Hi,
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.
Find
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
kayne
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.
Find
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
kayne
