PROBLEM 4.23 KNOWN:Dimensions, shape factor, and thermal conductivity of square rod with drilled interior hole. Interior and exterior convection conditions. FIND:Heat rate and surface temperatures. SCHEMATIC:ASSUMPTIONS:(1) Steady-state, two-dimensional conduction, (2) Constant properties, (3) Uniform convection coefficients at inner and outer surfaces. ANALYSIS:The heat loss can be expressed as ,1,2conv,1cond(2D)conv,2TTqRRR∞∞−=++where ()( )112conv,111RhD L50W mK0.25m 2m0.01273K Wππ−−==⋅×××=11cond(2D)RSk8.59m 150W m K0.00078K W
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