Short circuit current for transformers in serial....

[vranac]

Hello everybody...
Let say we have two transformers (please don't bother with transformers data, they are only to get the pricipal on how to calculate):
1. 20/10kV, uk=8%, S=40MVA, YNd1
2. 10/6.6kV, uk=6%, S=20MVA, DYn11
If we would like to make stability test of 20/10kV transformer on way that we make 3ph SC on 6.6kV side, and inject 400V on 20kV side, how much current we will get on 3ph SC on 6.6kV side, and how much current we will have on injection point on 20kV side?
I get 115A on 6.6kV side, and 38,3A on injection point on 20kV side. Am I close to the correct values?

Things that I would like to know:
-how to calculate impedance of transformer in this case. Are they calulate on 20kV base as we have injection on this side.?
-does and, if does, how vector group of transformers influent on calculations?

Thank you all....

 

[waross]

Transformer impedance is not calculated.
Transformer impedance is determined by factory test at the time of manufacture.
Transformer impedance or per unit impedance indicates the voltage as a fraction of rated voltage that, when applied to the transformer primary, will cause rated current to flow in the short circuited secondary.
In North America the pu value is reported at base KVA.
Example:
20/10kV, uk=8%, S=40MVA, 40,000,000 VA / 10,000 V = 4000 Amps.
Available Short Circuit Current = 4000 Amps / .08 pu Impedance voltage = 50,000 Amps.
With 400 Volts applied to the 20,000 Volt side the current in the 10,000 Volt side will be:
20,000V / 400V = 50:1
50,000 Amps / 50 = 1000 Amps on the 10,000 Volt winding.
Back feeding may introduce slight errors due to slightly different I²R losses.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[vranac]

thx waross for your reply...
You confused me a little bit with your pu calculation, but I manage to get it.
But, If I calculate the current on 10 000V, I get 577A. Should you take in to concideration sqrt(3)?
And, one more thing, if you will find time to calculate to the end...the thing I would like to Know;
-current on 6.6kV
-current on 20kV during short circuit on 6.6kV....
Maybe you missed it in my question...I am interesting in situation when we injecting voltage/current through two transformers connected in series.

 

[rcw retired EE]

I used the MVA method to calculate a 3 phase bolted 6.6 kV fault fed through the 40MVA and 20MVA transformers:

MVAsc1 = 40MVA/.08 = 500 MVA. MVAsc2 = 20MVA/0.06 = 333 MVA. MVAsc = 1/((1/MVAsc1)+(1/MVAsc2)) = 1/ (1/500 + 1/333) = 200 MVA short circuit with 20 kV applied at the 20 kV terminals.

MVAsc = 200 MVA. Isc @ 6.6 kV = 17,495A. But since your driving voltage is 0.4kv/20kV = 0.02 of 20 kV, the 6.6 kV side's current during your test will be 17,495 x 0.02 =350 Amps.

The 20 kV side current will be 6.6kV/20kV x 350A = 115 Amps. This neglects any losses in the transformers or the impedance of your 400 volt test source.

Vector groups have no bearing on the 3-phase current magnitudes.

Remember, Power In = Power Out. 400V x 115 A x 1.732 = 80 kVA. 80 kVA on the 6.6 kV side will be = 80 kVA/ (400V x 1.732 x (6.6kV/20kV)) = 80 kVA /( 132V x 1.732) = 350A.

Your numbers don't match up since the 6.6 kV side kVA is not equal to the 20 kV side kVA.

The test will work with a 100 kVA, 400V source supplying about 115 A, 3 phase into a three phase short circuit of 350A.

 

[waross]

Yes, I should have used root 3. Most of my impedance calculations have been for single phase transformers.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[vranac]

thx for detail calculation. Now it's more clear to me.
So, if we have transformes connected on this way, we can calculate currents on both sides without taking into concideration
10kV voltage level, and without thinking that we have transformers with different MVA?
And that means, on 10kV we will get 231A....

Thank you very much...This helps me a lot.

 

[vranac]

usually I calculate like this (for may example), but I had some doubts so thanky all for your help:
Xt1=(sqr(20kV)/40MVA)*(8/100)=0.8ohm
Xt2=(sqr(20kV)/20MVA)*(6/100)=1.2ohm
Xtotal=Xt1+Xt2=2ohm
-for injection of 400V on 20kV: I=400/(sqrt(3)*Xtotal)=115A
-and then recalculate this on 10kV: 20kv/10kv=2 >>>> I=2*115=230A
-and on 6.6kV: 10kV/6.6kV=1.51 >>>> I=1.51*230=348A

 

[prc]

vranac, your calculation is also perfectly correct. There is one more method of calculation ie pu system.

 

[waross]

The pu impedance is given. Your calculation converts pu impedance to an absolute impedance value in Ohms.
The impedance in ohms does not change for a given transformer.
However the pu value is given at the base KVA rating. If the transformer is re-rated to a different base KVA, the pu impedance will change to reflect the changed KVA base while the impedance in Ohms stays the same.
The impedance tests are conducted with the transformer at normal full load working temperature.
If the test is done on a cold transformer, particularly one with a relatively high temperature rise and relatively low X:R ratio, the error may be noticeable.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[prc]

Those who want to know more about the calculation method used by Wilson, refer to the IEEE paper with a catchy title "Short Circuit ABC-Learn it an hour, use it anywhere, memorize no formula" by Moon H Yuen ,IEEE Transactions on Industry Applications Vol1A -10 No2 March/April,1974 Pages 261-272. This paper also gives a comparison of the three different ways of calculation. For an in depth study, refer to the book "Short Circuit Currents"-Juergen Schlabbch, The Institution of engineering and Technology, UK,2005