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 1 of 4.
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)
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)
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.
(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)
Anuj said:
1 year ago
@All.
We have to find the cold coil outlet temp with the help of the given data and assume that Q rejected by the hot fluid is equal to the heat gained by the cold fluid, then we will get 36.8.
So, the right answer will be 37 °C for counter-current flow.
We have to find the cold coil outlet temp with the help of the given data and assume that Q rejected by the hot fluid is equal to the heat gained by the cold fluid, then we will get 36.8.
So, the right answer will be 37 °C for counter-current flow.
(1)
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)
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)
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)
Anon said:
2 years ago
how did u guys get 35.75 deg C as outlet temp of oil?
(1)
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.
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