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

02098 714 lbh rate of evaporation 1500 714 1429

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Unformatted text preview: e at the surface of the wet solid is higher than the wet-bulb temperature by a few degrees, thus reducing slightly the temperature driving force for heat transfer from the hot air to the wet surface of the solid. However, the overall heat-transfer coefficient for heating from both the top and bottom is increased considerably. The result is a significantly higher drying rate. From Eqs. (18-34) and (18-35), the drying-rate flux in the constant-rate drying period, Rc, in the absence of heat-transfer to the tray bottom and radiation to the wet surface of the solid is: Rc = h (Tg − Tv ) ∆H vvap = M dry air k y ( H v − H d ) (1) where, here, the subscript v refers to the wet-bulb temperature, w. Assuming the area for heat transfer to the top and bottom of the tray is the same, then, if convection to the uninsulated bottom of the tray is taken into account, with subsequent conduction through the tray thickness and the wet solid, then h in (1) is replaced by an overall coefficient: U = hc ,t + 1 1 t t + t + ws hc ,b kt kws (2) Assume that hc,t at the top = hc,b at the bottom = hc. Assume that the thermal resistance of the metal tray thickness, tt / kt , is negligible compared to the thermal resistance of the wet solid, tws / kws , where t is thickness and k is thermal conductivity. Then (2) becomes: Exercise 18.35 (continued) U = hc + 1 1 tws + hc kws (3) Now U in (3) is greater than h in (1). Therefore, the ratio U / (kyCsMB) > h / (kyCsMB) , the psychrometric ratio, and the temperature of the evaporating moisture will be higher than the wetbulb temperature of the hot air, thus reducing the temperature driving force. If we also account for radiation from the bottom of a tray to the tray below, the overall heat transfer coefficient for heat transfer to the tray below will be even higher than U, because an additional term for a radiation coefficient will be added to the right-hand side of (3). Therefore, the temperature of the evaporating moisture will be further increased. Exercise 18.36 Subject: Tunnel drying of raw cotton. Given: Dry 30 lb/h of raw cotton (dry basis) at 70oF and a moisture content of 100% (dry basis) in a tunnel dryer with a countercurrent flow of 1,800 lb/h of air (dry basis) entering at 200oF, 1 atm, and a relative humidity of 10%. The cotton will exit at 150oF with a moisture content of 10% (dry basis). Specific heat of dry cotton = 0.35 Btu/lb-oF. Assumptions: Adiabatic drying conditions. Specific heat of dry air = 0.24 Btu/ lb- oF. Specific heat of steam = 0.45 Btu/lb-oF. Find: (a) The rate of evaporation of moisture. (b) The outlet temperature of the air. (c) The rate of heat transfer. Analysis: (a) The rate of evaporation of moisture = 30(1.00 – 0.10) = 27 lb/h (b) The air must supply sensible heat and latent heat. Assume that moisture evaporation occurs at the wet-bulb temperature of the air. From Figure 18.17, this is 117oF. Therefore, heat the wet cotton from 70oF to 117oF, evaporate the 27 lb/h of moisture at 117oF, heat (or cool) the evaporated moisture to the final gas temperature, and heat...
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