Finding Iphase from Sequence Current Problem?

I was working this problem from a PE example exam and I was  stuck on how the solution to the problem for find the phase current was formulated. I understand the vector property but how was this equation formed to find the phase current has me stuck. thank you for any help on this I have  attached the  problem and the solution .

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I follow everything except why he multiplies the impedances times two

a4u2fear said:

I follow everything except why he multiplies the impedances times two

Click to expand...

I have not looked at the question in depth, but looking at that particular line equation it is because the PU impedance values are presumably the same for Negative and Positive Sequence circuits. 

So the (Postie Sequence Gen + Positive Sequence Trans ) x 2  =  (Pos + Neg) + (Pos + Neg)

The question is asking for a double line fault. Double line faults are calculated with positive and negative sequence impedances. The question states that Z1=Z2 for all components (generator and transformer).

There is no need to convert the impedances to a common base since the generator is 1.0 MVA, 13.8 kV and the transformer is 1.0 MVA, 13.8 kV (on the high side). However the fault occurs on the low side of the XFMR which means we will need a base current.

So, the total positive impedances would be (0.15+0.08) = 0.23 pu.

Since Z1=Z2 (as stated in the problem) then the total negative impedances would be 0.23 pu.

We are working in per-unit so the full voltage is 1.0 pu.

The current at Bus 2 would be (1.0 pu - votlage at full load)/(Z1+Z2) - Total impedance for positive and negative sequences.

Therefore, the totalcurrent (in per-unit) would be (1/(0.23+0.23)) = 2.17 pu.

To convert the per unit value to the actual value then we need the base. Remember in per unit - actual value/base value. So, actual value = per unit value x base value.

The base value on the low side of the transformer is the transformer's full load current:

1.0 MVA/(1000kV * root(3)) = 577 A.

577A * 2.17 pu = 1,252 A (for a single line) which is I1.

I1 = (1/3) IA * (1-a) where (1-a) is the vector property root (3) < -30 degrees.

So, Ia = 1252A * (3/root (3) < -30 degrees)) = 2,168A which is the short circuit current.