Appendix

Emissivity

Emissivity is defined as the ratio of radiance of a given body to that of a blackbody or "perfect radiator". As no physically realizable material is a perfect blackbody everyday objects will have emissivities <1.00.

The following material explains how emissivity is accounted for with the Compix thermal imaging systems.

When a thermal image is acquired radiant power is measured. An emissivity assumption, often unity (1.00), is used to convert the radiant power to a temperature value. Converting the temperature readings made based on one emissivity setting to a new set of temperatures based on a different emissivity is basically a two-step process. First, the original emissivity setting is used to compute the corresponding radiant power density. Secondly, the radiant power density is converted back to temperature using the newly specified emissivity factor.

Due to round off errors in the array data, converting temperature to power density and then back to temperature will not produce the same image as the original.

The concept of emissivity and how it affects measurements made by the PC2000 and represented by WinTES95 is illustrated through the detailed example which follows. This example first shows why an object at 50°C having an emissivity of 50% would appear to be at only 36.5°C if the emissivity setting were left at 100%. To see how this results we need to use Stefan's law which gives the radiant energy R as,

R = e*k*T4

where T is the temperature expressed in degrees Kelvin, e is the emissivity, and k is a conversion constant depending on the units used for T and R. For T in degrees Kelvin and R in watts per square meter, k is 5.6699*10-8. The arithmetic can be simplified if temperature is expressed in hundreds-of-Kelvin-degrees and R is in mW/cm2 making k have the numerical value of 0.56699.

Applying this to a blackbody of 50°C one may calculate as follows:

TH = (273 + 50)/100 = 3.23 where, TH is the temperature converted into hundreds-of-Kelvin-degrees, and 273 is the value of 0°C on the Kelvin scale

TH4 = 3.234 = 108.845

k*TH4 = .56699 * 108.845 = 61.714mW/cm2

Saying it was a blackbody is equivalent to saying e = 1, Hence

R = 1 * k * TH4 = 61.714mW/cm2

Now, this 61mW/cm2 is the so-called Blackbody Radiation, and is not in general the same as the total or net radiation. At the same time that the 50°C object is radiating due to its surface temperature, it is also absorbing radiation from its surroundings. If the surroundings are at an average effective temperature of 21°C, the incident radiation can also be calculated from Stefan's law since the cumulative effect of the surroundings are the same as a blackbody (or cavity) radiator:

TH4 = [ (273 + 21)/100 ]4 = 2.944 = 74.712

R = 1 * k * TH4 = .56699 * 74.712 = 42.361mW/cm2

Thus, the Net Radiant Power, NRP, for the 50°C object is

NRP = R_radiated-R_incident = 61.714 - 42.361 = 19.353mw/cm2

Note: if the above calculations were applied on the basis of a 50°C surroundings then the resultant NRP would be zero, a result which defines thermal equilibrium by describing a condition whoere the object is neither gaining nor losing energy.

Now, suppose the 50°C object of interest has only 0.50 emissivity. That means the object is only 50% efficient at radiating its internal heat. It also means that it absorbs only 50% of the radiation it receives from its surroundings and, unless it is transparent, it therefore reflects 50% of the radiation from the surroundings.

Using the results of the preceding calculation,

RBB-50 = 61.714mW/cm2 Blackbody radiant power for 50°C and RAMB = 42.361mw/cm2 21°C-Ambient surroundings' radiant power

the Net Radiant Power for the 50%-efficient 50°C object can be easily calculated:

NRP = ( Generated ) + (Reflected ) - ( Incident ) NRP = 0.5 * 61.714 + 0.5 * 42.361 - 42.361 = 9.678mW/cm2

In general, (ignoring the case of transparent objects) the NRP is given as

NRP = e * RBB + (1 - e) * RAMB - RAMB

Simplifying by combining terms yields

NRP = e * ( RBB - RAMB )

How does the PC2000 respond to this 50% emissivity, 50°C object? To find the answer it is necessary to ask "what temperature would a blackbody have to be at in order to have the same Net Radiant Power?" In the 50% emissivity case the NRP was computed to be 9.678mW/cm2. If that NRP were produced by a blackbody, what would the temperature of that blackbody need to be? That involves solving the equation,

NRP = 1.00 * ( RBB - RAMB ) = 9.678

giving

RBB = 9.678 + RAMB = 9.678 + 42.361 = 52.039

Stefan's law can now be applied, solving for TH ,

TH = Fourth-Root of [ R / (1.00 * k) ]

TH = Fourth-Root of [ 52.039 / (.56699) ]

TH = Fourth-Root of [91.781]

TH = 3.095

ToC = 309.5 - 273 = 36.5°C

Thus, for an emissivity setting of 100%, the indicated temperature of a 50% emissivity object at 50°C would be 36.5°C. If the setting for emissivity was changed to 50% (either when the original image was acquired, or afterwards using the emissivity function of WinTES95) the 50% emissivity object would then indicate 50°C as expected.

Suppose instead that an object of unknown temperature having 50% emissivity indicates a temperature of 50°C with an emissivity setting of 100%. Because the indicated temperature with emissivity of 100% is 50°C, from previous calculations the NRP must be 19.353 mW/cm2.

Using NRP = e * ( RBB - RAMB )

and solving for RBB

RBB = 19.353 / 0.5 + 42.361 = 81.067mW/cm2

Applying Stefan's law for the equivalent blackbody that radiates 81mW/cm2,

TH = Fourth-Root of [ R / (1.00 * k) ]

TH = Fourth-Root of [ 81.067 / (.56699) ]

TH = Fourth-Root of [142.977]

TH = 3.458 T°C = 345.8 - 273 = 72.8°C

 WinTES95 does not inhibit the user from specifying unrealistic factors in making conversions. Be aware that making units or emissivity conversions change the data and measurement parameters in ways that cannot be exactly reversed, and in extreme cases data can be drastically truncated or changed.

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