ME3122-HX-2014

Prof Andrew Tay ME3122 Heat Transfer

13. Heat Exchangers

A heat exchanger (HX) is a device which enables the

continuous transfer of heat between two fluid streams

separated by a solid wall.

Heat exchangers are used in numerous engineering systems

such as power generation, air conditioning, chemical

industries, food industries, aeroplanes, cars, etc. They are

usually classified according to flow arrangement and type of

construction.

1


13.1 Basic Types of Heat Exchangers

(i) Concentric tube or double-pipe heat exchanger.

One fluid flows through the inner tube, the other fluid flows through the

annulus. Both fluid streams traverse the exchanger once only and this

arrangement is called a single-pass arrangement.

If both fluids flow in the same direction, HX is called a parallel-flow

type. If both fluids flow in opposite directions, HX is called a counter-

flow type.

Parallel-flow Counter-flow

2


(ii) Cross-flow heat exchangers

When the two fluid flow paths along the heat transfer surface are at right

angles, the heat exchanger is of the cross-flow type.

(a) Both fluids unmixed

Each fluid is unmixed as it flows through the HX. The temperature of the

cross-flowing fluid is not uniform laterally and varies in x and y

directions.

3


(b) One unmixed, the other fully mixed

Temperature of the cross-flowing fluid is laterally uniform and varies in

the x direction only.

4


(iii) Shell-and-tube heat exchangers

In order to increase the effective heat transfer area per unit volume,

commercial HXs provide for multiple passes through the tubes; fluid

outside tubes in the shell may be routed back and forth by means of

baffles. The diagram above shows a shell and tube HX with one tube

passes and one shell pass in a cross-counter-flow arrangement.

One fluid flows in the tubes while the other flows in a shell enclosing

the tubes.

5


(a) One shell pass and two tube passes.

(b) Two shell passes and four tube passes.

6


Baffle Types

7


(iv) Compact heat exchangers

Compact heat exchangers have very large area per unit volume.

8


13.2 Overall Heat Transfer Coefficient

oo

w

iiooii AhR

AhAUAU

1111

Heat transfer between two fluids A and B across a tube wall is

Overall heat transfer coefficient, U is given by

(13.4)

(13.3)

(13.2)

(13.1)

q

ii Ah

1

wR

Thermal Resistance Network

oo

w

ii

BA

AhR

Ah

TTq

11

TA

hi

ho

TB

q

k

ro ri

oo Ah

1

resistance thermal walltube

2

ln

kL

rrR iow

where Ui , Uo are based on Ai , Ao , respectively.

BAoioi TTAUq //

From eqns (13.1) & (13.3):

where L = length of tube, and i/o refers to

inside/outside surface of the smaller tube.

9


oo

w

ii

BA

AhR

Ah

TTq

11

For a double-pipe heat exchanger,

B

1 2

A (13.5)

(13.1)

Hence from eqn (13.4):

Rw is usually negligible for thin metallic tubes since ro ri

ln

= 0, and k is high.

oi hhU

111

UUUAAA oioi ; since

Furthermore, if one of the fluids is boiling or condensing, where h is

very high,

kL

rrR iow

2

ln (13.2)

oi horhU

10


Approximate Values of Overall Heat Transfer Coefficients Physical Situation U (W/m2K)

Brick exterior wall, plaster interior, uninsulated 2.55

Frame exterior wall, plaster interior, uninsulated 1.42

Frame exterior wall, plaster interior, rock-wool insulation 0.4

Plate-glass window 6.2

Double plate-glass window 2.3

Steam condenser 1100-5600

Feedwater heater 1100-8500

Freon 12 condenser with water coolant 280-850

Water-to-water heat exchanger 850-1700

Finned-tube heat exchanger, water in tubes, air across tubes 25-55

Water-to-oil heat exchanger, steam in tubes, air over tubes 110-350

Steam to light fuel oil 170-340

Steam to heavy fuel oil 56-170

Steam to kerosene or gasoline 280-1140

Finned-tube heat exchanger, steam in tubes, air over tubes 28-280

Ammonia condenser, water in tubes 850-1400

Alcohol condenser, water in tubes 255-680

11


13.3 Fouling Factors

Over time the surfaces of the tube may be pitted

due to corrosion and/or covered with deposits,

leading to increased thermal resistance to heat

transfer between the two fluids, A and B, which

can be accounted for by Fouling Factors,

B A

ooo

fo

w

i

fi

ii

BA

AhA

RR

A

R

Ah

TTq

11

(13.6)

.and fofi RR

q

ii Ah

1

wR

Thermal Resistance Network

oo Ah

1

o

fo

A

R

i

fi

A

R

12

Prof Andrew Tay ME3122 Heat Transfer 13


There are generally two types of analysis performed on heat exchangers.

13.3 Heat Exchanger Analysis

(a) Heat Exchanger Design

In this analysis, usually the inlet temperatures of the heating and

cooling fluids are known, and the heat exchanger type and area is

required to be determined in order to bring the outlet temperature of

one of the fluids to a specified value. All the inlet and outlet fluid

temperatures are known or can be easily determined. For this

analysis, the Log Mean Temperature Difference (LMTD) Method

is best used.

(b) Heat Exchanger Performance/Rating

In this analysis, a heat exchanger is available and one is required to

determine its performance under various flow conditions. For this

analysis, the Effectiveness-NTU Method is best used.

14


13.4 Log Mean Temperature Difference Method

As can be seen, T between the hot and cold fluid streams is not

constant over the HX area. Hence, the rate of heat transfer between

the two fluid streams is

where T is some suitable mean temperature difference.

(13.7) TUAq

2 2

T

A

Thi

Tho

Tco

Tci

T1 T2

T

A

Thi

Tho

Tci

Tco T1

T2

Parallel Flow Counter Flow 1 1

The figure below illustrates the variations in fluid temperature for

both parallel and counter flow heat exchangers.

T T

15


Heat transfer across

elemental area dA is

Consider a parallel flow HX:

ccchhh dTcmdTcmdq

hh

hcm

dqdT

cc

ccm

dqdT

cchh

chcmcm

dqTdTTd

11)(

(13.8)

Parallel Flow Heat Exchanger

T

A

Thi

Tho

dq

Tco

Tci

dA

T T1 T2

dAcmcm

UT

Td

cchh

11Eliminating dq (13.11)

TUdATTUdAdq ch Now, dq is also given by (13.10)

(13.9)

16


cchh cmcmUA

T

T

11ln

1

2

Integrating

(13.16)

(13.15)

(13.14)

(13.13)

(13.12)

1212 cccchhhh TTcmTTcmq

21 hh

hhTT

qcm

12 cc

ccTT

qcm

q

TT

q

TTUA

T

T cchh 1221

1

2ln

lm

chch TUAT/T

TTUA

T/T

TTTTUAq

12

12

12

1122

lnln

The total heat transfer as given by the energy balance is

or and

Substituting and gives

Log Mean Temp. Difference (LMTD)

17


Special Operating Conditions

pcmC Rate,Capacity (13.17)

(a) Hot fluid condensing: Th is constant and Ch is infinite.

(b) Cold fluid evaporating: Tc is constant and Cc is infinite.

(c) Ch = Cc : T is constant.

18


For other heat exchanger systems, the heat transfer is calculated using a

correction factor F applied to the LMTD for a counter flow double-

pipe system with the same hot and cold fluid temperatures. Thus

lmTUAFq

The above recognizes the fact that the counter flow double-pipe heat

exchanger is the most efficient flow arrangement.

(13.18)

13.5 Correction Factor Charts

Correction Factor, F, values are available for various heat exchanger

systems, usually in the form of charts.

19


11

12

tT

ttP

12

21

tt

TT

2112 ttcmTTcm tpSp

Sp

tp

cm

cm

In the Correction Factor charts provided, the parameters P and R are

defined as

and R =

where t : tube side; T : shell side;

1 : inlet and 2 : outlet

R is also =

Correction Factor Charts

and because

(13.20)

(13.19)

20


F can be seen as a correction factor for a heat exchanger, and is a

measure of the deviation of Tlm from the counter flow case. Also

Note:

When a phase change occurs, as in condensation or boiling

(evaporation), the fluid remains at essentially constant temperature in

the heat exchanger. Therefore,

P = 0 (t2 = t1)

or R = 1 (T2 = T1)

and F = 1 for boiling or condensation

values of P : 0 < P < 1

values of R : 0 R

For all cases, F 1.0

21


Fig. 1: Correction Factor for Heat Exchanger with

One Shell Pass and Two (or Multiples of Two) Tube

Passes.

1 : in

2 : out

22


Fig. 2: Correction Factor for Heat Exchanger

with Two Shell Passes and Four (or Multiples of

Four) Tube Passes.

23


Fig. 3: Correction Factor for Single Pass Cross-

Flow Heat Exchangers with the Shell Side Fluid

Mixed, and the Other Fluid Unmixed.

24


Fig. 4: Correction Factor for a Single Pass Cross-

Flow Heat Exchanger with Both Fluids Unmixed.

25


Example

Determine the heat transfer surface area required for a heat exchanger

constructed from 25.4 mm OD tube to cool 25000 kg/h of ethyl

alcohol solution (cp = 3.822 kJ/kgK ) from 65.6 C to 39.4 C using

22727 kg/hr of water available at 10 C. Assume that the overall heat

transfer coefficient based on the outer tube area is 568 W/m2K and

consider each of the following arrangements

a) parallel-flow double-pipe HX

b) counter-flow double-pipe HX

c) Shell and tube HX with 2 shell passes and 18 multiples of 4 tube

passes, alcohol flowing through the shell and water through the

tubes, and

d) Cross-flow HX, one tube pass and one shell pass, shell side mixed.

26


C36.1

102422727439665822325000

2

2

1221

c

c

cccchhhh

T

T....

TTcmTTcmq

W692778

4.396.65822.32500021

hhhh TTcmq

212

12

m65.9318.5568

692778

C18.51065.6/36.139.4ln

1065.636.139.4

ln

LMTDq/UA

T/T

TTLMTD

Solution:

Energy balance for determining the outlet temperature of water for each of the

4 arrangements:

Heat transfer rate,

a) Parallel flow double-pipe HX:

m826025409365 ./.D/AL (Too long, needs multiple tube passes.)

27


TUAq

2m3541529568

692778.

.A

b) Counter flow double-pipe HX:

We note that = = 26542 W/K (special case) and in this case temperature difference T between hot and cold fluids is

constant, and T = 65.6 36.1 = 29.5 C

m518025403541 ./.D/AL

Too long still, needs multiple tube passes.

28


47.0106.65

101.36

11

12

tT

ttP

112

21

tt

TT

cm

cmR

s

t

lmTUAFq

c) Shell and tube HX with multiple shell and tube passes:

From F vs P chart (Fig. 2, multiple of 4 tube passes), F = 0.97;

So A = 41.35 / 0.97 = 42.63 m2

There are n = 18 4 = 72 tube segments: A = DL n L = A / (72 D ) = 7.45 m

For 72 tube segments of 25.4 mm OD tubes, L of 7.45 m is meaningful

as compared with parallel-flow and counter-flow double-pipe cases.

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d) Crossflow HX, shell side mixed arrangement:

Again P = 0.47 R = 1.0

From chart F = 0.88

Hence A = 41.35 / 0.88 = 46.99 m2

Corresponding L is physically reasonable if number of tube

passes is sufficiently large.

30


13.6 Effectiveness-NTU Method

The LMTD method is convenient when Tci, Tco, Thi, Tho are known or

easily determined. When either Tis or Tos need to be determined from

q, U or A, a trial & error procedure is needed. The -NTU method is

more convenient to use for this situation. Also, it provides a basis for

comparing various heat exchangers.

The effectiveness is defined as

* actual rate of heat lost or gained by either fluid

** rate of heat gained by one fluid if it were to undergo a change in

temperature equal to maximum temperature difference present in

the heat exchanger.

**

*

ratetransferheatpossibleMaximum

ratetransferheatActual (13.21)

31


cihi TTcmq minmax

The fluid which attains this maximum change in temperature is the

one with the smaller/minimun value of capacity rate, ie .

(13.22)

mincm

This can be reasoned out as follows. If the temperature change for the

fluid with the minimum capacity rate is and that for the other fluid is , then for energy balance,

From the above equation,

BA TcmTcmq maxmin (13.23)

BA TT

Hence the fluid which can undergo a temperature change equal to the

maximum temperature difference in the HX, must be the one with the

minimum capacity rate . mincm

We define the Capacity Rate Ratio, max

min

max

min

C

C

cm

cmCr

(13.24)

32


cihi

cico

cihicc

cicoccc

cihi

hohi

cihihh

hohihhh

TT

TT

TTcm

TTcm

TT

TT

TTcm

TTcm

For parallel flow heat exchanger:

T

A

Thi

Tho

Tci

The subscript on in the above equations denotes which fluid has the minimum value of .

(13.25)

(13.26)

Tco

mincm

2 1

33


Thi

Tho

2 1

Tco

Tci

T

A

For counter flow heat exchanger:

cihi

cico

cihicc

cicoccc

cihi

hohi

cihihh

hohihhh

TT

TT

TTcm

TTcm

TT

TT

TTcm

TTcm

(13.28)

(13.27)

HXindifferenceetemperaturMaximum

fluidimumminT )(

In general,

(13.29)

The minimum fluid is always the one experiencing the larger T in the

HX and the maximum T is always (Thi Tci).

34


For the parallel flow HX, we had

(13.30)

Derivation of Effectiveness of Heat Exchangers

cchh cmcmUA

T

T

11ln

1

2

if the cold fluid is the minimum fluid.

(13.12)

chcihi

coho

CCUA

TT

TTei

11exp..

r

ch

c

c

CC

UA

C

C

C

UA1exp1exp

cocirhihocicochohih TTCTTTTCTTC .Now

Substituting for Tho into LHS of eqn (13.30), we get

cihi

cocicocircihi

cihi

cococirhi

cihi

coho

TT

TTTTCTT

TT

TTTCT

TT

TT

rcihi

cocir C

TT

TTC

1111

Equating the above with the RHS of eqn (13.30), we get

rrc CCCUA 1/1/exp1 (13.31)

35


If the hot fluid is the minimum fluid, it can be shown that the expression

for the effectiveness is the same as eqn (13.30), except that Cc and Ch are interchanged. Hence, the following expression will cater for both cases:

r

r

C

CCUA

1

1/exp1 min (13.32)

Similarly, for the counter flow HX, it can be shown that

rr

r

CNTUC

CNTU

1exp1

1exp1 (13.35)

nits)Transfer U of(Number / min NTUCUA

since it indicates the size of the HX. Hence,

(13.33)

r

r

C

CNTU

1

1exp1 (13.34)

36

Prof Andrew Tay ME3122 Heat Transfer 2013 Prof Andrew Tay ME4225 Industrial Heat Transfer 37


13.7 -NTU Charts for Heat Exchangers

Expressions for (NTU, Cr), have been obtained for other heat exchanger systems.

Often, these are presented in graphical form (-NTU charts) which are more convenient to use.

39


-NTU Method for Boilers and Condensers

0 and

0 if finite, is As

max

minmaxm

m

C

CCCC

.C,TTCq

r

m

If one fluid is boiling or condensing, the fluid temperature remains

constant and the other fluid is the fluid.

-NTUe- 1Hence from eqn (13.34): (13.37)

(13.36)

Why?

43

-NTUcihiminmax e-TTCqq 1 and (13.38)

Eqns (13.36) to (13.38) apply for all types of heat exchangers if one of

the fluids is boiling or condensing.


Example

Water at the rate of 68 kg/min is heated from 35C to 75C by an oil

having a specific heat of 1.90 kJ/kgK. The oil enters the counter

flow double-pipe heat exchanger at 110C and leaves at 75C. The

overall heat transfer coefficient U is 320 W/m2K. Calculate the heat

exchanger area.

Calculate the water exit temperature if the mass flow rate of water is

reduced to 40 kg/min with other parameters unchanged. Calculate

also the total heat transfer rate under the new conditions.


ccchhh TcmTcm

min.mh kg/97170751101900

3575418068

J/min.K3249001711900 hhcm

J/min.K284240418068 cccm

533.075

40

35110

3575

cihi

cico

TT

TT

Solution

a) Energy balance to heat exchanger

Capacity rates:

So water is the minimum capacity rate fluid.

87.0324900

284240

max

min C

C

UA/Cmin = 1.05

A = 1.05284240/(32060) = 15.5 m2

From chart. NTU = 1.05 .


b) = = 40 kg/min = 40 4180 = 167200 J/min.K = 2787 W/K

Now NTU = UA/Cmin = (320 15.5)/2787 = 1.78

5150324900

167200

max

min .C

CCr

and q = 2787(91.3 - 35) = 156.8 kW

C391

35110

35

.T

T

TT

TT

co

co

cihi

cico

Hence from chart, 750.


Example

A shell and tube heat exchanger with one shell pass and four tube

passes has 4.83 m2 of heat-transfer area. The overall heat transfer

coefficient is 320 W/m2K. It is proposed to cool a stream of oil (cp

= 2.22 kJ/kgK) at 120C, flowing at 1.50 kg/s, with cooling water

available at 13C and a flow of 0.63 kg/s.

Determine the exiting temperature of the 2 streams.


W/K3330222050.1 oocm

795.03330

2646

max

min C

CCr 584.0

2646

83.4320

min

C

UANTU

350

131202646max.

q

q

q act

Solution

So water is the minimum fluid.

-NTU chart gives : 0.35

q = 99093 W

99093 = 3330 (120 To2)

To2 = 120 29.8 = 90.2 C

99093 = 2646(Tw2 13)

Tw2 = 50.45 C

W/K2646420063.0 wwcm

ME3122-HX-2014

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