The excitation voltage of synchronous motor refers to the DC supply given to the rotor to produce the required magnetic flux. The excitation voltage (Ef) of a synchronous motor can be determined for different power factors using complex algebra.
Let the supply voltage (V) be taken as the reference voltage. Thus,
$$\mathrm{V=V\angle0°=V+j0\:\:\:\:\:\:...(1)}$$
Then, the armature current at different power factors is given as follows,
For lagging power factor −
$$\mathrm{I_{a}=I_{a}\angle-φ=I_{a}cosφ-jI_{a}sinφ\:\:\:\:\:\:...(2)}$$
For unity power factor −
$$\mathrm{I_{a}=I_{a}\angle0°=I_{a}+j0\:\:\:\:\:\:...(3)}$$
For leading power factor −
$$\mathrm{I_{a}=I_{a}\angle+φ=I_{a}cosφ+jI_{a}sinφ\:\:\:\:\:\:...(4)}$$
Now, the excitation voltage of the synchronous motor is given by,
$$\mathrm{E_{f}=V-I_{a}Z_{S}\:\:\:\:\:\:...(5)}$$
Where,ZS is the synchronous impedance and is given by,
$$\mathrm{Z_{S}=R_{a}+jX_{S}\:\:\:\:\:\:...(6)}$$
$$\mathrm{E_{f}\angleδ=V\angle0°-(I_{a}=I_{a}\angle-φ)(R_{a}+jX_{S})}$$
$$\mathrm{\Longrightarrow\:E_{f}\angleδ=(V+j0)-(I_{a}cosφ-jI_{a}sinφ)(R_{a}+jX_{S})}$$
$$\mathrm{\Longrightarrow\:E_{f}\angleδ=(V+j0)-(I_{a}R_{a}cosφ+jI_{a}X_{S}cosφ-jI_{a}R_{a}sinφ+I_{a}X_{S}sinφ)}$$
$$\mathrm{\Longrightarrow\:E_{f}\angleδ=(V-I_{a}R_{a}cosφ-I_{a}X_{S}sinφ)-j(I_{a}X_{S}cosφ-I_{a}R_{a}sinφ)\:\:\:\:\:\:...(7)}$$
The magnitude of the excitation voltage at lagging power factor is given by,
$$\mathrm{\left|E_{f}\right|=\sqrt{(V-I_{a}R_{a}cosφ-I_{a}X_{S}sinφ)^{2}+(I_{a}X_{S}cosφ-I_{a}R_{a}sinφ)^{2}}\:\:\:\:\:\:...(8)}$$
The torque angle is given by,
$$\mathrm{δ=-tan^{-1}\left[\frac{I_{a}X_{S}cosφ-I_{a}R_{a}sinφ}{V-I_{a}R_{a}cosφ-I_{a}X_{S}sinφ}\right]\:\:\:\:\:\:...(9)}$$
As for unity power factor,
$$\mathrm{cosφ=1}$$
From Eqns.(8)&(9), we get,
$$\mathrm{\left|E_{f}\right|=\sqrt{(V-I_{a}R_{a})^{2}-(I_{a}X_{S})^{2}}\:\:\:\:\:\:...(10)}$$
$$\mathrm{δ=-tan^{-1}\left(\frac{I_{a}X_{S}}{V-I_{a}R_{a}}\right)\:\:\:\:\:\:...(11)}$$
$$\mathrm{\left|E_{f}\right|=\sqrt{(V-I_{a}R_{a}cosφ+I_{a}X_{S}sinφ)^{2}+(I_{a}X_{S}cosφ+I_{a}R_{a}sinφ)^{2}}\:\:\:\:\:\:...(12)}$$
Torque angle,
$$\mathrm{δ=-tan^{-1}\left[\frac{I_{a}X_{S}cosφ+I_{a}R_{a}sinφ}{V-I_{a}R_{a}cosφ+I_{a}X_{S}sinφ}\right]\:\:\:\:\:\:...(13)}$$
A 1500 kVA, 11000 V, 3-phase star connected synchronous motor has an armature resistance and synchronous reactance per phase of 4 Ω and 50 Ω respectively.
Determine the excitation EMF per phase and the angular retardation of the rotor when fully loaded at 0.8 power factor lagging.
Supply voltage per phase,
$$\mathrm{V=\frac{11000}{\sqrt{3}}= 6351V}$$
The armature current is,
$$\mathrm{kVA_{3φ}=\frac{\sqrt{3}V_{L}I_{a}}{1000}}$$
$$\mathrm{\therefore\:I_{a}=\frac{(kVA)_{3φ}\:\times\:1000}{\sqrt{3}V_{L}}=\frac{1500\times\:1000}{\sqrt{3}\:\times\:11000}= 78.7A}$$
At 0.8 power factor lagging −
$$\mathrm{cosφ = 0.8 \:then\: sinφ = 0.6}$$
The magnitude of the excitation voltage at lagging power factor is
$$\mathrm{\left|E_{f}\right|=\sqrt{(V-I_{a}R_{a}cosφ-I_{a}X_{S}sinφ)^{2}+(I_{a}X_{S}cosφ-I_{a}R_{a}sinφ)^{2}}}$$
$$\mathrm{\left|E_{f}\right|=\sqrt{[6351-(78.7\:\times4\:\times0.8)-(78.7\:\times50\:\times0.6)]^{2}+[(78.7\:\times50\:\times0.8)-(78.7\:\times4\:\times0.6)]^{2}}}$$
$$\mathrm{\left|E_{f}\right|=4767.6V}$$
The angular retardation of the rotor is,
$$\mathrm{Torque angle,δ=-tan^{-1}\left[\frac{I_{a}X_{S}cosφ-I_{a}R_{a}sinφ}{V-I_{a}R_{a}cosφ-I_{a}X_{S}sinφ}\right]}$$
$$\mathrm{δ=-tan^{-1}\left(\frac{2959.12}{3738.16}\right)=-38.37°}$$
Therefore, the excitation EMF per phase and the angular retardation of the motor is,
$$\mathrm{E_{f}\angleδ=4767.6\angle-38.37°Volts\: per \:phase}$$