Short Circuit Calculation; Primary Current for Sec. Fault?

[RWKeng]

I am trying to understand how to calculate the fault current that will flow on the primary of a transformer when the fault is on the secondary.

An example would be as follows.

When a transformer is rated, 1000 kVA, 5.75%Z, 13.2 kV-208/120V.

If there was a fault of approx 9000A on the secondary is there a simple method to calculate the amout of current flowing on the primary.

 

[davidbeach]

Fault current on the primary will have the same relationship to fault current on the secondary as load current on the primary does to load current on the secondary. Both are the inverse of the voltage ratio.

 

[RWKeng]

David,

Thank you for the reply. So do I have this right. The impedance of the transformer is irrelevant and if the fault current of 9000A is on the secondary of the transformer the current flowing at the same time as the fault on the primary would be 9000 divided by the turns ratio of 13200/208 or 63 resulting in approx 142Amps.

Is this correct?

 

[dpc]

The transformer impedance is not relevant in converting secondary current to equivalent primary current.

However it is the transformer impedance that largely determines what the secondary fault current will be.

So your calculation is correct.

 

[tinfoil]

You didn't describe your transformer, and the transformer connection may come into play.

For example, you may be talking about a single-line-to-ground fault on a system suppled by a delta/wye transformer.

In that case, two different primary phases are going to furnish the fault current, at 1/sqrt(3) each of the magnitude that a wye-wye connection would generate under similar circumstances..

 

[cuky2000]

See similar information in the enclose link.

http://cuky2000.250free.com/SC_Calc_Dist_Tranf_1.jpg

Paraphrasing Yogi Berra, engineering is 90% technical and the other half is administrative.

 

[tommom]

A typical, quicky calc for secondary fault current is to divide KVA/(%z*(1-.075))= fault kva. fault kva/v*1.73 = secondary fault current as though primary were infinite bus. (the 1-.075 takes into account mfr tolerances). As long as the motor contribution is not significant, this should be a good estimate. As a quickie estimate, I add up all motor currents (anything over say 1Hp), multiply x 4 and add to fault current from transformer. If you are paralleling with a generator, you need to add its fault contribution as well.

Then, I sec fault*V secondary/V primary will give you the current flowing on the primary.

However, if the secondary fault is line to ground on a delta-wye transformer, that fault current will appear as circulating current on the delta primary side.

 

[davidbeach]

I don't think the OP was dealing with a generator, but since it was mentioned, paralleling a generator and a transformer fed by the grid produces a system with higher fault currents than the sum of the two alone. The 3-phase fault level is not significantly higher than the sum, but the SLG fault is significantly higher. Need to watch that one or it can get you in trouble.

 

[dpc]

David,

I guess I'm not following you. That seems to violate superposition. Can you give an example.

 

[davidbeach]

Having been bit by it, I know it's there. This deals with a solidly grounded 480V generator (2MW in the case at hand) to be paralleled with a utility service. Draw the SLG fault sequence diagram (physically or mentally). Now assign the following values: Utility Z1, Z2, Z0 - medium with Z0 a medium high (grounded-wye / grounded-wye service transformer) or medium low (delta / grounded-wye service transformer). Generator Z1, Z2 - high, Z0 - low (or even really low).

Utility alone fault current flows through three medium (+/-)impedances and produces a medium current. Generator alone the fault current flows through two high impedances and one low impedance and produces a low fault current. So far, so good. Now, put the two in parallel and what you have is Z1 and Z2 are the parallel combination of medium and high, or something less than medium, but Z0 is now the parallel combination of medium and low, or something even lower than low, so the fault current is quite high.

Thinking about the diagrams now, I guess the 3-phase would be the sum of the two, but the SLG is most definitely not. When you have a utility SLG value of about 30kA and a generator SLG in the high 20kA range and you put them in parallel and the analysis software spits out upward of 80kA fault current (I don't have the exact numbers at hand from that project but the general order of magnitude is right) it really grabs your attention. That was the only time I found the need to do a hand calc to verify what SKM was telling me, and by gum they were right and my 65kA gear was in trouble.

There are advantages to working with the utility on projects that they want to make happen at a reasonable cost (read utility funded) - I got out of that bind by getting them to provide real transformer impedances and real system impedances ahead of the service and that knocked my fault currents way down and 65kA was great plenty.

Gotta watch those generator Z0 values for solidly grounded machines; Z0 is basically just the DC resistance of the windings and that is very low.

 

[dpc]

David,

(I guess we've hijacked this thread.)

OK, I understand you now - yes, I've run into this myself for SLG faults.

regards,

dpc

 

[jghrist]

David,

The higher fault currents occur in the generators that are the subject of thread238-175448. Total Ø-grd fault current with both the utility source (with grd-wye secondary) and the generator connected is 80 kA on the 480 volt generator bus. Ø-grd fault current with the utility source alone would be 38 kA. Ø-grd fault current with the generator alone would be 30 kA. 30+38=80. Current through the generator is 43 kA with the utility source connected.

 

[davidbeach]

jghrist, yep those are the types of numbers I saw; a bit higher but the sum of the two together much higher than the sum of the two separately.

 

[jbond]

Hey, TinFoils post above seems to say:

For example, you may be talking about a single-line-to-ground fault on a system suppled by a delta/wye transformer.
In that case, two different primary phases are going to furnish the fault current, at 1/sqrt(3) each of the magnitude that a wye-wye connection would generate under similar circumstances..

Whereas Tommom's post says:

However, if the secondary fault is line to ground on a delta-wye transformer, that fault current will appear as circulating current on the delta primary side.

Are these 2 posts contradicting? Tinfoil seems to say that there will be an influence on the input to the primary side of the Tx, whereas Tommom seems to say that the influence is contained within the primary side and the external phase connections will see no difference?

 

[davidbeach]

You'll see fault current on the phase conductors on the primary side of the transformer.

 

[00123456]

Hi,
Basic calculation is;
I have assumed; it's 3-ph symmetrical fault current = 9000A
Primary voltage = 13.2kV
Secondary voltage = 208V

Through fault current refelcted on Primary side = (208*9000) / (13.2*1000) = 141.81 Ampere

Is this three phase TR or single phase? If it is three phase then what is the TR configurations (D-Y,YY,YD etc.)?
9000A is the three phase?

Thanks