Chapter 12 - Heat Exchanger
Chapter 12 - Heat Exchanger
Chapter 12 - Heat Exchanger
CHAPTER 12
HEAT EXCHANGER
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Lecturer: Dr. Phan Thanh Nhan
HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY – VNU HCM
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Lecturer: Dr. Phan Thanh Nhan
HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY – VNU HCM
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Lecturer: Dr. Phan Thanh Nhan
HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY – VNU HCM
Recuperators / regenerators
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Lecturer: Dr. Phan Thanh Nhan
HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY – VNU HCM
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Lecturer: Dr. Phan Thanh Nhan
HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY – VNU HCM
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Lecturer: Dr. Phan Thanh Nhan
HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY – VNU HCM
Q k.F.t k.F.t
F 7
Lecturer: Dr. Phan Thanh Nhan
HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY – VNU HCM
Q k.F.t k.F.t F
Q
F k.t
- k : overall heat transfer coefficient, assumed that is constant over the heat
transfer surface area F
- t : the mean temperature difference
define t ???
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Lecturer: Dr. Phan Thanh Nhan
HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY – VNU HCM
t max t min
t
t max
ln
t min
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Lecturer: Dr. Phan Thanh Nhan
HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY – VNU HCM
t t .t ng
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Lecturer: Dr. Phan Thanh Nhan
HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY – VNU HCM
t
t
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Lecturer: Dr. Phan Thanh Nhan
HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY – VNU HCM
t
t
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Lecturer: Dr. Phan Thanh Nhan
HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY – VNU HCM
Example: Double pipe heat exchanger with counter flow is working with:
+ The hot fluid is passed throught the inner tube of double pipe heat exchanger
with mass flow rate G1 = 180kg/min, specific heat capacity Cp1 = 3kJ/kgK,
inlet temperature t1’ = 95oC
+ The cold fluid enters the annular side at t2’ = 30oC and rise to t2’’ = 50oC
with a mass flow rate of G2 = 240kg/min, specific heat capacity Cp2 =
4,174kJ/kgK
+ The overall heat transfer coefficient k = 2400W/m2K
a) Determine heat transfer area of heat exchanger
b) If heat exchanger is swiched to parallel flow with the nochange all the required
parameters, calculate heat transfer area in this case and compare with the result
from (a)
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Lecturer: Dr. Phan Thanh Nhan
HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY – VNU HCM
Counter flow: G2 = 240kg/ph= 4kg/s Heat release from hot fluid:
Cp2 = 4,174kJ/kgK
Q1 = G1.Cp1.(t1’ – t1’’)
t2’ = 30oC
G1 = 180kg/ph = 3kg/s Heat absorb by cold fluid:
Cp1 = 3kJ/kgK Q2 = G2.Cp2.(t2’’ – t2’) =333,92kW
t1’ = 95oC
Energy balance:
Q = Q 1 = Q2
t1’’ = ?
t2’’ = 50oC G1.Cp1.(t1’ – t1’’) = G2.Cp2.(t2’’ – t2’)
t1’’=57,9oC
Log mean temperature difference
method:
t1’ = 95oC Q k.F.t 333,92kW
tmax =
45oC k = 2400 W/m2k = 2,4kW/m2K
t2’’ = 50oC tmax = 45oC
tmin = 27,9oC
t1’’ = 57,9oC
tmin =
27,9oC
Q k.F.t
t2’ = 30oC
F 3,8895(m2) 14
Lecturer: Dr. Phan Thanh Nhan
HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY – VNU HCM
Parralel flow:
Q k.F.t
F 5,1353(m2)
Result:
Counter flow: F = 3,8895m2
Parralel flow: F = 5,1353m2
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Lecturer: Dr. Phan Thanh Nhan
HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY – VNU HCM
f C, NTU
C min
C
C max
NTU k F F
C min
Apply the suitable chart (or use the right equation) to define
if know C and NTU, or define NTU if known and C
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Lecturer: Dr. Phan Thanh Nhan
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HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY – VNU HCM
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Lecturer: Dr. Phan Thanh Nhan
HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY – VNU HCM
Example: Consider a shell and tube heat exchanger construted from a 0,0254m
OD tube to cool 6,93kg/s of a 95% ethy alcohol solution (Cp = 3810J/kgK) from
66oC to 42oC, using 6,3kg/s of water available at 10oC (Cp = 4187J/kgK). In heat
exchanger, 72 tubes will be used. Assume that the overall coefficient of heat
transfer based on the outer tube area is 568W/m2K. Calculate the surface area and
the length of the heat exchanger for each of the following arrangements (apply the
-NTU method):
a) Parallel flow shell and tube heat exchanger
b) Counter flow shell and tube heat exchanger
c) Shell and tube heat exchanger with one shell pass and multiples of two tube
passes
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Lecturer: Dr. Phan Thanh Nhan
HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY – VNU HCM
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Lecturer: Dr. Phan Thanh Nhan