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Unformatted text preview: PROBLEM 4.5
KNOWN: Boundary conditions on four sides of a rectangular plate.
FIND: Temperature distribution.
SCHEMATIC: y q′′
s W T1 T1 0
0 x L
T1 ASSUMPTIONS: (1) Twodimensional, steadystate 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 θ = T – T∞, the differential equation and boundary
conditions are
∂ 2θ
∂ 2θ
+ 2 =0
∂x 2
∂y
θ(0, y) = 0 θ(L, y) = 0 θ(x,0) = 0 k ∂θ
∂y = q′′
s (1a,b,c,d) y=W The solution is identical to that in Section 4.2 through Equation (4.11),
∞
nπx
nπy
θ = ∑ Cn sin
sinh
L
L
n=1 (2) To determine Cn, we now apply the top surface boundary condition, Equation (1d).
Differentiating Equation (2) yields Continued…. PROBLEM 4.5 (Cont.)
∂θ
∂y = ∞ ∑ Cn n=1 y=W nπ
nπx
nπW
sin
cosh
L
L
L (3) Substituting this into Equation (1d) results in q′′
s
=
k ∞ ∑ A n sin n=1 nπx
L (4) where An = Cn(nπ/L)cosh(nπW/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
nπx
. Equation (4.16) then becomes
f(x) = q ′′/k , g n (x) = sin
s
L
L An = q′′
nπx
s
∫ sin L dx
k0
L ∫ sin
0 2 nπx
dx
L = q′′ 2 (1) n+1 + 1
s
kπ
n Thus Cn = 2 q′′L
(1) n+1 + 1
s
k n 2 π 2 cosh(nπW/L) The solution is given by Equation (2) with Cn defined by Equation (5). (5) ...
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 Spring '08
 gough

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