Determination of Synchronous Motor Excitation Voltage

The excitation voltage of synchronous motor refers to the DC supply given to the rotor to produce the required magnetic flux. The excitation voltage (E_f) 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,Z_S is the synchronous impedance and is given by,

$$\mathrm{Z_{S}=R_{a}+jX_{S}\:\:\:\:\:\:...(6)}$$

Case 1 – Excitation Voltage for Lagging Power Factor

$$\mathrm{E_{f}\angleδ=V\angle0°-(I_{a}=I_{a}\angle-φ)(R_{a}+jX_{S})}$$

$$\mathrm{\Longrightarrow\:E_{f}\angleδ=(V+j0)-(I_{a}cosφ-jI_{a}sinφ)(R_{a}+jX_{S})}$$

$$\mathrm{\Longrightarrow\:E_{f}\angleδ=(V+j0)-(I_{a}R_{a}cosφ+jI_{a}X_{S}cosφ-jI_{a}R_{a}sinφ+I_{a}X_{S}sinφ)}$$

$$\mathrm{\Longrightarrow\:E_{f}\angleδ=(V-I_{a}R_{a}cosφ-I_{a}X_{S}sinφ)-j(I_{a}X_{S}cosφ-I_{a}R_{a}sinφ)\:\:\:\:\:\:...(7)}$$

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,

$$\mathrm{δ=-tan^{-1}\left[\frac{I_{a}X_{S}cosφ-I_{a}R_{a}sinφ}{V-I_{a}R_{a}cosφ-I_{a}X_{S}sinφ}\right]\:\:\:\:\:\:...(9)}$$

Case 2 – Excitation Voltage for Unity Power Factor

As for unity power factor,

$$\mathrm{cosφ=1}$$

From Eqns.(8)&(9), we get,

$$\mathrm{\left|E_{f}\right|=\sqrt{(V-I_{a}R_{a})^{2}-(I_{a}X_{S})^{2}}\:\:\:\:\:\:...(10)}$$

$$\mathrm{δ=-tan^{-1}\left(\frac{I_{a}X_{S}}{V-I_{a}R_{a}}\right)\:\:\:\:\:\:...(11)}$$

Case 3 – Excitation Voltage for Leading Power Factor

$$\mathrm{\left|E_{f}\right|=\sqrt{(V-I_{a}R_{a}cosφ+I_{a}X_{S}sinφ)^{2}+(I_{a}X_{S}cosφ+I_{a}R_{a}sinφ)^{2}}\:\:\:\:\:\:...(12)}$$

Torque angle,

$$\mathrm{δ=-tan^{-1}\left[\frac{I_{a}X_{S}cosφ+I_{a}R_{a}sinφ}{V-I_{a}R_{a}cosφ+I_{a}X_{S}sinφ}\right]\:\:\:\:\:\:...(13)}$$

Numerical Example

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.

Solution

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,

$$\mathrm{E_{f}\angleδ=4767.6\angle-38.37°Volts\: per \:phase}$$