DKK1493 HEAT TRANSFERDKK1493 HEAT TRANSFER
CHAPTER 4CHAPTER 4
HEAT EXCHANGERHEAT EXCHANGER
Heat exchanger type The Heat Exchanger Analysis: Log Mean
Temperature Difference (LMTD)
CONTENTSCONTENTS
It is expected that students will be able:to classify different types of heat exchangers to calculate the log mean temperature difference (LMTD) and determine correction factor for multi-passes heat exchanger.
TOPIC OUTCOMESTOPIC OUTCOMES
TYPE OF HEAT EXCHANGER
Double-pipe heat exchanger
Shell and Tube Heat Exchanger
1 shell pass I tube pass
(1-1 exchanger)
1 shell pass 2 tube passes
(1-2 exchanger)
Shell and tube exchangerShell inlet
Shell outlet
Tube inlet
Tube outlet
Shell inlet
Tube inlet
Tube outlet
Shell outlet
Cross-Flow Heat exchangers
One fluid mixed,one fluid unmixed
Both fluids unmixed
Gas flowGas flow
LOG MEAN TEMPERATURE DIFFERENCE (LMTD)
1'T
2'T1T
2T
Distance
T1
2T
Countercurrent
1T
1'T
2T
2'T
T2
T1
Distance
Co-current
T1
T’2
T2
T’1
T1
T’2
T2
T’1
Log mean temperature difference
Energy Balances and Heat Transfer Rate
Energy balance on hot fluid:
Energy balance on cool fluid:
Heat transfer rate:
ohhihp TTcmq
cicocp TTcmq
mTUAq
This Tlm holds for a double-pipe heat exchanger and 1-1 exchanger with 1 shell pass and 1 tube pass in parallel or counter-flow.
When the hot and cold fluids in a heat exchanger are in true counter-current flow or in co-current (parallel) flow,
LOG MEAN TEMPERATURE DIFFERENCE (LMTD)
2
1
21
1
2
12
lnlnTT
TT
TT
TTTlm
lmTUAq
Example
An oil which has a cpm = 2.30 kJ/kg.K is being cooled in a heat exchanger from 371.9 K to 349.7 K and flows inside the tube at a rate of 3630 kg/h. A flow of water enters at 288.6 K for cooling out and flows outside the tube at 319.1 K. Calculate the heat-transfer area if the U = 340 W/m2.K for each counter-current and co-current flow. Discuss your result.
(Ans: 2.66 m2; 2.87 m2)
For multiple-pass heat exchanger, the correction factor FT used to calculate the true mean temperature drop. (under assumption of counter-flow conditions)
Two dimensionless ratios are used as follows:
For counter-flow:
Correct mean temperature:
Heat transfer rate:
LOG MEAN TEMPERATURE DIFFERENCE (LMTD) CORRECTION FACTORS
ciho
cohi
cihocohilm
TTTT
TTTTT
ln
)(
Tlmm FTT
cico
hohi
TT
TTZ
cihi
cico
TT
TTY
moomii TAUTAUq
Correction factor FT for shell and tube heat exchanger with one shell pass and any multiple of two tube passes (2, 4, 6, etc, tube passes)
Correction factor FT for shell and tube heat exchanger with two shell pass and any multiple of four tube passes (4, 8, 12, etc, tube passes)
EXAMPLE 4.9-1: Temperature Correction Factor for a Heat Exchanger
A 1-2 heat exchanger containing one shell pass and two tubes passes heats 2.52 kg/s of water from 21.1 to 54.4oC by using hot water under pressure entering at 115.6 and leaving at 48.9oC. The outside surface area of the tubes in the exchanger is Ao = 9.30 m2
a) Calculate the mean temperature difference ΔTm in the exchanger and the overall heat-transfer coefficient Uo
b) For the same temperature but using a 2-4 exchanger, what would be the ΔTm ?
1T
2T
hiTC 15.61 o
C 9.48 ohoT
C 1.21 ociT
coTC 4.54 o
hot water
waterkg/s 2.52
kJ/kg.K 187.4waterpc
ExchangerHeat 21(a)
352.01.216.115
1.214.54
cihi
cico
TT
TTY
00.21.214.54
9.486.115
cico
hohi
TT
TTZ
1T
2T
hiTC 15.61 o
C 9.48 ohoT
C 1.21 ociT
coTC 4.54 o
hot water
waterkg/s 2.52
K 3.42
1.219.484.546.115
ln
1.219.484.546.115
ln2
1
21
TT
TTTlm
K 31.3C 31.3
)74.0(3.42 o
Tlmm FTT
0.74 4a,-4.9 Fig theFrom TF
(b) ExchangerHeat 42
W348200
)1.214.54)(4187(52.2
cicopm TTcmq
.K W/m1196)3.31(30.9
348200 2
mo
o TA
qU
94.0 4b,-4.9 Fig theFrom TF
K 39.8C 39.8
)94.0(3.42 o
Tlmm FTT
PROBLEM 4.9-2: Cooling Oil by Water in an Exchanger
Oil flowing at the rate of 5.04 kg/s (cpm = 2.09 kJ/kg.K) is cooled in a 1-2 heat exchanger from 366.5 K to 344.3 K by 2.02 kg/s of water entering at 283.2 K. The overall heat transfer coefficient Uo is 340 w/m2.K. Calculate the area required. (Hint: A heat balance must first be made to determine the outlet water temperature)