POWER FACTOR PROBLEM! HELP PLEASE ASAP!!

Discussion in 'Homework Help' started by mal1990p, Mar 9, 2009.

  1. mal1990p

    Thread Starter New Member

    Mar 9, 2009
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    Hi all, I have this problem regarding power factor correction. I was wondering if someone would tell me how I need to work it out. thanks;

    A single phase motor operating off a 400v, 50Hz supply is developing 10kw with an efficiency of 84% and a power factor of 0.7 lagging. Calculate;

    a. the input apparent power
    b. the active and reactive components of the current
    c. the reactive power (in kilovars)

    it is required to raise the supply power factor to 0.9 lagging. Determine the value of the capacitance required in parallel with the motor to achieve this correction

    if someone knows the solution please e-mail me on malachyp@gmail.com thanks.
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    Use the output power and efficiency to calculate the input electrical power and then use the calculated input electrical power and the power factor to calculate the apparent power.

    For the others have a look here:

    http://www.allaboutcircuits.com/vol_2/chpt_11/2.html
     
  3. mal1990p

    Thread Starter New Member

    Mar 9, 2009
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    Do you know how? can you tell me the method..i have no idea on how to do it as i have been given this question without ever being explained about this topic :S hence i have no clue about the equations that can be used and so on. thank you for your response
     
  4. mik3

    Senior Member

    Feb 4, 2008
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    I will tell you only the efficiency formula, for the others read the link.


    Input electrical power=output power/efficiency (not in percent)
     
  5. mal1990p

    Thread Starter New Member

    Mar 9, 2009
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    ok, can you do me another small favour..i will work it out in a couple of minutes then i will tell you my answers and the method, can you work it out yourself so we compare the answers? sorry for being irritating but you are of great help. thank you very much
     
  6. mal1990p

    Thread Starter New Member

    Mar 9, 2009
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    I worked out the first part;

    input power = output power/efficiency

    input power = 10,000w / 0.84
    input power = 11.9kW ........correct?

    Now for the second part..what does it mean by active and reactive components of the current..i have no idea how to calculate..i read the link but it doesnt help on this part does it :/
     
  7. mik3

    Senior Member

    Feb 4, 2008
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    First calculate the apparent power by using the power factor value.


    Real power=input electrical energy=Apparent power*power factor
     
  8. mal1990p

    Thread Starter New Member

    Mar 9, 2009
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    therefore 11.9kw = apparent power / 0.7 (power factor)

    hence apparent power = 8.33 Kva

    but then..how do i find the active and reactive components of the current??

    sry but this is not a subject i am good at..
     
  9. mal1990p

    Thread Starter New Member

    Mar 9, 2009
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    then...Apparent power = V I

    8.33Kva / 400v = I

    I = 20.83A, right? but which part did i find then the active or the reactive :S ?
     
  10. mik3

    Senior Member

    Feb 4, 2008
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    Apparent power=Real power/0.7=17KW

    S^2=P^2 + Q^2

    S=apparent
    P=real
    Q=reactive

    from this find Q
     
  11. mal1990p

    Thread Starter New Member

    Mar 9, 2009
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    ok, so for the first part of the answer i.e. part A. i found out that the input apparent power = 11.9k

    then for the second part i stil have not found the ACTIVE AND REACTIVE components of the current..i do not know what that means. i have done what you said i.e.

    APPARENT = REAL / 0.7 = 17kW (havent we found the input apparent power in part A?)
    then using the formula S^2 = P^2 + Q^2 i found that Q = 12.14 K var

    but where does this take me?
     
  12. mal1990p

    Thread Starter New Member

    Mar 9, 2009
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    if you can, add me on msn:we can talk better then this forum

    Note from moderator - you do not want an email address out in public to be harvested by a spambot. Use the PM feature. We do prefer that things get worked out in the public forums for the benefit of all.
     
    Last edited by a moderator: Mar 9, 2009
  13. mik3

    Senior Member

    Feb 4, 2008
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    I am sorry but I won't go on msn, I have other work to do.

    First you calculate the input real power (P) by using the output power and the efficiency.

    Then you calculate the apparent power (S) by using the value of (P) and the power factor.

    Then you use S^2=P^2 + Q^2 to find the reactive power (Q).
     
  14. rush2905

    New Member

    Mar 10, 2009
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    a) input apparent power:

    VA = 10 kW/0.7
    = 14285.7 kVA

    b) current:

    VA (output) = 14285.7 kVA x 0.84
    = 12 KVA

    I = 12 KVA/400V
    = 30 A

    c) reactive power (output):

    theta = cos ^-1 (0.7)
    = 45.57 deg.

    QL = 12 KVA x sin (45.57 deg)
    = 8569.71 VARS

    PF correction:

    Pout (before): 8.4 kW
    QL1 (before): 8569.71 VARS

    With PF correction:

    Theta = cos^-1 (0.9)
    = 25.84 deg.

    QL2 = 8.4 kW x tan (25.84 deg.)
    = 4068.31 VARS


    QL1 - QL2 = 4501.4 VARS

    XC = (V^2)/Q
    = (400^2)/4501.4 VARS
    = 35.54 Ohms

    C = 1/(2pie x f x XC)
    = 1/ (2pie x 50Hz x 35.54 Ohms)
    = 89.56 microF


    hope that helps. of course there are shorter ways, but i wanted to show you how i got everything. havent done that in a while. any other questions let me know. wait until you get into 3 phase. calculations are more fun.
     
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