Power Input of Synchronous Generator or Alternator

---

---

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

Let,

• V = Terminal voltage per phase
• E_f = Excitation voltage per phase
• I_a = Armature current
• δ = Load angle (between V and E_f )

By applying KVL in the circuit, we get,

$$\mathrm{E_{f} \:=\: V \:+\: I_{a}Z_{s} \:\:\:\dotso\: (1)}$$

$$\mathrm{\therefore\:I_{a} \:=\:\frac{E_{f} \:-\: V}{Z_{s}} \:\:\:\dotso\: (2)}$$

Where,

$$\mathrm{\text{Synchronous impedance, } \:Z_{s}\:=\:R_{a}\:+\:jX_{a}\:=\:Z_{s}\:\angle \theta_{z}\:\:\:\dotso\:(3)}$$

Also, for a synchronous generator the excitation voltage (E_f) leads the terminal voltage (V) by the load angle (δ). Thus,

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

Complex Power Input to the Alternator per Phase

The complex input power to an alternator per phase is given by,

$$\mathrm{S_{ig} \:=\: P_{ig} \:+\: jQ_{ig} \:=\: E_{f}{I^{*}_{a}}}$$

$$\mathrm{\Rightarrow \: S_{ig} \:=\: E_{f}\left(\frac{E_{f} \:-\: V}{Z_{s}}\right)^{*} \:=\: E_{f} \: \angle \delta \left(\frac{E_{f}\angle \delta \:-\: V \:\angle 0°}{Z_{s} \:\angle \theta_{z}} \right)^{*}}$$

$$\mathrm{\Rightarrow\:S_{ig} \:=\: E_{f}\:\angle \delta \left(\frac{E_{f}}{Z_{s}}\:\angle(\delta \:-\: \theta_{z}) \:-\: \frac{V}{Z_{s}}\:\angle -\theta_{z}\right)^{*}}$$

$$\mathrm{\Rightarrow\:S_{ig} \:=\: E_{f} \:\angle \delta \left(\frac{E_{f}}{Z_{s}}\:\angle(\theta_{z} \:-\: \delta) \:-\: \frac{V}{Z_{s}}\:\angle \theta_{z}\right)}$$

$$\mathrm{\Rightarrow\:S_{ig} \:=\: \frac{E^{2}_{f}}{Z_{s}}\:\angle\theta_{z} \:-\:\frac{VE_{f}}{Z_{s}}\:\angle(\theta_{z} \:+\: \delta)}$$

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

$$\mathrm{=\: \frac{E^{2}_{f}}{Z_{s}}(\cos \: \theta_{z} \:+\: j\:\sin\:\theta_{z })}$$

$$\mathrm{-\left[\frac{VE_{f}}{Z_{s}}\:\cos(\theta_{z} \:+\: \delta) \:+\: j\frac{VE_{f}}{Z_{s}}\:\sin(\theta_{z} \:+\: \delta)\right] \:\:\:\dotso\: (4)}$$

Real Power Input to the Alternator per Phase

Equating real parts of the eq. (4), we get the expression for real power input (P_ig) to the alternator,

$$\mathrm{P_{ig} \:=\: \frac{{E^{2}_{f}}}{Z_{s}}\:\cos\theta_{z}\:-\:\frac{VE_{f}}{Z_{s}}\:\cos(\theta_{z}\:+\:\delta)}$$

From the impedance triangle shown in Figure-2,

$$\mathrm{\cos\theta_{z}\:=\:\frac{R_{a}}{Z_{s}}\:\:and \:\:\theta_{z} \:=\: 90° \:-\: \alpha_{z}}$$

$$\mathrm{\therefore\:P_{ig} \:=\:\frac{E^{2}_{f}}{Z^{2}_{s}}R_{a}\:-\:\frac{VE_{f}}{Z_{s}}\:\cos(90°\:+\:\delta \:-\: \alpha_{z})}$$

$$\mathrm{\therefore\:P_{ig}\:=\:\frac{E^{2}_{f}}{Z^{2}_{s}}R_{a}\:+\:\frac{VE_{f}}{Z_{s}}\sin(\delta \:-\: \alpha_{z}) \:\:\:\dotso\: (5)}$$

Reactive Power Input to the Alternator per Phase

Equating imaginary parts of Eqn. (4), we obtain the reactive power input (Q_ig) to the alternator,

$$\mathrm{Q_{ig} \:=\:\frac{E^{2}_{f}}{Z_{s}}\sin\theta_{z}\:-\:\frac{VE_{f}}{Z_{s}}\sin(\theta_{z} \:+\: \delta)}$$

From the impedance triangle shown in Figure-2,

$$\mathrm{\sin\:\theta_{z}\:=\:\frac{X_{s}}{Z_{s}}\:\:and\:\:\theta_{z}\:=\: 90° \:-\: \alpha_{z}}$$

$$\mathrm{\therefore\:Q_{ig}\:=\:\frac{E^{2}_{f}}{Z^{2}_{s}}X_{s}\:-\:\frac{VE_{f}}{Z_{s}}\sin(90° \:+\: \delta \:-\: \alpha_{z})}$$

$$\mathrm{\Rightarrow\:Q_{ig}\:=\:\frac{E^{2}_{f}}{Z^{2}_{s}}X_{s}\:-\:\frac{VE_{f}}{Z_{s}}\cos(\delta \:-\: \alpha_{z}) \:\:\:\dotso\: (6)}$$

Note – The total mechanical power input to the alternator is

$$\mathrm{\text{Mechanical power input } \:=\: P_{ig} \:+\: \text{ rotational losses}}$$

Condition for Maximum Power Input to the Alternator per Phase

For the maximum power input to the alternator,

$$\mathrm{\frac{dP_{ig}}{d\delta}\:=\: 0\:\:and\:\:\frac{d^{2}P_{ig}}{{d\delta}^{2}}\:\lt\:0}$$

$$\mathrm{\therefore\:\frac{d}{d\delta}\:\left(\frac{E^{2}_{f}}{Z^{2}_{s}}R_{a} \:+\: \frac{VE_{f}}{Z_{s}}\:\sin(\delta \: - \: \alpha_{z})\right)\:= \:0}$$

$$\mathrm{0 \:+\:\frac{VE_{f}}{Z_{s}}\:\cos(\delta \:-\: \alpha_{z}) \:=\: 0}$$

$$\mathrm{\Rightarrow\:\cos(\delta \:-\: \alpha_{z}) \:=\: 0}$$

$$\mathrm{\Rightarrow\:\delta \:-\: \alpha_{z} \:=\: 90°}$$

$$\mathrm{\Rightarrow\:\delta \:=\: 90° \:+\: \alpha_{z} \:=\: 90° \:+\: (90° \:-\: \theta_{z}) \:=\: 180° \:-\: \theta_{z}}$$

Thus for maximum power input to the alternator,

$$\mathrm{\text{Load angle }(\delta) \:=\: 180° \:-\: \text{ Impedance angle }(\theta_{z}) \:\:\:\dotso\: (7)}$$

Hence, from Eqns. (5) and (7), the maximum power input to the alternator per phase is

$$\mathrm{P_{ig(max)} \:=\:\frac{E^{2}_{f}}{Z^{2}_{s}}R_{a} \:+\:\frac{VE_{f}}{Z_{s}} \:\:\:\dotso\: (8)}$$