# Example 5.2



## Wolverine (Oct 13, 2006)

A load is operating at 60Hz, 440V, 15kVA, and 10kW. Determine the capacitance of a capacitor necessary to correct the power factor to 0.95.

Bonus question: Determine the VA rating of the capacitor.


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## Art (Oct 13, 2006)

> A load is operating at 60Hz, 440V, 15kVA, and 10kW.  Determine the capacitance of a capacitor necessary to correct the power factor to 0.95.
> Bonus question: Determine the VA rating of the capacitor.


~ 7.9 KVAR

~ 108 uF

I assumed both pf's to be lagging...


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## Wolverine (Oct 14, 2006)

Show your work.

(okay, I admit, I'm just trying to trick you into showing me an alternate method to solving this problem. I don't like the method used in the study guide I lifted it from, nor an alternate graphical method I saw used elsewhere and wonder if there's an easier way.)


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## Art (Oct 14, 2006)

> Show your work.
> (okay, I admit, I'm just trying to trick you into showing me an alternate method to solving this problem.? I don't like the method used in the study guide I lifted it from, nor an alternate graphical method I saw used elsewhere and wonder if there's an easier way.)


Close? here goes...

draw the power triangle for the original situation...

arc cos 10/15, ~48.2 deg = pf 10/15 or .666

base = real power = 10W

hypot = complex power; real + reactive = 15KVA

rise = reactive power = sin 48.2 deg x 15 ~ 11.2 kvar

then superimpose the new 0.95 scenario over it...

angle =arc cos 0.95 ~ 18.2 deg

real power of 10 KW stays the same

complex equals 10kw/cos 18.2 = 10.5 kva

reactive = sine 18.2 x 10.5 ~ 3.3 kvar

subtract the reactive components, that's the vars you need to add in order to lower the pf: 11.2 - 3.3 = 7.9 kvar to be added

|Complex P| = V^2/Xc

Xc = 1/(2 Pi f C)

|Complex P| = 7.9 KVA

V = 440

f = 60

substitute and solve for C

C = 7900/(440^2 2 Pi 60) ~ 108 x 10^-6 or 108 uF


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## singlespeed (Oct 16, 2006)

C=[tan(cos^-1(pf_initial))-tan(cos^-1(pf_final))]/omega*V^2

C= 111 muF


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## Wolverine (Oct 19, 2006)

Answers are correct but here is a sweet little cheater formula that I happily discovered now that my little pea-brain is full one week out from the exam:

(Icorrecting) = Q/V - [P/(V*p.f.) * sqrt(1 - p.f.^2)]

where P &amp; Q are the beginning power components and p.f. is the new power factor desired.

(Icorrecting) then is the reactive current that will correct the p.f., old to new.

Then Xc = V/I,

C = 1/w*Xc,

and the capacitive kvar Qc = VI.

:drunk:

_Wake me up when October ends._


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