For an opaque surface, the absorptivity (α) , transitivity (τ) and reflectivity (ρ) are related by the equation
α + ρ = 1
When thermal radiation strikes any surface, it can interact in three primary ways: it can be absorbed by the surface, reflected off the surface, or transmitted through the surface. The portions of the incident radiation that undergo these processes are quantified by specific material properties.
The fundamental principle of energy conservation dictates that the sum of these fractions for any surface must always equal one, as all incident radiation must be accounted for. This general relationship is expressed by the equation:
\( \alpha + \tau + \rho = 1 \)
The question specifically refers to an opaque surface. An opaque surface is defined as a material that does not allow any thermal radiation to pass through it. In other words, for an opaque surface, the transmissivity (\( \tau \)) is effectively zero.
\( \tau = 0 \text{ (for an opaque surface)} \)
To find the relationship for an opaque surface, we substitute the condition \( \tau = 0 \) into the general energy conservation equation:
\( \alpha + 0 + \rho = 1 \)
This simplifies directly to the relationship:
\( \alpha + \rho = 1 \)
Therefore, for an opaque surface, the sum of its absorptivity and reflectivity always equals one. This means that an opaque surface either absorbs or reflects all the incident radiation.
Based on this derivation, the correct equation relating absorptivity (\( \alpha \)) and reflectivity (\( \rho \)) for an opaque surface is \( \alpha + \rho = 1 \).
If plate 1 is also a diffuse and gray surface with an emissivity value of 0.8, the net radiation heat exchange (in kW/m2) between plate 1 and plate 2 is
A gray body is defined such that
In a radiative heat transfer, a gray surface is one