PROBLEM 4.20 KNOWN:Dimensions, thermal conductivity and inner surface temperature of furnace wall. Ambient conditions. FIND:Heat loss. SCHEMATIC:ASSUMPTIONS:(1) Steady-state, (2) Uniform convection coefficient over entire outer surface of container, (3) Negligible radiation losses. ANALYSIS:From the thermal circuit, the heat loss is s,icond(2D)convTTqRR∞−=+where Rconv= 1/hAs,o= 1/6(hW2) = 1/6[5 W/m2⋅K(5 m)2] = 0.00133 K/W. From Equation (4.21), the two-imensional conduction resistance is dcond(2D)1RSk=where the shape factor S must include the effects of conduction through the 8 corners, 12 edges and 6
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This note was uploaded on 12/07/2010 for the course MAE Heat Trans taught by Professor Lee,j.s. during the Spring '10 term at Seoul National.