What is heat exchanger & its Functions Types of Heat Exchangers Compact Heat Exchangers Part of Fin Plate Heat Exchangers Advantages & Disadvantages of Fin Plate Exchangers Materials & Manufacturing Overall Heat transfer Coefficient & Fouling Factor LMTD Method Effectiveness - NTU Method

1. Design of Plate Fin
Heat Exchanger

2. Heat exchangers
• A Heat Exchanger is a device used to transfer heat
between one or more fluid.
• Minimum two fluid are required (S1 & S2)
• Heat transfer between S1 & S2 (Q).
• Temperature T1 & T2 (T1 ≠ T2 ).
• And two fluid are not allow to mix.

3. Application of heat exchanger
• Refrigeration
• Power plant (Boiler)
• Petroleum refining
• Automobile radiator
• In aircraft engine
• Air conditioning
• Water treatment

4. Applications of Heat Exchangers
Heat Exchangers
prevent car engine
overheating and
increase efficiency
Heat exchangers are
used in Industry for
heat transfer
Heat
exchangers are
used in AC and
furnaces

5. Types of Heat Exchanger
• Double tube heat exchanger
• Shell and Tube heat exchanger
• Plate heat exchanger
• Plate and shell heat exchanger
• Plate fin heat exchanger
• Spiral heat exchanger
• Helical coil heat exchanger

6. Plate fin heat exchanger
• Heat exchangers used in cryogenic applications need to have
very high effectiveness (0.95 or higher).
• In aircrafts where the demand on performance is not high, the
volume and weight of the heat exchanger should be kept at
minimum.
• For overcome these problem compact heat exchangers used
which provide large heat transfer area.
• Area to volume ratio greater than 700 𝑚2/ 𝑚3.
• Plate fin heat exchanger is compact heat exchanger.
• They are characterized by high effectiveness, compactness
(high surface area density), low weight and moderate cost.

8. Parts of pate fin HX
• Parting sheets: HX consisting of a block of alternating layers
of corrugated fins and flat separators known as parting sheets.
• Separating plates act as the primary heat transfer surfaces.
• Fins : The fins serve both as secondary heat transfer surface
and as mechanical support against the internal pressure
between layers.
• Steams exchange heat by flowing along the passages made by
the fins between the parting sheets.

9. • Side bars : The side bars prevent the fluid from spilling over
and mixing with the second fluid or leaking to outside.
• The fins and side bars are brazed with the parting sheets to
ensure good thermal link and to provide mechanical stability.
• Cap sheets: The first and the last sheets, called cap sheets, are
usually of thicker material than the parting sheets to support
the excess pressure over the ambient and to give protection
against physical damage.
• Headers: Each stream enters the block from its own header
via ports in the side-bars of appropriate layers and leaves in a
similar fashion.
• The header tanks are welded to the side-bars and parting sheets
across the full stack of layers

10. Advantages :
• Compactness : High overall heat transfer coefficient because
of large heat transfer area(area to volume ratio around 1000
𝑚2/𝑚3).
• Effectiveness: Ratio of actual heat transfer rate to maximum
heat transfer rate.
∈ =
𝑞 𝑎𝑐𝑡
𝑞 𝑚𝑎𝑥
𝑞 𝑚𝑎𝑥 = 𝐶 𝑚𝑖𝑛(𝑇ℎ,𝑖 − 𝑇𝑐,𝑖)
𝑞 𝑎𝑐𝑡= 𝐶ℎ(𝑇ℎ,𝑖 − 𝑇ℎ,𝑜) = 𝐶𝑐(𝑇𝑐,𝑜 − 𝑇𝑐,𝑖)
𝐶 𝑚𝑖𝑛 = min(𝐶ℎ , 𝐶𝑐)
• Very high thermal effectiveness more than 95% can be
obtained

11. • Temperature control : A close temperature approach
(temperature approach as low as 3K) is obtained for a heat
exchanger exchanging heat with single phase fluid streams.
• Flexibility : Changes can be made to heat exchanger
performance by utilizing a wide range of fluids and conditions
that can be modified to adapt to various design specifications.
• Multi stream operation is possible up to 10 streams.

12. Disadvantages :
• Difficulty in cleaning of passages, which limits its application
to clean and relatively non-corrosive fluids .
• Difficulty of repair in case of failure or leakage between
passages.
• Relatively high pressure drop due to narrow and constricted
passages .
• The rectangular geometry used puts a limit on operating range
of pressure and temperatures .

13. Flow Arrangement
• (1) Parallel flow (2) Counter flow (3) Cross flow
(4) Cross – counter flow.

14. Single Pass Multi pass

15. Fin Geometries
• (a) Plain rectangular (b) Plain trapezoidal
(c) wavy (d) offset strip fin

16. • (e) Louvered (f) perforated

17. Material
• Plate fin heat exchangers can be made in a variety of materials.
• Aluminium is preferred in cryogenic and aerospace
applications because of its low density, high thermal
conductivity and high strength at low temperature.
• The maximum design pressure for brazed aluminium plate fin
heat exchangers is around 90 bar.
• At temperatures above ambient, most aluminium alloys lose
mechanical strength.
.

18. • Stainless steels, nickel and copper alloys have been used at
temperatures up to 5000 C.
• The brazing material in case of aluminium exchangers is an
aluminium alloy of lower melting point,
• while that used in stainless steel exchangers is a nickel based
alloy with appropriate melting and good strength

19. Manufacturing
• The methods in common use are salt bath brazing and vacuum
brazing.
• In the salt bath process, the stacked assembly is preheated in a
furnace to about 550 ℃ and then dipped into a bath of fused
salt composed mainly of fluorides or chlorides of alkali metals.
• The molten salt works as both flux and heating agent,
maintaining the furnace at a uniform temperature.

20. • In the vacuum brazing process, no flux or separate pre-heating
furnace is required.
• The assembled block is heated to brazing temperature by
radiation from electric heaters and by conduction from the
exposed surfaces into the interior of the block.
• Many metals, such as aluminium, stainless steel, copper and
nickel alloys can be brazed satisfactorily in a vacuum furnace.

21. Overall heat transfer coefficient
• An essential requirement for heat exchanger design or
performance calculations.
• Contributing factors include convection and conduction
associated with the two fluids and the intermediate solid, as
well as the potential use of fins on both sides and the effects of
time-dependent surface fouling.
• Overall heat transfer coefficient related to total thermal and
fouling resistance.

22. • If temperature of hot fluid is 𝑇𝑎 and temperature of cold fluid
of is 𝑇𝑏 are known than heat transfer equation we can easily
write as given below.
Q = u*A*(𝑇𝑎-𝑇𝑏) = u*A*∆T
• Where Q = heat transfer coefficient (W)
A= area perpendicular to direction of heat flow (𝑚2)
u = overall heat transfer coefficient

23. • Overall heat transfer coefficient depend on total thermal
resistance of the system as given below,
u =
1
𝐴∗ ∑ 𝑅 𝑡ℎ
(W/𝑚2 𝐾)
u =
1
𝐴 [
1
ℎ 𝑖 𝐴
+
1
𝑘 𝐴
+
1
ℎ 𝑜 𝐴
]
u =
1
1
ℎ 𝑖
+
1
𝑘
+
1
ℎ 𝑜
• For metal value of k is very high, if we neglect 1/k term in
equation than,
u =
1
1
ℎ 𝑖
+
1
ℎ 𝑜

24. • Now u is depends on small value from both ℎ𝑖 and ℎ 𝑜. For
example, in a liquid to gas heat transfer convective heat
transfer is very small therefore u is control by this coefficient.
• To improve heat transfer coefficient on gas side we can
provide fins on gas side. Now total surface given by,
𝐴 𝑡𝑜𝑡𝑎𝑙 = 𝐴 𝑓𝑖𝑛 + 𝐴 𝑢𝑛𝑓𝑖𝑛𝑛𝑒𝑑
Where 𝐴 𝑓𝑖𝑛 = surface area of fins
𝐴 𝑢𝑛𝑓𝑖𝑛𝑛𝑒𝑑 = area of un-finned portion
• For short fin length we can take constant temperature
throughout the length because of high conductivity material
and effective surface area is given by above equation.

25. • For long fin there are temperature drop across the length
therefore effective surface area is given by below equation.
𝐴 𝑡𝑜𝑡𝑎𝑙 = 𝜂 𝑓𝑖𝑛 ∗ 𝐴 𝑓𝑖𝑛 + 𝐴 𝑢𝑛𝑓𝑖𝑛𝑛𝑒𝑑 (𝜂 𝑓𝑖𝑛 = fin
efficiency)
• Sometime overall efficiency 𝜂 𝑜 is used.
𝜂 𝑜 ∗ 𝐴 𝑡𝑜𝑡𝑎𝑙 = 𝜂 𝑓𝑖𝑛 ∗ 𝐴 𝑓𝑖𝑛 + 𝐴 𝑢𝑛𝑓𝑖𝑛𝑛𝑒𝑑
𝜂 𝑜 = 1 −
𝜂 𝑓𝑖𝑛
𝐴 𝑡𝑜𝑡𝑎𝑙
(1 - 𝜂 𝑓𝑖𝑛)
• We can write
𝑢𝑖 ∗ 𝐴𝑖 = 𝑢 𝑜 ∗ 𝐴 𝑜 =
1
∑ 𝑅 𝑡ℎ
=
1
1
ℎ 𝑖 𝐴 𝑖
+
1
𝑘 𝐴
+
1
ℎ 𝑜 (𝜂 𝑜∗ 𝐴 𝑡𝑜𝑡𝑎𝑙)

26. • Typical value of overall heat transfer coefficient are given
table,
[1] Ch.12 Heat Exchangers ,Fundamentals of Heat and Mass Transfer by M.Thirumaleshwar ,Pearson Education

27. Fouling Factor
• with time the surface become dirty because of scaling,
corrosion, deposits, etc. because of that the heat transfer
coefficient decrease.
• Reasons for fouling are given below,
1) deposits of finely divided particulates
2) Scaling or precipitation
3) corrosion
4) crystallization on surface by sub cooling
5) attachment of biological material
6) chemical reaction

28. • Effect of fouling is accounted for term called ‘Fouling factor’
or ‘dirt factor’, define as ,
𝑅𝑓 =
1
𝑢 𝑑𝑖𝑟𝑡𝑦
–
1
𝑢 𝑐𝑙𝑒𝑎𝑛
(𝑚2 𝐾/𝑊)
• Practically by finding out the overall heat transfer coefficient
for clean and dirt heat exchanger operating under same
condition.
• If we add fouling resistance in other thermal resistance than
we can write,
𝑢𝑖 ∗ 𝐴𝑖 = 𝑢 𝑜 ∗ 𝐴 𝑜 =
1
∑ 𝑅 𝑡ℎ
=
1
1
ℎ 𝑖 𝐴 𝑖
+
𝑅 𝑓𝑖
𝐴 𝑖
+
1
𝑘 𝐴
+
1
ℎ 𝑜 𝐴 𝑜
+
𝑅 𝑓𝑜
𝐴 𝑜
• Where 𝑅𝑓𝑖 and 𝑅𝑓𝑜 are fouling factor for hot fluid and cold
fluid surface.
• Fouling factor depends on flow velocity and operating
temperature; fouling decreases with increasing velocity and
decreasing temperature.

29. Fouling factors for industrial fluids (TEMA, 1988)
[1] Ch.12 Heat Exchangers ,Fundamentals of Heat and Mass Transfer by M.Thirumaleshwar ,Pearson Education

30. 1.LMTD method for Heat exchanger
analysis
• Assumptions made for this method :
1) u is considered as a constant for whole length
2) Fluids properties remain constant
3) No loss of heat to the surroundings.
4) Change in K.E. and P.E. are neglect.
5) Temperature of fluid remain constant across heat transfer
area.
• Temperature difference between two fluid is not constant
through out length, our aim is to find out appropriate ‘mean
temperature difference (∆Tm)’ .

31. Calculating U using Log Mean Temperature
coldhot dqdqdq =−=
ch TTT −=
ch dTdTTd −= )(
h
h
phh dTCmdq ..=
c
c
pcc dTCmdq ..=
Hot Stream :
Cold Stream:








−= c
pc
c
h
ph
h
Cm
dq
Cm
dq
Td
..
)(
dATUdq ..−=−








+−= c
pc
h
ph CmCm
dATUTd
.
1
.
1
...)(
 




 
+

−=



2
1
2
1
..
)( A
A
c
c
h
h
T
T
dA
q
T
q
T
U
T
Td
cc
c
hh
h
cm
Qd
dT
cm
Qd
dT



 −
=
−
=

32. 







−
=
1
2
12
ln
.
T
T
TT
AUq
Log Mean Temperature
( ) ( ) ( ) 2121
1
2 ...
ln cchhch TTTT
q
AU
TT
q
AU
T
T
−−−−=+−=







 







+−=



2
1
2
1
.
.
1
.
1
.
)( A
Ac
pc
h
ph
T
T
dA
CmCm
U
T
Td

33. CON CURRENT FLOW








−
=
1
2
12
ln
T
T
TT
TLn
73111 TTTTT ch −=−=
106222 TTTTT ch −=−=
COUNTER CURRENT FLOW
106122 TTTTT ch −=−=
73211 TTTTT ch −=−=
( ) ( )
Ln
c
pc
Ln
h
ph
TA
TTCm
TA
TTCm
U

−
=

−
=
.
..
.
.. 10763

T1
T2
T4 T5
T3
T7 T8 T9
T10
T6
Counter - Current Flow
T1 T2T4 T5
T6T3
T7
T8 T9
T10
Para llel Flow
Log Mean Temperature evaluation
T1
A
1 2
T2
T3
T6
T4 T6
T7
T8
T9
T10
Wall
∆T1
∆T2
∆ A
A
1 2

34. T1
A
1 2
T2
T3
T6
T4 T6
T7 T8
T9
T10
Wall
q = hh Ai Tlm
Tlm =
(T3 −T1) −(T6 −T2)
ln
(T3 −T1)
(T6 −T2)
q = hc Ao Tlm
Tlm =
(T1 −T7) −(T2 −T10)
ln
(T1 −T7)
(T2 −T10)

35. • Remember – 1 and 2 are ends, not fluids
• Same formula for parallel flow (but T’s are different)
•Counter flow has highest LMTD, for given T’s therefore smallest area for Q.
( )
( )
DifferenceeTemperaturMeanLogisLMTD
LMTD
/ln
2
1
putandmforSubstitute
12
12
222
111
c
UAQ
TT
TT
UAQ
ENDTTT
ENDTTT
ch
ch
=







−
=
−=
−=



LMTD equation derive above is for single pass parallel and cross flow
but , for cross flow and multi pass ‘correction factor’ is applied.
Q = u* A * F (LMTD)

36. 2.The effectiveness-NTU method for Heat
Exchanger analysis
• LMTD method used to determine area required, A (or size of
HX) using simple equation Q=u*A*(LMTD), when all four
end temperature are given.
• If two inlet temperature long with flow rate are given than exit
temperatures, heat transfer rate and overall heat transfer
coefficient easily calculated by NTU (Number of transfer
units) method.
• This method developed by Kays and London in1955.
• To find out performance of HX with different flow rate rather
than design flow rate.

37. Effectiveness of a heat exchanger (∈) :
∈ =
𝑄
𝑄 𝑚𝑎𝑥
where, Q = actual heat transferred in HX
𝑄 𝑚𝑎𝑥 = maximum heat transferred in HX
• Actual heat transfer rate in a heat exchanger is given by;
Q = 𝑚ℎ ∗ 𝐶 𝑝ℎ 𝑇ℎ1 − 𝑇ℎ2 = 𝐶ℎ (𝑇ℎ1 − 𝑇ℎ2)
or Q = 𝑚 𝑐 ∗ 𝐶 𝑝𝑐 𝑇𝑐2 − 𝑇𝑐1 = 𝐶𝑐 (𝑇𝑐2 − 𝑇𝑐1)
where 𝐶ℎ = capacity rate of the hot fluid
𝐶𝑐 = capacity rate of cold fluid
• Now 𝐶ℎ may be equal to 𝐶𝑐 or less than 𝐶𝑐 or greater than 𝐶𝑐.
• If 𝐶ℎ < 𝐶𝑐, we take 𝐶ℎ as 𝐶 𝑚𝑖𝑛.
• Instead, if 𝐶ℎ > 𝐶𝑐, we take 𝐶𝑐 𝑎𝑠 𝐶 𝑚𝑖𝑛.
• Capacity of other fluid in each case take as 𝐶 𝑚𝑎𝑥.

38. Capacity Ratio (C):
• Capacity ratio is defined as:
C =
𝐶 𝑚𝑖𝑛
𝐶 𝑚𝑎𝑥
Number of Transfer Units (NTU):
• NTU is a dimensionless number, defined as;
NTU =
𝑢∗𝐴
𝐶 𝑚𝑖𝑛
Where u = overall heat transfer coefficient
A = heat transfer area
• As NTU is larger the size of heat exchanger is larger.

39. Maximum possible heat transfer in a heat exchanger :
• Hot fluid temperature decrease from 𝑇ℎ1 𝑡𝑜 𝑇ℎ2.
• Cold fluid temperature increase from 𝑇𝑐1 𝑡𝑜 𝑇𝑐2.
• Maximum temperature difference is (𝑇ℎ1 − 𝑇𝑐1).
• Maximum heat transfer rate is depends on minimum capacity
rate.
𝑄 𝑚𝑎𝑥= 𝐶ℎ (𝑇ℎ1 − 𝑇𝑐1) (if 𝐶ℎ is minimum capacity rate, 𝐶 𝑚𝑖𝑛)
𝑄 𝑚𝑎𝑥= 𝐶𝑐 (𝑇ℎ1 − 𝑇𝑐1) (if 𝐶𝑐 is minimum capacity rate, 𝐶 𝑚𝑖𝑛)
• Generally we can write
𝑄 𝑚𝑎𝑥= 𝐶 𝑚𝑖𝑛 (𝑇ℎ1 − 𝑇𝑐1)

40. • Now we can write effectiveness as;
∈ =
𝑄
𝑄 𝑚𝑎𝑥
=
𝐶ℎ ( 𝑇ℎ1−𝑇ℎ2)
𝐶 𝑚𝑖𝑛 ( 𝑇ℎ1−𝑇𝑐1)
=
𝐶 𝑐 ( 𝑇𝑐2−𝑇𝑐1)
𝐶 𝑚𝑖𝑛 ( 𝑇ℎ1−𝑇𝑐1)
……..(a)
• If 𝐶ℎ < 𝐶𝑐 than 𝐶 𝑚𝑖𝑛 = 𝐶ℎ.
∈ =
( 𝑇ℎ1−𝑇ℎ2)
( 𝑇ℎ1−𝑇𝑐1)
• If 𝐶ℎ > 𝐶𝑐 than 𝐶 𝑚𝑖𝑛 = 𝐶𝑐.
∈ =
( 𝑇𝑐2−𝑇𝑐1)
( 𝑇ℎ1−𝑇𝑐1)
• Effectiveness of any heat exchanger can explained as function of
NTU and capacity ratio.
∈ = f(NTU,
𝐶 𝑚𝑖𝑛
𝐶 𝑚𝑎𝑥
)

41. Effectiveness, NTU and Capacity ratio relation for parallel flow:
• From LMTD method ;
ln(
𝑇ℎ2 − 𝑇𝑐2
𝑇ℎ1 − 𝑇𝑐1
) = -u*A ( 1
𝑚ℎ 𝐶 𝑝ℎ
+
1
𝑚 𝑐 𝐶 𝑝𝑐
) = -u*A ( 1
𝐶ℎ
+
1
𝐶 𝑐
)
𝑇ℎ2 − 𝑇𝑐2
𝑇ℎ1 − 𝑇𝑐1
= exp[-
𝑢∗𝐴
𝐶 𝑚𝑖𝑛
(1+
𝐶 𝑚𝑖𝑛
𝐶 𝑚𝑎𝑥
)] = exp[- NTU(1+
𝐶 𝑚𝑖𝑛
𝐶 𝑚𝑎𝑥
)]
• If we put 𝑇ℎ2 𝑎𝑛𝑑 𝑇𝑐2 from equation (a) than,
1- 𝜖 ∗ 𝐶 𝑚𝑖𝑛( 1
𝐶ℎ
+
1
𝐶 𝑐
) = exp[- NTU(1+
𝐶 𝑚𝑖𝑛
𝐶 𝑚𝑎𝑥
)]
• If we assume 𝐶ℎ > 𝐶𝑐 than 𝐶 𝑚𝑖𝑛 = 𝐶𝑐 and 𝐶 𝑚𝑎𝑥 = 𝐶ℎ.
𝜖 =
1−exp[− NTU(1+
𝐶 𝑚𝑖𝑛
𝐶 𝑚𝑎𝑥
)]
1+
𝐶 𝑚𝑖𝑛
𝐶 𝑚𝑎𝑥
=
1−exp[−𝑁𝑇𝑈 1+𝐶 ]
1+𝑐

42. Effectiveness, NTU and Capacity ratio relation for counter flow:
• From LMTD method,
ln(
𝑇ℎ2 − 𝑇𝑐1
𝑇ℎ1 − 𝑇𝑐2
) = -u*A ( 1
𝑚ℎ 𝐶 𝑝ℎ
−
1
𝑚 𝑐 𝐶 𝑝𝑐
) = -u*A ( 1
𝐶ℎ
−
1
𝐶 𝑐
)
• Assume that than 𝐶 𝑚𝑖𝑛 = 𝐶ℎ and 𝐶 𝑚𝑎𝑥 = 𝐶𝑐.
• If we put 𝑇ℎ2 𝑎𝑛𝑑 𝑇𝑐2 from equation (a) than, we get final
equation as given below,
𝜖 =
1−exp[−𝑁𝑇𝑈 1−𝐶 ]
(1−𝐶∗exp(−𝑁𝑇𝑈 1−𝑐 ))

43. Effectiveness – NTU relation for parallel flow

44. Effectiveness -NTU relation for counter flow

45. Effectiveness – NTU relation for cross flow

46. Thank You