Voltage during a short circuit condition

Status
Not open for further replies.
Grouch1980

Grouch1980

Senior Member
Location
New York, NY
Hey guys,
When you have a bolted 3 phase short circuit fault, say a few hundred feet into the building (doesn't matter where), the voltage where the fault is reduces significantly. (let's assume a 120/208 volt building).

My 2 questions are:
1. Does the voltage go down to zero at the fault location, or near zero? or can it still remain high, for example 20, 30, or 50 volts?
2. What happens to the voltage at the service entrance? Does the voltage there decrease in value to match that of the fault? or does the voltage at the service entrance stay at 120/208? I would assume it drops to whatever the voltage is at the fault location.

Thanks!
 
Voltage is measured between two points. So when you say "the voltage" at a point, you need to specify, or have an implicit agreement as to, the 0V reference point. Conductors have impedance, so when large currents are flowing through them, the voltage will change appreciably along the length of the conductor. [I'm just going to treat the impedances as resistances for my response, in reality they have some reactance which makes the math more complicated, but resistances just add up as single numbers.]

(1) So let's say you have a 120V perfectly constant voltage source and you earth one of the conductors and make that earthing point your 0V reference. Then if you have a dead short between those conductors some distance away, the behavior is just governed by Ohm's Law applied to the intervening conductors.

If the conductors are equal length and size, their impedances should be the same, and so the voltage at the dead short will be 60V. You just have two identical impedances in series, so the midpoint is at half the voltage. As you travel along the conductors from the earthing point (0V reference) to the dead short and back to the source, the voltage will start at 0V and rise proportionally to 60V at the short and 120V back at the source.

If the short itself has some impedance, then you now have 3 impedances in series. The voltage on each side of the short will be different, because of the short's impedance.

(2) For this question, the assumption of a perfectly constant voltage source is unrealistic. But you can treat the voltage source as a constant voltage in series with some "source" impedance. E.g. if the available fault current at the source is 12,000A, then that would mean the source impedance is 0.01 ohms, as 12,000A = 120V / 0.01 ohms.

So again you just need to apply Ohm's Law. If the short has an impedance of 0.03 ohms, and your conductors one way to the short have an impedance of 0.10 ohms, now the whole series circuit has an impedance of 0.24 ohms (source + one way conductor + short impedance + one way conductor). The current flow will be 120V / 0.24 ohms = 500A.

As the source impedance is 0.01 ohms, the source voltage will drop to 115V. Since we've defined the 0V reference to be our earthing point, that stays at 0V, and the unearthed source conductor is at 115V. As you travel from the 0V point to the short and back again, the voltage will begin at 0V, rise linearly to 50V at one side of the short, jump to 65V at the other side of the short (15V jump = 500A * 0.03 ohm short), and then rise linearly to 115V back at the source.

Cheers, Wayne
 
Last edited:
wwhitney said:
Voltage is measured between two points. So when you say "the voltage" at a point, you need to specify, or have an implicit agreement as to, the 0V reference point.
Click to expand...
I meant the voltage 'across' the short, across the point where the conductors are bolted together.

I have to go through what you wrote... it's a lot lol.
 
Grouch1980 said:
I meant the voltage 'across' the short, across the point where the conductors are bolted together.
Click to expand...
That's going to entirely depend on how good a short it is, i.e. what its impedance is. That impedance, plus all the other impedances in the circuit, will determine the current flow, and then the voltage across the short will be the impedance times the current.

For example, if your "short" is just a 1200W heater element, then the voltage drop across it will be basically the full line voltage, or 120V in the example.

Grouch1980 said:
I have to go through what you wrote... it's a lot lol.
Click to expand...
It's basically just all Ohm's law and could use some diagrams.

Cheers, Wayne
 
Grouch1980 said:
I meant the voltage 'across' the short, across the point where the conductors are bolted together.

I have to go through what you wrote... it's a lot lol.
Click to expand...
Voltage across a short? A “short” is either between two hot lines, or between a hot line and ground. In either case the voltage of the hot conductor(s) on either side of that short does not really change (until the protective device opens the circuit). The potential between the shorted elements changes by whatever resistance there is in the creation of that short. But the line voltage feeding the hot line(s) involved in the short is virtually unchanged.

If you were thinking of a BREAK in a single hot line (which is not necessarily a “short”) to where there is an arc across the gap, then the voltage across that gap actually goes UP because at some point in an AC circuit, the gap distance acts like a capacitor until the gap between the conductors exceeds the dielectric strength of the air (or whatever is there). This by the way happens every time you open a switch under load, which is why things like circuit breakers and contactors have what are called “arc chutes” to break up the arc and lower that voltage.
 
Grouch1980 said:
For number 1, I imagine the voltage at the fault is determined by any impedance at the fault.
Click to expand...
Your use of words doesn't add up, you know? When you described the event as a "bolted" fault, the picture that comes to mind is that there is no fault impedance.
You need to be clear on that first. Because, with no impedance whatsoever, the fault would be dependent on the source impedance and the voltage during the pre-fault condition. With zero impedance, you cannot expect a "voltage" (V=I x R).
 
Grouch1980 said:
I meant the voltage 'across' the short, across the point where the conductors are bolted together.

I have to go through what you wrote... it's a lot lol.
Click to expand...

By definition, there is zero Voltage across a bolted fault short circuit.

This really is a worst-case scenario model of a short circuit, since the short in question could be a short with a finite impedance. E.g. an arc fault, as is evident by the glow of light and UV, from the energy it removes from the electrical domain.

When we calculate short circuit current, we are looking at what current the source would cause to flow, if the only impediment to its flow were the internal impedance of the source and the impedance of the network of wires and transformers that deliver it. With the infinite bus model, we assume worst-case infinite fault current on the primary side of the transformer, such that the only impedances that matter to us, are the transformer impedance, and the impedance of the secondary wiring path.
 
It's basically all determined by the voltage divider formula:


So yes, the voltage across a bolted fault would be zero, because the resistance is zero.

For your questions 1 & 2, think of the voltage divider with 3 resistances - one for the fault (if its non-zero), one for the impedance between the service and the fault, and one for the source impedance (everything up to service entrance). The voltages will be divided per the ratio of the resistances.

So the voltage at the service entrance for a bolted fault will depend on the source impedance vs. the impedance to the fault. A high source impedance will drop the voltage more. A high feeder impedance and the voltage won't drop as much.

And if you want to be more exact for an AC circuit, substitute impedances for the resistances. And we are ignoring any transient effects.
 
If you go into the excruciating details, when you push two bare wires together (at right angles for simplification) the one thing that is totally unambiguoius is the current through the "short."
However, the voltage will depend on the exact point along each of the wires that you place your two voltage probes. If you were somehow able to sense the voltage inside the conductors you would see the voltage steadily dropping as the probes approach the surface where the two conductors are mashed together.
If it is a poor (at microscopic scale) connection there will still be a voltage across the interface. If you deform the metal surfaces into intimate contact (a "bolted" fault) the voltage will approach zero as you approach the interface.
 
Thanks everyone for clearing that up. Yes, I realize I mixed up my question when first asking about a bolted fault in post #1 (no impedance), and then mentioning the impedance across that fault in #2. Didn't mean to confuse things. Thanks for all the responses.
 
Last edited:
Status
Not open for further replies.
Top