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electricpete (Electrical) (OP)
7 Nov 04 22:25
This is a little bit of a spinoff from the IR for steam leaks question but I didn't want to clutter that one up.

Conservation of energy tells us that A+T+R=1 since A (absorption), T (transmission), and R (reflection) all represent what happens to incoming energy as fraction of total energy.

Now, it is often assumed that E = A.  I have a hard time understanding why it must be so.  There is no obvious reason from any conservation of energy argument I can see that compels E=A. I did read in CRC handbook that E=A is a good approximation for diffuse (vs specular) surfaces.   I cannot see any obvious reason for this.  Is there some particular reason that E=A for diffuse surfaces or is it just an empirically observed fact?

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Helpful Member!  IRstuff (Aerospace)
7 Nov 04 22:48
The emission must equal absorption, from conservation of energy, ASSUMING, thermal equilibrium.  This would be the case within a cavity blackbody.

If the object's temperature remains constant, then it can neither gain nor lose energy.  But Planck's law says that an object warmer than absolute zero must emit energy.  But, since its temperature in thermal equilibrium is constant, it MUST absorb the same amount of energy.

Given those conditions, we can see that actual value of emissivity doesn't matter, nor does its spectral characteristics.  A low emissivity object must have a low absorbance, again, because of thermal equilibrium.

Under thermal non-equilibrium conditions, the emissivity does not necessarily equal the absorbance spectrally, nor is that required.  For instance.  Consider the "standard NATO" tank that I encounter at work.  Its absorption is governed by sunlight with its Wien's law spectral peak around 0.55 microns and the camouflage paint applied, which is spectrally selective.  However, the blackbody temperature of the tank is usually a value that places the spectral peak arounf 8 microns, allowing the emissivity and absorbance to be unrelated for practical purposes.  

TTFN

electricpete (Electrical) (OP)
8 Nov 04 8:36
"The emission must equal absorption, from conservation of energy, ASSUMING, thermal equilibrium."

It would seem that this statement assumes the only source of heat transfer into the object is absorption and the only source of heat transfer out of the object is radiation.  Are these the intended assumptions?

In practical situations that I look at we have heat put in by electrical I^2*R losses (far more than is absorbed by the component by radiation from it's environment) and heat removed by conduction and convection cooling (far more than energy that is radiated for temperatures in the range 40 - 100C).  

In these situations it would seem there is reason to claim e = a based on conservation of energy. (agreed?) However, I am led to believe from the CRC handbook that it is still a reasonable approximation true for many cases, and it somehow tied to the diffuse or specular nature of the surface.  

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IRstuff (Aerospace)
8 Nov 04 12:49
Yes, you assume for the purposes of that radiative calculation that the objects in question are thermally isolated.

Howver, thermal equilibrium can be obtained in a number of ways; thermal equilibrium with non-zero conduction and convection does not require emissivity equal to absorbance.  The reason should be clear.  Conduction and convection require a thermal gradient.  This means that the object can be significantly hotter than its surroundings.  Therefore, emission and absorbance occur in different wavebands.

As mentioned above, in that case, the spectral characteristics of the object and its surroundings become important, and there is no practical reason that emissivity will equal absorbance, unless it was intentionally designed that way.

In general, emissivity in a waveband is essentially equal to the absorbance in the same waveband, but not across different wavebands.

TTFN

electricpete (Electrical) (OP)
8 Nov 04 18:04
I am struggling to find any basis for applying conservation of energy to a useful result... regardless of spectral considerations.

Let us assume greybodies -  no variation of emission on absorption with frequency.  Even then, if there is power added and removed by non-infrared sources (conduction, convection, I^2*R), then conservation of energy still provides no basis to make any conclusion regarding equivalence of e and a... correct?  After all, there are unspecified  elements of the energy conservation equation.

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electricpete (Electrical) (OP)
8 Nov 04 18:06
I will mention that I saw that conservation of energy (actually they called it KCL for Kirchoff's Current Law) listed as the justification for e = a  in an A.I.R.T. course material.  It has been bothering me ever since.

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jabonet (Mechanical)
9 Nov 04 12:03
I also have a problem with this E must equal A, no problem with the assumption that it happens only when there is thermal equilibrium, but its also that its true, only when you take in account the whole espectrum of radiation that is radiated to and from the object, let me explain:

in the case of detergent, there is a substance added to it that absorbs ultraviolet light (which we can't see) and emits visible light (which we can) the end result is that this detergent emits more light that we can see it absorbs, thats why its so much whiter, if we use black light, it will seem that it emits radiation (visible) that it doesn't absorbs, or at least that we can't see it absorbs.

in the case of thermal imagin, I can be in a room with many sources of very different spectra, if I stand next to the object and make an integration of all the field of view times the radiation detected by the camera, I will end up with what I think is the radiation that is hitting the object, now when I measure the object emitted radiation with the same camera, I see that its emiting more infrared that it should, but its temperature is not changing (is in thermal equilibrium), the problem is that the object is taking radiation in spectra I didn't accounted for like, visible, ultraviolet, and infrared that my camera can't detect, and is emiting in radiation that I can detect.

In this case a bolometer could work a little bit better, because its has a flatter spectra, I think...

on the other side, I don't think that there is so much energy coming from spectra other than Infrared as to change sensibly what the object emits, but in theory you should integrate over the whole spectra from very long radio waves to cosmic rays.

Hope I am not too wrong.
IRstuff (Aerospace)
9 Nov 04 12:35
You're answering a different question.  Obviously, if you're being radiated by the Sun, you're not going to be emitting like the Sun.

TTFN

JKEngineer (Chemical)
10 Nov 04 8:49
It sounds like the discussion is starting to confuse the material property, emissivity or emittance, with the energy being emitted.  They are not the same.  The energy emitted is a function of temperature of the object and its emissivity.  The equality between emissivity and absorptivity is the physical property, not the amount of energy being handled.  The thermal equilibrium argument, using only radistive transfer, is used to demonstrate that the physical properties, emissivity and absorptivity are the same.  They can continue to be the same when there is no thermal equilibrium, but the energy flows change.  
Jack

Jack M. Kleinfeld, P.E.  Kleinfeld Technical Services, Inc.
Infrared Thermography, Finite Element Analysis, Process Engineering
www.KleinfeldTechnical.com

electricpete (Electrical) (OP)
10 Nov 04 14:21
Jack - your comments are valued but in this case I think you have misunderstood the discussion. No-one here thinks emisssivity is a unit of power or energy that I can see.   I was seeking to understand any reason for concluding e=a , including use of energy balance justification for concluding that e=a.    

So far we have found some weaknesses in this energy balance "proof" which is cited by many people:
- IRStuff identified there are spectral considerations which complicate the simple analysis.
- Also it would seem to only apply only to systems at thermal equilibrium temperature where radiation is the predominant mode of heat transfer.

I am still trying to understand this energy balance so let us set those two issues aside by assuming we have a greybody rock in outerspace (no conduction/convection) in a constant thermal radiation field which reaches thermal equilibrium.  We are only interested in a and e as integrated number not function of wavelength.   Someone tell me how applying energy balance to this experiment proves a = e even under these idealized conditions.

Pin = Pout
where   Pin and Pout are energy per time.

Rock will cool to temperature T such that:
a*A*Ssun = e *A* Sblackbody = e*A* sigma * T^4.
where a and e are absorptivity and emissivity.
A = surface area... assume same surface area for absorption and emissivity for smiplicity.
Ssun is the thermal radiation flux  from the sun onto the rock.
Sblackbody is thermal radiation flux for blackbody at temperature T

so where it the proof that a = e?
why does anyone cite energy balance as proof that a=e?

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IRstuff (Aerospace)
10 Nov 04 16:25
You're doing it again.

The energy balance argument applies ONLY to thermal equilibrium and ONLY in limited situations.  ONCE you have established that e=a, then you're done.  

You don't, and can't demonstrate it under other conditions, nor should you want to or have to.  e and a are intrinsic properties of the material surface and do not care about thermal or any other equilibrium.

For a spectrally complicated material, it may not even be possible to prove or disprove.  e and a, and their rule of thumb was postulated for objects are nearly ideal blackbodies.

TTFN

electricpete (Electrical) (OP)
10 Nov 04 23:10
I have tried to make the limiting assumptions above.  It doesn't prove it.  all I got was
a*A*Ssun = e *A* Sblackbody = e*A* sigma * T^4
a*Ssun = e*sigma*T^4
proves nothing.

Let me say it the other way around.  Pick a rock with known e <> a and put it in outer space with no conduction or convection.  It will still reach thermal equilbrium... only perhaps at a different temperature than some rock with different characteristics including a=e.  So, we can draw no conclusions about a=e by observing that an object comes to thermal equilibrium.  Make all the assumptions you want... show me how to prove a=e by conservation of energy.

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IRstuff (Aerospace)
11 Nov 04 3:08
Thermal equilibrium meaning the source is at the same temperature as the blackbody.

TTFN

electricpete (Electrical) (OP)
11 Nov 04 10:20
Thanks for your comments.

So I understand the idealized scenario is an object in a vacuum surrounded by a blackbody at temperature Tbackground

Incoming power = a*A*sigma*Tbackground^4
Outgoing power = e*A*sigma*Tobject^4

Steady state is reached when incoming = outgoing

a*A*sigma*Tbackground^4 =  e*A*sigma*Tobject^4
a / e = (Tobject/Tbackground)^4

(For simplicity I will assume a and e do not change with temperature.)

If Tbackground is known, then Tobject is determined by ratio of a/e.

Saying that the PROOF that a=e comes from the ASSUMPTION that equilibrium Tobject=Tbackground seems like cheating. equilibrium Tobject is not something we are free to set arbitrarily, it is determined by the laws of physics (the equation above).
We might as well say that the PROOF of a=e comes from the ASSUMPTION that a=e.

I am not trying to be picky but just searching for some grain of logic as to why people say a=e is proved by conservation of energy.  I don't see it, no matter how idealized we make our assumptions.  Let me know if I have missed something.

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Helpful Member!  cev (Materials)
12 Nov 04 10:02
You are getting very close, electricpete. Alter your "idealized scenario" to just an isothermal cavity (without any enclosed object). Now you don't have to make any assumptions about T1 = T2, but you still get A=E.

....

Kirchoff's law (of emission) is often stated A=E, but that is oversimplified. In fact, the true statement is that absorption at a given wavelength, incidence angle, and temperature is equal to emission at the same wavelength, angle, and temperature:

A(wavelength, angle, Temperature) = E(wavelength, angle, Temperature)

Emissivity can vary quickly with changes in wavelength and angle, and less so to changes in temperature. Therefore, the generalization to E=A is not often valid.

Best,
CV
electricpete (Electrical) (OP)
12 Nov 04 12:04
Hi cev - thanks for joining the discussion.

I heard a lot about cavities. I thought I understood the basic idea that a cavity acts like a blackbody at the same temperature.  I roughly come up with this by saying that IF a the surface sees a background equivalent to a blackbody at Tcavity, then it's radiosity will be as as follows:

S = emitted + reflected
S = e*sigma*Tcavity^4 + rsigma*Tcavity
we do know from KCL that e+r=1 for opaque surfaces, so substitute r = 1-e (that part makes sense from KCL since r,e,and t are just fractions of a total).
S = e*sigma*Tcavity^4 + (1-e)sigma*Tcavity^4
S = e+(1-e)*sigma*Tcavity^4 = 1*sigma*Tcavity^4

Now seeing that this surface emits at blackbody temperature, I feel comfortable concluding that my initial IF might have been justified  ("IF a the surface sees a background equivalent to a blackbody at Tcavity")

So I believe an enclosed cavity has a flux equivalent to a blackbody.

Now let's go back and ask what fraction of incident energy is absorbed into the wall of a blackbody.  I know that there is an incident flux on the wall of the cavity:
IF a the surface sees a background equivalent to a blackbody at Tcavity
Sincident=sigma*Tcavity^4

I know the amount reflected from the wall of the cavity:
Sreflected=sigma*(1-e)*Tcavity^4

So where does the rest go?  It must be absorbed.  
Sabsorbed=Sincident-Sreflected=sigma*e*Tcavity^4

Looking at this result I see that e plays the role of a.

Now it seems to me that:
- e=a is a prerequisite for cavities to act like blackbodies of the same temperature.  If it were not true I would have a contradiction above.
- Therefore the observation that cavities act like blackbodies of the same temperature tends to prove e=a.

Does that sound right?  I have to think about it some more.

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cev (Materials)
12 Nov 04 20:27
- e=a is a prerequisite for cavities to act like blackbodies of the same temperature.

At first, I thought this was a silly statement, but after some thought I think it's interesting. It does seem like an incredible coincidence that a cavity treated with a coating of arbitrary reflectance acts exactly like a blackbody. If E!=A, this would not be true.

- Therefore the observation that cavities act like blackbodies of the same temperature tends to prove e=a.

A little bit of circular thinking now! Fortunately, this sort of circular thinking isn't a problem since it can be rigorously proven that E=A (in the angle-temperature-wavelength sense).


CV
electricpete (Electrical) (OP)
17 Nov 04 8:01
"...it can be rigorously proven that E=A (in the angle-temperature-wavelength sense."

That is interesting.  What is the nature of the proof?  Thx.

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mrpyro (Electrical)
24 Nov 04 15:12
Hi, Folks,
Sorry to be entering this discussion a little late, but I couldn't resist. The fact that e=a or epsilon=alpha for radiation is not something I believe is subject to dispute, it is most often subject to misunderstanding, especially in infrared thermography. Jack at JKEngineer, who I know, said it correctly and I think the discussion about thermal equillibrium has muddied the waters.

The "simple" reason is that it is true at all wavelengths for a given material or surface, or else the universe wouldn't work. There may be short enough times scales where it doesn't work, but that's not very often a problem in thermography. The fact that the values of e (and a) may be different at different wavelengths is what makes things a little confusing. In fact, they almost always vary by wavelength (some materials a lot, some a little) but the e and a are the same at each wavelength, almost by definition.

The argument to prove the fact is best done on a thought-experiment basis and one quickly understands the problem when explained in terms of total radiative heat transfer (most of the universe works that way). But in order for the total heat transfer explaination to work the spectral properties of an object have to be consistant, ie. e(Lambda-subn)=a(Lambda-subn).

The misunderstanding usually is encountered in thermography when one says emissivity or emittance and really means spectral band or spectral emissivity. Then you go and look it up a value in a table of total emissivities. That's the start and it often gets worse from there.

There's a group of pages on my website called the E-missivity Trail devoted to the issue and we can write more if anyone can convince me it's really needed. A good physics book will tell you the story better, but without the context on my site's pages. Check out: The E-missivity Trail http://www.temperatures.com/eindex.html for a starter and then maybe a more fruitful discussion will happen.

Happy Trails!
electricpete (Electrical) (OP)
24 Nov 04 23:27
mrpyro - thanks for trying. I found your website completely irrelevant to my question and a waste of my time.

All -

Let me clarify, correct, and consolidate my thought experiment from my previous message.

Given:
a = fraction of incoming energy which is absorbed.
r = fraction of incoming energy which is reflected.
t = fraction of incoming energy which is transmitted, assumed 0 for this discussion.
e = fraction of blackbody radiation which is emitted for a body at temperature T.

Note , from definitions of a,r, and assumption on t it is obvious a+r=1.  (sum of fractions of incoming energy must sum to 1).  However it is not immediately obvious a=e

Prove:  a = e

Thought experiment:  We have two very large identical parellel plates at the the same temperature located very close to each other (similar to a simple drawing of a capacitor).    Call them plates A and B

The radiosity (emitted plus reflected) leaving A toward B must equal the radiosity leaving B towards A.  Call it S.  

Total power entering plate A from the gap:
S_entering = a * S = (1-r) * S

Total energy leaving plate A into the gap (does not include reflected which never enters A):  
S_leaving = e*Sblackbody    (where Sblackbody = sigma*T^4).

By Symmetry, the net transfer of energy from A to B must be 0.  Therefore the energy entering  A must equal the energy leaving A:
S_entering = S_leaving
a * S  = e * Sblackbody

This is about as far as we can get with conservation of energy and symmetry... without knowing the relationship between S and Sblackbody we cannot draw conclusions about equality of a and e.

However, it is observed that cavities have radiosity emitted from their surfaces equivalent to a blackbody of the same temperature.   ie. we know S = Sblackbody.

Using this knowedge
a * S  = e * Sblackbody  becomes a*Sblackbody = e*Sblackbody
a = e

In other words, a=e is proved if and only if  we know that radiosity from the surface of a cavity is equal to a blackbody at the same temperature.   (examination of the thought experiment symmetry reveals a=e must apply to both spectral and total quantities, but only applies where emissions and absorption are at the same temperature.)

So I feel like I have come to a reasonable understanding as outlined above.  I always have more questions.  My questions now:
1 - It is an accepted fact that a cavity acts like a blackbody.  Is there any intuitive reason it should be so?  Any thought experiment (which does not rely on a=e... would be circular logic in light of the above "proof").
2 -  Does anyone have other thought experiments to prove a=e?  (any which do not rely on cavity acting like blackbody)

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