heat_3_82 - PROBLEM 3.82 KNOWN: Distribution of volumetric...

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Unformatted text preview: PROBLEM 3.82 KNOWN: Distribution of volumetric heating and surface conditions associated with a quartz window. FIND: Temperature distribution in the quartz. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligible radiation emission and convection at inner surface (x = 0) and negligible emission from outer surface, (4) Constant properties. ANALYSIS: The appropriate form of the heat equation for the quartz is obtained by  substituting the prescribed form of q into Eq. 3.39. d 2T α (1 − β ) q+ + -α x o e + =0 2 k dx Integrating, + (1 − β ) q+ o e-α x + C dT =+ 1 dx k T=− (1 − β ) q+ + e-α x + C x+C o kα 1 2 − k dT/dx) x=o = β q+ + o − k dT/dx) x=L = h Ⱥ T ( L ) − T∞ Ⱥ Ⱥ Ⱥ Boundary Conditions: Ⱥ (1-β ) Ⱥ − k Ⱥ q+ + + C1 Ⱥ = β q+ + o o Ⱥ k Ⱥ C1 = −q+ + / k o Hence, at x = 0: At x = L: Ⱥ (1-β ) Ⱥ Ⱥ (1-β ) Ⱥ − k Ⱥ q+ + e-α L + C1 Ⱥ = h Ⱥq+ + e-α L + C1L+C2 − T∞ Ⱥ o o Ⱥ k Ⱥ Ⱥ kα Ⱥ Substituting for C1 and solving for C2, q+ + C2 = o h Hence, q+ + o Ⱥ1 − (1 − β ) e-α L Ⱥ + q+ + + o(1-β ) e-α L + T . ∞ Ⱥ Ⱥ k Ⱥ Ⱥ kα T (x ) = + + (1 − β ) q+ o Ⱥe-α L − e-α x Ⱥ + q+ o kα Ⱥ Ⱥ Ⱥ Ⱥ k (L − x ) + q+ + Ⱥ o 1 − (1 − β ) e-α L Ⱥ + T . < Ⱥ ∞ Ⱥ Ⱥ h Ⱥ COMMENTS: The temperature distribution depends strongly on the radiative coefficients, α and β. For α → ∞ or β = 1, the heating occurs entirely at x = 0 (no volumetric heating). ...
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This note was uploaded on 01/24/2012 for the course MECHE 343 taught by Professor Professor during the Spring '11 term at Carnegie Mellon.

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