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

Thus x 2 l2 z x 1 l1 z 1 l2 ha tgi tw s h vap w

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Unformatted text preview: the final partially dry cotton to 150oF. Let the final temperature of the air = Tgo. From the steam tables, the heat of vaporization of water at 117oF = 1027.5 Btu/lb. Also, the humidity of the air from Figure 18.17 = 0.0528 lb H2O/lb dry air. Q transferred from the air to the wet, evaporating cotton = Q = 30(0.35)(150 – 70) + 27(1)(117 – 70) + 27(1027.5) + 3(1)(150 – 117) + 27(0.45)( Tgo – 117) = 29,950 + 12.15( Tgo – 117) = 28,530 + 12.15 Tgo Btu/h Q lost by the air = Q = [1,800(0.24) + 1,800(0.0528)(0.45)](200 - Tgo) = 474.8(200 - Tgo) = 94,960 – 474.8 Tgo Btu/h Q transferred from the air to the wet, evaporating cotton = Q lost by the air Therefore, 28,530 + 12.15 Tgo = 94,960 – 474.8 Tgo Solving, Tgo = 136.4oF (c) The rate of heat transfer = Q lost by the air = 94,960 – 474.8(136.4) = 30,200 Btu/h Exercise 18.37 Subject: Spray drying of an aqueous coffee solution. Given: A 25 wt% aqueous solution of coffee at 70oF. Air enters at 450oF and 1 atm with a humidity of 0.01 lb H2O/lb dry air. Final moisture content of the coffee solution is to be 5 wt% on the dry basis. Air exits at 200oF. Assumptions: Specific heat of coffee (solid or in solution) = 0.3 Btu/lb-oF. Dried coffee and heating air exit at the same temperature. Find: (a) The air rate in lb dry air/lb coffee solution. (b) The temperature of evaporation. (c) The heat-transfer rate in Btu/lb coffee solution. Analysis: Take as a basis: 100 lb of coffee solution containing 25 lb of coffee and 75 lb of water. Therefore, the amount of water evaporated = 75 – 25(0.05) = 73.75 lb (b) From a high-temperature humidity chart in Perry’s Handbook, Tw is approximately 120oF. From the steam tables, the heat of vaporization of water at 120oF = 1025.8oF. (a) Q transferred from the air to the evaporating coffee solution = Q = 25(0.30)( 200 – 70) + 75(1)(120 – 70) + 73.75(1025.8) +1.25(1)(200 – 120) + 73.75(0.45)(200 – 120) = 83,200 Btu Q lost by the air = Q = 83,200 = mdry air (0.24 + 0.45 H)(450 – 200) Solving, mdry air = 1,360 lb Therefore, the dry air rate for 100 lb of coffee solution = 1,360/100 = 13.6 lb (c) The heat transfer rate = 83,200/100 = 832 Btu/lb coffee solution Exercise 18.38 Subject: Flash drying of wet, pulverized clay particles. Given: 7,000 lb/h of wet, pulverized clay particles at 15oC and 1 atm with 27 wt% moisture (dry basis). Dry clay to a moisture content of 5 wt% (dry basis) with an exit temperature at the air wet-bulb temperature of 50oC. Cocurrent air enters at 525oC and exits at 75oC. Assumptions: Specific heat of dry clay = 0.3 Btu/lb-oF. Adiabatic drying conditions. Neglect moisture content of the entering air. Find: (a) The flow rate of air in lb/h (dry basis). (b) The rate of evaporation of moisture. (c) The heat-transfer rate in Btu/h. Analysis: Heat of vaporization at 50oC = 122oF = 1024.6 Btu/lb (b) mass flow rate of clay = 7000/1.27 = 5,512 lb/h mass flow rate of entering moisture in the clay = 7,000 – 5,512 = 1,488 lb/h moisture evaporated in dryer =...
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