Calculating temp increase from IR

[MacGyverS2000]

It has been entirely too long since I took thermo, but I'm not sure a more recent experience would have helped me anyway...

I'm trying to calculate the temperature rise of an area radiated with IR energy. I'm radiating a 0.005-0.006" diameter area (roughly circular, close enough for these purposes) with 70+ J/s at a wavelength of approximately 10.6um. For the purposes of the calculation, I'm going to assume all energy is absorbed by the material (little to no reflection) and does not have an opportunity to propagate to the rest of the material (though I'm happy to entertain thoughts on propagation as long as it stays 10,000 foot view level and doesn't launch into a treatise on partial differential equations)... hopefully this assumption removes material specific heat constants from the thought process.

Is this too little information to get a rough idea? In a nutshell, I'm trying to determine the increase in surface temperature over time as I radiate the substrate. I need to reach a minimum temp before the substrate surface will do what I need it to do.

Dan - Owner

http://www.Hi-TecDesigns.com

 

[IRstuff]

A bit too little. A long exposure involves the thermal conductivity and specific heat of the substrate, making it a much more complicated problem. Higher power, shorter pulses will keep the energy closer to the surface. We've got some calculations at work that we did with a pulsed laser; I can try and find that...

TTFN

FAQ731-376

 

[gwolf2]

We need the mass, material, and approximate dimension of the object which you want to heat.

70 watts feels like quite a lot of energy for such a small area and it is likely that your component will heat up quite quickly. If that is the case then as a first approximation you can ignore conduction losses and use this simple equation to see how much energy you need to get it up to the desired temperature:

h = m * Cp * (Tend-Tstart)

If you do this in Si then h is in J and H/70 is how long you need to heat it in seconds.

If you want to hold that temperature for any length of time, and I'm guessing of the order of 10-15 seconds or more then you will have to move from a loss-less system to a heat balance between 70 watts in vs conduction and re-radiation out.

Try the simple approach first, see what the scale of the problem is.

gwolf

 

[MacGyverS2000]

gw,

Actually, this is for a (relatively) short time period (on the order of milliseconds per spot, maybe tens of milliseconds). The application is to heat a thin material covering various substrates, from highly conductive metal to low conduction glass. While there is heat conduction into the substrate, I believe the (near) instantaneous application will put most of the heat into the thin covering material. The location is irradiated for a few milliseconds, the spot is moved to a partially overlapping location, and the process is repeated... heating of the substrate is incidental. Since the substrate itself is quite large (and possibly thick) compared to the covering material, the intent is to quickly heat the covering without regard to what it's attached to. Temps in the 1,500F+ range are desired.




IR,

A high-energy pulsed mode is not an option, but whatever info you have is bound to be useful. I can linger on the same spot as long as it takes to heat to those temps, but being able to do it quickly enough that the specific heat of the substrate becomes irrelevant is best (not to mention reducing processing time).


Dan - Owner

http://www.Hi-TecDesigns.com

 

[IRstuff]

What's the substrate's parameters?

TTFN

FAQ731-376

 

[gwolf2]

How thin is the film and what are it's material properties?

You need the thickness, specific heat capacity Cp and the density.

 

[MacGyverS2000]

The substrate is typically metal (aluminum, copper, stainless steel, etc.), glass, ceramic, and so on. The film is a glass frit added to a thickness of 0.001-0.002" (+/-).

Dan - Owner

http://www.Hi-TecDesigns.com

 

[IRstuff]

specific heat of glass = 0.2cal/gm-K
density = 2.6gm/cm^3
thk = .002in
diam = .006in

sp*(diam/2)^2*pi*thk*dens = 2 uJ/K

so for 70 W source --> 35MK/s

so, seems feasible.

TTFN

FAQ731-376

 

[MacGyverS2000]

Thanks, IR. Motion speeds for current systems are in the 1-8" per second, so if I can match those I'm happy.

Dan - Owner

http://www.Hi-TecDesigns.com