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Unformatted text preview: sychrometric chart and equations.
Given: From Example 18.1, air at 131oF with a humidity of 0.03 lb water/lb dry air, but with a
total pressure for this exercise of 1.5 atm.
Find: All of the psychrometric values listed below, using a psychrometric chart, steam tables,
and Table 18.3.
Analysis: (a) Molal humidity. From (186),
Molal humidity = Hm = 0.03(28.97/18.02) = 0.048 mol water/mol dry air
(b) Saturation humidity
From (187), using the vapor pressure of water at 131oF = 0.155 atm,
Saturation humidity, Hs = 18.02 ( 0.155 )
= 0.072 lb water/lb dry air
28.97 (1.5 − 0.155 ) (c) Relative humidity
Relative humidity, HR = pH 2 O
s
pH 2 O 0.048
(1.5)
1.048
=
= 0.443 or 44.3%
0.155 (d) Percentage humidity
From (189) using the saturation humidity computed in part (b), HP = 0.03
× 100% = 41.7%
0.072 (e) Humid volume
From (1810), humid volume us given by,
vH = 0.730 (131 + 460 )
RT 1
H
1
0.03
+
=
+
= 10.4 ft3/lb dry air
P M air M H2 O
1.5
28.97 18.02 (f) and (g) The small pressure change has a negligible effect.
Therefore, cs = 0.254 Btu/lb dry airoF
And H = 57.4 Btu/lb dry air Exercise 18.18
Subject: Humidity conditions for a mixture of nhexane and nitrogen.
Given: Gas at 70oF and 1.1 atm with a relative humidity for hexane of 25%.
Find: Humidity conditions given below.
Analysis: At 70oF, from Fig. 2.4, the vapor pressure of nhexane is 2.4 psia or 0.163 atm.
(a) Partial pressure of hexane.
From (188), p = psHR = 0.163(0.25) = 0.041 atm
(b) Humidity
From (185), H = M nC6 pnC6 ( M N2 P − pnC6 ) = 86.17
0.041
= 0.119 lb nC6/lb dry N2
28.02 (1.1 − 0.041) (c) Percentage humidity
Need saturation humidity, Hs, from (187), Hs = M nC6 ( s
p nC 6 M N2 P − p
nC
s 6 ) = 86.17
0.163
= 0.535 lb nC6/lb dry N2
28.02 (1.1 − 0.163) From (189), HP =(0.119/0.535) x 100% = 22.2%
(d) Mole fraction of nhexane
ynC6 = pnC6
P = 0.041
= 0.037
1.1 Exercise 18.19
Subject: Humidity conditions for a mixture of toluene and air.
Given: Gas at 180oF and 1 atm with a relative humidity for toluene of 15%.
Find: Humidity conditions given below.
Analysis: Use Fig. 18.20 for the airtoluene system at 1 atm.
Using Fig. 18.20: H = 0.21 lb toluene/lb dry air
Ts = 112oF To obtain the wetbulb temperature, which is not equal to the adiabatic saturation
temperature for the tolueneair system, a large extrapolation is needed, giving Tw =
approximately 118oF.
2
From Table 18.5, psychrometric ratio = N Le/ 3 = 1.908 Exercise 18.20
Subject: Wetsolid temperature profile in a continuous, adiabatic, directheat dryer
Given: Entering air at To = 180oF and 1 atm, with a relative humidity of 15%. Air flows
cocurrently to the waterwet solid, which maintains the wetbulb temperature of the entering air
(Ts = Tw). The air exits the dryer at 120oF. The temperature of the air in the dryer, Tg,
experiences an exponential decrease with distance through the dryer according to the relation, Tg = Ts + (To − Ts ) exp ( −0.1377 z ) (1) where, z = distance through the dryer in feet. Assumptions: Isobaric and adiabatic.
Find: Calculate and plot the var...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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