Power Triangle

Discussion in 'Homework Help' started by lisa3412, Sep 12, 2014.

  1. lisa3412

    Thread Starter Member

    Sep 6, 2014
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    hello!

    this is a question based on Power Triangle.

    My answer for Vs= 835.22 @ angle -130.

    Can somebody please verify this?
     
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    I probably did it wrong, but I got Vs= 368.53 volts @ 45°.
     
  3. subtech

    Senior Member

    Nov 21, 2006
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    Lisa, I think that is off by quite a bit.
    I would try figuring all the quantities related to the 30KVA load @ 240 Volts. (VL)
    Then do the same with the 40KW load.
    Combine the impedances (which are in parallel) of these two loads and insert that value in series with with the 0.1 ohm (resistive) and +j0.1 ohm (X). Then calculate the series circuit values.

    Hint:
    You can get a ballpark value for Vs by neglecting the angles and just using simple ohmic values.
     
    Last edited: Sep 12, 2014
  4. lisa3412

    Thread Starter Member

    Sep 6, 2014
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    what I did was that:

    For the 40kw load:
    We know that P=V*I*Power Factor
    V=240 @ angle=0
    So we can find out I2.

    Is this logic right?
    and since Power Factor is lagging, won't Current Angle be negative? that is -cos^-1(0.795)?
     
  5. shteii01

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    Feb 19, 2010
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    I noticed that it is 240 volts rms. I don't know off hand, but... Should you be using RMS value in your calculations or the Peak (Vpeak) value?

    240 volts rms is actually 339.4 volts peak. So if you use 240 volts rms, you will get one set of answers. If you use 339.4 volts peak, you get another set of answers.
     
  6. t_n_k

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    Mar 6, 2009
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    One must use the RMS value in this instance.
     
  7. t_n_k

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    Mar 6, 2009
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    Yes you are correct.
     
  8. shteii01

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    Feb 19, 2010
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    Ok, thank you, then my earlier solution is wrong.
     
  9. lisa3412

    Thread Starter Member

    Sep 6, 2014
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    So, t_n_k, is my solution correct? please can you verify. .I am having a tough time here!!!
     
  10. t_n_k

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    Mar 6, 2009
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    One step at a time.
    What values for the two individual load currents did you obtain?
     
  11. lisa3412

    Thread Starter Member

    Sep 6, 2014
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    OK! for I2 i got = 209.643 at an angle of -37.34 degrees
     
  12. t_n_k

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    Mar 6, 2009
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    That's correct.

    And the other.....
     
  13. lisa3412

    Thread Starter Member

    Sep 6, 2014
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    154.32 at an angle -25.84
     
  14. t_n_k

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    Wrong magnitude - correct phase angle. How did you work out the magnitude part?
     
  15. lisa3412

    Thread Starter Member

    Sep 6, 2014
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    Complex Power S= 30,000 VA
    Hence P= 30,000/0.9

    Current I1= (30,000/0.9)/(240V*0.9)
     
  16. lisa3412

    Thread Starter Member

    Sep 6, 2014
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    Sorry, Magnitude comes out to be 125
     
  17. t_n_k

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    Mar 6, 2009
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    That's incorrect.

    The current magnitude term is simply the total load VA (apparent power) divided by 240V.
     
  18. t_n_k

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    Mar 6, 2009
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    OK we are back in step.

    Now work out the total current from the source - i.e. I1 + I2
     
  19. lisa3412

    Thread Starter Member

    Sep 6, 2014
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    Ok. So, that comes out to be 125. Then Is= I1+I2
    Total Z= o.1+j0.1

    Vs= I/Z

    Is this correct?
     
  20. t_n_k

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    Mar 6, 2009
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    You've lost me.

    The total current is I1+I2 or 333A at about -33 degrees.

    The the line drop is Is*Zline. Add the line drop to 240V to get Vs.
     
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