Maximum Reactive Power for a Synchronous Generator or Alternator

Electronics & ElectricalElectronDigital Electronics

For a salient-pole synchronous generator or alternator, the per phase reactive power is given by,

$$\mathrm{𝑄_{1𝜑} =\frac{𝑉𝐸_{𝑓}}{𝑋_{𝑑}}cos\:𝛿 −\frac{𝑉^{2}}{2𝑋_{𝑑}𝑋_{𝑞}}\lbrace{(𝑋_{𝑑} + 𝑋_{𝑞} ) − (𝑋_{𝑑} − 𝑋_{𝑞})\:cos\:2\delta}\rbrace … (1)}$$

Where,

  • V is the terminal voltage per phase.

  • Ef is the excitation voltage per phase.

  • $\delta$ is the per phase angle between Ef and V.

  • Xd is the direct-axis synchronous reactance.

  • Xq is the quadrature-axis synchronous reactance.

For reactive power to be maximum,

$$\mathrm{\frac{𝑑𝑄_{1𝜑}}{𝑑\delta}= 0}$$

$$\mathrm{\Rightarrow\:\frac{𝑑}{𝑑\delta} \left(\frac{𝑉𝐸_{𝑓}}{𝑋_{𝑑}}cos\:\delta −\frac{𝑉^{2}}{2𝑋_{𝑑}𝑋_{𝑞}}\lbrace(𝑋_{𝑑} + 𝑋_{𝑞}) − (𝑋_{𝑑} − 𝑋_{𝑞}) cos\:2\delta\rbrace \right)= 0}$$

$$\mathrm{−\frac{𝑉𝐸_{𝑓}}{𝑋_{𝑑}}sin\:\delta −\frac{2𝑉^{2}}{2𝑋_{𝑑}𝑋_{𝑞}}(𝑋_{𝑑} − 𝑋_{𝑞})sin\:2\delta = 0}$$

$$\mathrm{\Rightarrow\:𝐸_{𝑓}\:sin\:\delta +\frac{𝑉}{𝑋_{𝑑}}(𝑋_{𝑑} − 𝑋_{𝑞})(2\:sin\:\delta\:cos\:\delta) = 0}$$

$$\mathrm{\Rightarrow\:cos\:\delta = −\frac{𝐸_{𝑓}𝑋_{𝑞}}{2\:𝑉(𝑋_{𝑑} − 𝑋_{𝑞} )}… (2)}$$

By putting the value of from eq. (2) in eq. (1), we have,

$$\mathrm{𝑄_{1𝜑\:𝑚𝑎𝑥} =\frac{𝑉𝐸_{𝑓}}{𝑋_{𝑑}}\left(−\frac{𝐸_{𝑓}𝑋_{𝑞}}{2\:𝑉(𝑋_{𝑑} − 𝑋_{𝑞} )}\right)−\frac{𝑉^{2}}{2𝑋_{𝑑}𝑋_{𝑞}}{(𝑋_{𝑑} + 𝑋_{𝑞})+\frac{𝑉^{2}}{2𝑋_{𝑑}𝑋_{𝑞}}}{(𝑋_{𝑑} - 𝑋_{𝑞})}(2 cos^{2}\:\delta − 1)}$$

$$\mathrm{\Rightarrow\:𝑄_{1𝜑\:𝑚𝑎𝑥}=−\frac{{𝐸^{2}_{𝑓}}𝑋_{𝑞}}{2\:𝑋_{𝑑}(𝑋_{𝑑}−𝑋_{𝑞})}-\frac{𝑉^{2}}{2𝑋_{𝑑}𝑋_{𝑞}}{(𝑋_{𝑑} + 𝑋_{𝑞})+\frac{𝑉^{2}}{2𝑋_{𝑑}𝑋_{𝑞}}}{(𝑋_{𝑑} - 𝑋_{𝑞})}\left(\frac{2{𝐸^{2}_{𝑓}}𝑋^{2}_{𝑞}}{4 𝑉^{2}(𝑋_{𝑑}− 𝑋_{𝑞})^{2}}-1\right)}$$

$$\mathrm{\Rightarrow\:𝑄_{1𝜑\:𝑚𝑎𝑥}=-\frac{𝑉^{2}}{2𝑋_{𝑑}𝑋_{𝑞}}\lbrace{(𝑋_{𝑑} + 𝑋_{𝑞})-(𝑋_{𝑑} - 𝑋_{𝑞})}\rbrace−\frac{𝐸^{2}_{𝑓}𝑋_{𝑞}}{2\:𝑋_{𝑑}(𝑋_{𝑑} − 𝑋_{𝑞})}+\frac{𝐸^{2}_{𝑓}𝑋^{2}_{𝑞}}{4 𝑋_{𝑑}(𝑋_{𝑑} − 𝑋_{𝑞} )}}$$

$$\mathrm{\Rightarrow\:𝑄_{1𝜑\:𝑚𝑎𝑥}=\frac{𝑉^{2}}{𝑋_{𝑑}}-\frac{{𝐸^{2}_{𝑓}}𝑋^{2}_{𝑞}}{4 𝑋_{𝑑}(𝑋_{𝑑} − 𝑋_{𝑞} )}… (3)}$$

Equation (3) gives the maximum value of the reactive power per phase for the salient-pole alternator.

Again, for a cylindrical rotor alternator,

$$\mathrm{𝑋_{𝑑} = 𝑋_{𝑞} = 𝑋_{𝑠}}$$

$$\mathrm{∴\:𝑄_{1𝜑} =\frac{𝑉𝐸_{𝑓}}{𝑋_{𝑠}}cos\:\delta −\frac{𝑉^{2}}{𝑋_{𝑠}}}$$

$$\mathrm{\Rightarrow\:𝑄_{1𝜑} =\frac{𝑉}{𝑋_{𝑠}}(𝐸_{𝑓}\:cos\:\delta − 𝑉) … (4)}$$

From Eq. (4), it can be seen that when $𝐸_{𝑓}\:cos\:\delta = 𝑉$ i.e. under normal excitation, then $𝑄_{1𝜑}$ = 0 and the alternator operates at unity power factor.

  • When 𝐸𝑓 cos $\delta$ > 𝑉, i.e., the alternator is over-excited, the reactive power is positive. Therefore, the alternator supplies reactive power to the busbars.

  • When 𝐸𝑓 cos $\delta$ < 𝑉, i.e., the alternator is under-excited, the reactive power is negative. Therefore, the alternator absorbs reactive power.

raja
Updated on 01-Oct-2021 07:20:39

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