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Thread: Calculating Real Power

  1. #11
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    Quote Originally Posted by LMAO View Post
    Just remember theta is always the angle between phase current and voltage. In other words theta is your impedance angle.
    Thanks this will help me out a lot.

    Correct me if I'm wrong,

    For a Balanced 3 Phase Delta System, since Line Voltage and Phase Voltage are equal, and if the the impedance is given, we can use the phase current for theta to find the Total Power for 3 Phase Load, which was equal to 2.448 kW.

    But, if we want to find the Single Phase Power we will use the line current angle, the power would be 0W? (Since the angle is 90 degrees)

  2. #12
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    Quote Originally Posted by BatmanisWatching1987 View Post
    Thanks this will help me out a lot.

    Correct me if I'm wrong,

    For a Balanced 3 Phase Delta System, since Line Voltage and Phase Voltage are equal, and if the the impedance is given, we can use the phase current for theta to find the Total Power for 3 Phase Load, which was equal to 2.448 kW.

    But, if we want to find the Single Phase Power we will use the line current angle, the power would be 0W? (Since the angle is 90 degrees)
    You had been told that the formula for power for every phase = phase voltage X phase current X cosine of the angle between the voltage and the current, period. That's not rocket science.

  3. #13
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    Quote Originally Posted by BatmanisWatching1987 View Post
    Thanks this will help me out a lot.

    But, if we want to find the Single Phase Power we will use the line current angle, the power would be 0W? (Since the angle is 90 degrees)
    Strange things show up when you look at the power factor of single phase loads connected to three phase sources.

    Say you have a perfect resistor connected line-line on a wye source. Clearly this resistor is using power and has a 0 degree current angle.

    But that 0 degree current angle is between the current flowing through the load and the l-l voltage applied to the load.

    Now look at the transformer source coils supplying the load. The L-N voltages are not in phase with the L-L voltages. This means that in the transformer coils there is a non-zero current angle.

    Back to your original calculation:

    The load has a current angle of 60 degrees. It has a 60 degree current angle relative to the applied voltage and must be consuming some real power.

    But one of the _source_ legs is seeing a 90 degree current angle, and is not supplying any real power to the load. If you do the calculation for the _other_ source leg, you should see a different result!

    -Jon

  4. #14
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    Quote Originally Posted by BatmanisWatching1987 View Post
    Thanks this will help me out a lot.

    Correct me if I'm wrong,

    For a Balanced 3 Phase Delta System, since Line Voltage and Phase Voltage are equal, and if the the impedance is given, we can use the phase current for theta to find the Total Power for 3 Phase Load, which was equal to 2.448 kW.

    But, if we want to find the Single Phase Power we will use the line current angle, the power would be 0W? (Since the angle is 90 degrees)
    I am not sure if I understand your question. Again, theta is you impedance angle; in this case (Zp=25<60 Ω) θ is 60 degrees. Where did you get 90 degrees? Load power always equals 3*|Iф|*|Vф|*cos(θ) where θ is the load impedance angle. Doesn't matter three phase Y, Δ or single phase. For single phase just remove the 3.
    "Because it's there!"
    George Mallory

  5. #15
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    Quote Originally Posted by LMAO View Post
    I am not sure if I understand your question. Again, theta is you impedance angle; in this case (Zp=25<60 Ω) θ is 60 degrees. Where did you get 90 degrees? Load power always equals 3*|Iф|*|Vф|*cos(θ) where θ is the load impedance angle. Doesn't matter three phase Y, Δ or single phase. For single phase just remove the 3.
    Just to amplify: the power through the load always depends on the voltage and current through the _load_, exactly as LMAO says.

    The original calculation converted a line-line value to a line-neutral value, an gave you the 90 degree phase angle...but this is not the current angle through the load, and can only reasonably be understood as the current angle seen by the source.

    -Jon

  6. #16
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    Deleted.
    Last edited by Sahib; Today at 04:04 AM.

  7. #17
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    Quote Originally Posted by BatmanisWatching1987 View Post
    Attachment 22295

    I am trying to solve Problem "B"

    I got Ia = 13.995∠-90°. (Which is the Line Current = IL)

    The power formula I am also using is, P =
    √3VLILcosθ

    What is the correct value of "
    θ" should I be using?

    In other problems, I always used the value of
    θ from the current angle.
    The lesson to be learned is the power factor angle to be used in problem(B) is the impedance angle of 60 degrees ie angle between phase voltage and phase current and not the angle between line voltage and line current.

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