Power Output of Synchronous Generator or Alternator

Electronics & ElectricalElectronDigital Electronics

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

  • $\delta$ = 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,

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

Also, for a synchronous generator the excitation voltage ($𝐸_{𝑓}$) leads the terminal voltage (V) by the load angle ($\delta$). Thus,

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

Complex Power Output of the Alternator per Phase

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

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

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

Therefore, the complex output power the synchronous generator is

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

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

Real Output Power per Phase of the Alternator

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

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

From the impedance triangle shown in Figure-2,

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

and

$$\mathrm{ 𝜃_{𝑧} = 90° − {α_{𝑧}}}$$

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

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

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($𝑄_{𝑜𝑔}$)

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

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

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

and

$$\mathrm{𝜃_{𝑧} = 90° − α_{𝑧}}$$

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

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

Condition for Maximum Power Output of the Alternator per Phase

For maximum power output of the alternator,

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

Therefore,

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

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

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

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

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

Hence for maximum power output of the alternator,

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

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

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

raja
Published on 18-Oct-2021 09:08:17
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