Net Radiation Between Surfaces
Net Radiation Between Surfaces
A hot surface loses heat by radiation at a net rate set by the difference of the fourth powers of the absolute temperatures.
The net rate is Q = ε·σ·A·(T_1⁴ − T_2⁴). where ε is emissivity, σ the Stefan–Boltzmann constant, A the area and T_1, T_2 the surface and surroundings temperatures.
Inputs
emis := 0.8=0.8
sigma_sb := 5.67e-8 W/(m^2·K^4)=5.67e-08 kg/K^4/s^3
A := 2 m^2=2 m^2
T_1 := 800 K=800 K
T_2 := 300 K=300 K
Apply the net radiation exchange formula.
Q := emis·sigma_sb·A·(T_1^4 - T_2^4)=36424.08 W
Results
The fourth-power law makes the hot surface dominate the exchange. Even with cool surroundings, an 800 K surface radiates strongly.
Assumptions & limitations
- Grey surface, constant emissivity.
- Surroundings much larger than the surface.
- Absolute temperatures.