Fault current question

[Matt Mckenzie]

how do I figure out the available fault current at the secondary side of a 150kva 480/208/120v transformer with 2% impedance?

(150,000/360.256)/.02 ? Availabel fault current at secondary = 20,818amps.....?

 

[don_resqcapt19]

That is correct, but assumes that you have infinite current available on the primary side.

 

[Julius Right]

I agree with don_resqcapt19, of course. There are some limitations as system impedance and may be, connected cables-at 480 V and/or at 208 V.
In IEEE 141/1993 IEEE Recommended Practice for Electric Power Distribution for Industrial Plants ch.4 Short-circuit current calculation sub.4.3.2 Type of short-circuit it is stated:
"In calculating the maximum short-circuit current, it is assumed that the short-circuit connection has zero impedance (is "bolted") with no current-limiting effect due to short-circuit itself. It should be recognized, however, that actual short-circuits often involve arcing, and variable arc impedance can reduce low-voltage short-circuit current magnitude appreciably.
In low-voltage systems, the minimum values of short-circuit current are sometimes calculated from known effects of arcing. Analytical studies indicate that the sustained arcing short-circuit currents, in per unit of bolted fault values, may be typically as low as:
a) 0.89 at 480 V and 0.12 at 208 for three-phase arcing."

 

[zog]

That is correct, but assumes that you have infinite current available on the primary side.

And does not include any other sources such as motor contribution.

 

[iwire]

And does not include any other sources such as motor contribution.

The OP asked for the transformers fault current, not the electrical systems fault current.

Wouldn't you need to know the transformers fault current before you went about looking at other factors?

 

[kingpb]

And does not include any other sources such as motor contribution.

From a power engineering standpoint, motor fault contribution on a 150KVA transformer is going to be negligible in comparison to the infinite bus through fault.

I_rated = 410A for 150KVA Xfmr
Max motor (75Hp?) = 221A, with 6 x FLA = 1326A

In comparison to 20.82kA; if the numbers are being used to determine equipment rating, then the contribution from motor would not change my selection of the bus; it would need to be over 22kA regardless.

Now, if the numbers are being used for relay setting or arc flash, well that's a whole other topic.

 

[LMAO]

From a power engineering standpoint, motor fault contribution on a 150KVA transformer is going to be negligible in comparison to the infinite bus through fault.

I_rated = 410A for 150KVA Xfmr
Max motor (75Hp?) = 221A, with 6 x FLA = 1326A

In comparison to 20.82kA; if the numbers are being used to determine equipment rating, then the contribution from motor would not change my selection of the bus; it would need to be over 22kA regardless.

Now, if the numbers are being used for relay setting or arc flash, well that's a whole other topic.

Correct, motor contribution should also be considered unless motors are driven by VFD. In that case only transformer is the source of fault current.

 

[Jraef]

From a power engineering standpoint, motor fault contribution on a 150KVA transformer is going to be negligible in comparison to the infinite bus through fault.

I_rated = 410A for 150KVA Xfmr
Max motor (75Hp?) = 221A, with 6 x FLA = 1326A

In comparison to 20.82kA; if the numbers are being used to determine equipment rating, then the contribution from motor would not change my selection of the bus; it would need to be over 22kA regardless.

Now, if the numbers are being used for relay setting or arc flash, well that's a whole other topic.

I understand that using 6x FLA is done to be extra conservative, and that ANSI recommends using 4x if nothing else is known about the motor, but I've never understood why it is even anything over 1x. Motor LRC being 6x is based upon the fact that the motor was NOT turning and the rotor fields were not providing impedance against the stator windings. When factoring motor contribution to a fault, it can only be contributed by motors that are ALREADY spinning when the fault happens and only for about 4 cycles until the residual magnetic fields in the stator colapse. So how can they contribute anything higher than their own FLA? As an induction generator, that would be their capacity as well. This is one of those little details that I took at face value way back when in school, but questioned later when there was nobody to ask. Anyone here care to enlighten me now?

 

[kingpb]

I understand that using 6x FLA is done to be extra conservative, and that ANSI recommends using 4x if nothing else is known about the motor, but I've never understood why it is even anything over 1x. Motor LRC being 6x is based upon the fact that the motor was NOT turning and the rotor fields were not providing impedance against the stator windings. When factoring motor contribution to a fault, it can only be contributed by motors that are ALREADY spinning when the fault happens and only for about 4 cycles until the residual magnetic fields in the stator colapse. So how can they contribute anything higher than their own FLA? As an induction generator, that would be their capacity as well. This is one of those little details that I took at face value way back when in school, but questioned later when there was nobody to ask. Anyone here care to enlighten me now?

Would not LRC current contribution be present if the fault occurs at the point in time when the DOL contactor closes to energize the motor?

 

[petersonra]

how do I figure out the available fault current at the secondary side of a 150kva 480/208/120v transformer with 2% impedance?

(150,000/360.256)/.02 ? Availabel fault current at secondary = 20,818amps.....?

where did you come up with 360.256?

 

[petersonra]

where did you come up with 360.256?

208 * sqrt(3)

I just did not recognize that "magic" number.

 

[winnie]

I understand that using 6x FLA is done to be extra conservative, and that ANSI recommends using 4x if nothing else is known about the motor, but I've never understood why it is even anything over 1x. [...] So how can they contribute anything higher than their own FLA? As an induction generator, that would be their capacity as well. [...]

Hmm. I don't know the answer, but here is my noodling.

Consider a separately excited synchronous generator feeding a unity power factor load at its full capability, with no other sources of power in the system. What sets the short circuit current of that system?

The short circuit current is set not by the full load rating of the generator, but by its impedance. If the short circuit current were the full load rating, then the thing would be some sort of constant current source, like a bicycle alternator.

At the instant of short circuit, I think you could consider the induction machine to be acting like a synchronous generator where the excitation has just been shorted and the magnetic field is starting to decay.

I think, that in general, the short circuit current contribution of an induction motor will be different from either the locked rotor current or the full load current.

-Jon

 

[Jraef]

Hmm. I don't know the answer, but here is my noodling.

Consider a separately excited synchronous generator feeding a unity power factor load at its full capability, with no other sources of power in the system. What sets the short circuit current of that system?

The short circuit current is set not by the full load rating of the generator, but by its impedance. If the short circuit current were the full load rating, then the thing would be some sort of constant current source, like a bicycle alternator.

At the instant of short circuit, I think you could consider the induction machine to be acting like a synchronous generator where the excitation has just been shorted and the magnetic field is starting to decay.

I think, that in general, the short circuit current contribution of an induction motor will be different from either the locked rotor current or the full load current.

-Jon

D'oh! :slaphead:

The motor is an INDUCTOR, so it IS storing energy and releasing it, even as the fields decay. I have no idea why I failed to see it until you brought up that issue of how a generator is able to deliver more current than it is rated for.

That's how we learn though... Thanks!

 

[iceworm]

Just a thought:
LAMO, Zog
How many 50hp+ motors do you think will be on a 150kva, 208V transformer?
I highly suspect the point is moot.


edit to add: Slow poster - adding to winnie's response

Jaref
I think this is stuff you already know - I don't think I have much enlightment:
The contribution has to do with the motor sub-transient reactance when acting as an Induction Generator. An excellent reference is a 1967 IEEE paper Procedure for determining short circuit values in secondary electrical distribution systems by Russell Ohlson. Which could be a bit tough to find with out IEexplore. Here is a link to a synopsis:
http://www.arcadvisor.com/reference.html#RotatingMachine

And, yeah, 4 X FLA is a good number. There is a good case for ignoring <50Hp. And there is a case for lumping <50hp Ohlson likes lumping. As I recall, ANSI likes ignore

worm

 

[Bugman1400]

Also, make sure that you recognize that resistive load is far different than a bolted fault where the "load" is mostly inductive. This means that the resistive load which requires an applied mechanical force (50HP) is replaced by an inductive load which requires little to no mechanical force. This is why the power factor for a fault can approach zero (90 degrees).

 

[zog]

Just a thought:
LAMO, Zog
How many 50hp+ motors do you think will be on a 150kva, 208V transformer?
I highly suspect the point is moot.

Doh! 150kVA, my mind saw 1500.

 

[don_resqcapt19]

Doh! 150kVA, my mind saw 1500.

That is because you work on the big stuff all the time...few of us do that.