Question

The outer surface of a grill hood is at 87 °c and the emissivity is 0.93. the heat transfer coefficient for convection between the hood and surroundings at 27 °c is 10 w/m2 /k. determine the net rate of heat transfer between the grill hood and the surroundings by convection and radiation, in kw per m2 of surface area.

Answer 1

The net flux, or rate of heat transfer per area, for convection and radiation are the following:

Convection:
Q/A = hΔT
Radiation:
Q/A = [εσ(T₁⁴ - T₂⁴)]/{1 - [A₁/A₂(1-ε)²]}
where
h is the heat transfer coefficient: 10 W/m²·K or 0.01 kW/m²·K
ε is the emissivity: 0.93
σ is the Stephen-Boltzmann constant: 5.67×10⁻¹¹ kW/m²·K⁴
A₁ = A₂ because they have the same surface area
T is absolute temperature in Kelvin: K = °C + 273

Thus,

Q/A = hΔT + [εσ(T₁⁴ - T₂⁴)]/{1 - [A₁/A₂(1-ε)²]}
Q/A = (0.01 kW/m²·K)(360 K - 300 K) + [0.93*5.67×10⁻¹¹ kW/m²·K⁴*(360⁴ - 300⁴)]/{1 - [1*(1-0.93)²]}
Q/A = 0.6 + 0.461
Q/A = 1.061 kW/m²