A couple fault current questions

[Ingenieur]

37kA @ 12.47kV sounds high unless you're near large generation or a large substation.

I also have never used PF as part of my fault calcs. Impedance Z% is the total of R%+jX% not P and Q.

sounds high 550 mva of fault

pf can usually be ignored
only influences offset from symmetrical
pf = cos(arctan x/r)
and it is related to S = P + jQ
in phasors v/z = i
so Z determines the i ph shift which is related to the pf by
S = v (i*)
in other words the reactive load X (of Z) determines pf arctan(Q/P)

only comes into play with transient and subtransient
symmetrical isvthe same other than ph angle

 

[Bugman1400]

So, you're saying that the X/R ratio determines the power factor of a circuit and a lower power factor percentage indicates a higher short circuit X/R ratio.....right?

 

[Ingenieur]

So, you're saying that the X/R ratio determines the power factor of a circuit and a lower power factor percentage indicates a higher short circuit X/R ratio.....right?

Basically yes
the ph ang or arctan(x/r) detrrmines the lead/lag ph ang of the current
Which determines the pf
ie cos of that ang is pf or cos(arctan x/r)

there are 3 stages of a fault
initial 1-2 cycle transient
next few Hz subtransient
steady state or symmetrical >4-6 cycles
roughly

what is normally calculated is the symmetrical
for 2 reasons
1 the peaking is usually small maybe 10-20% unless x/r is very high, ie pf low
2 most breakers don't operate until steady state 4 cycles or so

normally not factored in unless hv and very fast cb's and high x/r's

 

[Ingenieur]

Sub-transient-symmetrical
mixed them up above

The rate of decay of subtransient-transient-symmetrical is an natural exponential function
~ e ^(-t / x/r) or something like that
don't feel like looking it up lol

the higher the x/r the longer it takes to decay and the higher the current will be at 4-6 cycles (or whatever the system time constants determine) when the cb opens
ie higher fault current to be interrrupted

 

[Ingenieur]

Picture = 1000 words and a few equations ll

the durations vary whether on a large generator/transmission system vs a sc on a small (relative to supply) secondary load

 

[mivey]

And the only explanation how 37 kA would exist is that it is actually the available short-circuit amperes on the 12.47 kV side.

That is not what is happening here. Re-read the OP's posts. You made a turn a while back and need to get back on the paved road.

 

[Bugman1400]

As an exercise, assuming the 37kA fault is for a 3PH fault on the 480V side, the primary voltage is 12.47kV, what is the source impedance (Zs) and the 3PH fault current at the highside of the xfmr?

 

[topgone]

As an exercise, assuming the 37kA fault is for a 3PH fault on the 480V side, the primary voltage is 12.47kV, what is the source impedance (Zs) and the 3PH fault current at the highside of the xfmr?

Good points!
To deliver 37kA at 480 volts, the system impedance needs to be a negative ohms.:happysad:
This is gonna be quite boring for others:

Total impedance when delivering 37 kA = (480/1.732)/37,000 = 0.00749 ohms
Impedance of transformer (500 kVA) = 0.052 x (0.48^2)/0.5MVA = 0.0239616 ohms
And the system impedance must be = 0.00749 - 0.0239616 = -0.01647 ohms!​

 

[Ingenieur]

As an exercise, assuming the 37kA fault is for a 3PH fault on the 480V side, the primary voltage is 12.47kV, what is the source impedance (Zs) and the 3PH fault current at the highside of the xfmr?

assume
V 480
Ifs 37000 3 ph
sec Z = 480/(sqrt 3 x 37000) = 0.0075 ohm

reflected to primary (12460/480)^2 x 0.0075 = 5.05 ohm
Ifp ~ 12460/(sqrt 3 x 5.05) = 1425 A

 

[Ingenieur]

Good points!
To deliver 37kA at 480 volts, the system impedance needs to be a negative ohms.:happysad:
This is gonna be quite boring for others:

Total impedance when delivering 37 kA = (480/1.732)/37,000 = 0.00749 ohms
Impedance of transformer (500 kVA) = 0.052 x (0.48^2)/0.5MVA = 0.0239616 ohms
And the system impedance must be = 0.00749 - 0.0239616 = -0.01647 ohms!​

the 37000 is not based on the actual pu Z of 5.2%
that ohmic value is 0.0075 as you noted
as noted the Z is based on something else not the actual pu Z

the fault for the 5.2
Z = 0.052 x 480^2 / 500000 = 0.024 ohm
Ifs = 480/(sqrt 3 x 0.024) = 11574 as previously noted

system Z in this case will be 0 since only the xfmr is considered

 

[Ingenieur]

Which passes the sanity check
800 mva at the primary of a 0.5 mva xfmr
or
24 mva
???

looks like the 37000 was based on a pu Z of 1.63%

 

[Iron_Ben]

As an exercise, assuming the 37kA fault is for a 3PH fault on the 480V side, the primary voltage is 12.47kV, what is the source impedance (Zs) and the 3PH fault current at the highside of the xfmr?

The OP presented a 500 kva padmount in his initial post. In post #6 he showed us the nameplate, which indicates an impedance of 5.2%. My quick and dirty infinite bus calculation yields 11,500 amps at the low side terminals. This 37 ka number has never fit. It *can't* be the low side fault current, given 500 kva at 480Y/277 and 5.2%Z. And 37 ka on the high side is just about unthinkably high at 12.47 kv. That's 800 MVA or so. Our standard config at that distribution voltage was either a 7.5/10.5 or 10/14 MVA transformer with an impedance of 7 or 8%. 10 MVA at 7 % is 143 MVA, and that's at the substation transformer terminals, not at a customer 1/2 a mile or 10 miles away, where the line impedance will markedly reduce the available fault current. That 37 ka just doesn't jibe, given what's been presented.

 

[mivey]

I will again point out that the basis for the POCO 37 kA has not been specified. It would be unusual (except for arc flash study data) for the POCO to provide fault data for the exact transformer sitting in front of the OP.

 

[mivey]

Which passes the sanity check
800 mva at the primary of a 0.5 mva xfmr
or
24 mva
???

looks like the 37000 was based on a pu Z of 1.63%

There is no way to determine that. The fault data may even be for a 750 or 1000 kVA transformer. POCO practice probably includes some contingency analysis.

 

[Ingenieur]

There is no way to determine that. The fault data may even be for a 750 or 1000 kVA transformer. POCO practice probably includes some contingency analysis.

If at the primary 1000 ka at the secondary
might be difficult to find an of the shelf cb

pad mount 13.2-480 1.8-3.5%
http://www.madison-heights.org/depa...ment/docs/Available_Short_Circuit_Current.pdf

 

[Ingenieur]

PECO pg 3
https://www.peco.com/PartnersinBusi...Documents/04GeneralrequirementsRev5200808.pdf

500 kva at 480 wye
40 ka maximum at the 1.5%
24.06 ka at 2.5%
Z 1.5-2.5%

 

[mivey]

If at the primary 1000 ka at the secondary
might be difficult to find an of the shelf cb

1,000,000 amps? Where in the world did that come from? How far afield are we going to go here?

 

[mivey]

PECO pg 3
https://www.peco.com/PartnersinBusi...Documents/04GeneralrequirementsRev5200808.pdf

500 kva at 480 wye
40 ka maximum at the 1.5%
24.06 ka at 2.5%
Z 1.5-2.5%

My own data tops out at about 17 kA for a 750 kVA pad and 36 kA for a 750 kVA bank.

All of the data we can post might be interesting but until the OP hears from his POCO we don't know the basis for their 37 kA value.

 

[Ingenieur]

1,000,000 amps? Where in the world did that come from? How far afield are we going to go here?

Someone postulated the utility furnished 37000 figure was at the primary
that converts to 800 mva fault
on the secondary 1 million amps give or take
unlikely

it looks like the 37000 is a std figure for a 500 kva/480 based on a few utility data sheets
that computes to a z of 1.6% which on the low end of the data sheet ranges

 

[Ingenieur]

It most likely has primary fusing,
Ip rated 23 A

I would check that
this should limit let through well below 37000