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# Swing Equation in Power System

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## Swing Equation

The motion of a synchronous machine is governed by Newton’s law of rotation, which states that the product of the moment of inertia times the angular acceleration is equal to the net accelerating torque. Mathematically, this may be expressed as follows:

$\begin{matrix} J\alpha ={{T}_{a}}={{T}_{m}}-{{T}_{e}} & {} & \left( 1 \right) \\\end{matrix}$

Equation 1 may also be written in terms of the angular position as follows:

$\begin{matrix} J\frac{{{d}^{2}}{{\theta }_{m}}}{d{{t}^{2}}}={{T}_{a}}={{T}_{m}}-{{T}_{e}} & {} & \left( 2 \right) \\\end{matrix}$

where

J = moment of inertia of the rotor

Ta = net accelerating torque or algebraic sum of all torques acting on the machine

Tm = shaft torque corrected for the rotational losses including friction and windage and core losses

Te = electromagnetic torque

By convention, the values of Tm and Te are taken as positive for generator action and negative for motor action.

For stability studies, it is necessary to find an expression for the angular position of the machine rotor as a function of time t. However, because the displacement angle and relative speed are of greater interest, it is more convenient to measure angular position and angular velocity with respect to a synchronously rotating reference frame with a synchronous speed of${{\omega }_{sm}}$. Thus, the rotor position may be described by the following:

$\begin{matrix} {{\theta }_{m}}={{\omega }_{sm}}+{{\delta }_{m}} & {} & \left( 3 \right) \\\end{matrix}$

The derivatives of θm may be expressed as

\begin{align} & \begin{matrix} \frac{d{{\theta }_{m}}}{dt}={{\omega }_{sm}}+\frac{d{{\delta }_{m}}}{dt} & {} & \left( 4 \right) \\\end{matrix} \\ & \begin{matrix} \frac{{{d}^{2}}{{\theta }_{m}}}{d{{t}^{2}}}=\frac{{{d}^{2}}{{\delta }_{m}}}{d{{t}^{2}}} & {} & \left( 5 \right) \\\end{matrix} \\\end{align}

Substituting Eq. 5 into Eq. 2 yields

$J\begin{matrix} \frac{{{d}^{2}}{{\delta }_{m}}}{d{{t}^{2}}}={{T}_{a}}={{T}_{m}}-{{T}_{e}} & {} & \left( 6 \right) \\\end{matrix}$

Multiplying Eq. 6 by the angular velocity of the rotor transforms the torque equation into a power equation. Thus,

$J{{\omega }_{m}}\begin{matrix} \frac{{{d}^{2}}{{\delta }_{m}}}{d{{t}^{2}}}={{\omega }_{m}}{{T}_{a}}={{\omega }_{m}}{{T}_{m}}-{{\omega }_{m}}{{T}_{e}} & {} & \left( 7 \right) \\\end{matrix}$

Replacing ${{\omega }_{m}}T$by P and $J{{\omega }_{m}}$ by M, the so-called swing equation is obtained. The swing equation describes how the machine rotor moves, or swings, with respect to the synchronously rotating reference frame in the presence of a disturbance, that is, when the net accelerating power is not zero.

$M\begin{matrix} \frac{{{d}^{2}}{{\delta }_{m}}}{d{{t}^{2}}}={{P}_{a}}={{P}_{m}}-{{P}_{e}} & {} & \left( 8 \right) \\\end{matrix}$

where

M = Jω = inertia constant

Pa = Pm– Pe = net accelerating power

Pm = ωTm = shaft power input corrected for the rotational losses

Pe = ωTe = electrical power output corrected for the electrical losses

It may be noted that the inertia constant was taken equal to the product of the moment of inertia J and the angular velocity ωm, which actually varies during a disturbance. Provided the machine does not lose synchronism, however, the variation in ωm is quite small. Thus, M is usually treated as a constant.

Another constant, which is often used because its range of values for particular types of rotating machines is quite narrow, is the so-called normalized inertia constant H. It is related to M as follows:

$\begin{matrix} H=\frac{1}{2}\frac{M{{\omega }_{sm}}}{{{S}_{rated}}}{}^{MJ}/{}_{MVA} & {} & \left( 9 \right) \\\end{matrix}$

Solving for M from Eq. 9 and substituting into 8 yields the swing equation expressed in per unit. Thus,

$\frac{2H}{{{\omega }_{sm}}}\begin{matrix} \frac{{{d}^{2}}{{\delta }_{m}}}{d{{t}^{2}}}=\frac{{{P}_{a}}}{{{S}_{rated}}}=\frac{{{P}_{m}}}{{{S}_{rated}}}-\frac{{{P}_{e}}}{{{S}_{rated}}} & {} & \left( 10 \right) \\\end{matrix}$

It may be noted that the angle δm and angular velocity ωm in Eq. 10 are expressed in mechanical radians and mechanical radians per second, respectively. For a synchronous generator with p poles, the electrical power angle and radian frequency are related to the corresponding mechanical variables as follows:

\begin{matrix} \begin{align} & \delta \left( t \right)=\frac{p}{2}{{\delta }_{m}}\left( t \right) \\ & \omega \left( t \right)=\frac{p}{2}{{\omega }_{m}}\left( t \right) \\\end{align} & {} & \left( 11 \right) \\\end{matrix}

Similarly, the synchronous electrical radian frequency is related to synchronous angular velocity as follows:

$\begin{matrix} {{\omega }_{s}}=\frac{p}{2}{{\omega }_{sm}} & {} & \left( 12 \right) \\\end{matrix}$

Therefore, the per-unit swing equation of Eq. 10 may be expressed in electrical units and takes the form of Eq. 13.

$\frac{2H}{{{\omega }_{s}}}\begin{matrix} \frac{{{d}^{2}}\delta }{d{{t}^{2}}}={{P}_{a}}={{P}_{m}}-{{P}_{e}} & {} & \left( 13 \right) \\\end{matrix}$

Depending on the unit of the angle δ, Eq. 13 takes the form of either Eq. 14 or Eq. 15. Thus, the per-unit swing equation takes the form:

$\frac{H}{\pi f}\begin{matrix} \frac{{{d}^{2}}\delta }{d{{t}^{2}}}={{P}_{a}}={{P}_{m}}-{{P}_{e}} & {} & \left( 14 \right) \\\end{matrix}$

When δ is in electrical degrees, or

$\frac{H}{180f}\begin{matrix} \frac{{{d}^{2}}\delta }{d{{t}^{2}}}={{P}_{a}}={{P}_{m}}-{{P}_{e}} & {} & \left( 15 \right) \\\end{matrix}$

When δ is in electrical degrees.

When a disturbance occurs, an unbalance in the power input and power output ensues, producing a net accelerating torque. The solution of the swing equation in the form of the differential equation of (14) or (15) is appropriately called the swing curve δ (t).

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