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
32
37
45
50
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
36 comments Page 1 of 4.

Sovanpaul said:   1 decade ago
(m*cp*dT)oil = (m*cp*dT)water.

m of water = 10 kg/m3.

cp of water = 4.2 kj/kg K.

dT for water = 30 C.

m of oil = 40 kg/m3.

cp of oil = 2 kj/kg K.

dT for oil = (T-20).

From here T final for oil = 35.75°C.

And LMTD = 36.6°C.

Shan Rana said:   1 decade ago
Please correct it @Sovanpaul.

m mass flowrate of water = density x volumetric flowrate.
m mass flowrate of water = 1000 kg/m3 x 0.01 m3/min.
m mass flowrate of water = 10 kg/min.

NOT 10 kg/m3.

Similarly,

m mass flow rate of oil = 40 kg/min.

NOT 40 kg/m3.

Ankush said:   1 decade ago
Answer is 37. Since its a counter current flow.

Sanket said:   1 decade ago
LMTD is 36.73 = 37 (approx).

Jayesh said:   10 years ago
LMTD = ((80 - 35.7) - (30 - 20))/(ln((80 - 35.7)/(30 - 20)) = 23.

Answer is not listed in options.

Sandip said:   9 years ago
@JAYESH There is 50 instead of 30. So the ans is 36.66.

Hiten said:   9 years ago
((80 - 20) - (50 - 35.75))/(ln (60/14.25))

Hence it will be 31.82.

Hiten said:   9 years ago
31.82~32.

Ans is 32.

Lekhanandu said:   9 years ago
@ Hiten.

I think you have probably applied parallel flow logic but the question is in counter flow. So the answer would be 36.6 = 37 C.

Tushar said:   9 years ago
Parallel flow gives an answer of 32C but counter current gives 37C.


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