Chemical Engineering - Heat Transfer - Discussion
Discussion Forum : Heat Transfer - Section 1 (Q.No. 21)
21.
Hot water (0.01 m3 /min) enters the tube side of a counter current shell and tube heat exchanger at 80°C and leaves at 50°C. Cold oil (0.05 m3/min) of density 800 kg/m3 and specific heat of 2 kJ/kg.K enters at 20°C. The log mean temperature difference in °C is approximately
Discussion:
36 comments Page 2 of 4.
MADHAN said:
4 years ago
For counter current flow,
∇(t1) = hot water inlet (-) cold water outlet.
∇(t2) = hot water outlet (-) cold water inlet.
∇(t1) = hot water inlet (-) cold water outlet.
∇(t2) = hot water outlet (-) cold water inlet.
(1)
Jayesh said:
10 years ago
LMTD = ((80 - 35.7) - (30 - 20))/(ln((80 - 35.7)/(30 - 20)) = 23.
Answer is not listed in options.
Answer is not listed in options.
Mehtab Ali said:
3 years ago
For counter-current flow ---> LMTD = 36.66 ~ 37.
For parallel flow ----> LMTD = 31.82 ~ 32.
For parallel flow ----> LMTD = 31.82 ~ 32.
(4)
Enoch said:
8 years ago
Otherwise, the mass and Cp for water should be provided to find the solution.
SATHISH said:
5 years ago
For co-current flow answer is 32.
And for Countercurrent flow answer is 37.
And for Countercurrent flow answer is 37.
Faisal said:
8 years ago
Parallel flow gives an answer of 32C but the counter current gives 37C.
Hiten said:
9 years ago
((80 - 20) - (50 - 35.75))/(ln (60/14.25))
Hence it will be 31.82.
Hence it will be 31.82.
Tushar said:
9 years ago
Parallel flow gives an answer of 32C but counter current gives 37C.
Maha said:
4 years ago
For counter current (as mentioned in statement):
LMTD = 37.
LMTD = 37.
Riddhi said:
8 years ago
Yes, 36.6 is the correct answer for counter current flow.
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