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| General Electronics Chat Discussion forum for general chat about anything electronics related, including asking questions about material in the All About Circuits E-book, Worksheets, and Videos. |
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#1
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I am setting up a brief experiment that involves a switch an array on twelve 12V lead-acid batteries into a test inductor load at 150V with a limited 40A current.
This test will only be conducted for about 10 seconds and I dont want to spend too much time/effort worrying about current limiting methods. Hence, I was going to spec a high-wattage resistor around 4Ohm, connect it in series with the batteries and throw it in a bath of liquid nitrogen to keep it cool. Since the test will run for only 10 seconds, I thought I would get away with this. Anyone disagree or have any better suggestions? |
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#2
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That might end up with a steam explosion. You can get edge wound power resistors - http://www.ohmite.com/cgi-bin/showpa...powrrib_series - from any number of sources. 4 of them in series in air (1500 watt types) should do.
Place them in mineral oil for more heat transfer.
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First comes the hardware, then the software. |
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#3
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#4
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Heat = I*I*R = 40*40*5 = 8KW
Two 500W wire wound resistors in slow running water inside a container will easily cope with such wattage without issues. When I switch on my 2KW kettle for 10 seconds, the water inside is only barely warm. But with twelve 12V batteries you have 144V and a 5Ω resistor gives maximum 29A.
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L.Chung |
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#5
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I agree with L. Chung that your 500W resistors in water will be adequate for a 10 second test. In fact, I think you can just put them in a bucket of water and not worry about circulating the water. Liquid nitrogen would be way overkill, a hassle and a safety hazard and mineral oil would be unnecessarily messy and costly.
For electrical safety, use a plastic bucket and keep body parts out of the water and out of contact with any part of the circuit. If you are in an especially hard water area, use distilled water from the supermarket. Still much cheaper than mineral oil and no mess. But I don't think distilled water is necessary anyway because even with hard water the resistance through the water path will be MUCH higher than your test load resistance. 1 BTU raises 1 pound of water 1 degree F. 4 gallons of water (a convenient filling of a 5-gallon plastic bucket) at 8.34 pounds per gallon is about 33 pounds of water. 1 BTU is about 1000 Joules (watt-seconds). Your test will produce about 8KW for 10 seconds, or about 80,000 watt-seconds or about 80 BTU. 80 BTU/33 pounds of water = about 2.5 degrees F temp rise. Let us know what temperature rise you see (but avoid the temptation to feel the water temperature while the system is energized). awright |
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#6
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#7
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I assume just putting the resistor on the high voltage positive line in a bucket of water will be fine as long as the bucket is plastic and well insulated? Also, will I get away with unfiltered tap water? |
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#8
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Tap water in a plastic bucket will be fine. Do a little comparison for your comfort. Measure the resistance through your tap water between two bolts or metal tabs with about the same surface area as the exposed area of the terminals on your resistors and spaced about as far apart as the terminals on your resistors. Compare that reading with your 5 ohm test resistance. Will the water resistance have the slightest influence on your test resistance?
Even if the resistance was low enough to make a difference, it would only provide an additional "water resistor" in parallel with your test resistors. Here in Oakland, I measure about 100K to 250K Ohms between DVM test probes a couple of inches apart. No way will that have any affect at all on a 5 Ohm test resistor (except to keep it cool). awright |
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#9
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| 150v, 40a, current, limiting, technique |
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