Synchronous Machine Oscillations

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Synchronous Machine Oscillations

A synchronous machine under steady running conditions has a driving torque at every instant balancing its retarding torque. The retarding torque is developed by phase displacement (δ) between the axis of the stator and rotor poles. A mechanical rotary system possesses inertia and restoring torque that tends to restore its position when displaced, thus the system has a natural frequency of oscillations.

A synchronous machine operating in parallel with other machines or infinite busbars forms such an oscillatory system. Here, the restoring torque is due to the synchronizing torque that depends upon the phase displacement and opposes the displacement. The inertia in this system is due to the moment of inertia of the rotor and the prime mover.

Natural Time Period of Oscillations

Let

• J = Moment of inertia of rotating system (in kg m²)
• β = Load angle deviation from steady state position (in mech. radian)
• τ_syn = Synchronising torque coefficient (Nm per mech. radian)

If damping torque is neglected, then

$$\mathrm{J\frac{{d^{2}}\:\beta}{d{t^{2}}} \:=\: -\tau_{syn}\:\beta \:\:\:\dotso\: (1)}$$

The expression in eqn. (1) represents a single harmonic motion.

Also, the natural frequency of undamped oscillation is given by,

$$\mathrm{f \:=\:\frac{1}{2\pi}\sqrt{\frac{\tau_{syn}}{J}} \:\:\:\dotso\: (2)}$$

Therefore, the natural time period of oscillation is

$$\mathrm{T \:=\: \frac{1}{f} \:=\: 2\pi\sqrt{\frac{J}{\tau_{syn}}} \:\:\:\dotso\: (3)}$$

If I_a is the full-load current, then

$$\mathrm{\text{Reactance Voltage Drop } \:=\: I_{a}X_{s}}$$

$$\mathrm{\text{Per Unit Reactance, }\: X_{s \: (pu)} \:=\: \frac{I_{a}}{V_{ph}}X_{s}}$$

$$\mathrm{\therefore\:X_{s} \:=\: \frac{V_{ph}}{I_{a}}X_{s\:(pu)} \:\:\:\dotso\: (4)}$$

Hence, the short-circuit current is given by,

$$\mathrm{I_{sc} \:=\:\frac{V_{ph}}{X_{s}} \:=\: \frac{I_{a}}{X_{s\:(pu)}} \:\:\:\dotso\: (5)}$$

$$\mathrm{\Rightarrow\:\frac{I_{sc}}{I_{a}} \:=\: \frac{1}{X_{s\:(pu)}} \:\:\:\dotso\: (6)}$$

Since, the synchronising torque coefficient is given,

$$\mathrm{\tau_{syn} \:=\: \frac{3{V^{2}}_{ph}}{2\pi n_{s}X_{s}}\:\cdot\: p}$$

$$\mathrm{\Rightarrow\:\tau_{syn} \:=\:\frac{3{V^{2}}_{ph}}{2\pi n_{s}X_{s}}\:\cdot\: \left(\frac{f}{n_{s}}\right) \:\:\:\dotso\: (7)}$$

From Eqns. (5) & (7), we have,

$$\mathrm{\tau_{syn} \:=\:\frac{3{V}_{ph}I_{sc}f}{2\pi{n^{2}_{s}}} \:\:\:\dotso\: (8)}$$

Therefore, the time period of oscillation is

$$\mathrm{T\:=\:2\pi\sqrt{\frac{J}{\tau_{syn}}} \:=\: 2\pi \sqrt{\frac{J\:\cdot\:{2\pi{n^{2}_{s}}}}{3V_{ph}I_{sc}f}}}$$

$$\mathrm{\Rightarrow\:T \:=\: 9.093\:n_{s}\sqrt{\frac{J}{V_{ph}I_{sc}f}} \:\:\:\dotso\: (9)}$$

Now, the three-phase kVA is,

$$\mathrm{(kVA)_{3\phi} \:=\: \frac{3V_{ph}I_{a}}{1000}}$$

$$\mathrm{\Rightarrow\:V_{ph} \:=\: \frac{(kVA)_{3\phi} \:\times\: 1000}{3I_{a}}}$$

$$\mathrm{\therefore \: T \:=\: 9.093\:n_{s}\:\sqrt{\frac{J}{\left(\frac{1000}{3}\right)\:\cdot\:{(kVA)_{3\phi}}\:\cdot \: \left(\frac{I_{sc}}{I_{a}}\right)\:f}}}$$

$$\mathrm{\Rightarrow\:T \:=\: 0.498\:n_{s}\sqrt{\frac{J}{{(kVA)_{3\phi}}\:\cdot\:\left(\frac{I_{sc}}{I_{a}}\right)\:f}}}$$

$$\mathrm{\Rightarrow\:T\:=\:0.498\:n_{s}\:\sqrt{\frac{JX_{s\:(pu)}}{(kVA)_{3\phi}\:\cdot\:f}}\:\:\:\dotso\: (10)}$$

Equation (10) gives the natural time period of oscillation of a rotating system.

Numerical Example

A 6000 kVA, 3-phase, 11000 V, 50 Hz alternator runs at 1500 RPM, connected to constant voltage busbars. If the moment of inertia of the rotating system is 1.6 × 10⁵ kg-m² and the steady state short-circuit current is four times of the normal full-load current, find the natural time period of oscillation.

Solution

The per unit reactance is,

$$\mathrm{X_{s\:(pu)} \:=\:\frac{I_{a}}{I_{sc}} \:=\: \frac{1}{4} \:=\: 0.25}$$

Hence, the natural time of oscillation is,

$$\mathrm{T \:=\: 0.498 \: n_{s}\:\sqrt{\frac{JX_{s \:(pu)}}{(kVA)_{3\phi} \:\cdot\: f}}}$$

$$\mathrm{\Rightarrow\:T \:=\: 0.498 \:\times\: \left(\frac{1500}{60}\right)\:\times\:\sqrt{\frac{1.6\:\times\:10^{5} \: \times \: 0.25}{6000\:\times\:50}}}$$

$$\mathrm{\therefore \: T \:=\: 4.5404 \: seconds}$$