Motor Power Factor

Discussion in 'Homework Help' started by Three Phase, Jul 11, 2007.

  1. Three Phase

    Thread Starter Member

    Jul 11, 2007
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    A 240 Volt, 60Hz supply has a heater and a single-phase motor connected in parallel to it. The heater draws a current of 5.5 amps and the total line current is 35 amps at a power factor of 0.60. The motor power factor is 0.484. By the way, the answer to this question is 0.484 (pf). I just can't figure on what to calculate to arrive at that answer.
     
  2. recca02

    Senior Member

    Apr 2, 2007
    1,211
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    i dont get your question is .484 given or to be found out?
    the data seems incomplete to me.
     
  3. Three Phase

    Thread Starter Member

    Jul 11, 2007
    13
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    The power factor .484 is to be found out.
     
  4. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,373
    1,157
    What is the question?

    I see in input voltage [230V 60Hz], the Total I w/pf [35 A .. 0.6 pf], the heater's amps w/o a pf [5]. Then the next statement is the pf of the motor [0.454].

    What I didn't see was a question?
     
  5. Three Phase

    Thread Starter Member

    Jul 11, 2007
    13
    0
    A 240 volt, 60 Hz supply has a heater and a single-phase motor connected in parallel to it. The heater draws a current of 5.5 amps and the total line current is 35 amps at a power factor of 0.60. What is the power factor of the motor?
     
  6. recca02

    Senior Member

    Apr 2, 2007
    1,211
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    ok
    i have tried it with the little data that was given,
    my answer does not match ur's.
    however i will give the explanation try to correct me if i m wrong.
    we assume heater to be purely resistive since it must be a heat dissipating resistance.
    .6 is total pf right?
    so total active power supplied
    .6*35*240
    active power consumed by heater
    240*35*1.
    subtract and we get active power consumed by the motor.
    since the current draw is 29.5 by motor (35-5.5)
    apparent power consumed
    29.5*240
    pf= active/apparent.
    but it comes to abt .525 something.
    another approach wud be to actually find out the impedance using above idea but i think that wud yield the same answer and i m little lazy to find it out (maybe later).
     
  7. Three Phase

    Thread Starter Member

    Jul 11, 2007
    13
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    Thank you for your help, recca02.
     
  8. recca02

    Senior Member

    Apr 2, 2007
    1,211
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    oops sorry,
    240*5.5*1
    it was a typing error
    but that wont change the answer.
    someone with a correct answer?
     
  9. spar59

    Active Member

    Aug 4, 2007
    51
    0
    Here goes:-

    The total power factor is 0.6 and the total current drawn is 35A.
    Active current = Total current * Power factor = 35A * 0.6 = 21A.

    The active and reactive current are at 90 degrees to each other,
    therefore pythagoras' theorem applies.
    i.e. Total current squared = Active current squared + Reactive current squared.

    Hence:- Reactive current squared = Total current squared - Active current squared.
    In this case, Reactive current squared = (35 * 35) - (21 * 21) = 784.
    So the total Reactive current is 28A.

    The current drawn by the heater (taken to be purely resistive and hence entirely active power) is 5.5A

    The active current drawn by the motor is therefore equal to the total active current of 21A minus the heater current of 5.5A = 15.5A.
    The reactive curent of the motor is the entire reactive current of 28A.
    Total motor current squared = 15.5 squared + 28 squared = 1024.25
    Hence total motor current = 32A

    Power factor = Active current / Total current
    Hence the motor Power factor = 15.5A / 32A = 0.484 !

    Hope this helps, Steve.
     
  10. recca02

    Senior Member

    Apr 2, 2007
    1,211
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    gud answer,
    i wud like to be pointed where exactly i went wrong.
    hope three phase sees yer reply.
     
  11. spar59

    Active Member

    Aug 4, 2007
    51
    0
    Hi there.

    I reckon it started to go wrong here:-

    "subtract and we get active power consumed by the motor.
    since the current draw is 29.5 by motor (35-5.5)"

    The 35A is at 0.6 power factor (presumably lagging since the motor will possess inductive reactance). So the total current of 35A is at an angle of 53 degrees. The 5.5A drawn by the purely resistive heater is at unity power factor i.e. 0 degrees. You then arithmetically subtracted the two figures rather than considering it as a vectorial operation.

    Steve.
     
  12. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    got it.
    and i was so lost in getting what did i actually do wrong thanks.
     
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