CHAPTER 8

1. Calculate the specific volume of an air-vapor mixture in cubic meters per

kilogram of dry air at the following conditions t = 32 C , W = 0.016 kg kg ,
pt = 100 kPa .

Solution:
0.622 ps
W=
pt − ps
0.622 ps
0.016 =
100 − ps
ps = 2.508 kPa
pa = pt − ps = 100 − 2.508 = 97.492 kPa
RT
va = a =
(0.287 )(32 + 273) = 0.899 m3 kg
pa 97.492

2. Moist air at a dry bulb temperature of 25 C has a relative humidity of 50% when
the barometric pressure is 101.4 kPa. Determine (a) the partial pressures of water
vapor and dry air, (b) the dew point temperature, (c) the specific humidity, (d) the
specific volume, and (e) the enthalpy.

Solution:

At 25 C, pd = 3.169 kPa , hg = 2547.2 kJ kg ,

φ = 50%

(a) ps = φ pd = (0.50 )(3.169 ) = 1.5845 kPa

pa = pt − ps = 101.4 − 1.5845 = 99.82 kPa

(b) tdp = tsat at pd = 1.5845 = 13.7 C

0.622 ps 0.622(1.5845)
(c) W = = = 0.00987 kg kg
pt − ps 99.82

RTa (0.287 )(25 + 273)

(d) v a = = = 0.857 m 3 kg
pa 99.82

(e) h = c pt + Whg = (1.0062 )(25) + (0.00987 )(2547.2) = 50.30 kJ kg

 CHAPTER 8

3. Air at a temperature of 33 C has a relative humidity of 50%. Determine (a) the

wet bulb temperature, (b) the dew point temperature, (c) the humidity ratio, (d)
the enthalpy, and (e) the specific volume.

Solution:
At 33 C, pd = 5.075 kPa , hg = 2561.7 kJ kg ,

From psychrometric chart, at 33 C, RH = 50%

(a) t wb = 24.5 C
(b) tdp = 21.2 C
(c) W = 0.016 kg kg
(h) h = 74 kJ kg ,
(e) v = 0.887 m3 kg

4. How much heat is required to raise the temperature of 0.50 m3/s of air from 19 C
dry bulb and 15 C wet bulb to 36 C? What is the final dew point temperature?

Solution:
 CHAPTER 8

From psychrometric chart,

At 19 C DB and 15 C WB
h1 = 42 kJ kg
v1 = 0.84 m3 kg
V 0.50
m= 1 = = 0.595 kg s
v1 0.84

at 36 C, W2 = W1
h2 = 59 kJ kg

Q = m(h1 − h2 ) = 0.595(59 − 42 ) = 10.12 kW

tdp = tsat = 12.2 C

5. How much heat must be removed to cool 30 cu m per minute of air from 34 C dry
bulb and 18 C dew point to a wet bulb temperature of 19 C? What is the final
relative humidity?

Solution:
From psychrometric chart,
At 34 C DB and 18 C Dew Point
h1 = 67 kJ kg
v1 = 0.887 m3 kg
V& 30
m& = 1 = = 0.564 kg s
v1 (0.887 )(60 )

at 19 C, W2 = W1
 CHAPTER 8

h2 = 54 kJ kg
RH 2 = 83%

Q = m(h1 − h2 ) = 0.564(67 − 54 ) = 7.33 kW

RH 2 = 83%

6. How much heat and moisture must be added to 5 m3/minute of air at 21 C dry
bulb and 30% relative humidity to raise it to 37 C and 40% relative humidity?

Solution:

From psychrometric chart,

At 21 C , 30 % RH
h1 = 32.8 kJ kg
W1 = 0.0046 kg kg
v1 = 0.84 m3 kg
V& 15
m& = 1 = = 0.298 kg s
v1 (0.84 )(60 )
 CHAPTER 8

at 37 C, 40 % RH
h2 = 77.5 kJ kg
W2 = 0.0158 kg kg

heat added,
Q = m& (h2 − h1 ) = 0.298(77.5 − 32.8) = 13.32 kW

moisture added = m& (W2 − W1 ) = 0.298(0.0158 − 0.0046 ) = 0.00334 kg s

7. How much heat must be removed to cool 50 m3/min of air at 29 C dry bulb and 21
C wet bulb temperatures to 16 C dry bulb and 14 C wet bulb temperatures? How
much moisture was removed?

Solution:

From psychrometric chart,

At 29 C DB and 21 C WB
h1 = 61 kJ kg
W1 = 0.0124 kg kg
v1 = 0.87 m3 kg

At 16 C DB and 14 C WB
W2 = 0.0092 kg kg
h2 = 39.4 kJ kg

V&1 50
m& = = = 0.958 kg s
v1 (0.87 )(60 )

heat removed = m& (h1 − h2 ) = 0.958(61 − 39.4 ) = 20.69 kW

moisture removed = m& (W1 − W2 ) = 0.958(0.0124 − 0.0092 ) = 0.00307 kg s
 CHAPTER 8

8. Air at 32 C and 20 percent relative humidity is cooled and humidified by means

of an air washer until the relative humidity becomes 90%. How much moisture
was added per kg of dry air. What was the air washer efficiency and the dew point
temperature of the leaving air?

Solution:

From psychrometric chart,

At 1, 32 C and 20% RH
W1 = 0.0060 kg kg
At 2, 90% RH
W2 = 0.0116 kg kg

moisture added = W2 − W1 = 0.0116 − 0.0060 = 0.0056 kg kg

tdb1 − tdb2
air washer efficiency =
tdb1 − t wb
t wb = 17 C , tdb2 = 18.1 C
32 − 18.1
air washer efficiency = (100% ) = 92.7%
32 − 17

Dew point of leaving air = tdp2 = 16.5 C

9. A stream of outdoor air is mixed with a stream of return air in an air conditioning
system that operates at 101 kPa pressure. The flow rate of outdoor air is 2 kg/s,
and its condition is 35 C dry bulb temperature and 25 C wet bulb temperature.
The flow rate of return air is 3 kg/s, and its condition is 24 C and 50 percent
relative humidity. Determine (a) the enthalpy of the mixture, (b) the humidity
ratio of the mixture, and (c) the dry bulb temperature of the mixture.
 CHAPTER 8

Solution:

From psychrometric chart,

At 35 C DB, 25 C WB
ho = 76 kJ kg
Wo = 0.016 kg kg
at 24 C, 50% RH
hr = 47.8 kJ kg
Wr = 0.0094 kg kg

mo ho + mr hr (2)(76 ) + (3)(47.8)
(a) hm = = = 59.1 kJ kg
mm 5
m W + mrWr (2)(0.016) + (3)(0.0094 )
(b) Wm = o o = = 0.012 kg kg
mm 5
m h + mr hdbr (2 )(35) + (3)(24 )
(c) tdbm = o dbo = = 28.4 C
mdbm 5

10. An auditorium is to be maintained at 25 C dry bulb temperature and 50% relative

humidity. The supply air enters the auditorium at 17 C. The sensible and latent
heat loads are 150 kW and 61 kW, respectively. Determine the wet bulb
temperature, relative humidity, and volume flow rate of the supply air.

Solution:
 CHAPTER 8

QS = 150 kW , QL = 61 kW

At 25 C, 50% RH
h2 = 40 kJ kg
W2 = 0.0058 kg kg

(
QS = m& c p t db2 − tdb1 )
150 = m& (1.0062 )(25 − 17 )
m& = 18.63 kg s

QS + QL 150 + 61
h1 = h2 − = 40 − = 28.7 kJ kg
m& 18.63

QL = 2500m& (W2 − W1 )
61 = 2500(18.63)(0.0058 − W1 )

W1 = 0.0045 kg kg

From psychrometric chart,

At h1 = 28.7 kJ kg , W1 = 0.0045 kg kg

Then,
Wet bulb at 1, t wb1 = 10 C
Relative humidity, RH1 = 40%
v1 = 0.828 m 3 kg

V&1 = m& v1 = (18.63)(0.828) = 15.43 m 3 s

11. In a certain space to be air conditioned the sensible and latent heat loads are 20.60
kW and 6.78 kW, respectively. Outside air is at 33 C dry bulb and 24 C wet bulb
 CHAPTER 8

temperatures. The space is to be maintained at 25 C with a relative humidity not

exceeding 50%. All outside air is supplied with reheater. The conditioned air
enters at 18 C. Determine (a) the refrigeration load required, (b) the capacity of
the supply fan, and (c) the heat supplied in the reheater.

Solution:

From pyschrometric chart.

At 1, tdb1 = 33 C , t wb1 = 24 C
h1 = 72.2 kJ kg

At 4, tdb4 = 25 C , RH 4 = 50%
h4 = 50 kJ kg
W4 = 0.0099 kg kg

At 3, tdb3 = 18 C

(
QS = m& c p tdb4 − tdb3 )
20.60 = m& (1.0062 )(25 − 18)
m& = 2.925 kg s

QL = 2500m& (W4 − W3 )
 CHAPTER 8

6.78 = 2500(2.925)(0.0099 − W3 )
W3 = 0.0090 kg kg

at tdb3 = 18 C , W3 = 0.0090 kg kg
v3 = 0.83 m3 kg

At 2, with W2 = W3 = 0.0090 kg kg
h2 = 35 kJ kg

(a) Refrigeration load required = m& (h1 − h2 ) = 2.925(72.2 − 35) = 108.8 kW

(b) Capacity m& v3 = (2.925)(0.83) = 2.43 m3 s
(c) Heat supplied in the reheater = m& (h3 − h2 ) = 2.925(40.6 − 35) = 16.38 kW

12. An air conditioned auditorium with a capacity of 1000 persons is to be maintained

at 24 C dry bulb temperature and 55% relative humidity. The sensible and latent
heat loads are 115 kW and 42 kW, respectively. The conditioned air enters the
auditorium at 17 C. For proper ventilation, 40% of the supply air is fresh air and
the rest is recirculated air. Outside air is at 34 C and 50% relative humidity.
Determine (a) the volume flow rate of recirculated air, (b) the apparatus dew
point, and (c) the refrigeration load.

Solution:
 CHAPTER 8

m& o = 0.40m&
m& r = 0.60m&

At 4, tdb4 = 24 C , φ3 = 55%
h4 = 50.5 kJ kg
W4 = 0.0102 kg kg
v4 = 0.856 m 3 kg

At 1, tdb1 = 34 C , φ1 = 50%
h1 = 76.6 kJ kg

At 5, tdb5 = 0.40(34) + 0.60(24 ) = 28 C

h5 = 0.40(76.6) + 0.60(50.5) = 60.9 kJ kg

(
QS = m& c p tdb4 − tdb3 )
115 = m& (1.0062 )(24 − 17 )
m& = 16.33 kg s

QL = 2500m& (W4 − W3 )
42 = 2500(16.33)(0.0102 − W3 )
W3 = 0.0092 kg kg

QS + QL = m& (h4 − h3 )
115 + 42 = 16.33(50.5 − h3 )
h3 = 40.9 kJ kg

At 2, tdb2 = 12.6 C , W2 = W3 = 0.0092 kg kg

h2 = 36 kJ kg

(a) Volume flow rate of recirculated air = V&r = m& r v4 = (0.60 )(16.33)(0.856) = 8.39 m3 s
(b) Apparatus dew point = tdb2 = 12.6 C
(c) Refrigeration Load = = m& (h5 − h2 ) = 16.33(60.9 − 36 ) = 407 kW

13. A store to be maintained at 25 C and 50% relative humidity has a sensible heat
load of 18.90 kW and a latent heat load of 6.30 kW. Outside air is at 32 C dry
bulb and 23 C wet bulb temperatures. The conditioned air enters at 17 C. If 30%
of the supply air is fresh air and the bypass system is used, determine (a) the
refrigeration required, and (b) the volume of the bypass air at supply condition.
 CHAPTER 8

Solution:

tdb4 = 25 C , φ = 50%
QS = 18.90 kW
QL = 6.30 kW
tdb3 = 17 C
tdb1 = 32 C
t wb1 = 23 C
mo = 0.30m
mr + mb = 0.70m

From psychrometric chart

At 4, tdb4 = 25 C , φ = 50%
h4 = 50 kJ kg
 CHAPTER 8

W4 = 0.0098 kg kg
v4 = 0.858 m3 kg

(
QS = m& c p tdb4 − tdb3 )
18.90 = m& (1.0062 )(25 − 17 )
m& = 2.348 kg s

QL = 2500m& (W4 − W3 )
6.30 = 2500(2.348)(0.0098 − W3 )
W3 = 0.0087 kg kg

QS + QL = m& (h4 − h3 )
18.90 + 6.30 = 2.348(50 − h3 )
h3 = 39.3 kJ kg

At 3, W3 = 0.0087 kg kg , h3 = 39.3 kJ kg
v3 = 0.834 m3 kg

at 1, tdb1 = 32 C , t wb1 = 23 C
h1 = 68.1 kJ kg

To solve for tdb2 :

mb m
Let b = , c= c
m m

(1) b + c =1
tdb 3 = b(tdb 4 ) + c (tdb 2 )
(2) 17 = 25b + c(t db2 )
twb 3 = b(t wb 4 ) + c(twb 2 )
t wb 3 = 14 C , t wb4 = 17.8 C
t wb2 = tdb2
(3) 14 = 17.8b + c(tdb 2 )

(3) – (2) 17 − 14 = (25 − 17.8)b

b = 0.417
c = 1 − b = 1 − 0.417 = 0.583

Substitute in (2)
17 = 25b + c(t db2 )
 CHAPTER 8

t wb2 = tdb2 = 11.3 C

∴ m& c = cm& = 0.583(2.348) = 1.367 kg s

m& b = bm& = 0.417(2.348) = 0.979 kg s

m& o = 0.30m& = 0.30(2.348) = 0.704 kg s

m& r + m& b = 0.70m& = 0.70(2.348)
m& r + 0.979 = 0.70(2.348)
m& r = 0.665 kg s

∴ at 5,

m& r h4 + m& o h1 0.665(50 ) + 0.704(68.1)

h5 = = = 59.3 kJ kg
m& r + m& o 0.665 + 0.704

(a) Refrigeration load = m& c (h5 − h2 ) = 1.367(59.3 − 31.65) = 37.8 kW

(b) Volume of the bypass air at supply condition
= m& bv3 = 0.979(0.834) = 0.816 m3 s