amended mean log temperature difference
DESCRIPTION
unit operationTRANSCRIPT
Change in heat energy in a fluid stream
where, ΔTlm is the log-mean temperature difference
q = UA(ΔTlm )
ΔTlm = ΔT2 − ΔT1ln ΔT2
ΔT1
Log-Mean Temperature Difference Method Log-Mean Temperature Difference Method
ΔT1 = T1 − T’1
ΔT2 = T2 − T’2
For parallel flow For counter flow
T1
T’1
T2
T’2
T1
T’1
T2
T’2
Example 1: Counter flow heat exchanger
A liquid food (specific heat 4.0 kJ/kg.°C) flows in the inner pipe of a double-pipe heat exchanger. The liquid food enters the heat exchanger at 20°C and exits at 60°C. The flow rate of the liquid food is 0.5 kg/s. In the annular section, hot water at 90°C enters the heat exchanger and flows counter-currently at a flow rate of 1 kg/s. The average specific heat of water is 4.18 kJ/(kg.°C). Assume steady-state conditions.
Log-Mean Temperature Difference Method
1. Calculate the exit temperature of water.
2. Calculate log-mean temperature difference.
3. If the average overall heat transfer coefficient is 2,000 W/m2.°C and the diameter of the inner pipe is 5 cm, calculate the length of the heat exchanger.
Answer: Te = 70.9°C; ΔTlm = 39.5°C; L = 6.45 m
Log-Mean Temperature Difference Method
Log-Mean Temperature Difference Method
T1
T’1
T2
T’2
cp, food = 4.0 kJ/kg.°C
T’2 = 20°C
T’1 = 60°C.
mfood = 0.5 kg/s
T1 = 90°C
T2 = ?
mwater = 1 kg/s
cp, water = 4.18 kJ/kg.°C
U = 2,000 W/m2.°C
ID = 5 cm
.
.
1. Calculate the exit temperature of water.
Log-Mean Temperature Difference Method
q = !mHcpH (TH, inlet −TH, outlet ) = !mCcpC(TC, outlet −TC, inlet )(1)(4.18)(90 −T2 ) = (0.5)(4.0)(60 − 20)
(90 −T2 ) =(0.5)(4.0)(60 − 20)
4.18
T2 = 90 −(0.5)(4.0)(60 − 20)
4.18∴Exit temperature of water = 70.9°C
2. Calculate log-mean temperature difference.
Log-Mean Temperature Difference Method
ΔT1 = T1 −T '1ΔT1 = 90 − 60
ΔT1 = 30
ΔT2 = T2 −T '2ΔT2 = 70.9 − 20ΔT2 = 50.9
ΔTlm = ΔT2 − ΔT1ln ΔT2
ΔT1
ΔTlm = 50.9 − 30
ln 50.930
∴ΔTlm = 39.5°C
3. Calculate the length of the heat exchanger.
Log-Mean Temperature Difference Method
q =UAΔTlm
∴Length of exchanger = 6.45m
q =UπDiLΔTlm
L =!mcpΔT
UπDiΔTlm
L = (0.5)(4.0 ×103)(60 − 20)(2000)(3.142)(0.05)(39.5)
Repeat Example 1 for parallel-flow configuration? What can you conclude from two examples?
Log-Mean Temperature Difference Method Log-Mean Temperature Difference Method
T1
T’1
T2
T’2
cp, food = 4.0 kJ/kg.°C
T’2 = 60°C
T’1 = 20°C.
mfood = 0.5 kg/s
T1 = 90°C
T2 = ?
mwater = 1 kg/s
cp, water = 4.18 kJ/kg.°C
U = 2,000 W/m2.°C
ID = 5 cm
.
.
1. Calculate the exit temperature of water.
Log-Mean Temperature Difference Method
q = !mHcpH (TH, inlet −TH, outlet ) = !mCcpC(TC, outlet −TC, inlet )(1)(4.18)(90 −T2 ) = (0.5)(4.0)(60 − 20)
(90 −T2 ) =(0.5)(4.0)(60 − 20)
4.18
T2 = 90 −(0.5)(4.0)(60 − 20)
4.18∴Exit temperature of water = 70.9°C
2. Calculate log-mean temperature difference.
Log-Mean Temperature Difference Method
ΔT1 = T1 −T '1ΔT1 = 90 − 20
ΔT1 = 70
ΔT2 = T2 −T '2ΔT2 = 70.9 − 60ΔT2 = 10.9
ΔTlm = ΔT2 − ΔT1ln ΔT2
ΔT1
ΔTlm = 10.9 − 70
ln10.970
∴ΔTlm = 31.8°C
3. Calculate the length of the heat exchanger.
Log-Mean Temperature Difference Method
q =UAΔTlm
∴Length of exchanger = 8.01 m
q =UπDiLΔTlm
L =!mcpΔT
UπDiΔTlm
L = (0.5)(4.0 ×103)(60 − 20)(2000)(3.142)(0.05)(31.8)
Counter Flow Parallel Flow
Exit Temperate of Water 70.9°C 70.9°C
ΔTlm 39.5°C 31.8°C
Length of Heat Exchanger 6.45 m 8.01 m
Log-Mean Temperature Difference Method