Electricity to mechanical force calculation

AI Thread Summary
The discussion centers on calculating the mechanical force on a circuit breaker subjected to a 200kA fault current. Participants clarify that the force exerted is related to the square of the fault current and the physical characteristics of the breaker contacts. There is debate over the significance of electromagnetic repulsion in this context, with some asserting it is minimal compared to the need for stronger springs to prevent contact fusion. The conversation also touches on the nature of high-power distribution circuits, noting that 200kA is a common fault current for such systems. Understanding these dynamics is crucial for effective circuit breaker design and operation.
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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.
 
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You could start here. http://en.wikipedia.org/wiki/Circuit_breaker"
Low voltage (less than 1000 VAC) types are common in domestic, commercial and industrial application, include:
MCB (Miniature Circuit Breaker)—rated current not more than 100 A. Trip characteristics normally not adjustable. Thermal or thermal-magnetic operation. Breakers illustrated above are in this category.

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?
 
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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
 
m.s.j said:
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

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.
 
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.
 
negitron said:
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.
Hmm. Take a look at these circuit breaker pictures. Looks sigunificant to me.
BIGEYE said:
Image links:
http://tinyurl.com/9yvsw3"
http://tinyurl.com/7ou8pa"
http://tinyurl.com/8vqxkj"
http://tinyurl.com/7k85ze"
http://tinyurl.com/7wrvjk"
http://tinyurl.com/7s4v9p"
http://tinyurl.com/7vtvhl"
http://tinyurl.com/7z3yhq"
http://tinyurl.com/7r6g6a"
http://tinyurl.com/8ofrr3"
http://tinyurl.com/84y37h"
http://tinyurl.com/8tewxl"
http://tinyurl.com/92jcwu"
http://tinyurl.com/8ysw6t"
http://tinyurl.com/9aochw"
http://tinyurl.com/7tll6k"
http://tinyurl.com/8tufz4"
http://tinyurl.com/6v3be7"
http://tinyurl.com/7a8zxt"
http://tinyurl.com/92glox"
http://tinyurl.com/9scqhc"
http://tinyurl.com/85xmv6"
https://www.physicsforums.com/showthread.php?t=284671"
 
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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.
 
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.
 
dlgoff said:
The OP mentioned 200kA fault current.

Oops! I somehow missed that. Yeah, that's a fairly significant circuit. Way out of my league.
(Maybe a small nuke to disable it?)
 
  • #10
Danger said:
Oops! I somehow missed that. Yeah, that's a fairly significant circuit. Way out of my league.
(Maybe a small nuke to disable it?)


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.
 
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  • #11
Danger said:
Oops! I somehow missed that. Yeah, that's a fairly significant circuit. Way out of my league.
(Maybe a small nuke to disable it?)

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 existence? And why in the world would the OP be asking here?
 
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  • #12
negitron said:
That's a joke, right? Electromagnetic repulsion?

excuse me, but i think when persons just know a little of fact, they say jokey small talk.:wink:
 
  • #13
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.
 
  • #14
negitron said:
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.

OK, now that makes sense. 200K is the fault current, not the trip current, as I read it.
 

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