Calculating Losses for a Blackbody Heater and Infrared Sensor Geometry

[Natalie Johnson]

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.

 

[haruspex]

found it here
http://webserver.dmt.upm.es/~isidoro/tc3/Radiation View factors.pdf

Of the many cases listed at that link, which one do you think applies to your problem?

 

[Natalie Johnson]

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

 

[haruspex]

Hi. I know its a patch-disc setup

Why patch-disc rather than unequal discs? And which is the patch?

 

[Natalie Johnson]

Why patch-disc rather than unequal discs? And which is the patch?

Patch is the small sensor and the bb heater is the disc?

 

[haruspex]

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]

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.