# Power Flow Transfer Equations for a Synchronous Generator

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

## Let,

• π = Terminal voltage per phase

• $πΈ_{π}$ = Excitation voltage per phase

• $πΌ_{π}$ = Armature current

• $\delta$ = Load angle or angle between π and $πΈ_{π}$

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 ($πΈ_{π}$) leads the terminal voltage (V) by the load angle ($\delta$) of the machine. Thus,

$$\mathrm{π½ = π\angle0°\:\:and\:\:π¬_{π} = πΈ_{π}\angle \delta}$$

The synchronous impedance of the alternator is given by,

$$\mathrm{π_{π} = π _{π} + ππ_{π } = π_{π }\angleπ_{π§} … (1)}$$

Where, the angle ($π_{π§}$) is the impedance angle.

From the impedance triangle shown in Figure-3, $π_{π§}$ is given by,

$$\mathrm{π_{π§} = \tan^{-1} \left(\frac{π_{π }}{π _{π}} \right)… (2)}$$

And

$$\mathrm{α_{π§} = (90° − π_{π§} ) = \tan^{-1} \left(\frac{π _{π}}{π_{π }} \right)… (3)}$$

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

$$\mathrm{π¬_{π} = π½ + π°_{π}π_{π} … (4)}$$

$$\mathrm{∴\:π°_{π} =\frac{π¬_{π} − π½}{π_{π}}… (5)}$$

## 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 −

$$\mathrm{π_{ππ} = π_{ππ} + ππ_{ππ}}$$

$$\mathrm{=\frac{ππΈ_{π}}{π_{π }}cos(π_{π§} − \delta) +π\frac{ππΈ_{π}}{π_{π }}sin(π_{π§} − \delta)-\frac{π^{2}}{π_{π }}(coπ_{π§ }+ π\:sin\:π_{π§}) … (6)}$$

Real output power per phase of the alternator −

$$\mathrm{π_{ππ}=\frac{ππΈ_{π}}{π_{π }}sin(\delta + α_{π§}) −\frac{π^{2}}{π^{2}_{π }}π _{π} … (7)}$$

Reactive output power per phase of the alternator −

$$\mathrm{π_{ππ} =\frac{ππΈ_{π}}{π_{π }}cos(\delta + α_{π§}) −\frac{π^{2}}{π^{2}_{π }}π_{π } … (8)}$$

Complex input power to the alternator per phase −

$$\mathrm{π_{ππ} = π_{ππ} + ππ_{ππ}}$$

$$\mathrm{=\frac{πΈ^{2}_{π}}{π_{π }}(cos\:π_{π§} + π\:sin\:π_{π§}) −\left [\frac{ππΈ_{π}}{π_{π }} cos(π_{π§} + \delta) + π\frac{ππΈ_{π}}{π_{π }}sin(π_{π§} + \delta )\right ]… (9)}$$

Real power input to the alternator per phase −

$$\mathrm{π_{ππ} =\frac{πΈ^{2}_{π}}{π^{2}_{π }}π _{π} +\frac{ππΈ_{π}}{π_{π }}sin(\delta − πΌ_{π§} ) … (10)}$$

Reactive power input to the alternator per phase −

$$\mathrm{π_{ππ} =\frac{πΈ^{2}_{π}}{π^{2}_{π }}π_{π }−\frac{ππΈ_{π}}{π_{π }}cos(\delta − α_{π§} ) … (11)}$$

## Power Flow Equations for an Alternator with Armature Resistance Neglected

In practice, for a 3-phase alternator or synchronous generator $π _{π}$ < $π_{π }$ and hence the armature resistance ($π _{π}$) can be neglected in the power flow transfer equations. Therefore, when the armature resistance ($π _{π}$) is neglected, then the synchronous impedance is,

$$\mathrm{π_{π } = π_{π }\:\:and\:\:α_{π§} = 0}$$

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

$$\mathrm{π_{ππ} =\frac{ππΈ_{π}}{π_{π }}sin\:\delta … (12)}$$

$$\mathrm{π_{ππ} =\frac{ππΈ_{π}}{π_{π }}cos\:\delta −\frac{π^{2}}{π_{π }}… (13)}$$

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

$$\mathrm{π_{ππ} =\frac{ππΈ_{π}}{π_{π }}sin\:\delta = π_{ππ} … (14)}$$

$$\mathrm{π_{ππ} =\frac{πΈ^{2}_{π}}{π_{π }}−\frac{ππΈ_{π}}{π_{π }}cos\:\delta… (15)}$$

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