# Calculate the active and reactive power of the connected load after field increase in synchronous generator

A 3-phase synchronous generator connected in Y configuration has the following given parameters

$$\V_L=38kV, R_a=0.3\Omega,X_s=2\Omega, S=360MVA, cos{\phi}=0.83\$$

Where the parameters correspond to the line voltage, armature resistance, synchronous reactance, total complex power and power factor(lagging).

The following quantities are then calculated:
$$\ I_a=\frac{S}{\sqrt{3}V_L}=5469.6A\$$
with $$\ \phi=-33.9\$$
and $$\ E\angle\delta=V_{phase}+I_a\angle\phi(R_a+jX_s)=30.515kV \angle15.52 \$$

These numbers are correct and an indicative phasor diagram is drawn by hand The question now is:

The load connected to the generator has changed. In order to keep the demand and production in balance at the same frequency and same terminal voltage, the generator field current has increased 22% by the AVR. The corresponding power angle is now 12°.Calculate the active and reactive power of the connected load.

Here is where i get stuck, we know that $$\Esin{\delta}\$$ has to stay constant (right?) and i calculated the new excitation voltage by equating the old $$\Esin{\delta}\$$ with the new one. $$\E'=39.28kV\$$ but i cannot wrap my head around on how to implement the 22% field current increase, and whatever i did so far is wrong.

The correct answer is: 𝑃≈ 319 MW and 𝑄≈ 428.5 MVAr

Thanks for the help!

## 1 Answer

I think the key to your question is in your question.

and same terminal voltage

If you know the terminal voltage and the power angle (angle by which E leads terminal voltage) then using power transfer equation you should be able to calculate what you need.

$$P=\frac{EVsin\delta}{X}$$

and also

$$Q = \frac{E^2-EVcos\delta}{X}$$

where $$\\delta\$$ is the angle by which $$\E\$$ leads the terminal voltage.

But, since your circuit includes $$\R_a\$$, you need a version of the power transfer equation that includes resistance,

$$P=\frac{E^2cos\phi-EVcos(\delta+\phi)}{Z}$$ and also $$Q=\frac{E^2sin\phi-EVsin(\delta+\phi)}{Z}$$ where $$\Z = R_a + jX_s = Ze^{j\phi}\$$

You can derive these last two equations just like you derive the top two. Just let $$\R_a\$$ equal zero and these will devolve into the top two equations.