Ncees Problem #131

[banditxgbn]

Does anyone know how this equation in the solutions is derived

kvar = (Voperating/Vrated)2 X (KVAR rated)?

thank you

 

[Longlimbs]

Here is sort of how I thought it through.

kvar = 3*Vp*Ip*sin(cos-1(pf)) = 3*(Vp)2/Z*sin(cos-1(pf))

In the case of the capacitor it tells us what the kvar is for the rated voltage then it asks us for the kvar at a lower voltage. The only variable that changes is the voltage so you can then set a proportional equation

kvaroperating/kvarrated = (Voperating)2/(Vrated)2 = (Voperating/Vrated)2

-> kvaroperating = (Voperating/Vrated)2 * kvarrated

 

[Wheretostart]

My understanding is, since it only said "correcting the power factor" without specifying from one number to another, I can just assume XC=XL (for simplicity per phase). So, XC=(240/sqrt(3))**2/(110/3)=240**2/110=XL, then kvar that will be provided to the induction motor equals to kvar absorbed by inductive reactance, which means |kvar by XL|=|kvar by XC|=208**2/XC=208**2*110/240**2=82.6