How this calculation works
Radiative exchange between two surfaces depends on absolute temperatures and both surfaces’ emissivities. Using Celsius directly in a fourth-power law gives an incorrect result even when the temperature difference looks reasonable. This page models two large parallel diffuse-gray surfaces with view factor one and equal exchanging area. It returns signed net exchange, heat flux and the effective two-surface emissivity.
Q = area sigma(T1_K⁴−T2_K⁴)/(1/e1+1/e2−1)A worked example
For two black surfaces at one hundred and twenty degrees Celsius with common area two square metres, the calculation uses absolute temperatures 373.15 and 293.15 kelvin. The difference of their fourth powers multiplied by area and the Stefan–Boltzmann constant gives the modeled wattage. Setting both temperatures equal makes net radiation zero even though each surface still emits radiation individually.
Good to know
Convection, conduction and participating gases are omitted. A complete heat-loss estimate usually needs additional mechanisms and boundary conditions; this page isolates one radiative path. Real emissivity can depend on wavelength, direction, surface condition and temperature. This gray-surface idealization accepts fixed entered values and provides no material-property lookup or thermal safety certification.
Reference: NPTEL — exchange of radiation between gray surfaces (opens in a new tab) ↗
What this calculation answers
Radiative exchange between two surfaces depends on absolute temperatures and both surfaces’ emissivities. Using Celsius directly in a fourth-power law gives an incorrect result even when the temperature difference looks reasonable.
This page models two large parallel diffuse-gray surfaces with view factor one and equal exchanging area. It returns signed net exchange, heat flux and the effective two-surface emissivity.
The Counterflow or parallel heat exchanger NTU and LMTD calculator models two flowing streams linked by conductance. Large parallel-surface radiation instead uses absolute-temperature fourth powers and both emissivities.
Use the right input conventions
Enter temperatures in degrees Celsius; the calculator converts them to kelvin before applying the fourth powers. Temperatures must remain above absolute zero and emissivities must be positive and no greater than one.
Area is the common facing area in square metres. The idealized surfaces are large enough that edge losses to other surroundings are ignored, rather than treated as a finite plate with an automatically known view factor.
Interpret the result
Positive heat flow means energy travels from surface one to surface two. Reversing their temperatures changes the sign while keeping the magnitude unchanged when emissivities and area stay fixed.
The effective emissivity includes both surface resistances. It is not generally the arithmetic mean of entered emissivities, and reducing either surface emissivity reduces net exchange in this particular enclosure model.
Work through the example
For two black surfaces at one hundred and twenty degrees Celsius with common area two square metres, the calculation uses absolute temperatures 373.15 and 293.15 kelvin.
The difference of their fourth powers multiplied by area and the Stefan–Boltzmann constant gives the modeled wattage. Setting both temperatures equal makes net radiation zero even though each surface still emits radiation individually.
Where the model stops
Convection, conduction and participating gases are omitted. A complete heat-loss estimate usually needs additional mechanisms and boundary conditions; this page isolates one radiative path.
Real emissivity can depend on wavelength, direction, surface condition and temperature. This gray-surface idealization accepts fixed entered values and provides no material-property lookup or thermal safety certification.
A few questions, answered
Does zero net exchange mean neither surface emits?
No. Equal-temperature surfaces can each emit and absorb radiation while their net exchange cancels.
What changes in the comparison?
Set both emissivities to one half; the effective two-surface value becomes one third, so exchange is one third of the black-surface example.