Modern Engineering Thermodynamics

12.6 The Sling Psychrometer 423
p w
T WB1 = 4°C
φ = 100%
φ 1 = 80%
T DP1 = 2°C
p w1 = 700. N/m 2
5°C
T DB
T WB2 = 13°C
T
p WB1 = 4°C
w
T DP2 =T DP1 = 2°C
p w2 = p w1
= 700. N/m 2 State 2
State 1
5°C 25°C
T DB
φ = 100%
φ 1 = 80%
φ 2 = 20%
ω
ω 1 = 0.004 kg/kg
ω
ω 2 = ω 1
= 0.004 kg/kg
FIGURE 12.5
Example 12.7.
If a sling psychrometer is spun such that the air velocity over the wick is greater than 3.0 m/s, then the wet bulb
temperature is essentially equal to the adiabatic saturation temperature T 2 in Eq. (12.31). The following example
illustrates this point.
EXAMPLE 12.8
The wet and dry bulb temperatures measured in a dormitory room are 60.0°F and 70.0°F, respectively, when the barometric
pressure is 14.7 psia. Assuming that the wet bulb temperature is equal to the adiabatic saturation temperature, use
Eq. (12.31) to find the humidity ratio (ω) in the room and compare your answer with that obtained from the psychrometric
chart, Chart D.5.
Solution
Here, we have T WB = 60.0°F and T DB = 70.0°F. Then, from Table C.1a in Thermodynamic Tables to accompany Modern Engineering
Thermodynamics, we find
and
Then, Eq. (12.26a) gives
and, from Eq. (12.31), we get 6
h g1 = h g ð70:0°FÞ= 1092:0 Btu/lbm
h f g2 = h fg ð60:0°FÞ= 1059:6 Btu/lbm
h f 2 = h f ð60:0°FÞ= 28:1 Btu/lbm
p w3 = p sat ð60:0°FÞ¼0:2563 psia
ω 3 = 0:622ð0:2563Þ/14:7 ð − 0:2563Þ = 0:0110 lbm water per lbm of dry air
0:240ð60 − 70Þ + 0:0110ð1059:6Þ
ω 1 = = 0:00874 lbm water per lbm of dry air
1092:0 − 28:1
= 0:00874ð7000Þ = 61:2 grains of water per lbm of dry air
(Continued )