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are all of them grey bodies?

are all of them grey bodies?

are all of them grey bodies?

(OP)
hi,everybody:
What i'm trying to do on termography is to convert the lecture of temperature of a camera in emmited power.i think is possible to do it from  modified planck's law ,
emmited power=emmisivity*planck's constant*T**4
I need to know if it is a good aproximation ,in what materials can be used, and good tables of emmisivity.
¿could anybody tell me web sites where i can find theory about this?
 

RE: are all of them grey bodies?

To Termo,
Go to www.thermasearch.net and at the bottom of the page you will see a flashing "click here".  Click there and you will now see several site locations under their areas of coverage.  These include ir theory and tables (i.e. Omega's thermal book).  It is important that you understand that thermodynamics, imagery, equipment sensitivity/repeatability, ambient conditions and much more affect the developed E of a material and therefore a given E from one source may not agree with the E from another.  The most accurate method for obtaining the E of a target is to use the standard: ASTM E 1933-99a, "Standard Test Method for Measuring and Compensating for Emissivity Using Infrared Imaging Radiometers."  Of note is the typical case where an E given by a manufacturers table will generally be within +/- .02 of any one elses value.  Remember though that total radiant power error is a factor of one and equal to the sum of the loses expressed as E,R(reflectance),and T(transmittance), Therefore, do not exclude the other two predominant losses.  The equation you site is typically referred to as the Stefan-Boltzmann Law (using the Stefan-Boltzmann constant of 5.67 x 10 to the minus eight and expressed as Watts meter squared kelvin to the minus four power) and is reflective of what is called a Grey Body.  Good luck in your work.

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