NCEES probs 134 and 135

Hi all,

Okay looking at problems 134 and 135. I was able to solve these easy enough. My question...is there a reason why they solved it two different approaches in the Solutions?

Understanding leading, lagging, power triangle. neg vs pos, I used the general few formulas...

P=(V)(I)(pf) ....in 135 P was given

Qold=P (sin Theta old)

Qnew=P(sin Theta new)

Qadd=Qnew-Qold

instead of the SqRt of the Sum of the Squares.

I know there is more than one way to skin a cat, and I understand both approaches. Just wondering if there is a time (general rule) one is better used than the other? I may still discover it, but any input as always is appreciated.

John

jdd18vm said:

Hi all,
Okay looking at problems 134 and 135. I was able to solve these easy enough. My question...is there a reason why they solved it two different approaches in the Solutions?

Understanding leading, lagging, power triangle. neg vs pos, I used the general few formulas...

P=(V)(I)(pf) ....in 135 P was given

Qold=P (sin Theta old)

Qnew=P(sin Theta new)

Qadd=Qnew-Qold

instead of the SqRt of the Sum of the Squares.

I know there is more than one way to skin a cat, and I understand both approaches. Just wondering if there is a time (general rule) one is better used than the other? I may still discover it, but any input as always is appreciated.

John

Click to expand...

I am replying here without looking at the problem so I don't have all the details but looks like you did well. As long as you understand the concept and are comfortable with the way you attcacked it you are OK. You got it right and thats the bottom line.

There are no points for style in the test.

Keep it up JD. You are going to make it!!!!! :bio:

I took a look at the problems. I think you are correct in that they are trying to demonstrate that there are multiple ways to solve this type of problem. For me, choosing which method to use is decided by which one would get me to the correct answer in the shortest time. I tend to favor using S^2=P^2+Q^2 simply because it's faster for me and I'm less likely to fat finger a trig function on the calculator.

Good luck on the test.

Jim

jdd18vm said:

Hi all,
Okay looking at problems 134 and 135. I was able to solve these easy enough. My question...is there a reason why they solved it two different approaches in the Solutions?

Understanding leading, lagging, power triangle. neg vs pos, I used the general few formulas...

P=(V)(I)(pf) ....in 135 P was given

Qold=P (sin Theta old)

Qnew=P(sin Theta new)

Qadd=Qnew-Qold

instead of the SqRt of the Sum of the Squares.

I know there is more than one way to skin a cat, and I understand both approaches. Just wondering if there is a time (general rule) one is better used than the other? I may still discover it, but any input as always is appreciated.

John

Click to expand...

I cant get the correct answer for NCEES 134. Can you please explain how you solved it? 135 seems simple enough.

boltz said:

I cant get the correct answer for NCEES 134. Can you please explain how you solved it? 135 seems simple enough.

Click to expand...

The only trick with problem 134 is the voltage used will be a 480 Volt system.

Smotor=srt 3 * VL * IL = 1.732 * 480V * 34A = 28.27 kVA

P = S * pf = 28.27 * 0.75 = 21.20 kW

Qold= srt (S2 - P2) = 18.7 KVAR

Pnew = Pold = 21.20kW

Snew = Pnew / pf = 21.20kW / 0.9 = 23.56 KVA

Qnew = srt (23.56^2 - 21.20^2) = 10.26 KVAR

Qadd = Qnew - Qold = 10.26 - 18.7 = -8.4 kVAR

IFR_Pilot said:

The only trick with problem 134 is the voltage used will be a 480 Volt system.
Smotor=srt 3 * VL * IL = 1.732 * 480V * 34A = 28.27 kVA

P = S * pf = 28.27 * 0.75 = 21.20 kW

Qold= srt (S2 - P2) = 18.7 KVAR

Pnew = Pold = 21.20kW

Snew = Pnew / pf = 21.20kW / 0.9 = 23.56 KVA

Qnew = srt (23.56^2 - 21.20^2) = 10.26 KVAR

Qadd = Qnew - Qold = 10.26 - 18.7 = -8.4 kVAR

Click to expand...

Thank you IFR_Pilot

Are there different version of the NCEES or something? Questions 134 and 135 on my book are not about power factor correction. I actually have a question about 135 which is:

An electric generation facility uses a turbine-driven synchronous generator rated 3-phase, 150MVA, 13.8 kV. Per-unit (pu) reactances are X''d=0.15; X'd=0.25; Xd=Xq=1.20. Assume terminal voltage (Et)=1.0pu. For a transient stability study at rated MVA, rated voltage, and unity power factor, the internal voltage (pu) and reactance (pu) for the generator should be, respectively:

a. 1.01 and 0.15

b. 1.01 and 0.25

c. 1.03 and 0.25

d. 1.10 and 0.25

I actually don't understand the question. X'd and Xd that I know are from the conductor table use to find inductance/capacitance value. I'm not so sure what are X'dd and Xq.

The original question was from a previous version of the NCEES sample problems.

For the current version question, for motors and generators:

xd = steady state reactance

x'd = transient reactance

x''d = subtransient reactance

The first part of the problem is to select the proper impedance value for the type of fault study you are doing.

The second part is to determine the per-unit internal voltage required to maintain the terminal voltage at 1.0 p.u. at rated current.

Flyer_PE said:

The original question was from a previous version of the NCEES sample problems.
For the current version question, for motors and generators:

xd = steady state reactance

x'd = transient reactance

x''d = subtransient reactance

The first part of the problem is to select the proper impedance value for the type of fault study you are doing.

The second part is to determine the per-unit internal voltage required to maintain the terminal voltage at 1.0 p.u. at rated current.

Click to expand...

Are the questions for different version different? I would go find the older version to study as well if they are. Thanks.

The format of the test changed. The morning session in the old format was a breadth exam that wasn't specific to power/ECC/computer. A lot of the problems from the old version were carried over verbatim to to the new one. The problems from the old morning session were replaced with all power depth problems.

If you already have the current sample problem set, I doubt you would find the old book to be of much additional value for the current exam.

How come when doing the calculation for 134 (question 502 iin the newest version) real power, P, stays the same and the apparant power, S, doesn't change?

Real power drawn by the motor is not affected by the capacitor. Power factor correction only changes the reactive component. Since Q is being altered, the value for S will also change. They just don't bother calculating the new value for apparent power since it isn't necessary for solving the problem.

Thanks for the clarification!