dc electromagnet question?

I have constructed a DC electromagnet for a special use at my shop. Being a electrical contractor I thought this would be simple. (NOT) Made the
electromagnet, connected it to a 12vdc power supply at 5 amps and it works great with one problem, it starts to heat up quickly. My desired results was for it to run on a 80% duty cycle. Will resistors keep this heating problem from happening? DC voltage is not something I have dealt with indepth. Can anyone help me with this problem? snipped-for-privacy@cox.net Vick
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vick wrote:

A resistor in series will decrease the current, decrease the dissipation in the coil and decrease the strength of the electromagnet. If that is ok, then fine, stick a resistor in series or lower the voltage.
Otherwise, you need a way of getting rid of the 60W that you are dumping into the coil - or redesign the electromagnet. A fan or two (or four) might be an idea.
Another idea, that might be appropriate, is a resistor that can be switched out. So, the full strength of the electromagnet will be available when needed, but some form of automatic or manual switch reduces the current when the full strength isn't.
--
Sue




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Is there a way to reduce the wattage from the power supply or should I just get another power supply with less wattage? Thanks for your help!!!!!!!!!!!!!!!!! Vick

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vick wrote:

You can get a power supply with less *voltage*. That will reduce the current and hence the power dissipated - and the strength of the electromagnet.

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Being "positive" this time.
Rewind your coil with a finer gauge of wire but use more turns.
The higher resistance will reduce the current and hence the heating but as magnetic flux is the product of the number of turns and the current you can maintain the required strength of field/
For example, if you reduce the wire size from 1mm square to 0.5mm square and double the number of turns. The current, and hence the power dissipation, will be halved but you magnetic flux will remain virtually the same.
--
Stuart Winsor

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----------------------------

----- You indicate a doubling of the wire length and half the cross-section--resulting in 4 times the resistance. From this the current and losses will be 1/4 the original value but the force will also be approximately 1/4 (proportional to ^2). (unless the original setup was driving the iron into saturation ).
--

Don Kelly snipped-for-privacy@shawcross.ca
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Hmm! I must remember not to post at 20 mins past midnight, when I'm already half asleep, in future :-)
Your other post is of course the correct answer.
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Stuart wrote:

That was why, at a slightly earlier time, I settled for writing, "Or redesign the electromagnet" - without explaining how. :)
I thought, if he really wants to know, he can ask and I can answer - when the old brain cell isn't still swimming in cherry brandy..
--
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Nice!
Personally I prefer a single malt but cherry brandy's nice too.
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wrote:

---------------
You are not the only one to boob due to lack of sleep, surfeit of wine, or simple haste. I have done it more times than I want to count.
--

Don Kelly snipped-for-privacy@shawcross.ca
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Why (NI)^2?
MMF is just NI!!, as in ampere-turns!!
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----------------------------

-----------
mmf is NI but the mechanical force is dependent on (mmf)^2 . Check it out.
--

Don Kelly snipped-for-privacy@shawcross.ca
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Sorry about that.
A "senior moment," I guess.
Thanks again.
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No problem- join the crowd and be glad that they are not Roy or chairman moments.
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Stuart wrote:

I hate to be the wet blanket here, but winding a coil with smaller diameter wire will NOT change the heating if the field strength is kept the same! It will change the voltage required to give that heating.
As a hint: A coil wound with 950 feet of 20 gauge wire will weight about 3 pounds, run on 12 volts dissipating about 15 watts in the coil (think of a soldering pencil in there!) and produces a field of about 250 Gauss (without any iron present). The coil will draw about 1.2347 amps at 12 volts and have a resistance of roughly 9.7 Ohms. If you change the wire gauge keeping the coil size identical everything will remain the same except the voltage needed to get that field. Note that coil inductance will be greatly changed but that is not of concern here. (Inductance of the 20 ga version about .042 H.)
(Calculation of the above values left as an exercise for the interested student!)
You can only lower the heating by lowering the voltage drive which of course lowers the field strength of the magnet. Winding with finer wire but keeping the voltage the same will reduce heating but also the field. If you can get away with a lower field that is fine. Otherwise, the usual solution is to use lower voltage and heavier wire! Namely tiny copper tube with insulation on it. You run water through the tube to take the heat out of the coil. Except for dealing with water it makes a killer electromagnet. Takes a heavy current low voltage power supply though.
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Thanks again for your help. I will be applying your suggestions. You help me go back to school and you are a great teacher. Thanks again for you willingness to share your knowledge. Vick

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If you're an electrical contractor remind me never to employ you to do anything electrical. You clearly haven't a clue.
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Thanks for your help! It will be a do able solution hopefully. Being a electrical contractor I (believe it or not) specialize in design/build unique commerical electrical applications. I have dealt with dc only with systems that have already been design for special equipment. But, for me, this just did not compute in my brain compared to ac for what ever reasons. But thats again for your help. Its people like you helping people like me that make the world a better place. Vick
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<SNIP>
Glad to be of service - I trust you read my other post?
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----------------------------

----- If you use resistors, you will reduce the heating-but you will also reduce the force produced by the magnet in proportion. Using a lower voltage source will also do the same.- In that case increase the number of turns of wire to compensate -this will also add some resistance.
For example, twice the number of turns (same wire) doubles the resistance and halves the current resulting in half the heating for the same force(proportional to ampere-turns)- but a bigger magnet. Life consists of tradeoffs.
--

Don Kelly snipped-for-privacy@shawcross.ca
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