Calculation - Effect of emissivity error on delta-T between suspect an
Calculation - Effect of emissivity error on delta-T between suspect an
(OP)
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#t04052...
....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?
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#t04052...
....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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RE: Calculation - Effect of emissivity error on delta-T between suspect an
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
RE: Calculation - Effect of emissivity error on delta-T between suspect an
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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RE: Calculation - Effect of emissivity error on delta-T between suspect an
> 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
RE: Calculation - Effect of emissivity error on delta-T between suspect an
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....
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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RE: Calculation - Effect of emissivity error on delta-T between suspect an
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RE: Calculation - Effect of emissivity error on delta-T between suspect an
Jack
Jack M. Kleinfeld, P.E. Kleinfeld Technical Services, Inc.
Infrared Thermography, Finite Element Analysis, Process Engineering
www.KleinfeldTechnical.com
RE: Calculation - Effect of emissivity error on delta-T between suspect an
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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RE: Calculation - Effect of emissivity error on delta-T between suspect an
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
RE: Calculation - Effect of emissivity error on delta-T between suspect an
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
RE: Calculation - Effect of emissivity error on delta-T between suspect an
http://www.iriacenter.org/iriaweb.nsf/Planck.js?Op...
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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RE: Calculation - Effect of emissivity error on delta-T between suspect an
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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RE: Calculation - Effect of emissivity error on delta-T between suspect an
http://230nsc1.phy-astr.gsu.edu/hbase/quantum/radf...
This one gives 558 w/m^2 for the case above.
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RE: Calculation - Effect of emissivity error on delta-T between suspect an
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
RE: Calculation - Effect of emissivity error on delta-T between suspect an
Again the case 400k, e=1, in the band 8-14 um
The website
http://230nsc1.phy-astr.gsu.edu/hbase/quantum/radf...
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/plankbandfunct...
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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RE: Calculation - Effect of emissivity error on delta-T between suspect an
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
RE: Calculation - Effect of emissivity error on delta-T between suspect an
Thanks again.
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RE: Calculation - Effect of emissivity error on delta-T between suspect an
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RE: Calculation - Effect of emissivity error on delta-T between suspect an
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