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Active power of three-phase system

  1. Apr 3, 2016 #1
    1. The problem statement, all variables and given/known data
    Find active power of the given three-phase system.
    [itex]U_{RS}=U_{ST}=U_{TR}=220\sqrt 3 V[/itex]
    [itex]R=\omega L=\frac{1}{\omega C}=10\Omega[/itex]

    2. The attempt at a solution
    After reducing the circuit to one phase (see attachments), equivalent impedance is [itex]\underline{Z_e}=\sqrt{101}e^{-j84,3^{o}}[/itex].

    [tex]U_{RS}=U_{ST}=U_{TR}=U_{L}=\sqrt{3}U_R\Rightarrow U_R=220 V\Rightarrow \underline{U_R}=220e^{-j\pi/3}\Rightarrow \underline{I_R}=\frac{220}{\sqrt{101}}e^{j54,3^{o}}[/tex]

    Active power of each resistor is given by

    [tex]P_1=P_2=P_3=\mathfrak{R}(\underline{U_L}\underline{I_R^{*}})=\sqrt 3 U_RI_R\cos\angle(\underline{U_L}\underline{I_R})\Rightarrow P=3P_1[/tex]

    [tex]\cos\angle(\underline{U_L}\underline{I_R})\approx 0.099\Rightarrow P_1\approx 825.88 W\Rightarrow P=2477.64W[/tex]

    Is this correct?
     

    Attached Files:

  2. jcsd
  3. Apr 3, 2016 #2

    gneill

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    Staff: Mentor

    Can you show your calculations for the equivalent impedance?
     
  4. Apr 4, 2016 #3
    From circuit_2, [itex]\underline{Z_e}=X_C+(\frac{R}{3} || X_L)[/itex] where [itex]X_C=-j\frac{1}{\omega C},X_L=j\omega L[/itex].

    [tex]\underline{Z_e}=3-j9[/tex]

    I think this is the correct impedance.

    Question: Expect calculations, are the equations for powers, impedance, voltage and current in my original post set correctly?
     
    Last edited: Apr 4, 2016
  5. Apr 4, 2016 #4
    ATTEMPT EDITED (errors in calculation):

    After reducing the circuit to one phase (see attachment):


    Equivalent impedance is [itex]\underline{Z_e}=X_C+(R/3 || X_L)=-j10+(3+j)=3-j9[/itex]
    [tex]U_{RS}=U_{ST}=U_{TR}=U_L=\sqrt{3}U_R\Rightarrow \underline{U_R}=220e^{-j\pi/6}\Rightarrow \underline{I_R}=\sqrt{90}e^{-j71.56^{o}}[/tex]

    Active power of each resistor is given by:
    [tex]P_1=P_2=P_3=\mathfrak(\underline{U_R}\underline{I_R^{*}})=\sqrt{3}U_RI_R\cos\angle(\underline{U_L}\underline{I_R})[/tex]
    [tex]\cos\angle(\underline{U_L}\underline{I_R})\approx 0.75\Rightarrow P_1=165\sqrt{270}W\Rightarrow P=3P_1=495\sqrt{270}W[/tex]

    Question: Could someone check if equations are set correctly?
     
  6. Apr 4, 2016 #5

    gneill

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    Staff: Mentor

    Your equivalent impedance looks right now.

    How did you determine the potential across the resistor? In your equivalent circuit the capacitor is in series with the parallel combination of the resistor and inductor. Note that, since in a perfectly symmetric Y configuration the branches are essentially independent, you can ignore the phase angle of the source for a given branch if you're just looking for the real power in its load. Thus you have:

    upload_2016-4-4_8-10-16.png

    I put the magnitude of the voltage across the resistor at about 73 V, so whatever method you use to find it should find something similar.
     
  7. Apr 4, 2016 #6
    Why do we need voltage across the resistor?

    Total active power is given by [tex]P=3P_1=3\sqrt{3}U_RI_R\cos\angle(\underline{U_L}\underline{I_R})[/tex]

    To use this equation, we need to draw phase diagram and find [itex]\angle(\underline{U_L}\underline{I_R})[/itex] which is [itex]\approx 41.4^{o}[/itex].
     
  8. Apr 4, 2016 #7

    gneill

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    Staff: Mentor

    You don't necessarily need the voltage across the resistor. But it's a simple way to get to the power dissipated since P = V2/R. You should end up with a power (for one branch) of about 732/(10/3), or about 1600 W.

    In your previous attempt you wrote: ## \underline{I_R}=\sqrt{90}e^{-j71.56^{o}}##, but that looks more like the branch impedance value than the branch current.
     
  9. Apr 4, 2016 #8
    You are right, it should be [itex]\underline{I_R}=\frac{220}{\sqrt{90}}e^{j41.56^{o}}[/itex].

    But applying the formula [itex]P=3P_1=3\sqrt{3}U_RI_R\cos\angle(\underline{U_L},\underline{I_R})[/itex] doesn't give the result you get
    (I think that your result is correct, just don't know why it is not the same using complex domain).
     
  10. Apr 4, 2016 #9

    gneill

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    Staff: Mentor

    It must have something to do with the definitions of, or values of, your variables. Let's try a slightly different approach and see how it works out (It's a good idea to be able to check your work using different methods).

    If the branch source voltage is U = 220 V and the branch impedance is Z = (3 - j9) Ω, then the branch current is:

    ##I = \frac{U}{Z} = \frac{220}{3 - j9} = \left(\frac{22}{3} + j22 \right)~A##

    Then the complex power is p = VI* , so that:

    ##p = (220~V) \left(\frac{22}{3} - j22 \right) ~A = \left( \frac{4840}{3} - j4840 \right)~ VA ##

    The active power is the real component of the complex power, so P1 = 4840/3 W.

    What values are you using for UR and IR?


    edit: Fixed typo in power calculation (current value).
     
    Last edited: Apr 4, 2016
  11. Apr 4, 2016 #10
    I used [itex]U_R=220V,I_R=\frac{220}{\sqrt{90}},\cos\angle(\underline{U_L},\underline{I_R})=0.75[/itex].
     
  12. Apr 4, 2016 #11

    gneill

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    I have an idea of where your problem lies. I think you're applying a power formula for a Y-connected source with a Δ-connected load. That would explain the presence of the √3 before the voltage variable, converting the Y source voltage to a Δ line-to-line source.

    In this case you've already combined the loads into a single Y-connected load, and the given line-to-line voltage to a Y-source of 220 V, so the formula won't apply. What you have is:

    upload_2016-4-4_14-48-41.png
     
  13. Apr 4, 2016 #12
    Thanks for detailed explanations.
     
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