1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Adding KW, KVAR, and KVA

  1. Sep 12, 2011 #1
    1. The problem statement, all variables and given/known data

    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

    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,

    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?
     
  2. jcsd
  3. Sep 12, 2011 #2

    Zryn

    User Avatar
    Gold Member

    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?
     
  4. Sep 12, 2011 #3
    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?
     
  5. Sep 13, 2011 #4

    Zryn

    User Avatar
    Gold Member

    You mean S=sqrt[(60k^2)+(80k^2)]=100kVA, and yes this looks correct.
     
  6. Sep 13, 2011 #5
    yep, that's definitely what I meant, more of just a typo. Thank a lot for the help!!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Adding KW, KVAR, and KVA
  1. Motor power in Kw (Replies: 3)

  2. Kva rating calculation (Replies: 6)

Loading...