New Power NCEES #131 and #526

[Gerbera]

I’ve run across a couple of problems for which I feel that I am missing a mathematical foundation. For example #131:

A 3 phase capacitor rated at 240 V and 110 kVAr has been proposed for correcting the power factor of a 3 phase induction motor operating at 208 V. What will be the reactive power (kVAr) supplied to the motor?

Solution:

kVAr = (208/240)² * 110

The format of the solution is new voltage (208) divided by old voltage(240) squared times the old kVAr, to get the new kVAr.

I have seen this show up again in yet another problem #526:

% of Nameplate kVA - Total Losses (W)

0 - 420

100 - 2950

Maximum efficiency occurs when copper losses equal no load losses. The percentage of nameplate kVA which will result in transformer maximum efficiency is…?

Solution:

420W is no load losses

420 = (2950-420) * (k/100)², where k is the new value for % of nameplate. Same format: new divided by old squared, multiplied by “old” load losses (load losses at 100% efficiency). Except this time, we know the “new” value, which is 420 and is on the left side of the equal sign.

Perhaps the two are not at all identical and I am thinking about it all wrong.

 

[Flyer_PE]

Problem 131:

A capacitor is a passive component with an impedance value of Z.

Power is proportional to V2/Z

Changing the applied voltage does not affect the impedance.

VAR1= V12/Z

VAR2= V22/Z

VAR2 = (V2/V1)2

Hope this helps.

 

[jlozoya4]

Problem 131:

A capacitor is a passive component with an impedance value of Z.

Power is proportional to V2/Z

Changing the applied voltage does not affect the impedance.

VAR1= V12/Z

VAR2= V22/Z

VAR2 = (V2/V1)2

Hope this helps.

helps a lot, thanks!