Determination of Synchronous Motor Excitation Voltage

Electronics & Electrical Electron Digital Electronics

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,

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

Manish Kumar Saini

Published on 30-Oct-2021 06:59:41

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