# heat exchanger - suranaree university of...

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Heat exchangers are devices in which heat is transferred between two fluids at different temperatures without any mixing of fluids.

Heat exchanger type

Heat exchanger

1. Direct heat transfer type2. Storage type3. Direct contact type

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Heat exchanger

1. Direct heat transfer typeA 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

Concentric tube heat exchangers. (a) Parallel flow. (b) Counter flow.

hot fluid hot fluid

cold fluid

cold fluid

Heat exchanger2. Storage type heat exchangerA 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 exchanger3. 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.

Direct type heat exchanger1. Tubular 2. Plate3. Extended surface

Heat exchanger

Tubular heat exchanger1. 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 exchangerSeries 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 exchangerFins 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 arrangementThe three basic flow arrangements:

o Parallel flowo Counter flowo 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 (U) and fouling factor

Heat exchanger

In a heat exchanger, the heat is transferred by both convection and conduction.

q = UA (Ta –Tb)

U is overall heat transfer coefficient Tb

Ta

hohi

T1

T2

Heat exchanger

Tb

Ta

hohi

T1

T2

1ai

q T Th A

1 2q T Tk A

x

2 bo

q T Th A

a bq T T

U A

Conv.

Cond.

Conv.

Plate heat exchanger 1)

2)

3)

4)

Heat exchanger

a bi o

q q x q T Th A kA h A

a bq T T

U A

i o

q q q x qU A h A kA h A

1 1 1i o

xU h k h

Adding above three equations (1, 2, 3)

From (4)

Across a plain wall

21 h1

kb

h1

U1

Heat exchanger

r2

r1Tb

Ta

R i R a R o

T i To

To

Ti

condxR

kA

21

convo

Rh A

RT = Rconv1 + Rcond + Rconv2

11

convi

Rh A

Tubular heat exchanger

Heat exchanger

Ai = 2πriL

oolm

io

iiii Ah1

kA)rr(

Ah1

AU1

oo

i

lm

iio

ii rhr

krr)rr(

h1

U1

)r/rln()rr(rio

iolm

โดย

Based on inner area

oolm

io

iioo Ah1

kA)rr(

Ah1

AU1

Based on outter area

olm

oio

ii

o

o h1

krr)rr(

rhr

U1

Heat exchanger

oi h1

h1

U1

When the wall thickness of the tube is small and the thermal conductivity of the tube material is high, is as usually the case, the thermal resistance of the tube is negligible.

oi AA because

Heat exchanger

FoulingThe performance of heat exchangers usually deteriorate with time as a result of accumulation of deposits on heat transfer surfaces, representing additional resistance, called fouling.

(Log) Mean Temperature Difference

Heat exchanger

hot

Where:

cold

dA Th

Tc

TdAUdq ch TTT

Total heat transfer rate in heat exchanger TdAUq

Heat exchanger

If U is assumed to be a constant TdAUq

Define mean temperature difference

area

m TdAA1T

Thus:mTUAq

This is the basic performance equation for a direct transfer type heat exchanger

Parallel Flow

Heat exchanger

Assumption1. U is a constant2. Heat exchanger is adequately

insulated i.e. no heat loss to surrounding

Consider in elementary area dA (B.dx)

Heat exchanger

dxBTUdq

hphh dTCm

cpcc dTCm

ch TTT

ch dTdT)T(d

pccphh Cmdq

Cmdq

pccphh Cm1

Cm1dxBTU

Heat exchanger

L

0pccphh

T

T

dxUBCm1

Cm1

T)T(do

i

Where: i,ci,hi TTT o,co,ho TTT

UACm1

Cm1

TTln

pccphhi

o

UATTTTq1

i,co,co,hi,h

Heat exchanger

o

i

oi

TTln

TTUAq

This is the performance equation for a parallel-flow heat exchanger

mTUAq

Comparing with:

o

i

oim

TTln

TTT

Where:

Heat exchanger

For counter flowAssumption

1. U is a constant2. Heat exchanger is adequately

insulated i.e. no heat loss to surrounding

Consider an elementary area dA (B.dx)

Heat exchanger

dxB)TT(Udq ch

hphh dTCm

cpcc dTCm

ch TTT

ch dTdT)T(d

pccphh Cmdq

Cmdq

pccphh Cm1

Cm1dxBTU

Heat exchanger

L

0pccphh

T

T

dxUBCm1

Cm1

T)T(do

i

UACm1

Cm1

TTln

pccphhi

o

UATTTTq1

TTln i,co,co,hi,h

i

o

Where: o,ci,hi TTT i,co,ho TTT

Heat exchanger

o

i

oi

TTln

TTUAq

Th,i

Th,oTc,o

Tc,ioT

iT

mTUAq

Comparing with:

o

i

oim

TTln

TTT

Where:

Heat exchanger

pccphh cmcm

Special case of counter flow

mT

Then:i,co,co,hi,h TTTT

i,co,ho,ci,h TTTT Or

Substituting into the expression for , we get an indeterminate quantity

Heat exchanger

pTT

o

i

Define

Then:

Apply L’ Hopital’s rulepln

)1p(TlimT o1pm

oo

1pmT

p1

)1(TlimT

iom TTT

iT

oT

Heat exchangerCross flowCase 1: both fluids unmixed

Cold fluid

Hot fluid

Th,i

Th,o

Tc,i Tc,o

B

L

x

y

Heat exchanger

Cross flowCase 2: one fluid mixed, the other unmixed

Cold fluid

Hot fluid

Th,i

Th,o

Tc,i Tc,o

B

L

x

y

Th = f (x,y)Tc = f (x)

Heat exchanger

Cross flowCase -: Both fluid mixed

Cold fluid

Hot fluid

Th,i

Th,o

Tc,i Tc,o

B

L

x

y

Th = f (y)Tc = f (x)

Heat exchanger

Both Th and Tc are functions of x and y

Considering an elementary area dA (= dx dy)

dydx)TT(Udq ch

dydx)TT(UqB

0

L

0 ch

Comparing with mTUAq

dydx)TT(BL1T

B

0

L

0 chm

More complicated than before but it has been done.

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)

flowcounterwastarrengementheifm

flowcrossm

TT

F

If the bulk exit temperatures on the hot side and cold side are Th,o and Tc,o, then

i,co,ho,ci,hi,co,ho,ci,h

flowcounterm TT/TTlnTTTT

T

if the arrangement was encounter-flow

Heat exchanger

Mean temperature difference in cross flow

flowcounterwastarrengementheifm

flowcrossm

TT

F

For given values of Th,i; Th,o; Tc,i; Tc,o

is the highest amongst all flow arrangements flowcounterT

Therefore: 1F0

Heat exchanger

flowcrossmflowcross TUAq

flowcountermflowcross TUAFq

F is plotted as a function of two parameters, R and S i,2o,2

o,1i,1

TTTT

R

i,2i,1

i,2o,2

TTTT

S

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 = h2 = c

or 1 = c2 = 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 0R

The ratio of change in temperature of one of the fluid to the difference of inlet temperature of the two fluids 1S0

What is the parameter S?

Heat exchanger

i,2o,2

o,1i,1

TTTT

R

i,2i,1

i,2o,2

TTTT

S

i,1T

o,1T

i,2T o,2T

Both fluids unmixed cross flow heat exchanger

Heat exchangeri,1T

o,1T

i,2T o,2T

i,2i,1

i,2o,2

TTTT

S

i,2o,2

o,1i,1

TTTT

R

One fluids mixed and the other unmixed

mixed

unmixed

Heat exchanger

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Heat exchanger

The effectiveness - NTU methodGenerally, we encounter two type of problems:

Given:

hm

cm

o,hi,h T,T o,ci,c T,TUTwo fluids Find A?

Type 1

Given:hmi,hT i,cT

Two fluids A heat exchanger A

Type 2

cm Find Th,o; Tc,o?

Heat exchanger

Type 1

mTUAq

mTUqA

Type 2

mTUAq

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

maxqq

)TT()Cpm()TT(Cm

i,ci,hs

o,hi,hphh

)TT()Cpm()TT(Cm

i,ci,hs

i,co,cpcc

i,hT

i,cT

T

o,hT

i,cT

o,ci,h TT

Length of heat exchanger

Heat exchanger

Hence if , then pccphh CmCm spphh CmCm

i,ci,h

o,hi,h

TTTT

Hence if , then phhpcc CmCm sppcc CmCm

i,ci,h

i,co,c

TTTT

Note 1) The definition are equivalent when phhpcc CmCm 2) By definition 10

Heat exchanger

Effectiveness – parallel flowAssume spphh )Cm(Cm

i,ci,h

o,hi,h

TTTT

pcc

phh

pcc

phh

CmCm

1

CmCm

1

pcc

phh

i,ci,h

i,co,c

i,ci,h

o,hi,h

CmCm

1TTTT

TTTT

pcc Cmq i,co,c TT

pcc

phh

o,hi,h

i,co,c

CmCm

1

)TT()TT(

1

Heat exchanger

Effectiveness – parallel flow

pcc

phh

i,ci,h

o,co,h

CmCm

1TTTT

1

UACm1

Cm1exp

TTTT

pccphhi,ci,h

o,co,h

Derived earlier (slide 23)

UACm1

Cm1

TTln

pccphhi

o

Heat exchanger

pcc

phh

phhpcc

phh

CmCm

1Cm

UACmCm

1exp1

Substituting

Heat exchanger

If we had assumed initially sppcc )Cm(Cm

phh

pcc

pccphh

pcc

CmCm

1Cm

UACmCm

1exp1

, then

We combine the two expressions for

bp

sp

spbp

sp

CmCm

1Cm

UACmCm

1exp1

Heat exchanger

Define two new parameters

Capacity ratio (C)

max

min

bp

sp

CCor

)Cm()Cm(

C

Number of transfer unit (NTU)

minsp CUAor

)Cm(UANTU

Note: 1) Both are dimensionless2)3) NTU

10 0

Heat exchanger

Effectiveness – parallel flow

C1NTUC1exp1

Effectiveness – counter flow

NTUC11expC1

NTUC1exp1

Heat exchanger

Special casesCapacity rate (m.Cp) is infinite either on the hot side or the cold side

0C

For this solution, we obtain the relation

NTUexp1

Heat exchanger

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Heat exchanger

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Heat exchanger

Heat exchanger

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.

Example

Heat exchanger

LMTDSolution, Part 1:

K375T in,oil

K350T in,oil

K280T in,water

?T out,water K.Kg/J181,4C c,p

C/qTT and

TT Cq

ci,co,c

o,hi,hh

W125,26

)350375(090,25.0

K311

)181,42.0/(125,26280

Heat exchanger

• For parallel flow, 95Ti 39To

63)39/95ln(

3995T/Tln

TTToi

oiPFm,

• For counter flow, 64Ti 70To

67)70/64ln(

7064T/Tln

TTToi

oiCFm,

95Ti 39To

70To

64Ti

Heat exchanger

• For parallel flow,

• For counter flow,

2PF,mPF m66.1)TU/(qA

2CF,mCF m56.1)TU/(qA

Heat exchanger

Part 2:• Use -NTU method to determine the required NTU and heat transfer

area for parallel and counter flowSolution

K.Kg/J181,4C c,p

K/W104520905.0Cm phh

sppcc )Cm(K/W2.836181,42.0Cm

To determine the minimum heat capacity rate,

Heat exchanger• Then

W440,79)280375(2.836

)TT(Cq i,ci,hminmax

W26,125 TT Cq o,hi,hh • The effectiveness is

33.0440,79/125,26q/q max

With 8.0

045,12.836

)Cm()Cm(

Cbp

sp

Heat exchanger

Parallel flow Counter flow

2PFminPF m84.1U/NTUCA

2CFminCF m67.1U/NTUCA

0.55 0.50

Heat exchanger

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Heat exchanger

Other consideration in designing heat exchangers

1. Pressure drop on either sides2. Size restriction3. Stress consideration4. Servicing requirements5. Materials of construction6. System operation7. Cost