Chapter_3.pdf

3. Moist Air Properties and Conditioning Processes
3.1 Calculate values of humidity ratio, enthalpy, and specific volume for saturated air
at one standard atmosphere using perfect gas relations for temperatures of (a) 70 F
(20 C) and (b) 20 F (-6.7 C).
Solution:
(a) In English units, t = 70 F
Humidity Ratio:
Eq. (3-14b)
s
s
a
s
s
p
P
p
p
p
W
−
=
= 6219
.
0
6219
.
0
at t = 70 F, ps = 0.363 psia
P = 14.696 psia
363
.
0
696
.
14
363
.
0
6219
.
0
−
=
s
W = 0.01575 lbmv/lbma
Enthalpy:
Eq. (3-20a)
( )
t
W
t
i 444
.
0
2
.
1061
240
.
0 +
+
= Btu/lbma
( ) ( ) ( )( )
[ ]
70
444
.
0
2
.
1061
01575
.
0
70
240
.
0 +
+
=
i = 34.0 Btu/lbma
Specific Volume:
Ra = 53.352 ft-lbf/lbm-R
s
a
a
a
p
P
T
R
p
T
R
v
−
=
=
( )( )
( )( )
144
363
.
0
696
.
14
67
.
459
70
352
.
53
−
+
=
v = 13.69 ft3
/lbma
In SI units, t = 20 C
Humidity Ratio:
Eq. (3-14b)
s
s
a
s
s
p
P
p
p
p
W
−
=
= 6219
.
0
6219
.
0
at t = 20 C, ps = 0.00234 MPa = 2.34 kPa
P = 101.325 kPa
34
.
2
325
.
101
34
.
2
6219
.
0
−
=
s
W = 0.01407 kgv/kga
Enthalpy: Eq. (3-20b)

( )
t
W
t
i 86
.
1
3
.
2501
0
.
1 +
+
= kJ/kga
( ) ( ) ( )( )
[ ]
20
86
.
1
3
.
2501
01407
.
0
70
0
.
1 +
+
=
i = 55.7 kJ/kga
Specific Volume:
Ra = 287 J/kg.K
s
a
a
a
p
P
T
R
p
T
R
v
−
=
=
( )( )
( )( )
1000
34
.
2
325
.
101
15
.
273
20
287
−
+
=
v = 0.85 m3
/kga
(b) In English units, t = 20 F
Humidity Ratio:
Eq. (3-14b)
s
s
a
s
s
p
P
p
p
p
W
−
=
= 6219
.
0
6219
.
0
at t = 20 F < 32.02 F,
use ps at 32.02 F which is nearly equal by plotting on curve = 0.089 psia
P = 14.696 psia
089
.
0
696
.
14
089
.
0
6219
.
0
−
=
s
Enthalpy:
Eq. (3-20a)
( )
t
W
t
i 444
.
0
2
.
1061
240
.
0 +
+
= Btu/lbma
( ) ( ) ( )( )
[ ]
20
444
.
0
2
.
1061
0038
.
0
20
240
.
0 +
+
=
i = 8.7 Btu/lbma
Specific Volume:
Ra = 53.352 ft-lbf/lbm-R
s
a
a
a
p
P
T
R
p
T
R
v
−
=
=
( )( )
( )( )
144
089
.
0
696
.
14
67
.
459
20
352
.
53
−
+
=
v = 12.17 ft3
/lbma
In SI units, t = -6.7 C
Humidity Ratio:
Eq. (3-14b)
s
s
a
s
s
p
P
p
p
p
W
−
=
= 6219
.
0
6219
.
0
at t = -6.7 C < 0.01 C,
use ps at 0.01C which is nearly equal by plotting on curve = 0.00061 Mpa = 0.61 kPa

P = 101.325 kPa
61
.
0
325
.
101
61
.
0
6219
.
0
−
=
s
W = 0.0038 kgv/kga
Enthalpy:
Eq. (3-20b)
( )
t
W
t
i 86
.
1
3
.
2501
0
.
1 +
+
= kJ/kga
( ) ( ) ( )( )
[ ]
7
.
6
86
.
1
3
.
2501
0038
.
0
7
.
6
0
.
1 −
+
+
−
=
i = 2.8 kJ/kga
Specific Volume:
Ra = 287 J/kg.K
s
a
a
a
p
P
T
R
p
T
R
v
−
=
=
( )( )
( )( )
1000
61
.
0
325
.
101
15
.
273
7
.
6
287
−
+
−
=
v = 0.76 m3
/kga
3.2 The temperature of a certain room is 22 C, and the relative humidity is 50 percent.
The barometric pressure is 100 kPa. Find (a) the partial pressures of the air and
water vapor, (b) the vapor density, and (c) the humidity ratio of the mixtures.
Solution:
t = 22 C
φ = 50 % = 0.50
P = 100 kPa
(a) ps at 22 C = 2.672 kPa
s
v
p
p
=
φ ; s
v
p
p φ
= = (0.50)(2.672) = 1.336 kPa
v
a p
P
p −
= = 100 – 1.336 = 98.664 kPa
(b)
v
v
p
T
R
v =
Rv = 462 J/kg.K
( )( )
( )( )
1000
336
.
1
15
.
273
22
462 +
=
v = 102.065 m3
/kgv
(c)
v
v
p
P
p
W
−
= 6219
.
0
336
.
1
100
336
.
1
6219
.
0
−
=
W = 0.008421 kgv/kga

3.3 Compute the local atmospheric pressure at elevation ranging from sea level to
6000 ft (1830 m) in (a) inches of mercury and (b) kilopascals.
Solution:
(a) H = 6000 ft
Eq. (3-4)
P = a + bH
Table 3-2: H > 4000 ft
a = 29.42
b = -0.0009
P = 29.42 + (-0.0009)(6000) = 24.02 in. Hg.
(b) H = 1830 m
Eq. (3-4)
P = a + bH
Table 3-2: H > 1220 m
a = 99.436
b = -0.010
P = 99.436 + (-0.010)(1830) = 81.136 kPa.
3.4 Rework Problem 3.1 for an atmospheric pressure corresponding to an elevation of
(a) 5280 ft and (b) 1600 m.
Solution:
(a) H = 5280 ft
Eq. (3-4)
P = a + bH
Table 3-2: H > 4000 ft
a = 29.42
b = -0.0009
P = 29.42 + (-0.0009)(5280) = 24.668 in. Hg.
(b) H = 1830 m
Eq. (3-4)
P = a + bH
Table 3-2: H > 1220 m
a = 99.436
b = -0.010
P = 99.436 + (-0.010)(1600) = 83.346 kPa.

3.5 Compute the enthalpy of moist air at 60 F (16 C) and 80 percent relative humidity
for an elevation of (a) sea level and (b) 5000 ft (1525 m).
Solution:
(a) English units
ps at 60 F = 0.256 psia
φ = 80 % = 0.80
s
v p
p φ
= = (0.80)(0.256) = 0.2048 psia
At sea level, H = 0
Eq. (3-4)
P = a + bH
Table 3-2: H < 4000 ft
a = 29.92
P = a = 29.92 in. Hg = 101.325 kPa = 14.696 psia
2048
.
0
696
.
14
2048
.
0
6219
.
0
6219
.
0
−
=
−
=
v
v
p
P
p
Eq. (3-20a)
( )
t
W
t
i 444
.
0
2
.
1061
240
.
0 +
+
= Btu/lbma
( ) ( ) ( )( )
[ ]
60
444
.
0
2
.
1061
008789
.
0
60
240
.
0 +
+
=
i = 23.96 Btu/lbma
In SI units
ps at 16 C = 1.836 kPa
φ = 80 % = 0.80
s
v
p
p φ
= = (0.80)(1.836) = 1.469 psia
At sea level, H = 0
Eq. (3-4)
P = a + bH
Table 3-2: H < 4000 ft
a = 101.325
P = a = 101.325 kPa
469
.
1
325
.
101
469
.
1
6219
.
0
6219
.
0
−
=
−
=
v
v
p
P
p
W = 0.00915 kgv/kga
Eq. (3-20b)
( )
t
W
t
i 86
.
1
3
.
2501
0
.
1 +
+
= kJ/kga
( ) ( ) ( )( )
[ ]
16
86
.
1
3
.
2501
00915
.
0
16
0
.
1 +
+
=
i = 39.16 kJ/kga
(b) English units

ps at 60 F = 0.256 psia
φ = 80 % = 0.80
s
v p
p φ
= = (0.80)(0.256) = 0.2048 psia
At H = 5000 ft > 4000 ft
Eq. (3-4)
P = a + bH
Table 3-2: H > 4000 ft
a = 29.42
b = - 0.0009
P = 29.42 + (-0.0009)(5000) = 24.92 in. Hg = 12.24 psia
2048
.
0
24
.
12
2048
.
0
6219
.
0
6219
.
0
−
=
−
=
v
v
p
P
p
Eq. (3-20a)
( )
t
W
t
i 444
.
0
2
.
1061
240
.
0 +
+
= Btu/lbma
( ) ( ) ( )( )
[ ]
60
444
.
0
2
.
1061
010583
.
0
60
240
.
0 +
+
=
i = 25.91 Btu/lbma
In SI units
ps at 16 C = 1.836 kPa
φ = 80 % = 0.80
s
v
p
p φ
= = (0.80)(1.836) = 1.469 psia
At H = 1525 m > 1220 ,
Eq. (3-4)
P = a + bH
Table 3-2: H < 4000 ft
a = 99.436
b = - 0.010
P = 99.436 + (-0.010)(1525) = 84.186 kPa
469
.
1
186
.
84
469
.
1
6219
.
0
6219
.
0
−
=
−
=
v
v
p
P
p
W = 0.011045 kgv/kga
Eq. (3-20b)
( )
t
W
t
i 86
.
1
3
.
2501
0
.
1 +
+
= kJ/kga
( ) ( ) ( )( )
[ ]
16
86
.
1
3
.
2501
011045
.
0
16
0
.
1 +
+
=
i = 43.96 kJ/kga
3.6 The condition within a room is 70 F db, 50 percent relative humidity, and 14.696
psia pressure. The inside surface temperature of the window is 40 F. Will
moisture condense on the window glass?
Solution:
At 70 F db, ps = 0.363 psia
φ = 0.50

pv = 0.50 ( 0.363 psia ) = 0.1815 psia
at 0.1815 psia, t = 50.45 F
Since 40 F < 50.45 F , the moisture will condense on the window glass.
3.7 A duct has moist air flowing at a rate of 5000 ft3/min (2.36 m3/s). What is the
mass flow rate of the dry air, where the dry bulb temperature is 60 F (16 C), the
relative humidity is 80 percent and the pressure inside the duct corresponds to (a)
sea level, and (b) 6000 ft (1830 m).
Solution:
(a) English units
ps at 60 F = 0.2563 psia
s
v
p
p φ
= = (0.80)(0.2563) = 0.20504 psia
At sea level, P = 29.92 in. Hg = 14.696 psia
pa = P – pv = 14.696 – 0.20504 = 14.4910 psia
( )( )
( )( )
67
.
459
60
352
.
53
144
4910
.
14
+
=
=
T
R
p
a
a
ρ = 0.0753 lb/ft3
Q
& = 5000 ft3
/min
Q
ma
&
& ρ
= = (0.0753)(5000) = 376.5 lb/min
SI Units
ps at 16 C = 1.836 kPa
s
v p
p φ
= = (0.80)(1.836) = 1.4688 psia
At sea level, P = 101.325 kPa
pa = P – pv = 101.325 – 1.4688 = 99.8562
( )( )
( )( )
15
.
273
16
287
1000
8562
.
99
+
=
=
T
R
p
a
a
ρ = 1.2033 kg/m3
Q
& = 2.36 m3
/s
Q
ma
&
& ρ
= = (1.2033)(2.36) = 2.84 kg/s
(b) English units
ps at 60 F = 0.2563 psia
s
v
p
p φ
= = (0.80)(0.2563) = 0.20504 psia
At H = 6000 ft > 4000 ft
P = a + bH
a =29.42
b = - 0.0009
P = 29.42+ (-0.0009)(6000) = 24.02 in. Hg = 11.798 psia

pa = P – pv = 11.798 – 0.20504 = 11.593 psia
( )( )
( )( )
67
.
459
60
352
.
53
144
593
.
11
+
=
=
T
R
p
a
a
ρ = 0.06021 lb/ft3
Q
& = 5000 ft3
/min
Q
ma
&
& ρ
= = (0.06021)(5000) = 301.05 lb/min
SI Units
ps at 16 C = 1.836 kPa
s
v
p
p φ
= = (0.80)(1.836) = 1.4688 kPa
At H = 1830 m > 1220 m
P = a + bH
a = 99.436
b = - 0.010
P = 99.436 + (-0.010)(1830) = 81.136 kPa
pa = P – pv = 81.136 – 1.4688 = 79.667 kPa
( )( )
( )( )
15
.
273
16
287
1000
667
.
79
+
=
=
T
R
p
a
a
ρ = 0.96 kg/m3
Q
& = 2.36 m3
/s
Q
ma
&
& ρ
= = (0.96)(2.36) = 2.2656 kg/s
3.8 Compute the dew point for moist air at 80 F (27 C) and 50 percent relative
humidity for pressures corresponding to (a) sea level and (b) 5000 ft (1225 m).
Solution:
(a) English units
ps at 80 F = 0.507 psia
s
v
p
p φ
= = (0.50)(0.507) = 0.2535 psia
Dew Point = tdp = 59.68 F
SI units
ps at 27 C = 3.602 kPa
s
v p
p φ
= = (0.50)(3.602) = 1.801 kPa
Dew Point = tdp = 15.72 C
(b) H = 5000 ft (1225 m)
Since elevation does not affect dew point, the answers are the same as in (a).

3.9 A space is to be maintained at 70 F (21 C) dry bulb. It is estimated that the inside
wall surface temperature could be as low as 45 F (7 C). What maximum relative
and specific humidity can be maintained without condensation on the walls?
Solution:
English units
At 45 F, pv = 0.150 psia
At 70 F, ps = 0.363 psia
s
v p
p φ
=
( ) ( )
%
100
363
.
0
150
.
0
%
100 =
=
s
v
p
p
φ = 41.32 %
150
.
0
696
.
14
150
.
0
6219
.
0
6219
.
0
−
=
−
=
v
v
p
P
p
Maximum relative humidity = 41.32 %
Maximum specific humidity = 0.006413 lbmv/lbma
SI units
At 7 C, pv = 1.014 kPa
At 21 C, ps = 2.506 kPa
s
v
p
p φ
=
( ) ( )
%
100
506
.
2
014
.
1
%
100 =
=
s
v
p
p
φ = 40.46 %
014
.
1
325
.
101
014
.
1
6219
.
0
6219
.
0
−
=
−
=
v
v
p
P
p
W = 0.006287 kgv/kga
Maximum relative humidity = 40.46 %
Maximum specific humidity = 0.006287 kgv/kga
3.10 Outdoor air with a temperature of 40 F db and 35 F wb and with a barometric
pressure of 29 in. Hg is heated and humidified under steady-flow conditions to a
final temperature of 70 F db and 40 percent relative humidity. (a) Find the mass of
water vapor added to each pound mass of dry air. (b) If the water is supplied at 50
F, how much heat is added per pound mass of dry air?
Solution:
Solving for for W1 and i1 at Point 1
Using eq. (3-21d) and (3-14b) with its symbols.
At 35 F, 2
2 s
v
p
p = = 0.1013 psia

∗
2
fg
i = 1973.3 Btu/lbm
∗
w
i = 3.0 Btu/lbm
∗
2
t = 35 F
at 40 F , 1
v
i = 1078.5 Btu/lbm
P = 29 in Hg = 14.244 psia
1013
.
0
244
.
14
1013
.
0
6219
.
0
2
−
=
∗
s
Then
( )
∗
∗
∗
∗
−
+
−
=
w
v
fg
s
pa
i
i
i
W
t
t
c
W
1
2
2
1
2
1
( ) ( )( )
3
5
.
1078
3
.
1073
004454
.
0
40
35
24
.
0
1
−
+
−
=
( )
t
W
t
i 444
.
0
2
.
1061
24
.
0 +
+
= Btu/lbma
( ) ( ) ( )
[ ]
40
444
.
0
2
.
1061
003283
.
0
40
24
.
0
1 +
+
=
i = 13.14 Btu/lbma
Solving for W2 and i2 at point 2
At 70 F, ps = 0.363 psia
s
v
p
p φ
= = (0.40)(0.363 psia) = 0.1452 psia
P = 14.244 psia
1452
.
0
244
.
14
1452
.
0
6219
.
0
6219
.
0
2
−
=
−
=
v
v
p
P
p
( )
2
2
2
2 444
.
0
2
.
1061
240
.
0 t
W
t
i +
+
= Btu/lbma
( ) ( ) ( )
[ ]
70
444
.
0
2
.
1061
006408
.
0
70
240
.
0
2 +
+
=
i = 23.8 Btu/lbma
(a) Mass of water vapor added:
1
2
W
W
m
m
a
w
−
=
&
&
= 0.006408 – 0.003283 = 0.003125 lbmv/lbma
(b) At 50 F, iw = 18.1 Btu/lb
( ) w
a
w
a
i
m
m
i
i
m
q
&
&
&
&
−
−
= 1
2
= (23.8 – 13.14) – (0.003125)(18.1) = 10.3434 Btu/lbma
3.11 Air with a dry bulb temperature of 70 F and wet bulb temperature of 65 F is at a
barometric pressure of 29.92 in. Hg. Without making use of psychrometric chart,
find (a) the relative humidity of the air, (b) the vapor density, (c) the dew point,
(d) the humidity ratio, and (e) the volume occupied by the mixture associated with
a pound mass of dry air.
Solution:

1
t = 70 F
∗
2
t = 65 F
Solving for ∗
2
s
W , Eq. (3-14b)
2
v
p = 2
s
p at 65 F = 0.3095 psia
1
2 P
P = = 29.92 in Hg = 14.696 psia
3095
.
0
696
.
14
3095
.
0
6219
.
0
6219
.
0
2
2
2
2
−
=
−
=
∗
v
v
s
p
P
p
Solving for 1
W , Eq. (3-21c)
( )
∗
∗
∗
∗
−
+
−
=
w
v
fg
s
pa
i
i
i
W
t
t
c
W
1
2
2
1
2
1
∗
2
fg
i = fg
i at 65 F = 1056.5 Btu/lbm
∗
w
i = f
i at 65 F = 33 Btu/lbm
1
v
i = g
i at 70 F = 1091.7 Btu/lbm
( ) ( )( )
33
7
.
1091
5
.
1056
013379
.
0
70
65
24
.
0
1
−
+
−
=
Solving for 1
v
p , Eq. (3-14b)
1
1
1
1
6219
.
0
v
v
p
P
p
W
−
=
1
1
696
.
14
6219
.
0
012218
.
0
v
v
p
p
−
=
1
v
p = 0.2832 psia
at 70 F, 1
s
p = 0.363 psia
(a) Relative Humidity
363
.
0
2832
.
0
1
1
=
=
s
v
p
p
φ = 0.78 or 78 %
(b) Vapor Density
( )( )
( )( )
67
.
459
70
78
.
85
144
2832
.
0
+
=
=
T
R
p
v
v
ρ = 0.000898 lbmv/ft3
(c) Dew Point
At 1
v
p = 0.2832 psia
dp
t = 62.54 F
(d) Humidity Ratio

1
W
m
m
W
a
v
=
=
&
&
= 0.012218 lbmv/lbma
(e) Volume occupied by mixture per pound of mass of dry air.
( )( )
( )( )
144
2832
.
0
696
.
14
67
.
459
70
352
.
53
−
+
=
=
a
a
p
T
R
v = 13.62 ft3
/lbma
3.12 Air is cooled from 75 F db and 70 F wb until it is saturated at 55 F. Find (a) the
moisture removed per pound of dry air, (b) the heat removed to condense the
moisture, (c) the sensible heat removed, and (d) the total amount of heat removed.
Solution:
Use Figure 3-7
Determine state condition 1, 75 F db, 70 F wb
1
t = 75 F
∗
= 2
t
twb = 70 F
2
2 s
v p
p = at 70 F = 0.363 psia
2
P = 14.696 psia
363
.
0
696
.
14
363
.
0
6219
.
0
6219
.
0
2
2
2
2
−
=
−
=
∗
v
v
s
p
P
p
( )
∗
∗
∗
∗
−
+
−
=
w
v
fg
s
pa
i
i
i
W
t
t
c
W
1
2
2
1
2
1
∗
2
fg
i = fg
i at 70 F = 1053.7 Btu/lbm
∗
w
i = f
i at 70 F = 38Btu/lbm
1
v
i = g
i at 75F = 1093.85 Btu/lbm

( ) ( )( )
38
85
.
1093
7
.
1053
01575
.
0
75
70
24
.
0
1
−
+
−
=
W = 0.014581 Btu/lbma
( )
1
1
1
1 444
.
0
2
.
1061
240
.
0 t
W
t
i +
+
= Btu/lbma
( ) ( ) ( )
[ ]
75
444
.
0
2
.
1061
014581
.
0
75
240
.
0
1 +
+
=
i = 33.96 Btu/lbma
Determine state condition2
2
t = 55 F
2
s
p = 0.217 psia
2
2
2
2
6219
.
0
s
s
p
P
p
W
−
=
217
.
0
696
.
14
217
.
0
6219
.
0
2
−
=
( )
2
2
2
2 444
.
0
2
.
1061
240
.
0 t
W
t
i +
+
= Btu/lbma
( ) ( ) ( )
[ ]
55
444
.
0
2
.
1061
009321
.
0
55
240
.
0
2 +
+
=
i = 23.32 Btu/lbma
Determine state condition 3
1
3 t
t = = 75 F
2
3 W
W = = 0.009321 lbmv/lbma
( )
3
3
3
3 444
.
0
2
.
1061
240
.
0 t
W
t
i +
+
= Btu/lbma
( ) ( ) ( )
[ ]
75
444
.
0
2
.
1061
009321
.
0
75
240
.
0
3 +
+
=
i = 28.20 Btu/lbma
(a) Moisture removed, Eq. (3-29)
2
1
W
W
m
m
a
w
−
=
&
&
= 0.014581 – 0.009321 = 0.00526 Btu/lbma
(b) Heat removed to condense the moisture, Eq. (3-33)
3
1
i
i
m
q
a
l
−
=
&
&
= 33.96 – 28.20 = 5.76 Btu/lbma
(c) Sensible heat removed
2
3
i
i
m
q
a
s
−
=
&
&
= 28.20 – 23.32 = 4.88 Btu/lbma
(d) Total amount of heat removed
a
l
a
s
a
m
q
m
q
m
q
&
&
&
&
&
&
+
= = 4.88 + 5.76 = 10.64 Btu/lbma
3.13 The dry bulb and thermodynamic wet bulb temperature are measured to be 75 F
and 62 F, respectively, in a room. Compute the humidity ratio relative humidity
for the air at (a) sea level and (b) 5000 ft (1225 m).

Solution:
Use only English units as temperature are given in English units.
(a) At sea level, P = 29.92 in Hg = 14.696 psia
Eq. (3-14b)
∗
2
t = 62 F
2
2
2
2
6219
.
0
v
v
s
p
P
p
W
−
=
∗
2
v
p = 2
s
p at 62 F = 0.2774 psia
2774
.
0
696
.
14
2774
.
0
6219
.
0
2
−
=
∗
s
W = 0.0119865 lbmv/lbma
Eq. (3-21d)
( )
∗
∗
∗
∗
−
+
−
=
w
v
fg
s
pa
i
i
t
W
t
t
c
W
1
2
2
1
2
1
∗
2
fg
i = fg
i at 62 F = 1058.18 Btu/lbm
∗
w
i = f
i at 62 F = 30 Btu/lbm
1
v
i = g
i at 75F = 1093.85 Btu/lbm
( ) ( )( )
30
85
.
1093
18
.
1058
011965
.
0
75
62
24
.
0
1
−
+
−
=
W = 0.008969 lbmv/lbma – ans.
Solving for 1
v
p :
1
1
1
696
.
14
6219
.
0
v
v
p
p
W
−
=
1
1
696
.
14
6219
.
0
008969
.
0
v
v
p
p
−
=
1
v
p = 0.20893 psia
1
s
p = v
p at 75 F = 0.435 psia
435
.
0
20893
.
0
1
1
1
=
=
s
v
p
p
φ = 0.48 or 48 % - ans.
(b) H = 5000 ft = 1225 m
P = a + bH
Table 3-2. H > 4000 ft
a = 29.42
b = - 0.0009
P = 29.42 + (-0.0009)(4000) = 25.82 in Hg = 12.682 psia

2774
.
0
682
.
12
2774
.
0
6219
.
0
6219
.
0
2
2
2
2
−
=
−
=
∗
v
v
s
p
P
p
( )
∗
∗
∗
∗
−
+
−
=
w
v
fg
s
pa
i
i
t
W
t
t
c
W
1
2
2
1
2
1
( ) ( )( )
30
85
.
1093
18
.
1058
013907
.
0
75
62
24
.
0
1
−
+
−
=
W = 0.010900 lbmv/lbma – ans.
Solving for 1
v
p :
1
1
1
696
.
14
6219
.
0
v
v
p
p
W
−
=
1
1
682
.
12
6219
.
0
010900
.
0
v
v
p
p
−
=
1
v
p = 0.218448 psia
1
s
p = v
p at 75 F = 0.435 psia
435
.
0
218448
.
0
1
1
1
=
=
s
v
p
p
φ = 0.5022 or 50.22 % - ans.
3.14 To what temperature must atmospheric air at standard sea level pressure be cooled
to be saturated with a humidity ratio of 0.001 lbv/lba ? What is the temperature if
the pressure is 5 atmospheres?
Solution:
At standard sea level pressure
W = 0.001 lbmv/lbma
s
s
p
p
W
−
=
696
.
14
6219
.
0
s
s
p
p
−
=
696
.
14
6219
.
0
001
.
0
s
p = 0.0236 psia
Use Table A-1a, ≈
t 32.02 F – ans.
At P = 5 atm = 73.48 psia
Solving for 1
v
p :
s
s
p
p
W
−
=
=
48
.
73
6219
.
0
001
.
0
s
p = 0.118 psia
Use Table A-1a, interpolation, t = 39 F – ans.

3.15 Complete Table 3-3 using the ASHRAE psychrometric chart for (a) sea level, (b) 5000 ft
(1500 m) elevation; (c) compare parts (a) and (b).
Table 3-3 Psychrometric Properties for Problem 3-15.
State
Point
Dry
Bulb,
F(C)
Wet
Bulb,
F(C)
Dew
Point,
F(C)
Humid.
Ratio, w,
lbv/lba
(kgv/kga)
Enthalpy,
i,
Btu/lba
(kJ/kga)
Rel.
Humid.,
percent
Spec.
Vol, v,
ft3
/lba
(m3
/kga)
1 85(29) 60(16)
2 75(24) 50(10)
3 30(70) 60
4 70(21) 0.01143
5 82(28) 50(116)
Solution:
(a) Use Chart 1a.
State
Point
Dry Bulb,
F(C)
Wet Bulb,
F(C)
Dew
Point,
F(C)
Humid.
Ratio, w,
lbv/lba
(kgv/kga)
Enthalpy,
i,
Btu/lba
(kJ/kga)
Rel.
Humid.,
percent
Spec. Vol, v,
ft3
/lba
(m3
/kga)
1 85(29) 60(16) 40.7(4.83) 0.0054 26.3(21) 21 13.844(0.8643)
2 75(24) 60(16) 50(10) 0.0077 26.4(21) 41.4 13.640(0.852)
3 74.7(23.7) 65(18.3) 59.9(15.5) 0.0110 30(70) 60 13.70(0.855)
4 88.9(31.6) 70(21) 60.9(16.1) 0.01143 33.9(79) 39.1 14.078(0.879)
5 98.3(36.8) 85.6(29.7) 82(28) 0.0239 50(116) 60 14.6(0.912)

State
Point
Dry Bulb,
F(C)
Wet Bulb,
F(C)
Dew
Point,
F(C)
Humid.
Ratio, w,
lbv/lba
(kgv/kga)
Enthalpy,
i,
Btu/lba
(kJ/kga)
Rel.
Humid.,
percent
Spec. Vol, v,
ft3
/lba
(m3
/kga)
1 85(29) 60(16) 45(7.2) 0.0076 28.8(67) 24.7 16.70(1.043)
2 75(24) 59(15) 50(10) 0.0093 27.9(65) 41.4 16.44(1.026)
3 71.2(21.8) 61.4(16.3) 56.6(13.7) 0.118 30(70) 60 16.39(1.023)
4 103(39.5) 70(21) 55.7(13.2) 0.01143 37.4(87) 21.1 17.36(1.084)
5 N/A N/A 82(28) N/A 50(116) N/A N/A
(c) Comparison of (a) and (b)
State
Point
Dry
Bulb,
F(C)
Wet
Bulb,
F(C)
Dew
Point,
F(C)
Humid.
Ratio, w,
lbv/lba
(kgv/kga)
Enthalpy,
i,
Btu/lba
(kJ/kga)
Rel.
Humid.,
percent
Spec.
Vol, v,
ft3
/lba
(m3
/kga)
1 85(29) 60(16) a < b a < b a < b a < b a < b
2 75(24) a > b 50(10) a < b a < b a = b a < b
3 a > b a > b a > b 30(70) 60 a < b
4 a > b 70(21) a > b 0.01143 a < b a > b a < b
5 NA NA 82(28) NA 50(116) NA NA

3.16 To save energy, the environmental conditions in a room are to be regulated so that the dry
bulb temperature will be greater than or equal to 78 F (24 C) and the dew point will be
less than or equal to 64 F (17 C). Find the maximum relative humidity that can occur for
standard barometric pressure.
Solution:
Maximum relative humidity exist at tdb = 78 F (24 C), twb = 64 F (17 C), since increasing the dry
bulb constant dew point or decreasing the dew point at constant dry bulb will decrease relative
humidity.
From Chart 1a.
φmax = 46.6 %.
From Chart 1b.
φmax = 49.7 %.
3.17 A chilled water cooling coil receives 2.5 m3/s of air at 25 C db, 20 C wb. It is necessary
for the air to leave the coil at 13 C db, 12 C wb. Assume sea level pressure.
(a) Determine the SHF and the apparatus dew point.
(b) Compute the total and sensible heat transfer rates from the air.
Solution:
Chart 1b
State 1, tdb1 = 25 C, twb1 = 20 C
i1 = 57.26 kJ/kg
W1 = 0.01261 kgv/kga
v1 = 0.862 m3
/kg
State 2, tdb2 = 13 C, twb2 = 12 C
i2 = 34.08 kJ/kg
W2 = 0.00832 kgv/kga
Q
& = 2.5 m3
/s
862
.
0
5
.
2
1
=
=
v
Q
ma
&
& = 2.9 kg/s

(a) From chart 1b, SHF = 0.528 and td = 6.45 C
(b) Solving for total and sensible heat transfer rates;
( )
2
1
i
i
m
q a
−
= &
&
q
& = (2.9)(57.26 – 34.08) = 67.22 kW
( )
SHF
q
qs
&
& = = (67.22)(0.528) = 35.5 kW

3.18 The dry bulb and wet bulb temperatures are measured to be 78 F and 65 F, respectively,
in an air duct. Make use of psychrometric Charts 1a and 1b to find enthalpy, specific
volume, humidity ratio, and relative humidity in (a) English units and (b) SI units.
Solution:
(a) English units, Chart 1a
tdb = 78 F
twb = 65 F
Then,
i = 29.96 Btu/lbma
v = 13.773 ft3
/lbma
φ = 50 %
(b) SI units, Chart 1b
tdb = 25.56 C
twb = 18.33 C
Then,
i = 51.73 kg/kga
v = 0.860 m3
/kg
W = 0.01022 kgv/kga
φ = 50 %
3.19 The air in Problem 3.18 is heated to temperature of 110 F. Make use of Charts 1a and 1b
and compute the heat transfer rate if 4000 ft3
/min (1.9 m3
/s) is flowing at state 1, in (a)
English units and (b) SI units.
Solution:
(a) English units, Chart 1a.
2
t = 110 F
2
i = 37.79 Btu/lbma,
1
2 W
W =
1
v = 13.773 ft3
/lbma
1
i = 29.96 Btu/lbma
1
t = 78 F
Q
& = 4000 m3
/min

( )( )
773
.
13
60
4000
1
=
=
v
Q
ma
&
& = 17,425 lbma/hr
( )
1
2
t
t
c
m
q p
a
s
−
= &
&
s
q
& = (17,425)(0.245)(110 – 78) = 136,612 Btu/hr
or
( )
1
2 i
i
m
q a
s −
= &
&
s
q
& = (17,425)(37.79 – 29.96) = 136,525 Btu/hr
then s
q
& = 136,525 Btu/hr
Chart 1a:

2
t = 110 F = 43.33 C
2
i = 69.96 kJ/kga
1
2 W
W =
1
v = 0.860 m3
/kga
1
i = 51.73 kJ/kga
1
t = 25.56 C
Q
& = 1.9 m3
/s
860
.
0
9
.
1
1
=
=
v
Q
ma
&
& =2.21 kga/s
( )
1
2
t
t
c
m
q p
a
s
−
= &
&
s
q
& = (2.21)(1.02)(43.33 –25.56) = 40 kW
or
( )
1
2
i
i
m
q a
s
−
= &
&
s
q
& = (2.21)(69.96 – 51.73) = 40.3 kW
then s
q
& = 40.15 kW
Chart 1b:

3.20 The air in Problem 3.18 is cooled to 50 F (10 C) dry bulb and 90 percent relative
humidity. Make us of Charts 1a and 1b to compute the total heat transfer rate, the sensible
heat transfer, and the sensible heat factor (SHF) if 4000 ft3
/min (1.9 m3
/s) is flowing at
state 1 in (a) English units and (b) SI units.
Solution:

2
t = 50 F, 2
φ = 90 %
2
i = 19.46 Btu/lbma
Q
& = 4000 ft3
/min
1
v = 13.773 ft3
/lbma
1
i = 29.96 Btu/lbma
1
t = 78 F
( )( )
773
.
13
60
4000
1
=
=
v
Q
ma
&
& = 17,425 lb/hr
( )
2
1 i
i
m
q a −
= &
& = (17,425)(29.96 – 19.46) = 182,963 Btu/hr
( )
2
1
t
t
c
m
q p
a
s
−
= &
& = (17,425)(0.245)(78 – 50) = 119,536 Btu/hr
863
,
182
536
,
119
=
=
q
q
SHF s
&
&
= 0.653
2
t = 10 C, 2
φ = 90 %
2
i = 27.34 kJ/kga
Q
& = 1.9 m3
/s
1
v = 0.860 m3
/kga
1
i = 51.73 kJ/kga
1
t = 25.56 C
860
.
0
9
.
1
1
=
=
v
Q
ma
&
& = 2.21 kga/s
( )
2
1 i
i
m
q a −
= &
& = (2.21)(51.73 – 27.37) = 53.90 kW
( )
2
1
t
t
c
m
q p
a
s
−
= &
& = (2.21)(1.02)( 25.56 – 10) = 35.08 kW
90
.
53
08
.
35
=
=
q
q
SHF s
&
&
= 0.65

3.21 Air at 100 F (38 C) db and 65 F (18 C) wb is humified adiabatically with steam. The
steam supplied contains 20 percent moisture (quality of 0.80) at 14.7 psia (101.3 kPa). If
the air is humidified to 60 percent relative humidity, what is the dry bulb temperature of
the humidified air? Assume sea level pressure.
Solution:
Schematic of humidifying device
Using English units.
1
db
t = 100 F
1
wb
t = 65 F
x = 0.80
Chart 1a.
1
i = 29.79 Btu/lbma
1
For steam, Table A-1a, 14.7 psia, x = 0.8
f
i = 180.2 Btu/lbm
g
i = 1150.4 Btu/lbm
( )
f
g
f
w
i
i
x
i
i −
+
=
( )
2
.
180
4
.
1150
80
.
0
2
.
180 −
+
=
w
i = 956.4 Btu/lbm
W
i
i
W
W
i
i
w
∆
∆
=
=
−
−
1
2
1
2
= 956.4 Btu/lbm

At point 2, 2
φ = 60 %, draw a line from state 1 parallel to that determine from the protractor, then
locate 2 at the intersection of 2
φ = 60 % to determine 2
t = 91.6 F.
2
t = 91.6 F
Using SI units.
1
db
t = 38 C
1
wb
t = 18 C
x = 0.80
Chart 1b.
1
i = 50.38 kJ/kga
1
W = 0.00473 kgv/kga
For steam, Table A-1b, 101.3 kPa, x = 0.8
f
i = 419.1 kJ/kg
g
i = 2675.7 kJ/kg
( )
f
g
f
w
i
i
x
i
i −
+
=
( )
1
.
419
7
.
2675
80
.
0
1
.
419 −
+
=
w
i = 2224.4 kJ/kg
W
i
i
W
W
i
i
w
∆
∆
=
=
−
−
1
2
1
2
= 2224.4 kJ/kg

At point 2, 2
φ = 60 %, draw a line from state 1 parallel to that determine from the protractor, then
locate 2 at the intersection of 2
t = 33.1 C.
2
t = 33.1 C

3.22 Air at 84 F (29 C) db and 60 F (16 C) wb is humidified with the thr dry bulb temperature
remaining constant. Wet steam is supplied for humidification at 20 psia (138 kPa). What
quality must the steam have (a) to provide saturated air and (b) to provide air at 70
percent relative humidity? Assume sea level pressure.
Solution:
(a) Using English units
1
db
t = 84 F
1
wb
t = 60 F
2
db
t = 84 F
2
wb
t = 84 F
W
i
W
W
i
i
iw
∆
∆
=
−
−
=
1
2
1
2
Chart 1a.
1
i = 26.3 Btu/lbma
1
2
i = 48.23 Btu/lbma
2
00559
.
0
02556
.
0
3
.
26
23
.
48
−
−
=
w
i = 1098.1 Btu/lb
For 20 psia steam , Table A-1a
f
i = 196.2 Btu/lb
g
i = 1156.2 Btu/lb
2
.
196
2
.
1156
2
.
196
1
.
1098
1
−
−
=
−
−
=
f
g
f
w
i
i
i
i
x = 0.9395 or 93.95 %
Using SI units
1
db
t = 29 C

1
wb
t = 16 C
2
db
t = 29 C
2
wb
t = 29 C
W
i
W
W
i
i
iw
∆
∆
=
−
−
=
1
2
1
2
Chart 1b.
1
i = 44.57 kJ/kga
1
W = 0.00603 kgv/kga
2
i = 94.59 kJ/kga
2
W = 0.02562 kgv/kga
00603
.
0
02562
.
0
57
.
44
59
.
94
−
−
=
w
i = 2253.4 kJ/kg
For 138 kPa steam , Table A-1b
f
i = 456.4 kJ/kg
g
i = 2689.5 kJ/kg
4
.
456
5
.
2689
4
.
456
4
.
2553
1
−
−
=
−
−
=
f
g
f
w
i
i
i
i
x = 0.939 or 93.9 %
(b) Using English units
1
db
t = 84 F
1
wb
t = 60 F
2
db
t = 84 F
2
φ = 70 %
W
i
W
W
i
i
iw
∆
∆
=
−
−
=
1
2
1
2
Chart 1a.
1
i = 26.3 Btu/lbma
1
2
i = 39.57 Btu/lbma
2
00559
.
0
01767
.
0
3
.
26
57
.
39
−
−
=
w
i = 1098.5 Btu/lb
For 20 psia steam , Table A-1a
f
i = 196.2 Btu/lb
g
i = 1156.2 Btu/lb
2
.
196
2
.
1156
2
.
196
5
.
1098
1
−
−
=
−
−
=
f
g
f
w
i
i
i
i
x = 0.94 or 94 %

Using SI units
1
db
t = 29 C
1
wb
t = 16 C
2
db
t = 29 C
2
φ = 70 %
W
i
W
W
i
i
iw
∆
∆
=
−
−
=
1
2
1
2
Chart 1b.
1
i = 44.57 kJ/kga
1
W = 0.00603 kgv/kga
2
i = 74.37 kJ/kga
2
W = 0.01770 kgv/kga
00603
.
0
02562
.
0
57
.
44
37
.
74
−
−
=
w
i = 2253.6 kJ/kg
For 138 kPa steam , Table A-1b
f
i = 456.4 kJ/kg
g
i = 2689.5 kJ/kg
4
.
456
5
.
2689
4
.
456
6
.
2553
1
−
−
=
−
−
=
f
g
f
w
i
i
i
i
x = 0.939 or 93.9 %
3.23 Air at 38 db and 20 C and 101.325 kPa is humidified adiabatically with liquid water
supplied at 60 C, in such proportions that the mixture has a relative humidity of 80
percent. Find the dry bulb temperature of the mixture.
Solution:
Use chart 1b.

2
φ = 80 %
w
i at 60 C of water = 251.2 kJ/kg
W
i
i
W
W
i
i
w
∆
∆
=
=
−
−
1
2
1
2
= 251.2 kJ/kg
At point 2, 2
φ = 80 %, draw a line from state 1 parallel to that determined in the protractor then
locate state 2 at the intersection of 2
t = 22.86 C.
2
t = 22.86 C

3.24 It is desired to heat and humidify 2000 cfm (1.0 m3
/s) of air from an initial state defined
by a temperature of 60 F (16 C) dry bulb and relative humidity of 30 percent to a final
state of 110 F (43 C) dry bulb and 30 percent relative humidity. The air will first be
heated by a hot water coil, followed by adiabatic humidification using saturated vapor at
5 psig (34.5 kPa). Using the psychrometric chart, find the heat transfer rate for the
heating coil and the mass flow rate of the water vapor, and sketch the processes on a
skeleton chart showing pertinent data. Use (a) English units and (b) SI units. Assume sea
level pressure.
Solution:
Schematic of a heating and humidifying device
(a) English units
Q
& = 2000 cfm
At state 1,
1
db
t = 60 F
1
φ = 30 %
1
i = 17.97 Btu/lbma
1
1
v = 13.165 ft3
/lbma
State 2,
2
db
t = 110 F
2
φ = 30 %
2
i = 44.96 Btu/lbma
2
Table A-1a,
Steam at 5 psig saturated
w
i = 1155.9 Btu/lbm

Humidification process
χ
χ
χ W
W
i
i
w
i
−
−
=






∆
∆
2
2
= 1155.9 Btu/lb
Draw a line parallel to that determined from the protractor then locate state χ at 1
W
W =
χ =
0.00328 lbmv/lbma line.
Then,
χ
db
t = 107.47 F
χ
i = 29.425 Btu/lbma
( )( )
165
.
13
60
2000
1
=
=
v
Q
ma
&
& = 9115 lb/hr
( )
1
t
t
c
m
q p
a
−
= χ
&
& = (9115)(0.245)(107.47 – 60) = 106,009 Btu/hr
or
( )
1
i
i
m
q a
−
= χ
&
& = (9115)(29.425 – 17.97) = 104,412 Btu/hr
use
q
& = 105,210 Btu/hr
( )
1
2 W
W
m
m a
w
−
= &
& = (9115)(0.01672 – 0.00328) = 122.5 lbm/hr
SI units
Q
& = 1.0 m3
/s
At state 1,
1
db
t = 16 C
1
φ = 30 %
1
i = 24.62 kJ/kga
1
W = 0.00337 kgv/kga
1
v = 0.8236 m3
/kg
State 2,
2
db
t = 43 C
2
φ = 30 %
2
i = 85.40 kJ/kg.K
2
W = 0.01635 kgv/kga

Table A-1a,
Steam at 34.5 kpag saturated
w
i = 2689 kJ/kg

Humidification process
χ
χ
χ W
W
i
i
w
i
−
−
=






∆
∆
2
2
= 2689 kJ/kg
Draw a line parallel to that determined from the protractor then locate state χ at 1
W
W =
χ =
0.00337 kgv/kga line.
Then,
χ
db
t = 41.57 C
χ
i = 50.51 kJ/kga
8236
.
0
0
.
1
1
=
=
v
Q
ma
&
& = 1.2142 kg/s
( )
1
t
t
c
m
q p
a
−
= χ
&
& = (1.2142)(1.02)(41.57 –16) = 31.7 kW
or
( )
1
i
i
m
q a
−
= χ
&
& = (1.2142)(50.51 – 24.62) = 31.4 kW
use
q
& = 31.6 kW
( )
1
2 W
W
m
m a
w
−
= &
& = (1.2142)(0.01635 – 0.00337) = 0.01576 kg/s

3.25 Air at 40 F (5 C) db and 35 F (2 C) wb is mixed with warm air at 100 F (38 C) db
and 77 F (25 C) wb in the ratio of 2 lbm (kga) cool air to 1 lbm (kga) of warm air.
(a) Compute the humidity ratio and enthalpy of the mixed air, and (b) find the
humidity ratio and enthalpy using psychrometric chart 1.
Solution:
English units
(a)
1
db
t = 40 F
1
wb
t = 35 F
2
db
t = 100 F
2
wb
t = 77 F
2
1
a
a
m
m
&
&
= 2 lbm / 1 lbm = 2
Chart 1a
1
1
i = 12.66 Btu/lbma
2
2
i = 40.32 Btu/lbma
Mixed Air:
Eq. (3-44a)
2
1
2
1
2
1
3
1 a
m
a
m
a
m
a
m
i
i
i
&
&
&
&
+
+
=
( )( ) ( )
2
1
32
.
40
66
.
12
2
3
+
+
=
i = 21.88 Btu/lbma
Eq. (3-44b)
2
1
2
1
2
1
3
1 a
m
a
m
a
m
a
m
W
W
W
&
&
&
&
+
+
=
( )( ) ( )
2
1
01476
.
0
00315
.
0
2
3
+
+
=
Using chart 1a.
3
i = 22 Btu/lbma

3
W = 0.007 lbmv/lbma

SI units
(a)
1
db
t = 5 C
1
wb
t = 2 C
2
db
t = 38 C
2
wb
t = 25 C
2
1
a
a
m
m
&
&
= 2 kga / 1 kga = 2
Chart 1b
1
W = 0.00316 kgv/kga
1
i = 12.95 kJ/kga
2
W = 0.01462 kgv/kga
2
i = 75.79 kJ/kga
Mixed Air:
Eq. (3-44a)
2
1
2
1
2
1
3
1 a
m
a
m
a
m
a
m
i
i
i
&
&
&
&
+
+
=
( )( ) ( )
2
1
79
.
75
95
.
12
2
3
+
+
=
i = 33.90 kJ/kga
Eq. (3-44b)
2
1
2
1
2
1
3
1 a
m
a
m
a
m
a
m
W
W
W
&
&
&
&
+
+
=
( )( ) ( )
2
1
01462
.
0
00316
.
0
2
3
+
+
=
W = 0.00698 kgv/kga
Using chart 1b
3
i = 34 kJ/kga
3
W = 0.007 kgv/kga

3.26 Air at 10 C db and 5 C wb is mixed with air at 25 C db and 18 C wb in a
steady-flow process at standard atmospheric pressure. The volume flow rates
are 10 m3
/s and 6 m3
/s, respectively. (a) Compute the mixture conditions. (b)
Find the mixture conditions using Chart 1b.
Solution:
Chart 1b.
State 1, 1
db
t = 10 C, 1
wb
t = 5 C, 1
Q
& = 10 m3
/s
1
i = 18.58 kJ/kga
1
W = 0.0034 kgv/kga
1
v = 0.8065 m3
/kga
State 2, 2
db
t = 25 C, 2
wb
t = 18 C, 2
Q
& = 6 m3
/s
2
i = 50.70 kJ/kga
2
W = 0.0100 kgv/kga
2
v = 0.858 m3
/kga
8054
.
0
10
1
1
1 =
=
v
Q
ma
&
& = 12.40 kg/s
858
.
0
6
2
2
2 =
=
v
Q
ma
&
& = 6.99 kg/s
(a) Eq. (3-44a)
( )
99
.
6
40
.
12
1
70
.
50
58
.
18
99
.
6
40
.
12
1
2
1
2
1
2
1
3
+
+






=
+
+
=
a
a
a
a
m
m
i
i
m
m
i
&
&
&
&
= 30.16 kJ/kga
Eq. (3-44b)
( )
99
.
6
40
.
12
1
0100
.
0
0034
.
0
99
.
6
40
.
12
1
2
1
2
1
2
1
3
+
+






=
+
+
=
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
= 0.0058 kgv/kga
Then 3
db
t = 15.4 C, 3
wb
t = 10.4 C

(b)
99
.
6
40
.
12
13
32
____
____
2
1
=
=
a
a
m
m
&
&
= 1.774;
99
.
6
40
.
12
40
.
12
12
32
____
____
3
1
+
=
=
a
a
m
m
&
&
= 0.64;

99
.
6
40
.
12
99
.
6
12
13
____
____
3
2
+
=
=
a
a
m
m
&
&
= 1.774
Then, chart 1b. 3
db
t = 15.4 C, 3
wb
t = 10.4 C
3.27 Rework Problem 3.26, using Chart 1b, assuming that the mixture condition
can be computed on the basis of the volume flow rates rather than mass flow
rate. What is the percent error in the mixture enthalpy and humidity ratios?
Solution:
Chart 1b
State 1, 1
db
t = 10 C, 1
wb
t = 5 C, 1
Q
& = 10 m3
/s
1
i = 18.58 kJ/kga
1
W = 0.0034 kgv/kga
State 2, 2
db
t = 25 C, 2
wb
t = 18 C, 2
Q
& = 6 m3
/s
2
i = 50.70 kJ/kga
2
W = 0.0100 kgv/kga
3
5
6
10
13
32
____
____
2
1
=
=
=
Q
Q
&
&
;
8
5
6
10
10
12
32
____
____
3
1
=
+
=
=
Q
Q
&
&
;
8
3
6
10
6
12
13
____
____
3
2
=
+
=
=
Q
Q
&
&
Then, chart 1b. 3
db
t = 15.5 C, 3
wb
t = 10.5 C
( )
6
10
1
70
.
50
58
.
18
6
10
1
2
1
2
1
2
1
3
+
+






=
+
+
=
Q
Q
i
i
Q
Q
i
&
&
&
&
= 30.63 kJ/kga
Eq. (3-44b)
( )
6
10
1
0100
.
0
0034
.
0
6
10
1
2
1
2
1
2
1
3
+
+






=
+
+
=
Q
Q
W
W
Q
Q
W
&
&
&
&
= 0.0059 kgv/kga
% error in enthalpy = ( )
%
100
16
.
30
16
.
30
63
.
30 −
= 1.56 %
% error in humidity ratio = ( )
%
100
0058
.
0
0058
.
0
0059
.
0 −
= 1.73 %

3.28 ngiA zone in a building has a design cooling load of 136,000 Btu/hr (40 kW),
of which 110,000 Btu/hr (32 kW) is sensible cooling load. The space is to be
maintained at 76 F (24 C) dry bulb temperature and 50 percent relative
humidity. Locate the space condition line on Chart 1a and 1b.
Solution:
Chart 1a, English Units
q
& = 136,000 Btu/hr
s
q
& = 110,000 Btu/hr
db
t = 76 F
φ = 50 %
000
,
136
000
,
110
=
=
q
q
SHF s
&
&
= 0.81
Chart 1a:

Chart 1b, SI Units
q
& = 40 kW
s
q
& = 32 kW
db
t = 24 C
φ = 50 %
40
32
=
=
q
q
SHF s
&
&
= 0.80
Chart 1b:

3.29 Refer to Problem 3.28, and assume that the air can be supplied to the space at
55 F (13 C). Compute the volume flow rate of the air required in (a) English
units and (b) SI units.
Solution:
(a) Chart 1a
q
& = 136,000 Btu/hr
1
db
t = 76 F
1
φ = 50 % , SHF = 0.81
1
i = 28.74 Btu/lbma, 1
v = 13.71 ft3
/lb
2
db
t = 55 F, Chart 1a, SHF = 0.81
2
i = 22.39 Btu/lbma
39
.
22
74
.
28
000
,
136
2
1 −
=
−
=
i
i
q
ma
&
& = 21,417 lbma/hr
1
v
m
Q a
&
& = = (21,417 lb/hr)(1 hr / 60 min)(13.71 ft3
/lb) = 4,894 ft3
/min

(b) Chart 1b
q
& = 40 kW
1
db
t = 24 C
1
φ = 50 % , SHF = 0.80
1
i = 47.80 kJ/kga , 1
v = 0.854 m3
/kga
2
db
t = 13 C, Chart 1b, SHF = 0.80
2
i = 33.78 kJ/kga
78
.
33
8
.
47
40
2
1 −
=
−
=
i
i
q
ma
&
& = 2.853 kga/s
1
v
m
Q a
&
& = = (2.853 kga/s (0.854 m3
/kga) = 2.437 m3
/s

3.30 Rework Problem 3.28 using Charts 1Ha and 1Hb for (a) 5000 ft and (b) 1500
m elevation, respectively.
Solution:
(a) Chart 1Ha, English Units, 5000 ft
q
& = 136,000 Btu/hr
s
q
& = 110,000 Btu/hr
db
t = 76 F
φ = 50 %
000
,
136
000
,
110
=
=
q
q
SHF s
&
&
= 0.81
Chart 1Ha:

(b) Chart 1Hb, SI Units, 1500 m
q
& = 40 kW
s
q
& = 32 kW
db
t = 24 C
φ = 50 %
40
32
=
=
q
q
SHF s
&
&
= 0.80
Chart 1Hb:

3.31 Rework Problem 3.29 using Charts 1Ha and 1Hb for (a) 5000 ft and (b) 1500
m elevation, respectively.
Solution:
(a) 5000 ft, Chart 1Ha
q
& = 136,000 Btu/hr
1
db
t = 76 F
1
φ = 50 % , SHF = 0.81
1
i = 30.92 Btu/lbma, 1
v = 16.53 ft3
/lb
2
db
t = 55 F, Chart 1a, SHF = 0.81
2
i = 24.57 Btu/lbma
57
.
24
92
.
30
000
,
136
2
1 −
=
−
=
i
i
q
ma
&
& = 21,417 lbma/hr
1
v
m
Q a
&
& = = (21,417 lb/hr)(1 hr / 60 min)(16.53 ft3
/lb) = 5,900 ft3
/min

(b) 1500 m, Chart 1Hb
q
& = 40 kW
1
db
t = 24 C
1
φ = 50 % , SHF = 0.80
1
i = 52.58 kJ/kga , 1
v = 1.027 m3
/kga
2
db
t = 13 C, Chart 1b, SHF = 0.80
2
i = 38.56 kJ/kga
56
.
38
58
.
52
40
2
1 −
=
−
=
i
i
q
ma
&
& = 2.853 kga/s
1
v
m
Q a
&
& = = (2.853 kga/s (1.027 m3
/kga) = 2.930 m3
/s

3.32 A meeting hall is to be maintained at 25 C db and 18 C wb. The barometric
pressure is 101.3 kPa. The space has a load of 58.6 kW sensible and 58.6 kW
latent. The temperature of the supply air cannot be lower than 18 C db. (a)
How many kilograms per second of air must be supplied? (b) What is the
required wet bulb temperature of the supply air? (c) What is the sensible heat
ratio?
Solution: Chart 1b
1
db
t = 25 C
1
wb
t = 18 C
2
db
t = 18 C
s
q
& = 58.6 kW
l
q
& = 58.6 kW
p
c = 1.02 kJ/kg.C
(a) ( )
2
1 t
t
c
m
q p
a
s −
= &
&
( )( )
18
25
02
.
1
6
.
58 −
= a
m
&
a
m
& = 8.2073 kg/s
(b) l
s
q
q
q &
&
& +
= = 58.6 + 58.6 = 117.2 kW
( )
2
1
i
i
m
q a
−
= &
&
at 1
db
t = 25 C , 1
wb
t = 18 C
1
i = 50.71 kJ/kga
( )( )
2
71
.
50
2073
.
8
2
.
117 i
−
=
2
i = 36.43 kJ/kga
Then, 2
wb
t = 12.92 C
(c)
2
.
117
6
.
58
=
=
q
q
SHF s
&
&
= 0.50
3.33 A space is to be maintained at 72 F (22 C) and 30 percent relative humidity
during the winter months. The sensible heat loss from the space is 500,000
Btu/hr (146 kW), and latent heat loss due to infiltration is 50,000 Btu/hr (14.6
kW). Construct the condition line on (a) Chart 1a and (b) 1b.
Solution:
(a) Chart 1a
s
q
& = 500,000 Btu/hr

l
q
& = 50,000 Btu/hr
000
,
50
000
,
500
000
,
500
+
=
+
=
=
l
s
s
s
q
q
q
q
q
SHF
&
&
&
&
&
= 0.909
Chart 1a:

(b) Chart 1b
s
q
& = 146 kW
l
q
& = 14.6 kW
6
.
14
146
146
+
=
+
=
=
l
s
s
s
q
q
q
q
q
SHF
&
&
&
&
&
= 0.909
Chart 1b:

3.34 What is the maximum relative humidity to prevent condensation on pipes
carrying water at 50 F through a room that has an air temperature of 70 F?
Solution:
dp
t = 50 F
db
t = 70 F
From Chart 1a, φ = 49 %
3.35 Outdoor air at 95 F (35 C) db and 79 F (26 C) wb and at a barometric pressure
of 29.92 in Hg (101 kPa) is cooled and dehumidified under steady conditions
until it becomes saturated at 60 F (16 C). (a) Find the mass of water condensed
per pound of dry air. (b) If the condensate is removed at 60 F (16 C), what
quantity of heat is removed per pound (kilogram) of (16 C) dry air?
Solution:
English units:
1
db
t = 95 F
1
wb
t = 79 F
2
db
t = 60 F
2
wb
t = 60 F
1
i = 42.44 Btu/lbma
2
i = 26.46 Btu/lbma
1
2
(a) 2
1 W
W
W −
=
∆ = 0.0178 – 0.0111 = 0.0067 lbmv/lbma
(b) w
i at 60 F = 28 Btu/lb
Eq. (3.30)
( ) ( ) w
a
i
W
W
i
i
m
q
2
1
2
1 −
−
−
=
&
&
( ) ( )( )
28
0067
.
0
46
.
26
44
.
42 −
−
=
a
m
q
&
&
a
m
q
&
&
= 15.79 Btu/lbma

SI units:
1
db
t = 35 C
1
wb
t = 26 C
2
db
t = 16 C
2
wb
t = 16 C
1
i = 80.20 kJ/kga

2
i = 44.85 kJ/kga
1
W = 0.0175 kgv/kga
2
W = 0.0114 kgv/kga
(a) 2
1 W
W
W −
=
∆ = 0.0175 – 0.0114 = 0.0061 lbmv/lbma
(b) w
i at 60 F = 67.08 kJ/kg
Eq. (3.30)
( ) ( ) w
a
i
W
W
i
i
m
q
2
1
2
1 −
−
−
=
&
&
( ) ( )( )
08
.
67
0061
.
0
85
.
44
20
.
80 −
−
=
a
m
q
&
&
a
m
q
&
&
= 34.94 kJ/kga

3.36 Moist air enters a refrigeration coil at 89 F db and 75 F wb at a rate of 1400
cfm. The apparatus dewpoint temperature of the coil is 55 F. If 3.5 tons of
refrigeration is available, find the dry bulb temperature of the air leaving the
coil. Assume sea level pressure.
Solution: Chart 1a.
1
db
t = 89 F
1
wb
t = 75 F
1
Q
& = 1400 cfm
d
t = 55 F
q
& = (12000)(3.5) = 42,000 Btu/hr
at 1
db
t = 89 F, 1
wb
t = 75 F
1
i = 38.46 Btu/lbma
1
v =14.172 ft3
/lbma
( )( )
172
.
14
60
1400
1
1
=
=
v
Q
ma
&
& = 5,927 lb/hr
( )
2
1
i
i
m
q a
−
= &
&
( )( )
2
46
.
38
927
,
5
000
,
42 i
−
=
2
i = 31.37 Btu/lbma
From Chart 1a, d
t = 55 F, 2
i = 31.37 Btu/lbma
2
db
t = 73.18 F

3.37 Saturated steam at a pressure of 25 psia is sprayed into a stream of moist air.
The initial condition of the air is 55 F db and 45 F wb temperature. The mass
rate of airflow is 2000 lbma/min. Barometric pressure is 14.696 psia.
Determine (a) how much steam (in lbm/min) must be added to produce a
saturated air condition and (b) the resulting temperature of the saturated air.
Solution:
a
m
& = 2000 lbma/min
Steam at 25 psia, w
i = 1160.6 Btu/lbm
Eq. (3-40)
W
i
i
W
W
i
i
w
∆
∆
=
=
−
−
1
2
1
2
at 55 F db, 45 F wb, Chart 1a
1
i = 17.62 Btu/lbma
1
With
W
i
∆
∆
= 1160.6 Btu/lb, use protractor on Chart 1a
Then saturated air leaving condition:
2
db
t = 57.2 F
2
i = 24.60 Btu/lbma
2
(a) ( )
1
2
W
W
m
m a
s
−
= &
&
( )( )
0041
.
0
0101
.
0
2000 −
=
s
m
& = 12 lbm/min
(b) 2
db
t = 57.2 F

3.38 Saturated water vapor at 100 C is used to humidify a stream of moist air. The
air enters the humidifier at 13 C db and 2 C wb at a a flow rate of 2.5 m3
/s.
The pressure is 101.35 kPa. Determine (a) the mass flow rate of the steam
required to saturate the air and (b) the temperature of the saturated air.
Solution: Chart 1b
w
i at 100 C of water vapor = 2675.7 kJ/kg
at 13 C db, 2 C wb
1
v = 0.81 m3
/kga
1
W = 0.0000 kgv/kga
1
i = 12.97 kJ/kga
1
Q
& = 2.5 m3
/s
81
.
0
5
.
2
1
1
=
=
v
Q
ma
&
& = 3.0864 kg/s
w
i
W
i
=
∆
∆
= 2675.7 kJ/kg
From chart 1b
2
W = 0.0103 kgv/kga
2
i = 40.5 kJ/kga
2
db
t = 14.47 C
(a) ( )
1
2
W
W
m
m a
s
−
= &
&
( )( )
0000
.
0
0103
.
0
0864
.
3 −
=
s
m
& = 0.0318 kg/s
(b) 2
db
t = 14.47 C

3.39 Moist air at 70 F db and 45 percent relative humidity is recirculated from a
room and mixed with outdoor air at 97 F db and 83 F wb. Determine the
mixture dry bulb and wet bulb temperatures if the volume of recirculated air is
three times the volume of outdoor air. Assume sea level pressure.
Solution:
At 70 F db, 45 % relative humidity, Chart 1a
1
1
i = 24.47 Btu/lbma
1
v = 13.5 ft3
/lbma
At 70 F db, 45 % relative humidity, Chart 1a
2
2
i = 46.88 Btu/lbma
2
v = 14.51 ft3
/lbma
2
1
2
1
2
1
3
1
a
a
a
a
m
m
i
i
m
m
i
&
&
&
&
+
+
=
2
1
2
1
2
1
3
1
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
+
+
=
2
1 3Q
Q &
& =
1
Q
& is the recirculated air flow rate.
2
Q
& is the outdoor air flow rate
2
1 3Q
Q &
& =
2
2
1
1
3
v
m
v
m a
a
&
&
=
( )
51
.
14
5
.
13
3
3
2
1
2
1
=
=
v
v
m
m
a
a
&
&
= 2.7912
( )( )
7912
.
2
1
88
.
46
47
.
24
7912
.
2
3
+
+
=
i = 30.38 Btu/lbma
( )( )
7912
.
2
1
0214
.
0
007
.
0
7912
.
2
3
+
+
=
Then,
3
db
t = 77.3 F, 3
wb
t = 65.54 F

3.40 A structure has a calculated cooling load of 10 tons, of which 2.5 tons is latent
load. The space is to be maintained at 76 F db and 50 percent relative
humidity. Ten percent by volume of the air supplied to the space is outdoor air
at 100 F db and 50 percent relative humidity. The air is supplied to the space
cannot be less than 56 F db. Assume sea level pressure, and find (a) the
minimum amount of air supplied to the space in cfm, (b) the amounts of return
air and outdoor air in cfm, (c) the conditions and volume flow rate of the air
entering the cooling coil, and (d) the capacity and SHF for the cooling coil.
Solution:
q
& = 10 tons = 120,000 Btu/hr
l
q
& = 2.5 tons = 30,000 Btu/hr
l
s
q
q
q &
&
& −
= = 120,000 – 30,000 = 90,000 Btu/hr
Chart 1a
3
db
t = 76 F
3
φ = 50 %
0
db
t = 100 F
0
φ = 50 %
2
db
t = 56 F
2
0 10
0 Q
Q &
& .
=
2
3 90
0 Q
Q &
& .
=
9
2
0
Q
Q
&
& =
at state 0, 0
db
t = 100 F, 0
φ = 50 %
0
i = 47.08 Btu/lbma
0
o
v = 14.58 ft3
/lbma
at state 3, 3
db
t = 76 F, 3
φ = 50 %
3
i = 28.74 Btu/lbma
3
3
v = 13.71 ft3
/lbma












=
=
58
14
71
13
9
1
0
3
3
0
3
0
.
.
v
Q
v
Q
m
m
a
a
&
&
&
&
= 0.1045
at state 1,
( )( )
1045
0
1
74
28
08
47
1045
0
1
3
0
3
0
3
0
1
.
.
.
.
+
+
=
+
+
=
a
a
a
a
m
m
i
i
m
m
i
&
&
&
&
= 30.48 Btu/lbma

( )( )
1045
0
1
0096
0
0209
0
1045
0
1
3
0
3
0
3
0
1
.
.
.
.
+
+
=
+
+
=
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
= 0.01067 lbmv/lbma
1
db
t = 78.5 F
1
wb
t = 65.75 F
(a) ( )
2
3 t
t
c
m
q p
a
s −
= &
&
p
c = 0.245 Btu/lb-F
( )( )
56
76
245
.
0
000
,
90 −
= a
m
&
a
m
& = 18,367 lbma/hr
(b) 0
3 a
a
a
m
m
m &
&
& +
=
3
3 1045
0 a
a
a m
m
m &
&
& .
+
=
3
1045
1
367
18 a
m
&
.
, =
3
a
m
& = 16,629 lbma/hr return air.
( )( )( )
71
13
60
1
629
16
3
3
3 .
,
=
= v
m
Q a
&
& = 3,800 ft3
/min
( )( )=
=
= 629
16
1045
0
1045
0 3
0 ,
.
. a
a m
m &
& = 1,738 lbma/hr
( )( )( )
58
14
60
1
738
1
0
0
0 .
,
=
= v
m
Q a
&
& = 422 ft3
/min
(c) 1
db
t = 78.5 F, 1
wb
t = 65.75 F
1
v = 13.80 ft3
/lbma
1
i = 30.48 Btu/lbma
a
a
m
m &
& =
1
= 18,367 lbma/hr
( )( )( )
80
13
60
1
367
18
1
1
1 .
,
=
= v
m
Q a
&
& = 4,224 ft3
/min
(d) ( )
2
1 i
i
m
q a
c −
= &
&
For 2
i :
( )
2
3 i
i
m
q a −
= &
&
( )( )
2
74
.
28
367
,
18
000
,
120 i
−
=
2
i = 22.21 Btu/lbma, 2
db
t = 56 F
Then
( )
2
1 i
i
m
q a
c −
= &
&
( )( )
21
22
48
30
367
18 .
.
, −
=
c
q
& = 151,895 Btu/hr = 12.66 Tons
( )
2
1 t
t
c
m
q p
a
cs −
= &
&
( )( )( )
0
56
5
78
245
0
367
18 .
.
.
, −
=
cs
q
& = 101,248 Btu/hr

895
151
248
101
,
,
=
=
c
cs
c
q
q
SHF = 0.667

3.41 Rework Problem 3-40 for an elevation of 5000 feet.
Solution: Chart 1Ha
q
& = 10 tons = 120,000 Btu/hr
l
q
& = 2.5 tons = 30,000 Btu/hr
l
s
q
q
q &
&
& −
= = 120,000 – 30,000 = 90,000 Btu/hr
Chart 1a
3
db
t = 76 F
3
φ = 50 %
0
db
t = 100 F
0
φ = 50 %
2
db
t = 56 F
3
0
10
.
0 Q
Q &
& =
at state 0, 0
db
t = 100 F, 0
φ = 50 %
0
i = 51.95 Btu/lbma
0
o
v = 17.64 ft3
/lbma
at state 0, 3
db
t = 76 F, 3
φ = 50 %
3
i = 30.92 Btu/lbma
3
3
v = 16.53 ft3
/lbma
( ) 





=
=
64
.
17
53
.
16
10
.
0
0
3
3
0
3
0
v
Q
v
Q
m
m
a
a
&
&
&
&
= 0.09371
at state 1,
( )( )
09371
.
0
1
92
.
30
95
.
51
09371
.
0
1
3
0
3
0
3
0
1
+
+
=
+
+
=
a
a
a
a
m
m
i
i
m
m
i
&
&
&
&
= 32.72 Btu/lbma
( )( )
09371
.
0
1
0116
.
0
0253
.
0
09371
.
0
1
3
0
3
0
3
0
1
+
+
=
+
+
=
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
= 0.01277 lbmv/lbma
1
db
t = 78.05 F
1
wb
t = 64.73 F

(a) ( )
2
3
t
t
c
m
q p
a
s
−
= &
&
p
c = 0.245 Btu/lb-F
( )( )
56
76
245
.
0
000
,
90 −
= a
m
&
a
m
& = 18,367 lbma/hr
(b) 0
3 a
a
a
m
m
m &
&
& +
=
3
3
09371
.
0 a
a
a
m
m
m &
&
& +
=
3
09371
.
1
367
,
18 a
m
&
=
3
a
m
( )( )( )
53
.
16
60
1
794
,
16
3
3
3 =
= v
m
Q a
&
& = 4,627 ft3
/min
( )( )
794
,
16
09371
.
0
09371
.
0 3
0
=
= a
a
m
m &
& = 1,574 lbma/hr
( )( )( )
64
.
17
60
1
574
,
1
0
0
0 =
= v
m
Q a
&
& = 463 ft3
/min
(c) 1
db
t = 78.05 F, 1
wb
t = 64.73 F
1
v = 16.62 ft3
/lbma
1
i = 32.72 Btu/lbma
a
a
m
m &
& =
1 = 18,367 lbma/hr
( )( )( )
62
.
16
60
1
367
,
18
1
1
1 =
= v
m
Q a
&
& = 5,088 ft3
/min
(d) ( )
2
1
i
i
m
q a
c
−
= &
&
For 2
i :
( )
2
3
i
i
m
q a
−
= &
&
( )( )
2
92
.
30
367
,
18
000
,
120 i
−
=
2
i = 24.39 Btu/lbma, 2
db
t = 56 F
Then
( )
2
1
i
i
m
q a
c
−
= &
&
( )( )
39
.
24
72
.
32
367
,
18 −
=
c
q
& = 152,997 Btu/hr = 12.75 Tons
( )
2
1 t
t
c
m
q p
a
cs −
= &
&
( )( )( )
0
.
56
05
.
78
245
.
0
367
,
18 −
=
cs
q
& = 99,223 Btu/hr
997
,
152
223
,
99
=
=
c
cs
c
q
q
SHF = 0.649

3.42 A building has a calculated cooling load of 410 kW. The latent portion of the
load is100 kW. The space is to be maintained at 25 C db and 50 percent
relative humidity. Outdoor air is at 38 C and 50 percent relative humidity, and
10 percent by mass of the air supplied to the space is outdoor air. Air is to be
supplied to the space at not less than 18 C. Assume sea level pressure and find
(a) the minimum amount of air supplied to the space in m3
/s, (b) the volume
flow rates of the return air, exhaust air, and outdoor air, (c) the condition and
volume flow rates of the air entering the cooling coil, and (d) the capacity,
apparatus dew point, bypass factor, and SHF of the cooling coil.
Solution:
q
& = 410 kW
l
q
& = 100 kW
l
s
q
q
q &
&
& −
= = 410 kW – 100 kW = 310 kW
Chart 1b
State 3,
3
db
t = 25 C
3
φ = 50 %
3
W = 0.0099 kgv/kga
3
i = 50.31 kJ/kga
3
v = 0.858 m3
/kga
State 0,
0
db
t = 38 C
0
φ = 50 %
0
W = 0.0211 kgv/kga
0
i = 92.30 kJ/kga
0
v = 0.911 m3
/kga
1
2
0
10
.
0
10
.
0 a
a
m
m
m =
=
&
State 2,
2
db
t = 18 C
State 1,
9
1
10
.
0
10
.
0
1
1
1
3
0
=
−
=
a
a
a
a
a
m
m
m
m
m
&
&
&
&
&
( )
9
1
1
31
.
50
30
.
92
9
1
1
3
0
3
0
3
0
1
+
+






=
+
+
=
a
a
a
a
m
m
i
i
m
m
i
&
&
&
&
= 54.51 kJ/kga
( )
9
1
1
0099
.
0
0211
.
0
9
1
1
3
0
3
0
3
0
1
+
+






=
+
+
=
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
= 0.01102 kgv/kga

Chart 1b, 1
v = 0.863 m3
/kg, 1
db
t = 26.30 C, 1
wb
t = 19.19 C, 1
φ = 51.47 %
(a)
410
310
=
=
q
q
SHF s
= 0.756
Chart 1b, 2
db
t = 18 C
2
φ = 69.9 %
2
i = 40.87 kJ/kg
2
W = 0.0090 kgv/kga
2
v = 0.837 m3
/kga
( ) ( )( )
18
25
02
.
1
310
2
3
2
−
=
−
=
t
t
c
q
m
p
s
a
&
& = 43.4 kg/s
(b) 2
1 a
a
m
m &
& = = 43.4 kg/s
( )( )
4
.
43
1
.
0
1
.
0 1
0
=
= a
a
m
m &
& = 4.34 kg/s
( )( )
4
.
43
9
9 0
3
=
= a
a
m
m &
& = 39.06 kg/s
Return air:
( )( )
858
.
0
06
.
39
3
3
3
=
= v
m
Q a
&
& = 33.5 m3
/s
Exhaust air:
( )( )
911
.
0
34
.
4
0
0
0
=
= v
m
Q a
&
& = 3.96 m3
/s
Outdoor air = Exhaust air
( )( )
911
.
0
34
.
4
0
0
0
=
= v
m
Q a
&
& = 3.96 m3
/s
(a) Condition of air entering the coil
1
db
t = 26.30 C, 1
wb
t = 19.19 C, 1
φ = 51.47 %, 1
v = 0.863 m3
/kg
Volume flow rate;
( )( )
863
.
0
4
.
43
1
1
1
=
= v
m
Q a
&
& = 37.45 m3
/s
(b) Capacity ( ) ( )( )
87
.
40
51
.
54
4
.
43
2
1
1
−
=
−
= i
i
m
q a
c
&
& = 592 kW
Apparatus dew point , Chart 1b;
d
t = 7.2 C
By-pass factor
2
.
7
3
.
26
2
.
7
18
1
2
−
−
=
−
−
=
d
d
t
t
t
t
b = 0.566
SHF of cooling coil,
c
cs
c
q
q
SHF
&
&
=
( ) ( )( )( )
18
3
.
26
02
.
1
4
.
43
2
1
1
−
=
−
= t
t
c
m
q p
a
cs
&
&
cs
q
& = 368 kW
592
368
=
=
c
cs
c
q
q
SHF
&
&
= 0.622

3.43 Rework Problem 3-42 for an elevation of 1500 m
Solution:
q
& = 410 kW
l
q
& = 100 kW
l
s
q
q
q &
&
& −
= = 410 kW – 100 kW = 310 kW
Chart 1Hb
State 3,
3
db
t = 25 C
3
φ = 50 %
3
W = 0.0119 kgv/kga
3
i = 55.40 kJ/kga
3
v = 1.032 m3
/kga
State 0,
0
db
t = 38 C
0
φ = 50 %
0
W = 0.0254 kgv/kga
0
i = 103.46 kJ/kga
0
v = 1.099 m3
/kga
1
2
0
10
.
0
10
.
0 a
a
m
m
m =
=
&
State 2,
2
db
t = 18 C
State 1,
9
1
10
.
0
10
.
0
1
1
1
3
0
=
−
=
a
a
a
a
a
m
m
m
m
m
&
&
&
&
&
( )
9
1
1
4
.
55
46
.
103
9
1
1
3
0
3
0
3
0
1
+
+






=
+
+
=
a
a
a
a
m
m
i
i
m
m
i
&
&
&
&
= 60.21 Btu/lbma
( )
9
1
1
0119
.
0
0254
.
0
9
1
1
3
0
3
0
3
0
1
+
+






=
+
+
=
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
= 0.0133 lbmv/lbma
Chart 1Hb, 1
v = 1.038 m3
/kg, 1
db
t = 26.30 C, 1
wb
t = 18.79 C, 1
φ = 51.5 %
(a)
410
310
=
=
q
q
SHF s
= 0.756

Chart 1Hb, 2
db
t = 18 C
2
wb
t = 14.54 C
2
i = 45.96 kJ/kg
2
W = 0.0110 kgv/kga
2
v = 1.006 m3
/kga
96
.
45
4
.
55
310
2
3
2
−
=
−
=
i
i
q
m s
a
&
& = 43.4 kg/s
(b) 2
1 a
a
m
m &
& = = 43.4 kg/s
( )( )
4
.
43
1
.
0
1
.
0 1
0
=
= a
a
m
m &
& = 4.34 kg/s
( )( )
4
.
43
9
9 0
3
=
= a
a
m
m &
& = 39.06 kg/s
Return air:
( )( )
032
.
1
06
.
39
3
3
3
=
= v
m
Q a
&
& = 40.31 m3
/s
Exhaust air:
( )( )
099
.
1
34
.
4
0
0
0
=
= v
m
Q a
&
& = 4.77 m3
/s
( )( )
099
.
1
34
.
4
0
0
0
=
= v
m
Q a
&
& = 4.77 m3
/s
1
db
t = 26.30 C, 1
wb
t = 18.79 C, 1
φ = 51.5 %, 1
v = 1.038 m3
/kg
Volume flow rate;
( )( )
038
.
1
4
.
43
1
1
1
=
= v
m
Q a
&
& = 45.05 m3
/s
(b) Capacity ( ) ( )( )
96
.
45
21
.
60
4
.
43
2
1
1
−
=
−
= i
i
m
q a
c
&
& = 619 kW
Apparatus dew point , Chart 1Hb;
d
t = 8.9 C
By-pass factor
9
.
8
3
.
26
9
.
8
18
1
2
−
−
=
−
−
=
d
d
t
t
t
t
b = 0.523
c
cs
c
q
q
SHF
&
&
=
( ) ( )( )( )
18
3
.
26
02
.
1
4
.
43
2
1
1
−
=
−
= t
t
c
m
q p
a
cs
&
&
cs
q
& = 368 kW

619
368
=
=
c
cs
c
q
q
SHF
&
&
= 0.595

3.44 A building has a total heating load of 200,000 Btu/hr. The sensible heat factor
for the space is 0.8. The space is to be maintained at 72 F db and 40 percent
relative humidity. Outdoor air at 40 F db and 20 percent relative humidity in
the amount of 1000 cfm is required. Air is supplied to the space at 120 F db.
Find (a) the conditions and amount of air supplied to the space, (b) the
temperature rise of the air through the furnace, (c) the amount of water at 50 F
required by the dehumidifier, and (d) the capacity of the furnace. Assume sea
level pressure.
Solution: Chart 1a
State 0,
0
db
t = 40 F
0
φ = 20 %
0
i = 10.72 Btu/lbma
0
0
v = 12.611 ft3
/lbma
State 3,
3
db
t = 72 F
3
φ = 40 %
3
i = 24.58 Btu/lbma
3
3
v = 13.54 ft3
/lb
State 2,
2
db
t = 120 F
Chart 1a, SHF = 0.8
2
2
i = 39.28 Btu/lbma
2
wb
t = 76.18 F
2
φ = 12.8 %
q
& = 200,000 Btu/hr
( ) ( )( )
000
,
200
8
.
0
=
= q
SHF
qs
&
& = 160,000 Btu/hr
( )
3
2
2
i
i
m
q a
−
= &
&
( ) 58
.
24
28
.
39
000
,
200
3
2
2
−
=
−
=
i
i
q
ma
&
& = 13,606 lbma/hr
0
Q
& = 1000 ft3
/min

( )( )
61
.
12
60
1000
0
0
0 =
=
v
Q
ma
&
& = 4,758 lbma/hr
2
3 a
a
m
m &
& = = 13,606 lbma/hr
( )
606
,
13
758
,
4
1
58
.
24
72
.
10
606
,
13
758
,
4
1
3
0
3
0
3
0
1
+
+






=
+
+
=
a
a
a
a
m
m
i
i
m
m
i
&
&
&
&
= 20.99 Btu/lbma
( )
606
,
13
758
,
4
1
0067
.
0
0010
.
0
606
,
13
758
,
4
1
3
0
3
0
3
0
1
+
+






=
+
+
=
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
= 0.0052 lbmv/lbma
1
db
t = 63.8 F
At 50 F , w
i = 18.1 Btu/lb
For x – 2,
w
x
x
i
W
W
i
i
W
i
=
−
−
=
∆
∆
2
2
= 18.1
Draw a line parallel to the protractor line from 2 – x, x intersecting line of constant
humidity ratio from 1.
1
W
Wx
= = 0.0052 lbmv/lbma
x
t = 139 F
x
i = 39.20 Btu/lbma
(a) Air supplied to the space.
2
db
t = 120 F, 2
φ = 12.8 %, 2
wb
t = 76.18 F
2
a
m
& = 13,606 lbma/hr
(b) Temperature rise of the air through the furnace.
1
t
tx
− = 139 – 63.8 = 75.2 F
(c) Amount of water required by the humidifier.
( ) ( )( )
0052
.
0
0094
.
0
606
,
13
2
2
−
=
−
= x
a
w
W
W
m
m &
& = 57.2 lb/hr
(d) Capacity of the furnace.
( ) ( )( )
99
.
20
20
.
39
606
,
13
1
2
−
=
−
= i
i
m
q x
a
F
&
& = 247,765 Btu/hr

3.45 Rework Problem 3.44 for an elevation of 5000 ft.
Solution: Chart 1Ha
State 0,
0
db
t = 40 F
0
φ = 20 %
0
i = 10.96 Btu/lbma
0
0
v = 15.16 ft3
/lbma
State 3,
3
db
t = 72 F
3
φ = 40 %
3
i = 26.10 Btu/lbma
3
3
v = 16.31 ft3
/lb
State 2,
2
db
t = 120 F
Chart 1a, SHF = 0.8
2
2
i = 40.80 Btu/lbma
2
wb
t = 73.50 F
2
φ = 12.2 %
q
& = 200,000 Btu/hr
( ) ( )( )
000
,
200
8
.
0
=
= q
SHF
qs
&
& = 160,000 Btu/hr
( )
3
2
2
i
i
m
q a
−
= &
&
( ) 10
.
26
80
.
40
000
,
200
3
2
2
−
=
−
=
i
i
q
ma
&
& = 13,606 lbma/hr
0
Q
& = 1000 ft3
/min
( )( )
16
.
15
60
1000
0
0
0 =
=
v
Q
ma
&
& = 3,958 lbma/hr
2
3 a
a
m
m &
& = = 13,606 lbma/hr

( )
606
,
13
958
,
3
1
10
.
26
96
.
10
606
,
13
958
,
3
1
3
0
3
0
3
0
1
+
+






=
+
+
=
a
a
a
a
m
m
i
i
m
m
i
&
&
&
&
= 22.69 Btu/lbma
( )
606
,
13
958
,
3
1
00805
.
0
00125
.
0
606
,
13
958
,
3
1
3
0
3
0
3
0
1
+
+






=
+
+
=
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
= 0.00645 lbmv/lbma
1
db
t = 65.23 F
At 50 F , w
i = 18.1 Btu/lb
For x – 2,
w
x
x
i
W
W
i
i
W
i
=
−
−
=
∆
∆
2
2
= 18.1
Draw a line parallel to the protractor line from 2 – x, x intersecting line of constant
humidity ratio from 1.
1
W
Wx
= = 0.00645 lbmv/lbma
x
t = 139.52 F
x
i = 40.73 Btu/lbma
(a) Air supplied to the space.
2
db
t = 120 F, 2
φ = 12.2 %, 2
wb
t = 73.50 F
2
a
m
& = 13,606 lbma/hr
(b) Temperature rise of the air through the furnace.
1
t
tx
− = 139.52 – 65.23 = 74.29 F
(c) Amount of water required by the humidifier.
( ) ( )( )
00645
.
0
0104
.
0
606
,
13
2
2
−
=
−
= x
a
w
W
W
m
m &
& = 53.74 lb/hr
(d) Capacity of the furnace.
( ) ( )( )
69
.
22
73
.
40
606
,
13
1
2
−
=
−
= i
i
m
q x
a
F
&
& = 245,452 Btu/hr

3.46 The system of Problem 3.40 has a supply air fan located just downstream of
the cooling coil. The total power input to the fan is 2.0 hp. It is also estimated
that heat gain to the supply duct system is 500 Btu/hr. Rework Problem 3.40
taking the fan and duct system heat gain into account. Make a sketch of the
processes.
Solution:
q
& = 10 tons = 120,000 Btu/hr
l
q
& = 2.5 tons = 30,000 Btu/hr
l
s
q
q
q &
&
& −
= = 120,000 – 30,000 = 90,000 Btu/hr
Chart 1a
3
db
t = 76 F
3
φ = 50 %
0
db
t = 100 F
0
φ = 50 %
2
db
t = 56 F
2
1 ′
−
′
q
& = 2.0 hp (2545 Btu/hp.hr) = 5,090 Btu/hr
2
2 −
′
q
& = 500 Btu/hr
3
0
10
.
0 Q
Q &
& =
at state 0, 0
db
t = 100 F, 0
φ = 50 %
0
i = 47.08 Btu/lbma
0
o
v = 14.58 ft3
/lbma
at state 3, 3
db
t = 76 F, 3
φ = 50 %
3
i = 28.74 Btu/lbma
3
3
v = 13.71 ft3
/lbma
( ) 





=
=
58
.
14
71
.
13
10
.
0
0
3
3
0
3
0
v
Q
v
Q
m
m
a
a
&
&
&
&
= 0.094
at state 1,
( )( )
094
.
0
1
74
.
28
08
.
47
094
.
0
1
3
0
3
0
3
0
1
+
+
=
+
+
=
a
a
a
a
m
m
i
i
m
m
i
&
&
&
&
= 30.32 Btu/lbma

( )( )
094
.
0
1
0096
.
0
0209
.
0
094
.
0
1
3
0
3
0
3
0
1
+
+
=
+
+
=
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
= 0.0106 lbmv/lbma
1
db
t = 78.0 F
1
wb
t = 65.5 F
(a) ( )
2
3
t
t
c
m
q p
a
s
−
= &
&
p
c = 0.245 Btu/lb-F
( )( )
56
76
245
.
0
000
,
90 −
= a
m
&
a
m
& = 18,367 lbma/hr
(b) 0
3 a
a
a
m
m
m &
&
& +
=
3
3
094
.
0 a
a
a
m
m
m &
&
& +
=
3
094
.
1
367
,
18 a
m
&
=
3
a
m
( )( )( )
71
.
13
60
1
789
,
16
3
3
3 =
= v
m
Q a
&
& = 3,836 ft3
/min
( )( )
789
,
16
094
.
0
094
.
0 3
0
=
= a
a
m
m &
& = 1,578 lbma/hr
( )( )( )
58
.
14
60
1
578
,
1
0
0
0 =
= v
m
Q a
&
& = 384 ft3
/min
(c) 1
db
t = 78.0 F, 1
wb
t = 65.5 F
1
v = 13.78 ft3
/lbma
1
i = 30.32 Btu/lbma
a
a
m
m &
& =
1 = 18,367 lbma/hr
( )( )( )
78
.
13
60
1
367
,
18
1
1
1 =
= v
m
Q a
&
& = 4,218 ft3
/min
(d) ( )
2
2
1
2
2 ′
−
′ −
= i
i
m
q a
&
&
( )
1
2
1
2
1 ′
′
′
−
′ −
= i
i
m
q a
&
&
( )
1
2
1
2
2
2
1 ′
−
′
′
−
′ −
=
+ i
i
m
q
q a
&
&
&
( )
2
3
1
i
i
m
q a
−
= &
&
( )
1
3
1
2
2
2
1 ′
−
′
′
−
′ −
=
+
+ i
i
m
q
q
q a
&
&
&
&
( )( )
1
74
.
28
367
,
18
000
,
120
500
090
,
5 ′
−
=
+
+ i
1′
i = 21.90 Btu/lbma
( ) ( )( )
90
.
21
32
.
30
367
,
18
1
1
−
=
−
= ′
i
i
m
q a
c
&
& = 12.9 tons
( )
1
2
2
2
2
1 ′
−
′
′
−
′ −
=
+ t
t
c
m
q
q p
a
&
&
&
( )( )( )
1
56
245
.
0
367
,
18
500
5090 ′
−
=
+ t
1′
t = 54.76 F

( )
1
1 ′
−
= t
t
c
m
q p
a
cs
&
&
( )( )( )
76
.
54
0
.
78
245
.
0
367
,
18 −
=
cs
q
& = 104,578 Btu/hr
650
,
154
578
,
104
=
=
c
cs
c
q
q
SHF = 0.676

3.47 The system of Problem 3.42 has a supply air fan located just downstream of
the coil and a return air fan just upstream of the mixing box. The power input
to the supply fan is 18 kW, and the power input to the return fan is 12 kW.
Rework Problem 3.42 taking the fan power into account. Make a sketch of the
system processes on a skeleton psychrometric chart.
Solution:
q
& = 410 kW
l
q
& = 100 kW
l
s
q
q
q &
&
& −
= = 410 kW – 100 kW = 310 kW
Chart 1b
State 3,
3
db
t = 25 C
3
φ = 50 %
3
W = 0.0099 kgv/kga
3
i = 50.31 kJ/kga
3
v = 0.858 m3
/kga
State 0,
0
db
t = 38 C
0
φ = 50 %
0
W = 0.0211 kgv/kga
0
i = 92.30 kJ/kga
0
v = 0.911 m3
/kga
1
2
0
10
.
0
10
.
0 a
a
m
m
m =
=
&
State 2,
2
db
t = 18 C
State 1,
9
1
10
.
0
10
.
0
1
1
1
3
0
=
−
=
a
a
a
a
a
m
m
m
m
m
&
&
&
&
&
Solving for State 3′ ;
( )
3
3
3
3
3 t
t
c
m
q p
a −
= ′
′
−
&
&
( ) ( )
18
25
02
.
1
310
2
3
2
−
=
−
=
t
t
c
q
m
p
s
a
&
& = 43.4 kg/s
2
1 a
a
m
m &
& =
( )
4
.
43
9
.
0
9
.
0
10
.
0 1
1
1
3
=
=
−
= a
a
a
a
m
m
m
m &
&
&
& = 39.06 kg/s
( )
3
3
3
3
3 t
t
c
m
q p
a −
= ′
′
−
&
&
( ) ( )( )( )
25
02
.
1
06
.
39
12 3
−
= ′
t

3′
t = 25.3 C
at 3′ , 3′
t = 25.3 C, 3
3
W
W =
′ = 0.0099 kgv/kga
Chart 1b, 3′
i = 50.7 kJ/kga
For State 1;
( )
9
1
1
7
.
50
30
.
92
9
1
1
3
0
3
0
3
0
1
+
+






=
+
+
=
a
a
a
a
m
m
i
i
m
m
i
&
&
&
&
= 54.86 kJ/kga
( )
9
1
1
0099
.
0
0211
.
0
9
1
1
3
0
3
0
3
0
1
+
+






=
+
+
=
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
= 0.01102 kgv/kga
Chart 1b, 1
v = 0.864 m3
/kg, 1
db
t = 26.57 C, 1
wb
t = 19.3 C, 1
φ = 50.8 %
(a)
410
310
=
=
q
q
SHF s
= 0.756
Chart 1b, 2
db
t = 18 C
2
φ = 69.9 %
2
i = 40.87 kJ/kg
2
W = 0.0090 kgv/kga
2
v = 0.837 m3
/kga
( ) ( )( )
18
25
02
.
1
310
2
3
2
−
=
−
=
t
t
c
q
m
p
s
a
&
& = 43.4 kg/s
(b) 2
1 a
a
m
m &
& = = 43.4 kg/s
( )( )
4
.
43
1
.
0
1
.
0 1
0
=
= a
a
m
m &
& = 4.34 kg/s
( )( )
4
.
43
9
9 0
3
=
= a
a
m
m &
& = 39.06 kg/s
Return air:
( )( )
858
.
0
06
.
39
3
3
3
=
= v
m
Q a
&
& = 33.5 m3
/s
Exhaust air:
( )( )
911
.
0
34
.
4
0
0
0
=
= v
m
Q a
&
& = 3.96 m3
/s
( )( )
911
.
0
34
.
4
0
0
0
=
= v
m
Q a
&
& = 3.96 m3
/s

1
db
t = 26.57 C, 1
wb
t = 19.3 C, 1
φ = 50.8 %, 1
v = 0.864 m3
/kg
Volume flow rate;
( )( )
864
.
0
4
.
43
1
1
1
=
= v
m
Q a
&
& = 37.5 m3
/s
(a) Capacity
( )
2
1
1 ′
−
= i
i
m
q a
c
&
&
Solve for 2′;
( )
2
2
2
2
2 ′
−
′ −
= t
t
c
m
q p
a
&
&
( )( )( )
2
18
02
.
1
4
.
43
18 ′
−
= t
2′
t = 17.6 C
at 2′, 2′
t = 17.6 C, 2
2 W
W =
′ = 0.0090 kgv/kga
2′
i = 40.57 kJ/kga
( ) ( )( )
57
.
40
86
.
54
4
.
43
2
1
1
−
=
−
= ′
i
i
m
q a
c
&
& = 620 kW
Apparatus dew point , Chart 1b;
d
t = 8.8 C
By-pass factor
8
.
8
57
.
26
8
.
8
6
.
17
1
2
−
−
=
−
−
= ′
d
d
t
t
t
t
b = 0.495
c
cs
c
q
q
SHF
&
&
=
( ) ( )( )( )
6
.
17
57
.
26
02
.
1
4
.
43
2
1
1 −
=
−
= ′
t
t
c
m
q p
a
cs
&
&
cs
q
& = 397 kW
620
397
=
=
c
cs
c
q
q
SHF
&
&
= 0.64

3.48 A large warehouse located in Denver, Colorado (elevation = 5000 ft or 1500
m), is to be conditioned using an evaporative cooling system. Assume that the
space is to be maintained at 80 F (27 C) and 50 percent relative humidity by a
100 percent outdoor air system. Outdoor design conditions are 91 F (33 C) db
and 59 F (15 C) wb. The cooling load is estimated to be 100 tons (352 kW)
with a sensible heat factor of 0.9. The supply air fan is locatedjust downstream
of the spray chamber and is estimated to require a power input of 20 hp (15
kW). Determine the volume flow rate of air to the space, and sketch the
processes on a skeleton psychrometric chart in (a) English units and (b) SI
units.
Solution:
(a) Chart 1Ha, English units
State 2;
2
db
t = 80 F
2
φ = 50 %
2
i = 33.73 Btu/lbma
2
State 0;
0
db
t = 91 F
0
wb
t = 59 F
0
i = 28.0 Btu/lbma
0
State 1;
0
1
i
i = = 28.0 Btu/lbma
q
& = 100 tons = 1,200,000 Btu/hr
( ) ( )( )
9
.
0
000
,
200
,
1
=
= SHF
q
qs
&
& = 1,080,000 Btu/hr
( )
1
2 ′
−
= t
t
c
m
q p
a
s
&
& = 1,080,000 Btu/hr
but 1
1 ′
−
q
& = 20 hp (2545 Btu/hp-hr) = 50,900 Btu/hr
( )
1
1
1
1 t
t
c
m
q p
a −
= ′
′
−
&
& = 50,900 Btu/hr
By trial and error method,
Try 1′
t = 61 F
( )( )
61
80
245
.
0 −
= a
s
m
q &
& = 1,080,000 Btu/hr
a
m
& = 232,009 lb/hr
( )( )( )
1
1
1 61
245
.
0
009
,
232 t
q −
=
′
−
& = 50,900 Btu/hr
1
t = 60.10 F
locate at Chart 1Ha, 1
i = 28.0 Btu/lbma, 1
1 ′
=W

with 1′
W = 0.0125 lbmv/lbma and SHF = 0.9
at 1′
t = 61 F,
1′
v = 16.09 ft3
/lbma
( )( )( )
09
.
16
60
1
009
,
232
1 =
= ′
v
m
Q a
&
& = 62,217 ft3
/min

(b) Chart 1Hb, SI units
State 2;
2
db
t = 27 C
2
φ = 50 %
2
i = 61.34 kJ.kga
2
W = 0.0134kgv/kga
State 0;
0
db
t = 33 C
0
wb
t = 15 C
0
i = 47.10 kJ/kga
0
W = 0.0054 kgv/kga
State 1;
0
1
i
i = = 47.1 kJ/kga
q
& = 352 kW
( ) ( )( )
9
.
0
352
=
= SHF
q
qs
&
& = 316.8 kW
( )
1
2 ′
−
= t
t
c
m
q p
a
s
&
& = 316.8 kW
From Chart 1Hb, sketch of the process with SHF = 0.9, SI units equivalent for
temperatureas given in the problem will not intesect inside the chart as shown.
Therefore, use the answer in (a) converted to SI units.
( ) 















=
sec
60
min
1
281
.
3
1
min
/
217
,
62
3
3
ft
m
ft
Q
& = 29.36 m3
/s

3.49 The summer design conditions for Tuczon, Arizona, are 102 F (39 C) dry bulb
and 66 F (19 C) wet bulb temperature. In Shreveport, Louisiana, the design
conditions are 96 F (36 C) dry bulb and 76 F (24 C) wet bulb temperature.
What is the lowest air temperature that can theoretically be attained in an
evaporative cooler for these design conditions in each city?
Solution:
English units, Chart 1a.
In Tuczon, Arizona, 1
db
t = 102 F, 1
wb
t = 66 F,
1
2 i
i = saturated
then 2
t = 65.623 F
In Shreveport, Louisiana, 1
db
t = 96 F, 1
wb
t = 76 F,
1
2 i
i = saturated
then 2
t = 75.783 F
SI units, Chart 1b.
In Tuczon, Arizona, 1
db
t = 39 C, 1
wb
t = 19 C,
1
2 i
i = saturated
then 2
t = 18.836 C
In Shreveport, Louisiana, 1
db
t = 36 C, 1
wb
t = 24 C,
1
2 i
i = saturated
then 2
t = 23.892 C
3.50 Consider a conventional cooling system designed for use at high elevation
(5000 ft or 1500 m). The space, with SHF = 0.7, is to be maintained at 75 F
(24 C) db and 40 percent relative humidity, and outdoor design conditions are
100 F (38 C) and 10 percent relative humidity. Outdoor air is to be mixed with
return air such that it can be cooled sensibly to 50 F (10 C), where it crosses
the condition line. The air is then supplied to the space. Sketch the processes
on Chart 1Ha or 1Hb, and compute the volume flow rate of the supply air and
percent outdoor per ton of cooling load, in (a) English units and (b) SI units.
Solution:
(a) English units, Chart 1Ha.
At state 4, 4
db
t = 75 F, 4
φ = 40%
4
i = 27.77 Btu/lbma
4
4
v = 16.43 ft3
/lbma

At state 1 1
db
t = 100 F, 1
φ = 10%
1
i = 29.43 Btu/lbma
1
1
v = 17.08 ft3
/lbma
At state 3, From Chart 1Ha, 3
db
t = 50 F, SHF = 0.7
2
3
W
W = = 0.0065 lbmv/lbma,
3
i = 19.02 Btu/lbma
3
v = 15.60 ft3
/lbma
At state 2, From Chart 1Ha, 2
W = 0.0065 lbmv/lbma, along the line connecting the
state 1 and 4.
2
db
t = 90 F
2
i = 28.78 Btu/lbma
( )
3
2
i
i
m
q a
−
= &
& = 1 ton = 12,000 Btu/hr
( ) 000
,
12
02
.
19
78
.
28 =
−
a
m
&
a
m
& = 1,230 lb/hr per ton
( )( )( )
60
.
15
60
1
230
,
1
3
3 =
= v
m
Q a
&
& = 320 ft3
/min-tons
Percentage Outdoor Air.
4
1
4
1
4
1
2
1
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
+
+
=
( )
4
1
4
1
1
0089
.
0
0049
.
0
0065
.
0
a
a
a
a
m
m
m
m
&
&
&
&
+
+
=
4
1
a
a
m
m
&
&
=1.5
( )
%
100
1
5
.
1
5
.
1
%
4
1
1
3
1
+
=
+
=
=
a
a
a
a
a
m
m
m
m
m
Outdoor
&
&
&
&
&
= 0.6 (100 %) = 60 %

(b) SI units, Chart 1Hb.
At state 4, 4
db
t = 24 C, 4
φ = 40%
4
i = 46.81 kJ/kga

4
W = 0.0089 kgv/kga
4
v = 1.023 m3
/kga
At state 1 1
db
t = 38 C, 1
φ = 10%
1
i = 50.86 kJ/kga
1
W = 0.0049 kJ/kga
1
v = 1.065 m3
/kga
At state 3, From Chart 1Hb, 3
db
t = 10 C, SHF = 0.7
2
3
W
W = = 0.0065 kgv/kga,
3
i = 26.41 kJ/kga
3
v = 0.971 m3
/kga
At state 2, From Chart 1Hb 2
W = 0.0065 kgv/kga, along the line connecting the state 1
and 4.
2
db
t = 32.4 C
2
i = 49.22 Btu/lbma
( )
3
2
i
i
m
q a
−
= &
& = 1 ton = 3.517 kW
( ) 517
.
3
41
.
26
22
.
49 =
−
a
m
&
a
m
& = 0.1542 kga/s per ton
( )( )
971
.
0
1542
.
0
3
3
=
= v
m
Q a
&
& = 0.1497 m3
/s -tons
Percentage Outdoor Air.
4
1
4
1
4
1
2
1
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
+
+
=
( )
4
1
4
1
1
0089
.
0
0049
.
0
0065
.
0
a
a
a
a
m
m
m
m
&
&
&
&
+
+
=
4
1
a
a
m
m
&
&
=1.5
( )
%
100
1
5
.
1
5
.
1
%
4
1
1
3
1
+
=
+
=
=
a
a
a
a
a
m
m
m
m
m
Outdoor
&
&
&
&
&
= 0.6 (100 %) = 60 %

3.51 A space heating system is designed as shown in Fig. 3.21 for a large zone in a
building. Under design conditions for Kansas City, Missouri, air enters the
preheat coil at 6 F (-14 C) and essentially 0 percent relative humidity. The
outdoor air is heated to 60 F (16 C) and mixed with return air. It is then heated
and humidified in a separate process to 105 F (40 C) and 30 percent relative
humidity (RH) for supply to the space. Saturated vapor at 2.0 psig is used for
humidification. Twenty-five percent of the supply air is outdoor air by mass.
The total space heating load is 500,000 Btu/hr (145 kW), and the space design
conditions are 70 F (21 C) and 30 percent RH. Sketch the psychrometric
processes, and compute the supply air volume flow rate, the heat transfer rates
in both coils, and the steam flow rate in (a) English units, (b) SI units.
Solution: It is not Fig. 3.21 but Fig. 3.23
At state 3, 3
db
t = 70 F, 3
φ = 30 %
3
i = 21.90 Btu/lbma
3
At state 2, 2
db
t = 105 F, 2
φ = 30 %
2
i = 41.15 Btu/lbma
2
2
v = 14.56 ft3
/lbma
1
2
0
25
.
0
25
.
0 a
a
a
m
m
m &
&
& =
=
′
1
2
0
2
3
75
.
0
75
.
0 a
a
a
a
a
m
m
m
m
m &
&
&
&
& =
=
−
= ′
75
.
0
25
.
0
3
0
=
′
a
a
m
m
&
&
At state 0′, 0′
db
t = 60 F, 0′
φ = 0 %
0′
i = 14.4 Btu/lbma

0′
( )
75
.
0
25
.
0
1
90
.
21
4
.
14
75
.
0
25
.
0
1
3
0
3
0
3
0
1
+
+






=
+
+
=
′
′
′
a
a
a
a
m
m
i
i
m
m
i
&
&
&
&
= 20.03 Btu/lbma
( )
75
.
0
25
.
0
1
0047
.
0
0000
.
0
75
.
0
25
.
0
1
3
0
3
0
3
0
1
+
+






=
+
+
=
′
′
′
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
= 0.00353 lbmv/lbma
at state 1, 1
db
t = 67.42 F
state 1′, saturated steam at 2.0 psig, w
i = 1152.8 Btu/lb
W
i
i
W
W
i
i
w
∆
∆
=
=
−
−
′
′
1
2
1
2
= 1152.8 Btu/lb
From Chart 1a, intersection of line 1
2 ′
− parallel to
W
i
∆
∆
= 1152.8 Btu/lb and line
1
1 ′
− .
1′
db
t = 102.97 F
1
1 W
W =
′ = 0.00353 lbmv/lbma
1′
i = 28.62 Btu/lbma
( )
3
2
2
2
i
i
m
q a
a
−
= &
&
( )
90
.
21
15
.
41
000
,
500 2
−
= a
m
&
2
a
m
& = 25,974 lbma/hr
( )
974
,
25
25
.
0
25
.
0 2
0
=
=
′ a
a
m
m &
& = 6,494 lbma/hr
Supply air volume flow rate,
( )( )( )
56
.
14
60
1
974
,
25
2
2
2 =
= v
m
Q a
&
& = 6,303 ft3
/min
Heat transfer rate on pre-heat coil
( ) ( )( )( )
6
60
245
.
0
494
,
6
0
0
0
−
=
−
= ′
′ t
t
c
m
q p
a
p
&
& = 85,016 Btu/hr
Heat transfer rate on heating coil
( ) ( )( )( )
42
.
67
97
.
102
245
.
0
974
,
25
1
1
2
−
=
−
= ′ t
t
c
m
q p
a
h
&
& = 226,227 Btu/hr
Steam flow rate
( ) ( )( )
00353
.
0
0144
.
0
974
,
25
1
2
2
−
=
−
= ′
W
W
m
m a
v
&
& = 282 lbm/hr

At state 3, 3
db
t = 21 C, 3
φ = 30 %
3
i = 32.84 kJ/kga
3
W = 0.0046 kgv/kga

At state 2, 2
db
t = 40 C, 2
φ = 30 %
2
i = 76.01 kJ/kga
2
W = 0.0139 kgv/kga
2
v = 0.907 m3
/kga
1
2
0
25
.
0
25
.
0 a
a
a
m
m
m &
&
& =
=
′
1
2
0
2
3
75
.
0
75
.
0 a
a
a
a
a
m
m
m
m
m &
&
&
&
& =
=
−
= ′
75
.
0
25
.
0
3
0
=
′
a
a
m
m
&
&
At state 0′, 0′
db
t = 16 F, 0′
φ = 0 %
0′
i = 16.10 kJ/kga
0′
W = 0.0000 kgv/kga
( )
75
.
0
25
.
0
1
84
.
32
10
.
16
75
.
0
25
.
0
1
3
0
3
0
3
0
1
+
+






=
+
+
=
′
′
′
a
a
a
a
m
m
i
i
m
m
i
&
&
&
&
= 28.66 kJ/kga
( )
75
.
0
25
.
0
1
0046
.
0
0000
.
0
75
.
0
25
.
0
1
3
0
3
0
3
0
1
+
+






=
+
+
=
′
′
′
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
= 0.00346 kgv/kga
at state 1, 1
db
t = 19.75 F
state 1′, saturated steam at 2.0 psig (13.8 kPag) , w
i = 2681 kJ/kg
W
i
i
W
W
i
i
w
∆
∆
=
=
−
−
′
′
1
2
1
2
= 2681 kJ/kg
From Chart 1a, intersection of line 1
2 ′
− parallel to
W
i
∆
∆
= 2681 kJ/kg and line 1
1 ′
− .
1′
db
t = 38.9 C
1
1 W
W =
′ = 0.00346 kgv/kga
1′
i = 48.02 kJ/kga
( )
3
2
2
2
i
i
m
q a
a
−
= &
&
( )
54
.
32
01
.
76
145 2
−
= a
m
&
2
a
m
& = 3.12 kga/s
( )
12
.
3
25
.
0
25
.
0 2
0
=
=
′ a
a
m
m &
& = 0.78 kgas
Supply air volume flow rate,
( )( )
907
.
0
12
.
3
2
2
2
=
= v
m
Q a
&
& = 2.83 m3
/s

Heat transfer rate on pre-heat coil
( ) ( )( ) ( )
[ ]
14
16
02
.
1
78
.
0
0
0
0
−
−
=
−
= ′
′ t
t
c
m
q p
a
p
&
& = 23.9 kW
Heat transfer rate on heating coil
( ) ( )( )( )
75
.
19
9
.
38
02
.
1
12
.
3
1
1
2 −
=
−
= ′ t
t
c
m
q p
a
h
&
& = 61 kW
Steam flow rate
( ) ( )( )
00346
.
0
0139
.
0
12
.
3
1
2
2
−
=
−
= ′
W
W
m
m a
v
&
& = 0.0326 kg/s

3.52 A space is to be maintained at 78 F (26 C) and 68 F (20 C) wb. The cooling
system is a variable-air-volume (VAV) type where the quantity of air supplied
and the supply air temperature are controlled. Under design conditions, the
total cooling load is 150 tons (530 kW) with a sensible heat factor of 0.6, and
the supply air temperature is 60 F (16 C). At minimum load, about 18 tons (63
kW) with SHF of 0.8, the air quantity may be reduced no more than 80 percent
by volume of the full design value. Determine the supply air conditions for
minimum load. Show all the conditions on a psychrometric chart for (a)
English units and (b) SI units. Assume sea level pressure.
Solution:
(a) English units
Chart 1a,
At state 1
1
db
t = 78 F
1
wb
t = 68 F
1
i = 32.34 Btu/lbma
1
1
v = 13.82 ft3
/lbma
At state 2, SHF = 0.6
2
db
t = 60 F
2
wb
t = 57.8 F
2
i = 25.14 Btu/lbma
2
2
v = 13.30 ft3
/lbma
( )
2
1
2
i
i
m
q a
−
= &
&
( )
14
.
25
34
.
32
000
,
800
,
1 2
−
= a
m
&
2
a
m
& = 250,000 lbma/hr
( )( )( )
30
.
13
60
1
000
,
250
2
2
2 =
= v
m
Q a
&
& = 55,417 ft3
/min
At minimum load condition:
q
& = 18 tons = 216,000 Btu/hr
( ) ( )
417
,
55
2
.
0
2
.
0
8
.
0
1 2
2
2 =
=
−
=
′ Q
Q
Q &
&
& = 11,084 ft3
/min
Since space is to be maintained at 78 F db and 68 F wb.
At state 1′= state 1
1′
db
t = 78 F

1′
wb
t = 68 F
1′
i = 32.34 Btu/lbma
1′
1′
v = 13.82 ft3
/lbma
A state 2′, SHF = 0.8, Chart 1a
( ) ( )( )
8
.
0
000
,
216
=
= SHF
q
qs
&
& = 172,800 Btu/hr
( ) ( )
2
1
2
2
2
1
2 ′
′
′
′
′
′ −








=
−
= t
t
c
v
Q
t
t
c
m
q p
p
a
s
&
&
( )( ) ( )( ) ( )( ) 800
,
172
78
245
.
0
040
,
665
78
245
.
0
60
084
,
11
2
2
2
2
=
−








=
−






= ′
′
′
′
t
v
t
v
qs
&
2
2 9429
.
0
55
.
73 ′
′ −
= t
v
or
( ) ( )
2
1
2
2
2
1
2 ′
′
′
′
′
′ −








=
−
= i
i
v
Q
i
i
m
q a
s
&
&
( )( ) ( ) ( ) 000
,
216
34
.
32
040
,
665
34
.
32
60
084
,
11
2
2
2
2
=
−








=
−






= ′
′
′
′
i
v
t
v
qs
&
2
2 3248
.
0
34
.
32 ′
′ −
= v
i
From psychrometric chart, satisfying above equation, Chart 1a., State 2
2′
db
t = 63.75 F, 2′
v = 13.44 ft3
/lbma
2′
i = 28.0 Btu/lbma, 2′
wb
t = 62.35 F
2′
W = 0.01174 lbmv/lbma, 2′
φ = 92.7 %
Therefore the supply air condition is 63.75 F db and 62.35 F wb or 92.7 % RH.

(b) SI units
Chart 1b,
At state 1
1
db
t = 26 C
1
wb
t = 20 C
1
i = 57.23 kJ/kga
1
W = 0.01220 kgv/kga
1
v = 0.864 m3
/kga

2
db
t = 16 C
2
wb
t = 14.37 C
2
i = 40.23 kJ/kga
2
W = 0.00955 kgv/kga
2
v = 0.8317 m3
/kga
( )
2
1
2
i
i
m
q a
−
= &
&
( )
23
.
40
23
.
57
530 2
−
= a
m
&
2
a
m
& = 31.18 kga/s
( )( )
8317
.
0
18
.
31
2
2
2
=
= v
m
Q a
&
& = 25.93 m3
/s
q
& = 63 kW
( ) ( )
93
.
25
2
.
0
2
.
0
8
.
0
1 2
2
2 =
=
−
=
′ Q
Q
Q &
&
& = 5.186 kga/s
Since space is to be maintained at 26 C db and 20 C wb.
At state 1′ = State 1
1′
db
t = 26 C
1′
wb
t = 20 C
1′
i = 57.23 kJ/kga
1′
W = 0.01220 kgv/kga
1′
v = 0.864 kgv/kga
A state 2′, SHF = 0.8, Chart 1b
( ) ( )( )
8
.
0
63
=
= SHF
q
qs
&
& = 50.4 kW
( ) ( )
2
1
2
2
2
1
2 ′
′
′
′
′
′ −








=
−
= t
t
c
v
Q
t
t
c
m
q p
p
a
s
&
&
( )( ) 4
.
50
26
02
.
1
186
.
5
2
2
=
−








= ′
′
t
v
qs
&
2
2 105
.
0
729
.
2 ′
′ −
= t
v
or
( ) ( )
2
1
2
2
2
1
2 ′
′
′
′
′
′ −








=
−
= i
i
v
Q
i
i
m
q a
s
&
&

( ) 63
23
.
57
186
.
5
2
2
=
−








= ′
′
i
v
qs
&
2
2 148
.
12
23
.
57 ′
′ −
= v
i
From psychrometric chart, satisfying above equation, Chart 1b., State 2′
2′
db
t = 17.99 C, 2′
v = 0.840 m3
/kga
2′
i = 47.0 kJ/kga, 2′
wb
t = 16.74 C
2′
W = 0.0114 kgv/kga, 2′
φ = 88 %
Therefore the supply air condition is 17.99 C db and 16.74 C wb or 88 % RH.

3.53 Rework Problem 3.52 for an elevation of 5000 feet (1500 m).
Solution:
(a) English units
At design condition:
Chart 1Ha.
At state 1
1
db
t = 78 F
1
wb
t = 68 F
1
i = 35.70 Btu/lbma
1
1
v = 16.69 ft3
/lbma
2
db
t = 60 F
2
wb
t = 59.23 F
2
i = 28.35 Btu/lbma
2
2
v = 16.06 ft3
/lbma
( )
2
1
2
i
i
m
q a
−
= &
&
( )
35
.
28
70
.
35
000
,
800
,
1 2
−
= a
m
&
2
a
m
& = 244,899 lbma/hr
( )( )( )
06
.
16
60
1
899
,
244
2
2
2 =
= v
m
Q a
&
& = 65,551 ft3
/min
q
& = 18 tons = 216,000 Btu/hr
( ) ( )
551
,
65
2
.
0
2
.
0
8
.
0
1 2
2
2 =
=
−
=
′ Q
Q
Q &
&
& = 13,110 ft3
/min
Since space is to be maintained at 78 F db and 68 F wb.
At state 1′= state 1
1′
db
t = 78 F
1′
wb
t = 68 F
1′
i = 35.70 Btu/lbma
1′
1′
v = 16.69 ft3
/lbma

A state 2′, SHF = 0.8, Chart 1a
( ) ( )( )
8
.
0
000
,
216
=
= SHF
q
qs
&
& = 172,800 Btu/hr
( ) ( )
2
1
2
2
2
1
2 ′
′
′
′
′
′ −








=
−
= t
t
c
v
Q
t
t
c
m
q p
p
a
s
&
&
( )( ) ( )( ) ( )( ) 800
,
172
78
245
.
0
600
,
786
78
245
.
0
60
110
,
13
2
2
2
2
=
−








=
−






= ′
′
′
′
t
v
t
v
qs
&
2
2 1153
.
1
99
.
86 ′
′ −
= t
v
or
( ) ( )
2
1
2
2
2
1
2 ′
′
′
′
′
′ −








=
−
= i
i
v
Q
i
i
m
q a
s
&
&
( )( ) ( ) ( ) 000
,
216
70
.
35
600
,
786
70
.
35
60
110
,
13
2
2
2
2
=
−








=
−






= ′
′
′
′
i
v
t
v
qs
&
2
2 2746
.
0
70
.
35 ′
′ −
= v
i
From psychrometric chart, satisfying above equation, Chart 1Ha., State 2
2′
db
t = 63.46 F, 2′
v = 16.214 ft3
/lbma
2′
i = 31.22 Btu/lbma, 2′
wb
t = 62.83 F
2′
W = 0.01467 lbmv/lbma, 2′
φ = 96.8 %
Therefore the supply air condition is 63.46 F db and 62.83 F wb or 96.8 % RH.

(a) SI units
Chart 1Hb,
At state 1
1
db
t = 26 C
1
wb
t = 20 C
1
i = 64.83 kJ/kga
1
W = 0.0152 kgv/kga
1
v = 1.040 m3
/kga

2
db
t = 16 C
2
wb
t = 15.124 C
2
i = 47.83 kJ/kga
2
W = 0.01254 kgv/kga
2
v = 1.0014 m3
/kga
( )
2
1
2
i
i
m
q a
−
= &
&
( )
83
.
47
83
.
64
530 2
−
= a
m
&
2
a
m
& = 31.18 kga/s
( )( )
0014
.
1
18
.
31
2
2
2
=
= v
m
Q a
&
& = 31.22 m3
/s
q
& = 63 kW
( ) ( )
22
.
31
2
.
0
2
.
0
8
.
0
1 2
2
2 =
=
−
=
′ Q
Q
Q &
&
& = 6.244 kga/s
Since space is to be maintained at 26 C db and 20 C wb.
At state 1′ = State 1
1′
db
t = 26 C
1′
wb
t = 20 C
1′
i = 64.83 kJ/kga
1′
W = 0.0152 kgv/kga
1′
v = 1.040 kgv/kga
A state 2′, SHF = 0.8, Chart 1b
( ) ( )( )
8
.
0
63
=
= SHF
q
qs
&
& = 50.4 kW
( ) ( )
2
1
2
2
2
1
2 ′
′
′
′
′
′ −








=
−
= t
t
c
v
Q
t
t
c
m
q p
p
a
s
&
&
( )( ) 4
.
50
26
02
.
1
244
.
6
2
2
=
−








= ′
′
t
v
qs
&
2
2 1264
.
0
2855
.
3 ′
′ −
= t
v
or
( ) ( )
2
1
2
2
2
1
2 ′
′
′
′
′
′ −








=
−
= i
i
v
Q
i
i
m
q a
s
&
&
( ) 63
83
.
64
244
.
6
2
2
=
−








= ′
′
i
v
qs
&

2
2 09
.
10
83
.
64 ′
′ −
= v
i
From psychrometric chart, satisfying above equation, Chart 1Hb., State 2′
2′
db
t = 17.99 C, 2′
v = 1.0113 m3
/kga
2′
i = 54.63 kJ/kga, 2′
wb
t = 17.18 C
2′
W = 0.01442 kgv/kga, 2′
φ = 92.8 %
Therefore the supply air condition is 17.99 C db and 17.18 C wb or 92.8 % RH.

3.54 A 50-ton constant-volume space air-conditioning system uses face and bypass
and water temperature control. At the design condition the space is to be
maintained at 77 F (25 C) db and 50 percent relative humidity with 55 F (13
C) db supply air at 90 percent relative humidity. Outdoor air is supplied at 95
F (35 C) db, 60 percent relative humidity with a ratio of 1 lbm (kgm) to 5 lbm
(kgm) return air. A part-load condition exists where the total space load
decreases by 50 percent and the SHF increases to 90 percent. The outdoor air
condition changes to 85 F (29 C) db and 70 percent relative humidity. Assume
sea level pressure.
(a) At what temperature must the air be supplied to the space under part-
load condition?
(b) If the air leaving the coil has a dry-bulb temperature of 60 F (15 C),
what is the ratio of the air bypassed to that flowing through the coil?
(c) What is the apparatus dew point temperature for both the design and
part-load conditions?
(d) Show all the processes on a psychrometric chart.
Solution:
In English units, Chart 1a
c
q
& = 50 tons = 600,000 Btu/hr
5
1
3
0
=
a
a
m
m
&
&
= 0.2
at state 0, 0
db
t = 95 F, 0
φ = 60 %
0
i = 46.6 Btu/lbma
0
at state 3, 3
db
t = 77 F, 3
φ = 50 %
3
i = 29.4 Btu/lbma
3
at state 2, 2
db
t = 55 F, 2
φ = 90 %
2
i = 22.2 Btu/lbma
2
at state 1,
( )( )
2
.
0
1
4
.
29
6
.
46
2
.
0
1
3
0
3
0
3
0
1
+
+
=
+
+
=
a
a
a
a
m
m
i
i
m
m
i
&
&
&
&
= 32.27 Btu/lbma
( )( )
2
.
0
1
0099
.
0
0216
.
0
2
.
0
1
3
0
3
0
3
0
1
+
+
=
+
+
=
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
= 0.01185 lbmv/lbma
From chart 1a, 1
db
t = 80.3 F, d
t = 49.5 F

( )
2
1
2
i
i
m
q a
c
−
= &
&
( )
2
.
22
27
.
32
000
,
600 2
−
= a
m
&
2
a
m
& = 59,583 lbma/hr
Space load
( )
2
3
2
i
i
m
q a
−
= &
&
( )( )
2
.
22
4
.
29
583
,
59 −
=
q
& = 428,998 Btu/hr
( )
2
3
2 t
t
c
m
q p
a
s −
= &
&
( )( )( )
55
77
245
.
0
483
,
59 −
=
s
q
& = 321,152 Btu/hr
998
,
428
152
,
321
=
=
q
q
SHF s
&
&
= 0.7486
At part-load condition
q
& = (0.5)(428,998) = 214,499 Btu/hr
SHF = 0.90
state 0′, 0′
db
t = 85 F, 0′
φ = 70 %
0′
i = 40.5 Btu/lbma
0′
state 3′ , 3′
db
t = 77 F, 3′
φ = 50 %
3′
i = 29.4 Btu/lbma
3′
(a) Since volume is constant, 2
2 a
a
m
m &
& =
′ = 59,583 lbma/hr
( )
SHF
q
qs
&
& = = (214,499)(0.90) = 193,049 Btu/hr
( )
2
3
2 ′
′ −
= t
t
c
m
q p
a
s
&
&
( )( )( )
2
77
245
.
0
583
,
59
049
,
193 ′
−
= t
2′
t = 63.78 F
(b) Let b = bypass and c
t = coil leaving temperature
c
c
t
t
t
t
b
−
−
=
′
′
1
2
still,
5
1
3
0
=
′
′
a
a
m
m
&
&
= 0.2
( )( )
2
.
0
1
4
.
29
5
.
40
2
.
0
1
3
0
3
0
3
0
1
+
+
=
+
+
=
′
′
′
′
′
′
′
a
a
a
a
m
m
i
i
m
m
i
&
&
&
&
= 31.25 Btu/lbma

( )( )
2
.
0
1
0099
.
0
0183
.
0
2
.
0
1
3
0
3
0
3
0
1
+
+
=
+
+
=
′
′
′
′
′
′
′
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
= 0.0113 lbmv/lbma
then, Chart 1a, 1′
t = 78.61 F
60
6
.
78
60
78
.
63
1
2
−
−
=
−
−
=
′
′
c
c
t
t
t
t
b = 0.203
(c) For design condition, Chart 1a, d
t = 49.5 F
For part-load condition, Chart 1a, d
t ′ = 52.16 F
(d) Psychrometric Chart

SI units, Chart 1b
c
q
& = 50 tons = 176 kW
5
1
3
0
=
a
a
m
m
&
&
= 0.2
at state 0, 0
db
t = 35 C, 0
φ = 60 %
0
i = 90.19 kJ/kga
0
W = 0.0215 kgv/kga
at state 3, 3
db
t = 25 C, 3
φ = 50 %
3
i = 50.51 kJ/kga
3
W = 0.0099 kgv/kga
at state 2, 2
db
t = 13 C, 2
φ = 90 %
2
i = 34.25 kJ/kga
2
W = 0.0084 kgv/kga
at state 1,
( )( )
2
.
0
1
31
.
50
19
.
90
2
.
0
1
3
0
3
0
3
0
1
+
+
=
+
+
=
a
a
a
a
m
m
i
i
m
m
i
&
&
&
&
= 56.96 kJ/kga
( )( )
2
.
0
1
0099
.
0
0215
.
0
2
.
0
1
3
0
3
0
3
0
1
+
+
=
+
+
=
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
= 0.0118 lbmv/lbma
From chart 1b, 1
db
t = 26.67 C, d
t = 10 C
( )
2
1
2
i
i
m
q a
c
−
= &
&
( )
25
.
34
96
.
56
176 2
−
= a
m
&
2
a
m
& = 7.75 kga/s
Space load
( )
2
3
2
i
i
m
q a
−
= &
&
( )( )
25
.
34
31
.
50
75
.
7 −
=
q
& = 124.5 kW
( )
2
3
2
t
t
c
m
q p
a
s
−
= &
&
( )( )( )
13
25
02
.
1
75
.
7 −
=
s
q
& = 94.86 kW
5
.
124
86
.
94
=
=
q
q
SHF s
&
&
= 0.762
q
& = (0.5)(124.5) = 62.25 kW
SHF = 0.90

state 0′, 0′
db
t = 29 C, 0′
φ = 70 %
0′
i = 74.4 kJ/kga
0′
W = 0.01771 kgv/kga
state 3′ , 3′
db
t = 25 C, 3′
φ = 50 %
3′
i = 50.31 kJ/kga
3′
W = 0.0099 kgv/kga
2 a
a
m
m &
& =
′ = 7.75 kga/s
( )
SHF
q
qs
&
& = = (62.25)(0.90) = 56.03 kW
( )
2
3
2 ′
′ −
= t
t
c
m
q p
a
s
&
&
( )( )( )
2
25
02
.
1
75
.
7
03
.
56 ′
−
= t
2′
t = 17.91 C
c
c
t
t
t
t
b
−
−
=
′
′
1
2
still,
5
1
3
0
=
′
′
a
a
m
m
&
&
= 0.2
( )( )
2
.
0
1
31
.
50
4
.
74
2
.
0
1
3
0
3
0
3
0
1
+
+
=
+
+
=
′
′
′
′
′
′
′
a
a
a
a
m
m
i
i
m
m
i
&
&
&
&
= 54.33 kJ/kga
( )( )
2
.
0
1
0099
.
0
01771
.
0
2
.
0
1
3
0
3
0
3
0
1
+
+
=
+
+
=
′
′
′
′
′
′
′
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
= 0.0112 kgv/kga
then, Chart 1b, 1′
t = 25.67 C
13
67
.
25
13
91
.
17
1
2
−
−
=
−
−
=
′
′
c
c
t
t
t
t
b = 0.273
(c) For design condition, Chart 1b, d
t = 10 C
For part-load condition, Chart 1b, d
t ′ = 10.9 C

3.55 Rework Problem 3.54 for an elevation of 5000 ft.
Solution:
In English units, Chart 1Ha
c
q
& = 50 tons = 600,000 Btu/hr
5
1
3
0
=
a
a
m
m
&
&
= 0.2
at state 0, 0
db
t = 95 F, 0
φ = 60 %
0
i = 51.58 Btu/lbma
0
W = 0.026 lbmv/lbma
at state 3, 3
db
t = 77 F, 3
φ = 50 %
3
i = 31.6 Btu/lbma
3
W = 0.012 lbmv/lbma
at state 2, 2
db
t = 55 F, 2
φ = 90 %
2
i = 24.07 Btu/lbma
2
W = 0.010 lbmv/lbma
at state 1,
( )( )
2
.
0
1
6
.
31
58
.
51
2
.
0
1
3
0
3
0
3
0
1
+
+
=
+
+
=
a
a
a
a
m
m
i
i
m
m
i
&
&
&
&
= 34.93 Btu/lbma
( )( )
2
.
0
1
012
.
0
026
.
0
2
.
0
1
3
0
3
0
3
0
1
+
+
=
+
+
=
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
= 0.0143 lbmv/lbma
From chart 1Ha, 1
db
t = 80.2 F, d
t = 50.25 F
( )
2
1
2
i
i
m
q a
c
−
= &
&
( )
07
.
24
93
.
34
000
,
600 2
−
= a
m
&
2
a
m
& = 55,249 lbma/hr
Space load
( )
2
3
2
i
i
m
q a
−
= &
&
( )( )
07
.
24
6
.
31
249
,
55 −
=
q
& = 416,025 Btu/hr
( )
2
3
2 t
t
c
m
q p
a
s −
= &
&
( )( )( )
55
77
245
.
0
249
,
55 −
=
s
q
& = 297,792 Btu/hr

025
,
416
792
,
297
=
=
q
q
SHF s
&
&
= 0.716
q
& = (0.5)(416,025) = 208,013 Btu/hr
SHF = 0.90
state 0′, 0′
db
t = 85 F, 0′
φ = 70 %
0′
i = 44.69 Btu/lbma
0′
W = 0.022 lbmv/lbma
state 3′ , 3′
db
t = 77 F, 3′
φ = 50 %
3′
i = 31.6 Btu/lbma
3′
W = 0.012 lbmv/lbma
2 a
a
m
m &
& =
′ = 55,249 lbma/hr
( )
SHF
q
qs
&
& = = (208,013)(0.90) = 187,212 Btu/hr
( )
2
3
2 ′
′ −
= t
t
c
m
q p
a
s
&
&
( )( )( )
2
77
245
.
0
249
,
55
212
,
187 ′
−
= t
2′
t = 63.17 F
c
c
t
t
t
t
b
−
−
=
′
′
1
2
still,
5
1
3
0
=
′
′
a
a
m
m
&
&
= 0.2
( )( )
2
.
0
1
6
.
31
69
.
44
2
.
0
1
3
0
3
0
3
0
1
+
+
=
+
+
=
′
′
′
′
′
′
′
a
a
a
a
m
m
i
i
m
m
i
&
&
&
&
= 33.78 Btu/lbma
( )( )
2
.
0
1
012
.
0
022
.
0
2
.
0
1
3
0
3
0
3
0
1
+
+
=
+
+
=
′
′
′
′
′
′
′
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
= 0.0137 lbmv/lbma
then, Chart 1Ha, 1′
t = 78.2 F
60
2
.
78
60
17
.
63
1
2
−
−
=
−
−
=
′
′
c
c
t
t
t
t
b = 0.174
(c) For design condition, Chart 1Ha, d
t = 50.25 F

For part-load condition, Chart 1Ha, d
t ′ = 52.54 F
SI units, Chart 1Hb
c
q
& = 50 tons = 176 kW

5
1
3
0
=
a
a
m
m
&
&
= 0.2
at state 0, 0
db
t = 35 C, 0
φ = 60 %
0
i = 101.55 kJ/kga
0
W = 0.026 kgv/kga
at state 3, 3
db
t = 25 C, 3
φ = 50 %
3
i = 55.40 kJ/kga
3
W = 0.012 kgv/kga
at state 2, 2
db
t = 13 C, 2
φ = 90 %
2
i = 38.52 kJ/kga
2
W = 0.0101 kgv/kga
at state 1,
( )( )
2
.
0
1
40
.
55
55
.
101
2
.
0
1
3
0
3
0
3
0
1
+
+
=
+
+
=
a
a
a
a
m
m
i
i
m
m
i
&
&
&
&
= 63.09 kJ/kga
( )( )
2
.
0
1
012
.
0
026
.
0
2
.
0
1
3
0
3
0
3
0
1
+
+
=
+
+
=
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
= 0.0143 lbmv/lbma
From chart 1Hb, 1
db
t = 26.5 C, d
t = 9.89 C
( )
2
1
2
i
i
m
q a
c
−
= &
&
( )
52
.
38
09
.
63
176 2
−
= a
m
&
2
a
m
& = 7.163 kga/s
Space load
( )
2
3
2
i
i
m
q a
−
= &
&
( )( )
52
.
38
4
.
55
163
.
7 −
=
q
& = 120.9 kW
( )
2
3
2
t
t
c
m
q p
a
s
−
= &
&
( )( )( )
13
25
02
.
1
163
.
7 −
=
s
q
& = 87.68 kW
9
.
120
68
.
87
=
=
q
q
SHF s
&
&
= 0.725
q
& = (0.5)(120.9) = 60.45 kW
SHF = 0.90
state 0′, 0′
db
t = 29 C, 0′
φ = 70 %

0′
i = 83.68 kJ/kga
0′
W = 0.0214 kgv/kga
state 3′ , 3′
db
t = 25 C, 3′
φ = 50 %
3′
i = 55.4 kJ/kga
3′
W = 0.012 kgv/kga
2 a
a
m
m &
& =
′ = 7.163 kga/s
( )
SHF
q
qs
&
& = = (60.45)(0.90) = 54.41 kW
( )
2
3
2 ′
′ −
= t
t
c
m
q p
a
s
&
&
( )( )( )
2
25
02
.
1
163
.
7
41
.
54 ′
−
= t
2′
t = 17.55 C
c
c
t
t
t
t
b
−
−
=
′
′
1
2
still,
5
1
3
0
=
′
′
a
a
m
m
&
&
= 0.2
( )( )
2
.
0
1
4
.
55
68
.
83
2
.
0
1
3
0
3
0
3
0
1
+
+
=
+
+
=
′
′
′
′
′
′
′
a
a
a
a
m
m
i
i
m
m
i
&
&
&
&
= 60.11 kJ/kga
( )( )
2
.
0
1
012
.
0
0214
.
0
2
.
0
1
3
0
3
0
3
0
1
+
+
=
+
+
=
′
′
′
′
′
′
′
a
a
a
a
m
m
W
W
m
m
W
&
&
&
&
= 0.0136 kgv/kga
then, Chart 1b, 1′
t = 25.6 C
13
6
.
25
13
55
.
17
1
2
−
−
=
−
−
=
′
′
c
c
t
t
t
t
b = 0.241
(c) For design condition, Chart 1Hb, d
t = 9.89 C
For part-load condition, Chart 1Hb, d
t ′ = 11.4 C

3.56 A condition exists where it is necessary to cool and dehumidify air from 80 F
db and 67 F wb to 60 F db and 54 F wb.
(a) Discuss the feasibility of doing this in one process with a
cooling coil. (Hint: Determine the apparatus dewpoint
temperature for the process.)
(b) Describe a practical method of achieving the required process,
and sketch it on a psychrometric chart.
Answer:
(a) As shown on the psychrometric chart (Chart 1a), it is not feasible to cool and
dehumidify this in one process with a cooling coil only, since there is no
apparatus dewpoint temperature reached.
(b) An additional reheat coil will be feasible by cooling the air to a dewpoint of
49.7 F then reheated to 60 F. See Chart 1a.

3.57 Consider one zone of a dual-duct conditioning system. Conditions in the zone
are to be maintained at 75 F (24 C) and 50 percent relative humidity (RH). The
cold deck air is at 51 F (11 C) and 90 percent RH, while the hot deck air is
oputdoor air at 90 F (32 C) and 30 percent RH. The sensible heat factor for the
zone is 0.65. Assume sea level pressure. In what proportion must the warm
and cold air be mixed to satisfy the space condition? If the total space load is
60 tons (210 kW), what is the total volume flow rate of air supplied to the
zone? Sketch the states and process on a psychrometric chart. Use (a) English
units and (b) SI units.
Solution:
(a) English units
Chart 1a
At state 2, 2
db
t = 75 F, 2
φ = 50 %
2
i = 28.2 Btu/lbma
2
At state 1c, c
db
t 1
= 51 F, c
1
φ = 90 %
c
i1
= 20.0 Btu/lbma
c
W1
= 0.0072 lbmv/lbma
At state 1h, h
db
t 1
= 90 F, h
1
φ = 30 %
h
i1
= 31.4 Btu/lbma
h
W1
= 0.009 lbmv/lbma
From the psychrometric chart 1a, intersection of line with SHF = 0.65, from state 1 to
state 2, and line from state 1c to state 1h.
1
db
t = 62.2 F
1
wb
t = 55.4 F
1
φ = 65 %
1
i = 23.8 Btu/lbma
1
1
v = 13.32 ft3
/lbma
State 1:
h
a
c
a
h
c
h
a
c
a
m
m
i
i
m
m
i
1
1
1
1
1
1
1
1
&
&
&
&
+
+
=

( )
h
a
c
a
h
a
c
a
m
m
m
m
1
1
1
1
1
4
.
31
0
.
20
8
.
23
&
&
&
&
+
+
=
h
a
c
a
m
m
1
1
&
&
= 2.0
or
h
a
c
a
h
c
h
a
c
a
m
m
W
W
m
m
W
1
1
1
1
1
1
1
1
&
&
&
&
+
+
=
( )
h
a
c
a
h
a
c
a
m
m
m
m
1
1
1
1
1
009
.
0
0072
.
0
0078
.
0
&
&
&
&
+
+
=
h
a
c
a
m
m
1
1
&
&
= 2.0
or
c
a
h
a
m
m
1
1
&
&
= 0.5
Total volume flow rate:
q
& = 60 tons = 720,000 Btu/hr
( )
1
2
i
i
m
q a
−
= &
&
( )
8
.
23
2
.
28
000
,
720 −
= a
m
&
a
m
& = 163,636 lbma/hr
( )( )( )
32
.
13
60
1
636
,
163
1 =
= v
m
Q a
&
& = 36,327 ft3
/min

(b) SI units
Chart 1b
At state 2, 2
db
t = 24 C, 2
φ = 50 %
2
i = 48 kJ/kga
2
W = 0.0093 kgv/kga
At state 1c, c
db
t 1
= 11 C, c
1
φ = 90 %
c
i1
= 29.5 kJ/kga
c
W1
= 0.0072 kgv/kga
At state 1h, h
db
t 1
= 32 C, h
1
φ = 30 %
h
i1
= 55 kJ/kga
h
W1
= 0.009 kgv/kga
From the psychrometric chart 1b, intersection of line with SHF = 0.65, from state 1 to
1
db
t = 17 C
1
wb
t = 13.1 C
1
φ = 65 %
1
i = 38 kJ/kga
1
W = 0.0078 kgv/kga
1
v = 0.832 m3
/kga
State 1:
h
a
c
a
h
c
h
a
c
a
m
m
i
i
m
m
i
1
1
1
1
1
1
1
1
&
&
&
&
+
+
=
( )
h
a
c
a
h
a
c
a
m
m
m
m
1
1
1
1
1
55
5
.
29
38
&
&
&
&
+
+
=
h
a
c
a
m
m
1
1
&
&
= 2.0
or

h
a
c
a
h
c
h
a
c
a
m
m
W
W
m
m
W
1
1
1
1
1
1
1
1
&
&
&
&
+
+
=
( )
h
a
c
a
h
a
c
a
m
m
m
m
1
1
1
1
1
009
.
0
0072
.
0
0078
.
0
&
&
&
&
+
+
=
h
a
c
a
m
m
1
1
&
&
= 2.0
or
c
a
h
a
m
m
1
1
&
&
= 0.5
q
& = 210 kW
( )
1
2
i
i
m
q a
−
= &
&
( )
38
48
210 −
= a
m
&
a
m
& = 21 kga/s
( )( )
832
.
0
21
1
=
= v
m
Q a
&
& = 17.5 m3
/s

3.58 Rework Problem 3.57 for an elevation of 5000 ft (1500 m).
Solution:
(a) English units
Chart 1Ha
At state 2, 2
db
t = 75 F, 2
φ = 50 %
2
i = 30.2 Btu/lbma
2
At state 1c, c
db
t 1
= 51 F, c
1
φ = 90 %
c
i1
= 21.6 Btu/lbma
c
W1
= 0.0086 lbmv/lbma
At state 1h, h
db
t 1
= 90 F, h
1
φ = 30 %
h
i1
= 33.7 Btu/lbma
h
W1
= 0.0108 lbmv/lbma
From the psychrometric chart 1Ha, intersection of line with SHF = 0.65, from state 1
to state 2, and line from state 1c to state 1h.
1
db
t = 58.0 F
1
wb
t = 52.8 F
1
φ = 73.3 %
1
i = 23.8 Btu/lbma
1
W = 0.009 lbmv/lbma
1
v = 15.98 ft3
/lbma
State 1:
h
a
c
a
h
c
h
a
c
a
m
m
i
i
m
m
i
1
1
1
1
1
1
1
1
&
&
&
&
+
+
=
( )
h
a
c
a
h
a
c
a
m
m
m
m
1
1
1
1
1
7
.
33
6
.
21
8
.
23
&
&
&
&
+
+
=
h
a
c
a
m
m
1
1
&
&
= 4.5
or

h
a
c
a
h
c
h
a
c
a
m
m
W
W
m
m
W
1
1
1
1
1
1
1
1
&
&
&
&
+
+
=
( )
h
a
c
a
h
a
c
a
m
m
m
m
1
1
1
1
1
0108
.
0
0086
.
0
009
.
0
&
&
&
&
+
+
=
h
a
c
a
m
m
1
1
&
&
= 4.5
or
5
.
4
1
1
1
=
c
a
h
a
m
m
&
&
= 0.22222
q
& = 60 tons = 720,000 Btu/hr
( )
1
2
i
i
m
q a
−
= &
&
( )
8
.
23
2
.
30
000
,
720 −
= a
m
&
a
m
& = 112,500 lbma/hr
( )( )( )
98
.
15
60
1
500
,
112
1 =
= v
m
Q a
&
& = 36,327 ft3
/min

(b) SI units
Chart 1Hb
At state 2, 2
db
t = 24 C, 2
φ = 50 %
2
i = 53 kJ/kga
2
W = 0.0112 kgv/kga
At state 1c, c
db
t 1
= 11 C, c
1
φ = 90 %
c
i1
= 33.7 kJ/kga
c
W1
= 0.0088 kgv/kga
At state 1h, h
db
t 1
= 32 C, h
1
φ = 30 %
h
i1
= 59.9 kJ/kga
h
W1
= 0.0117 kgv/kga
From the psychrometric chart 1b, intersection of line with SHF = 0.65, from state 1 to
1
db
t = 16.2 C
1
wb
t = 12.5 C
1
φ = 68.3 %
1
i = 40 kJ/kga
1
W = 0.0095 kgv/kga
1
v = 0.997 m3
/kga
State 1:
h
a
c
a
h
c
h
a
c
a
m
m
i
i
m
m
i
1
1
1
1
1
1
1
1
&
&
&
&
+
+
=
( )
h
a
c
a
h
a
c
a
m
m
m
m
1
1
1
1
1
8
.
59
7
.
33
40
&
&
&
&
+
+
=
h
a
c
a
m
m
1
1
&
&
= 3.15
or

h
a
c
a
h
c
h
a
c
a
m
m
W
W
m
m
W
1
1
1
1
1
1
1
1
&
&
&
&
+
+
=
( )
h
a
c
a
h
a
c
a
m
m
m
m
1
1
1
1
1
0117
.
0
0088
.
0
0095
.
0
&
&
&
&
+
+
=
h
a
c
a
m
m
1
1
&
&
= 3.15
or
c
a
h
a
m
m
1
1
&
&
= 0.32
q
& = 210 kW
( )
1
2
i
i
m
q a
−
= &
&
( )
40
53
210 −
= a
m
&
a
m
& = 16.15 kga/s
( )( )
997
.
0
15
.
16
1
=
= v
m
Q a
&
& = 16.1 m3
/s

Chapter_3.pdf

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Chapter_3.pdf