Complex Imaginary Test 4 Problem 9

[K_Nova]

I've attached problem #9 from Complex Imaginary Test #4, with the solution. Can someone please explain to me why the 208V does not have a negative 30 degree phase angle when the problem is converted to single phase.

The solution simply shows 120V at 0 degree. Thanks, any help would be great.

 

[apoollo13]

I just took that fourth CI exam, and was wondering the same thing. BUMP.

 

[Wael]

it has been included in the PF angle, since theta = phase Voltage angle - I angle.

 

[DK PE]

Maybe think of it this way. They chose the line-neutral voltage of 120 at 0 and that set the current relative that at -31 degrees as the power factor of the load is voltage across it to current through it. So the current has to lag the voltage which is at 0 degrees by 31. Now just think of it as a phasor diagram. The internal voltage has to be the 120 at 0 + the drop across the stator resistance and syn reactance. Does that make sense? Note that if you figured out the magnitude correctly that is only solution.

 

[apoollo13]

http://www.uotechnology.edu.iq/dep-eee/lectures/3rd/Electrical/Machines%202/II_SG.pdf

There's an example on page 10 that has a problem/solution exactly like the solution given by complex imaginary.

I think I get it. Since you're given the current on a Wye connection, the line current = phase current. So for example, the impedance on leg AN of a Wye connection is found by the phase voltage divided by the phase current, with the resulting phasor angle of θv- θi. Since we're given the power factor of a Wye connection, we already know the difference between the phase voltage and the phase current and thus we do not need to change it.

Does that sound right? How would we hangle a Delta connected system with a given power factor? I assume we would then only divide the line current by root 3 and not change the power factor angle. Is this correct?

 

[Wael]

For me, when dealing with impedance, this tells me to deal with (phase values instead of line values)

. Then, when I am given the power factor, regardless of the configuration (delta or wye), it takes care of the angle difference of phase angle values (θv- θi) which is the same angle of impedance.

so to answer your question above, yes i would do the same

 

[K_Nova]

Thank you for the answers, that helped a lot.

 

[apoollo13]

Also, I think it has to do with the fact that you're setting the reference voltage as the internal phase voltage of the generator so the rest of the values must have phase values (or adjusted accordingly by shifting the phase).

If you look at question 111 in the NCEES you can see a problem where you're asked to find Vab (which is the reference voltage). You treat the delta like a "black box" and make a phase to neutral connection and shift the voltage phasor by 30 degrees. You add that to the line current times the impedance. That value is then converted BACK to a line to line value by multiplying by root 3 (it doesn't show it, but I'm pretty sure it would re-adjust the phase angle too).

If you have a copy of the Kaplan PE exam, the first question is another good refresher on this topic.

Thanks to you guys too... this would have been a very unfortunate mistake to make on the exam.

 

[apoollo13]

From an instructor:

In #111, a reference voltage was defined. In your problem it is not. The problem is incompletely stated, so phasor answer really is inappropriate.