Homework Help: Adding KW, KVAR, and KVA

1. Sep 12, 2011

ridemx

1. The problem statement, all variables and given/known data

3 loads are connected to a source (1000Vrms, 60hz, 1 phase) in parallel

Find total S, P, Q, PF

2. Relevant equations

S=apparent (VA)
P=real (W)
Q=reactive (VAR)

S^2=P^2+Q^2
PF=P/S

3. The attempt at a solution

Using the above equations, I found the S,P,Q for all 3 loads as follows,

S=given=125kva
P=(125kva)(.28)=35kw
Q=sqrt[(125kva^2)-(35kw^2)]=120kvar

S=sqrt[(10kw^2)+(40kvar^2)]=41.23kva
P=given=10kw
Q=given=40kvar

S=sqrt[(0^2)+(15kva^2)]=15kva
P=given=15kva
Q=0 since resistive

I added up all the P's Q's and S's to get

S=181.23kva
P=60kw
Q=160kvar

but then I checked myself using S^2=P^2+Q^2 and it doesn't work out

sqrt[(60kw^2)+(160kvar^2)]=170.9kva not 181.23kva

where did I screw up, can you not just add the loads?

2. Sep 12, 2011

Zryn

If you have an inductive load and you draw the Power Factor Triangle for it, and then you want to improve the Power Factor by adding in a capacitor, what will happen to the Power Factor Triangle?

3. Sep 12, 2011

ridemx

Adding a capacitor to and inductive load, it will decrease total Q, I guess I never thought of it that way for some reason, but in that case would the Q's be

which gives a total of 80kvar, so:

circuit totals are
P=60kw
Q=80Kvar
S=sqrt[(60^2)+(80^2)]=100VA

Is this at all correct?

4. Sep 13, 2011

Zryn

You mean S=sqrt[(60k^2)+(80k^2)]=100kVA, and yes this looks correct.

5. Sep 13, 2011

ridemx

yep, that's definitely what I meant, more of just a typo. Thank a lot for the help!!!