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Modern Engineering Thermodynamics

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12.8 Psychrometric Enthalpies 429<br />

or<br />

h # 3 = ð _m a1/ _m a3 Þh # 1 + ð _m a2/ _m a3 Þh # 2 (12.34)<br />

If the states of the inlet flow streams are known, then Eqs. (12.33) and (12.34) allow the calculation of two<br />

independent thermodynamic properties (ω 3 and h # 3 ) that fix the state of the outlet flow stream. The following<br />

example illustrates this type of problem.<br />

EXAMPLE 12.13<br />

Suppose 2000. ft 3 /min of air at 14.7 psia, 50.0°F, ϕ = 80.0% is<br />

adiabatically and aergonically mixed with 1000. ft 3 /min of air<br />

at 14.7 psia, 100.°F, and ϕ = 40.0%. Determine the dry bulb<br />

temperature and the relative humidity of the outlet mixture.<br />

Solution<br />

First, draw a sketch of the system (Figure 12.10).<br />

The unknowns are the dry bulb temperature and the relative<br />

humidity of the outlet mixture.<br />

∀ 1 = 2000. ft 3 /min<br />

T DB1 = 50.0°F<br />

φ 1 = 80.0%<br />

∀ 2 = 1000. ft 3 /min<br />

T DB2 = 100.°F<br />

φ 2 = 40.0%<br />

1<br />

2<br />

Q = 0 W = 0<br />

Flow stream<br />

mixing<br />

3<br />

T DB2 = ?<br />

φ 3 = ?<br />

At 50.0°F, p w1 = ϕ 1 p sat (50.0°F) = 0.800(0.178) = 0.142 psia.<br />

Then,<br />

FIGURE 12.10<br />

Example 12.13.<br />

Similarly,<br />

v a1 = R a T m /p a1<br />

= ½53:34 ft .lbf/ ðlbm.RÞŠð50:0 + 459:67 RÞ/½ð14:7 − 0:142 lbf/in 2 Þð144 in 2 /ft 2 ÞŠ<br />

= 13:0ft 3 /ðlbm dry airÞ<br />

p w2 = ϕ 2 p sat ð100:°FÞ = 0:400ð0:9503Þ = 0:380 psia<br />

and<br />

Also, since _m a = _V a/v a ,<br />

v a2 = 53:34ð100: + 459:67Þ/½ð14:7 − 0:380Þð144ÞŠ<br />

= 14:5ft 3 /ðlbm dry airÞ<br />

_m a1 = ð2000: ft 3 /minÞ/½13:0ft 3 /ðlbm dry airÞŠ = 154 lbm dry air/min<br />

_m a2 = ð1000: ft 3 /minÞ/½14:5ft 3 /ðlbm dry airÞŠ = 69:0 lbm dry air/min<br />

and, using the conservation of mass applied to the air,<br />

_m a3 = _m a1 + _m a2 = 154 + 69:0 = 223 lbm dry air/min<br />

Then, from the psychrometric chart (Chart D.5), we find 9<br />

ω 1 = ½44 grains of water vapor/ðlbm dry airÞŠ/ð7000: grains/lbmÞ<br />

= 0:0063 lbm water vapor/ðlbm dry airÞ<br />

ω 2 = ½115 grains of water vapor/ðlbm dry airÞŠ/ð7000: grains/lbmÞ<br />

= 0:0164 lbm water vapor/ðlbm dry airÞ<br />

h # 1<br />

h # 2<br />

= 19 Btu/ðlbm dry airÞ<br />

= 42 Btu/ðlbm dry airÞ<br />

From the water conservation of mass equation, Eq. (12.33), we can now calculate<br />

ω 3 = ð154/223Þð0:0063Þ + ð69:0/223Þð0:0164Þ<br />

= 0:0094 lbm water vapor/ðlbm dry airÞ<br />

= 0:0094 ð7000:Þ = 66 grains of water vapor/ðlbm dry airÞ<br />

and the resulting energy rate balance equation, Eq. (12.34), gives<br />

h # 3<br />

= ð154/223Þð19Þ + ð69:0/223Þð42Þ<br />

= 26 Btu/ðlbm dry airÞ<br />

(Continued )

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