Chapter4-4
4.4 Steady-State Temperature in a Disk
Figure 4.4-1 A thin circular plate with insulated top and bottom surfaces
The two-dimensional Laplace equation in polar coordinate can describe the steady-state temperature distribution u(r,) in a thin disk as shown in Figure 4.4-1.
+ = 00 < ra, 0 < 2(4.4-1)
The boundary condition for the disk at the circumference is
u(a,) = f(), for 0 < 2
u(a,) and f() must be 2periodic. We assume that u(r,) can be separated into R(r), a function of r alone, and T(), a function of alone.
u(r,) = R(r) T()(4.4-2)
Differentiating the partial derivative, we get
= R = R
= T = T
In terms of R(r) and T(), Eq. (4.4-1) becomes
T+ + R= 0(4.4-3)
Multiply Eq. (4.4-3) by to obtain
= + = n2(4.4-4)
Since the RHS of Eq. (4.4-4) depends on r only, and the LHS depends on only, they must equal to a constant n2. The constant must be chosen so that the solution is 2periodic in the -direction. The ODE on is
= n2T T = ancos(n) + bnsin(n), n = 0, 1, 2, 3, ...
The ODE on r is
+ = n2r2+ r n2R = 0
The equation for R is an Euler equation that has the following solution
R = r
The constant can be determined by substituting R = r and its derivatives into the original ODE
( 1)r + r n2r = 0
2 + n2 = 0 = n
The solution for R is written in the dimensionless form
R(r) = rn + r-n R(r) = c1 + c2, n = 1, 2, 3, ...
The case for n = 0 must be solved separately
+ = 0 r+ = 0 = 0
= dR = R = ln r +
The solution for the case n = 0 is finally
R = c1 + c2ln
Since the temperature and hence R must be finite at r = 0, c2 = 0. The constant c2will have a non zero value for temperature distribution outside a disk or on an annular region as shown in Figure 4.4-2.
Figure 4.4-2 The shaded areas indicate the region where 2u = 0 is solved.
The solution for R with the value of 1 for c1 is
R(r) = , n = 0, 1, 2, 3, ...
The product solution is
un(r,) = [ancos(n) + bnsin(n)]
Superposing these solutions, we get the general solution for the temperature distribution in a disk
u(r,) = [ancos(n) + bnsin(n)]
The coefficients an and bn can be obtained from the boundary condition at r = a.
u(a,) = f() = a0 + [ancos(n) + bnsin(n)]
The coefficients a0, an, and bn are the Fourier coefficients of the boundary function f().
a0 =
an =
bn =
1