4.5
KNOWN: PROBLEM Boundary conditions on four sides of a rectangular plate. FIND: Temperature distribution. SCHEMATIC:
y
q s
W
T1
T1
0 0 T1 L
x
ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties. ANALYSIS: This problem differs from the one solved in Section 4.2 only in the boundary condition at the top surface. Defining = T T, the differential equation and boundary conditions are 2 2 + 2 =0 x 2 y
(0, y) = 0 (L, y) = 0 (x,0) = 0 k y = q s
y=W
(1a,b,c,d)
The solution is identical to that in Section 4.2 through Equation (4.11), nx ny = Cn sin sinh L L n=1 To determine Cn, we now apply the top surface boundary condition, Equation Differentiating (1d). Equation (2) yields
(2)
Continued....
PROBLEM 4.5 (Cont.)
y
=
y=W
n=1
Cn
n nx nW sin cosh L L L
(3)
Substituting this into Equation (1d) results in
q s = k
n=1
A n sin
nx L
(4)
where An = Cn(n/L)cosh(nW/L). The principles expressed in Equations (4.13) through (4.16) still apply, but now with reference to Equation (4) and Equation (4.14), we should choose nx . Equation (4.16) then becomes f(x) = q /k , g n (x) = sin s L
q nx s sin L dx k 0
L 0 L
An =
sin
2
nx dx L
=
q 2 (-1) n+1 + 1 s k n
Thus
Cn = 2
qL (-1) n+1 + 1 s k n 2 2 cosh(nW/L)
(5)
The solution is given by Equation (2) with Cn defined by Equation (5).