Heat exchanger - Suranaree University of...
Transcript of Heat exchanger - Suranaree University of...
Heat exchangers are devices in which heat is transferred between two fluids at different temperatures without any mixing of fluids.
Heat exchanger type
Heat exchanger
1. Direct heat transfer type2. Storage type3. Direct contact type
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Heat exchanger
1. Direct heat transfer type
A direct transfer type of heat exchanger is one in which the cold and hot fluids flow simultaneously through the device and heat is transferred through a wall separating the fluids
Concentric tube heat exchangers. (a) Parallel flow. (b) Counter flow.
hot fluid hot fluid
cold fluid
cold fluid
Heat exchanger
2. Storage type heat exchangerA direct transfer type of heat exchanger is one in which the heat transfer from the hot fluid and the cold fluid occur though a coupling medium in the form of porous solid matrix. The hot and cold fluids alternatively through the matrix. The hot fluid storing heat in it and the cold fluid extracting heat from it.
Heat exchanger
3. Direct contact type heat exchangerA direct transfer type of heat exchanger is one in which the two fluids are not separated. If heat is to be transferred between a gas and a fluid, the gas is either bubbled through the liquid or the liquid is sprayed in the form of droplets in the gas.
Direct type heat exchanger1. Tubular 2. Plate3. Extended surface
Heat exchanger
Tubular heat exchanger1. Concentric tube 2. Shell and tube
Concentric tube Shell and tube
The heat transfer area available per unit volume 100 -500 m2/m3
Heat exchanger
Plate heat exchangerSeries of large rectangular thin metal plates which are clamped together to form narrow parallel-plate channel.
The heat transfer area available per unit volume 100-200 m2/m3
Heat exchanger
Extended surface heat exchangerFins attached on the primary heat transfer surface with the object of increasing the heat transfer area.
The heat transfer area available per unit volume 700 m2/m3
Heat exchanger
Classification by flow arrangement
The three basic flow arrangements:
o Parallel flowo Counter flowo Cross flow
Heat exchanger
Parallel flow Counter flow
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Cross flow
Both fluids unmixed One fluid mixed and the other unmixed
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Overall heat transfer coefficient (U) and fouling factor
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In a heat exchanger, the heat is transferred by both convection and conduction.
q = UA (Ta –Tb)
U is overall heat transfer coefficient Tb
Ta
hohi
T1
T2
Heat exchanger
Tb
Ta
hohi
T1
T2
1ai
qT T
h A
1 2
qT T
k A
x
2 bo
qT T
h A
a b
qT T
U A
Conv.
Cond.
Conv.
Plate heat exchanger 1)
2)
3)
4)
Heat exchanger
a bi o
q q x qT T
h A kA h A
a b
qT T
U A
i o
q q q x q
U A h A kA h A
1 1 1
i o
x
U h k h
Adding above three equations (1, 2, 3)
From (4)
Across a plain wall
21 h
1
k
b
h
1
U
1
Heat exchanger
r2
r1Tb
Ta
R i R a R o
T i To
To
Ti
cond
xR
kA
2
1conv
o
Rh A
RT = Rconv1 + Rcond + Rconv2
1
1conv
i
Rh A
Tubular heat exchanger
Heat exchanger
Ai = 2πriL
oolm
io
iiii Ah
1
kA
)rr(
Ah
1
AU
1
oo
i
lm
iio
ii rh
r
kr
r)rr(
h
1
U
1
)r/rln(
)rr(r
io
iolm
โดย
Based on inner area
oolm
io
iioo Ah
1
kA
)rr(
Ah
1
AU
1
Based on outter area
olm
oio
ii
o
o h
1
kr
r)rr(
rh
r
U
1
Heat exchanger
oi h
1
h
1
U
1
When the wall thickness of the tube is small and the thermal conductivity of the tube material is high, is as usually the case, the thermal resistance of the tube is negligible.
oi AA because
Heat exchanger
FoulingThe performance of heat exchangers usually deteriorate with time as a result of accumulation of deposits on heat transfer surfaces, representing additional resistance, called fouling.
(Log) Mean Temperature Difference
Heat exchanger
hot
Where:
cold
dA Th
Tc
TdAUdq ch TTT
Total heat transfer rate in heat exchanger
TdAUq
Heat exchanger
If U is assumed to be a constant TdAUq
Define mean temperature difference
area
m TdAA
1T
Thus:
mTUAq
This is the basic performance equation for a direct transfer type heat exchanger
Parallel Flow
Heat exchanger
Assumption1. U is a constant2. Heat exchanger is adequately
insulated i.e. no heat loss to surrounding
Consider in elementary area dA (B.dx)
Heat exchanger
dxBTUdq
hphh dTCm
cpcc dTCm
ch TTT
ch dTdT)T(d
pccphh Cm
dq
Cm
dq
pccphh Cm
1
Cm
1dxBTU
Heat exchanger
L
0pccphh
T
T
dxUBCm
1
Cm
1
T
)T(do
i
Where: i,ci,hi TTT
o,co,ho TTT
UACm
1
Cm
1
T
Tln
pccphhi
o
UATTTTq
1i,co,co,hi,h
Heat exchanger
o
i
oi
TT
ln
TTUAq
This is the performance equation for a parallel-flow heat exchanger
mTUAq
Comparing with:
o
i
oim
TT
ln
TTT
Where:
Heat exchanger
For counter flowAssumption
1. U is a constant2. Heat exchanger is adequately
insulated i.e. no heat loss to surrounding
Consider an elementary area dA (B.dx)
Heat exchanger
dxB)TT(Udq ch
hphh dTCm
cpcc dTCm
ch TTT
ch dTdT)T(d
pccphh Cm
dq
Cm
dq
pccphh Cm
1
Cm
1dxBTU
Heat exchanger
L
0pccphh
T
T
dxUBCm
1
Cm
1
T
)T(do
i
UACm
1
Cm
1
T
Tln
pccphhi
o
UATTTTq
1
T
Tln i,co,co,hi,h
i
o
Where:o,ci,hi TTT
i,co,ho TTT
Heat exchanger
o
i
oi
TT
ln
TTUAq
Th,i
Th,oTc,o
Tc,ioT
iT
mTUAq
Comparing with:
o
i
oim
TT
ln
TTT
Where:
Heat exchanger
pccphh cmcm
Special case of counter flow
mT
Then:
i,co,co,hi,h TTTT
i,co,ho,ci,h TTTT Or
Substituting into the expression for , we get an indeterminate quantity
Heat exchanger
pT
T
o
i
Define
Then:
Apply L’ Hopital’s rulepln
)1p(TlimT o
1pm
oo
1pm T
p1
)1(TlimT
iom TTT
iT
oT
Heat exchanger
Cross flowCase 1: both fluids unmixed
Cold fluid
Hot fluid
Th,i
Th,o
Tc,i Tc,o
B
L
x
y
Heat exchanger
Cross flowCase 2: one fluid mixed, the other unmixed
Cold fluid
Hot fluid
Th,i
Th,o
Tc,i Tc,o
B
L
x
y
Th = f (x,y)Tc = f (x)
Heat exchanger
Cross flowCase -: Both fluid mixed
Cold fluid
Hot fluid
Th,i
Th,o
Tc,i Tc,o
B
L
x
y
Th = f (y)Tc = f (x)
Heat exchanger
Both Th and Tc are functions of x and y
Considering an elementary area dA (= dx dy)
dydx)TT(Udq ch
dydx)TT(UqB
0
L
0 ch
Comparing with mTUAq
dydx)TT(BL
1T
B
0
L
0 chm
More complicated than before but it has been done.
Heat exchanger
The integration of the three cases of cross flow has been done numerically. The results are presented in the form of a correction factor (F)
flowcounterwastarrengementheifm
flowcrossm
T
TF
If the bulk exit temperatures on the hot side and cold side are Th,o and Tc,o, then
i,co,ho,ci,h
i,co,ho,ci,hflowcounterm TT/TTln
TTTTT
if the arrangement was encounter-flow
Heat exchanger
Mean temperature difference in cross flow
flowcounterwastarrengementheifm
flowcrossm
T
TF
For given values of Th,i; Th,o; Tc,i; Tc,o
is the highest amongst all flow arrangements flowcounterT
Therefore: 1F0
Heat exchanger
flowcrossmflowcross TUAq
flowcountermflowcross TUAFq
F is plotted as a function of two parameters, R and S
i,2o,2
o,1i,1
TT
TTR
i,2i,1
i,2o,2
TT
TTS
Heat exchanger
Subscripts 1 and 2 correspond to the two fluids
For case: 1 (both fluids unmixed) and case 3 (both fluids mixed)
It is immaterial which subscript corresponds to the hot side and which to the cold side.
For case: 1 and case 3
Subscripts 1 = h2 = c
or 1 = c2 = h
Heat exchanger
However for case 2, care must be taken to see that the mixed fluid has subscript 1
What is the parameter R?
The ratio of change of temperature of the two fluids 0R
The ratio of change in temperature of one of the fluid to the difference of inlet temperature of the two fluids 1S0
What is the parameter S?
Heat exchanger
i,2o,2
o,1i,1
TT
TTR
i,2i,1
i,2o,2
TT
TTS
i,1T
o,1T
i,2T o,2T
Both fluids unmixed cross flow heat exchanger
Heat exchanger
i,1T
o,1T
i,2T o,2T
i,2i,1
i,2o,2
TT
TTS
i,2o,2
o,1i,1
TT
TTR
One fluids mixed and the other unmixed
mixed
unmixed
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The effectiveness - NTU methodGenerally, we encounter two type of problems:
Given:
hm
cm
o,hi,h T,T o,ci,c T,T
UTwo fluids Find A?
Type 1
Given:
hm
i,hT i,cT
Two fluids A heat exchanger A
Type 2
cm
Find Th,o; Tc,o?
Heat exchanger
Type 1
mTUAq
mTU
qA
Type 2
mTUAq
We will need a trial and error approach to solve this type of problem i.e. assuming Th,o
??
Trial and error can be avoided if we adopt the alternative method called the effectiveness -NUT
Heat exchanger
Effectiveness of a heat exchanger =Rate of heat transfer in heat exchanger
Maximum possible heat transfer rate
maxq
q
)TT()Cpm(
)TT(Cm
i,ci,hs
o,hi,hphh
)TT()Cpm(
)TT(Cm
i,ci,hs
i,co,cpcc
i,hT
i,cT
T
o,hT
i,cT
o,ci,h TT
Length of heat exchanger
Heat exchanger
Hence if , then pccphh CmCm
spphh CmCm
i,ci,h
o,hi,h
TT
TT
Hence if , then phhpcc CmCm
sppcc CmCm
i,ci,h
i,co,c
TT
TT
Note 1) The definition are equivalent when phhpcc CmCm
2) By definition 10
Heat exchanger
Effectiveness – parallel flow
Assume spphh )Cm(Cm
i,ci,h
o,hi,h
TT
TT
pcc
phh
pcc
phh
Cm
Cm1
Cm
Cm1
pcc
phh
i,ci,h
i,co,c
i,ci,h
o,hi,h
Cm
Cm1
TT
TT
TT
TT
pcc Cm
q i,co,c TT
pcc
phh
o,hi,h
i,co,c
Cm
Cm1
)TT(
)TT(1
Heat exchanger
Effectiveness – parallel flow
pcc
phh
i,ci,h
o,co,h
Cm
Cm1
TT
TT1
UACm
1
Cm
1exp
TT
TT
pccphhi,ci,h
o,co,h
Derived earlier (slide 23)
UACm
1
Cm
1
T
Tln
pccphhi
o
Heat exchanger
pcc
phh
phhpcc
phh
Cm
Cm1
CmUA
Cm
Cm1exp1
Substituting
Heat exchanger
If we had assumed initiallysppcc )Cm(Cm
phh
pcc
pccphh
pcc
Cm
Cm1
CmUA
Cm
Cm1exp1
, then
We combine the two expressions for
bp
sp
spbp
sp
Cm
Cm1
CmUA
Cm
Cm1exp1
Heat exchanger
Define two new parameters
Capacity ratio (C)
max
min
bp
sp
C
Cor
)Cm(
)Cm(C
Number of transfer unit (NTU)
minsp C
UAor
)Cm(
UANTU
Note: 1) Both are dimensionless2)3) NTU
10 0
Heat exchanger
Effectiveness – parallel flow
C1
NTUC1exp1
Effectiveness – counter flow
NTUC11expC1
NTUC1exp1
Heat exchanger
Special cases
Capacity rate (m.Cp) is infinite either on the hot side or the cold side
0C
For this solution, we obtain the relation
NTUexp1
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Situation:
Light lubricating oil (Cp=2090 J/kg-K) is cooled with water in a small heat exchanger.
Oil flow = 0.5 kg/s, inlet T = 375 K
Water flow = 0.2 kg/s, inlet T = 280 K
Part 1:
If desired outlet temperature of the oil is 350 K, and you know the estimated overall heat transfer coefficient, U = 250 W/m²-K, from manufacturer’s data for this type of heat exchanger
Find: Required heat transfer area for a parallel flow heat exchanger and compare to the area needed for a counter flow heat exchanger.
Example
Heat exchanger
LMTD
Solution, Part 1:K375T in,oil
K350T in,oil
K280T in,water
?T out,water K.Kg/J181,4C c,p
C/qTT
and
TT Cq
ci,co,c
o,hi,hh
W125,26
)350375(090,25.0
K311
)181,42.0/(125,26280
Heat exchanger
• For parallel flow, 95Ti 39To
63)39/95ln(
3995
T/Tln
TTT
oi
oiPFm,
• For counter flow, 64Ti 70To
67)70/64ln(
7064
T/Tln
TTT
oi
oiCFm,
95Ti 39To
70To
64Ti
Heat exchanger
• For parallel flow,
• For counter flow,
2PF,mPF m66.1)TU/(qA
2CF,mCF m56.1)TU/(qA
Heat exchanger
Part 2:• Use -NTU method to determine the required NTU and heat transfer
area for parallel and counter flowSolution
K.Kg/J181,4C c,p
K/W104520905.0Cm phh
sppcc )Cm(K/W2.836181,42.0Cm
To determine the minimum heat capacity rate,
Heat exchanger
• Then
W440,79)280375(2.836
)TT(Cq i,ci,hminmax
W26,125 TT Cq o,hi,hh
• The effectiveness is
33.0440,79/125,26q/q max
With 8.0
045,1
2.836
)Cm(
)Cm(C
bp
sp
Heat exchanger
Parallel flow Counter flow
2PFminPF m84.1U/NTUCA
2CFminCF m67.1U/NTUCA
0.55 0.50
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Other consideration in designing heat exchangers
1. Pressure drop on either sides2. Size restriction3. Stress consideration4. Servicing requirements5. Materials of construction6. System operation7. Cost