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

# 752 from 2 of example 1813 ln x avg x xo x 8 2

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Unformatted text preview: e heat-transfer coefficient, h, use (3) or (4) for through-circulation in a packed bed, in Table 18.6, where the choice depends on the value of the bed Reynolds number, N Re = d pG (2) µ where, from Example 18.8, dp is the diameter of a sphere of the same surface area as the particle. From (1) of Example 18.8, with particle diameter, D, and particle length, L, πD 2 πd = πDL + 2 4 2 p D2 or d p = DL + 2 1/ 2 1 3 0.252 = + 48 2 1/ 2 = 0.354 in = 0.00897 m The superficial mass velocity = G = vρ, with v = 1.75 m/s = 1.75(3600) = 6,300 m/h For the gas density, calculate the humid volume. The humidity of the air, from a high-temperature humidity chart in Perry’s Handbook = H = 0.10 lb water/lb dry air. Exercise 18.29 (continued) From (18-10), 0.730 ( 320 + 460 ) 1 0.10 vH = + = 24.7 ft3/lb dry air 28.97 18.02 (1) 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 = 6,300(0.714) = 4,500 kg/h-m2 From Example 18.8, take the air viscosity = µ = 2 x 10-5 kg/m-s or 0.072 kg/m-h Therefore, bed Reynolds number, from (2) is, N Re = 0.00897 ( 4,500 ) 0.072 = 561 Therefore, (3) in Table 18.6 applies: h = 0.151 G 0.59 45000.59 = 0.151 = 149 W/m2-K or 26.3 Btu/h-ft2-oF 0.41 dp 0.00897 0.41 From (1), Rc = 26.3 ( 320 − 140 ) 1014.1 = 4.67 lb/h-ft2 From (18-42), the time for the constant-rate drying period is, m tc = s ( X o − X c ) (2) ARc Compute the mass of bone-dry solid in one tray = ms Tray volume = 1(0.5)(0.06) = 0.03 m3 For a porosity of 0.5, ms = 1,600(0.03)(0.5) = 24 kg = 53 lb of dry solid The moisture contents are given on a free-moisture-content basis. The area, A, in (2) is the total area of the particles in the bed on the tray. Volume of one extrusion = πD2L/4 = 3.14(0.25)2(0.375)/4 = 0.0184 in3 or 3.02 x 10-7 m3 The extrusions on one tray fill a volume of 0.015 m3 (50% of the tray volume) Therefore, the number of extrusions = 0.015/3.02 x 10-7 = 49,750 The surface area of one extrusion is, πD 2 = 3.14[(0.25)(0.375) + 0.5(0.25)2] = 0.393 in2 = 0.00273 ft2 πDL + 2 4 Therefore, total surface area of the extrusions on a tray = 49,750(0.00273) = 136 ft2 Substituting into (2), tc = 53 (1.10 − 0.70 ) = 0.0334 h or 2 min 136(4.67) Exercise 18.29 (continued) (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 = 0.154 h or 9.2 min ln (136)(4.67) 0.05 (3) Exercise 18.30 Subject: Tray drying in the constant- and falling-rate periods Given: Takes 5 hours to dry a wet solid in a tray from 36 to 8 wt% on dry basis. Critical and equilibrium moisture contents are 15 and 5 wt% dry basis, respectively. Negligible preheat time. Straight-line falling-rate period. Assumptions: Drying-air conditions stay constant. Cake does not shrink as it dries. Find: Drying time for same we...
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