calculate the energy transfer rate in w/m² across 6 in wall of firebrick with a temperature difference across the wall of 55°C. the thermal conductivity of the firebrick is 0.65 BTU/hr ft °F.
L=6inches(2.54cm/(1inch))=15.24cmL=6 inches(2.54cm/(1 inch))=15.24cmL=6inches(2.54cm/(1inch))=15.24cm
∆T=50℃∆T=50℃∆T=50℃
k=0.65Btu/hr−ft℉((1W/cm℃)/(57.79Btu/hr−ft℉))=65/5779W/cm℃k= 0.65 Btu/hr-ft℉((1W/cm℃)/(57.79Btu/hr-ft℉))= 65/5779 W/cm℃k=0.65Btu/hr−ft℉((1W/cm℃)/(57.79Btu/hr−ft℉))=65/5779W/cm℃
qk/A=k/L∆Tq_k/A=k/L ∆Tqk/A=k/L∆T
qk/A=(65/5779W/cm℃)/15.24cm×50℃q_k/A= ( 65/5779 W/cm℃)/15.24cm × 50℃qk/A=(65/5779W/cm℃)/15.24cm×50℃
qk/A=0.0369W/〖cm〗2=369W/m2q_k/A=0.0369 W/〖cm〗^2 = 369 W/m^2qk/A=0.0369W/〖cm〗2=369W/m2
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