Project ForgeHeat Transfer
Convection & Radiation
GATE 2016
Expert
An infinitely long furnace of 0.5 m × 0.4 m cross-section has all-black surfaces. Top and bottom walls (sides 1,3) are at 927°C; side walls (sides 2,4) are at 527°C. The view factor F1-2 is 0.26. The net radiation heat loss or gain on side 1 (in W/m) is __________. (σ = 5.67×10⁻⁸ W/m²-K⁴)
24530 W/m(± 3 accepted)
Explanation
By symmetry F1-2=F1-4=0.26, and summation gives F1-3=1-2(0.26)=0.48. Since T3=T1, the F1-3 term contributes nothing. Q1 = A1·σ·(T1⁴-T2⁴)·(F1-2+F1-4) = 0.5×5.67e-8×(1200⁴-800⁴)×0.52 ≈ 24,531 W/m (per unit length, T1=1200K, T2=800K).
Source: GATE ME 2016 Set-1 (2016)
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