Heat Exchanger.ppt

8/13/2019 Heat Exchanger.ppt

1/104

1

MEC551THERMAL ENGINEERING

4 0 Heat Exchangers


2/104

2

4.0 HEAT EXCHANGER

4.1 Types of Heat Exchanger: Shell and

Tube, Plate - Parallel Flow, Counter

Flow, Cross Flow.

4.2 Overall Heat Transfer Coefficient.4.3 Log-mean-temperature difference (LMTD)

method & correction

4.4 Effectiveness of NTU (e-NTU method4.5 Heat Exchanger Design and

Consideration


3/104


4/104

4

Heat Exchangers

Heat exchangers are typically classified according to

flow arrangement and type of construction.

The simplest heat exchanger is one in which the hotand cold fluids move in the same direction (parallel

flow) or opposite directions (counter flow) in a

concentric tube (or double pipe construction).


5/104

5

Heat Exchangers(Parallel-flow)

In the parallel-flow

arrangement, the hot

and cold fluids enter

the same end, flow in

the same direction, andleave at the same end.


6/104

6

Heat Exchangers(Counter-flow)

In the counter-flow

arrangement, the fluids

enter at opposite ends,

in opposite directions,

and leave at oppositeends.

out

in

cold fluid in

warmed

fluid out


7/104

7

Heat Exchangers(Cross-Flow)

Alternatively, the fluids

may move in cross-flow

(perpendicular) to one

another.

Fluid motion over them

may be mixed or

unmixed. The fluid is

unmixed, because fins

inhibit motion in a

direction that is

transverse to the main

flow direction (x)


8/104


9/104

9

Heat Exchangers(Shell and Tube Design)

These designs contain a large number of tubes

(packed in a shell) and heat transfer takes place as

one fluid flows inside the tubes while the other fluid

flows outside the tubes through the shell.


10/104

10

Heat Exchangers(Shell and Tube Design)

Baffles are commonly placed in the shell to force the

shell-side fluid to flow across the shell to enhance

heat transfer and to maintain a uniform spacing

between the tubes.


11/104

11


12/104

12

Heat Exchangers(Shell and TubeMulti-pass designs)

Shell and tube heat

exchangers are further

classified by how many shell

and tube passes are involved.

One-shell pass and two-tube

passes devices

Heat exchangers in which all the

tubes make 1 U-turn in the shell

are called

Two-shell passes and four-

tube passes devices


13/104

13

Heat Exchangers(Plate and Frame)

This is an

innovative design

which consists of

a series of plates

with corrugatedflat flow passages.

Hot and cold fluids

flow in alternative

passages, thus the

cold stream is

surrounded by two

hot streams.


14/104

14

Heat Exchangers(Terminology)

Heat exchangers are often given specific names to

reflect the specific application for which they are

used:

Condenser

A heat exchanger in which one of the fluids is cooled

and condenses as it flows through the heat exchanger.

Boiler

A heat exchanger in which one of the fluids absorbs heat

and vaporizes.

Space radiator

A heat exchanger that transfers heat from the hot fluid to

the surrounding space by radiation.


15/104

15

4.2 Overall Heat TransferCoefficient U,W/m2.oC


16/104

16

Double-pipe Heat Exchanger

A heat exchanger typically

involves two flowing fluids

separated by a solid wall.

Heat is transferred1. from the hot fluid to the wall

by convection,

2. through the wall by

conduction,3. from the wall to the cold fluid

again by convection.


17/104

17

Inner tube (dimensions)

DiDo

L

t

Ao= DoL

Ai= D

iL

t = (Do- Di)/2 = Ro- Ri

Outer surface area

inner surface area

Wall thickness


18/104

18

Thermal resistances of

the tube wall

Rwall= ln (Do/Di)2 kL

inner surface

Ri= 1/hiAi ,

outer surface

Ro= 1/hoAo


19/104

19

Double-pipe Heat Exchanger

The thermal resistance of the

network is thus:

oo

DD

ii

totAhkLAh

R io

1

2ln1

So the thermal resistance in the path of heat flow from hot tocold fluid in a heat exchanger includes:

i. Skin resistance associated with the flow

ii. Scale resistance from wall fouling (to be discussed later)

iii. Thermal resistance of wall material


20/104

20

Overall Heat Transfer Coefficient

The heat transfer is therefore:

Where U = heat transfer coefficient

Note:

TAUTAUTUA

R

TQ ooii

oioi

ooii

AAunlessUU

AUAU


21/104

21


In this equation, U is called the overall heat transfer

coefficient.

tot

ooiis

RAUAUAU

111


22/104

22

Since:

where Amis called the logarithmic mean area


) ) ) ) )

) )

) mAA

LR

LR

RRL

io

LR

LR

tthickness

ioR

R

D

D

wall

Ak

t

k

t

k

t

kLRR

RR

kLkLR

iAoA

io

i

o

io

i

o

i

o

i

o

-

-

-

-

ln

2

2ln

2

2

2

,

11

2

ln

2

ln

2

ln

)i

o

A

A

io

m

AAA

ln

-


23/104


24/104


25/104


26/104

26


For a thin tube:

1

11

0ln1;0

-

oi

oi

i

o

i

o

hhUU

D

Dso

D

Dt


27/104

27


When the tubes are finned on one side to enhance the

heat transfer, the total heat transfer surface area on

the finned side becomes:

unfinnedfintotals AAAA

SurfaceArea of fins

SurfaceArea of Unfinned portion


28/104

28


For short fins of high thermal conductivity, we can use

this total area in the convection resistance relation:

Since in this case the fins will be isothermal.

Otherwise we should determine the effective surfacearea from:

)unfinnedfinsconv

AAhAhR

11

finfinunfinneds AAA


29/104

29

Effect of Fouling

The performance of heat exchangers

usually deteriorates with time as a result

of the accumulation of deposits on heat

transfer surfaces.

The layer of deposits represents

additional resistance to heat transfer

and causes Q to decrease.

The net effect is represented by thefouling factor (Rf) which is a measure of

the thermal resistance introduced by

fouling.


30/104

30

Effect of Fouling

If we define the fouling factorson the inside and the

outside surface to be Rfiand Rfo, respectively. Then the

total thermal resistance (R) with fouling and the overall

heat transfer coefficient becomes:

Generally, Uois low for a fluid with low k-values (for example

gases and oils).

1

0

1ln2

1

11

-

o

fo

i

oo

i

ofi

i

o

i

o

oo

fo

mi

fi

ii

hRD

D

k

D

D

DRD

D

hU

hAA

R

Ak

t

A

R

hAR


31/104


32/104

32

Overall Heat Transfer Coefficient(Example 11.1)

Properties of water at 45C (Table A-9):

Properties of oil at 80C (Table A-16):

sm

waterCmW

water

waterm

kg

water

k 2

3

610602.0637.0

91.3Pr990

-

sm

oilCmW

oil

oilm

kg

oil

k 2

3

6105.37138.0

490Pr852

-


33/104

33


Since the thickness is small:

The hydraulic diameter for a circular tube is the diameter of the

tube itself, Dh, water = D = 0.02 m

Therefore the mean velocity (Vm) for water is:

oi hhU

111

)

) )

sm

m

kg

s

kg

waterhwater

water

cwater

water

waterm

m

Dm

AmV

61.102.0990

5.02

4

1

2

,41

3


34/104

34


The Reynolds number for the water is:

Since Re > 4,000, which is the Recritfor a pipe, the flow of water is

turbulent. The Nusselt number is thus:

) )490,53

10602.0

02.061.1Re

26

,

-

s

m

sm

water

waterhwaterm

water

mDV

) ) 6.24091.3490,53023.0

PrRe023.0

4.08.0

4.08.0

waterwaterwater

hwater

k

DhNu


35/104

35


Then:

)Cm

WCmW

water

waterh

waterwater

m

NuD

kh

2663,76.24002.0

637.0

,


36/104

36


Now we repeat the analysis for oil.

The mean velocity of the oil is:

m

DDD iooilh

01.002.003.0

,

-

-

) )

) ) sm

m

kg

s

kg

iooil

oil

coil

oil

oilm

m

DD

m

A

m

V

39.202.003.0852

8.0222

41

2241,

3

-

-


37/104

37


So the Reynolds number for oil is:

Recritfor oil in a pipe is higher, so the flow of oil is laminar.

The Nusselt number can be found from Table 13-3 (in text) for

Dt/D

s= 0.667 as:

) )637

105.37

01.039.2Re

26

,,

-

s

m

sm

oil

oilhoilm

oil

mDV

45.5oilNu


38/104

38


Therefore:

since

Then the overall heat transfer coefficient (U) for this heat

exchanger becomes:

)Cm

WCmW

oil

oilh

oiloil

mNu

D

kh

22.7545.5

01.0

138.0

,

Cm

W

CmW

CmW

oi hh

U

2

22

5.74

2.75

1

663,7

1

1

11

1

oi hhU

111


39/104

39

Effect of Fouling(Example 11.2)

Example 11.2 A double-pipe

(shell and tube) heat exchanger

is constructed of a stainless

steel inner tube with inner

diameter (Di) of 1.5 cm and

outer diameter (Do) of 1.9 cm.and an outer shell of inner

diameter 3.2 cm.

For the convection values and

fouling factors given, determine

the thermal resistance of the

heat exchanger and heat

transfer coefficients Ui and Uo.


40/104

40


The areas are:

The thermal resistance is:

)ooo

ofD

D

i

if

ii

ooiis

AhA

R

kLA

R

Ah

AUAUAUR

i

o

1

2

ln1

111

,,

) )

) )

2

2

0597.01019.0

0471.01015.0

mmmLDA

mmmLDA

oo

ii


41/104

41


Therefore:

)

) )

) ) ) )

)

W

C

WC

Cm

W

WCm

CmW

WCm

Cm

W

ooo

ofD

D

i

if

ii

mmm

m

m

mm

AhA

R

kLA

R

AhR i

o

0532.0

001396.000168.00025.000849.002654.0

0597.01200

1

0597.0

0001.0

11.152

015.0

019.0ln

0471.00004.0

0471.08001

1

2

ln1

22

22

,,

2

2

2

2


42/104

42


Note that 19% of the total resistance in this case is due to fouling

and about 5% of it is due to the steel tube separating the two

fluids. The rest of the 76% is due to convective resistances on

the two sides of the inner tube.

Now know the thermal resistances, the overall heat transfercoefficients can be found:

) )

) ) CmW

WC

o

o

Cm

W

WC

i

i

mARU

mARU

2

2

3150597.00532.0

11

3990471.00532.0

11

2

2


43/104

43

4.3 Log Mean TemperatureDifference Method


44/104

44

Analysis of Heat Exchangers

The first law of

thermodynamics requires

that the heat transfer to

the cold fluid must be

equal to the heat transferfrom the hot fluid.

Where the subscripts cand h

stand for cold and hot.

)

)outhinhphh

incoutcpcc

TTCmQ

TTCmQ

,,

,,

--

-


45/104

45


In heat exchanger analysis it is convenient to combine

the product of the mass flow and the specific heat of a

fluid into a single quantity called the heat capacity

rate.

)

)outhinhhincoutcc

TTCQ

TTCQ

,,

,,

--

-

phhh cmC pccc cmC


46/104

46


Consider an incremental area of the heat exchanger surface (as

shown for either a counter-flow or parallel flow heat exchanger).

Parallel Flow Counter-flow

To

dA

Hot

Cold

Th,in

Tc,inTc,out

TL

To

Th,in

Tc,out

Th,out

TL

Th,out

Tc,in

Hot

Cold

dA


47/104

47


The heat transfer over the area (dA) can be expressed in three

ways:

1) The heat flow between hot fluids:

2) The heat gain by the cold fluid:

3) The heat given up by the hot fluid:

TdAUQd

cc dTCQd

hh dTCQd -

Log Mean Temperature


48/104

48

Log Mean Temperature

Difference Method

However, since the temperature difference (T)

between the hot and cold fluids varies with distance

along the heat exchanger. It is convenient to use a

mean temperature difference (Tm) between the hot

and cold fluids, so that:

mTUAQ


49/104

49

LMTD Method

Consider a parallel double-

pipe heat exchanger. The

heat transfer on each fluid in

the differential section can

be expressed as:

cc

hh

dTCQ

dTCQ

-


50/104


51/104

51

LMTD Method

The rate of heat transfer in the differential section can

be expressed as:

Substituting this into the previous equations gives:

) dATTUQ ch -

)

--

--

chch

ch

CCdAU

TTTTd 11


52/104

52

LMTD Method

Integrating from the inlet to the heat exchanger to its

outlet, we get:

)

--

- L

ch

L

ch

ch

dACCUTT

TTd

00

11

+ For parallel-flow

- For counter-flow


53/104

53

LMTD Method

Solving the integral for parallel flow

(where T1=Th,in-Tc,inand T2=Th,out-Tc,out) we get:

-

-

-

-

-

hc

h

ch

c

ch

ch

ch

T

incinh

T

outcouth

CC

C

CC

CAU

CC

CCAU

T

T

CCAU

TT

TT

11ln

11ln

1

2

,,

,,

1

2


54/104

54

LMTD Method

)

--

-

---

hinhouth

incoutc

cincoutc

inhouth

CTT

TT

CTT

TTAUT

T 11ln,,

,,

,,

,,

1

2

Since:

Then:

)inhouth

incoutc

c

h

incoutc

inhouth

h

c

TT

TT

C

Cand

TT

TT

C

C

,,

,,

,,

,,

--

-

-

--


55/104

55

LMTD Method

Solving:

) )

-

----

-

--

QTTAU

Q

TTTTAU

Q

TT

Q

TTAU

T

T

incinhoutcouth

incoutcinhouth

12

,,,,

,,,,

1

2ln


56/104


57/104

57

LMTD Method

The log mean temperature difference method (LMTD) can be usedfor both parallel and counter-flow heat exchangers, if the

following terms are used for T:

Parallel Heat Exchangers Counter-flow Heat Exchangers

Correction factors


58/104

58

Correction factorsfor LMTD Method

The LMTD method is strictly limited to single pass parallel andcounter-flow heat exchangers only.

However, similar relations are also developed for cross-flow

and multi-pass shell and tube heat exchangers, but these are

very complicated expressions.

In these cases it is more convenient to relate the equivalent

temperature difference relation for the counter-flow case as:

flowcounterLMTDm TFT

- ,

Log Mean Temperature Difference

for counter-flow heat exchanger

Correction factor

Correction factors


59/104

59


Correction factors (F) for several common

configurations are available from standard charts.

Figure 11-18 (in text) shown on the next slide, lists

some for common shell and tube cross-flow heatexchangers. The correction factor used for this is:

) )

sideshellp

sidetubep

cm

cm

tt

TTR

tT

ttP

-

-

-

-

12

21

11

12


60/104

LMTD Method


61/104

61

LMTD Method(Example 6.3)

Example 6.3 Alcohol is to be cooled at a rate of 0.2kg/s from 75C to 35C in a counter-flow heatexchanger. Cooling water enters the heat exchanger

at 12C at a rate of 0.16 kg/s. The convectivecoefficient between alcohol and the tube wall andwater is 0.34 kW/(m2K) and between the tube walland water is 0.225 kW/(m2K). The tube may beassumed to be thin. The specific heat for the alcohol

is 2.52 KJ/(kgK) and water is 4.187 KJ/(kgK). Calculate the capacity ratio (C), effectiveness (),

and heat exchanger surface area (As).

LMTD Method


62/104

62


Th,in= 75C

Tc,out= ?

Tc,in

= 12C

Th,out= 35C

TTh,in

Tc,out

Th,out

Tc,in

T1

T2

Km

kWo

Km

kWi

h

h

2

2

225.0

340.0


63/104

LMTD Method


64/104

64


Energy balance:

) ) )

) ) )

CCT

CC

CT

TcmTcm

QQ

outc

Kkg

kJ

s

kg

outcKkgkJ

s

kg

alcoholphhwaterpcc

lossgain

-

-

1.4267.0

16.2012

357552.22.0

12187.416.0

,

,

LMTD Method


65/104

65


The effectiveness is thus:

Cminis used for Qmaxbecause the fluid with the smallerheat capacity rate will experience the largesttemperature change and thus be the first toexperience the maximum temperature, at which pointheat transfer stops.

)

) ) )

635.01275

3575

,,m in

,,

m ax

-

-

-

-

CC

CC

TTC

TTC

Q

Q

incinh

outhinhh

e

LMTD Method


66/104

66


The temperature differences are:

Therefore the LMTD is:

CCCTTT

CCCTTT

outcinh

incouth

--

--

9.321.4275

231235

,,1

,,2

) ) CCCTTTC

CT

Tm -

-

7.27

ln9.3223

ln9.32

2312

1

2

LMTD Method


67/104

67


For a thin tube (as given) then rori, the overall heattransfer coefficient is:

Finally, the surface area is:

Km

kW

oi

i

hh

U

--

21354.0

225.0

1

34.0

111 11

) )2

495.02737.271354.0

16.20

2

mK

kW

TU

Q

A

Km

kW

ms

LMTD Correction


68/104

68

LMTD Correction(Example 6.4)

Example 6.4 Cooling water (Cpc=4.187 kJ/(kgK))flows through a two-shell pass, four-tube pass heat

exchanger at the rate of 2 kg/s and temperatures at

entry of 20and exit at 80C. Hot oil enters throughthe shell side of the heat exchanger at 140C andleaves at 90C. Calculate the heat exchanger surfacearea if the overall heat transfer coefficient (U) is 300

W/(m2C).

LMTD Correction


69/104

69


Th,in= 140C

Th,out= 90C

Tc,out= 80C

Tc,in= 20C

LMTD Correction


70/104

70


The LMTD is:

) )C

CC

CC

CCCC

T

T

TTTm

-

-

---

-

9.64

80140

2090ln

801402090

ln2

1

21

LMTD Correction


71/104

71


Parameters to use the LMTD correction chart:

833.02080

90140

5.020140

2080

,,

,,

,,

,,

-

-

-

-

-

-

-

-

CC

CC

TT

TTR

CC

CC

TT

TTP

incoutc

inhouth

incinh

incoutc

LMTD Correction


72/104

72


From chart (Figure 13-18 in text) the value F= 0.97

0.97

LMTD Correction


73/104

73


Therefore the surface area (As) is:

) ) )

) ) )23

,,

6.269.6497.010300

2080187.42

2

mC

CC

TFU

TTCm

A

Cm

kW

KkgkJ

s

kg

m

incoutcpcc

s

-

-

-



74/104

74


Two different design tasks:1) Specified:

- the temperature change in a fluid stream, and

- the mass flow rate.

Required:

- the designer needs to select a heat exchanger.

2) Specified:

- the heat exchanger type and size,

- fluid mass flow rate,

- inlet temperatures.

Required:

- the designer needs to predict the outlet temperatures and heat transferrate.

Two methods used in the analysis of heat exchangers:

the log mean temperature difference (or LMTD)

best suited for the #1,

the effectivenessNTUmethod

best suited for task #2.

LMTD


75/104

75

LMTD

)1 2

1 2ln

s lm

lm

Q UA T

T TT

T T

-


76/104

76

Correction factors


77/104

77


The LMTD method is strictly limited to single pass parallel andcounter-flow heat exchangers only.

However, similar relations are also developed for cross-flow

and multi-pass shell and tube heat exchangers, but these are

very complicated expressions.

In these cases it is more convenient to relate the equivalent

temperature difference relation for the counter-flow case as:

flowcounterLMTDm TFT

- ,

Log Mean Temperature Difference

for counter-flow heat exchanger

Correction factor


78/104

78

4.4 The EffectivenessNTU Method

The NTU Method


79/104

79

The -NTU Method

This method is useful when the LMTD method cannotbe determined. The required data may instead be

determined from the -NTU method charts of several

common configurations

Definitions:

The heat capacity of the cold stream is given by:

The heat capacity of the hot stream is given by:

pccc cmC

phhh cmC

The NTU Method


80/104

80

The -NTU Method

The capacity ratio is defined as:

ch

c

h

ch

h

c

CCifC

CC

CCif

C

CC

;

;


81/104

The NTU Method


82/104

82

The -NTU Method

The maximum temperature difference possible in aheat exchanger is:

The heat transfer in the heat exchanger will reach its

maximum value when:

1) The cold fluid is heated to the inlet temperature of the hot

fluid.

2) The hot fluid is cooled to the inlet temperature of the cold

fluid.

incinh TTT ,,m ax -


83/104

The NTU Method


84/104

84

The -NTU Method

Therefore, the maximum possible heat transfer rate inheat exchanger is:

)incinh TTCQ ,,minmax -

hc CorCofsmallertheisCwhere m in:

The NTU Method


85/104

85

The -NTU Method

Therefore the heat exchanger effectiveness for a

counter-flow heat exchanger is:

)

) )incinh

outhinhh

incinh

incoutcc

TTC

TTC

TTC

TTC

,,m in

,,

,,m in

,,

-

-

-

-

e

e

ratetransferheatpossibleMaximum

ratetransferheatActual

Q

Q

max

e

The NTU Method


86/104

86

The -NTU Method

NTUNumber of Heat Transfer Unit

The NTU is a measure of physical size of the heat

exchanger; the larger NTU, the larger heat exchanger

size. It is defined as the ratio of the heat capacity ofthe heat exchanger to the minimum heat capacity of

the flow.

)minmin p

ss

CmAU

CUANTU

The -NTU Method


87/104

87

(Step-by-step Procedure)

The -NTU Method Procedure

1) Determine Cminand Cmaxfrom of the streams

and calculate the capacity ratio C.

max

min

C

Cc

pcm


88/104

The -NTU Method


89/104

89


3) Charts

Knowing , NTU, and C use

the chart (Figure 11-26 in the

text) to determine U and As

in question.

OR from standard derivation

Tables 11-4 & 11-5

The -NTU Method


90/104

90


4) Compute the heat transfer rate:

)incinh TTCQ ,,min - e

The -NTU Method


91/104

91


5) Calculate the outlet temperature:

c

incoutc

h

inhouth

C

QTT

C

QTT

-

-

,,

,,

The -NTU Method


92/104

92

(Example 6.5)

Example 6.5 A single, pass counter-flow shell-and-tube heat exchanger is used to cool hot mercury

(Cph=1.37 kJ/(kgK)) from 110C to 70C flowing at arate of 1 kg/s with water (Cpc=4.187 kJ/(kgK)) enteringat 30C and flowing at a rate of 0.2 kg/s. Calculatethe heat transfer surface area required and the exit

temperature of the water if the overall heat transfer

coefficient (U) is 250 kW/(m2K).

The -NTU Method


93/104

93

30C

70C110C

(Example 6.5)

The -NTU Method


94/104

94

(Example 6.5)

The temperature difference (T2) is:

The heat capacities are:

CCC

TTT incouth

-

-

403070

,,2

) ) sK

kJKkg

kJs

kg

pccc

sKkJ

KkgkJ

s

kg

phhh

CmC

CmC

836.0187.42.0

37.137.11

The -NTU Method


95/104

95

(Example 6.5)

Since Ch> Cc, therefore:

61.0

37.1

836.0

836.0

370.1

max

min

min

max

C

CC

CC

CC

sK

kJ

c

sKkJ

h

The -NTU Method


96/104

96

(Example 6.5)

The total heat transfer (Q) is:

The maximum heat transfer is:

) ) ) )

kW

K

TTCQ

KkgkJ

outhinhh

8.54

2737027311037.1

,,

-

--

--

) ) ) )

kW

K

TTCQ

sKkJ

incinh

9.66

27330273110836.0

,,m inm ax

-

-

The -NTU Method


97/104

97

(Example 6.5)

The heat exchanger effectiveness is:

Now consult the -NTU chart (Figure 11-26 in text)

82.0

9.66

8.54

max

kW

kW

Q

Q

e

The -NTU Method


98/104

98

(Example 6.5)

0.82

2.7

From the chart:

NTU = 2.7

The -NTU Method


99/104

99

(Example 6.5)

Therefore the surface area (As) is:

The exit temperature of the water is thus:

) 2min 03.9250

836.07.2

2

mU

CNTUA

Km

kW

sKkJ

s

)

) CorKKkW

TCQT

TTCQ

sK

kJ

inc

c

outc

incoutcc

-

6.956.36827330836.0

8.54

,,

,,


100/104

100

4.5 Heat Exchanger DesignConsiderations

Design Considerations


101/104

101


Some design considerations are:

Heat transfer rate

This is the most important quantity. A heat exchanger

must be capable of transferring heat at a specified rate in

order to achieve the desired temperature change of thefluid at a specified mass flow rate.

Cost

Budgetary limitations always play an important role.

Operating and maintenance costs are also a factor.

Pumping Power

The fluids are normally forced by pumps or fans which

require a pump with associate electrical costs.



102/104

102


Size and weight Normally the smaller and lighter, the better. This is

especially true for automotive and aerospace industries.

Type

The type of heat exchanger depends on therequirements, fluids involved, size and weight limitations

etc.

Materials

The materials in the heat exchanger construction may bean important consideration, especially if thermal

stresses are an issue.

End Of Heat Exchanger SectionC C


103/104

103

End Of HeatExchanger SectionC C

End Of LecturesC C


104/104

End Of LecturesC C

Study hard for your finals

I hope you all do well

Documents

Heat Exchanger.ppt