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

# 1 wt dry basis drying time data plotted below

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Unformatted text preview: for the falling-rate period Analysis: (a) The constant-rate drying flux, from (18-34) is, Rc = h (Tg − Tw ) vap ∆H w (1) Tg = Gas temperature = 160oC = 320oF, Tw = drying temperature = 60oC = 140oF v From the steam tables, at 140oF, ∆H wap = 1014.1 Btu/lb For the heat-transfer coefficient, h, use (18-38) in Table 18.6: h = 0.0204 G0.8, where h is in W/m2-K and G is in kg/h-m2 G = vρ, v = 3.5 m/s = 3.5(3600) = 12,600 m/h The humidity of the air, from a high-temperature humidity chart in Perry’s Handbook = H = 0.10 lb water/lb dry air. Use the humid volume to get the air density. From (18-10), vH = 0.730 ( 320 + 460 ) (1) 1 0.10 + = 24.7 ft3/lb dry air 28.97 18.02 or 24.7/1.10 = 22.5 ft3/lb moist air or 1.40 m3/kg moist air Therefore, ρ = 1/1.40 = 0.714 kg/m3 Therefore, G = vρ = 12,600(0.714) = 9,000 kg/h-m2, which is within the range of (1) in Table 18.6. Therefore, h = 0.0204(9,000)0.8 = 29.7 W/m2-K or 5.24 Btu/h-ft2-oF Exercise 18.28 (continued) From (1), Rc = 5.24 ( 320 − 140 ) From (18-42), 1014.1 = 0.93 lb/h-ft2 ms ( Xo − Xc ) ARc Compute the mass of bone-dry solid in one tray = ms Tray volume = 1(0.5)(0.03) = 0.015 m3 Therefore ms = 1,600(0.015) = 24 kg = 53 lb of dry solid The moisture contents are given on a free-moisture-content basis. The tray area = A = 0.5(1) = 0.5 m2 = 5.38 ft2 Substituting into (2), 53 tc = (1.10 − 0.70 ) = 4.24 h 5.38(0.93) tc = (2) (b) 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 Xc = critical free-moisture content = 0.70 lb water/lb dry solid X = final free-moisture content = 0.05 lb water/lb dry solid Therefore, from (3), tf = 53 ( 0.70 ) 0.70 ln = 19.6 h (5.38)(0.93) 0.05 (3) Exercise 18.29 Subject: Batchwise drying of extrusions 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). The cake is extruded into cylindrical-shaped pieces measuring ¼-inch in diameter by 3/8-inch long, which are placed in trays that are 6 cm high by 1 m long by 0.5 m wide. Drying air at 160oC and 1 atm, with a wet-bulb temperature of 60oC, passes through the bed of pieces at a superficial velocity of 1.75 m/s. Falling-rate period follows the type curve in Fig. 18.31a, based on free-moisture content. Bed porosity is 50%. Assumptions: Neglect sensible heat to increase the temperature of the evaporated moisture from 60oC to 160oC. Drying-air conditions stay constant as the air through the pieces on the trays. Cake does not shrink as it dries. Find: (a) Drying time for the constant-rate period. (b) Drying time for the falling-rate period. Analysis: (a) The constant-rate drying flux, from (18-34) is, Rc = h (Tg − Tw ) (1) vap ∆H w Tg = Gas temperature = 160oC = 320oF, Tw = drying temperature = 60oC = 140oF v From the steam tables, at 140oF, ∆H wap = 1014.1 Btu/lb For th...
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