Calculation - Effect of emissivity error on delta-T between suspect an 2

[electricpete]

Subject: Calculation - Effect of emissivity error on delta-T between suspect and faulty connections.

Let's say we have two connections, A (suspect faulty connection) and B (reference normal connection).
Both have emissivity e_actual.
We set our camera to assign emissivity e_assumed for all points (camera setting does not match actual).

Connection A has actual temperature Ta_actual and emits radiosity Sa = sigma * e_actual*Ta_actual
Camera indicates Ta_indicated to produce the same radiosity based on assumed emisisivity
Sa = sigma * e_assumed*Ta_indicated^4 = sigma * e_actual*Ta_actual^4
Solving for Ta_indicated:
Ta_indicated = (e_actual/e_assumed)^0.25 * Ta_actual

Likewise we can find indicated temperature for connection B
Tb_indicated = (e_actual/e_assumed)^0.25 * Tb_actual


The indicated differential temperature between A and B dTab_indicated is given as
dT_ab_indicated = Ta_indicated-Tb_indicated = (e_actual/e_assumed)^0.25 * (Ta_actual-Tb_actual)
Expressing this in terms of the actual differential temperature (dTab_actual), we have
dT_ab_indicated = (e_actual/e_assumed)^0.25 * dTab_actual
Or equivalently.
dT_ab_actual = (e_assumed/e_actual)^.25 * dTab_indicated

Differential temperatures in C are equivalent to differential temperatues in K, so we can work direclty in C without needing to convert to absolute temperature K.

If indicated differential temperature is 44C using assumed (camera) emissivity of 1 but actual emissivity is 0.5, we have
dT_ab_actual = (e_assumed/e_actual)^.25 * dTab_indicated
dT_ab_actual = (1/0.5)^.25 * 44 C
dT_ab_actual = 52.3 C

The above seems right to me. But if I look at the 4/5/04 tip of the week at irinfo (How Delta T's Understate Priorities – Pt. 1) at the following address....

http://www.irinfo.org/tip_of_week_2004.html#t04052004

....I see they have analysed this exact situation and conclude the actual differential temperature is 70C.

So, either the tip of the week is wrong or I am wrong. If I had to bet, I would say that I must be wrong, but I can't find any error above. Can anyone shed some light?

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[IRstuff]

One obvious issue is that what Boltzman's equation describes is the TOTAL radiance, but your thermal camera is usually only sensitive to 8-12 um radiation.

As the innate temperatue increases, the total radiation increases with T^4, but what you collect in the thermal camera does not vary by as much as the T^4 relationship would predict.

One way around such problems is to use a "two-color", e.g., MWIR/LWIR camera that records radiation from both bands. What you would see in that case is that the MWIR radiation will increase substantially more than the radiation in the LWIR band.

TTFN

 

[electricpete]

Thanks. You are right ours is long-wave... I guess most modern uncooled cameras fall in that category.

I will have to think about whether I can estimate a solution using the integral of energy over that band. Sounds a little tougher.


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[IRstuff]

You can probably get there by doing:

> Calibrate your emissivities of typical materials that you're likely to view with the IR camera. I'd look to having an accurate thermocouple or thermistor and putting the material at a known temperature and then seeing what temperature the camera reports. Brute force twiddling can get you to the correct emissivity for the material.

> Develop a model that will allow you to input the deltaT and calibrated emissivity and have it kick back the correct deltaT. Mathcad using a solve block can handle that.

> There used to be a freeware blackbody emission calculator that would give total and in-band radiation and flux for blackbodies called Blackbody Calculator and was given out by Integrated Sensors, Inc. from Goleta, CA. The copyright on my copy is 1989, so it may not be available anymore.

TTFN

 

[electricpete]

Thanks IRStuff - your comments have been invaluable.

I did the integration numerically using Maple. Developed a curve of in-band radiation (watts/m) vs Kelvin temperature. Calcs and curve are here.

http://reliability-magazine.com/pub/PlankLongWave.PDF

Also in that file I analyzed the example case. Still didn't work out quite right. I am curious if anyone can check the reasonability of the curve.

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[electricpete]

I will re-try the calc with 1-5 micron range (short wave) and see if it matches what was published in that article

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[JKEngineer]

I did not review the calculation in detail, but it appears that you are neglecting the reflected component of radiation. If the body is not transmissive, then the radiation coming from it is the sum of emitted and reflected energy. The reflected energy depends on the ambient or background temperature. Note that this is not the air temperature, it is the temperature of objects that reflect from the surface, typically considered to be "behind" the camera.
Jack

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

http://www.KleinfeldTechnical.com

 

[electricpete]

Thanks Jack.

Assuming my graph gives blackbody radiation in long-wave band as a function of temperature...call it Sbb(T)

Then would the full expression for an opaque item at temperature Tspecimen with background temperature Tbackground be given by the following?

S = e* Sbb(Tspecimen) + (1-e) * S(Tbackground)

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[IRstuff]

That's essentially correct. However, the error, even with poor emissivities, is usually low, particularly with high temperatures. That's because the the energy from reflection is from a lower temperature and thus contributes less than from a higher temperature reflection.

We usually have problems in that regard, measuring a blackbody at say, 25º, but seeing reflections from the operator, which is hotter than the blackbody.

Another possibility is that the uncooled camera is "seeing" more or less of the ray bundle than it should. Classical cooled thermal detectors use what's called a "re-imaging" optic to project an "image" of the system f-stop onto a cold shield. The cold shield becomes the limiting aperture of the system and is positioned with some accuracy. On some uncooled cameras, conventional optics are used, and the f-stop of the system may not provide sufficient baffling to produce a clean ray bundle. We've had problems in this regard as well, when testing with a pinhole, it's possible to get >100% apparent transmission, because the detector can see more of the test blackbody than it's supposed to.

TTFN

 

[IRstuff]

btw... I can't see anything wrong with your calculations, but the flux at the 400K end are a bit high. I'm getting about 556 W/m^2 from both the ISI program and Mathcad, and it looks likearound 562 on your graph. A minor nitnoid, to be sure, but annoying, nonetheless.

Another thing comes to mind. Most thermal imagers are two-point calibrated, e.g., both for offset and gain. However, no thermal sensor that we've seen is actually linear. There is an infamous "W" curve, that describes the response of the sensor after calibration. The two points on the "W" correspond to the two calibration temperatures. So, how the camera is corrected for non-uniformity can affect the apparent response. I'm not sure whether imaging radiometers account for that or how.

TTFN

 

[electricpete]

I found an infrared calculator here:

http://www.iriacenter.org/iriaweb.nsf/Planck.js?OpenPage

I put in the band 8 - 14 um and evaluated 400K.

The result was 0.0462 W/(cm² * sr)

OK, I can convert from cm^2 to m^2 with factor of 1E4 giving

462 W/(m² sr)

Somewhere around 100 off of our result. But what is "sr"? Can someone explaion it?





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[electricpete]

Whoops, I'm sorry. For 400k, the above result 462 W/(m² sr) is shown as total energy.

In the band 8-14 um it shows 176.9 watts / (m^2 * sr).

That is about exactly a factor of pi off from what IRStuff posted. I am familiar with the term steradian as solid angle but I don't quite see what the relationship is here.

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[electricpete]

I found another calculator

http://230nsc1.phy-astr.gsu.edu/hbase/quantum/radfrac.html#c1

This one gives 558 w/m^2 for the case above.

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[IRstuff]

sr is indeed a steradian and is involved in the distribution of radiated energy in the hemisphere above the emitter. The assumption here is that the emitter is Lambertian, e.g., uniformly diffuse.

I forget the exact process, but the integrals work out such that the energy emitted per unit solid angle comes out to the total energy divided by pi steradians. The derivation is usually found in most texts convering radiometry.

TTFN

 

[electricpete]

Going back to the difference in computed emissions at 400k as computed by the various programs.

Again the case 400k, e=1, in the band 8-14 um

The website

http://230nsc1.phy-astr.gsu.edu/hbase/quantum/radfrac.html#c1

computes 558.00 watts/m^2

I redid my calculation using numberical integration by trapezoidal rule with a variable number of steps.
Program is here:

http://reliability-magazine.com/pub/plankbandfunction.PDF

My results (please excuse the large number of decimal places which are not intended to reflect actual precision of the calculation)
1 step - 557.309151489089
2 steps - 562.879349079766
5 steps - 562.926504744547
10 steps 562.867887147106
100 steps 562.844927391277
1000 steps 562.844688753207
10000 steps 562.844686365965


From this one might conclude that the website calculator used only one step? (I doubt it, but it's the only explanation I can come up with at the moment.)

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[IRstuff]

I thought I had vetted your constants against the results, but apparently I messed up the first time. Your c2 comes out to be 1.4338 cm*K, while my c2 comes out to be 1.4388 cm*K. That changes the result from 555.82 to 562.83.

I'm using c2:h*c/k, with h:6.62606876E-34*J*s, c:2.99792458E+8 m/s and k:1.3806503E-23 J/K

My c2 is calculated on-line in Mathcad, but yours appears to have been calculated elsewhere. Can you verify your c2?

TTFN

 

[electricpete]

Aha. Thanks, you figured it out. I must have had dyslexia when I typed that constant. I tried to doublecheck by solving again from "scratch" in the 2nd program, but I got lazy and cut/pasted my constants and carried my error into the 2nd program (duh). Maybe that's a lesson learned for next time I'm troubleshooting my programs.
Thanks again.

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[electricpete]

btw - I saw a curve of typical relative sensitivity of long-wave bolometers or whatever you call them. It was close to 1 between 8 and 10 and then started decreasing toward 0.4 at 14. So my square-shaped assumption of equal sensitivity between 8 and 14 probably won't be very accurate at predicting camera behavior.

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[IRstuff]

That can be very involved exercise. Not only might there be variability in responsivity, there is also variability in spectral emissivity.

This is often coupled with transmission issues, particularly at the 7 um end. Even a couple of feet at 7 um will knock the transmission down significantly. Even just below 8 um, there's a bunch messy stuff.

Oh, and don't forget that humidity can drastically affect the performance of LWIR.

TTFN