automatic power factor corrector

Discussion in 'Physics' started by edennabo, Dec 11, 2011.

  1. edennabo

    Thread Starter New Member

    Dec 9, 2011
    dear sir

    please be inform you that our total load capacity of our transformer in bank is 3 @ 500 kva . if im going to put an automatic power factor corrector, what capacity of the said equipment that im going to use

    thank you and best regards
  2. polychromeuganda

    New Member

    Dec 12, 2011
    I can't convince myself I have all the information to answer your question. Here what I can figure out...

    You want to perform automatic power factor correction.

    You have either 1 or 3 transformers rated at 500kVA in the environment it is or will be used in. this is either the total capacity of number of tansformers connected to a variety of loads, a pre-connected 3-phase transformer with a 500kVA total power rating (167kVA/core), serving a load, or you may have 3 500kVA transformers to be interconnected to form a single 3-phase transfomer rated 1,500kVA max.

    IF... the load is already safely attached to the secondary side of the transformer,

    ...then we can assume the total load is less than the transformer capacity (either 500kVA or 1500kVA depending on the above) If the entire load might be reactive, you need 500k (1500k) VARS. Extreme power factors do happen with long transmission lines, and as transient events for a variety of causes, but electric customers (other than residential) are usually heavily penalized for, and thus in practice required to have, a load power factor magnitude less than 0.8 at the point of connection. That would mean that uncorrected load was <400kW (1200kW) and <100k (300k) VARS. If your course hasn't covered this yet, just say 500kVARS at 0PF, 100kVARS max for |PF| < 0.8 (1500kVARS at 0PF, 300kVARS max for |PF| < 0.8

    IF... the load is not yet connected to the transformer, and the idea is to use the entire transformer capacity for the real part of the load by cancelling the reactive part, then the real part of the load is <= 500kW (or 1500kW depending on the above)

    the load's power factor is unstated. if it is zero, you have to supply infinite k VARS, which is an appropriate answer only if this was a trick question.

    If you are expected to assume the load |PF| < 0.8, then the reactive parts was < 125k VARS (375k VARS depending on above)

    ... please try to provide the whole problem when asking your question. I assume that's why no one else answered, and I don't think I'll it again now that I see how slow it is to type this kind of answer.
  3. edennabo

    Thread Starter New Member

    Dec 9, 2011
    thank you sir for the answer of my question may God bless you always.

    best regards eden