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Dr. JITENDRA B. CHAUDHARI,
Assistant Professor
DEPARTMENT OF MECHANICAL ENGINEERING
R. N. G. PATEL INSTITUTE OF TECHNOLOGY-RNGPIT, Bardoli
[Formerly known as FETR]
Email: jbc.fetr@gmail.com, M: 960 156 4684
HEAT TRANSFER
SUBJECT CODE: 2151909 | B.E. 5th SEMESTER
Topic Title: Overall Heat Transfer Coefficient And LMTD
M. Tech, (Turbo-Machines) Ex. Design Engineer, Hindustan Aeronautics Ltd. HAL
Ph.D., (Combustion Technology) Ex. Production Engineer, Suzlon Energy Ltd.
Lecture: 03
A Typical Heat Exchanger
Hot Fluid flows
through inner
tube
Cold Fluid flows
through outer
tube
Heat transfer
takes place from
Hot fluid to cold
fluid through
inner tube
thickness
Hot Fluid In
Hot Fluid Out
Cold Fluid In
Cold Fluid Out
A Typical Heat Exchanger
Q
Q
Hot Fluid flows
through inner tube
Cold Fluid flows
through outer tube
Heat transfer takes
place from Hot fluid
to cold fluid through
inner tube thickness
Inlet Section Outlet Section
ΔTin
ΔTout
LMTD Special Cases
1. In case of Counter Flow HX
Which is Indeterminate, So applying L Hospital’s Rule
co
T
hi
T
ho
T
ci
T
 
0
0
ln








o
i
o
i
m
T
T
T
T
T
o
i T
T 


o
T

i
T

ci
ho
o
co
hi
i
T
T
T
T
T
T






 
o
i
T
T
o
i
o
i
T
T
T
T
T
T
o
i














1 ln
o
i
m T
T
T 



 o
i
m T
UA
T
UA
T
UA
Q 





LMTD Special Cases
Inlet Outlet Inlet Outlet
CONDENSER EVAPORATOR
 
o
i
o
i
m
T
T
T
T
T







ln
co
h
o
ci
h
i
T
T
T
T
T
T






i
T

c
ho
o
c
hi
i
T
T
T
T
T
T






o
T

o
T

i
T

co
T
h
T
ho
T
h
T
ci
T
hi
T
c
T
c
T
• Determine
experimentally by
testing HX in clean
and dirty condition
Determination of Factor
• TEMA-
clean
dirty
f
U
U
R
1
1


The Tubular Exchanger Manufacturers
Association (also known as TEMA)
is an association of fabricators of
shell and tube type heat
exchangers.
TEMA has established and maintains a
set of construction standards for
heat exchangers, known as the
TEMA Standard..
TEMA was founded in 1939, and is
based in Tarrytown, New York
Overall Heat Transfer Co-efficient- U















0
1
1
1
h
k
L
h
U
i















0
1
1
1
h
k
L
h
U i
• Overall HT Coefficient
Ignoring wall thickness,
• Considering
Wall Thickness
• Considering
Scaling effect
OR
fouling factor

























i
i
o
i
o
o
o
o h
r
r
r
r
k
r
h
U
1
ln
1
1

























o
o
i
i
o
i
i
i h
r
r
r
r
k
r
h
U
1
ln
1
1
Ao = 2лroL,
Ai = 2лriL
Ao = Ai = A



































i
i
o
fi
i
o
i
o
o
fo
o
o h
A
A
R
A
A
r
r
kL
A
R
h
U
1
ln
2
1
1



































o
o
i
fo
o
i
i
o
i
fi
i
i h
A
A
R
A
A
r
r
kL
A
R
h
U
1
ln
2
1
1
m
T
UA
Q 

Steps to Solve LMTD
Area of HX?
No of Tubes?
Temperature?
Mass of Fluid?
Rate of Heat Transfer?
Type of HX?
 
 
fg
f
ci
co
pc
c
ho
hi
ph
h
h
m
Q
T
T
C
m
Q
T
T
C
m
Q





dl
n
A 

Counter
co
T
hi
T
ho
T
ci
T
Parallel
co
T
hi
T
ho
T
ci
T
Inlet Outlet
CONDENSER
co
T
ci
T
h
T h
T
Inlet Outlet
EVAPORATOR
hi
T
ho
T
C
T
C
T
 
o
i
o
i
m
T
T
T
T
T







ln
m
T
UA
Q 

Steps to Solve LMTD
Counter
co
T
hi
T
ho
T
ci
T
Parallel
co
T
hi
T
ho
T
ci
T
co
ho
o
ci
hi
i
T
T
T
T
T
T






i
T
 o
T

i
T

o
T

ci
ho
o
co
hi
i
T
T
T
T
T
T






 
o
i
o
i
m
T
T
T
T
T







ln
Steps to Solve LMTD
Parallel
co
T
hi
T
ho
T
ci
T
co
ho
o
ci
hi
i
T
T
T
T
T
T






i
T
 o
T

1. Find all 4 temperatures
   
ci
co
pc
c
ho
hi
ph
h T
T
C
m
T
T
C
m
Q 



2. Find LMTD
 
o
i
o
i
m
T
T
T
T
T







ln
m
T

3. Find Heat Transfer
m
T
UA
Q 

dl
n
A 

4. No of Tubes
Careful about Units
   
ci
co
pc
c
ho
hi
ph
h T
T
C
m
T
T
C
m
Q 



kW
s
kJ
K
kgK
kJ
s
kg


*
*
m
T
UA
Q 

K
m
K
m
W
W *
* 2
2

U
C ph
LMTD-Ex-01
• In a counter flow heat exchanger, 2.78 kg/s of an oil having specific
heat of 2.095 kJ/kg K is cooled from 800C to 500C by 2.095 kg/s of
water entering at 250C.
• Determine the heat exchanger area. Take U=300 W/m2K and Cp
=4.187 kJ/kg K.
s
kg
m
s
kg
m
C
h
/
095
.
2
/
78
.
2


kgK
kJ
C
kgK
kJ
C
pc
ph
/
186
.
4
/
095
.
2

 C
T
C
T ho
hi
0
0
50
,
80 

?
,
250

 co
ci T
C
T
K
m
W
U 2
/
300

m
T
UA
Q 
 ?

A
 
 
ci
co
pc
c
ho
hi
ph
h
T
T
C
m
Q
T
T
C
m
Q




LMTD-Ex-01
• In a counter flow heat exchanger, 2.78 kg/s of an oil having specific heat of 2.095 kJ/kg K
is cooled from 800C to 500C by 2.095 kg/s of water entering at 250C.
• Determine the heat exchanger area. Take U=300 W/m2K and Cp =4.187 kJ/kg K.
   
ci
co
pc
c
ho
hi
ph
h T
T
C
m
T
T
C
m
Q 



Heat Lost by hot fluid
Heat Gain by Cold fluid
C
T
C
T ho
hi
0
0
50
,
80 

?
,
250

 co
ci T
C
T
   
25
*
186
.
4
*
095
.
2
50
80
*
095
.
2
*
78
.
2 

 co
T
C
Tco
0
8
.
43

LMTD-Ex-01
  27
.
30
25
2
.
36
ln
25
2
.
36



 m
T
m
T
UA
Q 

C
T
C
T ho
hi
0
0
50
,
80 

C
T
C
T co
ci
0
0
8
.
43
,
25 

Counter
co
T
hi
T
ho
T
ci
T
i
T

o
T

8
.
43
80



 co
hi
i T
T
T
25
50



 ci
ho
o T
T
T
  kW
T
T
C
m
Q ho
hi
ph
h 7
.
174



27
.
30
*
*
300
1000
*
174 A

2
16
.
19 m
A 
LMTD-Ex-02
• Stream of hot and cold fluid running through parallel flow heat
exchanger, are 0.2 kg/s and 0.5 kg/s respectively. The inlet
temperature of hot and cold fluid are 750C & 200C, respectively. The
exit temperature of hot fluid is 450C, If overall heat transfer
coefficient on both side are 650 W/m2K.
• Determine the heat exchanger area. Cp =4.187 kJ/kg K.
s
kg
m
s
kg
m
C
h
/
5
.
0
/
2
.
0


kgK
kJ
C
kgK
kJ
C
pc
ph
/
187
.
4
/
187
.
4

 C
T
C
T ho
hi
0
0
45
,
75 

?
,
200

 co
ci T
C
T
c
h h
h
U
1
1
1


m
T
UA
Q 
 ?

A
Parallel
co
T
hi
T
ho
T
ci
T  















o
i
o
i
m
T
T
T
T
T
ln
   
ci
co
pc
c
ho
hi
ph
h T
T
C
m
T
T
C
m
Q 



   
ci
co
c
ho
hi
h T
T
m
T
T
m 


   
20
5
.
0
45
75
2
.
0 

 co
T
C
Tco
0
32

i
T
 o
T

55
20
75 




 ci
hi
i T
T
T
13
32
45 




 co
ho
o T
T
T   C
Tm
0
11
.
29
13
55
ln
13
55










c
h h
h
U
1
1
1


m
T
UA
Q 

Parallel
co
T
hi
T
ho
T
ci
T
 
kW
T
T
C
m
Q ho
hi
ph
h
122
.
25
)
45
75
(
*
187
.
4
*
2
.
0





i
T
 o
T

c
h h
h
U
1
1
1


m
T
UA
Q 

K
m
W
U 2
325

650
1
650
1
1


U
18
.
29
*
*
325
1000
*
122
.
25 A

m
T
UA
Q 

2
65
.
2 m
A 
LMTD-Ex-3
• Saturated steam at 120OC is condensed with the help of water (inlet
temperature 200C, outlet temperature 900C, Cp =4.187 kJ/kg K) in
surface condenser. The overall heat transfer coefficient U=1800
W/m2K .
• Determine the heat exchanger area & mass of condensate.
• Take hfg =2200 kJ/kg and mass of water=16.67 kg/min.
min
/
67
.
16
/
?
kg
m
s
kg
m
C
h

 C
T
T
T h
ho
hi
0
120


C
T
C
T co
ci
0
0
90
,
20 

K
m
W
U 2
/
1800

m
T
UA
Q 
 ?

A Inlet Outlet
CONDENSER
co
T
ci
T
h
T h
T
 
ci
co
pc
c
fg
stem T
T
C
m
h
m
Q 


Inlet Outlet
CONDENSER
co
T
ci
T
h
T h
T
• Saturated steam at 120OC is condensed with the help of water (inlet temperature 200C, outlet
temperature 900C, Cp =4.187 kJ/kg K) in surface condenser. The overall heat transfer
coefficient U=1800 W/m2K .
• Determine the heat exchanger area & mass of condensate.
• Take hfg =2200 kJ/kg and mass of water=16.67 kg/min.
 
o
i
o
i
m
T
T
T
T
T







ln
m
T
UA
Q 

100
20
120 




 ci
h
i T
T
T
30
90
120 




 co
h
o T
T
T
• Oil 0.27 kg/s, Cp =2.09 kJ/kg, cool from 80OC to 40OC using coolent
with flow rate of 0.27 kg/s, Cp =4.18 kJ/kg at 300C.
• Give your choice for HX.
• Determine the heat exchanger area .
• Take U =24 W/m2K
ho
co T
T 
Parallel
co
T
hi
T
ho
T
ci
T
Counter
co
T
hi
T
ho
T
ci
T
To check given HX is
Counter or Parallel
Compare Hot &
Cold Exit Temp.
IF,
HX is Counter Flow Type
ho
co T
T 
IF,
HX is Parallel Flow Type
THANK YOU

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Worked Example_LMTD.pptx

  • 1. Dr. JITENDRA B. CHAUDHARI, Assistant Professor DEPARTMENT OF MECHANICAL ENGINEERING R. N. G. PATEL INSTITUTE OF TECHNOLOGY-RNGPIT, Bardoli [Formerly known as FETR] Email: jbc.fetr@gmail.com, M: 960 156 4684 HEAT TRANSFER SUBJECT CODE: 2151909 | B.E. 5th SEMESTER Topic Title: Overall Heat Transfer Coefficient And LMTD M. Tech, (Turbo-Machines) Ex. Design Engineer, Hindustan Aeronautics Ltd. HAL Ph.D., (Combustion Technology) Ex. Production Engineer, Suzlon Energy Ltd. Lecture: 03
  • 2. A Typical Heat Exchanger Hot Fluid flows through inner tube Cold Fluid flows through outer tube Heat transfer takes place from Hot fluid to cold fluid through inner tube thickness Hot Fluid In Hot Fluid Out Cold Fluid In Cold Fluid Out
  • 3. A Typical Heat Exchanger Q Q Hot Fluid flows through inner tube Cold Fluid flows through outer tube Heat transfer takes place from Hot fluid to cold fluid through inner tube thickness Inlet Section Outlet Section ΔTin ΔTout
  • 4. LMTD Special Cases 1. In case of Counter Flow HX Which is Indeterminate, So applying L Hospital’s Rule co T hi T ho T ci T   0 0 ln         o i o i m T T T T T o i T T    o T  i T  ci ho o co hi i T T T T T T         o i T T o i o i T T T T T T o i               1 ln o i m T T T      o i m T UA T UA T UA Q      
  • 5. LMTD Special Cases Inlet Outlet Inlet Outlet CONDENSER EVAPORATOR   o i o i m T T T T T        ln co h o ci h i T T T T T T       i T  c ho o c hi i T T T T T T       o T  o T  i T  co T h T ho T h T ci T hi T c T c T
  • 6. • Determine experimentally by testing HX in clean and dirty condition Determination of Factor • TEMA- clean dirty f U U R 1 1   The Tubular Exchanger Manufacturers Association (also known as TEMA) is an association of fabricators of shell and tube type heat exchangers. TEMA has established and maintains a set of construction standards for heat exchangers, known as the TEMA Standard.. TEMA was founded in 1939, and is based in Tarrytown, New York
  • 7. Overall Heat Transfer Co-efficient- U                0 1 1 1 h k L h U i                0 1 1 1 h k L h U i • Overall HT Coefficient Ignoring wall thickness, • Considering Wall Thickness • Considering Scaling effect OR fouling factor                          i i o i o o o o h r r r r k r h U 1 ln 1 1                          o o i i o i i i h r r r r k r h U 1 ln 1 1 Ao = 2лroL, Ai = 2лriL Ao = Ai = A                                    i i o fi i o i o o fo o o h A A R A A r r kL A R h U 1 ln 2 1 1                                    o o i fo o i i o i fi i i h A A R A A r r kL A R h U 1 ln 2 1 1
  • 8. m T UA Q   Steps to Solve LMTD Area of HX? No of Tubes? Temperature? Mass of Fluid? Rate of Heat Transfer? Type of HX?     fg f ci co pc c ho hi ph h h m Q T T C m Q T T C m Q      dl n A  
  • 9. Counter co T hi T ho T ci T Parallel co T hi T ho T ci T Inlet Outlet CONDENSER co T ci T h T h T Inlet Outlet EVAPORATOR hi T ho T C T C T   o i o i m T T T T T        ln m T UA Q  
  • 10. Steps to Solve LMTD Counter co T hi T ho T ci T Parallel co T hi T ho T ci T co ho o ci hi i T T T T T T       i T  o T  i T  o T  ci ho o co hi i T T T T T T         o i o i m T T T T T        ln
  • 11. Steps to Solve LMTD Parallel co T hi T ho T ci T co ho o ci hi i T T T T T T       i T  o T  1. Find all 4 temperatures     ci co pc c ho hi ph h T T C m T T C m Q     2. Find LMTD   o i o i m T T T T T        ln m T  3. Find Heat Transfer m T UA Q   dl n A   4. No of Tubes
  • 12. Careful about Units     ci co pc c ho hi ph h T T C m T T C m Q     kW s kJ K kgK kJ s kg   * * m T UA Q   K m K m W W * * 2 2  U C ph
  • 13. LMTD-Ex-01 • In a counter flow heat exchanger, 2.78 kg/s of an oil having specific heat of 2.095 kJ/kg K is cooled from 800C to 500C by 2.095 kg/s of water entering at 250C. • Determine the heat exchanger area. Take U=300 W/m2K and Cp =4.187 kJ/kg K. s kg m s kg m C h / 095 . 2 / 78 . 2   kgK kJ C kgK kJ C pc ph / 186 . 4 / 095 . 2   C T C T ho hi 0 0 50 , 80   ? , 250   co ci T C T K m W U 2 / 300  m T UA Q   ?  A
  • 14.     ci co pc c ho hi ph h T T C m Q T T C m Q     LMTD-Ex-01 • In a counter flow heat exchanger, 2.78 kg/s of an oil having specific heat of 2.095 kJ/kg K is cooled from 800C to 500C by 2.095 kg/s of water entering at 250C. • Determine the heat exchanger area. Take U=300 W/m2K and Cp =4.187 kJ/kg K.     ci co pc c ho hi ph h T T C m T T C m Q     Heat Lost by hot fluid Heat Gain by Cold fluid C T C T ho hi 0 0 50 , 80   ? , 250   co ci T C T     25 * 186 . 4 * 095 . 2 50 80 * 095 . 2 * 78 . 2    co T C Tco 0 8 . 43 
  • 15. LMTD-Ex-01   27 . 30 25 2 . 36 ln 25 2 . 36     m T m T UA Q   C T C T ho hi 0 0 50 , 80   C T C T co ci 0 0 8 . 43 , 25   Counter co T hi T ho T ci T i T  o T  8 . 43 80     co hi i T T T 25 50     ci ho o T T T   kW T T C m Q ho hi ph h 7 . 174    27 . 30 * * 300 1000 * 174 A  2 16 . 19 m A 
  • 16. LMTD-Ex-02 • Stream of hot and cold fluid running through parallel flow heat exchanger, are 0.2 kg/s and 0.5 kg/s respectively. The inlet temperature of hot and cold fluid are 750C & 200C, respectively. The exit temperature of hot fluid is 450C, If overall heat transfer coefficient on both side are 650 W/m2K. • Determine the heat exchanger area. Cp =4.187 kJ/kg K. s kg m s kg m C h / 5 . 0 / 2 . 0   kgK kJ C kgK kJ C pc ph / 187 . 4 / 187 . 4   C T C T ho hi 0 0 45 , 75   ? , 200   co ci T C T c h h h U 1 1 1   m T UA Q   ?  A
  • 17. Parallel co T hi T ho T ci T                  o i o i m T T T T T ln     ci co pc c ho hi ph h T T C m T T C m Q         ci co c ho hi h T T m T T m        20 5 . 0 45 75 2 . 0    co T C Tco 0 32  i T  o T  55 20 75       ci hi i T T T 13 32 45       co ho o T T T   C Tm 0 11 . 29 13 55 ln 13 55           c h h h U 1 1 1   m T UA Q  
  • 18. Parallel co T hi T ho T ci T   kW T T C m Q ho hi ph h 122 . 25 ) 45 75 ( * 187 . 4 * 2 . 0      i T  o T  c h h h U 1 1 1   m T UA Q   K m W U 2 325  650 1 650 1 1   U 18 . 29 * * 325 1000 * 122 . 25 A  m T UA Q   2 65 . 2 m A 
  • 19. LMTD-Ex-3 • Saturated steam at 120OC is condensed with the help of water (inlet temperature 200C, outlet temperature 900C, Cp =4.187 kJ/kg K) in surface condenser. The overall heat transfer coefficient U=1800 W/m2K . • Determine the heat exchanger area & mass of condensate. • Take hfg =2200 kJ/kg and mass of water=16.67 kg/min. min / 67 . 16 / ? kg m s kg m C h   C T T T h ho hi 0 120   C T C T co ci 0 0 90 , 20   K m W U 2 / 1800  m T UA Q   ?  A Inlet Outlet CONDENSER co T ci T h T h T
  • 20.   ci co pc c fg stem T T C m h m Q    Inlet Outlet CONDENSER co T ci T h T h T • Saturated steam at 120OC is condensed with the help of water (inlet temperature 200C, outlet temperature 900C, Cp =4.187 kJ/kg K) in surface condenser. The overall heat transfer coefficient U=1800 W/m2K . • Determine the heat exchanger area & mass of condensate. • Take hfg =2200 kJ/kg and mass of water=16.67 kg/min.   o i o i m T T T T T        ln m T UA Q   100 20 120       ci h i T T T 30 90 120       co h o T T T
  • 21. • Oil 0.27 kg/s, Cp =2.09 kJ/kg, cool from 80OC to 40OC using coolent with flow rate of 0.27 kg/s, Cp =4.18 kJ/kg at 300C. • Give your choice for HX. • Determine the heat exchanger area . • Take U =24 W/m2K ho co T T  Parallel co T hi T ho T ci T Counter co T hi T ho T ci T To check given HX is Counter or Parallel Compare Hot & Cold Exit Temp. IF, HX is Counter Flow Type ho co T T  IF, HX is Parallel Flow Type