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Maximum Reactive Power for a Synchronous Generator or Alternator
For a salient-pole synchronous generator or alternator, the per phase reactive power is given by,
$$\mathrm{π_{1π} =\frac{ππΈ_{π}}{π_{π}}cos\:πΏ −\frac{π^{2}}{2π_{π}π_{π}}\lbrace{(π_{π} + π_{π} ) − (π_{π} − π_{π})\:cos\:2\delta}\rbrace … (1)}$$
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,
$$\mathrm{\frac{ππ_{1π}}{π\delta}= 0}$$
$$\mathrm{\Rightarrow\:\frac{π}{π\delta} \left(\frac{ππΈ_{π}}{π_{π}}cos\:\delta −\frac{π^{2}}{2π_{π}π_{π}}\lbrace(π_{π} + π_{π}) − (π_{π} − π_{π}) cos\:2\delta\rbrace \right)= 0}$$
$$\mathrm{−\frac{ππΈ_{π}}{π_{π}}sin\:\delta −\frac{2π^{2}}{2π_{π}π_{π}}(π_{π} − π_{π})sin\:2\delta = 0}$$
$$\mathrm{\Rightarrow\:πΈ_{π}\:sin\:\delta +\frac{π}{π_{π}}(π_{π} − π_{π})(2\:sin\:\delta\:cos\:\delta) = 0}$$
$$\mathrm{\Rightarrow\:cos\:\delta = −\frac{πΈ_{π}π_{π}}{2\:π(π_{π} − π_{π} )}… (2)}$$
By putting the value of from eq. (2) in eq. (1), we have,
$$\mathrm{π_{1π\:πππ₯} =\frac{ππΈ_{π}}{π_{π}}\left(−\frac{πΈ_{π}π_{π}}{2\:π(π_{π} − π_{π} )}\right)−\frac{π^{2}}{2π_{π}π_{π}}{(π_{π} + π_{π})+\frac{π^{2}}{2π_{π}π_{π}}}{(π_{π} - π_{π})}(2 cos^{2}\:\delta − 1)}$$
$$\mathrm{\Rightarrow\:π_{1π\:πππ₯}=−\frac{{πΈ^{2}_{π}}π_{π}}{2\:π_{π}(π_{π}−π_{π})}-\frac{π^{2}}{2π_{π}π_{π}}{(π_{π} + π_{π})+\frac{π^{2}}{2π_{π}π_{π}}}{(π_{π} - π_{π})}\left(\frac{2{πΈ^{2}_{π}}π^{2}_{π}}{4 π^{2}(π_{π}− π_{π})^{2}}-1\right)}$$
$$\mathrm{\Rightarrow\:π_{1π\:πππ₯}=-\frac{π^{2}}{2π_{π}π_{π}}\lbrace{(π_{π} + π_{π})-(π_{π} - π_{π})}\rbrace−\frac{πΈ^{2}_{π}π_{π}}{2\:π_{π}(π_{π} − π_{π})}+\frac{πΈ^{2}_{π}π^{2}_{π}}{4 π_{π}(π_{π} − π_{π} )}}$$
$$\mathrm{\Rightarrow\:π_{1π\:πππ₯}=\frac{π^{2}}{π_{π}}-\frac{{πΈ^{2}_{π}}π^{2}_{π}}{4 π_{π}(π_{π} − π_{π} )}… (3)}$$
Equation (3) gives the maximum value of the reactive power per phase for the salient-pole alternator.
Again, for a cylindrical rotor alternator,
$$\mathrm{π_{π} = π_{π} = π_{π }}$$
$$\mathrm{∴\:π_{1π} =\frac{ππΈ_{π}}{π_{π }}cos\:\delta −\frac{π^{2}}{π_{π }}}$$
$$\mathrm{\Rightarrow\:π_{1π} =\frac{π}{π_{π }}(πΈ_{π}\:cos\:\delta − π) … (4)}$$
From Eq. (4), it can be seen that when $πΈ_{π}\:cos\:\delta = π$ i.e. under normal excitation, then $π_{1π}$ = 0 and the alternator operates at unity power factor.
When πΈπ cos $\delta$ > π, i.e., the alternator is over-excited, the reactive power is positive. Therefore, the alternator supplies reactive power to the busbars.
When πΈπ cos $\delta$ < π, i.e., the alternator is under-excited, the reactive power is negative. Therefore, the alternator absorbs reactive power.
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