Session 8
Multi-Dimensional, Steady-State,
Linear Heat Conduction: Analytical Methods
Ernesto Gutierrez-Miravete
Spring, 2002
1 Steady State Heat Conduction
If the temperature field inside a solid body does not change with time
then steady state conditions prevail. Under steady state conditions the
term ¶H/¶t = 0 and the heat equation becomes
where k is the thermal conductivity, T is temperature, g is the
rate of internal heat generation and r and t are, respectively,
position and time.
The steady state heat equation must be solved subject to appropriate
boundary conditions. The boundary conditions may consist of
- the temperature being specified (T = TS),
- the heat flux being specified
(q = - k ÑT ·n = qS), or
- a combination of temperature and heat flux being specified
(-k ÑT ·n = h (T - T¥)).
at each boundary in the system.
The corresponding forms of the steady state heat conduction equation in
Cartesian, cylindrical and spherical coordinates are, respectively
|
|
|
¶
¶x
|
(k |
¶T
¶x
|
) + |
¶
¶y
|
(k |
¶T
¶y
|
) + |
¶
¶z
|
(k |
¶T
¶z
|
) + g(x,y,z) = 0 |
|
| |
|
|
|
|
1
r
|
|
¶
¶r
|
(k r |
¶T
¶r
|
) + |
1
r2
|
|
¶
¶f
|
(k |
¶T
¶f
|
) + |
¶
¶z
|
(k |
¶T
¶z
|
) + g(r,f,z) = 0 |
|
| |
|
and
|
|
|
1
r2
|
|
¶
¶r
|
(k r2 |
¶T
¶r
|
) + |
1
r2 sinq
|
|
¶
¶q
|
(k sinq |
¶T
¶q
|
) + |
1
r2 sin2 q
|
|
¶
¶f
|
(k |
¶T
¶f
|
) + g(r,f,q) = 0 |
|
| |
|
2 One-dimensional Steady State Problems
The one dimensional forms of the steady state heat equation are particularly
simple and readily solvable in many cases as they reduce to second order
ordinary differential equations.
The corresponding forms of the steady state heat conduction equation in
Cartesian, cylindrical and spherical coordinates are, respectively
|
|
|
d
d x
|
(k |
d T
d x
|
) + g(x) = 0 |
|
| |
|
|
|
|
1
r
|
|
d
d r
|
(k r |
d T
d r
|
) + g(r) = 0 |
|
| |
|
and
|
|
|
1
r2
|
|
d
d r
|
(k r2 |
d T
d r
|
) + g(r) = 0 |
|
| |
|
2.1 Steady State Conduction in a Slab
This section contains solutions to a couple of simple but important problems.
Consider first the problem of steady state heat conduction in a slab of
thickness L, without internal heat generation (g = 0) and with
fixed temperatures at its boundaries T(0) = T1 and T(L) = T2.
Furthermore, assume the thermal conductivity is constant.
Integrating the heat equation twice and introducing the given boundary
temperatures yields
The heat flux through the slab is
|
|
| q = - k |
dT
dx
|
= - k |
T2 - T1
L
|
|
|
| |
|
and the rate of energy transfer (heat flow) through an area A of slab is
|
|
| Q = q A = - k A |
T2 - T1
L
|
= - |
T2 - T1
Rslab
|
|
|
| |
|
where Rslab = L/k A is the thermal resistance of the slab.
Exercise. Derive the above equation.
As a second problem consider the same slab as before but now let heat be generated internally at a constant rate g per unit volume.
Furthermore, assume the slab
exchanges heat by convection with an environment at temperature T¥.
In this case the solution is
|
|
| T(x) = T¥ + |
g L2
2 k
|
[1 - ( |
x
L
|
)2] + |
g L
h
|
|
|
| |
|
Exercise. Derive the above equation.
The concept of thermal resistance introduced above is used
to determine the steady state heat flow in more complex situations.
For instance, consider the composite slab made up of two slabs 1 and 2,
of respective thicknesses L1 and L2. The thermal contact
between the slabs is perfect. Assume heat is
exchanged by convection at x = 0 with an environment at T¥1
(heat transfer coefficient h1),
and at x = L = L1 + L2 with an environment
at T¥2 (heat transfer coefficient h2).
Furthermore, let the convective thermal resistances be defined as
and
The steady state heat flow is simply given by
|
|
| Q = |
T¥1 - T¥2
Rconv,1 + Rslab1 + Rslab2 + Rconv,2
|
|
|
| |
|
2.2 Steady State Conduction in a Cylinder
Consider a hollow cylinder (inner radius a, outer radius b). Let
the temperatures at A and b be fixed as T(a) = T1 and
T(b) = T2. Assume g = 0 and constant k. The solution to
this problem is
|
|
| T(r) = T1 + |
T2 - T1
ln(b/a)
|
ln( |
r
a
|
) |
|
| |
|
The heat flux through the cylindrical wall is
|
|
| q = - k |
dT
dr
|
= - k |
T2 - T1
r ln(b/a)
|
|
|
| |
|
and the rate of energy transfer through an area A = 2 pr l
of shell is
|
|
| Q = q A = - k (2 pr l) |
T2 - T1
r ln(b/a)
|
= - |
T2 - T1
Rcyl
|
|
|
| |
|
where l is the axial length and
Rcyl = ln(b/a)/2 pl k is the
thermal resistance of the cylindrical shell.
Exercise. Derive the above equation.
The concept of thermal resistance can also be
applied to the determination of the steady state heat flow
in composite cylindrical shells. Consider a hollow cylinder consisting
of two concentric layers. let the inner radius be r1, the
radius at the joint bewtween layers r2 and the outer radius r3.
The heat flux is simply given by
|
|
| Q = |
2 pl (T¥1 - T¥2)
1/(h1 r1) + ln(r2/r1)/ k + ln(r3/r2)/ k + 1/(h2 r3)
|
|
|
| |
|
The above expression can be used to determine the
critical thickness of cylindrical insulation.
Exercise. Derive an expression for the
critical thickness of cylindrical insulation.
Consider now a solid cylinder (radius b), inside which heat is generated
at constant rate g per unit volume while its boundary exchanges heat with
an environment
at T¥. The solution to this problem is
|
|
| T(r) = T¥ + |
g b2
4 k
|
[1 - ( |
r
b
|
)2 + |
2 k
h b
|
] |
|
| |
|
Exercise. Derive the above equation.
Now, assume a cladding is placed around the cylinder but that no heat
is generated in the cladding. Let the conductivities and radii of the
inner cylinder and cladding be k1, r1 and k2, r2.
The following expression allows the
calculation of the temperature drop between the axis of the
cylinder (Tc) and that of the surrounding fluid,
|
|
| Tc - T¥ = |
g r12
4 k1
|
+ |
g r12
2 k2
|
ln( |
r2
r1
|
) + |
g r12
2 r2 h
|
|
|
| |
|
In practice, the contact between the cladding and the inner cylinder is
not perfect and and additional term accounting for this effect must
be added to the right hand side of the above equation.
As a final example consider the problem of one-dimensional steady-state conduction in a hollow cylindrical shell subject to a boundary condition
of the second kind on the inner surface (r = r1) and
of the first kind on the outside surface (r = r2), i.e.
|
|
|
1
r
|
[ |
¶
¶r
|
( r k |
¶T
¶r
|
)] = 0 |
|
| |
|
subject to
and
This problem has an analytical solution given by
|
|
| T(r) = T2 - |
q1 r1
k
|
ln( |
r
r2
|
) |
|
| |
|
Exercise. Derive the above equation.
2.3 Steady State Conduction in a Sphere
Consider a hollow spherical shell (inner radius a, outer radius b). Let
the temperatures at a and b be fixed as T(a) = T1 and
T(b) = T2. Assume g = 0 and constant k. The solution to
this problem is
|
|
| T(r) = T1 + |
T2 - T1
1/b - 1/a
|
( |
1
r
|
- |
1
a
|
) |
|
| |
|
The heat flux through the spherical shell is
|
|
| q = - k |
dT
dr
|
= k |
T2 - T1
r2 (1/b - 1/a)
|
|
|
| |
|
and the rate of energy transfer through an area A = 4 pr2 of shell is
|
|
| Q = q A = k (4 pr2) |
T2 - T1
r2 (1/b - 1/a)
|
= |
T2 - T1
Rsph
|
|
|
| |
|
where Rsph = (b-a)/4 pab k is the
thermal resistance of the cylindrical shell.
Exercise. Derive the above equation.
Consider now a solid sphere (radius b), inside which heat is generated
at constant rate g while its boundary exchanges heat with en environment
at T¥. The solution to this problem is
|
|
| T(r) = T¥ + |
g b2
4 k
|
[1 - ( |
r
b
|
)2 + |
2 k
h b
|
] |
|
| |
|
3 Steady State Conduction in Fins and Spines
Fins and spines are extended surface structural elements commonly used
to enhanced the efficiency of heat exchange surfaces.
Fins are thin and flat while spines are pointy.
Consider a fin or spine protruding from a flat wall.
At the base of the fin x = 0 and the fin stretches towards
x ® ¥. The fin/spine has a nonuniform
cross sectional area A(x) and its perimeter is P(x).
As energy travels through the fin, part of it gets dissipated
by convection though the exposed extended surface while the rest
continues travelling along the fin.
Let the heat transfer coefficient be h and the surrounding
environment temperature T¥.
A steady state heat balance on a small
fin/spine element of size Dx gives
but
|
|
| Q(x+Dx) - Q(x) = |
d Q
dx
|
Dx = |
d
dx
|
(- k A(x) |
dT
dx
|
) Dx |
|
| |
|
and
|
|
| Qconv = h P(x) (T - T¥) Dx |
|
| |
|
Combination of the above yields
|
|
|
d
dx
|
( A(x) |
dT
dx
|
) - |
h
k
|
P(x) (T - T¥) = 0 |
|
| |
|
If the fin has constant cross section A and perimeter P and
T¥ = 0, the above differential equation becomes
which has the general solution
|
|
| T(x) = C1 e- m x + C2 em x = C3 sinh(m x) + C4 cosh(m x) |
|
| |
|
where m = hP/kA. If the temperature at the base of the fin/spine
is Tb and also the fin/spine is very long,
T(x ® ¥) » Tinfty = 0 so that
4 Steady State Temperature in a Rectangular Plate
At steady state the temperature
satisfies Laplace's equation
From Green's theorem, Laplace's equation requires that
which states that under steady state conditions the boundary heat flux
cannot be chosen arbitrarily but must average zero.
Also, from Green's theorem, if T1 and T2 are two solutions
of a steady state problem whose values coincide at the boundary
|
|
|
ó
õ
|
|
ó
õ
|
|
ó
õ
|
V
|
[ Ñ(T2 - T1]2 d t = 0 |
|
| |
|
so that T2 - T1 = constant. The constant is zero when the problem
involves only prescribed temperatures at the boundary (Dirichlet problem)
and can be nonzero when normal derivatives of T at the boundary are
specified (Neumann problem).
Consider steady state heat conduction in a thin rectangular plate
of width 1 and height 1. The edges
x = 0, x = 1 and y = 0 are maintained at T = 0 while at the edge y = 1,
T(x,1) = f(x). No heat flow along the z direction perpendicular to
the plate. The required temperature T(x,y) satisfies
To find T by separation of variables we assume the a particular
solution can be represented as a product of two functions each
depending on a single coordinate, i.e.
substituting into Laplace's equation gives
|
|
| - |
1
X
|
|
d2 X
dx2
|
= |
1
Y
|
|
d2 Y
dy2
|
= k2 |
|
| |
|
where k2 is a constant and this is true
since the LHS is a function of x alone while the RHS a function of
y alone. The constant is selected as k2 in order to obtain a
proper Sturn-Liouville problem for X (with real eigenvalues).
With the above the original PDE problem has been transformed into a
system of two ODE's, i.e.
subject to X(0) = X(1) = 0 and
subject to Y(0) = 0.
The solution for X(x) is
with eigenvalues
for n = 1,2,3,....
The solution of Y(y) is
so that the particular solution desired is
|
|
| Tn = |
¥
å
n = 1
|
an sin(n px) sinh(n py) |
|
| |
|
The an's are determined by making the above satisfy the
nonhomogeneous condition at y = 1, i.e.
|
|
| f(x) = |
¥
å
n = 1
|
[ an sinh(n p) ] sin(n px) |
|
| |
|
which is a Fourier sine series representation of f(x) with
coefficients
|
|
| cn = an sinh(n p) = 2 |
ó
õ
|
1
0
|
f(x) sin(n px) dx |
|
| |
|
so that the final solution is
|
|
| T(x,y) = |
¥
å
n = 1
|
cn sin(n px) |
sinh(n py)
sinh(n p)
|
|
|
| |
|
As a particular example consider the case when f(x) = 1. In this case
|
|
| cn = 2 |
ó
õ
|
1
0
|
f(x) sin(n px) dx = |
-2 [cos(n p) - 1]
n p
|
= |
ì
ï
ï
í
ï
ï
î
|
|
|
|
|
| |
|
Therefore, as long as f(x) is representable in terms of Fourier series,
the obtained solution coverges to the desired solution. Note also
that the presence of homogeneous conditions at x = 0, x = l made
feasible the determination of the required eigenvalues.
5 Steady State Temperature in a Circular Annulus
Consider now the problem of determining the steady state temperature in a
thin annular plate where the temperatures are specified at the inner
and outer radii r1 and r2 as
Laplace's equation in this case is
|
|
| Ñ2 T = |
¶2 T
¶r2
|
+ |
1
r
|
|
¶T
¶r
|
+ |
1
r2
|
|
¶2 T
¶q2
|
= 0 |
|
| |
|
Assume now a particular solution is of the form
Substituting leads to the following two ODEs
where the separation constant as been chosen as k2 in order to
obtain periodic trigonometric functions as the solutions for Q.
The general solution for R is
whereas that for Q is
|
|
| Q = |
ì
í
î
|
|
|
Ck cos(k q) + Dk sin(k q); k ¹ 0 |
|
|
|
|
|
| |
|
The periodicity requirement is satisfied by taking k = n, with
n = 1,2,3,....
The particular single valued solution Tp is then
|
|
|
Tp = (a0 + b0 lnr) + |
¥
å
n = 1
|
[(an rn + bn r-n) cos(n q) |
| | + (cn rn + dn r-n) sin(n q) ] |
|
| |
|
Finally, the desired solution must satisfy the stated boundary conditions.
Substituting f1 and f2 leads to Fourier series representations
and the relationships
|
|
| a0 + b0 lnr1 = |
1
2 p
|
|
ó
õ
|
2 p
0
|
f1(q) d q |
|
| |
|
|
|
| a0 + b0 lnr2 = |
1
2 p
|
|
ó
õ
|
2 p
0
|
f2(q) d q |
|
| |
|
|
|
| an r1n + bn r1-n = |
1
p
|
|
ó
õ
|
2 p
0
|
f1(q) cos(n q) d q |
|
| |
|
|
|
| an r2n + bn r2-n = |
1
p
|
|
ó
õ
|
2 p
0
|
f2(q) cos(n q) d q |
|
| |
|
|
|
| cn r1n + dn r1-n = |
1
p
|
|
ó
õ
|
2 p
0
|
f1(q) sin(n q) d q |
|
| |
|
|
|
| cn r2n + dn r2-n = |
1
p
|
|
ó
õ
|
2 p
0
|
f2(q) sin(n q) d q |
|
| |
|
A special case of interest is steady heat conduction in a disk
(r1 = 0) of radius r2 = a subject to a temperature f(q)
at its boundary. In this case, the solution is
|
|
| T(r,q) = A0 + |
¥
å
n = 1
|
( |
r
a
|
)n [An cos(n q) + Cn sin(n q)] |
|
| |
|
with
|
|
| A0 = |
1
2 p
|
|
ó
õ
|
2 p
0
|
f(q) d q |
|
| |
|
|
|
| An = |
1
p
|
|
ó
õ
|
2 p
0
|
f(q) cos(n q) d q |
|
| |
|
|
|
| Cn = |
1
p
|
|
ó
õ
|
2 p
0
|
f(q) sin(n q) d q |
|
| |
|
for n = 1,2,.... Note that the temperature at the center of the disk
is simply the average value of the boundary temperature.
Poisson's integral formula allows direct determination of the temperature
T(r,q) from the values of it at its boundary as follows
|
|
| T(r,q) = |
1
2 p
|
|
ó
õ
|
2 p
0
|
|
a2 - rr
a2 - 2 a r cos(q- y) + r2
|
T(a,y) d y |
|
| |
|
Another important special case is that of computing the temperature
field around a circular hole (r2 ® ¥)
of radius r1 = a subject to a temperature distribution
T(a,q) = f(q) at the hole boundary. In this case
|
|
| T(r,q) = A0 + |
¥
å
n = 1
|
( |
a
r
|
)n [Bn cos(n q) + Dn sin(n q)] |
|
| |
|
with
|
|
| A0 = |
1
2 p
|
|
ó
õ
|
2 p
0
|
f(q) d q |
|
| |
|
|
|
| Bn = |
1
p
|
|
ó
õ
|
2 p
0
|
f(q) cos(n q) d q |
|
| |
|
|
|
| Dn = |
1
p
|
|
ó
õ
|
2 p
0
|
f(q) sin(n q) d q |
|
| |
|
for n = 1,2,.... Note that the temperature at infinity
is simply the average value of the boundary temperature.
6 Steady State Temperature in a Sphere
Consider now the problem of steady state heat conduction in a sphere
(radius a) incorporating symmetry in the azimuthal direction
(i.e. solution independent of q).
Laplace's equation in this case is
|
|
|
¶
¶r
|
(r2 |
¶T
¶r
|
) + |
1
siny
|
|
¶
¶y
|
(siny |
¶T
¶y
|
) = 0 |
|
| |
|
The temperature at the surface of the sphere is prescribed as
As before, a particular solution of the form Tp = R Y is
assumed leading to
|
|
|
1
siny
|
|
d
d y
|
(siny |
d Y
d y
|
) + k2 Y = 0 |
|
| |
|
The general bounded solution for R is
while the general solution for Y, finite at y = 0, p
in terms of Legendre polynomials
is obtained by making k2 = n(n+1) giving
Therefore the particular solution obtained is
|
|
| T(r,y) = |
¥
å
n = 0
|
cn ( |
r
a
|
)n Pn(cosy) |
|
| |
|
The coefficients cn are determined from the imposed boundary
condition as
|
|
| cn = |
2n+1
2
|
|
ó
õ
|
p
0
|
f(y) Pn(cosy) sinyd y |
|
| |
|
7 Steady State Temperature in a Brick
Consider now the problem of steady state heat conduction in a
rectangular parallelepiped (length l1, width l2,
height d) subject to zero temperature on all five faces except that at
z = d which is maintained at T(x,y,d) = f(x,y).
Laplace's equation is
|
|
|
¶2 T
¶x2
|
+ |
¶2 T
¶y2
|
+ |
¶2 T
¶z2
|
= 0 |
|
| |
|
Assume a particular solution is of the form
Two separation constants are needed in this case giving
and
Note that homogeneous conditions occur at both boundaries in both the
x and y directions leading to
|
|
| Xm = Am sin(k1 x) = Am sin( |
m px
l1
|
) |
|
| |
|
|
|
| Yn = Bn sin(k2 y) = Bn sin( |
n py
l2
|
) |
|
| |
|
and
where kmn2 = k12 + k22.
The desired particular solution is then
|
|
| Tp = |
¥
å
m = 1
|
|
¥
å
n = 1
|
amn sin( |
m px
l1
|
) sin( |
n py
l2
|
) sinh(kmn z) |
|
| |
|
where the coefficients amn must be determined by incorporating the
nonhomogeneous condition. This leads to a double Fourier series representation
and the coefficients
|
|
| amn = |
1
sinh(kmn d)
|
|
4
l1 l2
|
|
ó
õ
|
l1
0
|
|
ó
õ
|
l2
0
|
f(x,y) sin( |
m px
l1
|
) sin( |
n py
l2
|
) dy dx |
|
| |
|
File translated from TEX by TTH, version 2.34.
On 24 Apr 2002, 11:58.