voltage drop-affects?

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jflynn

Senior Member
Would voltage drop affect a breaker from opening during a short circuit or fault condition?
The breaker in question is a 1P- 277V- 20AMP GE TEY,thanks....
 

Mayimbe

Senior Member
Location
Horsham, UK
Would voltage drop affect a breaker from opening during a short circuit or fault condition?
At short circuit, the voltage at the breaker is near 0, thats the reason of a "short" circuit.

So voltage drop doesnt seem to affect, since voltage is "0".

In fault condition you have to be more specific, there are several types of faults conditions. phase to phase, phase to ground, two phase to ground, open phase, and so on.
 

bob

Senior Member
Location
Alabama
Voltage drop is related to the circuit impedance. It is possible that the fault impedance is such that is limits the fault current to a level that is not high enough to cause the breaker to trip.
 

hurk27

Senior Member
Resistance/impedance is current limiting, so yes if it doesn't allow enough current to flow in a fault it can keep a breaker from opening, but with out the specifics of the circuit, this can not be chiseled in stone.

This is because even with a voltage drop, a bolted fault can still produce very high currents in a circuit to open a breaker, one of the reasons smaller conductors for EGC's are allowed above a #8 wire size.
 

Charlie Bob

Senior Member
Location
West Tennessee
tell me if i'm wrong but i guess we're talking about a relatively small amount of voltage drop, ( 3% to 10%, and tha;s a high value as it is), so if you have a low impedance path for the fault current to travel ,as a properly sized EGC, the ground fault current should be way high enough to open the OCP.
The same with a short circuit, in a grounded system.
 

macmikeman

Senior Member
I keep hearing on various forums how you can short out the end of a 1,000 ft spool of 14-2 and not get a 15 A breaker to trip. I never attempted this spetacular feat of tempting the transformer myself. I just wonder if there is anything to it.
 

SG-1

Senior Member
Would voltage drop affect a breaker from opening during a short circuit or fault condition?
The breaker in question is a 1P- 277V- 20AMP GE TEY,thanks....

This breaker is designed with a long time delay thermal trip.

Here is a link to the Time Current Curve http://www.geindustrial.com/publibrary/checkout/Time Current Curves|GES6237B|generic

If you have an idea of the fault current you can determine the trip time.
For 200A of fault it will open in 2 seconds.
For 40A of fault current it will take 50 seconds.
For 30A of fault current 350 seconds.
For 28A of fault current 1000 seconds.
All times are worse case.

It is really not the volt drop, but the impedance of the circuit limiting the amount of fault current.

Or, bad breaker ?
 

iwire

Moderator
Staff member
Location
Massachusetts
we're talking about a relatively small amount of voltage drop, ( 3% to 10%, and thats a high value as it is)

Your talking about the normal running load voltage drop.

Under short circuit conditions the voltage drop will increase dramatically as the load increases dramatically. But ultimately it is about impedance.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
091002-0617 EST

macmikeman:

The resistance of #14 cooper at 68 deg F. is 2.525 ohms. Double this and the loop resistance is 5.05 ohms. Assume 120 V to the 2000 ft loop and the current is 23.8 A. Now answer your question.

.
 

Mayimbe

Senior Member
Location
Horsham, UK
Your talking about the normal running load voltage drop.

Under short circuit conditions the voltage drop will increase dramatically as the load increases dramatically. But ultimately it is about impedance.

???:confused:

How does that happen? How does the load increase??? If it is under short circuit.

By load you mean current??

The amount of short circuit current is inverse proportional to the impedance, agree. But Voltage Drop?? Why??
 

jflynn

Senior Member
[
In fault condition you have to be more specific, there are several types of faults conditions. phase to phase, phase to ground, two phase to ground, open phase, and so on.[/QUOTE]



The condition from the information I received (I was not directly involved) is a- phase to ground condition.
 

jflynn

Senior Member
Resistance/impedance is current limiting, so yes if it doesn't allow enough current to flow in a fault it can keep a breaker from opening, but with out the specifics of the circuit, this can not be chiseled in stone.


Circuit wiring consists of:
Est length-240'-275'
#12 THHN CU-in the form of 12/2-MC cable.
277v load-Est to be 4-6amps of flourescent lighting.
 

ELA

Senior Member
Occupation
Electrical Test Engineer
This is an interesting thing to think about.

In many breakers it does take some small level of voltage to power the instantaneous trip coil. So it seems like- in theory -that a short circuit with low enough impedance between it and the breaker could limit the available voltage to such a low value that there is not enough voltage present to trip the instantaneous coil?

Of course the thermal overload, which does not require nearly as much voltage (more current dependent), would eventually trip the breaker provided the short circuit current was above its trip levels.
 
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Mayimbe

Senior Member
Location
Horsham, UK
The condition from the information I received (I was not directly involved) is a- phase to ground condition.

If it is a phase to ground, then as I said before, the voltage would be 0 V. Thats asuming that it was a solid to ground fault, a fault with very low impedance. If it was a fault with some impedance, then the voltage phase to ground at fault point would be a few volts.

In the not faulted phases the current will be zero, because the current found a new easy-to-flow path and it will tend to flow on the faulted phase.

So, my final answer is NO, the voltage drop had nothing to do with the breaker functions. Since the breaker doesnt care about the voltage, it cares about the amount of current flowing through it.

If you were a breaker would you care about the voltage drop??;)
 
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macmikeman

Senior Member
091002-0617 EST

macmikeman:

The resistance of #14 cooper at 68 deg F. is 2.525 ohms. Double this and the loop resistance is 5.05 ohms. Assume 120 V to the 2000 ft loop and the current is 23.8 A. Now answer your question.

.

Thanks Gar, that does the trick. Like I said, I never attempted to try it out myself to see, I considered the very idea of it pretty stupid actually. When you think about it, 23.8 amps on a 15 would of course trip the breaker, but such a load is not all that uncommon after hacks get their hands on a lighting circuit......
 

hurk27

Senior Member
Resistance/impedance is current limiting, so yes if it doesn't allow enough current to flow in a fault it can keep a breaker from opening, but with out the specifics of the circuit, this can not be chiseled in stone.


Circuit wiring consists of:
Est length-240'-275'
#12 THHN CU-in the form of 12/2-MC cable.
277v load-Est to be 4-6amps of flourescent lighting.

At 275' your circuit would produce a bolted fault current of 2505.6 amps, way more than enough to open that breaker in less time it would take to damage the conductors.
At 240' it would produce a 2871 amp bolted fault.


This is an interesting thing to think about.

In many breakers it does take some small level of voltage to power the instantaneous trip coil. So it seems like- in theory -that a short circuit with low enough impedance between it and the breaker could limit the available voltage to such a low value that there is not enough voltage present to trip the instantaneous coil?

Of course the thermal overload, which does not require nearly as much voltage (more current dependent), would eventually trip the breaker provided the short circuit current was above its trip levels.


The voltage is only at zero at the point of the fault, not at the breaker, the resistance of the conductors, between the breaker and the point of fault will only lower the voltage to the point of the voltage drop of the applied fault current, it will be the voltage drop of the conductors ahead of this breaker, that will determine the voltage at the breaker.
 
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ELA

Senior Member
Occupation
Electrical Test Engineer
The voltage is only at zero at the point of the fault, not at the breaker, the resistance of the conductors, between the breaker and the point of fault will only lower the voltage to the point of the voltage drop of the applied fault current, it will be the voltage drop of the conductors ahead of this breaker, that will determine the voltage at the breaker.

As I said in theory;
and I also stated low impedance between breaker and fault (like directly at breaker output), thus volts at breaker are
near zero.
Resistance upstream is higher than <near zero ohm fault> and drops nearly all the voltage.
 

SG-1

Senior Member
Thanks Gar, that does the trick. Like I said, I never attempted to try it out myself to see, I considered the very idea of it pretty stupid actually. When you think about it, 23.8 amps on a 15 would of course trip the breaker, but such a load is not all that uncommon after hacks get their hands on a lighting circuit......

It will trip the breaker in approximately 70 to 200 seconds if the temperature of the wire is kept constant while the breaker is timing. I am using the curve for a 15A GE TEY type breaker that is mentioned above. A more common breaker would be faster. Even if it were twice as fast many would terminate the experiment before the breaker had time to trip and thus reach a "faulty" conclusion.
 
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