Calculating Losses for a Blackbody Heater and Infrared Sensor Geometry

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SUMMARY

The discussion focuses on calculating thermal losses for an infrared sensor positioned above a blackbody heater. The sensor, with an area of 0.1 mm², is located 0.1 m above a disc-shaped blackbody heater with a diameter of 0.05 m. Key equations include the emittance from a blackbody, represented as σT⁴, and the view factor formula, r² / (d² + A²), where r is the radius of the heater and d is the distance to the detector. Participants emphasize the need to correctly apply the view factor concept to determine the irradiance received by the sensor.

PREREQUISITES
  • Understanding of blackbody radiation and thermal emittance
  • Familiarity with view factors in radiative heat transfer
  • Knowledge of basic geometry related to circular discs
  • Proficiency in applying the Stefan-Boltzmann law
NEXT STEPS
  • Study the application of view factors in thermal radiation calculations
  • Learn about the Stefan-Boltzmann law and its implications for blackbody radiation
  • Research the derivation and application of formulas for unequal disc geometries
  • Explore practical examples of infrared sensor applications in thermal analysis
USEFUL FOR

Students and professionals in thermal engineering, physicists studying radiative heat transfer, and engineers designing infrared sensor systems will benefit from this discussion.

Natalie Johnson
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Homework Statement


At room temperature, an infrared sensor with an area 0.1mm2 is suspended 0.1m above a disc that is a blackbody heater of diameter 0.05m . The infrared detector absorbs all thermal infrared radiation (0.2 - 100 microns) produced by the blackbody heater. What are the losses for the geometry. You may assume the detector acts as a blackbody.

Homework Equations


Emittance from blackbody = σT4
Irradiance of infrared at detector = σT4

The Attempt at a Solution


All I have found is the view factor for this geometry is
r2 / ( d2 + A2)
Where r is radius of blackbody heater and d is the distance from the detector.
And I found it here
http://webserver.dmt.upm.es/~isidoro/tc3/Radiation View factors.pdf

I have read that irradiance is 1/r2 drop off but I don't understand how to implement this. Can someone share knowledge please.
 
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haruspex said:
Of the many cases listed at that link, which one do you think applies to your problem?
Hi. I know its a patch-disc setup which on arrangements gives the equation I wrote in the attempt. But I am not sure if this is correct, I just found this information on a google search. What are the losses for something like this? Its quite difficult
 
Natalie Johnson said:
Hi. I know its a patch-disc setup
Why patch-disc rather than unequal discs? And which is the patch?
 
haruspex said:
Why patch-disc rather than unequal discs? And which is the patch?
Patch is the small sensor and the bb heater is the disc?
 
Natalie Johnson said:
Patch is the small sensor and the bb heater is the disc?
Ok, but the formula at the link gives you the fraction of the emissions from the patch that would be intercepted by the disc. Your problem is the other way around.
 
haruspex said:
Ok, but the formula at the link gives you the fraction of the emissions from the patch that would be intercepted by the disc. Your problem is the other way around.
I assume you are still stuck.

If source 1, area A1, illuminates area 2, and the view factor is F12, then the light reaching 2 is σA1F12. But light paths are all reversible, so if we switch source and sink then the light reaching 1 from 2 is the same.
This leads to the general equation A1F12=A2F21.
Use that to find the formula for illuminating the patch from the disc.

By the way, I am almost sure that the formula at that link for two unequal discs is wrong, but I do not know what it should be.
 
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