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 4 of 4.
Sandip said:
9 years ago
@JAYESH There is 50 instead of 30. So the ans is 36.66.
Jayesh said:
10 years ago
LMTD = ((80 - 35.7) - (30 - 20))/(ln((80 - 35.7)/(30 - 20)) = 23.
Answer is not listed in options.
Answer is not listed in options.
Sanket said:
1 decade ago
LMTD is 36.73 = 37 (approx).
Ankush said:
1 decade ago
Answer is 37. Since its a counter current 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.
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.
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