Calculating Losses for a Blackbody Heater and Infrared Sensor Geometry

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Homework Help Overview

The discussion revolves around calculating thermal losses in a system involving a blackbody heater and an infrared sensor. The setup includes an infrared sensor positioned above a blackbody heater, with specific dimensions provided for both components. The problem is situated within the context of thermal radiation and view factors.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the view factor for the geometry and discuss the implications of irradiance drop-off. There are questions about the applicability of formulas found online and the correct interpretation of the patch-disc setup.

Discussion Status

The discussion is ongoing, with participants sharing insights about the view factors and questioning the appropriateness of the formulas referenced. Some guidance has been offered regarding the relationship between the areas and the view factors, but no consensus has been reached on the correct approach or formula to use.

Contextual Notes

Participants note potential confusion regarding the definitions of the components in the setup, specifically the roles of the patch and the disc. There is also mention of uncertainty about the accuracy of external resources referenced in the 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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