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 2 of 4.

Rana Rohan said:   6 years ago
Counter-current:

T1 = T_Hot_In - T_Cold_Out.
= 80.00 - 50.00,
= 30.00.

T2 = T_Hot_Out - T_Cold_In.
= 36.00 - 20.00,
= 16.00.

LMTD= (T1-T2)/ ln(T1/T2).
= (30-16)/ ln(30/16),
= 22.27.

Am I right?
(1)

Ritik said:   7 years ago
Thanks for your answer @Shan Rana.

Hariom said:   7 years ago
Yes, 37 is the answer.

Monil said:   7 years ago
How to put t2=35.54?

Please explain.

Zubair Farooq said:   7 years ago
Log mean Temp Diff = T2-T1/ln(T2/T1).

Tekena said:   7 years ago
The answer is 38.22~ 38.

Mujahid Yusuf Barde said:   7 years ago
The answer is 36.5 = 37 ° considering counterflow.

Riddhi said:   8 years ago
Yes, 36.6 is the correct answer for counter current flow.

Faith said:   8 years ago
Here, m*Cp*dT (hotwater) = m*Cp*dT (Cold Oil).

(1000 kg/m3)(0.01m3/min)(4.184kJ/kg.K)(80C-50C) = (800kg/m3)(0.05m3/min)(2kJ/kg.K)(T2-(273.25+20))

T2 = 308.69K = 35.54C (Substitute this T2 for cold oil in LMTD formula for counter current flow)

LMTD = (Th1-Tc2) - (Th2 - Tc1) / ln ((Th1-Tc2) / (Th2 - Tc1)).
= (80-35.54) - (50-20) / ln ((80-35.54) / (50-20)).

LMTD = 36.76 C or approx. 37C.
(3)

Danish said:   8 years ago
37 (for counter flow).

32 (for parallel flow).


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