
Transformer Primary Injection Testing Calculation
Originally Posted by
slts1991
Thanks for the input. Here are some answers
Original Question: A primary injection is to be performed on a transformer 138kV/25kV, 2500kva, 8%Z using a 480V generator connected to the high side with the low side grounded. The expected current in the high side is?
7. You did not specify the xfmr connection but assuming it is Delta/Wye my calculation is 3007.12 A primary
Didn't read all the answers but this one jumped out and just slapped me as very high. I think that if no three phase connection is given one assumes a single phase transformer. If I tell you I am standing here with a 480V to 120V transformer in my hand you would assume one transformer with a turns ratio of 4 to 1 and a primary voltage of 480 volts.
Ok  so the full load current at 138,000V and 2,500,000 VA is 18.116 A. At 8%Z with a shorted secondary you would multiply that by 12.5 so at rated voltage you have 226.45 amps Isc. But you are only applying 480V so you would get, proportionately, 0.787 AMPS. Generally, if I specify that a transformer is (for example) 10%Z that means that with the secondary shorted it will take 10% of the rated primary voltage to get full load current. www.cttestset.com

Originally Posted by
Vernon
Ok  so the full load current at 138,000V and 2,500,000 VA is 18.116 A. At 8%Z with a shorted secondary you would multiply that by 12.5 so at rated voltage you have 226.45 amps Isc. But you are only applying 480V so you would get, proportionately, 0.787 AMPS. Generally, if I specify that a transformer is (for example) 10%Z that means that with the secondary shorted it will take 10% of the rated primary voltage to get full load current.
www.cttestset.com
Hi Vernon, can you explain why we multiply by 12.5 and why we don't use three phase voltage in the calculation? Thank you.

Originally Posted by
Vernon
Didn't read all the answers but this one jumped out and just slapped me as very high. I think that if no three phase connection is given one assumes a single phase transformer. If I tell you I am standing here with a 480V to 120V transformer in my hand you would assume one transformer with a turns ratio of 4 to 1 and a primary voltage of 480 volts.
Ok  so the full load current at 138,000V and 2,500,000 VA is 18.116 A. At 8%Z with a shorted secondary you would multiply that by 12.5 so at rated voltage you have 226.45 amps Isc. But you are only applying 480V so you would get, proportionately, 0.787 AMPS. Generally, if I specify that a transformer is (for example) 10%Z that means that with the secondary shorted it will take 10% of the rated primary voltage to get full load current.
www.cttestset.com
I can follow all the math except where the 12.5 number came from. Can anyone fill me in on what I missed?

Originally Posted by
Vernon
Didn't read all the answers but this one jumped out and just slapped me as very high. I think that if no three phase connection is given one assumes a single phase transformer. If I tell you I am standing here with a 480V to 120V transformer in my hand you would assume one transformer with a turns ratio of 4 to 1 and a primary voltage of 480 volts.
Ok  so the full load current at 138,000V and 2,500,000 VA is 18.116 A. At 8%Z with a shorted secondary you would multiply that by 12.5 so at rated voltage you have 226.45 amps Isc. But you are only applying 480V so you would get, proportionately, 0.787 AMPS. Generally, if I specify that a transformer is (for example) 10%Z that means that with the secondary shorted it will take 10% of the rated primary voltage to get full load current.
www.cttestset.com
I see it now...
(kVA x 100)/(kV x %Z)
Therefore,
(2,500,000VA x 100)/(138,000V x 8) = 226.449A
But at 480VAC vice 138,000VAC so,
138,000/480 = 287.5
So dividing the shortcircuit primary amps by the voltage ratio difference:
226.449A/287.5 = 0.787A

Originally Posted by
ddd675
7. A primary injection is to be performed on a transformer 138kV/25kV, 2500kva, 8%Z using a 480V generator connected to the high side with the low side grounded. The expected current in the high side is?
Assuming that the answer from the practice tests is correct, I believe I have finally satisfied myself with a reasonable solution to this problem.
 Assume the xfmr is singlephase since no other information is given.
 Using SC calculation, we find that the secondary fault current in a SC would be 1,250A.
 Knowing the secondary voltage and current, we can solve for the reactance in the secondary coil, giving 20 ohms @ 60Hz (another assumption).
 Changing gears; we now have a known secondary impedance since the secondary is grounded (20 ohms).
 With the fixed turns ratio of 5.52, we find that applying 480 to the primary gives us 86.9V on the secondary.
 Apply Ohm's Law to find secondary current over our 20 ohm impedance (4.35A).
 Again, use the TR of 5.52 to find what level of primary current will cause 4.35A in the secondary.
With all these assumptions, I come up with 0.787A of primary current.
This reference pointed me in the right direction. Until now, all the solutions I have seen on here were as clear as mud. https://www.eaton.com/ecm/groups/pub...wp009001en.pdf
Last edited by young_dad; November 25, 2019 at 07:13 PM.

Originally Posted by
roberts311
I can follow all the math except where the 12.5 number came from. Can anyone fill me in on what I missed?
It just popped into my head the other night: 100% ÷ 8% = 12.5. Vernon multiplies by 12.5 rather than dividing by 0.08.

A primary injection is to be performed on a DY transformer rated 138kV/25kV, 2500kva, 8%Z using a 480V generator connected to the high side with the low side grounded. What is the expected current on the high side?
0.262A
0.455A
0.787A
1.36A

acs69
Originally Posted by
mmacdonell
A primary injection is to be performed on a DY transformer rated 138kV/25kV, 2500kva, 8%Z using a 480V generator connected to the high side with the low side grounded. What is the expected current on the high side?
0.262A
0.455A
0.787A
1.36A
Answer: 0.455A

what is the point of primary injection of a transformer?
i understand the math, but have never seen or heard of this done. hopefully this is just an old method to something we currently do in its place for testing.

I got 1.364A on the high side.
So using a power base of 2500kVA and a voltage base of 138kv, the actual impedance seen by the high side would be
(.08)*(138kV^2)/(2500kVA) = 609.41 Ohms seen by the high side.
percent imp*(Base Voltage Squared)/(Base KVA Squared)
the equivalent circuit would be a 480V source to a three 609.41 impedances connected in a Delta.
the single phase equivalent using ohms law would be 480V = (Line Current/Square Root of 3)* 609.41 Ohm
Or
I = 480/609.41*1.732 = 1.364A
If I use a Y connection. I would get the .45476A.
If I solve for phase current instead of line current, I'd get .7876A
Did anyone else get this answer?
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