 123, 171,
Packed column, 145,
Phase rule, 121,
Plate column, 123, 135,
Ponchon-Savarit method, 131,q-line, 129,
Raoult’s law, 121, Reboiler 123,
Rectifying section, 127, Reflux ratio, 130,
Relative volatility, 122, 128,
Reynolds numbers, 137, (tray), 149, (packed)
Stripping factor, 135, 147, Stripping section, 127,
Tray model, 136, Two-film theory, 135, 146,
Vapor-liquid equilibrium constant, 121,
Downcomer, 124, (see Tray)
Eddy
Eddy diffusivity, 196,
Eddy scales, 204, Length scale, 204, Time scale, 206,
Elbow, 32, 79, (see Pipe fittings)
Electrolytic reaction, 164, 166,
Electrochemical reaction method, 165~168, 223,
Electrolytic cell, 164, Limiting current, 165,
Mass transfer measurement, 165,
Velocity-gradient measurement, 167,
Energy spectrum, 200, (see Turbulence)
Enthalpy balance, 156, 160, (see Humidification)
Equations of change
Differential balances (Microscopic balance), 37,Continuity, 37 Energy, 42, Mass, 43,
Momentum, 38,
Application (momentum): Circular pipe flow, 46,
Rotating cylinders, 49, Non-Newtonian, 52, 53,
Application (energy): Circular pipe, 56, Hollow
cylinder, 60,
Application (mass): Liquid jet, 65,
Turbulent flows: motion, 194, energy, 194, mass, 194,
Equivalent diameter, 81,
Definition, 81, Hydraulic diameter, 81,
Non-circular channel, 81, Shell side 94,
Equivalent length, 77,
Eulerian viewpoint, 3, 204,
Evaporation 174, (see Boiling)
Evaporative cooling, 159,
Operating line, 161,
Wet-bulb temperature, 154,
F-factor, 137 (plate), 150 (packed), (see Distillation)
Fick’s law of diffusion, 12, 44, 106,
Flooding, 117, (see Packed column)
Flow work, 25, (see Macroscopic energy balance)
Fluid flows
Body force, 4, surface force, 4, static pressure, 4Force balance, 23, 24, Tube flow, 23,
Bend, 23, Convergent nozzle, 24,
Reynolds number, 16, 18,
Laminar, Newtonian fluids, 17,
Non-Newtonian fluids, 52, 53,
Velocity profile in pipe flow, 16, 48,
Turbulent, Energy spectrum, 200,
Intensity, 71, 193,
Kinetic energy, 193, in pipe flow, 193,
Reynolds stress, 195,
Scale of turbulence, 203,

Transport flux, 195,
Turbulent fluctuations, 18, 70, 84, 192,
Velocity distribution, 205,
Flux 5, (cf. rate)
Definition, 5, Heat, 10, Mass, 11, Molar, 11,Momentum, 8, Ionic mass flux, 166,
Turbulent: heat, 195, mass, 195, momentum, 195,
Fouling, 88, (see Heat exchanger)
Fourier’s law of heat conduction, 9,Free convection, 226,
Boundary layer theory, 226,
Convective heat transfer, 228, Grashof number, 228,
Heat loss, 229, Heat transfer correlations, 228,
Nusselt number, 228,
Friction factor/Friction loss, 73,
Definition, 73, 74, Drag coefficient, 82,
Fanning, 49, 74, Friction loss, 26,
Friction loss factor, 77,of laminar pipe flow,74, 75,
for noncircular channels, 81, 96,
for heat exchangers, 96,
for dry packed column, 118,
for turbulent boundary layer, 220,
Froude number, 236, (see Agitation)
Grashof number, 228, (see Free convection)
Hagen-Poiseuille equation, 48,
Heat conduction, 9,
Steady, 60, Unsteady, 62,
Heat exchangers, 92,
Engineering design, 96,
Double-pipe, 89, 93,
Design, 89, Heat transfer 93,
LMTD (logarithmic mean overall temperature
difference), 89,
Overall heat transfer, 87, 88,Heat transfer pipe layout, 94,
Shell-and-Tube 92,
Baffles, 94, Correction factor, 96, Design, 96,
Overhead condenser, 171,
Shell-side heat transfer, 94,
Shell-side pressure drop, 96,
True temperature difference, 95,
Tube-side heat transfer, 93,
Heat transfer
in boiling, 174,in circular pipe flow, 56, 93,
Circular cylinder in cross flow, 102,
Coefficient (definition), 59, 85, 86, Correlation, 171,
Pool boiling heat transfer, 176,
in condensation, 171, in distillation, 149,
Condensation heat transfer, 170,
Free convection, 226, Heat loss, 229,
Henry’s law, 68, 108,
Hot-wire anemometry, 71, 103, 193,
HETP (height equivalent to a theoretical plate),
147, 149, (see Packed column distillation)
HTU (height of a transfer unit), 113, 146,
(see Packed column absorption & distillation)
Humidification, 152,
Adiabatic cooling line, 157, Enthalpy balance, 153,
Humidity, 157, Lewis relation, 156, Operating line,
160, Water-cooling, 159, Wet-bulb temperature,
154,
Hydrostatics, 4Ideal plate/ideal stage, 125,

245
Impinging flow, 220,
Stream function analysis, 220,
Impinging jet flow, 222, Heat transfer, 222,
Instability
Boundary layer, 218, Circular pipe flow, 17,
Rotating coaxial cylinders, 51,
Integral equation, 2117, (see Boundary layer)
Interphase transfer
Momentum transfer, 70, 73,
Heat or enthalpy transfer, 85, 143, 145,
Mass transfer, 106, 146, 155,
Ionic mass transfer, 163,
Electrode reaction, 163, 164,
Electric double layer, 165,
Limiting current, 165,
Molar-flux of ions, 163,
Supporting electrolyte, 165,
Jet
Impinging jet, 222, Liquid, 65,
j-factor
j-factor for heat transfer, 87, 93, 108, 149, 216,
j-factor for mass transfer, 108, 149, 156, 217, 220,in boundary layer, 217, 220,
Lagrangian viewpoint, 3, 205,
Laminar flow, 16,
in boundary layer, 212,
in circular pipe, 16, 47,
in rotating coaxial cylinders, 50,
Length Reynolds number, 216,, (see Boundary layer)
Lewis number, 13, 156, (see Humidification)
Limiting current, 165, (see Ionic mass transfer)
LMTD (Logarithmic Mean Temperature Difference),
89, 90, (see Heat exchangers)
Macroscopic balanceof mass, 20, of momentum, 22, of thermal energy, 24,
31, of mechanical energy, 26, of individual component,
34,
Mass-averaged, 11, (cf. Molar-averaged)
Mass balancein humidification, 152, 153,
in packed column, 111,
in plate column, 133,
Mass transfer
Coefficient (definition),106, 109, 114,
correlations, 114~116,
interphase transfer, 106,
Model for gas absorption, 110, 113,
Penetration theory, 67,
Model for distillation, 133 (plate), 145~147) (packed),
McCabe-Thiele method, 128, (see Distillation)
Mechanical energy
Balance, 26, Bernoulli equation, 26, 77,
Mechanical energy loss, 26, 77, Pipe fittings, 77,
Microscopic balance (Differential balance), 32,
of energy, 42, of mass, 37, 43, of momentum 38,
Mixing length theory, 196, 197,
Eddy diffusivity, 196,
Molecular or molar
Diffusion, 12, Fick’s law, 12, 44, 106,
Flux, 5, Molar-average velocity, 11,
Natural convection (see Free convection)
Navier-Stokes equation, 38, 41,
Newton’s law of viscosity, 6,
Non-Newtonian fluid, 14,
Bingham model, 14, 52,

For circular pipe flow, 52,
Power law model, 14, 53, (Ostwald-de Waele Model)
Yield stress, 14, 52,
NTU (number of transfer unit),
114, (see Packed column), 117, (see Plate column)
Nucleate boiling, 175,
Nusselt numberof laminar pipe flow, 56, 93,of turbulent pipe flow, 87, 93,of free convection, 228,
One-seventh power law, 18, 219,
in boundary layer, 219, in tube flow, 18,
Operating line
in gas absorption, 112, (see Absorption)in distillation, 127, 128, (see Distillation)
in water cooling, 161,
Overall heat-transfer coefficient, 88, 90, 149,
Overall mass-transfer coefficientof absorption, 110,of distillation, 146,
Packed column
Absorption, 109,
Design, column diameter,116,
Pressure drop, 116 (wet), 118 (dry),
Column height, 113, HTU,114, 115,
Distillation, 145, Flooding, 117,Heat or enthalpy transfer, 146,
Limiting liquid-gas ratio, 113,
Mass transfer, 114, correlations, 114~116,
Packings, 110,
Random packings, 110, Structured packings, 111,
Penetration theory, 67,
PipingEquivalent length, 77,
Fittings, 77, Friction loss factor, 77,
Pipeline, 32, 79, Valves, 78,
Pitot tube, 29,
Plate column
Bubble-cap plate, 124, Sieve plate, 124,
Downcomer, 124,
Reboiler, 123, Overhead condenser, 171,
Ponchon-Savarit method, 131,
Power law model, 14, (see non-Newtonian)Ostwald-de Waele model, 53,
Power number, 235, (see Agitation)
Prandtl number, 13, (definition)
Pressure
Static pressure, 4,
Pressure drop
Circular pipe, 49 (laminar), 74 (turbulent),
Packed column, 116 (wet), 118 (dry),
Shell-and-tube exchanger, 96,
Pumps (see Mechanical energy balance)
Power requirement, 26, 33,
Raoult’s law, 121, (see Distillation)
Rate (cf. flux)
Definition, 4,Rate of momentum,22, energy, 25, mass, 21,
Rectifying section, 127, (see Distillation)
Reflux and reflux ratio, 130, (see Distillation)
Relative volatility, 122,
Resistance
Thermal resistance, 62
Mass transfer resistance, 67, 110,
Reynolds number

246
Definition, 16, 18,
in agitation, 235,
in boundary-layer flow, 213,
(Length Reynolds number)in circular pipe flow, 16,
in coaxial rotating cylinder, 51,
in heat exchangers(tube-side, 93, shell-side, 94, 96,
double-tube, 93)
in packed column flow, 149, in plate column, 137,
Length Reynolds number, 149, 213, 216, 218,
Reynolds stresses, 195, (cf. turbulent momentum flux)
Roughness
Relative roughness in pipes, 76,
Schmidt number, 13, (definition)
Shaft work, 25, (see Macroscopic energy balance)
Shear stress, 6, 7,
ShellsShell-and-tube exchangers, 92,
Shell-side heat transfer, 94,
Shell-side pressure drop, 96,
True temperature difference, 95,
Sherwood number, 115, (definition)
SI units (Systeme International), 3,
Similarity
among molecular transports of momentum,
energy, and mass, 13,
Simultaneous transfer,
143 (distillation), 152 (humidification),
Chilton-Colburn analogy, 217,
Mass and enthalpy, 143, 152,
SphereDrag force, 82, Heat transfer, 104,
Stage-by-stage calculation, 128,
( see McCabe-Thiele method)
Stagnation point,
Heat transfer, 222, Impinging jet, 222,
Stanton number, 59, (definition)
Stream function, 211, application, 226,
Stripping factor, 135, 148, ( see Distillation)
Stripping section, 127, (see Distillation)
Superficial velocity, 118, 137, 149, (see Packed column)
Taylor vortex flow, 51, Taylor number, 51,
Tensor, 3,
ThermalConductivity (Fourier’s law), 8,
Diffusivity, 8,
Resistance 62,
Time-averaged, 84, 192,
TorqueRotating cylinder, 51, Agitator, 234,
Transfer Units (see HTU, packed column)
Transition
Laminar to turbulent flow, 17,
Boundary layer flow, 218,

Transport intensity, flux and rate, 4Tray, 124, (see Distillation)
Crossflow tray, 124, Dual-flow tray, 124,
Bubble-cap, 124, Sieve, 124,
Downcomer, 124,
Plate efficiency/ Murphree efficiency, 136,
Turbine impeller, 233,
Turbulence
Correlation functions, spatial/temporal, 203,
Energy spectrum, 200, Intensity, 71, 193,
Generation, 17, Kinetic energy, 193,
Length/time scale, 203, Scale of turbulence, 203,
Sizes of eddies, 203, Structure, 200,
Viscous dissipation, 203,
Turbulent flow
Boundary-layer flow, 211,
Circular tube flow, 17,
Fluctuation, 18, 70, Intensity, 71, 193,
Turbulent core, 206,
Turbulent flux of heat and mass, 195,
Turbulent transport, 194,
Two-film theory, 67,
Vapor-liquid equilibrium constant,(see Distillation)
Valves, 78,
Check, 79, Diaphragm, 78,
Gate, 78, Globe, 28, 79,
Vector, 3,
Unit vector, 21, Vector equation, 38, 41, 43, 44,
VelocityF-factor, 137, 150, (see Distillation)
Friction velocity, 206,
Velocity fluctuation, 18, 70, 84, 192,
Velocity profile: 1/7th power law, 18, 219,
Mass-average velocity, 11 (see Diffusivity)
Molar-average velocity, 11 (see Diffusivity)
Non-Newtonian, 53, 54,
Superficial velocity, 118, 137, 149,
Universal velocity profile, 207,
Viscous sublayer, 206, buffer layer, 206, turbulent
core, 206,
Viewpoints
Eulerian, 3, 204, Lagrangian, 3, 205,
Viscosity
Definition (Newton’s law), 7, Kinematic, 8,
Volumetric transfer coefficient
Absorption, 113, Distillation, 146~149,
VortexCellular vortex, 51, (Taylor vortex)
Water-cooling, 159, (Packed bed)
Enthalpy balance, 161,
Operating line, 161, Tie line, 161,
Wet-bulb temperature, 154, (see Humidification),Yield stress, 14, 52, (see non-Newtonian fluid)