A 10-m- long counter current-flow heat exchanger is being used to heat a liquid
ID: 1532616 • Letter: A
Question
A 10-m- long counter current-flow heat exchanger is being used to heat a liquid food from 20 to 80 degree C. The heating medium is oil, which enters the heat exchanger at 150 degree C and exits at 60 degree C. The specific heat of the liquid food is 3.9 kJ/(kg K) The overall heat transfer coefficient based on the inside area is 1000 W/(m^2 K). The inner diameter of the inside pipe is 7 cm. Estimate the flow rate of the liquid food Determine the flow rate of the liquid food if the heat exchanger is operated in a concurrent-flow mode for the same conditions of temperatures at the inlet and exit from the heat exchanger.Explanation / Answer
(a) Flow rate in countercurrent
Calculate Log Mean Temperature Difference (LMTD)
LMTD = (delta T1 - delta T2) / ln(delta T1/delta T2)
delta T1 = T Hot in - T Cold out
= 150.00 - 80.00
= 70.00
delta T2 = T Hot out - T Cold in
= 60.00 - 20.00
= 40.00
LMDT = 70 - 40/ In(70/40)
= 53.608 degree C
Calculate the heat transfer
Q = U*Ar*LMDT
= 1000*(pi*d*L)*53.608
= 1000*(pi*0.07*10)*53.608
= 117890.14 W = 1.178*10^5 W
Mass flow rate of the liquid food
Q = m*Cp*dT
= m*Cp*(Tco - Tci)
1.178*10^5 = m*3.9*1000*(80-20)
m = 0.5038 kg/s
(b) Flow rate in concurrent
Calculate Log Mean Temperature Difference (LMTD)
LMTD = (delta T1 - delta T2) / ln(delta T1/delta T2)
delta T1 = T Hot in - T Cold in
= 150.00 - 20.00
= 130.00
delta T2 = T Hot out - T Cold in
= 80.00 - 60.00
= 20.00
LMDT = 130 - 20/ In(130/20)
= 58.767 degree C
Calculate the heat transfer
Q = U*Ar*LMDT
= 1000*(pi*d*L)*58.767
= 1000*(pi*0.07*10)*58.767
= 129235.38 W = 1.292*10^5 W
Mass flow rate of the liquid food
Q = m*Cp*dT
= m*Cp*(Tco - Tci)
1.292*10^5 = m*3.9*1000*(80-20)
m = 0.5522 kg/s
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