# NCEES 502



## Blink (Sep 5, 2013)

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?

I know how to solve it once I find the magnitude of S. My question is why does the solution say:

S=SQR(3)*V*I

I am having trouble understanding where the SQR(3) comes from. Any clarification helps. Thanks


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## Flyer_PE (Sep 5, 2013)

It's the 3-phase power equation.

S=sqrt3*V*I

Where V is Line-Line


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## Blink (Sep 5, 2013)

Just figured it out. Thank you for verifying.


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## Ship Wreck PE (Sep 5, 2013)

The problem states 3-phase. If it were 480 volt single phase (only two hot lines) you would not use square root of three, but with the 3rd additional

hot conductor, it is logical for the additional power..


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## GTOShoota (Oct 12, 2013)

I'm currently struggling with this problem as well. I understand what they're asking for, but in the solution, I don't know where they're getting Qnew = P[tan(cos^-1(theta)) and Qold = [sin(cos^-1(pf))]*[sqrt(3)*V*I]

Any help would be greatly appreciated!


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## Ship Wreck PE (Oct 12, 2013)

Their answer is hard to follow. I do the problem by drawing a right triangle and using the a^2 + b^2 = c^2.

Start off by finding VA. 34 x 480 x sq rt 3 = 28.2

That goes on hyp of triangle.

Multiply VA x power factor to get Watts

28.2 x .75 = 21.2

To find vars take sq root(28.2^2 -21.2^2=18.6

Now find new VA using Watts / new PF

21.2/.9=23.6

Now find new var

Sq rt( 23.6^2-21.2^2=10.3)

Take first Vars - second vars and that is answer

18.6-10.3=8.3

It is a lot easier than it sounds. Just use and label triangle as you go.


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## iahim (Oct 12, 2013)

GTOShoota said:


> I'm currently struggling with this problem as well. I understand what they're asking for, but in the solution, I don't know where they're getting Qnew = P[tan(cos^-1(theta)) and Qold = [sin(cos^-1(pf))]*[sqrt(3)*V*I]
> 
> Any help would be greatly appreciated!


Don't try to solve it one step, like they show it in the solution. Go step by step:

1. Calculate Smotor using V and I

2. Calculate Pmotor using Smotor and motor pf

3. Calculate Qmotor using Pmotor and Smotor

4. Calculate Sdesired, using Pmotor and desired pf

5. Calculate Qdesired using Sdesired and Pmotor

6. Calculate deltaQ from Qdesired and Qmotor.


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## iahim (Oct 12, 2013)

Ship Wreck said:


> Their answer is hard to follow. I do the problem by drawing a right triangle and using the a^2 + b^2 = c^2.
> 
> Start off by finding VA. 34 x 480 x sq rt 3 = 28.2
> 
> ...


You beat me by a few seconds ...


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## GTOShoota (Oct 12, 2013)

Thanks Guys! When I saw tan in the answer, I figured I might have to use a triangle, but wasn't sure exactly how. This was much easier when approached in the manner!


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## apoollo13 (Oct 14, 2013)

http://www.geindustrial.com/publibrary/checkout/GEP-974G?TNR=Application%20and%20Technical|GEP-974G|PDF

Page 26-41 should be brought to the exam, especially page 33. It makes this question infinitely easier.


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## Redskinsdb21 (Oct 14, 2013)

CI website has some really good videos of where the sqrt(3) comes from


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## GTOShoota (Oct 14, 2013)

That's extremely informative on this topic. Thanks Apoollo!


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## apoollo13 (Oct 14, 2013)

Also, page 30 with the formulas has the equation you use for the NCEES question where the capacitor is rated differently than the system itself.


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## GTOShoota (Oct 14, 2013)

I haven't had the opportunity to see any engineering economy questions, but is it related to this? Cost of an electrical grid and improving efficiency?


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## apoollo13 (Oct 14, 2013)

engineering economy is more geared towards time value of money problems. if you haven't gone over that yet, google it and do some practice problems. energy rates and KWH is still a good thing to know, though.


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## daw4888 (Oct 15, 2013)

I would get the time value of money tables and thrown them in. Just Google "present and future value table".


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## GTOShoota (Oct 16, 2013)

Thanks!


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## DJMB (Dec 8, 2020)

Can anyone tell me why Q(old)'s formula does not work for Q(new) ? 

Q(old) = sqrt3*V*I*sin(theta) = sqrt3*480*34*sin(cos^-1(.75)) = 18.7 kVAR

Qnew = sqrt3*480*34*sin(cos^-1(.9)) =12.3 kVAR

Qnew = P*tan(cos^-1(.9)) = 21.2*tan(cos^-1(.9))= 10.26 kVAR


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## kris7o2 (Jun 22, 2021)

Another way to solve this problem:

formula: P (tan (Phi1) - tan (Phi2) )

S=Sqrt(3)VI* = Sqrt(3) x 480 x 34 Ang (acos(.75))
|S| = Sqrt(3) x 480 x 34 = 28.3 kVA

P = PF x S = 28.3 kVA x .75 = 21.2 kW

=> 21.2 kW x (tan(-acos(.75) - tan( - acos(.9))
= -8.4290 kVAR

I used Phi1 and Phi2, both negative above because this was a lagging power factor


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