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)
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)
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.
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.
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.
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.
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)
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.
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.
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.
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)
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