Power Output of Synchronous Generator or Alternator

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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,

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

$$\mathrm{∴\:I_{a} \:=\:\frac{E_{f} \:−\: V}{Z_{s}}\:\:…\:\: (2)}$$

Where,

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

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 Output of the Alternator Per Phase

$$\mathrm{S_{og} \:=\: P_{og} \:+\: jQ_{og} \:=\:V{I^{*}_{a}}}$$

$$\mathrm{\Rightarrow\:S_{og} \:=\: V \left( \frac{E_{f} \:−\: V}{Z_{s}}\right)^{∗}\:=\:V\:\angle 0° \left(\frac{E_{F}\angle \delta − V \angle 0°}{Z_{} \angle _{z}} \right)^{∗}}$$

$$\mathrm{\Rightarrow\:S_{og} \:=\: V \:\angle 0° \left(\frac{E_{F}}{Z_{}}\:\angle(\delta \:−\: _{z}) \:−\: \frac{V}{Z_{}}\:\angle − _{z}\right)^{∗}\:=\:\frac{VE_{F}}{Z_{}}\:\angle(_{z} \:−\: \delta) \:−\:\frac{V^{2}}{Z_{}}\:\angle _{z}}$$

Therefore, the complex output power the synchronous generator is

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

$$\mathrm{=\: \frac{VE_{F}}{Z_{}}cos(_{z} \:-\: \delta) \:+\: j\frac{VE_{F}}{Z_{}}sin(_{z} \:-\: \delta) \:- \:\frac{V^{2}}{Z_{}} \:(cos\:_{z} + j\:sin\:_{z }) \:\:…\:\: (4)}$$

Real Output Power Per Phase of the Alternator

Equating real parts of Eqn. (4), we get the real output power ($P_{og}$)

$$\mathrm{P_{og} \:=\:\frac{V{E_{F}} }{Z_{}}cos(_{z} \:−\: \delta) \:−\:\frac{V^{2}}{Z_{}}cos\:_{z}}$$

From the impedance triangle shown in Figure-2,

$$\mathrm{cos\:_{z}\:=\:\frac{R_{a}}{Z_{}}}$$

and

$$\mathrm{ _{z} \:=\: 90° \:− \:{α_{z}}}$$

$$\mathrm{∴\:\:P_{og}\: =\: \frac{VE_{F}}{Z_{}}cos(90° \:-\: \delta \:+\: α_{z}) \:- \:\frac{V^{2}}{Z^{2}_{}}R_{a}}$$

$$\mathrm{\Rightarrow\:P_{og} \:=\:\frac{VE_{F}}{Z_{}}sin(\delta \:+\: α_{z}) \: -\:\frac{V^{2}}{Z^{2}_{}}R_{a} \:\:…\:\: (5)}$$

The power ($P_{og}$) is also known as electrical power developed by the alternator.

Reactive Output Power Per Phase of the Alternator

Equating imaginary parts of the eq. (4), we get the reactive output power($Q_{og}$)

$$\mathrm{Q_{og} \:=\:\frac{VE_{F}}{Z_{}}(_{z} \:-\: \delta) \:−\:\frac{V^{2}}{Z_{}}sin\:_{z}}$$

From the impedance triangle shown in Figure-2, we get,

$$\mathrm{sin\:_{z} \:=\:\frac{X_{}}{Z_{}}}$$

and

$$\mathrm{_{z} \:=\: 90° \:−\: α_{z}}$$

$$\mathrm{∴\:Q_{og} \:=\:\frac{VE_{F}}{Z_{}}sin(90° \:-\: \delta \:+\: α_{z} )\:−\:\frac{V^{2}}{Z^{2}_{}}X_{ }}$$

$$\mathrm{\Rightarrow\:Q_{og} \:=\:\frac{VE_{F}}{Z_{}}cos(\delta \:+ \:α_{z})\:−\:\frac{V^{2}}{Z^{2}_{}}X_{ } \:\:…\:\: (6)}$$

Condition for Maximum Power Output of the Alternator Per Phase

For maximum power output of the alternator,

$$\mathrm{\frac{P_{og}}{}\:=\: 0\:\:and\:\:\frac{^{2}P_{og}}{{\delta}^{2}} \:<\: 0}$$

Therefore,

$$\mathrm{\frac{}{\delta}\left[\frac{VE_{F}}{Z_{}}sin(\delta \:+ \:α_{z})\:-\:\frac{V^{2}}{Z^{2}_{}}R_{a} \right] \:=\: 0}$$

$$\mathrm{\Rightarrow\:\frac{VE_{F}}{Z_{}}cos(\delta \:+\: α_{z})\:-\:0 \:=\: 0}$$

$$\mathrm{\Rightarrow\: cos(\delta \:+\:α_{z}) \:=\: 0}$$

$$\mathrm{\Rightarrow\:\delta \:+\: α_{z} \:=\: 90°}$$

$$\mathrm{\Rightarrow\:\delta \:=\: 90° \:-\: α_{z} \:=\: _{z} \:\:…\:\: (7)}$$

Hence for maximum power output of the alternator,

$$\mathrm{Load\:angle() \:=\: Impedance \:angle(_{z})}$$

Thus, from Eqns. (5) and (7), the maximum power output of the alternator is given by,

$$\mathrm{P_{og(a)} \:=\:\frac{VE_{F}}{Z_{}} \:-\:\frac{V^{2}}{Z^{2}_{}}R_{a}\:\:…\:\: (8)}$$