heat transfer in fins

1. Heat Transfer Third Year
Conduction-ConvectionSystems
In heat-exchanger applications a finned-tube arrangement might be used to remove heat from
a hot liquid. The heat transfer from the liquid to the finned tube is by convection. The
extended-surface problem is a simple application of a conduction-convection systems.
Consider the one-dimensional fin exposed to a surrounding fluid
shown in Figure below. The temperature of the base of the fin is T0.
at a temperature T∞ as
Let P = perimeter = 2(Z + t)
The energy balance on an element of the fin of thickness dx is
 dT d 2
T 
dT
  kA  kA  dx  hPdx(T T ),
dx dx
 dx 
dx2
 

2. Heat Transfer Third Year
d
dT
let   T  T ,  
dx dx
d 2
T d 2


dx2
dx2
Let m2
= hP/kA
2
d
 m2
  0
dx2
(D-Operator)
D2
1  0  D   1
 The roots of the equation above are:
m1  1, m2  1
The general solution is
Several cases may be considered:
CASE1: The fin is very long.
at x  0 T  T0   0
at x   T  T   0
B.C.1
B.C.2
By applying the boundary conditions on the general equation above, we get
CASE2: The end of the fin is insulated so that dT/dx = 0 at x = L.
at x  0 T  T0   0
at x  L dT / dx  0 d / dx  0
B.C.1
B.C.2
From B.C.1 0  C1  C2  C1  0  C2
dT
dx
mL mL
From B.C.2  mC1e  mC2 e  0
Subs. B.C.1 in B.C.2, we get

3. Heat Transfer Third Year
emL
emL
 
 0
 0
 
C  C
2 1 0
emL
 emL
emL
 emL
Subs. C1&C2 in the general equation, we obtain
CASE3: The fin is of finite length and loses heat by convection from its end.
at x  0 T  T0   0
B.C.1
 k
d
dx
 h
B.C.2 at x  L
From B.C.1 0  C1  C2  C1  0  C2
 k mC e mL
 mC emL
 hC e mL

C
emL

From B.C.2 1 2 1 2
By applying C1&C2 in the general equation, we get,
CalculationofHeatLostbytheFin
The heat lost by the fin can be calculated either by
or by

4. Heat Transfer Third Year
FORCASE1:
 d
dx
d
dx
mx mx
 m0 e ,  m0
 e ,
0 x0
hP
 kA( m0 )  kAm0 , For q  kA 0 

kA
q  hPkA 0
FORCASE2:
cosh[m(L  x)] d   me mx
memx


,  0  
2mx
 
2mx
0 cosh mL

dx  1  e 1 e 
emL
 e mL


d
1 1 
 0 m   0 m  
mx mL
 2mx 
2mx
 1 e 1 e dx e  e
  
x0
mL
 mL


 q  kA m 
e e
 
 emx
 emL 
0
 
q  hPkA 0 tanh mL
FORCASE3:
The heat flow for this case is
d
dx