problem4-05 - PROBLEM 4.5 KNOWN: Boundary conditions on...

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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) 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 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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problem4-05 - PROBLEM 4.5 KNOWN: Boundary conditions on...

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