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

# 072 kgm h therefore bed reynolds number from 2 is n re

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Unformatted text preview: , X* = 0.04 lb H2O/lb dry cotton Therefore, the initial free-moisture content = 0.95 – 0.04 = 0.91 lb H2O/lb dry cotton and the free-moisture content at the end of the constant-rate drying period = 0.40 – 0.04 = 0.36. Therefore, the moisture evaporated from the cotton on one tray = 69.4(0.91 – 0.36) = 38.2 lb H2O This moisture is evaporated from an area = 1.5 m2 or 1.5/0.30482 = 16.2 ft2 During this period, the temperature of the cotton will be at the wet-bulb temperature of the air, which from Figure 18.17 is 118.7oF. The heat transfer from the air will have to provide latent heat to evaporate the moisture at 118.7oF and sensible heat to increase the temperature of the evaporated moisture to 160oF. From the steam tables, the heat of vaporization = 1026.5 Btu/lb and the specific heat of steam = 0.45 Btu/lb-oF. Thus, the heat that must be transferred from the air = Q = 38.2[1026.5 + 0.45(160 – 118.7)] = 39,900 Btu/tray Note that if the sensible heat is neglected, Q = 39,200 Btu/tray (just 1.8% less) Now, compute the rate of heat transfer to each tray from: q = hA∆T, where A = 16.2 ft2 and ∆T = 160 – 118.7 = 41.3 oF Obtain h from (1) of Table 18.6: h = 0.0204 G0.8, where h is in W/m2-K and G is in kg/h-m2 Exercise 18.27 (continued) G = 500 lb/h-ft2 or 2,440 kg/h-m2 , which is just within the range of the correlation. Therefore, h = 0.0204(2,440)0.8 = 10.5 W/m2-K or 1.85 Btu/h-ft2-oF and q = 1.85(16.2)(41.3) = 1,240 Btu/h Therefore, the drying time for the constant-rate period is, Q/q = 39,900/1,240 = 32.2 h = tc (c) For the falling-rate period based on the curve in Figure 18.31a, (18-41) applies: tf = ms X c X c ln ARc X The rate of drying per unit area in the constant-rate period is, Rc = 38.2 /[32.2(16.2)] = 0.0732 lb H2O/h-ft2 ms = mass of bone-dry solid per tray = 69.4 lb dry cotton Xc = critical free-moisture content = 0.40 – 0.04 = 0.36 lb water/lb dry solid X = final free-moisture content = 0.10 – 0.04 = 0.06 lb water/lb dry solid Therefore, from (1), tf = 69.4 ( 0.36 ) 0.36 ln = 37.7 h (16.2)(0.0732) 0.06 (d) The total drying time = 1.0 + 32.2 + 37.7 = 70.9 h or almost 3 days (1) Exercise 18.28 Subject: Batchwise drying of slabs of filter cake in trays. Given: Filter cake of bone-dry density of 1,600 kg/m3 with an initial free-moisture content of 110% (dry basis). Critical free-moisture content = 70% (dry basis) and final free-moisture content is to be 5% (dry basis). Trays are 1 m long x 0.5 m wide x 3 cm deep. Drying takes place only from the top. Drying air at 160oC and 1 atm, with a wet-bulb temperature of 60oC, passes across the trays at 3.5 m/s. Falling-rate period follows the type curve in Fig. 18.31a, based on free-moisture content. Assumptions: Assume filter cake fills the trays. Neglect sensible heat to increase the temperature of the evaporated moisture from 60oC to 160oC. Drying-air conditions stay constant as the air passes over the trays. Cake does not shrink as it dries. Find: (a) Drying time for the constant-rate period. (b) Drying time...
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