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4.53. Heat-Transfer Area and Use of Log Mean Temperature Difference. A reaction mix-ture having a cpmfrom 377.6 K to 344,3 K. Cooling water at 288.8 K is available and the flow rate is4536 kg/h. The overall U, is 653 W/m2.K(a) For counterflow, calculate the outlet water temperature and the areathe exchanger.(b) Repeat for cocurrent flow.= 2.85 kJ/kg-K is flowing at a rate of 7260 kg/h and is to be cooled13.2-9Ao ofAns. (a) T 325.2 K, A5.43 m2 (b) A6.46 m2

Question
4.53. Heat-Transfer Area and Use of Log Mean Temperature Difference. A reaction mix-
ture having a cpm
from 377.6 K to 344,3 K. Cooling water at 288.8 K is available and the flow rate is
4536 kg/h. The overall U, is 653 W/m2.K
(a) For counterflow, calculate the outlet water temperature and the area
the exchanger.
(b) Repeat for cocurrent flow.
= 2.85 kJ/kg-K is flowing at a rate of 7260 kg/h and is to be cooled
13.2-9
Ao of
Ans. (a) T 325.2 K, A
5.43 m2 (b) A
6.46 m2
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4.53. Heat-Transfer Area and Use of Log Mean Temperature Difference. A reaction mix- ture having a cpm from 377.6 K to 344,3 K. Cooling water at 288.8 K is available and the flow rate is 4536 kg/h. The overall U, is 653 W/m2.K (a) For counterflow, calculate the outlet water temperature and the area the exchanger. (b) Repeat for cocurrent flow. = 2.85 kJ/kg-K is flowing at a rate of 7260 kg/h and is to be cooled 13.2-9 Ao of Ans. (a) T 325.2 K, A 5.43 m2 (b) A 6.46 m2

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Step 1

Specific heat of reaction mixture cpm = 2.85 kJ/kg.K

Mass flow rate of reaction mixture mh = 7260 kg/h = 2.0167 kg/s

Initial temperature of the mixture Th1= 377.6 K

Final temperature of the mixture after cooling Th2 = 344.3 K

Temperature of the cooling water Tc1= 288.8 K

Mass flow rate of the cooling water mc = 4536 kg/h = 1.26 kg/s

The value of overall heat transfer coefficient Uo = 653 W/m2K

Step 2

 The outlet water temperature and area Ao of the heat exchanger for counter flow can be calculated as follows.

T377.6K
T 344.3K
h2
T
T 288.8K
.1
Now, from the energy balance equation
х Сp, x (Tи — Тз) - т, хс х (Т. -Т.)
2.0167 x 2.85 x(377.6-344.3) = 1.26 x 4.187x(T, - 288.8)
191.3949 5.27 x ( T-288.8)
191.3949
T2-288.8
5.27
T2 325.08 K
T2 325.1K
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T377.6K T 344.3K h2 T T 288.8K .1 Now, from the energy balance equation х Сp, x (Tи — Тз) - т, хс х (Т. -Т.) 2.0167 x 2.85 x(377.6-344.3) = 1.26 x 4.187x(T, - 288.8) 191.3949 5.27 x ( T-288.8) 191.3949 T2-288.8 5.27 T2 325.08 K T2 325.1K

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Step 3

The value of LMTD (Log Mean Temperature ...

LMTD(e)= -6
In
(344.3-288.8)- (377.6-325.1)
(344.3-288.8)
In
|(377.6-325.1)
55.5 52.5
e
0.05556
e=53.995 K
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LMTD(e)= -6 In (344.3-288.8)- (377.6-325.1) (344.3-288.8) In |(377.6-325.1) 55.5 52.5 e 0.05556 e=53.995 K

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