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0.007 0.00113 = 0.00097 cm/s, which is about 7% lower than in Example 17.9.
Thus, here the crystal growth rate is still controlled by surface reaction, but mass transfer does
make a small contribution with about 14% of the resistance. (1) Exercise 17.22
Subject: Heat transfer area of a cooling crystallizer.
Given: Feed of 2,000 kg/h of 30 wt% Na2SO4 in water at 40oC; cooled to a temperature at
which 50% of the sulfate is crystallized as the decahydrate. Overall heat transfer coefficient = 15
Btu/hft2oF. Specific heat of solution = 0.80 cal/goC. Coolant is chilled water in countercurrent
flow, entering at 10oC and exiting so as to give a logmean temperature driving force of 10oC.
Assumptions: Equilibrium and no heat loss.
Find: Heat transfer area in m2.
Analysis: First compute the mass balance around the crystallizer.
Feed is 600 kg/h of Na2SO4 and 1,400 kg/h of water.
From Table 17.5, at 40oC, the solubility of 48.8 g/100 g of water.
Since the feed contains 42.9 g/100 g of water, the feed is below solubility
MW of Na2SO4 = 142.1
MW of Na 2SO 4 10H 2 O = 322.2
Therefore, the crystals are 142.1/322.2 = 0.441 mass fraction Na2SO4
Na2SO4 crystallized = 0.5(600) = 300 kg/h
Water of crystallization = (0.559/0.441)300 = 380 kg/h
Therefore, have exiting, 680 kg/h of crystals and a solution of 300 kg/h Na2SO4
and 1,400 – 380 = 1,020 kg/h of water. This gives a solubility of (300/1,020)100 = 29.4 g/100 g
From Table 17.5, the temperature corresponding to that solubility is, by interpolation,
approximately 25oC.
Now compute the heat transfer rate by an enthalpy balance.
From Table 17.5, the heat of solution at 25oC = +18,700 cal/mol of compound
Therefore, the heat of crystallization is exothermic at 18,700 kcal/kmol.
Since the final temperature is also 25oC, assume a thermodynamic path of cooling the
feed from 40oC to 25oC and then crystallizing at 25oC.
Therefore, in cgs units,
Q to be transferred to chilled water = 2,000(1,000)(0.80)(40 – 25) +
(680/322.2)(18,700)(1,000) = 24,000,000 + 39,500,000 = 63,500,000 cal/h
Now apply the heat transfer rate equation, Q = UA ∆TLM in SI units
Q = 63,500,000(4.185) = 266,000 kJ/h
U = 15(20.44) = 307 kJ/hm2K
∆TLM = 10oC = 10 K
Therefore, heat transfer area = A = 266,000/[307(10)] = 87 m2 Exercise 17.23
Subject: Number of cooling crystallizer units
Given: Two tons (4,000 lb) per hour of Na 3 PO 4 12H 2 O to be produced in a cooling crystallizer
from a saturated aqueous solution entering at 40oC and exiting at 20oC. Countercurrent chilled
water enters at 10oC and exits at 25oC. Average specific heat of the solution = 0.80 cal/goC.
Overall heat transfer coefficient = 20 Btu/hft2oF.
Assumptions: Equilbrium and no heat loss. Linear temperature profile on both sides of the heat
transfer surface.
Find: (a) The tons/h of required feed solution.
(b) Heat transfer area in ft2.
(c) The number of crystallizer units needed, if each has 30 ft2 of heat transfer surface.
Analysis:
(a) First, compute the mass balance aroun...
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 Spring '11
 Levicky
 The Land

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