transformer available fault current

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ericsherman37

ericsherman37

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I don't have an actual practical application for this; it was brought up at class recently and I realized I've never done these calculations before.

So, a hypothetical situation: a 50 kVA transformer, 208Y/120 V secondary, 2.8% impedance per nameplate.

How do I calculate the fault current available at the terminals? What about at a disconnect 50 feet away?
 
Here is what is throwing me off. The transformer impedance is a percent - but of what? Here is my calculation thus far:

P = I x E
50,000 kVA = I x (208V x 1.732) <-- for three phase
I = about 139 amps

Now at this point I want to divide the amps by the impedance, but I'm confused about the units. How does the impedance percentage relate to the amps? If I just ignore that and use the numerical values I get this:

139 / .028 = about 4960

If that final number is amps, it seems right, but I'm just kinda confused what the impedance percentage is all about. I would think that it is a dimensionless number (it's not labelled as ohms, just a percentage) so then it would follow that I could divide the amps by the impedance percentage and have the answer still be in amps. But if someone could elucidate this I would appreciate it.
 
The 2.8% is a per unit value based on the impedance. You've calculated the fault current that could come out of the transformer if the available fault current from the primary side was infinite (which you know is not possible due to utility impedance) called an infinite primary calculation.
 
Okay, so the calculated amperage can be divided by the impedance percentage without violating mathematical rules because the percent is based on a per unit value of impedance - it's not ohms. I think I am beginning to grasp that.

How would I calculate the available fault current from the same transformer at some point down the line? I figure I would need information on wire size, length, material, etc... how do I factor all of that in?
 
get a hold of Buss Fuse Plan Review .
They give you all the formulars and impedance factors for conductor configuration in RMC, PVC, and I think AL.

The % of Z is sometimes considered as follows

It is the % of Pri Voltage needed to cause Sec FLA with a bolted fault. at the Sec lugs.
 
ron said:
The 2.8% is a per unit value based on the impedance. You've calculated the fault current that could come out of the transformer if the available fault current from the primary side was infinite (which you know is not possible due to utility impedance) called an infinite primary calculation.
Click to expand...

I agree with you Ron except for the formula. The per unit method rule of thumb I have always used is:
I (available fault current) = KVA
%Z
I = 50
.028
I = 1786 A

The equation came out goofy but the current equals the KVA devided by the % Z as a fraction: 1786 A.

Tony
 
cpal said:
get a hold of Buss Fuse Plan Review .
They give you all the formulars and impedance factors for conductor configuration in RMC, PVC, and I think AL.
Click to expand...

So the fault current calculation to an arbitrary point in the electrical system differs with the type of raceway your wires are in?
 
tonytonon said:
I agree with you Ron except for the formula. The per unit method rule of thumb I have always used is:
I (available fault current) = KVA
%Z
I = 50
.028
I = 1786 A

The equation came out goofy but the current equals the KVA devided by the % Z as a fraction: 1786 A.

Tony
Click to expand...

Clearly I'm no expert at this but something seems weird about that. I think that if you divided your kVA units by the % number you'd wind up with an answer still in kVA, not amps. The units don't cancel.
 
tonytonon said:
I agree with you Ron except for the formula. The per unit method rule of thumb I have always used is:
I (available fault current) = KVA
%Z
I = 50
.028
I = 1786 A

The equation came out goofy but the current equals the KVA devided by the % Z as a fraction: 1786 A.

Tony
Click to expand...

The rule of thumb method does not address the reality that the utility will not provide a infinet amount of power at the primary.That being said the rule of thumb method ( that I am familure with) is a good place to start when deciding if you need to increase the rating of gear or OCP.

Assume the % of Z as a whoel or mixed number.

A 75kVA 2%Z 480/208/120 trans has a sec fla of 208A. What is the availble fault current with at the sec terminals.

Devide 100 by the 2 (as a whole number) = 50

50 is your multiplier.
Available short circuit current is 50 X 208 = 10,400A.


This probably the most current thai trans will deliver.

Don't forget these are rough calcualtions and do not consider other circuit dynaics.

The further away from the you go the less the fault current each foot of conductor has a value of resistance which will reduce the point to point fault value.

Theos calculation asre not difficult but there are several step to perform a point to int calculation. As I mentioned in a previous post Buss Fuse (that I know of) has a neat program in their Plan Review EPR Booklet.


If you have access to Delmar Pub their Commercial Wiring Text has two chapters with exercise that will provide all the information needed unless you are studing to become a engineer.
 
ericsherman37 said:
Clearly I'm no expert at this but something seems weird about that. I think that if you divided your kVA units by the % number you'd wind up with an answer still in kVA, not amps. The units don't cancel.
Click to expand...

As cpal said "It is the % of Pri Voltage needed to cause Sec FLA with a bolted fault at the Sec lugs"
So the % number is percent voltage and the units cancel.
 
Last edited:
ericsherman37 said:
Here is what is throwing me off. The transformer impedance is a percent - but of what? Here is my calculation thus far:

P = I x E
50,000 kVA = I x (208V x 1.732) <-- for three phase
I = about 139 amps

Now at this point I want to divide the amps by the impedance, but I'm confused about the units. How does the impedance percentage relate to the amps? If I just ignore that and use the numerical values I get this:

139 / .028 = about 4960

If that final number is amps, it seems right, but I'm just kinda confused what the impedance percentage is all about. I would think that it is a dimensionless number (it's not labelled as ohms, just a percentage) so then it would follow that I could divide the amps by the impedance percentage and have the answer still be in amps. But if someone could elucidate this I would appreciate it.
Click to expand...
The 0.028 figure you used (rather than the percentage) is just a number.
Dimentionless, as you say.
It thus doesn't change the units used in your calculation.
FWIW, I agree with your calculations - assuming you don't really mean 50,000 kVA

:wink:
 
ericsherman37 said:
If I just ignore that and use the numerical values I get this:

139 / .028 = about 4960
Click to expand...

Really this is 139A x(100%/2.8%) = Isc

I know it is mathmaticly the same but look at it this way for understanding what I am about to say.

ericsherman37 said:
If that final number is amps, it seems right, but I'm just kinda confused what the impedance percentage is all about.
Click to expand...

Ever notice how the %Z is stamped on the name plate and nto laser etched like the rest of the info? Thats because it is a test result. (I havent seen this done in a while and it is sunday Am so I am doing this off the top of my head.)

Here is what they do, they take your transformer at the factory, and short out the secondary, around that short (A bus bar normally) they put a CT so they can monitor current flow through the short. On the primary side they attach a variable voltage source, they slowly raise voltage and monitor current in the secondary short, when that current reached the rated full load current of the transformer (139A in your example) they stop raising primary voltage. The % of rated primary voltage that created the rated full load secondary current in the short is your %Z.

All it is really is a ratio. Now put this transformer in service, and use the % of rated primary voltage that is actually being used (usually 100%, but could vary), divide by the %Z to get your multiplier, and multiply that by your 139A.
 
zog said:
Really this is 139A x(100%/2.8%) = Isc

I know it is mathmaticly the same but look at it this way for understanding what I am about to say.



Ever notice how the %Z is stamped on the name plate and nto laser etched like the rest of the info? Thats because it is a test result. (I havent seen this done in a while and it is sunday Am so I am doing this off the top of my head.)

Here is what they do, they take your transformer at the factory, and short out the secondary, around that short (A bus bar normally) they put a CT so they can monitor current flow through the short. On the primary side they attach a variable voltage source, they slowly raise voltage and monitor current in the secondary short, when that current reached the rated full load current of the transformer (139A in your example) they stop raising primary voltage. The % of rated primary voltage that created the rated full load secondary current in the short is your %Z.

All it is really is a ratio. Now put this transformer in service, and use the % of rated primary voltage that is actually being used (usually 100%, but could vary), divide by the %Z to get your multiplier, and multiply that by your 139A.
Click to expand...

Zog,
That is how I remember the test was performed. Thanks for the refresher course!;)

Tony
 
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