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Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

# 23 total moisture content xt xin 050 lb waterlb dry

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Unformatted text preview: iation of the moisture-evaporation 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 wet-bulb temperature = Tw = 114.5oF In the dryer, the solid will follow the adiabatic-saturation temperature, which for the airwater system is the same as the entering wet-bulb 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 2 4 6 8 10 12 14 16 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: Wet-bulb temperature of high-temperature air Given: Air at 1,000oF with a humidity of 0.01 kg H2O/kg dry air. Assumptions: Wet-bulb temperature = adiabatic-saturation temperature for air-water system Find: The wet-bulb 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 (18-24), ∆H svap Ts = T − (Hs – H) ( cs )in (1) From (18-11), cs = 0. 24 + 0.45(0.01) = 0.245 Btu/lb-oF 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 (18-7), 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 (18-7), 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 (18-7), 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 (18-7), 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 process-flow 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 process-flow diagram "# \$...
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