# NCEES Power #502



## tootie87 (Apr 4, 2010)

The problem reads as follows:

A 3-phase, 460-V, 25-hp induction motor draws 34A at 0.75 lagging power factor from a 480-V source. The reactive power required to correct the power factor to 0.90 lagging is most nearly?

The solution finds P to be 21.7 and I'm not sure how this is calculated. Any help is appreciated, thanks so much!


----------



## pelaw (Apr 4, 2010)

You have to draw the power triangle with P1, Q1, and S1. You are only adjusting Q up and down. P1 remains. As you adjust Q based on the angle, (the angle goes from 41.4 deg at 0.75 pf to 25.8 deg at .9 pf) the total power S will be adjusted. But P, real power, remains. Then the answer in kVAR is difference between the Q1 at 0.75 and Q2 at 0.9 pf.


----------



## Flyer_PE (Apr 4, 2010)

The magnitude of S = sqrt(3)*V*I = sqrt(3)*480*34 = 28.3 kVA

P = pf*S = 0.75*28.3 = 21.2 kW

This problem is also discussed in this thread.


----------



## BamaBino (Oct 1, 2010)

tootie87 said:


> The problem reads as follows:
> A 3-phase, 460-V, 25-hp induction motor draws 34A at 0.75 lagging power factor from a 480-V source. The reactive power required to correct the power factor to 0.90 lagging is most nearly?
> 
> The solution finds P to be 21.7 and I'm not sure how this is calculated. Any help is appreciated, thanks so much!


The 25-hp is 18.64 kW. Can't that number be used to verify the correct solution?

Thanks.


----------



## cableguy (Oct 1, 2010)

To me, the big trick in this question is using the correct voltage. It's a 460 volt motor, but a 480 volt at 34 amp source. If you use 460 volts, you get it wrong.


----------



## eng787 (Oct 2, 2010)

I think P can be calculated directly as sqrt(3)[email protected]= SQRT(3)x480x34x.75=21.2


----------



## kris7o2 (May 20, 2021)

Hi, I tried to use the formula from the NCEES study guide, 5.1.3


And upon doing this I get

S= sqrt(3) * 480 *34 ang (acos(.75)) = 28.3 ang (41.4 deg) kVA
Preal = S * .75 = 21.2 kW

21.2 kW * (tan(acos(.75))-tan(acos(.90))) = 8.4290 kVAR

Moreover, my answer is positive and matches the method of how I used to solve problem 14 in the PE final Exam by Justin Kauwale

Why is the kVAR positive when the NCEES solution manual says it should be negative?


----------



## akyip (May 20, 2021)

Power correction capacitors and capacitive loads deliver reactive power Q. Typically a load absorbs power, so it has a positive sign in front of the power value (whether it is in VA, W, or VAR). But when a load delivers power instead, it should have a negative sign. This is part of typical source and load conventions for circuit analysis.

SOURCE AND LOAD POWER CONVENTIONS FOR CIRCUIT ANALYSIS:

When a source delivers power, its power sign is positive or +.

When a source absorbs power, its power sign is negative or -.

When a load absorbs power, its power sign is positive or +.

When a load delivers power, its power sign is negative or -.

In the case of power correction, capacitors/capacitive loads are delivering reactive power Q (VAR). So really the VAR value should be negative to denote the capacitive load delivering reactive power.

(Though I should mention that most or at least half of the questions I have seen involving power factor correction like this typically just ask for the absolute value or magnitude of the amount of reactive power Q being delivered for power factor correction. So the negative sign is usually omitted in the final answer.)


----------

