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Finding power consumed given a load

  1. Oct 26, 2016 #1
    1. The problem statement, all variables and given/known data
    The power consumed by 2 3-phase parallel loads are:
    Load 1: 12 kVA with power factor 0.7 lagging
    Load 2: 8 kW with power factor 0.8 lagging

    A)Find the complex power consumed by Load 1 and Load 2.
    B)Find the total complex power consumed by these two loads
    C)Determine the combined load power factor

    2. Relevant equations
    S = V * I*
    S = sqrt(P^2 + Q^2)
    P = V * I

    3. The attempt at a solution
    A)
    Load 1: Q = 12kVA, cos(θ) = 0.7, sin(θ) = 0.71414284272
    S = Q/sin(θ) = 16803.36VA

    Load 2: P = 8kW, cos(θ) = 0.8
    S = P/cos(θ) = 8kW/0.8 = 10000VA

    B)
    Total Complex = S1 + S2 = 26803.36VA

    Am I doing this correctly?
     
  2. jcsd
  3. Oct 26, 2016 #2

    cnh1995

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    This is not Q. Check the unit.
    Right.
     
  4. Oct 26, 2016 #3
    Is the complex power consumed by load 1 simply 12kVA?
     
  5. Oct 26, 2016 #4

    gneill

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    Staff: Mentor

    Complex power is expressed as a complex number, with both real and imaginary parts. In the power triangle, the components of that number are represented by the real and reactive power, while the apparent power is the hypotenuse (and thus the magnitude of the complex power). Just like vectors, you can't add them by adding their magnitudes. You have to add the components separately.

    In this problem I would argue that the 12 kVA of first load is not the reactive power but rather the apparent power (i.e. the magnitude of the complex power). The second load has units of watts, so that can only be the real power and you've treated it as such.

    So I would revisit the first load and assume that the given value is the apparent power. Then I would sum the real and reactive components of the loads to find the combined power.
     
  6. Oct 26, 2016 #5

    cnh1995

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    Yes. Unit of Q is VAR or kVAR.
     
  7. Oct 26, 2016 #6
    So complex power consumed by Load1 = 12kVA
    Complex power consumed by Load2 = 10kVA

    P/S = cos(θ)
    Load 1:
    P/12k = 0.7
    P = 8400W
    S = sqrt(P^2 + Q^2)
    12000 = sqrt(8400^2 + Q^2)
    144000000 = 8400^2 + Q^2
    Q = 8569.7
    Load 1 = 8400W + 8569.7VAR

    Load2:
    10000VA = sqrt(8000^2 + Q^2)
    100000000 = 64000000 + Q^2
    Q = 6000
    Load 2 = 8000W + 6000VAR

    Combined = 16400W + 14569.7VAR

    Stotal = 21937.1VA

    P/S = cos(θ)
    16400/21937.1 = cos(θ)
    PF = cos(θ) = 0.7476

    correct?
     
  8. Oct 26, 2016 #7

    cnh1995

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    Looks good.
     
  9. Oct 27, 2016 #8
    I forgot the last part:
    Find the capacitance of a 3-phase capacitor to be connected in parallel with the load to raise the total power factor to 0.95 lagging.
    Consider
    a) the capacitor is connected in delta
    b) The capacitor is connected in Y.
    Load line-line voltage is 207.85 V.

    What role does delta vs Y connection play? The only example in my notes did not mention how the capacitor would be connected. The example in my notes also gave the circuit frequency, which is not given for this problem.
     
  10. Oct 27, 2016 #9

    cnh1995

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    The capacitor bank supplies reactive power to the load and improves the power factor. This is called reactive power compensation.
    Current through the capacitors will depend on their connection (star or delta). Also, frequency of the supply is used to calculate the capacitance value from the capacitive reactance obatined from reactive power calculations.
     
  11. Oct 27, 2016 #10
    How would I do power factor correction without a given frequency?
     
  12. Oct 27, 2016 #11

    cnh1995

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    You can take the standard power frequency. It's either 50Hz or 60Hz, depending on your country.
     
  13. Oct 27, 2016 #12
    I have:
    current power factor = 0.74
    desired power factor = 0.95 lagging
    Load line-line voltage: 207.85

    How do calculate the capacitance required? I don't understand the delta/Y part. My notes have something like this:

    V = 120V<60, I = 10<25
    S = 1200<35
    P1 = 982.98
    Q1 = 688.3
    desired power factor: 0.95
    current power factor: 0.82
    arccos(0.95) = 18.19 degrees
    Q2 = tan(18.19) * P1 = 323
    Qchange = Q1-Q2 = 365.3
    Zcapacitor = (120*120)/365.3 = 39.42 ohm
    C = 1/(w(Zcapacitor)) =67.27uF
     
  14. Oct 27, 2016 #13

    cnh1995

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    So how much reactive power should be supplied by the capacitor bank? It will be the difference between current reactive power (with pf=0.74) and reactive power with the desired power factor (0.95). Star or delta connection determines the capacitor current. For the line voltage of 207.85V (you sure this is line-line and not line to neutral?), capacitor current will be 207.85/Zc for delta and 207.85/√3Zc for star.
     
  15. Oct 27, 2016 #14
    Prompt says load line-to-line voltage
     
  16. Oct 27, 2016 #15

    cnh1995

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    Ok..That's 120V line to neutral, USA's standard distribution voltage.
    Well, how much is the difference between the reactive powers?
     
  17. Oct 27, 2016 #16
    Q1 = 14569.7
    new pf = 0.95
    new Q = Q2 = 5390.46
    Q change = 9179.24
    Zcapacitor = 120^2/9179.24
    Zcapacitor = 1.57ohm

    For same capacitance, Delta = Star/3, correct?

    Are these correct:
    Zcapacitor = 120^2/(3*9179.24)
    and for Y/star:
    Zcapacitor = 120^2/(9179.24)
    ?
     
    Last edited: Oct 27, 2016
  18. Oct 27, 2016 #17

    cnh1995

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    I can't double check your calculations right now. You have the difference in reactive powers as Qc. This is the power supplied by the capacitor bank.
    Qc=√3VLIL
    From this, you can find the line current IL. For star connection, IL=Iph=I, hence, you can use Qc=3I2Xc and calculate Xc per phase.
    For delta connection, you have VL=Vph=V, hence, Qc=3V2/Xc. You can get Xc from this and then C=1/(ωXc).
     
  19. Oct 27, 2016 #18
    I don't have I, how do I find Xc for star?
     
  20. Oct 27, 2016 #19

    cnh1995

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    I=Qc/(√3VL).
    Then,
    ∴Xc=Qc/3I2.
    And
    C=1/(ω*Xc).
     
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