# New NCEES Power Sample Exam 517 and 522



## swooda2 (Mar 18, 2009)

In problem 517, the synchronous motor is said to have a leading pf of .7, and the solution shows the pf angle to be negative. Then for the the induction motor, which is said to have a lagging pf, they show the angle of this machine to be positive. This actually agrees with the EE Reference Manual (from PPI), page 27-13.

But in problem 522, a synchronous machine is said to have a lagging pf (current lagging voltage), and the solution shows this as a negative angle, which is the opposite of what was done in 517.

What am I missing here?

thanks.


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## niurou (Mar 19, 2009)

swooda2 said:


> In problem 517, the synchronous motor is said to have a leading pf of .7, and the solution shows the pf angle to be negative. Then for the the induction motor, which is said to have a lagging pf, they show the angle of this machine to be positive. This actually agrees with the EE Reference Manual (from PPI), page 27-13.
> But in problem 522, a synchronous machine is said to have a lagging pf (current lagging voltage), and the solution shows this as a negative angle, which is the opposite of what was done in 517.
> 
> What am I missing here?
> ...



if it is lagging PF, and voltage is at 0 degree, current should be at negative angle and vise-versa.

but for apparent power, S=I*V ,I* is the complex conjugate of current. so if I is negative angle, I*is at positive angle, so S is at positve angle.


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## jcreit (Mar 29, 2009)

niurou said:


> if it is lagging PF, and voltage is at 0 degree, current should be at negative angle and vise-versa.but for apparent power, S=I*V ,I* is the complex conjugate of current. so if I is negative angle, I*is at positive angle, so S is at positve angle.


In the solution why do we have to deduct the angle (-18.19) first by 90 degree.

Thanks


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## niurou (Mar 29, 2009)

jcreit said:


> In the solution why do we have to deduct the angle (-18.19) first by 90 degree.
> Thanks



Z=R + jX

X is the reactance.

Z is the impedance

R is the resistance

for motor:

V= E + ZI = E + jXI (R is 0 or negligible)

since

j= 1 at 90 degree

I= 1.05 at -18.19 degree.

j*I= 1.05 at (-18.19 + 90) degree.

jXI = XjI= 0.9 * 1.05 at (-18.19 + 90) degree.


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## jcreit (Mar 30, 2009)

Thanks a lot


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