Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani
The conduction shape factor
q  kSToverall
The values of (S) have been worked out for several geometries and are summarized in
Table 3-1.For a three-dimensional wall, as in a furnace, separate shape factors are
used to calculate the heat flow through the edge and corner sections. When all the
interior dimensions are greater than one-fifth of the wall thickness,
S wall 
A
L
Sedge = 0.54D
Scorner = 0.15L
Where A = area of wall
L = wall thickness
D = length of edge
Example :-A horizontal pipe 15 cm in diameter and 4 m long is buried in the earth at
a depth of 20 cm. The pipe-wall temperature is 75℃, and the earth surface calculate
the heat lost by the pipe?
Solution. We may calculate the shape factor for this situation using the equation given
in Table 3-1. Since D < 3r.
S
2 (4)
2L

 15.35m
1
cosh ( D / r ) cosh 1 (20 / 7.5)
The heat flow is calculated from
q  kST  (0.8)(15.35)(75  5)  859.6W [2933Btu / h]
Example :-A small cubical furnace 50 by 50 by 50 cm on the inside is constructed of
fireclay brick [ k = 1.04 W/m·℃] with a wall thickness of 10 cm. The inside of the
furnace is maintained at 500℃, and the outside is maintained at 50℃. Calculate the
heat lost through the walls.
Solution. We compute the total shape factor by adding the shape factors for the walls,
edges, and corners:
A (0.5)(0.5)
S 
 2.5m
Walls:
L
0.1
S  0.54 D  (0.54)(0.5)  0.27m
Edges:
Corners:
S = 0.15L = (0.15)(0.1) = 0.015m
There are six wall sections, twelve edges, and eight corners, so that the total shape
factor is
S = (6)(2.5) + (12)(0.27) + (8)(0.015) = 18.36 m
And the heat flow is calculated as
q = kSΔT = (1.04)(18.36)(500-50) = 8.592 kW [29,320 Btu/h]
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani
Numerical method of analysis:- Consider a two-dimensional body that is
to be divided into equal increments in both the x and y directions, as shown in
Figure. The nodal points are designated as shown,the (m) locations indicating
the (x) increment and the n locations indicating the y increment.We wish to
establish the temperatures at any of these nodal points within the body, using

d 2T d 2T q
 2  0
2
dy
k
equation [ dx
] as a governing condition. Finite differences are
used to approximate differential increments in the temperature and space
coordinates; and the smaller we choose these finite increments. In order to solve
the above equation numerically .we will change each differential term in numeric
form. So, the first temperature derivative can be written as.
T
 Tm,n 
 T 
 m1,n
 x 

x
  m1/ 2,n
 inthe (x) nodels points
 T 
Tm,n  Tm1,n 

 

x
 x  m1/ 2,n

 T 
T
T 
 m ,n1 m ,n 
 
y
 y  m ,n1/ 2

inthe (y) nodels points

Tm ,n  Tm ,n1 


T



 y  m ,n1/ 2

y
Then ,the second derivative in x-direction can be wrote
T 
T 



x

T
 m1/ 2,n x  m1/ 2,n Tm1,n  Tm1,n  2Tm,n


x 2  m,n
x
(x) 2
in the (y) direction the second derivative can be wrote as;
2
 T

y 2  m,n
2
T 
T 


y  m,n1/ 2 y  m,n1/ 2
y

Tm,n1  Tm,n1  2Tm,n
(y ) 2

q
In the case of two dimensional heat conduction with no heat generation (  0 ),we
k
can apply the above result s ,then
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani
 2T  2T

0
x 2 y 2
Tm1,n  Tm1,n  2Tm,n Tm,n1  Tm,n1  2Tm,n

0
(x) 2
(y ) 2
If Δx = Δy, then
Tm1,n  Tm1,n  2Tm,n Tm,n1  Tm,n1  2Tm,n

 0 * (x) 2
2
2
(x)
(y )
Tm+1,n+Tm-1,n+Tm,n+1+Tm,n-1-4Tm,n = 0……………..(***)
Or
Tm,n 
Tm1,n  Tm1,n  Tm,n1  Tm,n1
4
Equation (***) states that the net heat flow in to any node is zero at steady state
conditions. When heat generation takes in account, then

Tm1,n  Tm1,n  2Tm,n Tm,n1  Tm,n1  2Tm,n q

 0
(x) 2
(y ) 2
k
and for square mesh Δx=Δy

q * (x) 2
Tm1,n  Tm1,n  Tm,n1  Tm,n1  4Tm,n 
 0(* * **)
k
Equation (****) can be applied for all internals nodes. The number of equations equal
to numbers of nodes, the equations can be solved simultaneously.
If the solid exposed to the some convention boundary condition, the temperature
calculated by energy balance.
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani
Consider steady heat transfer in an L-shaped solid body whose cross
section is given in Figure. Heat transfer in the direction normal to the
plane of the paper is negligible, and thus heat transfer in the body is twodimensional. The thermal conductivity of the body is k=15 W/m · °C, and
heat is generated in the body at a rate of g· =2 *106 W/m3. The left
surface of the body is insulated, and the bottom surface is maintained at a
uniform temperature of 90°C.The entire top surface is subjected to
convection to ambient air at T=25°Cwith a convection coefficient of h
=80 W/m2 · °C, and the right surface is subjected to heat flux at a
uniform rate of q·R =5000 W/m2. The nodal network of the problem
consists of 15 equally spaced nodes with Δx=Δy = 1.2 cm, as shown in
the figure. Five of the nodes are at the bottom surface, and thus their
temperatures are known. Obtain the finite difference equations at the
remaining nine nodes and determine the nodal temperatures by solving
them?