one wattmeter and two wattmeter method

Discussion in 'Homework Help' started by Ibbyy, Jun 22, 2013.

  1. Ibbyy

    Thread Starter New Member

    Jun 22, 2013
    1
    0
    Using phase currents, phase voltages and line voltages, I worked out the single phase power for each phase by doing V * I * pf. = power in watts, doing this for all three phases i added them (W1+W2+W3) to give the three phase power.

    Now I need to know how to calculate the total power for both of these in the one wattmeter method (balanced and unbalanced load) and the two wattmeter method (balanced and unbalanced load). Any help is appreciated
     
  2. francisB.

    New Member

    Aug 31, 2013
    2
    0
    for the two wattmeter method balanced load, you will have a high and low meter reading that correspond to the value of bar angle of your line current. for low wattmeter readin=(Vline)(Iline)cos(30+angle of current), for high reading just replace the addition sign to minus sign. i hope it will help :)
     
  3. francisB.

    New Member

    Aug 31, 2013
    2
    0
    just add the two reading to have your total power
     
  4. donpetru

    Active Member

    Nov 14, 2008
    186
    25
    The method of the two wattmeters (see P1 and P2 in the schematic below).
     
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