The equation governing the rotor dynamics

\[J\dfrac{d^{2}\theta}{dt^{2}}=T_{m}-T_{e}=T_{a}\] where \[\begin{array}{cc} J & \mbox{total moment of inertia of the rotor mass in}~kg-m^{2}\\ T_{m} & \mbox{mechanical torque supplied by the prime mover in}~N-m\\ T_{e} & \mbox{electrical torque output of the alternator in}~N-m\\ \theta & \mbox{angular position of the rotor in }~rad \end{array}\]

\[J\dfrac{d^{2}\theta}{dt^{2}}=T_{m}-T_{e}=T_{a}\]

Neglecting the losses, the difference between \(T_m\) and \(T_e\) gives the net accelerating torque \(T_a\) .

In the steady state, \(T_m = T_e\), and hence \(T_a = 0\).

During this period the rotor will move at synchronous speed \(\omega_s\) in rad/s.

The angular position \(\theta\) is measured with a stationary reference frame.

To represent it with respect to the synchronously rotating frame, we define \[\theta=\omega_{s}t+\delta\] where \(\delta\) is the angular position in rad with respect to the synchronously rotating reference frame.

Defining the angular speed of the rotor as \[\begin{aligned} \omega_{r} & =\dfrac{d\theta}{dt}\\ \omega_{r}-\omega_{s} & =\dfrac{d\delta}{dt} \end{aligned}\]

Therefore, the rotor angular speed is equal to the synchronous speed only when \(d\delta/dt = 0\)

We can therefore term \(d\delta/dt\) as the error in speed.

Then, \[\begin{aligned} J\dfrac{d^{2}\delta}{dt^{2}} & =T_{m}-T_{e}=T_{a}\\ \Rightarrow J\omega_{r}\dfrac{d^{2}\delta}{dt^{2}} & =P_{m}-P_{e}=P_{a} \end{aligned}\] where \(P_m\), \(P_e\) and \(P_a\) are the mechanical, electrical and accelerating power in MW.

We now define a normalized inertia constant as \[\begin{aligned} H&=\dfrac{\mbox{Stored Kinetic energy at synchronous speed in mega-joules}}{\mbox{Generator MVA rating}}\\ &=\dfrac{J\omega_{s}^{2}}{2G_{rated}} \end{aligned}\]

Substituting for \(J\) we get \[\left[2H\dfrac{G_{rated}}{\omega_{s}^{2}}\right]\omega_{r}\dfrac{d^{2}\delta}{dt^{2}}=P_{m}-P_{e}=P_{a}\]

In steady state, the machine angular speed is equal to the synchronous speed and hence we can replace \(\omega_r\) in the above equation by \(\omega_s\).

Note \(P_m\), \(P_e\) and \(P_a\) are given in MW.

Therefore dividing them by the generator MVA rating, \(G_{rated}\) we can get these quantities in per unit.

Hence dividing both sides by \(S_{rated}\) we get \[\dfrac{2H}{\omega_{s}}\dfrac{d^{2}\delta}{dt^{2}}=P_{m}-P_{e}=P_{a}~\text{per-unit}\]

Equation describes the behaviour of the rotor dynamics and hence is known as the .

The angle \(\delta\) is the angle of the internal emf of the generator and it dictates the amount of power that can be transferred. This angle is therefore called the .

Let, \[\begin{aligned} G_{mach} & = \mbox{machine rating (base)}\\ G_{system} & = \mbox{system base} \end{aligned}\]

Then, \[\begin{aligned} \dfrac{G_{mach}}{G_{system}}\left(\dfrac{H_{mach}}{\pi f}\dfrac{d^{2}\delta}{dt^{2}}\right) & =\left(P_{m}-P_{e}\right)\dfrac{G_{mach}}{G_{system}}\\ \Rightarrow\dfrac{H_{system}}{\pi f}\dfrac{d^{2}\delta}{dt^{2}} & =\left(P_{m}-P_{e}\right) \end{aligned}\]

where \[H_{system}=H_{mach}\left(\dfrac{G_{mach}}{G_{system}}\right)\]

Consider the swing equations of two machines on a common system base. \[\begin{aligned} \dfrac{H_{1}}{\pi f}\dfrac{d^{2}\delta_{1}}{dt^{2}} & =P_{m1}-P_{e1}\\ \dfrac{H_{2}}{\pi f}\dfrac{d^{2}\delta_{2}}{dt^{2}} & =P_{m2}-P_{e2} \end{aligned}\]

Since the machine rotors swing together (coherently or in unison) \[\delta_{1} = \delta_{2} = \delta\]

Adding \[\dfrac{H_{eq}}{\pi f}\dfrac{d^{2}\delta}{dt^{2}}=P_{m}-P_{e}\] where \[\begin{aligned} P_{m} & =P_{m1}+P_{m2}\\ P_{e} & =P_{e1}+P_{e2}\\ H_{eq} & =H_{1}+H_{2} \end{aligned}\]

The two machines swinging coherently are thus reduced to a single machine. The equivalent inertia can be written as \[H_{eq}=\dfrac{H_{1,mach}G_{1,mach}}{G_{system}}+\dfrac{H_{2,mach}G_{2,mach}}{G_{system}}\]

The above results can be easily extendable to any number of machines swinging coherently.