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

120 vs mw 00519 na avagadros number 6022 x 1026

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Unformatted text preview: quid H liquid + mcrystals H crystals Substituting given flow rates and Figure 17.10 enthalpies into (1) gives, 5,870(-3) + 10,500(-44) + Qin = 1,490(1,099) + 11,160(-43) + 3,720(-148) Solving, Qin = 1,087,000 Btu/h (1) Exercise 17.15 Subject: Rate of heat removal from a cooling crystallizer Given: Feed of saturated oxalic acid in water at 100oC. From Example 17.4, solution is cooled to 10.6oC to crystallize 95% of the oxalic acid. Assumptions: No heat losses from the crystallizer. Find: Required rate of heat removal. Analysis: Use cgs units. Assume a thermodynamic path of cooling to 10.6oC, followed by crystallization to the dihydrate. From Example 17.4, using a basis of 100 grams of water in the feed, the oxalic acid in the feed is 84.4 grams from Table 17.7. Also, the amount of oxalic acid remaining in the solution at 10.6C is 4.2 g. Therefore, 84.4 – 4.2 = 80.2 g of oxalic acid is crystallized. MW of oxalic acid = 90.0. Therefore the mols of oxalic acid crystallized = 80.2/90 = 0.891 mols From Perry’s Handbook, the estimated feed specific heat = 0.76 cal/g-oC. Therefore, to cool the solution, must remove: (100 + 84.4)(0.76)(100 – 10.6) = 12,530 cal/100 g water in feed Also, the heat of exothermic crystallization = 8,580 cal/mol Therefore, must remove an additional 0.891(8,580) = 7,650 cal/100 g water in feed The total amount of heat that must be removed is: 12,530 + 7,650 = 20,180 cal/100 g water in the feed Exercise 17.16 Subject: Effect of crystal size on solubility Given: Crystals of KCl, BaSO4, and sucrose at 25oC , together with values of their interfacial tensions in water, σs,L of 0.028, 0.13, and 0.01 J/m2, respectively. Assumptions: Spherical crystals Find: Effect of crystal size, say from 0.01 to 10 microns on the solubility in water Analysis: From (17-16), ln From a handbook, Component KCl BaSO4 Sucrose Use SI units in (1). KCl: MW 74.6 233.42 342.30 4v σ c = s s,L cs υRTD p Density, kg/m3 1,980 4,499 1,588 (1) vs, m3/kmol 0.0376 0.0519 0.216 υ 2 2 1 Solubility at 25oC from Table 17.5 by interpolation = cs = 35.5 g/100 g water Results from Example 17.7 are: Dp, microns 0.01 0.10 1.00 10.00 Dp , m 0.00000001 0.0000001 0.000001 0.00001 c/cs 1.0085 1.00085 1.000085 1.0000085 c, g/100 g water 35.80 35.53 35.50 35.50 BaSO4: Calculate solubility from solubility product in Table 17.6 assuming value of 0.87 x 10-7 at 18oC equals that at 25oC. () K c = ( cBa ++ ) cSO= = ( cBaSO4 ) = 0.87 × 10−10 2 4 Solving, ( cBaSO4 ) = 0.93 x 10-5 mol/L water or cs = 0.93 x 10-5 (0.1)(233.42) = 2.2 x 10-4 g/100 g water From (1), ln 4 ( 0.0519 ) ( 0.13) c = cs 2 ( 8315 )( 298 ) ( D p in m ) (2) Exercise 17.16 (continued) Solving (2) for a range of crystal size, Dp, microns 0.01 0.10 1.00 10.00 Sucrose: Dp , m 0.00000001 0.0000001 0.000001 0.00001 c/cs 1.716 1.055 1.0054 1.00054 c, g/100 g water 0.00038 0.00023 0.00022 0.00022 Solubility at 25oC from Table 17.7 by interpolation = cs = 212 g/100 g water From (1), ln 4 ( 0.0216 ) ( 0.01) c = cs 1( 8315 )( 298 ) ( D p in m ) (3) So...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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