Power factor correction problem

Discussion in 'Homework Help' started by testing12, Apr 11, 2011.

  1. testing12

    Thread Starter Member

    Jan 30, 2011
    80
    2
    Hello ,
    I had to solve the following problem:
    [​IMG]

    I did this problem using admittance. (since there in parallel) I simply took the impedance above took Y= (1/Z)

    and
    G (susceptance) = 2(pi)fC

    i got C = 312 microfarads. My question is how did they do part b in the solution manual. Im not understanding that. Please advise thank you.
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I wonder if it shouldn't be

    C=\frac{Ptan(\theta_1-\theta_2)}{\omega V_{rms}^2}

    where

    \theta=(\theta_1-\theta_2)

    is the angular difference between supply voltage and current

    and

    p.f.=cos(\theta_1-\theta_2)=cos(\theta)

    In any case that formula should be derived or stated somewhere in the text from which the problem is quoted.

    It's easy enough to show that

    Motor Reactive Power, Sm

    S_m=VIsin(\theta)=VIcos(\theta)\frac{sin(\theta)}{cos(\theta)}=P_mtan(\theta)

    Where Pm is the motor real power.

    The capacitive reactive power, Sc is

    S_c=\frac{V^2}{X_c}=V^2\omega C

    And assuming Sm=Sc for complete compensation.

    Leads to ..

    C=\frac{P_mtan(\theta)}{\omega V^2}
     
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