Power Developed by Salient Pole Synchronous Motor

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Power Developed by Salient Pole Synchronous Motor

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

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

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

$$\mathrm{V \:=\: V\:\angle -\delta \:=\:V\:\cos\delta \:-\: jV\:\sin\delta \:\:\:\:\dotso\: (2)}$$

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

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

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

$$\mathrm{S_{1\phi}\:=\:(V\:\cos\delta\:-\:jV\:\sin\delta)\:(I_{q}\:+\:jI_{d}) \:\:\:\:\dotso\: (4)}$$

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

$$\mathrm{I_{q}\:=\:\frac{V\:\sin\delta}{X_{q}} \:\:\:\:\dotso\: (5)}$$

$$\mathrm{I_{d}\:=\:\frac{E_{f}\:-\:V\:\cos\delta}{X_{d}} \:\:\:\:\dotso\: (6)}$$

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

$$\mathrm{S_{1\phi}\:=\:(V\:\cos\delta\:-\:jV\:\sin\delta)\left(\frac{V\sin\delta}{X_{q}}\:+\:j\frac{E_{f}\:-\:V\:\cos \delta}{X_{d}}\right)}$$

$$\mathrm{\Rightarrow\:S_{1\phi}\:=\:\left(\frac{v^{2}}{X_{q}}\sin\delta\cos\delta\:+\:\frac{VE_{f}}{X_{d}}\sin\delta \:-\: \frac{v^{2}}{X_{d}}\sin\delta\cos\delta\right) \:+\: j\left(\frac{VE_{f}}{X_{d}}\cos\delta\:-\:\frac{V^{2}}{X_{d}} \cos^{2}\delta\:-\:\frac{V^{2}}{X_{d}}\sin^{2}\delta\right)}$$

$$\mathrm{\Rightarrow\:S_{1\phi}\:=\:\left[\frac{VE_{f}}{X_{d}}\sin\delta\:+\:\frac{V^{2}}{2}\left(\frac{1}{X_{q}} \:-\: \frac{1}{X_{d}}\right)\sin 2\delta\right] \:+\: j\left[\frac{VE_{f}}{X_{d}}\cos\delta\:-\:\frac{v^{2}}{2X_{d}}(1\:+\:\cos 2\delta)\:-\:\frac{v^{2}}{2X_{d}}(1\:-\:\cos 2\delta)\right]}$$

$$\mathrm{\Rightarrow\:S_{1\phi}\:=\:\left[\frac{VE_{f}}{X_{d}}\sin\delta \:+\: \frac{V^{2}}{2}\left(\frac{1}{X_{q}} \:-\: \frac{1}{X_{d}}\right)\sin 2\delta\right] \:+\: j\left[\frac{VE_{f}}{X_{d}}\cos\delta \:-\: \frac{v^{2}}{2X_{d}X_{q}} \left\{(X_{d} \:+\: X_{q}) \:-\: (X_{d}\:-\:X_{q})\cos 2\delta\right\}\right] \:\:\:\:\dotso\: (7)}$$

Also,

$$\mathrm{S_{1\phi} \:=\:P_{1\phi}\:+\:jQ_{1\phi} \:\:\:\:\dotso\: (8)}$$

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

$$\mathrm{P_{1\phi}\:=\:\frac{VE_{f}}{X_{d}}\sin\delta\:+\:\frac{V^{2}}{2}\left(\frac{1}{X_{q}}\:-\: \frac{1}{X_{d}} \right)\:\sin 2\delta \:\:\:\:\dotso\: (9)}$$

Thus, the total real power for three phases is,

$$\mathrm{P_{3\phi}\:=\:3P_{1\phi}\:=\:\frac{3VE_{f}}{X_{d}}\sin\delta\:+\:\frac{3V^{2}}{2}\left(\frac{1}{X_{q}} \:-\: \frac{1}{X_{d}}\right)\:\sin 2\delta \:\:\:\:\dotso\: (10)}$$

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,

$$\mathrm{Q_{1\phi} \:=\:\frac{VE_{f}}{X_{d}}\:\cos\delta\:-\:\frac{V^{2}}{2X_{d}\:X_{q}} \:[(X_{d} \:+\: X_{q}) \:-\: (X_{d} \:-\:X_{q})\:\cos 2\delta] \:\:\:\:\dotso\: (11)}$$

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

$$\mathrm{Q_{3\phi}\:=\:3Q_{1\phi}\:=\:\frac{3VE_{f}}{X_{d}}\:\cos\delta \:-\: \frac{3V^{2}}{2X_{d} \:X_{q}} \:[(X_{d} \:+\: X_{q})\:-\: (X_{d}\:-\: X_{q})\:\cos 2\delta] \:\:\:\:\dotso\: (12)}$$

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= 8,Xq = 4. 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,

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

$$\mathrm{I_{d} \:=\: I_{a}\:\cos\delta \:=\:1574.64\: \times \: \cos 27.5° \:=\: 1396.72 \:A}$$

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

$$\mathrm{E_{f} \:=\: V\:\cos 2\delta \:+\: I_{d}X_{d}}$$

$$\mathrm{\Rightarrow \: E_{f} \:=\: \frac{11000}{\sqrt{3}}\:\cos 27.5° \:+\:(1396.72\:\times\:8)\:=\:16807.2\:V}$$

The real power for three phases is,

$$\mathrm{P_{3\phi}\:=\:3P_{1\phi}\:=\:\frac{3VE_{f}}{X_{d}}\:\sin\delta \:+\: \frac{3V^{2}}{2} \left(\frac{1}{X_{q}} \:-\: \frac{1}{X_{d}}\right)\:\sin 2\delta}$$

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

$$\mathrm{\Rightarrow\:P_{3\phi}\:=\:18483194.48\:+\:6195200.78 \:=\: 24678395.26\:W}$$

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

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