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Power Correction Unity

  1. Nov 4, 2016 #1
    1. The problem statement, all variables and given/known data
    Not sure if this is the correct place for Electrical Engineering Homework help but here goes

    A single-phased motor connected across a 240-V source at 50Hz as shown in the figure has a power factor of 1.0, I = 20A, I1 = 25 A.

    Find the capacitance required to give a unity power factor when connected in parallel with the load as shown.

    HPjEffT.jpg

    2. Relevant equations

    Apparent Power S = VRMSIRMS
    Angular Frequency ω = 2πf
    cosθ = 1.0
    Reactive Power Q = Ssinθ = VRMSIRMSsinθ
    Active Power P = Scosθ = VRMSIRMScosθ
    ZRL = R + ωLi
    Zequiv = ZCZR/[ZC + ZR]

    QC = 1/ωC
    3. The attempt at a solution

    I think I am misunderstanding the question because if it is already at unity, wouldn't that make θ = 0° meaning QL = 0 VAR?

    If we want this at unity, QL = QC, correct? But I am getting 0 when the answer is allegedly non-zero.

    The only thing I was able to find was P = (240 V)(20 A) = 4800 W and I think I2 = 5 A which is entering the node. Meaning I1 = I + I2. This I2 is an RMS value, correct?
     
  2. jcsd
  3. Nov 4, 2016 #2

    cnh1995

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    It can't be already at unity pf. Every ac motor has a lagging pf, which is the result of its magnetizing current. So, unity pf condition can be achieved only after putting a capacitor in parallel with the motor.
    Yes and rms values aren't added algebraically.
     
  4. Nov 4, 2016 #3
    Ok, so when the problem says that it has a power factor of 1.0, they are talking about after you correct it with the capacitor not before? The wording made it seem as if it is already at unity which made no sense to me. So basically, I have to find the power factor before the capacitor is added.

    Ok so I have multiply the rms values by √2 to find the peak values and then add the currents using phasors in rectangular form aka complex numbers?
     
  5. Nov 4, 2016 #4

    cnh1995

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    That would complicate the problem. Start with the phasor diagram of the circuit. Using elementary geometry, you can find I2. Once you know I2, you can find capacitive reactance Xc and capacitance C.
     
  6. Nov 4, 2016 #5
    Hmm ok so I drew the diagram on a graph and got that IC = √[IL2 - I2] which gave me IC = 15 A

    Using the voltage source, which is in parallel with the capacitor, I found reactance Xc which gave me C = 1.19mF. Did I make a mistake somewhere?
     
  7. Nov 4, 2016 #6

    cnh1995

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    Right.
    I am getting a different value for C. How much is the capacitive reactance?
     
  8. Nov 4, 2016 #7
    Zc = Vc/Ic = (240 < 0°)/(15 < 90°) = 2.67 < -90° which means Xc = -2.67i = -i/(ωC)

    Edit: Nvm I see that I must've mistyped in my calculator. Zc = 16 < -90°

    Now I got C = 199 μF
     
  9. Nov 4, 2016 #8

    cnh1995

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    Right.
     
  10. Nov 4, 2016 #9
    Thanks, I really appreciate it.
     
  11. Nov 4, 2016 #10

    cnh1995

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    You're welcome!:smile:
     
  12. Nov 4, 2016 #11
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