What are the Output Power of a Synchronous Motor?

---

---

Consider a synchronous motor is operating at lagging power factor. The voltage equation of a synchronous motor is given by,

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

Where,

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

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

$$\mathrm{\Rightarrow \: I_{a} \:=\: \frac{V \angle 0° \:-\: E_{f} \:-\: \delta}{Z_{S}\angle \theta_{Z}} \:=\: \frac{V}{Z_{S}}\:\angle -\theta_{Z} \:-\: \frac{E_{f}}{Z_{S}} \: \angle -(\delta \:+\: \theta_{Z})}$$

$$\mathrm{\therefore \: I^{*}_{a} \:=\: \frac{V}{Z_{S}} \: \angle \theta_{Z} \:-\: \frac{E_{f}}{Z_{S}} \: \angle(\delta \:+\: \theta_{Z}) \:\:\:\:\dotso\: (3)}$$

Complex Power Output per Phase of a Synchronous Motor

The complex power output of a synchronous motor is given by,

$$\mathrm{S_{o} \:=\: E_{f}I^{*}_{a} \:=\: P_{o} \:+\: jQ_{o} \:\:\:\: \dotso\: (4)}$$

$$\mathrm{\Rightarrow\:S_{o} \:=\: E_{f} \: \angle -\delta \left(\frac{V}{Z_{S}} \:\angle \theta_{Z} \:-\: \frac{E_{f}}{Z_{S}} \: \angle(\delta \:+\: \theta_{Z})\right)}$$

$$\mathrm{\Rightarrow \: S_{o} \:=\: \left(\frac{VE_{f}}{Z_{S}}\:\cos(\theta_{Z} \:-\: \delta) \:+\: j \: \frac{VE_{f}}{Z_{S}} \:\sin(\theta_{Z} \:-\: \delta)\right) \:-\: \left(\frac{E^{2}_{f}}{Z_{S}} \: \cos \theta_{Z} \:+\: j\: \frac{E^{2}_{f}}{Z_{S}}\:\sin \theta_{Z}\right)}$$

$$\mathrm{\therefore\:S_{o}\:=\:\left(\frac{VE_{f}}{Z_{S}}cos(\theta_{Z} \:-\: \delta) \:-\: \frac{E^{2}_{f}}{Z_{S}} \: \cos\theta_{Z}\right) \:+\:j\left(\frac{VE_{f}}{Z_{S}} \: \sin(\theta_{Z} \:-\:\delta)\:-\: \frac{E^{2}_{f}}{Z_{S}} \: \sin\theta_{Z}\right) \:\:\:\:\dotso\: (5)}$$

Real Power Output per Phase of the Synchronous Motor

By equating the real part of equation(5), we get the real power output of the synchronous motor, i.e.,

$$\mathrm{P_{o}\:=\:\frac{VE_{f}}{Z_{S}}\:\cos(\theta_{Z} \:-\:\delta)\:-\:\frac{E^{2}_{f}}{Z_{S}}\:\cos\theta_{Z}}$$

$$\mathrm{\because\:\cos\theta_{Z}\:=\:\frac{R_{a}}{Z_{S}}}$$

$$\mathrm{\therefore\:P_{o}\:=\:\frac{VE_{f}}{Z_{S}}\:\cos(\theta_{Z} \:-\:\delta)\:-\:\frac{E^{2}_{f}}{Z^{2}_{S}}R_{a} \:\:\:\:\dotso\: (6)}$$

But,

$$\mathrm{\theta_{Z}\:=\:(90°\:-\:\alpha_{Z});\:\cos(\theta_{Z}\:-\:\delta)\:=\:\cos(90°\:-\:\delta\:+ \: \alpha_{Z}) \:=\: \sin(\delta\:+\:\alpha_{Z})}$$

$$\mathrm{\therefore\:P_{o}\:=\:\frac{VE_{f}}{Z_{S}}\:\sin(\delta\:+\:\alpha_{Z})\:-\:\frac{E^{2}_{f}}{Z^{2}_{S}}R_{a} \:\:\:\:\dotso\: (7)}$$

Reactive Power Output per Phase of the Synchronous Motor

By equating the imaginary part of Equation(5), we obtain the reactive power output of the synchronous motor, i.e.,

$$\mathrm{Q_{o}\:=\:\frac{VE_{f}}{Z_{S}}\:\sin(\theta_{Z}\:-\:\delta)\:-\:\frac{E^{2}_{f}}{Z_{S}}\:\sin\theta_{Z}}$$

$$\mathrm{\because\:\sin\theta_{Z}\:=\:\frac{X_{S}}{Z_{S}}}$$

$$\mathrm{\therefore\:Q_{o}\:=\:\frac{VE_{f}}{Z_{S}}\:\sin(\theta_{Z}\:-\:\delta)\:-\:\frac{E^{2}_{f}}{Z^{2}_{S}}\:R_{a} \:\:\:\:\dotso\: (8)}$$

But,

$$\mathrm{\theta_{Z} \: = \:(90° \:-\: \alpha_{Z}); \:\sin(\theta_{Z} \:-\: \delta) \:=\: \sin(90°\:-\:\delta\:+\: \alpha_{Z}) \:=\:\cos(\delta\:+\:\alpha_{Z})}$$

$$\mathrm{\therefore \: Q_{o} \:=\: \frac{VE_{f}}{Z_{S}}\:\cos(\delta\:+\:\alpha_{Z})\:-\:\frac{E^{2}_{f}}{Z^{2}_{S}} \: X_{S} \:\:\:\:\dotso\: (9)}$$

Also, for a synchronous motor, the output power available at the shaft is given by,

$$\mathrm{P_{sh}\:=\:P_{o}\:-\:Rotational \: losses \:\:\:\:\dotso\: (10)}$$

Where,p_o is the mechanical power (or gross power) developed by the motor. The rotational losses include core losses, friction and windage losses.