Power Developed by Salient Pole Synchronous Motor

Electronics & ElectricalElectronDigital Electronics

The complex power input per phase of the synchronous motor is,

$$\mathrm{S_{1φ}=VI^{*}_{a}\:\:\:\:\:\:...(1)}$$

Taking excitation voltage (Ef) as the reference phasor, then,

$$\mathrm{V=V\angle-δ=V\:cosδ-jVsinδ\:\:\:\:\:\:...(2)}$$

$$\mathrm{I_{a}=I_{q}-jI_{d}}$$

$$\mathrm{\therefore\:I^{*}_{a}=I_{q}+jI_{d}\:\:\:\:\:\:...(3)}$$

Hence, from Eqns.(1),(2)&(3), we get,

$$\mathrm{S_{1φ}=(V\:cosδ-jVsinδ)(I_{q}+jI_{d})\:\:\:\:\:\:...(4)}$$

For a salient-pole synchronous motor, the quadrature-axis current and direc taxis current are given by,

$$\mathrm{I_{q}=\frac{Vsinδ}{X_{q}}\:\:\:\:\:\:...(5)}$$

$$\mathrm{I_{d}=\frac{E_{f}-Vcosδ}{X_{d}}\:\:\:\:\:\:...(6)}$$

Substituting the values of Iq and Id in Eqn.(4), we have,

$$\mathrm{S_{1φ}=(V\:cosδ-jVsinδ)\left(\frac{Vsinδ}{X_{q}}+j\frac{E_{f}-Vcosδ}{X_{d}}\right)}$$

$$\mathrm{\Longrightarrow\:S_{1φ}=\left(\frac{v^{2}}{X_{q}}sinδcosδ+\frac{VE_{f}}{X_{d}}sinδ-\frac{v^{2}}{X_{d}}sinδcosδ\right)+j\left(\frac{VE_{f}}{X_{d}}cosδ-\frac{V^{2}}{X_{d}}cos^{2}δ-\frac{V^{2}}{X_{d}}sin^{2}δ\right)}$$

$$\mathrm{\Longrightarrow\:S_{1φ}=\left[\frac{VE_{f}}{X_{d}}sinδ+\frac{V^{2}}{2}\left(\frac{1}{X_{q}}-\frac{1}{X_{d}}\right)sin2δ\right]+j\left[\frac{VE_{f}}{X_{d}}cosδ-\frac{v^{2}}{2X_{d}}(1+cos2δ)-\frac{v^{2}}{2X_{d}}(1-cos2δ)\right]}$$

$$\mathrm{\Longrightarrow\:S_{1φ}=\left[\frac{VE_{f}}{X_{d}}sinδ+\frac{V^{2}}{2}\left(\frac{1}{X_{q}}-\frac{1}{X_{d}}\right)sin2δ\right]+j\left[\frac{VE_{f}}{X_{d}}cosδ-\frac{v^{2}}{2X_{d}X_{q}}\left\{(X_{d}+X_{q})-(X_{d}-X_{q})cos2δ\right\}\right]\:\:\:\:\:\:...(7)}$$

Also,

$$\mathrm{S_{1φ}=P_{1φ}+jQ_{1φ}\:\:\:\:\:\:...(8)}$$

Comparing Eqns.(7)&(8), we get the real power per phase in watts,

$$\mathrm{P_{1φ}=\frac{VE_{f}}{X_{d}}sinδ+\frac{V^{2}}{2}\left(\frac{1}{X_{q}}-\frac{1}{X_{d}}\right)sin2δ\:\:\:\:\:\:...(9)}$$

Thus, the total real power for three phases is,

$$\mathrm{P_{3φ}=3P_{1φ}=\frac{3VE_{f}}{X_{d}}sinδ+\frac{3V^{2}}{2}\left(\frac{1}{X_{q}}-\frac{1}{X_{d}}\right)sin2δ\:\:\:\:\:\:...(10)}$$

The first term on right-hand side of Eqn.(10) is called the excitation power and the second term is called the reluctance power. Also, the reactive power per phase in VARs is,

$$\mathrm{Q_{1φ}=\frac{VE_{f}}{X_{d}}cosδ-\frac{V^{2}}{2X_{d}X_{q}}[(X_{d}+X_{q})-(X_{d}-X_{q})cos2δ]\:\:\:\:\:\:...(11)}$$

Hence, the total reactive power for three phases in VARs is,

$$\mathrm{Q_{3φ}=3Q_{1φ}=\frac{3VE_{f}}{X_{d}}cosδ-\frac{3V^{2}}{2X_{d}X_{q}}[(X_{d}+X_{q})-(X_{d}-X_{q})cos2δ]\:\:\:\:\:\:...(12)}$$

Equations (10)&(12) indicate the real power and reactive power of the salient pole synchronous motor. The torque angle (δ) is negative for the salient pole synchronous motor.

Numerical Example

A 30 MVA, 3-phase, star-connected, 11 kV, 8-pole, 50 Hz salient-pole synchronous motor has reactance of Xd= 8Ω,Xq = 4Ω. The torque angle is 27.5°. At full load, unity power factor and rated voltage determine the real power of the motor.

Solution

Armature current,

$$\mathrm{I_{a}=\frac{S_{3φ}}{\sqrt{3}V}=\frac{30\:\times\:10^{6}}{\sqrt{3}\:\times\:11\times\:10^{3}}= 1574.64A}$$

$$\mathrm{I_{d}=I_{a}cosδ=1574.64\:\times\:cos27.5° = 1396.72 A}$$

Hence, the excitation voltage per phase of the motor is,

$$\mathrm{E_{f}=Vcos2δ+I_{d}X_{d}}$$

$$\mathrm{\Longrightarrow\:E_{f}=\frac{11000}{\sqrt{3}}cos27.5° + (1396.72\:\times\:8) = 16807.2V}$$

The real power for three phases is,

$$\mathrm{P_{3φ}=3P_{1φ}=\frac{3VE_{f}}{X_{d}}sinδ+\frac{3V^{2}}{2}\left(\frac{1}{X_{q}}-\frac{1}{X_{d}}\right)sin2δ}$$

$$\mathrm{\Longrightarrow\:P_{3φ}=\left[\frac{3\:\times\:11000\:\times\:16807.2}{\sqrt{3}\:\times\:8}sin 27.5°\right]+\left[\left(\frac{3}{2}\right)\times\left(\frac{11000}{\sqrt{3}}\right)^{2}\:\times\:\left(\frac{1}{4}-\frac{1}{8}\right)sin(2\:\times\:27.5)\right]}$$

$$\mathrm{\Longrightarrow\:P_{3φ}=18483194.48\:+\:6195200.78 = 24678395.26W}$$

$$\mathrm{P_{3φ}=24678.39 kW = 24.678MVA}$$

raja
Updated on 30-Oct-2021 07:40:43

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