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.
Tekena said:
7 years ago
The answer is 38.22~ 38.
Zubair Farooq said:
7 years ago
Log mean Temp Diff = T2-T1/ln(T2/T1).
Monil said:
7 years ago
How to put t2=35.54?
Please explain.
Please explain.
Hariom said:
7 years ago
Yes, 37 is the answer.
Ritik said:
7 years ago
Thanks for your answer @Shan Rana.
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?
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)
Rana Rohan said:
6 years ago
Given answer is right for co-current.
Temidayo Gideon said:
6 years ago
Here, 37 is the correct answer.
(1)
Urvish Makwana said:
6 years ago
The given answer is wrong as by using counter flow we will get 37 as answer but if we use co current than we will get 32 as the answer.
(1)
Kishan said:
6 years ago
LMTD = (ΔT1 - ΔT2) / ln(ΔT1/ΔT2).
Counter flow,
ΔT1 = T_Hot_In - T_Cold_Out.
= 80.00 - 35.75.
= 44.25.
ΔT2 = T_Hot_Out - T_Cold_In.
= 50.00 - 20.00.
= 30.00.
LMTD = 36.66.
For parallel flow,
ΔT1 = T_Hot_In - T_Cold_In.
= 80.00 - 20.00.
= 60.00.
ΔT2 = T_Hot_Out - T_Cold_Out.
= 50.00 - 35.75
= 14.25.
LMTD = 31.82.
Counter flow,
ΔT1 = T_Hot_In - T_Cold_Out.
= 80.00 - 35.75.
= 44.25.
ΔT2 = T_Hot_Out - T_Cold_In.
= 50.00 - 20.00.
= 30.00.
LMTD = 36.66.
For parallel flow,
ΔT1 = T_Hot_In - T_Cold_In.
= 80.00 - 20.00.
= 60.00.
ΔT2 = T_Hot_Out - T_Cold_Out.
= 50.00 - 35.75
= 14.25.
LMTD = 31.82.
(11)
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