Download ( ) ( ) q mc T T mc T T = - = - x x

Design of a Tubular Heat Exchanger
Design of a Tubular Heat Exchanger
We will use the following assumptions:
1. Heat transfer is under steady-state conditions.
2. The overall heat-transfer coefficient is constant throughout the length of pipe.
3. There is no axial conduction of heat in the metal pipe.
4. The heat exchanger is well insulated.
Change in heat energy in a fluid stream, if its temperature changes from T1 to T2, is expressed
as:
 = mass flow rate of a fluid (kg/s),
where m
cp = specific heat of a fluid (kJ/kgC),
temperature change of a fluid is from some inlet temperature T1 to an exit temperature T2.
Hot fluid, H, enters the heat exchanger at location (1) and it flows through the inner pipe, exiting
at location (2).
Its temperature decreases from TH, inlet to TH,exit .
Fluid C, is a cold fluid that enters the annular space between the outer and inner pipes of the
tubular heat exchanger at location (1) and exits at location (2).
Its temperature increases from TC,inlet to TC,exit.
Conducting an energy balance, the rate of heat transfer between the fluids is:
q  m H c pH (TH ,inlet  TH ,exit )  m C c pC (TC ,exit  TC ,inlet )
where cpH is the specific heat of the hot fluid (kJ/kgºC),
cpC is the specific heat of the cold fluid (kJ/kgºC),
m H is the mass flow rate of the hot fluid (kg/s),
m C is the mass flow rate of the cold fluid (kg/s).
The preceding equation has limited application
1
Fluid c
Insulation
Fluid h
Fluid h
Fluid c
2
1
T
Th, inlet
dTh
dq
Th, exit
Tc, exit
dTc
Tc, inlet
A
Consider a thin slice of the heat exchanger, the rate of heat transfer, dq, from fluid H to
fluid C may be expressed as:
where T is the temperature difference between fluid H and fluid C.
The temperature difference, T, between the two fluids H and C is
For a small differential ring element as shown in Figure, using energy balance for the hot stream
H
and, for cold stream C
2
dq  m C c pC dTC
In Eq , dTH is a negative quantity, therefore we added a negative sign to obtain positive value for
dq. Solving for dTH and dTC, we obtain
dTH  
dq
m H c pH
and
dTC 
dq
m C c pC
Then subtracting
 1
1
dTH  dTC  d (TH  TC )  dq 

 m c

 H pH mC c pC



or
 1
d (TH  TC )
1
 U 


(TH  TC )
 m H c pH m C c pC

 dA

Integrating
ln
 1
1
 UA 


 TC ,inlet )
 m H c pH m C c pC
(TH ,exit  TC ,exit )
(TH ,inlet



Noting that
we get
ln
 1
T2
1
 UA 


T1
 m H c pH m C c pC



Substituting
T
T
T
 T 
T

ln  2   UA  H ,inlet H ,exit  C ,exit C ,inlet 
q
q


 T1 
Rearranging terms
3
 T 
UA
ln  2   
 (TH ,inlet  TC ,inlet )  (TH ,exit  TC ,exit ) 
q
 T1 
Rearranging terms,
q  UA( Tlm )
where
Tlm 
T2  T1
T
ln 2
T1
Tlm is called the log mean temperature difference.
A Counter-Current Heat Exchanger
Fluid c
Insulation
Fluid h
Fluid h
Fluid c
2
1
T
Th, inlet
Tc, exit
dq
Th, exit
Tc, inlet
A
4
Class Problem
Calculate the log mean temperature difference for a heat exchanger with the following data, the
temperature of hot stream decreases from 150C to 50C, while the temperature of cold stream
increases from 40C to 80C.
Consider another case when temperature of hot stream decreases from 150C to 50C and cold
stream temperature increases from 40C to 45C.
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