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Electricity to mechanical force calculation

  1. Jun 8, 2009 #1
    I am trying to understand the mechanical force exerted on a breaker when it is subjected to 200kA fault current. In other words is there a formula which gives the force (newton) value when we key in the value of the current. Please guide me.
     
  2. jcsd
  3. Jun 8, 2009 #2
    You could start here. http://en.wikipedia.org/wiki/Circuit_breaker" [Broken]
    There are few types. I've long been curious as to how they operate. Actually, that goes for just about anything hiding inside a case. Got a hammer and a dead circuit breaker?
     
    Last edited by a moderator: May 4, 2017
  4. Jun 11, 2009 #3
    The repulsion forces of circuit breaker electrical contacts are related to square of through fault current (peak amount) and some physical situation of contacts. For more information you can refer to General electrical riddle NO.37 from http://electrical-riddles.com
     
  5. Jun 16, 2009 #4

    negitron

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    That's a joke, right? Electromagnetic repulsion? Please. While there are certainly repulsive forces at work, they are absolutely insignificant in this application; the reason for stronger springs on high-interrupt breakers is simply to get the contacts apart faster so they don't fuse.
     
  6. Jun 16, 2009 #5

    Danger

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    There are 3 main types of breakers. The simplest ones use a bimetallic strip similar to that found in a thermostat. Excess current heats it to the point where it bends and thus breaks contact. Others have an electromagnet that will pull the contacts apart when the current overage increases the magnetic field to a particular level. Yet another type has a small explosive charge that blows the contacts apart when triggered.
     
  7. Jun 16, 2009 #6

    dlgoff

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    Hmm. Take a look at these circuit breaker pictures. Looks sigunificant to me.
    https://www.physicsforums.com/showthread.php?t=284671"
     
    Last edited by a moderator: May 4, 2017
  8. Jun 16, 2009 #7

    negitron

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    Well, okay, technically those photos are the end result of electromagnetic forces (as are nearly every event we can observe) at work. But, not in the more conventional sense that's being discussed here; those look like the result of an arc blast--and a fairly significant one, at that.
     
  9. Jun 16, 2009 #8

    dlgoff

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    The OP mentioned 200kA fault current. I assumed it had to do with power generation/transmission; which is what m.s.j was thinking I think.
     
  10. Jun 16, 2009 #9

    Danger

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    Oops! I somehow missed that. Yeah, that's a fairly significant circuit. Way out of my league.
    (Maybe a small nuke to disable it?)
     
  11. Jun 17, 2009 #10

    Somehow I missed that too, but I learned a few things from wilipedia out of it.

    A home with 22KW @ 220V service would draw 50 Amps on each 120 VAC half.So the main breakers in a house can handle about 50 Amps.
     
    Last edited: Jun 17, 2009
  12. Jun 17, 2009 #11
    Ya. I missed that too somehow. I think the OP is confused.

    The service to a house might be about 125 Amps. 125 Amps on each 120 VAC leg for 30K Watts. The OP is talking 1600 times that.

    A line supplying 1600 amps wouldn't be 120 VAC. That would be a waste of copper. So whatever sort of breaker, would have to blow out an arc with a few thousand volts behind it when opened.

    Maybe the question was intended to be about 200 amp.

    Say, we are talking about a 16 Megawatt plus line. Is there such a thing in existance? And why in the world would the OP be asking here?
     
    Last edited: Jun 17, 2009
  13. Jun 17, 2009 #12
    excuse me, but i think when persons just know a little of fact, they say jokey small talk.:wink:
     
  14. Jun 17, 2009 #13

    negitron

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    200 kA is not an uncommonly high instantaneous fault current for high-power distribution circuits. Note that a current of this magnitude will only flow for a few tens to hundreds of microseconds before the breaker trips. The equipment I deal with at work routinely has main breakers rated for 65 kA interrupt and higher.
     
  15. Jun 17, 2009 #14
    OK, now that makes sense. 200K is the fault current, not the trip current, as I read it.
     
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