Reflectivity, emissivity, and absorptivity of opaque materials

[rmjmu507]

Homework Statement

what is the relation between reflectivity, emissivity, and absorptivity for an opaque material?

if an opaque material's reflectivity is 0.075 for solar radiation in the visible range and 0.95 for radiation of a heated surface of 80 degrees Celsius in the far infrared range, what are the corresponding emissivity and absorptivity?


2. The attempt at a solution

For an opaque material where the transmitivity is equal to 0, the absorptivity + reflectivity must equal 1.

I know for non-opaque materials, the emissivity is equal to the absorptivity, but I do not know any such relationship for opaque materials...

For the second part, I am unable to find a starting point.

Any help is much appreciated.

 

[jlefevre76]

First, for any material,

\alpha ( \lambda, \theta, \phi ) = \epsilon ( \lambda, \theta, \phi )

Where \lambda, \theta, and \phi represent wavelength, incident angle, and radial angle respectively (basically, \theta and \phi determine what direction the incoming or outgoing ray is).

If we have a diffuse surface, angle is no longer part of the equation, and thus:

\alpha ( \lambda ) = \epsilon ( \lambda )

If we have a grey, non-diffuse surface, then:

\alpha ( \theta, \phi ) = \epsilon ( \theta, \phi )

If we have a diffuse, grey surface (a favorite assumption in radiative heat transfer):

\alpha = \epsilon

Now, if diffuse, grey, and opaque (transmissivity is zero), we can say

\rho + \epsilon = \rho + \alpha = 1

If not, we can at least say:

\rho ( \lambda, \theta, \phi ) + \alpha ( \lambda, \theta, \phi ) = \rho ( \lambda, \theta, \phi ) + \epsilon ( \lambda, \theta, \phi )

So, the exact assumption we make depends on whether we can assume diffuse, grey, etc. And emissiviy and absorptivity will be functions of different variables depending one which ones we can assume.

Hope this helps a bit!