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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/goC.
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 (1716), 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 107 at
18oC equals that at 25oC. () K c = ( cBa ++ ) cSO= = ( cBaSO4 ) = 0.87 × 10−10
2 4 Solving, ( cBaSO4 ) = 0.93 x 105 mol/L water or cs = 0.93 x 105 (0.1)(233.42) = 2.2 x 104 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.
 Spring '11
 Levicky
 The Land

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