Calculate Short Circuit Current

[vdubEE]

Another question revolving a Complex Imaginary problem. The problem statement is:

The problem I have is that they add the PU impedance of the transformer to the line (including the j) instead of just adding the two PU values together. I have solved it two other ways and got the same answer but wanted to make sure I was not wrong on just adding the two numbers and ignoring the j.

Again, any input is appreciated!

 

[Flyer_PE]

You're not really ignoring the j. The unwritten assumption is that the transformer impedance is nearly all reactive (Very high X/R value) and that the angles are both close enough to 90o that the resistance component of the transformer impedance will not materially affect the answer.

I would have the following critique of the problem though: They really should tell you the base MVA for the line impedance. it's probably safe to assume it's on a 20 MVA base due to the transformer but if you're going to use pu values, the base values need to be spelled out prominently somewhere.

 

[vdubEE]

Flyer,

So their solution is incorrect and the two solutions I came up with are correct? I understand what you said about the j and that is why I solved it like I did. Just wanted to make sure I was not missing a concept somewhere. Thanks!

I agree about the problem statement but it seems CI has quite a bit of problems with this.

 

[Insaf]

Another question revolving a Complex Imaginary problem. The problem statement is:

The problem I have is that they add the PU impedance of the transformer to the line (including the j) instead of just adding the two PU values together. I have solved it two other ways and got the same answer but wanted to make sure I was not wrong on just adding the two numbers and ignoring the j.

Again, any input is appreciated!

---------------------------------------

Base MVA =20

Base kV (low side)=22

Base Current= 20 * 10^3 /(sqrt(3)*22)=524.86 A

Now I(pu)= 1/(0.06+0.072)=7.58 pu

I(fault)= 7.58*524.86=3976 A.

Simply calculate the PU current and multiply by base current.

Thanks,