Per Unit Question 1

[fitzg7]

I'm working in the field right now, and graduated from a program that didn't dictate that I took a power class, which is where my job is. Needless to say that I plan on taking it in the fall. However, I'm working on a little project that dictates that I know, so I taught myself, but I'm having problem figuring the xfrmr per unit impedance. It's a 225kVA 12.5pri/208sec, with a 1.8%Z. I'm using 100MVA for a p base, and 208 for a V base, but I keep coming up with numbers that I know don't jive. I can't find a good resource on the net for calculating xfrmr impedance based on %Z, and I'm wondering if anyone has some input, or could tell me how to do this problem. I figured out the rest of the impedances of the circuit that I am working with, and need to add the transformer to that value in order to have an accurate indication as to what the short circuit current is going to be at the xfrmr versus at the load. Any help would be great. Thanks

 

[Skogsgurra]

You did the only mistake when you decided to use the non-existing 100 MVA as a base. The Z is referred to the actual transformer rating. And, if you are as smart as I think you are, you will use a single-phase circuit for the calculations. That will give you 225/3 = 75 kVA as the P base.

Then proceed as usual. Find rated secondary current, find voltage drop at full load (Z times secondary phase-neutral voltage). Divide voltage drop with current to get impedance in ohms. You will not get real and imagery parts. Only the absolute ohms value.

Gunnar Englund

http://www.gke.org

 

[fitzg7]

I'm using 100 MVA, becuase that is the utility standard, so the xfrmr would be .00225 p.u. My boss asked me to use 100MVA, noting the standard, and the fact that it would help me practice per unit conversion. So what would I do in that case. Thanks.

 

[dpc]

On transformer base (225 kVA), per unit Z is 0.018.

To convert to new base (100 MVA):

Zpu_new = Zpu_old [Vbase_old/Vbase_new]^^2 * [VAbase_new/VAbase_old]

Since your voltages are the same, you just multiply by the new base and divide by the old base, making sure to get your units right.

If you convert to ohms and then to per unit on the new base, you should get the same answer.

For the same transformer, the per unit impedance increases as the MVA base increases. Of course, the actual transformer doesn't know about per unit, so the actual ohmic impedance doesn't change.

So you would have 100,000/225 * 0.018.

xnet.rrc.mb.ca/janaj/E602%20Power%20Systems/Introduction%20to%20Per%20Unit%20Calculations.doc

 

[powerjunx]

dpc, suggested a case of 3 phase fault having infinite source. In this case, if the source of the subsytem is too large that is an infinite bus and cable impedance are negligible. The fault current is dictated by the size and the impedance of the transformer.

"So you would have 100,000/225 * 0.018."

Z_tx =((%IZ/100)*(MVA_base/MVA_tx)),
= ((1.8%/100)*(100MVA/225KVA)
= 100,000/225 * 0.018.

So 3 Phase fault current is;

:= (MVA_base*FLA_tx)/Z_tx.



To fitzg7,
"I'm using 100 MVA, becuase that is the utility standard, so the xfrmr would be .00225 p.u."

Considering the standard having utility available short circuit MVA (SCMVA),e.i.100 MVA. I think, fitzg, you include the utility in lieu of infinite bus assumption.
So your per unit impedance would be;

Z_utility = MVA_base/SCMVA,

Z_tx = ((%IZ/100)*(MVA_base/MVA_tx)), tx:transformer

Z_total = Z_utility + Z_tx

Then, your 3 Phase Fault current is;

:= (MVA_base*FLA_tx)/Z_total,

FLA_tx:transformer full load current


I bet your boss exactly knew the convenience and advantages of per unit calculation.