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Associate

Join Date: Apr 2011
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Correction Factor for Emissivity

10/09/2012 2:22 AM

I have IR Thermometer HT 826, but it's emmissivity is fixed at 0.95 by manufacturer. How we can measure surface temperature of black body (emmissivity 1.0) or other material who has different emmissivities. is there any chart available to convert one emmissivity reading to other.

In meter there is no facility to change emmissivity

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#1

Re: correction factor for Emissivity

10/09/2012 9:09 AM

There is no reliable mathmatical conversion of emmissivity that I trust. It is a function of the type of material, surface finish, cleanliness, etc. In practical situations, we almost always cross check an IR Thermometer against a thermocouple probe readout, and adjust emmissivity. You could do the same and create your own conversion chart, but you may be better off budgeting for another IR Thermometer where you can adjust emmissivity.

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#2

Re: correction factor for Emissivity

10/09/2012 12:31 PM

For materials of known emissivity like copper, i have used the standard value and have had good readings (~2-5% of Thermocouple readings)..

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Power-User

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#3

Re: Correction Factor for Emissivity

10/10/2012 6:25 PM

>is there any chart available to convert one emmissivity reading to other.

No chart, but you can calculate a corrected temperature.

An IR thermometer with a fixed emissivity of 0.95 will indicate lower temperatures than reality for objects with emissivities lower than 0.95.

In most commercial products, emissivity is a arithmetical divisor or multiplier factor.

If an object has an emissivity of 01.0, then it emits IR at only 10% of that which a black body would emit at that same temperature. The IR thermometer only "sees" 10% of the energy that a black body would produce at the same temperature.

IR thermometers with an emissivity correct factor solve an algebraic equation using the stated emissivity factor, (in the example, 0.10)

10% of what (x) = "energy seen"
0.10 * x = "energy seen"
x = "energy seen"/0.10

Dividing by 0.10 is the same arithmetically as multiplying by 10.

If the 'energy seen' for a object with a 0.10 emissivity factor = 95 Deg C (if it were a black body), then 95 Deg C/0.10 = 950 Deg C for a reading from an object with a 0.10 emissivity.

The IR thermometer is already adding about 5% as its compensation for its fixed 0.95 emissivity factor.

To correct for the fixed 0.95 emissivity of the IR thermometer, the fixed correction factor, 0.95, is divided by object's emissivity to get a correction multiplier factor: 0.95/object emissivity

For example:
For a black body with a 1.0 emissivity, the correction factor is 0.95/1.0 = 0.95

For an object with a 0.95 emissivity, the correction factor is 0.95/0.95 = 1.0; no correction necessary.

For an object with a 0.50 emissivity, the correction factor is 0.95/0.50 = 1.9

For an object with a 0.13 emissivity, the correction factor is 0.95/0.13 = 7.31

An instrument using a fixed emissivity of 0.95 might not have a sensitivity needed to accurately read the lower levels of IR radiation of objects with low emissivity.

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Participant

Join Date: Mar 2020
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#4
In reply to #3

Re: Correction Factor for Emissivity

03/30/2020 1:45 AM

Old thread but it is still showing up in search engines and the above answers are not correct.

The above answer has two problems.

Firstly is Celsius instead of Kelvin.

Secondly there is a power to four between the temperature and the instrument sensor output.

j* = εσT4

https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law

j* is what the instrument is measuring under ideal conditions. (watts per square meter)

ε is emissivity, a number between 0 and 1 (approx. 0.98 for human skin)

σ is a constant

T is temperature in Kelvin

So with the instrument set to 0.95 a simplified equation for conversion could be expressed as:

0.95σ(Tinstrument)4= εnewσ(Tcorrected)4

since σ appears on both sides we can substitute it with 1

0.95(Tinstrument)4= εnew(Tcorrected)4

Tcorrected = 4√((0.95(Tinstrument)4)/(εnew))

/Martin

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