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 3 of 4.
Faisal said:
8 years ago
Parallel flow gives an answer of 32C but the counter current gives 37C.
Enoch said:
8 years ago
Otherwise, the mass and Cp for water should be provided to find the solution.
Enoch said:
8 years ago
Using a density=1000kg/m3 and Cp=4187J/kg/K for water, heat lost by water = 20935W which is the same as heat gained by the oil hence the temp of the oil leaving the exchanger is 35.7°C corresponding to approximately log mean temperature difference of 37°C for counter flow.
Bhavin Patel said:
8 years ago
Q=m*Cp*DT for water Q=(60*0.01*1000)*(1)*(80-50), Q= 18000 Kcal/hr.
where [Cp in (Kcal/Kg°C), m in (Kg/hr) and Q in (Kcal/hr)].
For oil, Q=m*Cp*DT, 18000=(0.05*60*1000)*(2*0.4777)*(Tout-20), Tout=32.56°C.
DT(LMTD)= [(DT1-DT2)/ln(DT1/DT2)] where DT1=(80-32.56=47.44) & DT2=(50-20=30).
= [(47.44-30)/ln(47.44/30)].
= 38.056 ~ 38°C.
where [Cp in (Kcal/Kg°C), m in (Kg/hr) and Q in (Kcal/hr)].
For oil, Q=m*Cp*DT, 18000=(0.05*60*1000)*(2*0.4777)*(Tout-20), Tout=32.56°C.
DT(LMTD)= [(DT1-DT2)/ln(DT1/DT2)] where DT1=(80-32.56=47.44) & DT2=(50-20=30).
= [(47.44-30)/ln(47.44/30)].
= 38.056 ~ 38°C.
Nilesh said:
9 years ago
37 is the correct answer.
Ashwin said:
9 years ago
36.69 is the answer in the case of counter current flow.
Tushar said:
9 years ago
Parallel flow gives an answer of 32C but counter current gives 37C.
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.
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.
Hiten said:
9 years ago
31.82~32.
Ans is 32.
Ans is 32.
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.
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