hnc question

Discussion in 'Homework Help' started by redshaw, Jul 16, 2008.

  1. redshaw

    Thread Starter Member

    Jul 15, 2008
    12
    0
    could anyone help

    a single phase motor takes 50 A at a power factor of 0.6 lagging from a 240V, 50 Hz supply. Determin a.) the current taken by the capacitor in parrallel with the motor to correct the power factor to unity, and b.) the value of the supply current after power factor correction.

    i worked out the Ic by

    pf=cosO
    0=cos -1 (0.6)
    0=53.13

    true power=Vcoc0
    =240 cos 53.13
    144W

    Q=true power * tan 0
    Q=144 * tan 53.13
    Q=191.99 VA

    Ic=Q/V
    Ic=191.99/240
    Ic=0.799A

    but am struggling with the value of the supply current question b

    i have calculated Xc as 300.38 ohms and C as 0.106uF but dont know where to go from here

    can anyone help

    cheers
     
  2. redshaw

    Thread Starter Member

    Jul 15, 2008
    12
    0
    is the answer to b just simply

    supply corrected to power factor of 0.6 = 144W/(240V*0.6)

    =1A ?????????
     
  3. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    redshaw,

    I'll try.

    The apparent power is 50*240 = 12000 volt-amps

    The average power is 12000*.6 = 7200 watts

    The reactive capacitor power is sqrt(12000^2-7200^2) = 9600 vars

    The reactance of the capacitor is (240^2)/9600 = 6 ohms

    The reactive current is sqrt(9600/6) = 40 amps

    The current after correction is 7200/240 = 30 amps

    So correcting the power factor reduced the current load of the power supply by 20 amps. Ask if you have any questions.

    Ratch
     
  4. redshaw

    Thread Starter Member

    Jul 15, 2008
    12
    0
    thank you for your assistance
     
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