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3-1 Calculate the specific volume of an air-vapor mixture in cubic meters per kilogram of dry air when the following conditions prevail: t = 30 C, W = 0.015 kg/kg, and pt = 90 kPa.
Solution: Equation 3-4.
t s
a
a
a
p p
RT
p
R T
T = 30 C + 273 = 303 K
Ra = 287 J/kg.K Pt = 90 kPa = 90,000 Pa
Equation 3-
t s
s
p p
0.622p
W
s
s
90 p
0.622p
1.35 - 0.15ps = 0.622ps ps = 2.1193 kPa
p p
R T
t s
a
νννν = 0.99 m
3 /kg - - - Ans.
3-2. A sample of air has a dry-bulb temperature of 30 C and a wet-bulb temperature of 25 C. The barometric pressure is 101 kPa. Using steam tables and Eqs. (3-2), (303), and (3-5), calculate (a) the humidity ration if this air is adiabatically saturated, (b) the enthalpy of air if it is adiabatically saturated, (c) the humidity ratio of the sample using Eq. (3-5), (d) the partial pressure of water vapor in the sample, and (e) the relative humidity.
Solution: Eq. 3-2.
t s
s
p p
0.622p
W
Eq. 3-3. h = cpt + Whg
Eq. 3- h 1 = h 2 - (W 2 - W 1 )hf
h 1 = cpt 1 + Whg
hg1 at 30 C = 2556.4 kJ/kg t1 = 30 C cp = 1.0 kJ/kg.K
h 1 = (1)(30) + 2556.4W 1 h 1 = 30 + 2556.4W 1
h 2 = cpt 2 + Whg
hg 2 at 25 C = 2547.3 kJ/kg
t 2 = 25 C
cp = 1.0 kJ/kg.K
h 2 = (1)(25) + 2547.3W 2
h 2 = 25 + 2547.3W 2
hf at 25 C = 125.66 kJ/kg
Then:
h 1 = h 2 - (W 2 - W 1 )hf
30 + 2556.4W 1 = 25 + 2547.3W 2 - (W 2 - W 1 )(125.66)
5 = 2421.64W 2 - 2430.74W 1
But,
t s
s 2
p p
0.622p
W
ps at 25 C = 3.171 kPa
W 2 =
W 2 = 0.0201 kg/kg
5 = 2421.64(0.0201) - 2430.74W 1
W 1 = 0.018 kg/kg
(a) Humidity Ratio W 2 = 0.0201 kg/kg - - - Ans.
(b) h 2 = cpt 2 + W 2 hg 2
h 2 = (1)(25) + (0.0201)(2547.3) h 2 = 76.2 kJ/kg - - - Ans.
(c) Humidity Ratio W 1 = 0.018 kg/kg - - - Ans.
(d) ps
t s
s 1
p p
0.622p
W
s
s
101 p
0.622p
ps1 = 2.84 kPa ps1 = 2840 kPa - - - Ans.
Dew-Point = 17.5 C - - - Ans.
at 101 kPa
t s
s
p p
0.622p
ps = 2.3783 kPa
Dew-Point = 20.3 C - - - Ans.
3-5. A cooling tower is a device that cools a spray of water by passing it through a stream of air. If 15 m
3 /s of air is at 35 C dry-bulb and 24 C wet-bulb temperature and an atmospheric pressure of 101 kPa enters the tower and the air leaves saturated at 31 C, (a) to what temperature can this airstream cool a spray of water entering at 38 C with a flow rate of 20 kg/s and (b) how many kilograms per second of make-up water must be added to compensate for the water that is evaporated?
Solution: At 35 C dry-bulb, 24 C wet-bulb.
Fig. 3-1, Psychrometric Chart h 1 = 71.524 kJ/kg,
ν 1 = 0.89274 m^3 /kg W 1 = 0.0143 kg/kg At 31 C saturated, Table A-2. h 2 = 105 kJ/kg W 2 = 0.0290 kg/kg
Then; m = (15 m^3 /s) / (0.89274 m^3 /kg) = 16.8022 kg/s
(a) tw1 = 38 C mw = 20 kg/s cpw = 4.19 kJ/kg.K
mwcpw(tw1 - tw2) = m(h 2 - h 1 )
(20)(4.19)(38 - tw2) = (16.8022)(105 - 71.524)
tw2 = 31.3 C - - - Ans.
(b) Make-Up Water = mm
mm = m(W 2 - W 1 ) mm = (16.8022)(0.0290 - 0.0143)
mm = 0.247 kg/s - - - Ans.
3-6. In an air-conditioning unit 3.5 m^3 /s of air at 27 C dry-bulb temperature, 50 percent relative humidity, and
standard atmospheric pressure enters the unit. The leaving condition of the air is 13 C dry-bulb temperature and 90 percent relative humdity. Using properties from the psychrometric chart, (a) calculate the refrigerating capacity inkilowatts and (b) determine the rate of water removal from the air.
Solution:
At 27 C dry-buld, 5 Percent Relative Humidity h 1 = 55.311 kJ/kg,
ν 1 = 0.86527 m
3 /kg W 1 = 0.0112 kg/kg
At 13 C Dry-Bulb, 90 Percent Relative Humidity h 2 = 33.956 kJ/kg W 2 = 0.0084 kg/kg
m = (3.5 m
3 /s)/(0.86526 m
3 /kg) = 4.04498 kg/s
(a) Refrigerating Capacity = m(h 1 - h 2 ) = (4.04498)(55.311 - 33.956) = 86.38 kW - - - Ans.
(b) Rate of Water Removal = m(W 1 - W 2 ) = (4.04498)(0.0112 - 0.0084) = 0.0113 kg/s - - - Ans.
3-7. 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, (c) the dry-bulb temperature of the mixture from the properties determined in parts (a) and (b) and (d) the dry-bulb temperature by weighted average of the dry-bulb temperatures of the entering streams.
Solutions:
Use Fig. 3-1, Psychrometric Chart At 35 C Dry-Bulb, 24 C Wet-Bulb h 1 = 75.666 kJ/kg, m 1 = 2 kg/s W 1 = 0.0159 kg/kg
At 24 C Dry-Bulb, 50 Percent Relative Humidity h 2 = 47.518 kJ/kg, m 2 = 3 kg/s W 2 = 0.0093 kg/kg
(a)
hm
hm = 58.777 kJ/kg - - - Ans.
3-9. A winter air-conditioning system adds for humidification 0.0025 kg/s of saturated steam at 101 kPa pressure to an airflow of 0.36 kg/s. The air is initially at a temperature of 15 C with a relative humidity of 20 percent. What are the dry- and wet-bulb temperatures of the air leaving the humidifier?
Solution:
At 15 C Dry-Bulb, 20 Percent Relative Humidity h 1 = 20.021 kJ/kg W 1 = 0.0021 kg/kg
At 101 kPa steam, hfg = 2675.85 kJ/kg ms = 0.0025 kg/s m = 0.36 kg/s ms = m(W 2 - W 1 ) 0.0025 = 0.36(W 2 - 0.002) W 2 = 0.00894 kg/kg
m(h 2 - h 1 ) = mshg (0.36)(h2 - 20.021) = (0.0025)(2675.85) h 2 = 38.6 kJ/kg
Fig. 3-1, Psychrometric Chart W 2 = 0.00894 kg/kg h 2 = 38.6 kJ/kg
Dru-Bulb Temperature = 16.25 C Wet-Bulb Temperature = 13.89 C
3-10. Determine for the three cases listed below the magnitude in watts and the direction of transfer of sensible
heat [ using Eq. (3-8)], latent heat [ using Eq. (3-9)], and total heat [ using Eq. (3-14)]. the area is 0.15 m
2 and hc = 30 W/m^2 .K. Air at 30 C and 50 percent relative humidity is in contact with water that is at a temperature of (a) 13 C, (b) 20 C, and (c) 28 C.
Solution: Equation 3-8. dqs = hcdA(ti - ta) Equation 3-9. dqL = hDdA(Wi - Wa)hfg Equarion 3-14.
(h h )
c
hdA
dq i a
pm
c
t = −
At 30 C, 50% Relative Humidity ha = 63.965 kJ/kg = 63,965 J/kg Wa = 0.0134 kg/kg
(a) 13 C
dqs = hcdA(ti - ta) dqs = (30)(0.15)(13 - 30)
dqs = -76.5 W - - - Ans.
dqL = hDdA(Wi - Wa)hfg
Wi at 13 C = 0.00937 kg/kg from Table A-
hfg at 13 C = 2,470,840 J/kg
hD = hc / cpm
cpm = 1020 kJ/kg.K
hD = 30 / 1020 = 0.
dqL = (0.029412)(0.15)(0.00937 - 0.0134)(2,470,840)
dqL = -43.93 W - - - Ans.
hi at 13 C = 36,719 J/kg from Table A-
(h h )
c
hdA
dq i a
pm
c
t = − = −
dqt = -120.2 W - - - Ans.
(b) 20 C
dqs = hcdA(ti - ta)
dqs = (30)(0.15)(20 - 30)
dqs = -45 W - - - Ans.
dqL = hDdA(Wi - Wa)hfg
Wi at 20 C = 0.01475 kg/kg from Table A-
hfg at 20 C = 2,454,340 J/kg
hD = hc / cpm
cpm = 1020 kJ/kg.K
hD = 30 / 1020 = 0.
dqL = (0.029412)(0.15)(0.01475 - 0.0134)(2,454,340)
dqL = 14.62 W - - - Ans.
hi at 20 C = 57,544 J/kg from Table A-
(h h )
c
hdA
dq i a
pm
c
t = − = −
dqt = -28.33 W - - - Ans.
(c) 28 C
dqs = hcdA(ti - ta)
dqs = (30)(0.15)(28 - 30)
dqs = -9.0 W - - - Ans.
