Heat Lecture Exchanger (Double Pipe)

1

Heat Exchanger Effectiveness

( )lmTm TFUATUAQ ∆=∆=

( ) ( )( )( )22

11

2211

lnTTTT

TTTTTlm

−′−′

−′−−′=∆

To calculate Q, we need both inlet and outlet temperatures:

1T ′

2T ′

2T1T

What if the outlet temperatures are unknown? e.g. calculate the performance of a given heat exchanger.

© Faith A. Morrison, Michigan Tech U.

Procedure:1. guess outlet temperatures

2. calculate ∆Tlm, FT

3. calculate Q4. calculate Q from energy balance5. compare, adjust, repeat.

To calculate unknown outlet temperatures:

This tedious procedure can be simplified by the definition of heat-exchanger effectiveness, ε.


2

Heat Exchanger Effectiveness

Consider a counter-current double-pipe heat exchanger:

coT

ciT′

hoThiT

hiT

hoT

ciT

coT

distance along the exchanger


Energy balance cold side:

( ) ( )cicocoldpcoldin TTmCQQ −==,

Energy balance hot side:

( ) ( )hihohotphotin TTmCQQ −=−=,

( )( )

( )( ) h

c

hiho

cico

coldp

hotp

TT

TTTT

mC

mC

∆∆=

−−−=


3

Temperature profile in a double-pipe heat exchanger:

xeTT

TT0

11

α=−′−′

−

′′=

mCmCRU

ppˆ1

ˆ1

20 πα


** REVIEW OF LECTURE 7 ** REVIEW OF LECTURE 7 **

0

20

40

60

80

100

120

0 0.2 0.4 0.6 0.8 1

x/L

T, T

' (oC

)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1T’

T

Note that the temperature curves are only approximately linear.

KkWmC

KkWCm

mRL

mKWU

p

p

/3

/6

4.152

/300

'

2

=

=′=

=π

=−′−′ L

xL

eTT

TT 0

11

α

11 TT

TT

−′−′

© Faith A. Morrison, Michigan Tech U.© Faith A. Morrison, Michigan Tech U.

−

′′=

mCmCRLUL

ppˆ1

ˆ1

20 πα

** REVIEW OF LECTURE 7 ** REVIEW OF LECTURE 7 **

4

hiT

hoT

ciT

coT


hT∆

cT∆

Case 1: ( ) ( )hc

coldphotp

TT

mCmC

∆>∆> cold fluid = minimum fluid


We want to compare the amount of heat transferred in this case to the amount of heat transferred in a PERFECT heat exchanger.

cohi TT =

hoT

ciT


If the heat exchanger were perfect, Thi=Tco

∞=A

( ) ( )cihicoldpA TTmCQ −=∞=

this temperature difference only depends on inlet temperatures

cold side:


5

∞=≡

AQQε

Heat Exchanger Effectiveness, ε

( ) ( )cihicoldp TTmCQ −=⇒ εcold fluid = minimum fluid

if ε is known, we can calculate Q without iterations


hiT

hoT

ciT

coT


hT∆

cT∆

Case 2: ( ) ( )hc

coldphotp

TT

mCmC

∆<∆< hot fluid = minimum fluid


6

If the heat exchanger were perfect, Thi=Tco

∞=A

( ) ( )cihihotpA TTmCQ −=∞=

this temperature difference only depends on inlet temperatures

hot side:

hiT

ciho TT =coT



∞=≡

AQQεHeat Exchanger Effectiveness, ε

( ) ( )cihihotp TTmCQ −=⇒ εhot fluid = minimum fluid

if ε is known, we can calculate Q without iterations

( ) ( )cihip TTmCQ −=min

ε

in general,


7

But where do we get ε ?The same equations we use in the trial-and-error solution can be combined algebraically to give ε as a function of (mCp)min, (mCp)max.

This relation is plotted in Geankoplis, as is the relation for co-current flow.

( )( )( )

( )( )

( )( )( )

−

−=

−−

−−

min

min

min

max

min

min

1

max

min

1

1

1

p

p

p

p

p

p

mC

mC

mCUA

p

p

mC

mC

mCUA

emC

mC

eε

counter-current flow:


Geankoplis 4th ed., p299

counter-current co-current

note: Geankoplis’Cmin=(mCp)min

Heat Exchanger Effectiveness forDouble-pipe or 1-1 Shell-and-Tube Heat Exchangers


8

Final ExamCM 310

November 16, 1998

3. (25 points) Water flowing at a rate of 0.723 kg/s enters theinside of a countercurrent, double-pipe heat exchanger at 300K andis heated by an oil stream that enters at 385K at a rate of 3.2kg/s.The heat capacity of the oil is 1.89 kJ/kg K, and the average heatcapacity of water over the temperature range of interest is 4.192kJ/kg K. The overall heat-transfer coefficient of the exchanger is300 W/m2 K, and the area for heat transfer is 15.4 m2. What is thetotal amount of heat transferred?


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Heat Lecture Exchanger (Double Pipe)