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

Separation Process Principles 2n Seader& Henley Solutions Manual

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Unformatted text preview: r = 1.09 kJ/kg-K = ( CP ) g Density of the air = 0.942 kg/m3 = ρ g Bed cross-sectional area = 0.0929 m2 Volumetric flow rate of solid particles = 0.0283 m3/min Dry density of calcium carbonate = 1500 kg/m2 = ( ρb )ds Mass flow rate of calcium carbonate = 42.5 kg/min = 0.708 kg/s (dry basis) Solve this exercise using SI units, but with temperature in oC and time in sec Zone 1: H = 2 in = 0.0508 m, L1 = 8 ft = 2.44 m From (18-74), gas temperature leaving Zone 1 is Tgo = 37.8 + ( 76.7 − 37.8 ) exp − 0.188(395)(0.0508) = 50oC 0.942(1.09)(2) The moisture-content distribution is given by (18-75). Applying this equation at the end of Zone 1 gives the following, where x is distance along the bed, which equals 8 ft = 2.44 m and z is the height measured from the bottom of the bed for upward flow of air. Exercise 18.40 (continued) X 1 { L1 , z} = 0.30 1 − 2.44 ( 0.188 )( 395 )( 76.7 − 37.8 ) 0.00508 ( 2413)(1500 ) exp − 0.188 ( 395 ) z 0.942 (1.09 )( 2 ) z, m 0.0000 0.0127 0.0254 0.0381 0.0508 Using a spreadsheet, the following results are obtained: X, % 18.5 22.7 25.4 27.1 28.2 Zone 2: Use (18-78), with x measured at L1 and height given by (H – z), because of downward flow from z = H to z = 0. Thus, X 2 {L2 , z} = X 1 { L1 , z} 1 − L2 ha (Tgi − Tw ) S ∆H vap w ( ρb )ds exp − ha ( H − z ) ρ g ( CP ) g us (1) where L2 = length of zone 2 = 2.44 m and, again, z is measured from the bottom of the bed. Using a spreadsheet, the following results are obtained: z, m X, % 0.0000 17.4 0.0127 20.5 0.0254 21.5 0.0381 20.5 0.0508 17.4 Zone 3: Use (18-78), with x measured at L2 and height given by z because of upward flow from z. = H to z = 0. Thus, using an equation similar to (1), but with X3 in terms of X2. Using a spreadsheet, the following results are obtained: z, m X, % 0.0000 10.7 0.0127 15.6 0.0254 18.2 0.0381 18.6 0.0508 16.3 The final average moisture content is obtained from integrating the values of X at the end of zone 3 with (18-76), which gives Xavg = 16.5% or 0.165 lbH2O/lb dry solid, which is greater than the critical moisture content so that drying is in the constant-rate period as assumed. Exercise 18.41 Subject: Drying of extruded filter cake of calcium carbonate with a through-circulation belt dryer having 2 zones. Given: Calcium carbonate extruded into cylindrical pieces of 3/8-inch diameter and 1/2-inch length with an initial moisture content of 30% (dry basis) and a critical moisture content of 10% (dry basis). Belt dryer is 6-ft wide with two drying zones of 12-ft long each, running with a belt speed, S, of 1 ft/min (0.00508 m/s). Carbonate bed height on the belt is 2-inches with an external porosity of 50%. Air at 170oF (76.7oC) and 10% relative humidity flows through the bed at a superficial velocity, us,of 2 m/s, upward in the first zone and downward in zone 2. Assumptions: Negligible preheat period. Constant-rate drying at the wet-bulb temperature of the entering air. Find: Moisture content distribution with height at the end of each zone and the final average moisture content. Analysis: This exercise is exactly the same as Example 18.18 except that the diameter of the extrusion is 3/8 inch instead of 2/8 inch. Thus, the following preliminary calculations from Example 18.18 apply here: Entering air has a wet-bulb temperature of 110.2oF = 37.8oC = Tw Entering air has a humidity of 0.0265 lb H2O/lb dry air vap Heat of vaporization of water at 110.2oF = 2413 kJ/kg = ∆ H w Specific heat to the air = 1.09 kJ/kg-K = ( CP ) g Density of the air = 0.942 kg/m3 = ρ g Bed cross-sectional area = 0,0929 m2 Dry density of calcium carbonate = 1500 kg/m2 = ( ρb )ds Preliminary calculations: First, calculate the effective particle diameter, dp, the diameter of a sphere having the same particle surface area, with D = 3/8 inch, L = 4/8 πD 2 πd 3 = 2 + πDL p 4 Solving, dp = 0.506 inch = 0.013 m Now, compute the heat-transfer coefficient from Table 18.6. Need to calculate the bed Reynolds number to determine whether (3) or (4) applies. Using data in Example 18.8, NRe = (0.013)(1.96)/2x10-5 = 1,274 Therefore, (3) in Table 18.6 applies: h = 0.151(14,300 / 2 ) / ( 0.013) = 168 J/s-m2-K or 0.168 kJ/s-m2-K Now compute the extrusion surface area pe...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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