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

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
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Chapter10: Heat Exchangers
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Problem 10.17P
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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
Transcribed Image Text: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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