Power electronics

Discussion in 'Homework Help' started by Kayne, Aug 4, 2010.

  1. Kayne

    Thread Starter Active Member

    Mar 19, 2009

    I have attached the question that I have attempted. I am not confident that the answers are correct as I am having trouble with the angle (wt-15) component of the question. So I appreciate any guidance in solving this question.


    1.Active power supply to the load.

    Active Power = 3 x RMS of Van \times RMS of Fundamental Component of ia \times Cos of the phase shift between Van and the fundamental component of ia

    RMS of Van = 339/√2 = 240
    RMS of ia = 30/√2 = 21.2
    Cos of the angle is what I am unsure of because of the 9th harmonic has 30 deg and the rest 15.

    I think that I have to subtract -15 –(- 30) = -15 which is the same as the rest them I can use Cos 15 in the above formula.

    Active power = 3 \times 240 \times 21.2 \times Cos15 = 14744.3W

    2. Percentage harmonic distortion

    %THD = √((RMS 3rd Harmonic)^2+(RMS 5th Harmonic)^2+.......)

    %THD = \frac{(\sqrt{(5.53^2)+(2.1^2)+(0.707^2)+(0.35^2))}}{21.2} \time 100 \Rightarrow 19.7%

    3. The RMS value in

    For this part I understand that the phases are 120 degrees apart so the values of current would be the same just the angles would change. This is what I am not sure about doing.

    4. Displacement power factor

    Cos15 = 0.965

    5.Apparent Power supplied to the load

    3 x RMS of phase voltage x RMS of phase current

    3 \times 240 x 21.2 = 15264Va

    6.Power factor

    Active power/Apparent Power = 0.965

    Thanks for your time