Unformatted text preview: iation of the moistureevaporation temperature, Tv, and the gas
temperature as a function of z.
Analysis: From Fig. 18.17, humidity of the entering air = Ho = 0.052 lb water/lb dry air, and
the wetbulb temperature = Tw = 114.5oF
In the dryer, the solid will follow the adiabaticsaturation temperature, which for the airwater system is the same as the entering wetbulb temperature. Therefore, Tv = 114.5oF.
A spreadsheet is used to compute the gas temperature from (1), using Ts = Tw = 114.5oF. z, ft Tg, F 0
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18 180.0
164.2
152.3
143.2
136.3
131.0
127.0
124.0
121.7
120.0 Exercise 18.21
Subject: Wetbulb temperature of hightemperature air
Given: Air at 1,000oF with a humidity of 0.01 kg H2O/kg dry air.
Assumptions: Wetbulb temperature = adiabaticsaturation temperature for airwater system
Find: The wetbulb temperature for pressures of (a) 1 atm, (b) 0.8 atm, and (c) 1.2 atm
Analysis: Note that 1,000oF is way off the humidity chart of Fig. 18.17.
Therefore, use (1824),
∆H svap
Ts = T −
(Hs – H)
( cs )in (1) From (1811), cs = 0. 24 + 0.45(0.01) = 0.245 Btu/lboF
Because Hs and ∆H svap depend on the value of Ts, solve for Ts iteratively.
(a) P = 1 atm
Iteration 1: Assume Ts = 150oF
From steam tables, ∆H svap = 1008.2 Btu/lb and Ps = 3.72 psia = 0.253 atm
From (187), Hs = 18.02 0.253
= 0.211 lb H2O/lb dry air
28.97 1 − 0.253 Substituting into (1), gives Ts = 1000 − 1008.2
( 0.211 − 0.01) = 173o F (too high)
0.245 Iteration 2: Assume Ts = 152oF
From steam tables, ∆H svap = 1007 Btu/lb and Ps = 3.906 psia = 0.266 atm
From (187), Hs = 18.02 0.266
= 0.225 lb H2O/lb dry air
28.97 1 − 0.266 Substituting into (1), gives Ts = 1000 −
By extrapolation, Ts = Tw = 151oF 1007
( 0.225 − 0.01) = 116o F (too low)
0.245 (b) P = 0.8 atm
Iteration 1: Assume Ts = 140oF
From steam tables, ∆H svap = 1014 Btu/lb and Ps = 2.889 psia = 0.197 atm Exercise 18.21 (continued)
From (187), Hs = 18.02
0.197
= 0.203 lb H2O/lb dry air
28.97 0.8 − 0.197 Substituting into (1), gives Ts = 1000 − 1014
( 0.203 − 0.01) = 201o F (too high)
0.245 Based on interpolation of results for Part (a), Ts = Tw = 141oF
(c) P = 1.2 atm
Iteration 1: Assume Ts = 158oF
From steam tables, ∆H svap = 1003.5 Btu/lb and Ps = 4.519 psia = 0.3075 atm
From (187), Hs = 18.02
0.3075
= 0.214 lb H2O/lb dry air
28.97 1.2 − 0.3075 Substituting into (1), gives Ts = 1000 − 1003.5
( 0.214 − 0.01) = 164o F (too high)
0.245 Based on interpolation of previous results, Ts = Tw = 158oF Exercise 18.22
Subject: Drying of paper with two dryers and a recirculating air system
Given: Equipment and process flow conditions shown in the flowsheet below. Pressure is 1 atm
throughout the process.
Assumptions: Adiabatic drying conditions and no heat losses in the heat exchangers
Find: (a) Labeled processflow diagram
(b) lb water evaporated in each dryer per lb of dry air circulated
(c) lb water condensed in the second heat exchanger per lb of dry air circulated
Analysis: (a) Labeled processflow diagram "# $...
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 Spring '11
 Levicky
 The Land

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