PPI Practice Problem (Chapter 36 Problem 6)

[sam314159]

This problem is out of the "Distribution" practice set.

I can follow along with the solution and I get the same answers but when I try to solve the problem using a formula I got out of Gross's "Power System Analysis" book, I get a different answer. It works for the P part but I get 100 MVAR for Q.

The forumla is on page 230: (Eq 6.22d)

S = [(V*Ef)/Xd]*sin(delta) + j[SIZE=12pt][[/SIZE][(V*Ef)/(Xd)]*cos(delta) - (V^2)/Xd[SIZE=12pt]][/SIZE]

P = [(V*Ef)/Xd]*sin(delta)

= [(0.98*1.03)/0.08].sin(7.5)

= 1.647 pu = 329.38 MW

(This part seems to work)

Q = [(V*Ef)/(Xd)]*cos(delta) - (V^2)/Xd

= [(0.98*1.03)/(0.08)]*cos(7.5) - (0.98^2)/0.08

= 12.51 - 12.005 = 0.505 pu = 101 MVAR

I looked at Gross's derivation of Eq (6.22d) and it looks like he is using he same principle of the PPI solution so I am not sure what I can't get 150 MVAR.

Any idea what I am doing wrong?

 

[sam314159]

I just figured out what I was doing wrong.

The formula out of Gross's book was calculating VARs into Substation 2, 100 MVAR, and the problem was asking to calculate VARs out of Substation 1 which means I have to account for VARs dissipated in the transmission line which are approximately 49.8 MVARs (I^2*X).

So VARs out of Substation 1 = 100 MVAR + 49.8 MVAR = 149.8 MVAR which matches the answer in the book.