Heat exchanger
Heat exchangers are devices in which heat is transferred between two
fluids at different temperatures without any mixing of fluids.
Heat exchanger type
1. Direct heat transfer type
2. Storage type
3. Direct contact type
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
cold fluid
hot fluid
hot fluid
cold fluid
Concentric tube heat exchangers. (a) Parallel flow. (b) Counter flow.
Heat exchanger
2. Storage type heat transfer
A 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 exchanger
A 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.
Heat exchanger
Direct type heat exchanger
1. Tubular
2. Plate
3. Extended surface
Heat exchanger
Tubular heat exchanger
1. 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 exchanger
Series 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 exchanger
Fins 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 flow
o Counter flow
o Cross flow
Heat exchanger
Parallel flow
Counter flow
Heat exchanger
Cross flow
Both fluids unmixed
One fluid mixed and the
other unmixed
Heat exchanger
Overall heat transfer coefficient and fouling factor
Across a plain wall
1
1
b
1

 
U
h1 k h 2
Across a tubular surface
r  r 1
1
1
r

 i ln  o   i
U ri
hi
k  ri  r o h o
These equations are valid for clean surfaces.
Heat exchanger
Mean Temperature Difference
dA
hot
Th
Tc
cold
Where:
dq  U  TdA
 T  Th  Tc
Total heat transfer rate in heat exchanger
q 
 U  TdA
Heat exchanger
If U is assumed to be a constant
q  U   TdA
Define mean temperature difference
 Tm 
1
A
  TdA
area
Thus:
q  UA  T m
This is the basic performance equation for a direct transfer type heat
exchanger
Heat exchanger
Parallel Flow
Assumption
1. U is a constant
2. Heat exchanger is adequately
insulated i.e. no heat loss to
surrounding
Heat exchanger
Consider in elementary area dA (B.dx)
dq  U  T B dx
  m h C ph dT h
 m c C pc dT c
 T  Th  Tc
d (  T )  dT h  dT c
dq
dq
  
 
m h C ph
m c C pc

1
1
  U  T B dx  
 
m C
m c C pc
 h ph




Heat exchanger
 To

 Ti

d (T )
1
1
  
 
m C
T
m c C pc
 h ph
Where:

 UB


L
 dx
0
 Ti  T h ,i  T c ,i
 To  T h ,o  Tc ,o


  To 
1
1
 UA
    
 
ln 

m c C pc 
  Ti 
 m h C ph
1
  T h , i  T h , o  T c , o  T c , i  UA
q
Heat exchanger


 Ti   To
q  UA 

 Ti
ln

 To







This is the performance equation for a parallel-flow heat exchanger
Comparing with:
q  UA  T m
Where:
 Tm 
 Ti   To
 Ti
ln
 To
Heat exchanger
For counter flow
Assumption
1. U is a constant
2. Heat exchanger is adequately
insulated i.e. no heat loss to
surrounding
Heat exchanger
Consider an elementary area dA (B.dx)
dq  U ( T h  T c ) B dx
  m h C ph dT h
  m c C pc dT c
 T  Th  Tc
d (  T )  dT h  dT c
dq
dq
  
 
m h C ph
m c C pc

1
1
  U  T B dx  
 
m C
m c C pc
 h ph




Heat exchanger

d (T )
1
1
  
 
m C
m c C pc
T
 h ph

 UB



  To 
1
1
    
ln 
 
m C
m c C pc
  Ti 
 h ph

 UA


 To

 Ti
L
 dx
0
Where:
 Ti  T h ,i  T c ,o
 T o  T h ,o  T c ,i
  To 
1
   T h , i  T h , o  T c , o  T c , i  UA
ln 
q
  Ti 
Heat exchanger


 Ti   To
q  UA 

 Ti
ln

 To







Th,i
 Ti
Tc,o
Tc,i
Comparing with:
q  UA  T m
Where:
Th,o
 Tm 
 Ti   To
 Ti
ln
 To
 To
Heat exchanger
Special case of counter flow
m h c ph  m c c pc
Then:
T h ,i  T h ,o  T c ,o  T c ,i
Or
T h ,i  T c ,o  T h ,o  T c ,i
Substituting into the expression for
 Tm
, we get an indeterminate quantity
Heat exchanger
Define
  Ti 

  p

T
o 

Then:
 T m  lim
p1
 T e ( p  1)
ln p
Apply L’ Hopital’s rule
 T m  lim
p1
 T e (1)
  Te
1
p
 T m   Te   Ti
 Ti
 To
Heat exchanger
Cross flow
Case 1: both fluids unmixed
Hot fluid
Th,i
Tc,i
Tc,o
Cold fluid
B
y
x
Th,o
L
Heat exchanger
Both Th and Tc are functions of x and y
Considering an elementary area dA (= dx dy)
dq  U ( T h  T c ) dx dy
q 
B
 
0
L
( T h  T c ) dx dy
0
Comparing with
 Tm 
1
BL
q  UA  T m
B
 
0
L
0
( T h  T c ) dx dy
More complicated than before but it has been done.
Heat exchanger
Cross flow
Case 2: one fluid mixed, the other unmixed
Hot fluid
Th,i
Tc,i
Tc,o
Cold fluid
B
y
x
Th,o
Th = f (x,y)
Tc = f (x)
L
Heat exchanger
Cross flow
Case 2: one fluid mixed, the other unmixed
Hot fluid
Th,i
Tc,i
Tc,o
Cold fluid
B
y
x
Th,o
L
Th = f (y)
Tc = f (x)
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)
F
Tm cross flow
Tm if the arrengement was counter flow
If the bulk exit temperatures on the hot side and cold side are Th,o and Tc,o, then
Tm counter flow 
T  T   T
ln T  T  / T
h ,i
h ,i
c ,o
c ,o
h ,o
h ,o
 Tc ,i 
 Tc ,i 
Heat exchanger
Mean temperature difference in cross flow
F
Tm cross flow
Tm if the arrengement was counter flow
For given values of Th,i; Th,o; Tc,i; Tc,o
 T counter  flow
Therefore:
is the highest amongst all flow arrangements
0  F 1
q  UA Tm cross  flow
Heat exchanger
q  UA Tm cross  flow
q  UAFTm counter flow
F is plotted as a function of two parameters, R and S
R
T
T
 T1,o 
T
T
 T2,i 
1,i
2 ,o
S
2 ,o
1,i
 T2 ,i 
 T2,i 
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 = h
2=c
or
1=c
2=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
R0
The ratio of change in temperature of one of the fluid the fluids to the difference
of inlet temperature of the two fluids 0  S  1
Heat exchanger
T1,i
T2,i
T2,o
T1,o
R
T
T
1,i
2 ,o
 T1,o 
 T2 ,i 
Both fluids unmixed cross flow heat exchanger
S
T
T
2 ,o
1,i
 T2 ,i 
 T2 ,i 
Heat exchanger
T1,i
mixed
T2,i
T2,o
unmixed
R
T
T
1,i
2 ,o
T1,o
 T1,o 
 T2,i 
S
One fluids mixed and the other unmixed
T
T
2 ,o
1,i
 T2 ,i 
 T2 ,i 
Heat exchanger
The effectiveness - NTU method
Generally, we encounter two type of problems:
Given:
Type 1
Two fluids m h
T h ,i , T h ,o
Type 2
m c
U
Find A?
T c ,i , T c ,o
Given:
Two fluids A heat exchanger A
m h
T h ,i
m c
T c ,i
Find Th,o; Tc,o?
Heat exchanger
Type 1
q  UA  T m
A 
Type 2
?
q
U  Tm
?
q  UA  T m
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
T

q
Th ,i  Tc ,o
Th ,o
q max
Tc ,i
Length of heat exchanger
T h ,i
T h ,i


m h C ph ( T h , i  T h , o )
( m  Cp ) s ( T h , i  T c , i )
m c C pc ( T c , o  T c , i )
( m  Cp ) s ( T h , i  T c , i )
Heat exchanger
Hence if
m h C ph  m c C pc

Hence if
Note
T h ,i  T h ,o
T h ,i  T c ,i
m c C pc  m h C ph

, then m h C ph  m  C p s
, then m c C pc  m  C p s
T c ,o  T c ,i
T h ,i  T c ,i
1) The definition are equivalent when
2) By definition 0    1
m c C pc  m h C ph
Heat exchanger
Effectiveness – parallel flow
Assume

h

m C ph  ( m C p ) s

m h C ph 
1 




m
C
 T h ,i  T h ,o  
c
pc 

  

 

T
T
m
h
,
i
c
,
i

  1  h C ph 



m
c C pc 

 T  T h ,o   T c ,e  T c ,i 


   h , i
 T T 
T
T

c ,i 
c ,i 
 h ,i
 h ,i
q
 T c ,o  T c ,i
m c C pc

m h C ph
1 


m
c C pc





Heat exchanger
Effectiveness – parallel flow





T  Tc ,o
   1  h , o
T h ,i  T c ,i


m h C ph
1 

m c C pc





Derive earlier (slide 17)
 T h ,o  Tc ,o

 T T
c ,i
 h ,i


1
  exp   1


m C
m c C pc

 h ph

 UA


Heat exchanger
Substituting


 
m
C
   1  exp    1  h ph


m c C pc
 

If we had assumed initially

 UA




 
m c C pc  ( m  C p ) s


 
m
C
   1  exp    1  c pc


m h C ph
 





s
b




, then

 UA




 

m c C pc
1  

m h C ph


 UA




 

m C p
1 


m
Cp

We combine the two expressions for 

 
m C p

  1  exp    1 



m
Cp

 


m h C ph
1 


m
c C pc









s
b




Heat exchanger
Define two new parameters
Capacity ratio (C)
C
(m  C p )s
(m  C p ) b
or
C min
C max
Number of transfer unit (NTU)
NTU 
Note:
UA
(m  C p ) s
or
UA
C min
1) Both are dimensionless
2) 0    1
3) NTU  0
Heat exchanger
Effectiveness – parallel flow
1  exp  1  C NTU 

1  C 
Effectiveness – counter flow

1  exp  1  C NTU 
1  C exp  11  C NTU 
Heat exchanger
Special cases
Capacity rate (m.Cp) is infinite either on the hot side or the cold side
C 0
For this solution, we obtain the relation
  1  exp  NTU
Heat exchanger

Heat exchanger
Heat exchanger
Heat exchanger
Heat exchanger
Example
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.
Heat exchanger
LMTD
Toil,in  350 K
Solution, Part 1:
Toil,in  375 K
Cp,c  4,181 J / Kg.K
q  C h Th ,i  Th ,o 
and
Tc ,o  Tc ,i  q / C c
Twater,out  ?
Twater,in  280 K
 0.5  2,090  (375  350)
 26,125 W
 280  26,125 /(0.2  4,181)
 311 K
Heat exchanger
• For parallel flow,
Tm,PF 
 T i  95
To  39
Ti  To
95  39

 63
ln Ti / To ln(95 / 39)
• For counter flow,
Ti  64
 T i  95
To  39
To  70
Ti  64
Tm,CF 
Ti  To
64  70

 67
ln Ti / To ln(64 / 70)
To  70
Heat exchanger
• For parallel flow,
APF  q /(UTm,PF )  1.66 m2
• For counter flow,
ACF  q /(UTm,CF )  1.56 m2
Heat exchanger
Part 2:
• Use -NTU method to determine the required NTU and heat transfer
area for parallel and counter flow
Solution
Cp,c  4,181 J / Kg.K
To determine the minimum heat capacity rate,
mh Cph  0.5  2090  1045 W / K
mcCpc  0.2  4,181  836.2 W / K  (mCp )s
Heat exchanger
• Then
q max  Cmin (Th ,i  Tc,i )
 836.2  (375  280)  79,440 W
q  Ch Th,i  Th,o   26,125 W
• The effectiveness is
  q / qmax  26,125/ 79,440  0.33
With
C
( m C p ) s
( m C p ) b

836.2
 0.8
1,045
Heat exchanger
0.55
Parallel flow
APF  Cmin NTUPF / U  1.84 m2
ACF  CminNTUCF / U  1.67 m2
0.50
Counter flow
Heat exchanger
Other consideration in designing heat exchangers
1.
2.
3.
4.
5.
6.
7.
Pressure drop on either sides
Size restriction
Stress consideration
Servicing requirements
Materials of construction
System operation
Cost