GATE AG 2010 • Q43 MCQ • 2 Marks

Section 7: Dairy and Food Engineering

Heat and Mass Transfer

Subtopic: Steady state heat transfer in conduction, convection and radiation

Question Statement:

The heating surface of an oven has emissivity 0.7, area 0.1 m², maintained at 280°C. View factor of this surface w.r.t. a piece of bread of 0.01 m² area is 0.05. Bread has emissivity 0.3 and receives 10 W of energy through radiation from the heating surface (σ=5.67×10⁻⁸ W m⁻²K⁻⁴). The steady state bread surface temperature in °C is

Options:

  • (A) 137.2
  • (B) 118.5
  • (C) 97.3
  • (D) 84.5

Official Answer Key:

C

Step-by-Step Solution & Explanation

Two-surface radiation network resistance: R = (1−ε₁)/(ε₁A₁) + 1/(A₁F₁₂) + (1−ε₂)/(ε₂A₂) = (0.3)/(0.7×0.1) + 1/(0.1×0.05) + (0.7)/(0.3×0.01) = 4.286 + 200 + 233.3 = 437.6 m⁻². Q = σ(T₁⁴−T₂⁴)/R. Oven T₁ = 280+273.15 = 553.15 K. Solving 10 = 5.67×10⁻⁸(553.15⁴ − T₂⁴)/437.6 for T₂ gives T₂ ≈ 358.1 K = 97.3°C.

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