A Derivation of the Heat Conduction Equation

Ernesto Gutierrez-Miravete

September 19, 1994

1  The Thermal Energy Balance

Consider a small region inside a solid material which is being heated (or cooled). The molecules inside the region vibrate due to thermal excitation. A quantitative expression of molecular thermal excitation is provided by the specific internal energy u (i.e. internal energy per unit volume). The specific internal energy of a heated (or cooled) solid body changes when energy is added to or substracted from the body.

Consider the small region inside the body to be a tetrahedron with a vertex at the origin of a rectangular cartesian system of coordinates and with three of its sides along the three coordinate axes. Let ax, ay, az denote the areas of the tetrahedron faces perpendicular to the Ox, Oy, and Oz axes, respectively, and let ao be the area of the fourth face. Further, let [(no)\vec] the outward normal to the inclined face. Assume that energy enters (or exits) the tetrahedron through all faces. A simple energy balance (i.e. time rate of change of specific internal energy = total energy flux through all boundaries) for the tetrahedral element gives




u
t
dV =

f(x,y,z,
i
 
,t) dax +

f(x,y,z,
j
 
,t) day +


f(x,y,z,
k
 
,t) daz +

f(x,y,z,-
no
 
,t) dao
Here, the f's are the energy flows through the different faces of the element (i.e. energy per unit area per unit time). These flows are regarded as positive if they are in the direction of, respectively, [i\vec], [j\vec], [k\vec] or [(no)\vec].

The total volume of the tetrahedron is [1/3] h ao, where h is the perpendicular distance from the origin to the face of area ao, from the mean value theorem one can write

1
3
h ao
u
t
 
= ax
qx
 
+ ay
qy
 
+ az
qz
 
+ ao
f(x,y,z,-
no
 
,t)
 
(1)
where the bars denote mean values and
qx = f(x,y,z,
i
 
,t)
(2)
qy = f(x,y,z,
j
 
,t)
(3)
qz = f(x,y,z,
k
 
,t)
(4)
are used to simplify notation. Now, taking the limit as h 0 while maintaining [(no)\vec] constant, the term [1/3] h ao[`([(u )/(t)])] 0 and since
f(x,y,z,-
no
 
,t) = - f(x,y,z,
no
 
,t)
(5)
one obtains,
f(x,y,z,
no
 
,t) = qx cos(
no
 
,x) + qy cos(
no
 
,y) + qz cos(
no
 
,z)
(6)
where cos[((no)\vec],xi) are direction cosines of the outward normal to the face of area ao. Thus, the function f(x,y,z,[n\vec],t) is the projection of certain vector [q\vec] = (qx,qy,qz) onto the direction of the normal vector [(no)\vec]. The vector [q\vec], analogous to the mass flux (i.e. mass crossing an unit area per unit time) in fluid flow problems, is called the heat flux vector and represents energy transferred per unit time across an unit area.

Finally, for a small region of arbitrary shape, the energy balance equation is




u
t
dV = -


q
 
·
n
 
da
(7)
where [n\vec] is the normal to the region's boundary. The divergence (Green's) theorem can be applied to the right hand side of the above equation to give



u
t
dV = -


·
q
 
dV
(8)
Since this equation is valid for any volume, the integrands are equal and
u
t
= - ·
q
 
= - div (
q
 
)
(9)
This is the general form of the differential thermal energy balance equation or simply the energy equation.

2  The Heat Flux Vector and the Heat Conduction Equation

The temperature change at any point in a body is the result of heat propagation processes. The rate of temperature change with distance along a particular direction [n\vec] is

T
n
=
n
 
·T
(10)
where T is the temperature gradient vector. For an isotropic medium, if the temperature increases in a direction normal to a surface S, then the heat flow across S must be negative and proportional to the rate of temperature increase along that direction, (Fourier's law) i.e.

q
 
= - k T
(11)
where k is some positive scalar quantity which may depend on the medium and the temperature and is called the thermal conductivity. For small variations in temperature k may be assumed to depend on position only. Substitution of Fourier's law in the integral form of the energy equation gives



u
t
dV =


·(k T) dV
(12)
On the other hand, from thermodynamics we know that the time rate of change of specific internal energy is given by
u
t
= (rc T)
t
(13)
where r is the density and c is the specific heat. Combining the above with the fact that if the value of the integral is zero the argument must be zero, the integral form of the energy equation acquires the differential form
(rc T)
t
= ·(k T) = div( k T)
(14)
We will call this equation the heat conduction equation. Adopting rectangular cartesian coordinates (x,y,z) one obtains
(rc T)
t
=
x
k ( T
x
) +
y
k ( T
y
) +
z
k ( T
z
)
(15)
In cylindrical polar coordinates (r,q,z)
(rc T)
t
= 1
r

r
(r k ( T
r
)) + 1
r2

q
( k ( T
q
)) +
z
( k ( T
z
))
(16)
and in Spherical Coordinates (r,j,q)
(rc T)
t
= 1
r2

r
( r2 k ( T
r
)) + 1
r2 sinq

q
( sinq k ( T
q
)) + 1
r2 sinq2

j
( k ( T
j
))
(17)

Finally, if r, c and k are all assumed constant one obtains what is often also called the heat conduction equation:

T
t
= a2 T
(18)
where
a = k
rc
(19)
is the thermal diffusivity.

3  References

[1] S.L. Sobolev ``Partial Differential Equations of Mathematical Physics'', Dover Publications, New York, 1989, Sections 1.1 and 1.5.

[2] F.B. Hildebrand, ``Advanced Calculus for Applications'', 2nd. ed., Prentice-Hall, Inc., Englewood Cliffs, NJ, 1976. Chapter 6.


File translated from TEX by TTH, version 2.34.
On 10 Jan 2001, 11:53.

Updated: 2001-01-10, 11:53