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Cooling a resistor load at 6 to 10kW


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I have some resistors to cool that I presume are going to get rather hot!!

I am using 6x wirewound resistors in parallel to create a 3.75 Ohm load for a 150V source, hence generating 40A and 6kW. Later I may up the voltage to 200 so I need to spec the cooling for a max of 10kW. Each wirewound resistor is 28mm dia and 350mm long.

I was thinking of a open perspex box filled with some oil (maybe olive oil). Then creating a custom lid for the box out of a thermally conductive material that is an electrical insulator. Then mounting the resistors on to the bottom side of the lid on long legs so that they are completely submerged. On the top side of the lid I would put a large heatsink and fan.

The maximum length of time the power will flow through the load is 1 minute, so this will help reduce the overall specs.

If anyone likes this idea then my next challenge is to spec the size of the box, amount of oil and the size of the heatsink and fan. This is the tricky bit obviously and I'm not really sure where to start.

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I dont think your biggest problem will be getting the heat away. I am worried about the thermal resistance between the actual wire and the outside of the resistor. Maybe that will be your limiting factor. Use power resistors in aluminium heatsink (they will give that figure in their datasheet) and then mount everything on a big heatsink with a fan. Try to avoid oil, it is very messy. Consider adding more resistors in parallel to distribute the heat better.

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I concur with Alex's recommendation to avoid oil as a coolant on the basis of mess and cost.  Disposal of a tank of water is SO much cheaper and easier than a tank of oil.  Is this a long-term or a single-shot project?  Oil is used in transformers for reasons other than thermal efficiency, like corrosion, fouling, freezing, and boiling.

Heat capacity of oil (BTU/lb) is a little less than half that of water and it gets even worse in BTU/unit volume because oil is less dense than water.  I haven't investigated the topic, but I'd guess that heat transfer from an element to water is faster than to oil also due to viscosity.  That is, the convective flow of coolant past the element would be slower in oil than in water. 

Consider using electric water heater elements in a water tank as loads to completely eliminate the problem of electrical insulation of the resistance wire from the coolant.  Their resistance varies from about 10 ohms for a 240 volt 6 KW element  to about 20 ohms for a 3KW element.  Find out where all those rusted out water heaters go to die.  The elements are probably still usable and are probably just a waste disposal problem for plumbers.

6 KW for 1 minute is 100 watt-hours of energy.  1 KWH equals 3412 BTUs so your 1-minute test will generate 341.2 BTUs.  1 BTU is  the amount of energy that will raise 1 lb. of water 1 degree F.  If we assume that a reasonable temperature rise for your test is, say, 50 degrees F (from 55 to 105 degrees F) you would need about 7 pounds or less than 1 gallon of water.  You will have to have a 10 or 20 gallon tank just to have enough physical space for your water heater elements, so temperature rise is not a problem.

Sorry to use non-metric units.


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