0
\$\begingroup\$

Salient Pole Generator Mode leading PF

In the figure above we can see the location of two armature currents. In case 1, \$I_{a1}\$ is leading \$Vt\$ (terminal voltage) but lagging \$E_f\$ (excitation voltage). In case 2, it is obvious that \$I_{a2}\$ is leading both voltages.

I have seen many cases that have stated that case 2 is a leading pf. However, in a few references, I have seen people use case 1 as a leading pf. I would like to ask to make sure about which case is right when the machine is in generator mode with leading PF.

It is obvious the internal PF angle will have a different value in each case.

\$\endgroup\$
1
  • \$\begingroup\$ Yes it is obvious +Ve angle is leading \$\endgroup\$ – Tony Stewart EE75 Feb 20 '20 at 0:46
0
\$\begingroup\$

The load impedance is inductive and current lags behind voltage when the phase angle between voltage and current is positive.

$$ \phi = \phi_v - \phi_i > 0 \rightarrow Lagging$$

The load impedance is capactive and current leads voltage when the phase angle between voltage and current is negative.

$$ \phi = \phi_v - \phi_i < 0 \rightarrow Leading$$

Both cases can be correct as it depends if you've defined the rotation of your d-q plane to be clockwise or counterclockwise. As long as you're consistent you'll be fine. I've learnt it as counterclockwise. Therefore with Ef as the reference point (0°), Ia_1 is lagging Ef and leading Vt. Ia_2 is leading both Ef and Vt.

\$ Ia_1 = -30°, Ia_2 = 30°, Ef = 0°, Vt = -60° \$

\$ \phi = \phi_v - \phi_i = Ef - Ia_1 = 0° - (-30°) = 30° > 0 \rightarrow Lagging \$ \$ \phi = \phi_v - \phi_i = Ef - Ia_2 = 0° - 30° = -30° < 0 \rightarrow Leading \$

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.