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Help with simple Equation 1

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MatthewMachinist

Industrial
Jun 29, 2014
5
Hi Guys,

I am a machinist and not very good with equations. What I have is an old school pyrometer that has no on board emissivity compensation. I have found the equation to make the compensation but I can not get my head around the fraction in the calculation.

T rad = ε 1/4 T kin

So lets say I measure 1000 degrees C or 1273 Kelvin (T rad) and ε Emissivity compensation is 0.7, how can I solve this, is there a way to turn that fraction into a constant number?

Thanks and sorry for the math question.
Matt.
 
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So you just want to get T_kin ? You can rearrange the formula like this:

T_kin = (4*T_rad)/ε
 
Do you have a source for that equation? It doesn't look right. I think the equation should look like the last one below, where it's emissivity to the 4th power. Your emissivity of 0.7 results in its 4th power being 0.9147, so your corrected temperature should be ~1392 K, which makes sense, since the emissivity essentially is not allowing you to measure the full power of the heated surface. This equation is from Stefan-Boltzmann Law which says that radiance is proportional to the 4th power of the absolute temperature
emiss_jqh0tp.png


TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
 
Ok, thanks for the replies guys, the webpage that confused me is this one...

However I am happy to say that I have used the advice on this thread to get the correct answer on their sample equation so I am sorted. I have also worked it out the other way around... solving for both actual and measured.

Once I got my head around using my calculators power function (a chevron symbol on my calculator) and the square root to the power of x function I was sorted.

Cheers.
 
Your link shows the equation in the correct form; the 1/4 is the exponent of the emissivity. The T_kin is the kinetic temperature, which I called T_actual, while the T_rad is the radiance temperature, which I called T_measured

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
 
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