Project ForgeRadiative heat transfer is intended between the inner surfaces of two very large isothermal parallel metal plates. While the upper plate (plate 1) is a black surface and is the warmer one being maintained at 727°C, the lower plate (plate 2) is a diffuse and gray surface with an emissivity of 0.7 and is kept at 227°C. Assume the surfaces are sufficiently large to form a two-surface enclosure and steady-state conditions to exist. Stefan-Boltzmann constant = 5.67×10⁻⁸ W/m²K⁴. The irradiation (in kW/m²) for the upper plate (plate 1) is
- A2.5
- B3.6
- C17.0
- 19.5
Explanation
Since plate 1 is black, all radiation leaving it (Eb1=σT1⁴) reaches plate 2, so G2=σT1⁴=5.67e-8×(1000)⁴=56700 W/m². Plate 2's radiosity J2=ε2·σT2⁴+(1-ε2)G2=0.7×5.67e-8×(500)⁴+0.3×56700=2480.6+17010≈19490.6 W/m²≈19.5 kW/m². Since view factor F21=1, all of J2 reaches plate 1, so irradiation on plate 1, G1=J2≈19.5 kW/m².
Source: GATE ME 2009 (2009)
This question is part of a 15-day revision cycle — sign in and Project Forge will bring it back the day before you'd forget it.