Heat Exchanger
Tutorial-1
Question-1
The overall heat transfer coefficient of an automotive radiator that is a compact cross flow
water to air heat exchanger with both fluids (air and water) unmixed as shown in the following
figure is tested. The radiator has 35 tubes of internal diameter 0.5 cm and length 65 cm in a
closely spaced plate-finned matrix. Hot water enters the tubes at 90°C at the rate of 0.6 kg/s
and leaves at 70°C. Air flows across the radiator through the interfin spaces and is heated from
20°C to 35°C. Determine the overall heat transfer coefficient of this radiator based on the inner
surface area of the tubes. [Assumption: the specific heat capacity of water is 4.195 kJ/(kg.C)].
Question-1
Question-1: Solution
Using LMTD method, we have
Given
C p , w = 4.195 kJ
Di = 0.005 m
L = 65 m
Q = UAF ( LMTD )
kg .C
T1 = Th ,in − Tc ,out = 55 C
T2 = Th ,out − Tc ,in = 50 C
T1 − T2
= 52.46 C
ln ( T1 T2 )
Th ,in = 90 C
LMTD =
Th ,out = 70 C
A = n DL = 35    0.005  .65 = 0.36 m 2
Tc ,in = 20 C
Tc ,out = 35 C

t2 − t1 Th ,out − Th ,in
P=
=
= 0.3 
T1 − t1 Tc ,in − Th ,in

 F = 0.985
T −T T −T
R = 1 2 = c ,in c ,out = 0.75

t2 − t1 Th ,out − Th ,in

t = Tube side
T = Shell side
Question-1: Solution
Solution 4
LMTD = 52.46 C
Q = mC p , w (T ) = 0.6  4.195  (90 − 70) = 50.34kW
A = 0.36 m 2
Q
U=
= 2.73 kW m 2 .C
AF ( LMTD )
F = 0.985
Question-2
Hot oil is to be cooled by water in a 1-shell-pass and 8 tube passes heat exchanger. The tubes
are thin-walled and are made of copper with an internal diameter of 1.4 cm. The length of each
tube pass in the heat exchanger is 6 m, and the overall heat transfer coefficient is 300 W/m2.C.
Water flows through the tubes at the rate of 0.2 kg/s, and the oil through the shell at a rate of
0.3 kg/s. The water and the oil enter at the temperature of 30֯C and 100֯C, respectively.
Determine the rate of heat transfer in the heat exchanger and the outlet temperature of the
water and the oil.
The specific heat capacity of water and oil are assument to be
C p , water = 4.195 kJ
kg .C
and C p ,oil = 2.13 kJ
kg .C
Question-2: Solution
Di = 0.014 m
Leach −tube = 6 m
U = 300 W
m 2 .C
mw = 0.2 kg
s
mo = 0.3 kg
s
Tw,in = 30 C
To ,in = 100 C
Given → not enough info for LMTD method
 -NTU method will be implemented
Question-2: Solution
kg
 4.195 kJ
= 0.839 kW
kg .C
C
s
kg
Ch = moC p ,oil = 0.3  2.13 kJ
= 0.639 kW
kg .C
C
s
Cmin = Ch = 0.639 kW
C
C
0.639
c = min =
= 0.762

Cmax 0.836
 = 0.5
Qmax = Cmin (100 − 30) = 44.73 kW
Cc = mwC p , water = 0.2
Q =  Qmax → Need to find 
To find 
2
UA 300 W m .C  ( n DL )
NTU =
=
Cmin
Cmin
300 W m 2 .C  ( 8    0.014  6 )
NTU =
= 0.99
639
Question-2: Solution
c = 0.764

  = 0.5
NTU = 0.99 
Q =  Qmax = 0.5  44.73 = 22.37 kW
Q = mwC p , w ( T ) w = moC p ,o ( T )o
Tw,out = 56.7 C
To ,out = 65 C
Question-3
Steam in the condenser of a power plant is to be considered at a temperature of 30oC with
cooling water from a nearby lake, which enters the tubes of the condenser at 14oC and leaves at
22oC. The surface area of the tube is 45m2, and the overall heat transfer coefficient is 1500
W/m2C. Determine the mass flow rate of the cooling water needed and the rate of condensation
of the steam in the condenser.
Use
Q = UA ( LMTD ) and
kJ
kg
For condensation, the heat reject from
h fg at condenser = 2431
the process of steam to condensate water
can be calculated as multiplication of
mass and h fg at condenser (enthalpy difference
between phase changes).
Question-3: Solution
Using LMTD method, we have
Q = UA( LMTD )
T1 = Th ,in − Tc ,out = 30 − 22 = 8 C
T2 = Th ,out − Tc ,in = 30 − 14 = 16 C
LMTD =
LMTD = Tm =
T1 − T2
= 11.5 C
ln ( T1 T2 )
T1 − T2
T

ln  1


T

2
where
A = 45 m ( Given )
2
T1 = Th ,in − Tc ,out
T2 = Th ,out − Tc ,in
Q = UA ( LMTD )
W
2

45
m
11.54 C
2
mC
Q = 779.055 kW
Q = 1500
Q = mwc p , w ( T ) w = ms h fg
779.055 kW = mw  4.195
mw = 23.2 kg
s
kJ
(22 − 14) K
kg .K
Question-4
The overall heat transfer coefficient U of a circular fin heat exchanger in which a hot gas
flows across the finned tubes and oil flows through the tubes is given by
1
1
1
=
+
U t hg   o 
  g  ho


i.
Explain briefly how the efficiency on the gas side t of the finned tube is determined.
ii.
For the case of circular finned tube, surface CF-8.7-5/8J, determine the value of for the  o
flow of oil inside the finned tubes.
For surface CF-8.7-5/8J the following expression may be assumed:


 =  Ac A  =  rh
fr 

The symbols have their usual meanings. The data for surface CF-8.7-5/8J are given in the
following figure. Use the plate spacing of 1.848 in.
Question-4
Question-4: Solution
1
1
1
=
+
U t hg   o 
  g  ho


t = efficiency of finned tube
or overall surface efficiency
t  A = ( A − A f ) + ( A f   f )
t =
( A − Af )
t = 1 −
A
Af
A
(
+
1 − )
(
A
f
f
 f
A
)
Question-4: Solution


=
 =  rh 



  o (oil side) = 
 A   
 rh

 = A /V = 
=

 LAfr   rh 





Ac
 Arh
=
Afr  LA
 fr
2
( 5)
  Arh
 = 
  V

Ac
=

 Afr rh
2
D
= 2
4
8 4
Afr = S  1.35 = 1.848  1.35
Ac = 
rh = flow cross sectional area/wetter perimeter
D2

4 =D
rh =
D
4
D2

4
Ac
D
in 2
 o (oil side) =
=
=
= 0.787 3
Afr rh 1.848 1.35  D 1.848 1.35
in
4
Question-5
Figure shows the core geometry of a circular, finned-tube core structure (CF-7.0-5/8J)
for the design of a cross-flow heat exchanger for recovery for heat from the exhaust
gas of a micro-turbine. In operation, hot exhaust gas flows across the finned tube
while water flows through the tubes.
(i) Determine the frontal area, free flow area, and the total heat transfer area on the
gas side based on a tube length of 1 inch.
(ii) Determine the free flow area, the frontal area and the hydraulic radius for the
flow on the water side.
The tube outside diameter is 0.645 in or 16.38
mm, fin pitch is 7.0 per inch or 276 per m. The fin
thickness is 0.01 inch or 0.254 mm.
Question-5: Solution
Air Side:
S = 1.232"
d o = 0.645"
Frontal Area = Afr = S 1 = 1.232 in 2 
Free flow area = Ac = Agap + Awithin − fins

d
Agap =  S − 2  o
 2

Awithin − fins
  1.121 − d o
 − 2
2
 
 1.121 − d o
= 2
2

Ac = 0.5537in 2 

2

1
=
0.111
in




 
2


1
−
7

0.01
=
0.4427
in
(
)
 
  total fin thickness 
Question-5: Solution
A = total heat transfer area = Afin + Afin −tip + Aunfin
Afin =

1.121
(
4
2
)
− 0.6452 
7
no. of fins

2
two surfaces
Afin −tip =   1.121 0.01 7
Aunfin =   0.645  (1 − 0.01 7)
A = 11.37 in 2 
OilSide
Ac ,o =
Afr ,o
 5
2
2
=
0.3068
in
 
4 8
= 1.35 1.232 = 1.6632 in 2
rh = flow cross sectional area/wetter perimeter
D2
5

D
4 = = 8 = 0.15625 in
rh =
D
4
4
END
END