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3 Phase Balance Delta System - 30 Degree Phase Shift for Apparent Power

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    3 Phase Balance Delta System - 30 Degree Phase Shift for Apparent Power

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    With Problem 111
    I solved it by using the following formula
    S = √3∠-30*VL*IL* = √3∠-30 * (12.5∠0kV) * (70∠20) = 1515.54∠-10.

    But in the other problem posted, why isn't there a 30 Degree Phase shift as well.

    They are both 3 Phase Balanced Delta Loads.

    #2
    Originally posted by BatmanisWatching1987 View Post
    [ATTACH=CONFIG]22427[/ATTACH][ATTACH=CONFIG]22428[/ATTACH][ATTACH=CONFIG]22429[/ATTACH]

    With Problem 111
    I solved it by using the following formula
    S = √3∠-30*VL*IL* = √3∠-30 * (12.5∠0kV) * (70∠20) = 1515.54∠-10.

    But in the other problem posted, why isn't there a 30 Degree Phase shift as well.

    They are both 3 Phase Balanced Delta Loads.
    Where is the rest of the work?

    It matters if you are using I_phase vs. I_line.

    V_L in the problem is using V_LL voltage (perhaps the L they use is for "load"?). Maybe the I_L is I_phase.
    BB+/BB=?

    Comment


      #3
      Originally posted by BatmanisWatching1987 View Post
      [ATTACH=CONFIG]22427[/ATTACH][ATTACH=CONFIG]22428[/ATTACH][ATTACH=CONFIG]22429[/ATTACH]

      With Problem 111
      I solved it by using the following formula
      S = √3∠-30*VL*IL* = √3∠-30 * (12.5∠0kV) * (70∠20) = 1515.54∠-10.

      But in the other problem posted, why isn't there a 30 Degree Phase shift as well.

      They are both 3 Phase Balanced Delta Loads.
      I am pretty much sure this question was asked in a different thread and answered. When calculating 3 phase power (reactive, real or apparent) forget about the 30 degree shift between phase and line voltage/current. Only angle that matters is your load impedance angle which in this case is negative arc cos(0.85) = -31.8 = θ

      P=3XVφXIφXcos(θ)
      Q=3XVφXIφXsin(θ)
      S=3XVφXIφ
      "Because it's there!"
      George Mallory

      Comment


        #4
        Originally posted by BatmanisWatching1987 View Post
        [ATTACH=CONFIG]22427[/ATTACH][ATTACH=CONFIG]22428[/ATTACH][ATTACH=CONFIG]22429[/ATTACH]

        With Problem 111
        I solved it by using the following formula
        S = √3∠-30*VL*IL* = √3∠-30 * (12.5∠0kV) * (70∠20) = 1515.54∠-10.

        But in the other problem posted, why isn't there a 30 Degree Phase shift as well.

        They are both 3 Phase Balanced Delta Loads.
        This isn't at all helpful, but I'm so glad that I don't have to study for the Power PE Exam again! I was doing what you're doing this time last year. Good luck!

        Comment


          #5
          Originally posted by mivey View Post
          Where is the rest of the work?

          It matters if you are using I_phase vs. I_line.

          V_L in the problem is using V_LL voltage (perhaps the L they use is for "load"?). Maybe the I_L is I_phase.

          Here is my solution for Problem 111

          Click image for larger version

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