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- Synchronous Machines
- Introduction to 3-Phase Synchronous Machines
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- Working of 3-Phase Alternator
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- Discussion
Power Flow Transfer Equations for a Synchronous Generator
The circuit model of a cylindrical rotor synchronous generator is shown in Figure-1.

Let,
- V = Terminal Voltage per Phase
- Ef = Excitation Voltage per Phase
- Ia = Armature Current
- δ = Load Angle or Angle between V and Ef
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 (Ef) 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 (θz) is the impedance angle.

From the impedance triangle shown in Figure-3, θz 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 Ra < Xs and hence the armature resistance (Ra) can be neglected in the power flow transfer equations. Therefore, when the armature resistance (Ra) 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,
$$\mathrm{P_{ig} \:=\: \frac{VE_{f}}{X_{s}}\:\sin \delta \:=\: P_{og} \:\: \dotso \:(14)}$$
$$\mathrm{Q_{ig} \:=\: \frac{E^{2}_{f}}{X_{s}} \:-\: \frac{VE_{f}}{X_{s}}\: \cos \delta \:\: \dotso \:(15)}$$