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

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Unformatted text preview: al water content of 40 wt% on dry basis (Xin = 0.40). Dried in a tunnel dryer at 1 atm. Fibers brought to equilibrium with air at 25oC and a relative humidity of 30%. Equilibrium moisture content given in Figure 18.24. Find: The kg of moisture evaporated per kg of bone-dry fibers. Analysis: From Figure 18.24, the final moisture content = Xout = X* = 0.07 lb water/lb dry solid Therefore, evaporation = Xin - Xout = 0.40 – 0.07 = 0,33 lb water/lb dry solid Exercise 18.25 Subject: Slow drying of lumber using equilibrium moisture content Given: Initial moisture content = Xin = 0.50 lb water/lb dry lumber. Final air conditions of 25oC and 1 atm, with a relative humidity of 40%. Assumptions: Applicability of Figure 18.24, 10 – Lumber. Find: (a) Unbound moisture before drying (b) Bound moisture before drying (c) Free moisture before drying (d) lb moisture evaporated per lb of bone-dry wood Analysis: The definitions of moisture content are shown schematically in Figure 18.23. Total moisture content = XT = Xin = 0.50 lb water/lb dry lumber Figure 18.24 applies at the desired conditions of 25oC and 1 atm. Bound moisture content = XB = 0.306 lb water/lb dry lumber Equilibrium moisture content = X* = 0.075 lb water/lb dry lumber (a) Unbound moisture content = XT - XB = 0.500 – 0.306 = 0.194 lb water/lb dry lumber (b) Bound moisture content = XB = 0.306 lb water/lb dry lumber (c) Free moisture content = XT – X*= 0.500 – 0.075 = 0.425 lb water/lb dry lumber (d) Moisture evaporated is the free moisture = 0.425 lb water/lb dry lumber Exercise 18.26 Subject: Drying of cotton cloth in a closed air system Given: 50 lb of cotton cloth at 100oF and 1 atm with a total moisture content of 20 wt% (dry basis). Cloth is hung in a closed room containing 4,000 ft3 of air at 100oF and 1 atm with an initially wet-bulb temperature of 69oF. Air is kept at 100oF and equilibrium is achieved in the room. Assumptions: Equilibrium moisture content of cotton cloth at 100oF is given by Figure 18.25. Neglect the increase in air pressure. Air is not vented and no new air enters the room. Find: Final moisture content of the cotton cloth and the final relative humidity of the air. Also, determine the increase in air pressure. Analysis: The relative humidity of the air will increase, so an iterative solution is required. Initial composition of the wet cloth: % water in solid on a wet basis = 20/(20+100) = 0.1667 Therefore, the initial wet solid is: 0.1667(50) = 8.33 lb water 50 – 8.33 = 41.67 lb water-free cotton cloth Initial air conditions: From Figure 18.17, humidity = 0.008 lb water/lb dry air From (18-10), the humid volume is: vH = 0.730 (100 + 460 ) (1) 1 0.008 + = 14.29 ft3/lb dry air 28.97 18.02 Therefore, have 4,000/14.29 = 280 lb dry air Initial moisture in the air = 0.008(280) = 2.24 lb water Now, perform iterations to determine final equilibrium conditions for a room containing: 41.67 lb water-free cotton cloth 280 lb air on the dry basis 8.33 + 2.24 = 10.57 lb water If all the water were evapor...
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