Unformatted text preview: % & ! (b) The only place in the process where all conditions are fixed for the air is after the coolercondenser, HX2, where because of partial condensation of moisture, the air leaving the phase
separator is saturated at 60oF and 1 atm. Because the pressure is 1 atm. Fig. 18.17 can be used to
follow the process from that point. The process path is shown on a schematic of that figure on
the next page. The humidity values from Fig. 18.17 are as follows:
For Dryer 1, H in = 0.0110 lb/lb dry air, H out = 0.0325 lb/lb dry air
Therefore, the moisture evaporated = 0.0325 – 0.0110 = 0.0215 lb water/lb dry air
For Dryer 2, H in = 0.0325 lb/lb dry air, H out = 0.0478 lb/lb dry air
Therefore, the moisture evaporated = 0.0478 – 0.0325 = 0.0153 lb water/lb dry air
(c)
Using conditions from the humidity chart for HX2,
H in = 0.0478 lb/lb dry air, H out = 0.0110 lb/lb dry air
Therefore, the moisture condensed = 0.0478 – 0.0110 = 0.0368 lb water/lb dry air Exercise 18.22 (continued)
0.06 Humidity, lb water/lb dry air 0.05
Dryer 2
Dryer 2 0.04
Heater HX1 0.03
Dryer 1 0.02
Heater HX3 0.01 0.00
60 80 100 120 140 Temperature, F 160 180 200 Exercise 18.23
Subject: Dehumidification of air
Given: Air at 96oF, 1 atm, and 70% relative humidity. Dehumidify to 10% relative humidity.
Cooling water at 50oF is available
Find: A labeled processflow diagram for carrying out the dehumidification. Coolingwater
requirement in lb water per lb of dry air recirculated.
Analysis: One method of carrying out the dehumidification is to cool the air to 60oF (i.e. 10oF
above the entering coolingwater temperature). Assume the cooling water flows
countercurrently to the air in a heat exchanger, exiting at 60oF. This gives a temperaturedriving
force of about 20oF in the exchanger. The labeled processflow diagram is From Figure 18.17, the air entering the heat exchanger has a humidity of 0.0257 lb water per lb
dry air. At the exit conditions of the heat exchanger, the air is saturated at 60oF and 1 atm. From
Figure 18.17, the humidity is now 0.0113 lb water per lb dry air. To achieve a relative humidity
of 10%, using Figure 18.17 again, we must heat the air to 136 F.
To make an energy balance around the first exchanger, use (1812) with a To of 32oF.
From (1811), cs of the entering air = 0.24 + 0.45(0.0257) = 0.252 Btu/lb dry air – oF.
At 32oF, the heat of vaporization = 1075 Btu/lb water. Therefore, from (1812),
Enthalpy of air in = 0.252(96 – 32) + 1075(0.0257) = 43.8 Btu/lb dry air
For the exit air, cs = 0.24 + 0.45(0.0113) = 0.245 Btu/lb dry air – oF.
Therefore, Enthalpy of air out = 0.245(60 – 32) + 1075(0.0113) = 19.0 Btu/lb dry air
Therefore, the heat duty = 43.8 – 19.0 = 24.8 Btu/lb dry air. Since the cooling water temperature
change is 10oF, this amounts to 10 Btu/lb of water. Therefore, need 24.8/10 = 2.48 lb H2O/lb dry
air. Exercise 18.24
Subject: Drying of nitrocellulose fibers using equilibrium moisture content
Given: Nitrocellulose fibers with initial tot...
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 Spring '11
 Levicky
 The Land

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