Calculating Total Power and Power Factor for Parallel Loads in an AC Circuit

[ridemx]

Homework Statement

3 loads are connected to a source (1000Vrms, 60hz, 1 phase) in parallel
load 1:Inductive, 125kva, .28pf lagging
load 2:capacitive, 10kw, 40 kvar
load 3:resistive, 15kw

Find total S, P, Q, PF

Homework Equations

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

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

The Attempt at a Solution

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

Load 1:
S=given=125kva
P=(125kva)(.28)=35kw
Q=sqrt[(125kva^2)-(35kw^2)]=120kvar

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

load 3:
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?

 

[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?

 

[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

Load 1: 120kvar
Load 2: -40kvar
Load 3: 0

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?

 

[Zryn]

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

 

[ridemx]

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