Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani
Chapter three
Steady-State Conduction two dimensions
For steady state, two dimensional heat transfers by conduction with no
heat generation the general heat conduction equation reduced to
 2T  2T

 0......(3 - 1)
x 2 y 2
assuming constant thermal conductivity. The solution to this equation
may be obtained by analytical, numerical, or graphical techniques.The
objective of any heat-transfer analysis is usually to predict heat flow or
the temperature that results from a certain heat flow. The solution to
Equation (3-1) will give the temperature in a two-dimensional body as a
function of the two independent space coordinates x and y. Then the heat
flow in the x and y directions may be calculated from the Fourier
equations q x  kAx
q y  kAy
T
.....(3  2)
x
T
.........(3  3)
y
So if the temperature distribution inthe material is known, we may easily
establish the heat flow.
MATHEMATICAL ANALYSIS OF TWO-DIMENSIONAL HEAT
CONDUCTION
To solve Equation (3-1), the separation-of-variables method is used. The
essential point of this method is that the solution to the differential
equation is assumed to take a product form
T = XY where X = f(x)
2
Y = f(y)
dT
dX
dT
d2X
 Y.
 Y. 2 ..(3  4) and
dx
dx
dx 2
dx
2
2
dT
d Y
 Y. 2 ....(3  5)
2
dy
dy
by subsituting in controllin g differntial eqn.

1 d 2 X 1 d 2Y

.....(3  6)
X dx 2 Y dy 2
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani
Each side of Equation (3-6) is independent of the other because x and y
are independent variables. This requires that each side be equal to some
constant. We may thus obtain two ordinary differential equations in terms
of this constant .
1 d 2 X 1 d 2Y


 2
2
2
X dx
Y dy
This equation gives two differential equations
d2X
 2 X  0 …(3-7)
2
dx
d 2Y
 2Y  0 ….(3-8)
2
dy
Where λ2is called the separation constant. Its value must be determined
from the boundary conditions. Consider a rectangular plate shown in
figure ,three sides of plate maintained at constant temperature (T1). The
boundary conditions with a sine-wave temperature distribution impressed
on the upper edge of the plate. Thus
(1)T=T1
at y = 0
(2)T = T1
at x = 0
(3)T=T1
at x = W
(4) Tm  Tm sin
x
W
at y = H
Where (Tm) is the amplitude of the sine function.
The first step in solving the problem is examine the value of (λ2),there are
three possibilities (λ2=0, λ2<0,and λ2 >0) one of this value is acceptable
,and the other will be neglected.
For λ =0
2
d2X
0
dx 2
The solutions are X=(C1+C2x)
Y=(C3+C4y)
and
d 2Y
and
0
dy 2
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani
T= (C1+C2x) *(C3+C4y)…(3-9)
This function cannot fit the sine-function
2
boundary condition, so theλ =0 solution may be excluded
X  (C5e  x  C6 e x )
For  2 < 0
Y  (C7 cos y  C8 sin y )
T  (C5e x  C6 e x )(C7 cos y  C8 sin y ) .....(3 - 10)
Again, the sine-function boundary condition cannot be satisfied, so this
solution is excluded also.
X  (C9 cos x  C10 sin x)
For  > 0
Y  (C11e y  C12e y )
2
T  (C9 cos x  C10 sin x)(C11e y  C12e y )....(3  11)
Now, it is possible to satisfy the sine-function boundary condition; so we shall attempt
to satisfy the other conditions. The algebra is somewhat easier to handle when the
substitution is made as
  T  T1
The transform of the boundary conditions are
(1)θ = 0
(3)θ = 0
at y = 0
(2)θ = 0
at x = W
(4)θ =
Tm sin
Applying these conditions, we have
0  (C9 cos x  C10 sin x)(C11  C12 )...[ a]
0  C9 (C11e y  C12e y ).....[b]
0  (C9 cos W  C10 sin W )(C11e y  C12e y )....[c]
Tm sin
x
W
 (C9 cos x  C10 sin x)(C11e H  C12e H )...[ d ]
Accordingly
C11=-C12
and
From [c]
0  C10 C12 sin W (e y  e  y )
This requires that
SinλW = 0 …..(3-13)

n
W ……(3-14)
C9=0
at x = 0
x
W
at y = H ..(3-12)
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani
where n is an integer. The solution to the differential equation may thus
be written as a sum of the solutions for each value of n. This is an infinite
sum, so that the final solution is the infinite series.

  T  T1   Cn sin
n 1
nx
ny
sinh
.....(3  15)
W
W
Where the constants have been combined and the exponential terms
converted to the hyperbolic function. The final boundary condition may
now be applied:
Tm sin
x
W

  Cn sin
n 1
nx
nH
sinh
W
W
which requires that Cn = 0 for n > 1. The final solution is therefore
T  Tm
sinh( y / W )
x
sin( )  T1......(3  16)
sinh( H / W )
W
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani
Graphical analysis
Consider the two-dimensional system shown in Figure. The inside surface
is maintained at some temperatureT1, and the outer surface is maintained
atT2. To calculate the heat transfer, isotherms and heat-flow lanes have
been sketched to aid in these calculation .The isotherms and heat-flow
lanes form groupings of curvilinear figures like that shown inFigure. The
heat flow across this curvilinear section is given by Fourier’s law,
assuming unit depth of material.
if [Δx=Δy], the heat flow is proportional to the ΔT across the element
and, since this heat flow is constant, the ΔT across each element must be
the same within the same heat-flow lane. Thus the ΔT across an element
is given by
Where N is the number of temperature increments between the inner and
outer surfaces.
Where M is the number of heat-flow lanes. the number of temperature
increments between the inner and outer surfaces is a bout N=4, while the
number of heat flow lanes for the corner section may be estimated as
M=8.2. The total number of heat-flow lanes is four times this value, or
4×8.2=32.8. The ratio [M/N] is thus 32.8/4=8.2 for the whole wall
section.