Calculating number of cooling fins on a waveguide load 4

[microwavemark]

I need some help in ways of calculating the number and size of fins required to disperse heat generated by a load element inside a piece of wave guide.

The load element is silicon carbide absorbing 200W of power in copper or aluminium WR75 waveguide.

Any pointers would be very much appreciated

 

[logbook]

The problem shouldn't be that hard so long as you deliberately keep your calculations simple. Typically the fins are pretty much as small as you can make them by machining or extruding. Say 2mm thick as a starting point, depending on the overall size.

1) Please confirm that the heatsink is on the outside of the waveguide?
2) Is the heatsink force convection cooled (fan blown) or natural convection cooled (or dipped in water!)
3) How hot is the base of the heatsink allowed to get?

Let's do some simple sums. If the load dissipates 200W and you want a temperature rise on the load of not more than say 50 degC then you need a heatsink better than 50degC/200W= 0.25degC/watt. You can look in a distributer's catalog at 4degC/watt heatsinks and see how big they are. You may not find one as big as a 0.25degC/watt. This is a BIG heatsink and without fan cooling it may not be workable.

 

[microwavemark]

1. Yes, the heatsink is on the outside of the waveguide.
2. Is natural convection cooled
3. Say about 60 degC

I was thinking of using a set of individual fins mounted in the same plane as a flange would be, about 100mm by 100mm by 2mm, spaced along the guide at regular intervals.

The finished load would look something like this one .

Many thanks for your help

 

[IRstuff]

You need to work the numbers. Natural convection can be as low 4 W/m²ºC. If you design for a 50ºC delta temp and 300 W of dissipation, you'd need 1.5 m² of effective fin area.

Obviously, a higher HTC would yield different results.

TTFN

 

[biff44]

Well, it is important to not exceed the allowed operating temperature of the load material. If it is some sort of lossy ceramic material that doesn't melt until 1500 deg F, then you can get away with less heatsinking. If it is an elastomeric load material that will turn to carbon at high temperatures, then you need to get the heat out of the load by keeping the waveguide wall cooler.

In otherwords, the higher you are willing to allow the waveguide wall temperature to rise, the less heatsink area you need.

 

[Warpspeed]

There are a lot of variables here as others have already pointed out. But a very rough as guts initial guess based on normal free convective airflow, would be one square inch of double sided fin area per watt for about a 60C temperature rise.

That will give you a rough starting point. Another way is to look up the specifications of commercially available heatsinks of roughly similar fin size and spacing, and do a little reverse engineering.

For 200W, and a fin size of 100 x 100mm, I would start thinking about having at least a minimum of twelve to fifteen fins, spaced widely enough apart not to restrict free convective airflow.

Build up a prototype, and use a simple resistance heating element and a dc power supply to test with, and get some exact temperature rise figures. Then you will have a pretty good idea where to go from there.

 

[biff44]

You could just make them this big:

http://www.milpi.com/582main.html

 

[zappedagain]

IRStuff,

"Natural convection can be as low 4 W/m2ºC."

Is that from your experience or can I find that in a (thermodynamics) book sowmewhere?

Thanks,

Z

 

[IRstuff]

should be textbooks. There are also on-line calculators:

http://www.frigprim.com/online/natconv_heatsink.html

http://www.mhtl.uwaterloo.ca/cgi-bin/rect_hs

Caveat on the supposed conditions in the calculator.

TTFN