Power Flow Transfer Equations for a Synchronous Generator

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The circuit model of a cylindrical rotor synchronous generator is shown in Figure-1.

Let,

• V = Terminal Voltage per Phase
• E_f = Excitation Voltage per Phase
• I_a = Armature Current
• δ = Load Angle or Angle between V and E_f

Also, the phasor diagram of the alternator at lagging power factor is shown in Figure-2.

For an alternator or synchronous generator, the excitation voltage (E_f) leads the terminal voltage (V) by the load angle (δ) of the machine. Thus,

$$\mathrm{V \:=\: V \:\angle 0° \:\: and \:\: E_{f} \:=\: E_{f}\:\angle \delta}$$

The synchronous impedance of the alternator is given by,

$$\mathrm{Z_{s} \:=\: R_{a} \:+\: jX_{s} \:=\: Z_{s} \:\angle \theta_{z} \:\: \dotso \:(1)}$$

Where, the angle (θ_z) is the impedance angle.

From the impedance triangle shown in Figure-3, θ_z is given by,

$$\mathrm{\theta_{z}\:=\:\tan^{-1}\:\left(\frac{X_{s}}{R_{a}}\right) \:\: \dotso \:(2)}$$

And

$$\mathrm{\alpha_{z}\:=\:(90°\:-\:\theta_{z})\:=\:\tan^{-1}\left(\frac{R_{a}}{X_{s}}\right) \:\: \dotso \:(3)}$$

Now, by applying KVL in the circuit of Figure-1, we get,

$$\mathrm{E_{f} \:=\: V \:+\: I_{a}Z_{s} \:\: \dotso \:(4)}$$

$$\mathrm{\therefore\:I_{a} \:=\:\frac{E_{f} \:-\: V}{Z_{s}} \:\: \dotso \:(5)}$$

Power Flow Transfer Equations for an Alternator

The various power relation of the alternator, when the armature resistance is considered, are given as follows −

Complex power output per phase of the alternator −

$$\mathrm{S_{og} = P_{og} + jQ_{og}}$$

$$\mathrm{= \: \frac{VE_{f}}{Z_{s}} \: \cos(\theta_{z} \:-\: \delta) \:+\: j\frac{VE_{f}}{Z_{s}}\:\sin(\theta_{z} \:-\: \delta)\:-\:\frac{V^{2}}{Z_{s}}(cos \theta_{z} \:+\: j\:\sin \theta_{z}) \:\: \dotso \:(6)}$$

Real output power per phase of the alternator −

$$\mathrm{P_{og} \:=\: \frac{VE_{f}}{Z_{s}}\:\sin(\delta \:+\: \alpha_{z}) \:-\: \frac{V^{2}}{Z^{2}_{s}}R_{a} \:\: \dotso \:(7)}$$

Reactive output power per phase of the alternator −

$$\mathrm{Q_{og} \:=\: \frac{VE_{f}}{Z_{s}}\:\cos(\delta \:+\: \alpha_{z}) \:-\: \frac{V^{2}}{Z^{2}_{s}}\:X_{s} \:\: \dotso \:(8)}$$

Complex input power to the alternator per phase −

$$\mathrm{S_{ig} \:=\: P_{ig} \:+\: jQ_{ig}}$$

$$\mathrm{= \: \frac{E^{2}_{f}}{Z_{s}}\: (cos \theta_{z} \:+\: j\:\sin \theta_{z}) \:-\: \left[\frac{VE_{f}}{Z_{s}} \:\cos(\theta_{z} \:+\: \delta) \:+\: j\frac{VE_{f}}{Z_{s}}\:\sin(\theta_{z} \:+\: \delta)\right] \:\: \dotso \:(9)}$$

Real power input to the alternator per phase −

$$\mathrm{P_{ig} \:=\: \frac{E^{2}_{f}}{Z^{2}_{s}}\:R_{a} \:+\: \frac{VE_{f}}{Z_{s}}\:\sin(\delta \:-\: \alpha_{z}) \:\: \dotso \:(10)}$$

Reactive power input to the alternator per phase −

$$\mathrm{Q_{ig} \:=\: \frac{E^{2}_{f}}{Z^{2}_{s}}\:X_{s}\:-\:\frac{VE_{f}}{Z_{s}}\:\cos(\delta \:-\: \alpha_{z}) \:\: \dotso \:(11)}$$

Power Flow Equations for an Alternator with Armature Resistance Neglected

In practice, for a 3-phase alternator or synchronous generator R_a < X_s and hence the armature resistance (R_a) can be neglected in the power flow transfer equations. Therefore, when the armature resistance (R_a) is neglected, then the synchronous impedance is,

$$\mathrm{Z_{s} \:=\: X_{s} \:\: and \:\: \alpha_{z} \:=\: 0}$$

Therefore, the output power per phase of the alternator is,

$$\mathrm{P_{og} \:=\: \frac{VE_{f}}{X_{s}}\:\sin \delta \:\: \dotso \:(12)}$$

$$\mathrm{Q_{og} \:=\: \frac{VE_{f}}{X_{s}}\:\cos \delta \:-\: \frac{V^{2}}{X_{s}} \:\: \dotso \:(13)}$$

And, the input power to the alternator per phase is,

$$\mathrm{P_{ig} \:=\: \frac{VE_{f}}{X_{s}}\:\sin \delta \:=\: P_{og} \:\: \dotso \:(14)}$$

$$\mathrm{Q_{ig} \:=\: \frac{E^{2}_{f}}{X_{s}} \:-\: \frac{VE_{f}}{X_{s}}\: \cos \delta \:\: \dotso \:(15)}$$