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Fundamentals for Thermal
Environment Engineering
熱環境工学基礎
1st & 2nd partProfessor TANIMOTO, Jun
谷本 潤 教授

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Human – Environment – Social (HES)
SystemHolistic andBridging fororganic viewpointis requisite.

Environment

different spatial
scale systems

Building
Urban
Vegetation

Each of the three isinter-dependently
as well as interpermeably
connected.

Social system

Human

Bridging fordifferent physicalsystems

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Globe scale

Urban scale

Global warming

Urban Heat Island

106 105∞
Spatial scale [m]

104

103

102

The physical environmental
system surrounding us
extends wide range ofspatial scales.

Building scale
Ventilation for street
canyon & pollutants
dispersion

Room scale
Energyconservation;robust and efficient
controllability

101

Individual productivity

Human scale
100
10-1

10-2
-∞

Thermal comfort,
IAQ, visual &acoustic comfort

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Contents #1
This class provides you fundamental knowledge of
Building Physics (建築環境工学) and Building Service
Engineering (建築設備学). Both of those consist of solid
infrastructure for Built Environmental Engineering (生活圏環境工学) that deals with theoretical backgrounds
as well as down-to-earth applications to make our daily
environment safe, healthy, comfort and more productive.
First off, in the former part, Prof. Tanomoto gives an
overview for the theoretical template for Building
Physics. In the latter part, Prof. Tanimoto follows to makeyou understand what Building Service Engineering is asa whole.

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Contents #2- Building Physics; fundamental theory for heat transfer,ventilation, human comfort as well as building exterior
environment is primarily addressed in the lecture.
- This class is basically “chalk & talk”. Instead of
blackboard with chalk, I will use PowerPoint slides.
- Although you can DL slides through my web site, you’re
still requested to write down supplementary issues,
additional notes and whatever you think as important.
- Reviewal is expected to avoid over-night studying justbefore exam.
- The semester-end exam will score your final evaluation.
- You can DL materials for the class from the web site;
http://ktlabo.cm.kyushu-u.ac.jp/

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Fundamental theory for heat transfer #1
Conduction

Convection

Solid - solid

Solid surface- fluid

Radiation
surface - surface

Thermalboundary layer

wall

Direction of heat flux

Direction of heat flux

Net direction of heat flux causedby radiative heat exchange

Latent heat transfer by evaporation/ condensation
Xf
temperature
100 oC

Vapor boundary
layer

Xs

Thermal energy that you can observe as
temperature increasing is called sensitive heat.
Direction of evaporative cooling flux
time

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Fundamental theory for heat transfer #2
Conduction
Solid - solid

Convection

Radiation

surface - [Wsurfacem-1 K-1]
Solid surface-Thermal
fluid conductivity

q1
wall

ql
Thermalboundary layer

q1  q2
x2  x1>0

q2
Direction of heat flux

Direction of heat flux

x1

Fourier's law;Conductive heat fluxx2

qqcond  l
x

[W m-2]

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Fundamental theory for heat transfer #3
Thermal conductivity[W m-1 K-1] lSpecific heat
[J kg-1 K-1] Cp

Density [kg m-3]

Concrete (normal)

1.4

880

2200

Mortar

1.5

800

2000

Plaster

0.79

840

2000

Glass

1.0

750

2500

Tile

1.3

840

2400

Brick

0.64

840

1700

Asphalt

0.11

950

1000

Carpet

0.08

800

400

Wood (middle weight)

0.17

1300

500

Plywood

0.19

1300

550

Glass wool (24K)

0.042

840

24

Foam polystyrene

0.037

1300

28

Soil (sand)

0.9

1300

1600

Cooper

390

0.39

9000

Air space (well sealed)

R= 0.07 m2 K W-1

Thermal property
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Fundamental theory for heat transfer #4
Let us suppose a small cubic (dx, dy, dz), and evaluate theDxstored heat during dt.

Dz Dy

Outflow

Inflowdq
dydzdt
l
dx
z

Dx
y

q

d 
dq 
 l q dx dydzdt
dx 
dx 
dq
q  dx
dx
dq
dq
x dx

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Fundamental theory for heat transfer #5
How the stored heat in x direction can be expressed?

Dx

Outflow

Inflowdq
dydzdt
l
dx
z

Dx
y

q

Dz Dy

d 
dq 
 l q dx dydzdt
dx 
dx 
dq
q  dx
dx
dq
dq
x dx
2


q
(Stored) = (inflow) – (outflow) =l
[J]
dx

dy

dz

dt
2
x

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Fundamental theory for heat transfer #6
How it becomes if considering all 3 directions?

Dx

Outflow

Inflowdq
dydzdt
l
dx
z

Dx
y

q

Dz Dy

d 
dq 
 l q dx dydzdt
dx 
dx 
dq
q  dx
dx
dq
dq
x dx
2

  2q  2q  q (Stored) = l  2  2  2 dx  dy  dz  dt
y
z 
 x

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Fundamental theory for heat transfer #7
What this stored heat brings about as physics phenomenon?

Dx

Outflow

Inflowdq
dydzdt
l
dx
z

Dx
y

q

Dz Dy

d 
dq 
 l q dx dydzdt
dx 
dx 
dq
q  dxdx
dq
Specific heat [J kg-1 K-1]
dq
Density [kg m-3]
x dx
2

  2q  2q  q (Stored) = l  2  2  2 dx  dy  dz  dt  C p r  dq  dV
y
z 
 x

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Fundamental theory for heat transfer #8
Thus…

This physical property expressing thermal mass…
The physical property expressing how it efficiently transfer heat…

2

2

2

 q  q  q q
Cpr  l  2  2  2   l  Dq
t
y
z 
 x
Note this is analogous to Newton’s Equation of motion for aparticle.

vm a  m   ftThat is, of course…making
the particle accelerate.

Sum up all forces acting to theparticle. And what it happen
on the particle?

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Fundamental theory for heat transfer #9
Convection

Radiation
Convective
heat flux; (qs > qf )surface - surface

Solid surface- fluid

qconv  a c q s  q f [W m-2]

Convective heat transfer coefficient [W m-2 K-1]
Thermalboundary layer

Buckingham P Theory gives the universal relation of
Non-dimensional Numbers for convective heattransfer.
abc

Direction of heat flux

Convective heat transfer
coefficient [W m-2 K-1]

ac  
l
Thermal conductivity [W m K ]Nusselt number Nu 
-1-1

Representative velocity [m s-1]

Reynolds number Re 
Dynamic viscosity [m2 s-1]

Nu  k  Re  Pr  Gr

Representative
length [m]Prandtl number Pr 

Thermal diffusivity [m2 s-1]
coefficient of thermal
Inertia force
expansion [K-1]u Density [Kg m-1

Specific heat [J kg-1 K-1]

Grashof number

n Visors force

n
a

Gr 

Gravity 9.8 [m s-2]

n 

Cp  rlTemperature difference [K]g    Dq   3n2

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Fundamental theory for heat transfer #10
Free (Natural) convection

u

qf

qs

Dominant

Forced convectionqf

qs

Reynolds’s experiment (1883)
Turbulent flow

Laminar flow

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Fundamental theory for heat transfer #11
Modegeometry

Forced

Plate

Free

Verticalplate
Horizontal
plate

Laminar or turbulent/
upward or downward
Laminar
Turbulent
Laminar
Turbulent
Upwardheat flux
Downward
heat flux

Representative
lengthFetch from theedge
Length of thevertical-side

Laminar

Length of thehorizontal-side

Turbulent-Threshold oflaminar/turbulentRe<5*105
Re>5*105
Gr･Pr<109
Gr･Pr>109
Gr･Pr<2*107

Constants
inuniversal relationkab
0.664 1/2 1/3
0.037 4/5 1/3
0.56
0
1/4
0.13
0
1/3
0.54
0
1/4

c
0
0
1/4
1/3
1/4

Gr･Pr>2*107
-

0.14
0.27

1/3
1/4

0
0

1/3
1/4

Dynamic viscosity-6-1]]
×10 [m[m2ss-1

22
20
18
16

linear relation ;q (temperature) vs n (dynamic viscosity)

14
12
10
0

10

20

30

40

50

60

70

80

90

Air temperature [℃]

Dynamic viscosity of air varying with temperaturetheGenerated by Foxit PDF Creator © Foxit Software
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Fundamental theory for heat transfer #12
Electromagnetic wave

Radiation
surface - surface

Radiative heat flux from Facet 1 & 2

4

qrad _1  e 1sT1
4
qrad _ 2  e 2sT2

[W m-2]

Stephan – Boltzmann Constants  5.67 10 8

Net direction of heat flux caused by radiative
heat exchange

Cb  5.67[W m-2 K-4]

e : Emissivity [ND] (= Absorptance [Kirchhoff’s law])
Why sT4 ?
Plank’s law

I b (l ) 

3.741  108

l5 exp14388  l1T 1   1[W m-2 mm-1]

Stephan-Boltzmann’s lawRb   I b ( l )dl  sT 4 [W m-2]0Generated by Foxit PDF Creator © Foxit Software
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Fundamental theory for heat transfer #13
Let us evaluate net radiation flux from Facet 1
to Facet 2.
It must be a difference between absorbed
radiation by 2 out of emitted one from 1 andabsorbed radiation by 1 out of emitted one
from 2.
42
1
4

H12  e 2 F12 A1e 1sT1  e 1 F21 A2e 2sT2
8Stephan – Boltzmann Constant s  5.67  10

Cb  5.67
4[W m-2 K-4]

[10-8 ×W m-2 K-4]

 T1 
 T2 
 e 1e 2 A1 F12Cb   e 1e 2 A2 F21Cb  100  100 4

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Fundamental theory for heat transfer #14 2
Reciprocal Relation on Geometric View Factor

F12

A1 F12  A2 F21
 T1  4  T2  4 
H12  e 1e 2Cb A1 F12    100   100  

F211

  T1  4  T2  4 
 
 

100   100   A1 F12  e 1e 2Cb q1  q 2 
T1  T2
 Temperature coefficient
Emissivity…It’s around 0.9 – 3
0.95 except for metal or


T

T
/
2


glass surfaces.
 0.04 1 2This can be treated as a constant value.

a rad  0.9  0.9  5.67 1  4.6 [W m-2 K-1]
Radiative heat transfer coefficient

100



1
If it’s cold…Substitute
T=300 [K] for example.

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Point – Facet View Factor;
f=a/

a

1

b

Formulas of VFs based on analytic solution

1

b

半径=1
面積=a

a

c

Point – Facet VF

p
①
a
1
b

c

②

b
1  1
1
1 
1
tan
① f   tan x 
2 
1 y2
1  y 2 
2
c
c


1 
x
y
y
x 
1
1
tan
tan

③
④
② f  2 
2
2
2
1 x
1 y
1  y 2 
 1 xFacet – Facet VF

p
a

1
2


1 y2
1  1 1x
1
1 x 
 tan③ F    y tan x  y tan
2
y 
1 y


④

2  1  y 2F 
tan 1 y 1x2  y2 1 x2 
y 1 x2  y2
1 x2  y2xy
 2 xy ln 1  x 2 1  y 2  2 x ln x 2  y 2 1  y 2  2 y ln x 1  x 2  y 2  1
1 x2 1 y2 
1
1 x2
1
x
y

1
1
1

ln
tan

 tan x   tan y 
2
2
2
2
xy
x
y
1 y
1 x
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Chart for Point – Facet VF ①

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Chart for Point – Facet VF ②

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Chart for Facet – Facet VF ③

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Chart for Facet – Facet VF ④

b
a
d
X=a/d
Y=b/d

Y

X

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Fundamental theory for heat transfer #15
Mean Radiative Temperature
放射率=1

Point ofevaluation

表面温度=平均放射温度ｐｐSurfacetemperature [oC]pに入射する放射熱総量

Surface area[m2]

A hypothetic temperature for the evaluation point
surrounded by a uniform globe having e = 1 withhomogeneous surface temperature, which canemit same amount of radiative flux as the realasymmetric radiation field.

MRT 

4

p

4

[oC]


f

e

q

273
.
2

273
.
2
 p i i i

A q

[ C]
Average Surface temperature; AST 
A
i

i

o

i

If the closed space assumed is like a rectangular-box shape, MRT
approximates AST.

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Fundamental theory for heat transfer #16
Latent heat transfer by evaporation/ condensation
Xf

Vapor transfer flux

Xs

Vapor boundarylayerm  k X s  X f [kg m-2]

Mass transfer coefficient[kg m-2 (kg/kg’)-1]

Imperfectly wet surface
Direction of evaporative cooling flux

Xf
Heat flux brought by mass transfer

q evap  L  m[W m-2]

Vapor boundary
layer

Xs

Latent heat [J kg-1] water; 2500kJ kg-1 (at 0.6112kPa & 0oC) Perfectly wet surface
If the surface is perfectly wet like water; Saturated absolute humidity at surface temperature; qs.
Hererogeneous surface
Homogeneousm  k X sat s  X f
1- rwetsurface
If the surface is imperfectly wet like soil
Normalized area = 1
Represented by revap
Evaporative
ratio,or building materials;



q 



evaporation efficiencym  k revap  X sat q s   X f wetnessm  rwet  k X sat (qs )  X f 

Microscopic view

Macroscopic view

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Psychrometric chart

You can DL ET_AEE from the web-site of my lab.

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Fundamental theory for heat transfer #17
Analogy between mass transfer and convective heat transfer

qconv  a c q s  q f  [W m-2]
[kg m-2]m  k X s  X f 
Sherwood number Sh  h D  

DMass transfer coefficient [m s-1]

Schmidt number

Sc 

Nu  k  Re a  Pr b ･･･(1)abSh  k  Re  Sc ･･･(2)(6), (7) ⇔

k

Cp

Le

b 1

ac  
lPrandtl number

n

Pr 

D

 Nu   Pr     ･･･(4)(1)/(2) ⇔  Sh   Sc a
Le b 1(3), (4), (5) ⇔ hD 
C p r air

ac

Nu 

Mass diffusive
coefficient [m s-2]

Reynolds number Re b

Nusselt number

Representative
length [m]

n

n 

u 

a

n

a

Cp  rll･･･(3)

C p r humidair
Pr n / a D1 ･･･(5)
Sc n / D a Le
Lewis Number･･･(6)hD  k / r humidair ･･･(7)One can assume Le  1 under usual flow situation.

1.0 kＪ kg-1 K-1

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Fundamental theory for heat transfer #18
Heat flux under steady state for a multi-layer wall
Water

Water flow rate (flux) is always
same, wherever you measure.

A river with time-constant flow-> steady stateHightemperature

Heat

Low
temperature

Likewise, heat flux MUST be consistent, wherever you measure.

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Compositive thermal conductancel1

l2

Imagine Ohm's Law. You can add
two parts only in resistance unit.

Dx2

Dx1

1
Dx1

l1



→ R1+R2+R3 []Dx2Comprehensive thermal conductance(U-value) of a multi-layer wall

[W m-2K-1]

l2
qout [K]･･･

ain

aout [W m-2 K-1]

U

[W m-2 K-1]

1

l1

1
1
 

a out ik li Ck 1 a in

[W m-1 K-1]

1

lk-2 Air space lk

l1 [m] l2

li

l2

lk-2

Ck-1 lk[W m-2 K-1]

qin [K]

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Fundamental theory for heat transfer #19
lk-2 Air space lk

l1 [m] l2

q1 q

qout [K]aout

2[W m-2 K-1]

q3
l1

ain

･･･

l2

[W m-1 K-1]

qk-2

qk-1
lk-2

qk

[W m-2 K-1]

qk+

Ck-1 lk 1[W m-2 K-1]

qin [K]

Given qout and qin, let us solve each temperature; q1 ･･･ qk+1.
q  U  q out  q in  ･･･(1)1q out  q k  ･･･(2)
q
li
1
1
 
a out ik li C k 1
By solving Eqs. (1) & (2) simultaneously, you get qk. By way like
this, temperature of any arbitrary point in the wall can be solved.

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Fundamental theory for heat transfer #20
lk-2 Air space lk

l1 [m] l2

q1 q

qout [K]aout

2[W m-2 K-1]

q3
l1

ain

･･･

l2

[W m-1 K-1]

qk-2

qk-1
lk-2

qk

[W m-2 K-1]

qk+

Ck-1 lk 1[W m-2 K-1]

qin [K]

Given qout and qin, let us solve each temperature; q1 ･･･ qk+1.
q  U  q out  q in  ･･･(1)q  a out  q out  q1  ･･･(2)q  a in  q k 1  q in  ･･･(3)Both surface temperatures can be solved as below:U
q1  q out  q out  q in 

a out

q k 1  q in 

U

a in

 q out  q in 

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Fundamental theory for heat transfer #21
lk-2 Air space lk

l1 [m] l2

q1 q

qout [K]aout

in [%]
2[W m-2 K-1]

q3
l1

l2

[W m-1 K-1]

q k 1  q in 

U

a in

ain

･･･

qk-2

qk-1

qk

[W m-2 K-1]

qk+

lk-2

Ck-1 lk 1[W m-2 K-1]

qin [K] q out  q in 
inIf qk+1 < DPT, surface
condensation happens
on the indoor surface.

Dew Point
Temperature;
DPT

DPTin[oC]←qin [oC]qin

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Fundamental theory for heat transfer #22
OutdoorIndoorBy how in-wall condensation
DBT can be prevented?
DPT

condensation

DBT
DPT
Insulation

DBT
DPT

Vapor proof

Usual (Default) wallconstruction

Putting the insulation
layer at outdoor-side

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Fundamental theory for heat transfer #23
lk-2 Air space lk

l1 [m] l2

q1 q

qout [K]aout

2[W m-2 K-1]

q3
l1

qk-2

l2

aout

[W m-2 K-1]

qk

[W m-2 K-1]

qk+

Ck-1 lk 1[W m-2 K-1]

SAT：Sol-Air Temperatureqout [K]

qk-1
lk-2

[W m-1 K-1]

qSAT

ain

･･･

q SAT  q out 

･･･

aI

a out

qin [K]
Absorptance of solar
radiation [ND]
Incidental solar
radiation [W m-2]

ain
[W m-2 K-1]

qin [K]

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Fundamental theory for heat transfer #24
Heat bridge

Passive solar system

Direct gain
OutdoorIndoorlow temperaturehigh temperature

Surface condensation,
Mold

Trombe wall

Green house

Terminology

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Building exterior environment #1 Solar radiation
Pyranometer

Solar constant [W m-2]

Io
Celestial
hemisphere

Diffusion by aerosol
Sky → secondary illuminant
Sky solar radiation;none directional
Direct solar radiation;
directional
Normalplane

solar radiation
I SH ;Skymeasured on aholizontal plane [W m-2]

solar radiation
I DN ;Direct
measured on a normal
plane [W m-2]

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Building exterior environment #2 Solar radiation
Automobile chasing device forsolar position
A tube for only
detecting directcomponent

Shielding ring to kick out direct
element component.

Measuring device for direct solar radiation
Measuring device for sky solar radiation

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Building exterior environment #3 Solar radiation
Bouguer’s theoretical relation
Sky transmittanceIo

Air massm

I DN  I o  P  I o  P

Io

sin h

Aerosolattenuatesradiation.

m

I DN

h

Berlage’s relation …bad for cloudy days

1
1  P1/sin h
I SH  I o  sin h. ln P2
1  14
Nagata’s relation …bad for cloudy days







I SH  I o 0.66  0.322 sin h 0.5  0.4  0.3 P  sin h sin h  1  P1/sin hMatsuo’s relation …good for cloudy days as well as fine daysI  12. I  sin h1  P1/sin h 1  PSHo. ln P1  14

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Building exterior environment #4 Solar radiation
Practical engineering problems require decomposed solar radiation by direct
and sky elements, because you do need those two to evaluate total solar
radiation on an arbitrary plane as below;
Incidental angle
I D  I DN  cos i I  I    I 1  cosq
SH
sky
SH
 S
2
Sky

Sky view factor;
Walltiltingangleq sky Solar
altitude SolarhazimuthaS

a Wall azimuth

1  cosq
2

Groundq

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Building exterior environment #5 Solar radiation
Incidental angle
I D  I DN  cos i I  I    I 1  cosq
SH
sky
SH
 S
2

Thus, this only
depends on date &time and geographicallocation.
Walltiltingangleq

cos i  cosq sin h  sin q cos h cosa  a 
Depending on time,
Depending on date date and longitude Latitude Solar declination Time angle
sin h  sin  sin d  cos  cos d cos tSolar
cos d sin t
altitude Solarsin a hazimuthcosh
asin h sin   sin dS
cos a cos h cos a Wall azimuth

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Building exterior environment #6 Solar radiation
Time anglet  15  ST  E  12   L  Ls

Standard Time;

Equation of time;

JST in Japandifference between
true solar time andmean solar time

Longitude
Longitudeof the place of the placeofTimezone;E  0.1645 sin 2 B  0.1255 cos B  0.025 sin Bn  81B  360 n; Jan 1st = 1, ･･･, Dec 31st = 365
365
Solar declination[deg]

135 deg incase of JST(GMT+9)d = 0 deg on Equinox, d = -23.45 deg on Wintersolstice and d =23.45 deg on Summer solstice.

sin d  sin 23.45  sin B
Solar declination is caused by the situation that the
globe axis is not perpendicular to the rotation orbit.

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Building exterior environment #7 Solar radiation
Terminology #1

Perimeter zone

Interior zone

Perimeter-less = Preventing or at least lessening Skin Loadoccurrence
Air Flow
Window (AFW)
Air flow removes reconvection
component once
absorbed by blind.
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Building exterior environment #８ Sky radiation
Terminology #2
glazinga  r   1

Transmission; Each of proportions varies byincidental angle and glazing’s
characteristics.

OutdoorAfter absorbed by
the glazing, outflowing to outdoorside through theform of convection &radiation.

Absorption; aIndoorAfter absorbedby the glazing,out-flowing toindoor-side
through theform ofconvection &radiation.

raa, r,  [ND]

Refraction; r

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Building exterior environment #9 Solar radiation
Terminology #3

Heat gain by solar radiation entails “Delighting” .
Celestial hemisphere
Illuminance measured at an
outside horizontal plane Lo

Delighting factor;ref
Li Ldir

L
i
DLF 
 i
Lo
Lo

Indirect component
Illuminance measured at anindoor horizontal plane Li
Direct component

Evenness = Lmin / Lmax

Evenness
Distancefromwindow

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Plane angle & Solid angle

Definition of plane angle


q
r

r

半径=1

radian[rad]
Point – Facet View
Factor; f = a / →Projected area
ratio of solid angle

circleq
globe

Definition of solid angle


r

S
 2
r

面積=a

View
objectsteradian[sr]Projected areaon the globe

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Light source
Divergenceof light flux

Luminous intensity
Illuminance

Receiving facet

Iq
Luminance : L 
Luminous intensity
S cos qof viewing direction
At a complete diffusive
facet, where Iq=In cos q, L
must be, regardless of q;

Luminous intensity
of normal direction
Apparent area

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Building exterior environment #10 Solar radiation
Terminology #4
Atrium space for amenity &architectural design, but hard forenvironmental engineering…
Extra space
can help tostore overheat,
which loses
through large
glazing area.

Screen, Shield andShading…those help
solar gain decreasing.

If not Extra space, thethermal environment
at the top floor issignificantly
deteriorated.

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Building exterior environment #11 Sky radiation
According to Stephan-Boltzmann’s law, everything even aerosol “as longas having absolute temperature” emits long wave radiation…
Aerosol, dust,
small ice,particles

Average emissivity of sky

qsky  e sky  s  Tair
Cloud height constant; 0.45
(low height) – 0.82 (high)4

Average emissivity Cloud amount; 0 (clear sky)of clear sky– 10 (perfect cloudiness)CC
e sky  1  K   e scl  K
10 
10


e scl  0.711  0.56 10 2  DPTair  0.73 10 4  DPTair 2

Berdahl & Martin (1984)

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Building exterior environment #12
New York

Climograph

Tokyo

BDT [oC]

LondonMoscow

Psychrometric chart
RH [%]

Wet-bulb temperature [oC]; WBTRelative humidity [%]; RH, 
Absolute humidity [g/kg’]; X
Dew-point
temperature[oC]; DPT

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Building exterior environment #13 urban wind
Urban edge

Urban center
Urban
boundary layer
Building array= roughness

Urban
wind
profileSmoothsurfacez – intercept =roughness length; z0
Log law
100 z0 =slope = quotient
; 1/n
100

Indicatingreference

1/n=
1/4

z [m]

z/zref [ND]Power law
1/ n
 z 
U


U ref  z ref 
10

1/n
=1/8

1
U/Uref50.25 m10

z0 =5*10-4mU
1  z  ln 
U *   z0 
Karman constant= 0.4

0 1 2 3 4
-1
U/Uref Friction velocity [m s ]Generated by Foxit PDF Creator © Foxit Software
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Evaluation of human comfort #1
Physical stimulus
Thermal stress

Human sensation
Hot ⇔ Cold

Thermal comfort
Air Quality

Smelly⇔ Not smelly

Indoor Air Quality
Light

Glaring⇔ Not glaring

Visual comfort
Sound

Noisy ⇔ Calm

Acoustic Quality
Input signals

Output signals

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Evaluation of human comfort #2
Utility, which economists use as
their terminology, actually means
human sensation.

Weber-Fechner’s law

Weber – Fechner’s law indicatesutility function as well as humansensation can be approximated by alog-function.

感覚量Sensation

ΔSensation／ΔStimulus=k／Stimulus
Sensation=log（ Stimulus ）

0
0

刺激量
Stimulus

Human sensation digresses with increasing stimulus to be given.

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Evaluation of human comfort #3(1) Acoustic comfort

intensity [W m-2]

 I Sound intensity level [dB] IL  10  log

I 
 0

Minimum intensity
a human can hear
10-12 W m-2
There are two noise sources. Each of those two is respectively evaluated 60
dB. If those two sources are multiplied, how much is the superimposed noselevel?
2  60  120 [dB]

 I  ⇒ I  106 [W m-2]
60  10  log 12 
6


2
10

10
Thus, IL  10  log

[dB]


63
 1012 (2) Indoor Air Quality (IAQ)
Olf (stimulus) & Decipol (sensation) (Fanger, 1987)(2) Visual comfort

Luminance of the facet [cd m-2]

1.6 0.6 Solid angle seeing theBs   high luminance facet [sr]Glare Index GI  10  log 0.478
 BackgroundBb luminance [cd m-2]

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Evaluation of human comfort #4

Thermal comfort

Metabolic rate –External work

Temperature
Humidity
velocity

Air
Clothing

MRT

K
Heat balance equation
governing humanRadiation
thermal comfort: Metabolic Consumed Convection
Evaporation Conduction

M W  C  E  R  K

rateenergy forexternal work

Heat loss from a body

[W m-2]

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Evaluation of human comfort #4

Thermal comfort

Metabolic rate –External work

Temperature
Humidity
velocity

Air
Clothing

MRT

K
Heat balance equation
governing humanRadiation
thermal comfort: Metabolic Consumed Convection
Evaporation Conduction

M W  C  E  R  K

rateenergy forexternal work

Heat loss from a body

[W m-2]

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Evaluation of human comfort #4

Thermal comfort

Metabolic rate –External work

Temperature
Humidity
Air velocity
Clothing

MRT

K
Heat balance equation
governing humanRadiation
thermal comfort: Metabolic Consumed Convection
Evaporation Conduction

M W  C  E  R  K

rateenergy forexternal work

Heat loss from a body

[W m-2]

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Evaluation of human comfort #4

Thermal comfort

Metabolic rate –External work

Temperature
Humidity
Air velocity
Clothing

MRT

K
Heat balance equation
governing humanRadiation
thermal comfort: Metabolic Consumed Convection
Evaporation Conduction

M W  C  E  R  K

rateenergy forexternal work

Heat loss from a body

[W m-2]

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Evaluation of human comfort #4

Thermal comfort

Metabolic rate –External work

Temperature
Humidity
velocity

Air
Clothing

MRT

K
Heat balance equation
governing humanRadiation
thermal comfort: Metabolic Consumed Convection
Evaporation Conduction

M W  C  E  R  K

rateenergy forexternal work

Heat loss from a body

[W m-2]

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Evaluation of human comfort #4

Thermal comfort

Metabolic rate –External work

Temperature
Humidity
velocity

Air
Clothing

MRT

K
Heat balance equation
governing humanRadiation
thermal comfort: Metabolic Consumed Convection
Evaporation Conduction

M W  C  E  R  K

rateenergy forexternal work

Heat loss from a body

[W m-2]

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Evaluation of human comfort #4

Thermal comfort

Metabolic rate –External work

Temperature
Humidity
velocity

Air
Clothing

MRT

K
Heat balance equation
governing humanRadiation
thermal comfort: Metabolic Consumed Convection
Evaporation Conduction

M W  C  E  R  K

rateenergy forexternal work

Heat loss from a body

[W m-2]

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Evaluation of human comfort #4

Thermal comfort

Metabolic rate –External work

4 environmental factors
2 human factors

Temperature
Humidity
Air velocity
Clothing

MRT

K
Heat balance equation
governing humanRadiation
thermal comfort: Metabolic Consumed Convection
Evaporation Conduction

M W  C  E  R  K

rateenergy forexternal work

Heat loss from a body

[W m-2]

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Evaluation of human comfort #5

Thermal comfort

M W  C  E  R  K

C↑

R↓
R↑

By qroom↑, heatloss through C& R isdecreased.

HeaterHeat loss
through R might
be negative.

M W  C  E  R  K

R↓

C↑ R↑

0.1

0.6

1.0
2.0 Clo
1 Clo = ＝0.155 m2 K W-1

M W > C  E  R  K

C↑E↑
FannerFloor heating
By aconv↑,heat loss
through C &E ↑.
M W > C  E  R  K

C↑E↑
By qroom↓,heat loss
through C ↑.

1.0
M  W >> C  E  R  K

1.4

3 Met
1 Met = ＝58.2 W m-2
M  W > C  E  R  K

Cooler

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Evaluation of human comfort #6

Thermal comfort

ThermalSensation Vote

To evaluate thermal comfort it is necessary to take account 6 elements; 1)
DBP, 2) humidity, 3) MRT, 4) air velocity, 5) clothing and 6) metabolic rate.
(1) Globe Temperature (GT) < 1), 3)
Sensible temperature(, some extent 4))
0.5
Hot = +3
MRT

GT

0
.
24

v
 GT  DBT 
Experimental
Neutralresult=0(2) Operative Temperature (OT) ;(Gagge 1937) < 1), 3)
Cold = -3

OT  w  DBT  1  w  MRT

Sensible temperature(3) Effective Temperature (ET) ;(4) Corrected Effective Temperature(Yaglou & Houghten 1923)(CET) ; (Bedford 1935) < 1), 2), 3), 4)< 1), 2), 4)(5) ET*, SET* ; (Gegge 1971) < 1), 2), 3), 4), 5),6) ; considering not only insensitive but also
ordinal perspiration.

Thermal sensation index
PMV (Predicted Mean Vote) ;(Fanger 1970) < 1), 2), 3),4), 5), 6) ; considering only
insensitive perspiration. -3.0
cold

Slightly
cool

Slightly
warm

cool

warm
neutral+3.0

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Chart for ET and CET

修

正

有

効

温

度

WBT

湿球温度［℃］

ま

た

は

有

効

温

度

［

℃

］

DBT or GT

グローブ温度または乾球温度［℃］

Air 風速［m/s］
velocity

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Chart for ET and CET

温
効
有
正
修

Air 風速［m/s］
velocity

度

WBT

湿球温度［℃］

ま

た

は

有

効

温

度

［

℃

］

DBT or GT

グローブ温度または乾球温度［℃］

Solve ET when a room isfeatured with DBP = 25 oC,
RH = 50 % and air velocity
is 1.0 m s-1.

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Evaluation of human comfort #7

Terminology“Comfort and energy conservation” is not always conflicting.

Conventional system

Task & ambient system

Space air-conditioning & lighting

Space air-conditioning & lighting: minimum.
Task-system covers individual
preference.

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Ventilation #1

Air change rate; n [s-1]ventilation rate; Q [m3 s-1]
Air volume of a room; V [m3]
Q=n･V

-

+
Any pollutant never comes in.

Any pollutant never comes out.
Pollutant concentration
C [m3 m-3]

Seidel’s formula
Air volume V [m3]ventilation rate
Q [m3 s-1]

Q

The mass balance during Dt;

Pollutant generating rate k [m3 s-1]
(generated pollutant) Dt  k

C0Equilibrium k / Q
concentration– (removed pollutant by ventilation) Dt  Q  C= (stored pollutant that raising concentration) DC  V

V

t

dC
k 
k
 Q 
 k  Q  C  C t     C 0   exp  t 
dt
Q 
Q
 V 

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Ventilation #2 Air flow rate

Air flow coefficient [m3 s-1 (Pa)-1/n][m3 s-1]

Q

Openingarea

Q  a  Dp 

A [m ]
2

Q2

a 2 A2a 1 A1Q

Dp1

Dp

Q  Q1  Q2  a 1 A1[Pa] = [N m-2] =[kg m s-2 m-2] =[kg s-2 m-1]
Fluid density [kg m-3]
humid air  1.205

r

Dp

zeta

2

r

2

 Q  r  Q  r
Dp  Dp1  Dp2   a 1 A1  2  a 2 A2  22Q  r 22
a 2 A2   2
 1   1 
 1 
2
a
A



  





Dp
2

 aA 

 a 1 A1 

Remind calculation procedure of U-vale!

2

2

Pressure resistance coefficient   1 / a [ND]

Composition of openings
Q1

2

Flow coefficient [ND]

Equivalent opening area [m2]a 1 A1Laminar-like flow← →ordinal flow

Q a A

Dp [Pa]tandem

Quotient parameter;
1～2

In case or an ordinal opening;

Pressure difference

parallel1n

Dp  a 2 A2

2r

Dp  aA2

r a 2 A2 

Dp  aA  a 1 A1  a 2 A

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Compositive thermal conductancel1

l2

Imagine Ohm's Law. You can add
two parts only in resistance unit.

Dx2

Dx1

1
Dx1

l1



→ R1+R2+R3 []Dx2Comprehensive thermal conductance(U-value) of a multi-layer wall

[W m-2K-1]

l2
qout [K]･･･

ain

aout [W m-2 K-1]

U

[W m-2 K-1]

1

l1

1
1
 

a out ik li Ck 1 a in

[W m-1 K-1]

1

lk-2 Air space lk

l1 [m] l2

li

l2

lk-2

Ck-1 lk[W m-2 K-1]

qin [K]

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Ventilation #3

Wind pressure [Pa]U[m s-1]

Ventilation caused by wind pressure

Dp w

Dp w  C w r2Wind pressure coefficient [ND]
Physical unit of the right-hand side =[kg m-3] [m2 s-2 ] = [kg s-2 m-1] = [Pa]Kinetic energy ofunit volume air

U 2

Clashing with the wall
Pressurepushing the wall1
mU 2 ･[m-3] = [kg m2 s-2] ･[m-3] = [kg m-3] [m2 s-2 ] = [kg s-2 m-1]
2
-2
-2
= [kg m s m ] = [N m-2] = [Pa]

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Opening type Flow coefficient a
SimpleopeningBladeorificeBell
mouth

Projected
window

Swivelwindow

Venetian
blindWind pressure coefficient Cw

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Ventilation #4
Stack effectVentilation caused by temperature differencepoqi > qopi
outflow+

Density varied with temperature [kg m-3]

353.25353.25rTq  273.15
In-out pressure difference at the point
higher h [m] than the base base-level
where respective absolute pressures
are defined; Dp(h)

qo

qi

hn

Neutral zone
The height
where Dp is zero.

Dp = pi - po

inflow
Gravity [m s-2]

Dp

AbsolutepressureDph    pi _ base  r q i   h  g    po _ base  r q o   h  g 
At Neutral zone; Dp(hn) MUST be 0, then;h0   pi _ base  r q i   hn  g    po _ base  r q o   hn  g =

Baselevel

pi_base po_baseground-level hn 

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Ventilation #5

Ventilation caused by temperature difference

In-out pressure difference at the point higher h [m] than Neutral zone,of which absolute pressure is given p [Pa] that are for both in- and outbuilding (stack); Dp(h)

h
p

NZ

hn

p

Dph    p  r q i   h  g    p  r q o   h  g 

 Dph   h  r q o   r qi   g

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Ventilation #６
qi

qo

Ventilation caused by both temperature
difference as well as wind pressureqi > qo
NZ

h

→

0

po_basepi_baseRelative pressure tooutside at ground level pi_base-rel

Dph   pi _ baserel  r q i   h  g   0  r q o   h  g 
 pi _ base  rel  r q o   r q i   h  g
Dp h  > 0 ∴Def; (in-) – (out)Dp h  > 0
NZU[m s-1]Dph  r q o  2
pi _ base  rel r q o   r qi   h  g  Cw
U2

Dp h   0

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Ventilation #4

Ventilation network for multi-rooms system
Room #i

Room #1

Opening #1

#n
#4

#2

#m
Opening #j

#n-1
#2

#3

i
j

#n-2

i
j

Q a A

Equivalent opening areaof Opening #j at Room #i

#m-1

i
j



2

r q ij upstream



Dp ij

Pressure difference MUST
be considered both windand buoyancy effects.

This temperature MUST be the upstream
temperature of Opening #j at Room #i.

Num  oppening at Room iiFlow balance equation of Room i:jj 1Q  0Sing is important; either
inflow or outflow is defined
positive.

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Ventilation #5

Ventilation network for multi-rooms system

Room #1#2#3

Num  oppening at Room i
i
j
j 1

Q  0

#n
#4

#n-1
#n-2



Fi q i , pi ;q k , pk ;;q l , pl   0

Temperature of Room #i
Baseline pressure at height 0 of Room #i that isdefined to be relative to the absolute pressure ofoutdoor air at ground level.

 F1  0 F  0 n

All sets of q and p for the spaces
neighboring to Room #i.

By solving those n nonlinear equations simultaneously,

p1 ,, pn are fixed. The calculation procedureentails iteration process by applying Newton-Raphson
method.

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Acoustic environment

Sound -> Compression wave.

Refraction

Reflection

Diffraction

You can hear distantsound in winter.

Sound withhigh frequency
hardly happens
diffraction.

absorptancefrequency

Sound absorption by porous
furnishing materialSound absorption
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General; Life- cycle assessment
NOT initial cost minimum,
NOT cost minimum, even though relying on life-cycle cost,
BUT Life-cycle Impact (e.g., by CO2) should be considered.

Construction

Operation

Renovation

Lifetime of the building

Demolition

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建築設備・Building Service Engineering●空気調和・換気設備 Air Conditioning●給排水衛生設備
Water Supply, Drainage & Sanitary●電気設備
Electricity●防災・消火設備
Safety Control &Fire Fighting●自動制御・計装設備 Automatic Control & Instrumentation
Out of printed books from SHASE: Society of Heating, Air-conditioning andSanitary Engineering Japan

Basic knowledge for Air-Conditioning
and Sanitary Equipment

Practical knowledge of Planning for
Air-Conditioning Equipment

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冷暖房・Cooling and Heating vs. 空気調和・Air-conditioning
暖房

Heating

In Building: higher temp.
Outdoor: lower temp.
Heat:higher→lower

暖房

Heat loss
Conduction
Convection
Radiation
Ventilation(Infiltration)
Generated heat
Natural room temp.
Heat supply(Heating)Cooling

冷房

In Building: lower temp.
Outdoor: higher temp.
Heat:higher→lower

冷

Heat gain房

Conduction
Convection
Radiation
Ventilation(Infiltration)
Generated heat
Natural room temp.
Heat Extraction (Cooling)

熱は温度の高い方から低い方へ流れる。
Heat flows where the temperature is higher to where the temperatures is lower.

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冷暖房・Cooling and Heating vs. 空気調和・Air-conditioning
Promotion of Indoor Thermal Comfort
室内の熱的快適性向上のため、・室内空気の温湿度を制御→冷暖房 ・Control of indoor air temperature andhumidity →Cooling and Heating・室内空気温度のみならず、室内・Not only the control of indoor air
空気質（IAQ）も制御→空気調和

temperature and humidity but the control
of Indoor Air Quality(IAQ）→Air-Conditioning

Specified Buildings of Building Maintenance Law
ビル管理法で対象となる特定建築物
・ 建物の規模 ： 床面積 ３０００㎡以上
・ Building size： Floor area 床面積 more than ３０００㎡
・ 建物の用途 ： 興行場、集会所、図書館、 ・ Building Usage：Game center, Hall, libraryMuseum, Art museum, Arena,
博物館、美術館、遊技場、
Office build. Shop, Hotel
事務所、店舗 、旅館など
ビル管理法における室内空気環境の基準
温 度
１７～２８℃
相対湿度
４０～７０％気 流
０．５ｍ／秒 以下
浮遊粉塵量
０．１５ｍｇ／ｍ3 以下
二酸化炭素
１０００ｐｐｍ 以下
一酸化炭素
１０ｐｐｍ 以下
ホルムアルデヒド ０．１ｍｇ／ｍ3 以下（０．０８ｐｐｍ 以下）

Standard of Indoor Air Environment under B.M.L.
Temperature
１７～２８℃
Humidity
４０～７０％Air speed
less than 0.15m/s
Dust
less than 0.15mg/ｍ3
Carbon dioxide
less than １０００ｐｐｍ
Carbon Monoxide less than １０ｐｐｍ
Form-aldehyde
less than 0.1mg/ｍ3（less than ０．０８ｐｐｍ）

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Basic Knowledge of Ventilation/
換気の基礎知識・人体の呼吸
Respiration of Human Body
・１時間、１人当たりの呼吸量 →480 l/h
・O2の減少量 → 480 l/h×0.0493=24 l/h ・ Respiration ratio2 par 1h par person →480 l/h
・ Decrease of O → 480 l/h×0.0493=24 l/h
・CO2の増加量 → 480 l/h×0.0435=21 l/h ・ Increase of CO2 → 480 l/h×0.0435=21 l/h

１人当たりの換気量（新鮮空気量）(1)窒息しないために室内空気の酸素濃度を18％以上とする。
(2)新鮮空気の酸素濃度を20.95％とする(3)１人当たりの換気量をX l/hとしてX×(0.2095-0.18)= 24 l/h
X = 814 l/h(1)室内空気の二酸化炭素濃度を1000ppm(0.1％)以下とする。
(2)新鮮空気の二酸化炭素濃度を300ppm(0.03％)とする(3)１人当たりの換気量をX l/hとしてX×(0.001-0.0003)= 21 l/h
X = 30000 l/h = 30m3/h

Ventilation Demand（Fresh Air Demand）(1)Keep the Oxygen density of indoor air
less than 18% to prevent suffocation(2)Oxygen density of Fresh air is 20.95%(3)Put the ventilation volume par person as X l/h
X×(0.2095-0.18)= 24 l/h
X = 814 l/h(1)Keep the carbon dioxide density of indoor air
less than 1000ppm(0.1%)(2)Assume the carbon dioxide density of fresh
air as 300ppm(0.03％)(3)Put the ventilation volume par person as X l/h
X×(0.001-0.0003)= 21 l/h
X = 30000 l/h = 30m3/h

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典型的(古典的)な空気調和設備・
Basic(Classic) Air-conditioning System
Exhausted
Heat

Cooling
Tower
Supply Air

Heating Coil
Return Air

Cooling
Water

Cooling Coil(Cooling)
OutletPump

Inlet
Heat lord

Air
FilterRoom(Heating)

Outdoor Air

Fan
Humidifier

冷水温度

冷凍機出口 5～7℃
冷凍機入口 10～12℃
冷却水温度 冷却塔出口 28～30℃
冷却塔入口 33～35℃
Chilled Water Temp.

Chimney,Stack

At chiller outlet 5～7℃
At chiller inlet 10～12℃
Cooling Water Temp. At C.T. outlet 28～30℃
At C.T. inlet
33～35℃

Chilled Water
Pump

Refrigerator.
Chiller

Vapor
Condensed Water

BoilerBanner
Pump

Pump

Oil

Oil Tank

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Fig. 5.1 Composition and Flow of Air-Conditioning System

Heat Flow
Exhausted to
outdoorExhausted Energy

Exhausted to outdoor

Heating

From rooms

Cooling

To rooms
To rooms
To rooms
Air

Heat carrier

Terminal

Air Handling Unit

Heat Source Unit
EnergyBoiler,Chiller

From outdoor

Rooms
Outlet, Inlet
Terminal unit

Pump,
PipingHeat exchanger
Air filter

Fan,
Duct

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空調負荷計算・最大負荷計算 装置容量の決定：危険率・年間負荷計算 年間エネルギー使用量：部分負荷効率
Heat load calculation
・Maximum heat load calculation to decide unit capacity: Level of significance・Annual heat load calculation to estimate annual energy consumption: Partial load efficiency

Heat Load

Data Base

Input Data

Temperature difference between
Outdoor air and room air
Computer

ProgramBy Hand

Input Form

E.T.D. of heavy wall

E.T.D. of light wall
Design of the System,
Running Cost

Unit
Capacity

Fig. 3.1 Heat Load and Its Calculation

Fig. 3.6 Effective Temperature Difference
between Wall Surfaces under Solar heat Gain

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冷房負荷

Cooling Load
(1) Solar heat gain
(2) Transmitted heat through walls and roofs as thetemperature of outdoor is higher than that of indoor(3) Generated heat from illumination and human body(4) Outdoor air

Outdoor air

Solar
heat
gain

Solar heat gain
Transmitted heat
Generated heat

Generated
heat
Transmitted heatFig. 3.1 Component of Cooling Load

Outdoor air

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暖房負荷

Heating Load
(1)Transmitted heat caused by the temperature
difference between the outdoor and indoor(2) Heat loss caused by infiltration(3) Outdoor air

Outdoor air(heating)

Humidification

Transmitted heatTransmitted heatTransmitted heat

Infiltration(heating)

Fig. 3.12 Component of Heating Load (Office building except piping heat loss)

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2.1 Characteristics of the air[1] Dry air and vapor(1) Dry air and moist (wet ,moist) air
Moist air = Dry air+ Vapor(1～3%)

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(2) Saturated air and Partial vapor Pressure
・Amount of vapor contained in air is limited
according to the temperature and pressure・The air that contains as much as vapor (at thelimit)
is called saturated air or saturated moist air

・Partial pressure： Pressure of a certain gas
contained with mixed gas
・If the above mentioned certain gas is vapor andthe mixed gas is the air, the pressure of the vapor
is called ‘Partial vapor pressure’・ Partial vapor pressure is related to the amount ofvapor , Partial vapor pressure is one of the index
to denote how much the air contains vapor

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[2] Expression of vapor
Table 2.2 expression of vapor in the air← Absolute humidity
Mass ratio of vapor to dry air,if x=0 : Dry air←Partial vapor pressure
pw = pws : Saturated air
pw < pws :Unsaturated air
pw =0 : Dry air← Relative humidity
Percentage of partial vapor pressure to saturated partial
vapor pressure at the same temperature← Comparative absolute humidity or Percentage of saturation
Percentage of absolute humidity to saturated absolute
humidity← Dew point temperature
Temperature of saturated air which has the same partial
vapor pressure as the concenning air←Wet bulb temperature
Temperature of wet bulb which exposed to air(more than 5m/s)

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(1) Absolute humidity/Mass ratio of vapor to dry air/When the mass of moist air is (1+x)kg
the mass of dry air is
1 kg
the mass if vapor is
x kg/The unit of absolute humidity is kg/kg(DA)DA means dry Air
Moist air, Its absolute humidity is x kg
Open the bag,

Close the bag

Moist air mass: Xm kg
Dry air mass: Xd kg
Vapor mass: Xv kg
Absolute humidity
x= Xd / Xd

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(2) Partial vapor pressure
The relation between partial vapor pressure pw andabsolute humidity x is shown as Eq.2.1, supposing the
moist air is ideal gas ( regulated with Boyle-Charle’s law,sometime called perfet gas or complete gas.)

Where
P : Total pressure of moist air
pw : Partial vapor pressure
Ra : Gas constant of dry air
Rw : Gas constant of vapor
Supposing moist air is a complete gas, the partial pressureof saturated air, pws, is the same as the saturated vapor
pressure of water at the same temperature ps. Namely pws= ps and this supposition can be applied the practicaldesign of air-conditioning.
There are many proposal to calculate the saturated vapor
pressure, ex. The approximate expression of Antoine or
Goff-Gratch.
Table-2.3 and 2.4 show the saturated water vapor tables.

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Table 2.3 Saturated water vapor table (Temperature base)Temperature

Pressure

Specific volume

Specific enthalpy‘ : Liquid phase (Water), ‘”: Gas phase(Vapor)

Specific entropy

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Table 2.4 Saturated water vapor table (Pressure base, SI unit)

Temperature

Pressure

Specific volume

Specific enthalpy‘ : Liquid phase (Water), ‘”: Gas phase(Vapor)

Specific entropy

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As to the absolute humidity of saturated air xs, Eq.2.2 can
Be led from Eq.2.1

Here, pws is partial water vapor pressure of saturated moist
air.
Example 2-1 How about the absolute humidity of saturated
air xs, at standard atmospheric pressure (101.325kPa) andtemperature 25℃?

Solution : From table-2.3, saturated partial vapor presurer
pws =3.1660 kPa, therefore
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(3) Relative humidity
Percentage of partial vapor pressure of moist air against
the saturated partial vapor pressure at the same
temperature is called Relative Humidity :RH.
RH is often used to show the humidity relating humansensation. Relative humidity φ is given by Eq. 2.3.

Partial vapor pressure of saturated air is changed
according to the temperature. Therefore the relative
humilities at the same partial vapor pressure anddifferent temperature are another things.
(4) Comparative absolute humidity
Percentage of the absolute humidity of moist air against
the saturated absolute humidity xs at the same
temperature is called percentage of saturation orcomparative absolute humidity.
Comparative absolute humidity Φ is given by Eq. 2.4.

The difference between relative humidity φ andcomparative absolute humidity Φ becomes large
according to the temperature increase, but at thenormal temperature level or below than that thedifference is ignorable.

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(5) Dew point temperature
The temperature is higher, the moisture content is larger.
In the contrary, the temperature of moist air is getting
lower, the moist air reaches to the saturated state. The
moist air is cooled more over, at a certain temperaturesome portion of vapor is condensed and become dew.
This temperature is called dew point temperature.
Dew point temperature has a certain value against
partial vapor pressure, this partial vapor pressure is thesame as the saturated vapor pressure at the dew point.
Under constant pressure, dew point has a constantvalue against absolute temperature.
(6) Wet bulb temperature
Psychrometer (dry and wet bulb thermometer) is often
used to measure humidity. Dry bulb thermometer(normal thermometer) is used to measure airtemperature. In case of wet bulb thermometer, the dry
bulb is wrapped with cloth and the cloth is wetted withwater driven by capillary effect. Evaporation at the cloth
extract the heat from wet bulb and the wet bulb shows
the heat balanced state of heat transfer. This heatbalance temperature is called wet bulb temperature :WB. Air temperature measured with dry bulb
thermometer is called dry bulb temperature: DB.

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(5) Specific enthalpy
Enthalpy means the heat contained in fluid. The enthalpy
par unit mass is called specific enthalpy.
The enthalpy of moist air is normalized as its value of dry
air at temperature 0℃ is 0 and the specific enthalpy isdefined as the value of par 1kg dry air.
Specific enthalpy h [kJ/kg(DA)] at temperature t [℃] andabsolute humidity x [kg/kg(DA)] can be calculated using
Eq.2.5 under the assumption of ideal gas. The specificenthalpy of moist air is the sum of 1kg dry air value andvapor one.

Here, ha : Specific enthalpy of dry air [kJ/kg(DA)]hw : Specific enthalpy of vapor [kJ/kg(DA)]cpa : Specific heat at constant pressure of dry air[1.006kJ/kg(DA)・K]
cpw : Specific heat at constant pressure of vapor[1.805kJ/kg(DA)・K]
r0 : Evaporative latent heat of water at temperature
0℃ [2501.6 kJ/kg]

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Specific heat at constant pressure of moist air cp is theheat to change its temperature 1℃,namely the change of
specific enthalpy, is given by Eq.2.6(4) Specific volume
Specific volume v is the volume of moist air that contains
dry air 1kg and given by Eq.2.7 supposing ideal gas

Here, v : Specific volume [m3/kg(DA)]Ra, Rw : Gas constant of dry air and vapor(reffer Eq.2.1)
T : Absolute temperature [K]P : Pressure of moist air [kPa]
At normal atmospheric pressure, Eq.2.8 can be obtained
from Eq.2.7 where temperature unit is ℃.

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Dry air at 20℃ is called standard air, its density is 1.2
kg/m3. The specific volume of this standard air vo is 0.83
[m3/kg(DA)]. The ratio of real air-flow volume tostandardized air-flow volume is v/ vo = v/0.83, where v isthe specific volume of moist air. This ratio is nearly equal
1 at normal temperature. In case of the design of airconditioning system, which needs not excessive accuracy,
air-flow volume is treated as standard air. If you need theaccurate value, you can get it by multiplying v/0.83 tostandard air.

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2.2 Psychrometric Chart[1] Psychrometric chart
The state of moist air is expressed using pressure,temperature, humidity, specific volume and so on. If you
can get three state value such as pressure, temperatureand partial vapor pressure, you can get the other statevalues.
Commonly, the pressure is supposed as atmosphericpressure and the constant value. So if you can get two
state values, you can get the other state values.
Therefore, you can express the state of moist air using
these two values as the axes. This type of chart thatdefine the state of moist air is called psychrometric chart.
of moist air cp is the heat to change its temperature
1℃,namely the change of specific enthalpy, is given by
Eq.2.6
Usually, psychrometric chart uses specific enthalpy h andabsolute humidity x as tilted axes. This type of chart isused for the calculation of heat load of air-conditioning
and the analysis of the state of moist air. The other typesof chart, using dry bulb temperature t and absolute
humidity x (t-x chart) or dry bulb temperature t andspecific enthalpy h (t-h chart ) is utilized in the samemanner.

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[1] Psychrometric charta. h-x chart
Figure 2.1 is a h-x chart. In this figure, horizontal axis isabsolute humidity, tilted axis is specific enthalpy. You canfind dry bulb temperature, wet bulb temperature, absolute
humidity, partial vapor pressure, relative humidity, specificabsolute humidity, specific volume.
In this a psychrometric chart, adiabatic saturated
temperature is used as we bulb temperature .

Air with Mist

Unsaturated Air

Dry Air x=0

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In fig.2.4
t : Dry bulb temperature [℃]t’: Wet bulb temperature [℃]x: Absolute humidity [kg/kg(DA)]pw: Partial vapor pressure [kPa]Φ : Relative humidity [%]Φ : Specific absolute humidity [%]h : Specific enthalpy [kJ/kg(DA)]v : Specific volume [m3/kg(DA)]

Absolute humidity

Partial vapor pressure
Partial vapor pressure

Dew point
temperature

Dry bulb
temperature

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[Example 2-2]
The moist air, dry bulb temperature is 26℃
relative humidity is 50%
How about wet bulb temperature t’, dew point temperature t’’,absolute humidity x and specific enthalpy h?

湿り空気の加熱と冷却
・加熱：乾球温度は上昇、絶対湿度は一定、相対湿度は減少
・冷却：乾球温度は減少、絶対湿度は一定、相対湿度は増加
乾球温度=露点温度、相対湿度95～100%
・冷却：乾球温度は減少、絶対湿度は減少、
相対湿度は一定、

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Heating and Cooling of Moist Air・Heating: dry bulb temp. increases, absolute humidity isconstant, relative humidity decreases.
・Cooling: dry bulb temp. decreases, absolute humidity isconstant, relative humidity increases.
Dry bulb temp.=Dew-point temp.
Relative humidity is 95～100%
・Cooling: dry bulb temp. decreases,
absolute humidity decreases
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Cooling with Air Handling Unit
Supply air
Room airOutdoor air
Room airMixed airOutdoor air

Mixed airRoom airSupply air

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①乾球温度32℃、相対湿度70％の外気1000m3/hを空調機に導入する。室内空気を乾球温度26℃、相対湿度50％として、外気顕熱負荷、外気潜熱負荷、外気全熱負荷を求めよ。
Outdoor air (dry bulb temperature DB: 32℃, relative humidity φ: 70% ) is induced to the air handling unit (AHU.) Whenthe room air is kept at dry bulb temperature DB: 26℃, and the relative humidity φ: 50%. Get the sensible cooling heatload, latent cooling heat load and total cooling heat load of the outdoor air.
Outdoor air: Specific volume=0.894, W=1000/0.893=1119kg/h, Specific enthalpy=86 kJ/kg[DA]Room air: Specific enthalpy=54kJ/kg[DA].
Outdoor air sensible heat load=(60-54)x1119=6714kJ/h, Outdoor air latent heat load=(86-60)x1119=29094kJ/h,
Outdoor air total heat load=(86-54)x1119=35808kJ/h

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Supply air

Cooling CoilRoom airOutdoor air
Room airMixed airOutdoor air

Mixed airRoom airSupply air

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②①の外気と室内空気(乾球温度26℃、相対湿度60％) 環気3000m3/hを空調機内で混合する。この混合空気の乾球温度、相対湿度、絶対湿度、露点温度、比エンタルピーを求めよ。
The outdoor air mentioned above is mixed with the return air (dry bulb temperature DB: 26℃, relative humidity φ:50% ) from the room in AHU. Get the dry bulb temperature, relative humidity, absolute humidity, dew pointtemperature and specific enthalpy of the mixed air.
Room air: Specific volume=0.892, W=3000/.892=3363kg/h, Outdoor air: W=1119kg/h, 3363:1119=3:1
Mixed air: Dry bulb temperature=27.5 ℃, Relative humidity=57%, Absolute humidity x=0.0135kg/kg[DA],Dew point temperature=18.5 ℃, Specific enthalpy= 62kJ/kg[DA]Outdoor air

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③②の混合空気を空調機の冷却コイルで乾球温度13℃、相対湿度95％まで冷却する。除去顕熱量、除去潜熱量、除去全熱量と凝縮水量（kg/h）を求めよ。
The mixed air mentioned above ② is cooled with the cooling coil of AHU to the state of dry bulb temperature DB:
13℃ and relative humidity φ: 95%. Get extracted sensible heat, extracted latent heat, extracted total heat and theamount (weight) of condensed water
Mixed air: W=3363+1119=4482kg/h, Absolute humidity x=0.0135kg/kg[DA], Specific enthalpy= 62kJ/kg[DA]Supply air:Absolute humidity x=0.0090kg/kg[DA], Specific enthalpy= 36kJ/kg[DA]Extracted sensible heat=(51-36)x4482=67230kJ/h, Extracted latent heat =(62-51)x4482=49302kJ/h,
Extracted total heat =(62-36)x4482=116532kJ/h, Weight of condensed water=(0.0135-0.0090) )x4482=20.2kg/h

Mixed airSupply air

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④③の空気が室内に吹き出し、乾球温度26℃、相対湿度50％となる。室の顕熱、潜熱、全熱および室の顕熱比（SHF）を求めよ。
The cooled air mentioned above ③ blows into the room and become the state of dry bulb temperature DB: 26℃ andrelative humidity φ: 50%. Get the sensible cooling heat load, latent cooling heat load, total cooling heat load andsensible heat factor SHF of the room..
Room air: Specific enthalpy=54kJ/kg[DA], Supply air: Specific enthalpy= 36kJ/kg[DA]Sensible cooling heat load=(49-36)x4482=58266kJ/h, Latent cooling heat load =(54-49)x4482=22410kJ/h
Total cooling heat load= (54-36)x4482=80676kJ/h
Sensible heat factor SHF=(Sensible heat load)/(Total cooling heat load)=58266/80676=0.72

Room airSupply air

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⑤③で求めた空調機の装置容量(顕熱、潜熱、全熱)が①＋④と一致することを確かめよ。
Assure the capacity of AHU given by ③ is the same as the sum of ① and ④.
Unit Extracted sensible heat=(51-36)x4482=67230kJ/h
Extracted latent heat =(62-51)x4482=49302kJ/h,
Extracted total heat =(62-36)x4482=116532kJ/h
Outdoor air sensible heat load=(60-54)x1119=6714kJ/h Room Sensible heat load=(49-36)x4482=58266kJ/h (64980kJ/h)latent heat load=(86-60)x1119=29094kJ/h
Latent heat load =(54-49)x4482=22410kJ/h (51504kJ/h)
total heat load=(86-54)x1119=35808kJ/h
Total heat load= (54-36)x4482=80676kJ/h (116484kJ/h)

Outdoor air

Mixed airRoom airSupply air

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③’②の混合空気を空調機の冷却コイル(表面温度11℃、バイパスファクターBF: 0.125 or コンタクトファクター CF: 0.875)に通過させる。除去顕熱量、除去潜熱量、除去全熱量と凝縮水量（kg/h）を求めよ。
The mixed air given by ② goes through the cooling coil (coil surface temperature is 11℃, bypass factor BF: 0.125 orcontact factor CF: 0.875.) Get extracted sensible heat, extracted latent heat, extracted total heat and the amount(weight) of condensed water.
Supply air: Dry bulb temperature 13 ℃, x=0.0090kg/kg[DA], Specific enthalpy= 36kJ/kg[DA]Extracted sensible heat=(51-36)x4482=67230kJ/h, Extracted latent heat =(62-51)x4482=49302kJ/h,
Extracted total heat =(62-36)x4482=116532kJ/h, Weight of condensed water=(0.0135-0.0090) )x4482=20.2kg/h

Mixed air

Coil surface temp.
Supply air

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④’③’の空気が室内に吹き出す。室の冷房負荷の顕熱比（SHF）が0.6、室の乾球温度が26℃に制御されるとき、室内の相
対湿度と室の除去顕熱量、除去潜熱量、除去全熱量を求めよ。
The cooled air mentioned above ③’ blows into the room. When the sensible heat factor SHF of the room is 0.6 andthe room air is controlled as 26℃,Get the sensible cooling heat load, latent cooling heat load and total cooling heatload of the room
Room air: Specific enthalpy=5457kJ/kg[DA], Supply air: Specific enthalpy= 36kJ/kg[DA]Sensible cooling heat load=(49-36)x4482=58266kJ/h
Latent cooling heat load =(=5457-49)x4482=2241035856kJ/h
Total cooling heat load= (5457-36)x4482=8067694122kJ/h

Room air (reality)
Room air (designed)

Supply air

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Supply air

Heating CoilHeat with Air Handling Unit

Room airOutdoor air
Room airMixed air

Humidifier

Supply air
Heated air

Outdoor air
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⑥乾球温度4℃、相対湿度40％の外気1000m3/hと乾球温度20℃、相対湿度50％の室内空気３000m3/hを混合する。この混合空気の乾球温度、相対湿度、絶対湿度を求めよ。
Mixing the outdoor air (dry bulb temperature DB: 4℃, relative humidity φ: 40% and airflow volume1000m3/h ) andthe return air of the room(dry bulb temperature DB: 20℃, relative humidity φ: 50% and airflow volume 3000m3/h.)
Get the dry bulb temperature, relative humidity and absolute humidity.
Outdoor air: Specific volume=0.788 W=1000/0.788=1269kg/h
Room air: Specific volume=0.841, W=3000/0.841=3567kg/h, 3567:1269=2.8:1
Mixed air: Dry bulb temperature=15.8 ℃, Relative humidity=53%, Absolute humidity x=0.0057kg/kg[DA],Outdoor air

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⑦⑥の空気を乾球温度40℃まで加熱する。加熱に必要な熱量を求めよ。また、相対湿度と絶対湿度はいくつになるか。
Above mentioned mixed air is heated to dry bulb temperature DB=40℃, get the heating load, relative humidity andabsolute humidity.
Mixed air: Dry bulb temperature=15.8 ℃, W=1269+3567=4836kg/h, Specific enthalpy=31kJ/kg[DA]Heated air: Specific enthalpy=56kJ/kg[DA] Heating load=(56-31)x4836=120900kJ/h
Relative humidity=12%, Absolute humidity x=0.0057kg/kg[DA],Room airOutdoor air

Mixed airHeated air

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⑧⑦の空気を水噴霧で加湿し室内に供給する。この吹出空気により室内が乾球温度20℃、相対湿度50％に保たれる。吹
出空気の乾球温度は何度か。また、室の損失熱量と加湿に必要な水量（kg/h）を求めよ。
Heated air mentioned ⑦ is humidified with water spray and supplied to the room. This supply air keep the room atdry bulb temperature DB: 20℃ and relative humidity φ: 50%. How about the dry bulb temperature of the supply air?
How about the heat loss of the room and how much is the water to be sprayed?
heated air: Absolute humidity x=0.0057kg/kg[DA]Supply air: Dry bulb temperature=36.0 ℃, W=4836kg/h, Specific enthalpy=56kJ/kg[DA]Absolute humidity x=0.0064kg/kg[DA]Room air: Specific enthalpy=39kJ/kg[DA]Heat loss(heat load)=(56-39)x4836= 82212kJ/h
Water weight (0.0064-0.0057)x4836=3.4kg/h

Supply air
Heated air

Outdoor air
Room airMixed airGenerated by Foxit PDF Creator © Foxit Software
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典型的(古典的)な空気調和設備・
Basic(Classic) Air-conditioning System

Exhausted
Heat

熱源機器
Heat Source
Machinery
Supply Air

Cooling
Tower

Heating Coil
Return Air

Cooling
Water

Cooling Coil(Cooling)
OutletPump

Inlet
Heat lord

Air
FilterRoom(Heating)

Outdoor Air

Fan
Humidifier

冷水温度

冷凍機出口 5～7℃
冷凍機入口 10～12℃
冷却水温度 冷却塔出口 28～30℃
冷却塔入口 33～35℃
Chilled Water Temp.

Chimney,Stack

At chiller outlet 5～7℃
At chiller inlet 10～12℃
Cooling Water Temp. At C.T. outlet 28～30℃
At C.T. inlet
33～35℃

Chilled Water
Pump

Refrigerator.
Chiller

Vapor
Condensed Water

BoilerBanner
Pump

Pump

Oil

Oil Tank

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冷凍機
冷水温度

Refrigerator or Chiller
冷凍機出口 5～7℃
冷凍機入口 10～12℃

Chilled Water Temp.

At chiller outlet 5～7℃
At chiller inlet 10～12℃

Mechanical compression

Centrifugal (turbine)
Reciprocal
RotaryElectric motor

Absorption

Electric motor
Compressor

Compressor
Compressor

ロータリ, Rotary
ターボ, Turbine

レシプロ, Reciprocal

吸収, Absorption Chiller

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圧縮式冷凍機Mechanical compression Chiller

冷媒の相変化熱を利用して冷水を作る。
冷媒： CFC(Freon) オゾン層破壊効果のため、
モントリオール議定書（1987年）で禁止
HCFC, HFC, NH4, CO2
凝縮器のチューブ内で冷媒は冷却水に冷やされて凝縮（液化）、冷却は高温になる。
In the tube of condenser, refrigerant is cooled by
the cooling water, becomes liquid (condensation. )cooling water becomes high temperature.

Utilizing the phase-change heat of refrigerant, Chiller
makes chilled water.
Refrigerant: CFC(Freon) prohibited by Montreal protocol
for shake of Ozone layer depletion.
HCFC, HFC, NH4, CO2
b: 冷媒は高圧、低温液体
b: refrigerant is highpressure, lowtemperature liquid

Condenser

Cooling water, coolanta: 冷媒は高圧、高温ガス
a: refrigerant is highpressure, hightemperature gas

c: 冷媒は低圧、低温液体
c: refrigerant is low
pressure, lowtemperature liquid

Expansion valve

圧縮器内で冷媒は圧縮され高圧、高温になる。
In the compressor, refrigerant(gas) is compressed, become
high pressure and hightemperature

Chilled water

Evaporator

Compressord: 冷媒は低圧、低温ガス
d: refrigerant is low pressure, lowtemperature gas

蒸発器のチューブ内で冷媒は冷水から熱を奪って蒸発（気化）、冷水は熱を奪われて低温になる。
In the tube of evaporator, refrigerant extract the heatfrom the chilled water, becomes gas (evaporation. )chilled water becomes low temperature.

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Carnot cycle; Heat engine; transforming thermal energy into work
Reversible process

Pressure-volume graph
A Isothermal

Temperature-entropy diagram
isothermal

expansion
Adiabatic
compression

W

B

D

iso-entropy
Adiabatic
expansion

Isothermal
compressionC

Inverse Carnot cycle; Heat pump; transforming work into
thermal energy

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Fun
Electric motor
Upper water tank
Louver

Cooling water

Filler
Cooling tower

Outdoor air
Eliminator

Cooling water

Lower water tank
Supply water
Over flow
Condenser
Evaporator

Fig. 5.43 Cooling Tower
Compressor
Electricity

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Cooled bycooling water
Condenser

Compressor

Evaporator

Cooled byoutdoor air

Condenser

Evaporator
Heated bychilled water

Heated byroom air
直膨コイル： 冷媒が冷却コ
イルのチューブ内で直接蒸発する。
Direct expansion coil:
Refrigerant directly
evaporate in the tube ofcooling coil.

Compressor

Room Air Conditioner

Mechanical Compression Chiller

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Compressor
Condenser

Evaporator

Outdoor

Temperature order in Heating Cycle
High
Refrigerant in condenser
Indoor air

OutdoorHeat Flow
Outdoor air
Refrigerant in condenser

Cooling Cycle

Low
Compressor

Get the heat from where thetemperature is relatively low,
Supply the heat to where thetemperature is relatively high.

Condenser

Evaporator

IndoorHeat FlowFig 5.42 Heat Pump Air Conditioner

Indoor

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吸収式冷凍機Absorption Chiller

冷媒： 水
吸収剤：臭化リチュム

Refrigerant: Water
Absorbent: Lithium bromide

Absorption Chiller
Low pressurized (vacuumed)
Water vapor
Water
Water vapor

Water vapor

Chilled water

冷水
Evaporator
Absorber

Evaporator

Solvent of water and lithium
bromideSolvent absorbs vapor, thedensity of lithium bromide
becomes thin and finallycannot absorbs anymore

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Hot water vapor
Regenerator
Thin solvent

Steam

Thin solvent

Condenser

Regenerator
Steam

Evaporator

Cooling water

冷却水

Thick solvent
Water vapor

Heatexchanger

Evaporator
Water vapor

Absorber
Absorber

Solvent pump

Chilled water pumpThin solvent in absorber is pumped up
to regenerator.
The solvent in regenerator is boiled up
by steam and the solvent becomes
thick.
The thick solvent drops into absorberby gravity.

Regenerated hot water vapor moves to condenser.
In condenser, cooling water condenses the hotwater vapor to water.
The thin solvent is sent to regenerator and heated up.
The thick solvent (hot) is sent to the absorber where
the temperature is relatively low.
Therefore heat exchanger is effective.

Absorption chiller needs steam and cooling water to make chilled water.
How ever it needs not much amount of electricity.
Some absorption chiller which need not steam for regenerator called
double effect, direct fired absorption chiller

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空調方式の分類Classification of air conditioning system
熱搬送媒体
Heat
carrierAir

名称
Naming

熱源機器
Heat source machinery

定風量単一ダクト方式
Constant air volume single duct

冷房用：水冷式冷凍機
暖房用：温水ボイラー
蒸気ボイラー
冷暖両用：
ヒートポンプチラー
冷温水発生器

変風量単一ダクト方式
Variable air volume single duct
二重ダクト方式
Dual duct, Multi-zoon unit

中央式
CentralAir + Water

ダクト併用ファンコイルユニット
Fan coil unit with air handlingunit
インダクション（誘引）ユニット
Induction unit

Water

ファンコイルユニット
Fan coil unit

Cooling: Water heat
Source chiller
Heating: Hot water boiler
Vapor boiler
Cooling & heating:
Heat pump chiller
Double effect, direct
fired
absorption chiller

ルームエアコン
Room air conditioner
個別方式
Local

Refrigerant

パッケージエアーコン
Packaged air conditioner
パッケージターミナルユニット
Packaged terminal unit

適要Remarks

不要
Need not

冷房専用と冷暖兼用(ヒートポンプ)タイプがある。
空気熱源、水熱源がある。
Cooling only or heat pumptype.
Air heat source or Water
heat source.

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典型的(古典的)な空気調和設備・
Basic(Classic) Air-conditioning System

Exhausted
Heat

Cooling
Tower
Supply Air

Heating Coil
Return Air

Cooling
Water

Cooling Coil(Cooling)
OutletPump

Inlet
Heat lord

Air
FilterRoom(Heating)

Outdoor Air

Fan
Humidifier

冷水温度

冷凍機出口 5～7℃
冷凍機入口 10～12℃
冷却水温度 冷却塔出口 28～30℃
冷却塔入口 33～35℃
Chilled Water Temp.

Chimney,Stack

At chiller outlet 5～7℃
At chiller inlet 10～12℃
Cooling Water Temp. At C.T. outlet 28～30℃
At C.T. inlet
33～35℃

Chilled Water
Pump

Refrigerator.
Chiller

Vapor
Condensed Water

BoilerBanner
Pump

Pump

Oil

Oil Tank

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Thermostat

Outdoor air

定風量単一ダクト方式
・送風機の風量は一定
・室温の制御は送風温度を変えて行う
・サーモの指令で自動弁の開度が変化し、コイル通過水量が変わり、送風温度が
変化する。

Fan
Heating CoilBoiler

Cooling Coil Air filter

Chiller

Chilled water

Automatic valve
Thermostat

Fan

Cooling CoilConstant air volume single duct system
・Air flow volume of the fun is constant.
・Room temperature is controlled throughthe change of supply air temperature.
・The opening ratio of automatic valve ischanged with the signal of thethermostat. Water flow amount of thecoil is changed and the supply air
temperature is changed.

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North

外周部と内周部
Exterior zone and interior

Exterior zone
Meeting room

Office room
Interior zoneWest

Adjacent building
East

Central core

Adjacent building
Office room

Reception
room

Executive roomSouth

Adjacent building

Adjacent building

Radiation
Net
Conduction

Radiation
Conduction

Net

Exterior zone
Interior zone

Fig.5.8 Thermal characteristics of Exterior zone and interior zone
Fig.5.10 Fluctuation of space heat load facing to the south

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North

Interior zone
Central core
East

West
Exterior zoneSouth

Fig.5.7 Zooning
Return air

OutletFun

Fig.5.14 Stepped installation

Outdoor air(a) Single duct
Return air
Fun

Terminal unit
(heater)

Outdoor air(b) Terminal unit

Fig.5.15 Single duct and terminal unit

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Dumper
Thermostat

Fan

Heating Coil

Cooling CoilAir filter

Outdoor air

Fig.5.20 Variable air volume system
Sound proof plenum box

Sound insulator

Sound buffer

Dumper

Fig.5.21 Detail of terminal unit in variable air volume system

可変風量単一ダクト方式
・室温の制御は送風量を変えて行う
・サーモの指令でターミナルユニットのダンパーの開度が変化し、吹き出し風量が変化する。
Variable air volume single duct system
・Room temperature is controlled through the change ofsupply air volume.
・The opening ratio of dumper in terminal unit is changed withthe signal of the thermostat.

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Boilerwarm air

Air filter

Outdoor air

Cool airMixingdumperFan
Chiller
Heating Coilwarm airFig.5.22 Multi-zoon unit

マルチゾーンユニット方式
・空調機内で温風と冷風を同時に作り、各ゾーンのサーモスタットの指令で混合ダンパーの開度を変えて、各ゾーンに最適な送風温度に制御する。
・混合損失が生じ、省エネ性能に劣る。
Multi-zoon unit
・Warm air and cool air are made at the same time in the air
handling unit. The opening ratio of mixing dumpers ischanged with the signal of the thermostat in each zoon andsupply air temperature of each zoon is controlled.
・Energy loss because of mixing loss.

Fan

Cool air

Cooling CoilFig.5.23 Multi-zoon unit

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Cooling

Cooling

Cooling

Supply air

Thermostat

Chilled or hot water
Fun coil unit

Chiller

Supply pipe

coil

BoilerReturn pipe
2 pipe system

Cooling

Thermostat
Chiller

Heating

ファンコイルユニット方式
・通常、外気取入能力は無い。

Fun coil unit

Chilled water Supply pipe
Hot water supply pipe
3 pipe system
Cooling

Chiller

Boiler

Thermostat

Fan

Fig.5.24 Section of fun coil unit

Return pipe

BoilerReturn air

Heating

Fun coil unit

Chilled water Supply pipe
Chilled water return pipe
Hot water return pipe
Hot water supply pipe
4 pipe system

Fan coil unit
・Usually fan coil unit can notventilate the outdoor air.

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Primary air
Heating Coil Cooling Coil
Outdoor air
Air outlet

Fan
Fan coil unit

Air filter
BoilerAir exhaustion
Chiller orBoilerSecondary water (chilled or hot water)
Fig.5.25 Fan coil unit with air handling unit

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パッケージ型空調機、ルームエアコンの基本形
Basic pattern of packaged air-conditioner or room air-conditioner
Water cooled(Water heat source)

Outdoor unit

Indoor unit

Air cooled(Air heat source)

Air cooled(Air heat source)
Evaporator

Compressor Condenser

Evaporator

Fan

Condenser

Separate or split type, some unit has thecompressor in outdoor unit

All in one type

IndoorOutdoorCooling tower

IndoorOutdoor
OutdoorCooling waterpump

Wall through or window type
Water heat source cooler

Water cooled(Water heat source)

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Cooling tower
OutdoorIndoor

Cooling tower
Cooling waterpump

Heat exchanger forwater and refrigerant
BoilerIndoor unit

Water heat source heat pump
BoilerOutdoor unit
Indoor unit Indoor unit

Indoor unit
Water piping
Refrigerant piping
Cooling water
pump
Refrigerant piping

Air heat source heat pump multi type(one outdoor unit and plural indoor units)

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給水衛生設備Water supply system

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Supply water amount par person

Office building

Residential building

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Table 6 Minimum water pressure (head)
Name
Flush valve
FaucetNecessary water head

=7m Aq=70 kPa(minimum)=3m Aq=30 kPa

Auto-close faucet=7m Aq=70 kPa

Shower head
Instantaneous hot water supplier, large
Instantaneous hot water supplier, middle=7m Aq=70 kPa=7m Aq=70 kPa=7m Aq=70 kPa

Instantaneous hot water supplier, small=1m Aq=10 kPa (Low pressure type)

Water head1
Atmospheric pressure=1 kg/cm2f= 100,000 Pa/cm2 =100 kPa/cm2=760 mmHg/cm2=10 mAq/cm2=10,000 mmAqcm2

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Table 4 Water flow volume and pipe connecting size (mm)Toilet (flush valve)
Toilet (water tank)
Urinary (flush valve)
Urinary (water tank)
Urinary (water tank)
Sink (small)
Sink (large)
Faucet (small)Faucet (large)
Drinking fountain
Garden faucet
Bath tub (Japanese)
Bath tub (Western)

15l/time/ for10s
5l/time/ for 6s

Shower

Water amount(l/time)Usage rate
(time/hour)
Maximum flow
Rate (l/min)

Connecting pipe
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2.2 Water supply system
In normal buildings, next 4types of water supply
systems are popular.
Direct connection system Fig.1
Elevated tank system Fig.2
Pressure tank system Fig.3, Fig.4

Tank-less system Fig.5

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Property line

FaucetMeter
Stop valve

Branch water supply pipe Gate valve
Main water supply pipe

Fig.1 Direct conn system

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Elevated tank

Supplypipe
FaucetLiftingpipe

Liftingpump

Receiver tankFig.2

Elevated tank system

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FaucetPressure tank

Pump
Compressor

Supplypipe
Shallowwell

Receiver tankFig.4 Pressure tank system
Fig.3 Pressure tank system

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Pressurecontroller

Pressure
Signal unit

Motor
controller

Pump with
Variable speed motor

Receiver tank

Fig.5

Tank-less system using variable speed motor

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Elevated tankElevated tank

Suction tank

Suction tankElevated tank

Suction tank

Lifting pumpLifting pumpLifting pumpLifting pump

Fig.6

Water supply system for high-rise building