Maximum Reactive Power for a Synchronous Generator or Alternator

Salient-pole Synchronous Generator or Alternator

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

$$\mathrm{Q_{1\phi} \:=\: \frac{VE_{f}}{X_{d}} \: \cos \: \delta \:-\: \frac{V^{2}}{2X_{d}X_{q}}\lbrace{(X_{d} \:+\: X_{q}) \: -\: (X_{d} \:-\: X_{q})\:\cos\:2\delta}\rbrace \:\:\dotso\: (1)}$$

Where,

• V is the terminal voltage per phase.
• E_f is the excitation voltage per phase.
• δ is the per phase angle between E_f and V.
• X_d is the direct-axis synchronous reactance.
• X_q is the quadrature-axis synchronous reactance.

For reactive power to be maximum,

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

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

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

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

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

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

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

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

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

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

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

Cylindrical Rotor Alternator

Again, for a cylindrical rotor alternator,

$$\mathrm{X_{d} \:=\: X_{q} \:=\: X_{s}}$$

$$\mathrm{\therefore \:Q_{1\phi} \:=\:\frac{VE_{f}}{X_{s}}\:\cos\:\delta\:-\:\frac{V^{2}}{X_{s}}}$$

$$\mathrm{\Rightarrow\:Q_{1\phi} \:=\:\frac{V}{X_{s}}(E_{f}\:\cos\:\delta\:-\: V)\:\:\dotso\: (4)}$$

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

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