NCEES prob 522

Where is the voltage of 90 degrees we are subtracting from? I'm remembering why the other 2 times I studied my ass off and exam got canceled I got frustrated at the white book. I have worked 50 of these problems and never seen a random 90 degree subtraction coming in. Just like the voltage drop problems, zero issues till you look at the NCEES way and whatever way they want you to do things.  I worked this problem like solution, except my current angle I had as lagging 18.19, not 90-18.19. 

The extra 90 is to take into account for the j in 0.9j which is the synchronous reactance.

Yeah, this one is a little tricky.  The term reactance, gives you the "j", which gives you the 90 degrees. 

Motor:  Power flow from terminal to internal voltage.  

V_internal = V_terminal - ( 1.05 angle -18.19)(0.9j)

justin-hawaii said:

Yeah, this one is a little tricky.  The term reactance, gives you the "j", which gives you the 90 degrees. 

Motor:  Power flow from terminal to internal voltage.  

V_internal = V_terminal - ( 1.05 angle -18.19)(0.9j)

Click to expand...

Thank you, for my calculations I use the J instead of converting to the 90 degrees. Thank you!

Basically what the above two said.

In synchronous machines (synch motors and synch gens), the synchronous reactance is typically given as Xs, which is an inductive reactance (think jX). To properly represent it as an impedance phasor: jXs

There is also armature resistance Ra, which is the real component of the overall synchronous impedance Zs = Ra + jXs. Ra is usually small and negligible compared to synchronous reactance Xs.

So the overall synchronous impedance in synchronous machines is:

Zs = Ra + jXs

Ra is usually much smaller compared to Xs, so Zs = jXs usually.