Power System Stability

The equal area criterion (EAC) is a graphical method to check the transient stability of a single machine connected to an infinite bus (or of two machines) without solving the swing equation step by step. It states that the system is stable if, after a disturbance, the area of accelerating energy under the power-angle curve can be balanced by an equal area of decelerating energy.

Basis

For a machine connected to an infinite bus, the swing equation (in pu, resistance and damping neglected) is

$$\frac{2H}{\omega_s}\frac{d^2\delta}{dt^2} = P_m - P_e = P_a, \qquad P_e = P_{max}\sin\delta = \frac{EV}{X}\sin\delta$$

Multiplying both sides by $2\dfrac{d\delta}{dt}$:

$$\frac{2H}{\omega_s}\cdot 2\frac{d\delta}{dt}\frac{d^2\delta}{dt^2} = 2(P_m - P_e)\frac{d\delta}{dt} \;\Rightarrow\; \frac{d}{dt}\left(\frac{d\delta}{dt}\right)^2 = \frac{\omega_s}{H}(P_m - P_e)\frac{d\delta}{dt}$$

Integrating from the initial angle $\delta_0$ (where $d\delta/dt = 0$, the rotor runs at synchronous speed):

$$\left(\frac{d\delta}{dt}\right)^2 = \frac{\omega_s}{H}\int_{\delta_0}^{\delta}(P_m - P_e)\,d\delta$$

For the machine to be stable, the rotor angle must stop increasing at some maximum angle, i.e. $d\delta/dt = 0$ at $\delta_{max}$:

$$\int_{\delta_0}^{\delta_{max}}(P_m - P_e)\,d\delta = 0$$

This means the area under the $P_a$–$\delta$ curve must be zero: the accelerating area $A_1$ (where $P_m > P_e$) must be equal to the decelerating area $A_2$ (where $P_e > P_m$):

$$A_1 = \int_{\delta_0}^{\delta_c}(P_m - P_e)\,d\delta = A_2 = \int_{\delta_c}^{\delta_{max}}(P_e - P_m)\,d\delta$$

Power-angle diagram

 Pe
  ^          Pmax1 sin d (pre-fault)
  |           .-''-.
  |         .' .--. '.    Pmax3 sin d
  |        / .' A2 '. \   (post-fault)
 Pm-------+-+-------+--\------
  |      /|A1|        \ \
  |     / |..|.....     \ \  Pmax2 sin d
  +----+--+--+--------+--+--> d
       0  d0 dc     dmax 180
 A1: accelerating area, A2: decelerating area

Applying it to a fault (sudden disturbance)

1. Before the fault: the machine runs at $\delta_0$, where $P_m = P_{max1}\sin\delta_0$.
2. During the fault: the transfer reactance rises, the power curve drops to $P_{max2}\sin\delta$ ($P_{max2} = 0$ for a fault at the generator bus). Since $P_m > P_e$, the rotor accelerates and $\delta$ increases. The energy gained is area $A_1$.
3. Fault cleared at $\delta_c$: the post-fault curve $P_{max3}\sin\delta$ applies. Now $P_e > P_m$, so the rotor decelerates, but $\delta$ keeps increasing until the kinetic energy is given back. This is area $A_2$.
4. Stability test: the system is stable if $A_2$ can become equal to $A_1$ before $\delta$ reaches $\delta_{max} = 180^\circ - \sin^{-1}(P_m/P_{max3})$. The rotor then swings back and oscillates about the new operating point. If the available $A_2 < A_1$, the machine loses synchronism.

Critical clearing angle

The largest clearing angle for which $A_1 = A_2$ just holds (with $\delta$ reaching exactly $\delta_{max}$) is the critical clearing angle:

$$\cos\delta_{cr} = \frac{P_m(\delta_{max} - \delta_0) + P_{max3}\cos\delta_{max} - P_{max2}\cos\delta_0}{P_{max3} - P_{max2}}$$

For a fault at the generator terminals with no change in network ($P_{max2} = 0$, $P_{max3} = P_{max}$, $\delta_{max} = \pi - \delta_0$):

$$\delta_{cr} = \cos^{-1}\left[(\pi - 2\delta_0)\sin\delta_0 - \cos\delta_0\right]$$

and, since $P_e = 0$ during the fault, the critical clearing time is

$$t_{cr} = \sqrt{\frac{4H(\delta_{cr} - \delta_0)}{\omega_s P_m}}$$

Assumptions and limitation

• Mechanical input $P_m$ constant during the transient; damping and resistance neglected.
• Machine represented by a constant voltage behind transient reactance.
• Valid only for one machine against an infinite bus (or two machines); for multi-machine systems, numerical solution of swing equations is needed.

Example: with $P_m = 1$ pu and $P_{max} = 2$ pu, $\delta_0 = 30^\circ$; for a terminal fault $\delta_{cr} = \cos^{-1}[(\pi - 1.047)(0.5) - 0.866] = \cos^{-1}(0.181) = 79.6^\circ$.

Transient stability depends on how far the rotor angle swings after a large disturbance. From the equal area criterion, it improves by (a) reducing the accelerating area, i.e. reducing the time or size of the power imbalance, or (b) increasing the decelerating area, i.e. raising the power transfer capability $P_{max} = EV/X$. The main techniques are:

Short notes on the main ones

1. Fast clearing: if the fault is cleared before the critical clearing time, the rotor gains less kinetic energy. Modern breakers clear in about 2-3 cycles.
2. Reducing reactance: series compensation or a second circuit raises $P_{max}$, so for the same $P_m$ the operating angle $\delta_0$ is smaller and the stability margin larger.
3. Single-pole reclosing: keeps two healthy phases in service, so power transfer during the dead time is not zero.
4. Fast excitation: quickly boosting field voltage raises $E$ and the electrical output, absorbing the excess mechanical power.

Power system stability is the ability of a power system to return to a normal (synchronous) operating condition after being subjected to a disturbance.

• Steady state stability is the ability of the system to remain in synchronism when subjected to small and gradual changes in load (or generation). The maximum power that can be transmitted through a point without loss of synchronism under such slow changes is the steady state stability limit. For a machine connected to an infinite bus through reactance $X$:

$$P_{max} = \frac{EV}{X} \quad (\text{at } \delta = 90^\circ)$$

• Transient stability is the ability of the system to remain in synchronism after a large and sudden disturbance such as a fault, loss of a line, sudden loss of large load or generation. It is studied over the first swing (about 1 s). The maximum power that can be transmitted while the system remains stable after a specified large disturbance is the transient stability limit. It is always lower than the steady state limit and depends on the type, location and clearing time of the disturbance.

Comparison

Methods of improving transient stability

The most effective in practice are high-speed fault clearing with auto-reclosing, single-pole switching and fast-acting excitation.

Data: $|E| = 1.2$ pu, $|V| = 1.0$ pu, $P_m = 1.0$ pu. Generator reactance j0.25, parallel lines j0.5 and j0.4, j0.05 to the infinite bus. The fault P is on the j0.4 line next to breaker 2, i.e. practically at the receiving-end bus.

1. Pre-fault (both lines in)

$$\begin{aligned} X_1 &= 0.25 + (0.5\|0.4) + 0.05 = 0.25 + 0.2222 + 0.05 = 0.5222\ \text{pu} \\ P_{max1} &= \frac{EV}{X_1} = \frac{1.2 \times 1}{0.5222} = 2.298\ \text{pu} \\ \delta_0 &= \sin^{-1}\frac{P_m}{P_{max1}} = \sin^{-1}\frac{1}{2.298} = 25.80^\circ = 0.4502\ \text{rad} \end{aligned}$$

2. During fault

The fault at P is at the receiving bus, which is the only path from the generator to the infinite bus. That bus is at zero voltage, so no power reaches the infinite bus:

$$P_{max2} = 0$$

3. Post-fault (j0.4 line removed by breakers 1 and 2)

$$\begin{aligned} X_3 &= 0.25 + 0.5 + 0.05 = 0.80\ \text{pu} \\ P_{max3} &= \frac{1.2 \times 1}{0.8} = 1.5\ \text{pu} \\ \delta_{max} &= 180^\circ - \sin^{-1}\frac{1}{1.5} = 180^\circ - 41.81^\circ = 138.19^\circ = 2.4119\ \text{rad} \end{aligned}$$

4. Equal area criterion

 Pe
  ^           post-fault 1.5 sin(d)
  |            .--''''--.
  |         .'    A2      '.
 1|....+---+---------------+.... Pm = 1.0
  |    |A1 |                \
  |    |   |                 \
  +----+---+------------------+--> d
      d0   dc               dmax
 During fault Pe = 0, so A1 = Pm(dc - d0)

Accelerating area $A_1 = P_m(\delta_c - \delta_0)$ must equal decelerating area $A_2 = \int_{\delta_c}^{\delta_{max}}(P_{max3}\sin\delta - P_m)\,d\delta$. This gives

$$\cos\delta_{cr} = \frac{P_m(\delta_{max} - \delta_0) + P_{max3}\cos\delta_{max} - P_{max2}\cos\delta_0}{P_{max3} - P_{max2}}$$

Substituting ($P_{max2} = 0$):

$$\begin{aligned} \cos\delta_{cr} &= \frac{1.0\,(2.4119 - 0.4502) + 1.5\cos138.19^\circ}{1.5} \\ &= \frac{1.9617 + 1.5(-0.7454)}{1.5} = \frac{1.9617 - 1.1180}{1.5} = 0.5624 \\ \delta_{cr} &= \cos^{-1}(0.5624) = 55.78^\circ \end{aligned}$$

Answer: critical clearing angle $\delta_{cr} \approx 55.8^\circ$ (initial angle 25.8°). Breakers 1 and 2 must open before the rotor swings beyond 55.8°, otherwise the machine loses synchronism.

Power system stability is the ability of a power system to return to a normal (synchronous) operating condition after being subjected to a disturbance.

• Steady state stability is the ability of the system to remain in synchronism when subjected to small and gradual changes in load (or generation). The maximum power that can be transmitted through a point without loss of synchronism under such slow changes is the steady state stability limit. For a machine connected to an infinite bus through reactance $X$:

$$P_{max} = \frac{EV}{X} \quad (\text{at } \delta = 90^\circ)$$

• Transient stability is the ability of the system to remain in synchronism after a large and sudden disturbance such as a fault, loss of a line, sudden loss of large load or generation. It is studied over the first swing (about 1 s). The maximum power that can be transmitted while the system remains stable after a specified large disturbance is the transient stability limit. It is always lower than the steady state limit and depends on the type, location and clearing time of the disturbance.

Limits for a machine on an infinite bus

 Pe
  ^      steady state limit EV/X
  |          .-''-.
  |        .'      '.
  |------.'---------'.---- transient limit
  |     /             \    (lower; depends on fault)
  |    /               \
  +---+--------+--------+--> delta
      0       90       180

• Steady state limit: $P_{ss} = EV/X$; operation is stable only where $dP_e/d\delta > 0$, i.e. $\delta < 90^\circ$.
• Transient limit: the largest $P_m$ for which, after the given fault and clearing time, the equal area condition $A_1 = A_2$ can still be met.

Techniques for enhancing transient stability

Power system stability is the ability of a power system to return to a normal (synchronous) operating condition after being subjected to a disturbance.

• Steady state stability is the ability of the system to remain in synchronism when subjected to small and gradual changes in load (or generation). The maximum power that can be transmitted through a point without loss of synchronism under such slow changes is the steady state stability limit. For a machine connected to an infinite bus through reactance $X$:

$$P_{max} = \frac{EV}{X} \quad (\text{at } \delta = 90^\circ)$$

• Transient stability is the ability of the system to remain in synchronism after a large and sudden disturbance such as a fault, loss of a line, sudden loss of large load or generation. It is studied over the first swing (about 1 s). The maximum power that can be transmitted while the system remains stable after a specified large disturbance is the transient stability limit. It is always lower than the steady state limit and depends on the type, location and clearing time of the disturbance.

The limits are shown on the power-angle curve $P_e = \frac{EV}{X}\sin\delta$: the steady-state limit is its peak; the transient limit is the largest $P_m$ for which the equal area condition ($A_1 = A_2$) can still be satisfied after the given disturbance.

Factors affecting transient stability

From the swing equation $\frac{2H}{\omega_s}\frac{d^2\delta}{dt^2} = P_m - P_{max}\sin\delta$, any factor that lowers $P_{max}$ during or after the fault, or keeps $P_m - P_e$ large for longer, makes the system less stable.

Power system stability is the ability of a power system to return to a normal (synchronous) operating condition after being subjected to a disturbance.

• Steady state stability is the ability of the system to remain in synchronism when subjected to small and gradual changes in load (or generation). The maximum power that can be transmitted through a point without loss of synchronism under such slow changes is the steady state stability limit. For a machine connected to an infinite bus through reactance $X$:

$$P_{max} = \frac{EV}{X} \quad (\text{at } \delta = 90^\circ)$$

• Transient stability is the ability of the system to remain in synchronism after a large and sudden disturbance such as a fault, loss of a line, sudden loss of large load or generation. It is studied over the first swing (about 1 s). The maximum power that can be transmitted while the system remains stable after a specified large disturbance is the transient stability limit. It is always lower than the steady state limit and depends on the type, location and clearing time of the disturbance.

Methods of improving transient stability

Factors affecting transient stability

Steady state stability is the ability of a power system to remain in synchronism (return to its operating point) after small and gradual changes in load or generation. Its limit, for a machine connected to an infinite bus, is $P_{max} = EV/X$ at $\delta = 90^\circ$.

Transient stability is the ability of a power system to remain in synchronism after a large and sudden disturbance such as a short circuit, switching out of a line, or sudden loss of a large generator or load. It is judged over the first rotor swing.

Methods of improving transient stability

1. High-speed fault clearing using fast relays and circuit breakers, reducing the accelerating area.
2. Reducing transfer reactance by parallel lines, bundled conductors and series capacitor compensation, raising $P_{max} = EV/X$.
3. Single-pole switching and high-speed auto-reclosing, so healthy phases keep transferring power.
4. Fast-acting excitation systems with power system stabilisers, raising the internal emf during the disturbance.
5. Fast valving of steam turbines to cut mechanical input quickly.
6. Braking resistors at the generator bus to absorb excess power during the fault.
7. Generator tripping or load shedding to restore the power balance.
8. Higher system voltage, intermediate switching stations, and FACTS devices (SVC, STATCOM, TCSC).

Definitions

• Power system stability: the ability of a power system to stay in a state of operating equilibrium under normal conditions and to regain an acceptable equilibrium after a disturbance, with all synchronous machines remaining in synchronism.
• Steady state stability: ability to remain in synchronism for small, gradual changes in load; limit $P_{max} = EV/X$ at $\delta = 90^\circ$, with stable operation where $dP_e/d\delta > 0$.
• Dynamic stability: ability to remain stable for small disturbances over a longer time (several seconds to minutes) when the effects of control systems (AVR, governor, PSS) are included; it concerns damping of low-frequency oscillations.
• Transient stability: ability to remain in synchronism after a large, sudden disturbance (fault, loss of a line, loss of a generator); judged over the first swing (about 1 s).
• Voltage stability: ability of the system to maintain acceptable voltages at all buses after a disturbance; it is lost when reactive power demand cannot be met, leading to voltage collapse.

Equal area criterion for parallel lines

System: generator ($E$, $X_d'$) feeding an infinite bus ($V$) through two parallel lines of reactance $X_L$ each. A 3-phase fault occurs on one line and is cleared by opening the breakers at both ends of that line.

          +---[CB]--- X_L ---[CB]---+
 E--Xd'--(1)                       (2)--- V (inf. bus)
          +---[CB]--- X_L -x-[CB]---+
                           F

Three power-angle curves:

$$\begin{aligned} \text{Pre-fault: } & P_{e1} = \frac{EV}{X_d' + X_L/2}\sin\delta = P_{max1}\sin\delta \\ \text{During fault: } & P_{e2} = \frac{EV}{X_{tr}}\sin\delta = P_{max2}\sin\delta \quad (X_{tr} \text{ from star-delta reduction; } P_{max2}=0 \text{ if F is at bus 1 or 2}) \\ \text{Post-fault: } & P_{e3} = \frac{EV}{X_d' + X_L}\sin\delta = P_{max3}\sin\delta \end{aligned}$$

with $P_{max1} > P_{max3} > P_{max2}$.

Procedure:

1. Initial angle: $\delta_0 = \sin^{-1}(P_m/P_{max1})$.
2. During the fault, $P_m > P_{e2}$; the rotor accelerates from $\delta_0$ to the clearing angle $\delta_c$. Accelerating area: $A_1 = \int_{\delta_0}^{\delta_c}(P_m - P_{max2}\sin\delta)\,d\delta$
3. After clearing, the operating point jumps to the post-fault curve; $P_{e3} > P_m$, the rotor decelerates. Decelerating area: $A_2 = \int_{\delta_c}^{\delta_1}(P_{max3}\sin\delta - P_m)\,d\delta$
4. The system is stable if $A_2 = A_1$ is reached at some $\delta_1 < \delta_{max} = \pi - \sin^{-1}(P_m/P_{max3})$.
5. Setting $A_1 = A_2$ with $\delta_1 = \delta_{max}$ gives the critical clearing angle:

$$\cos\delta_{cr} = \frac{P_m(\delta_{max}-\delta_0) + P_{max3}\cos\delta_{max} - P_{max2}\cos\delta_0}{P_{max3} - P_{max2}}$$

 Pe
  ^          Pmax1 sin d (pre-fault)
  |           .-''-.
  |         .' .--. '.    Pmax3 sin d
  |        / .' A2 '. \   (post-fault)
 Pm-------+-+-------+--\------
  |      /|A1|        \ \
  |     / |..|.....     \ \  Pmax2 sin d
  +----+--+--+--------+--+--> d
       0  d0 dc     dmax 180
 A1: accelerating area, A2: decelerating area

Having two parallel lines is itself a stability aid: after one line trips, the other still carries power ($P_{max3} > 0$), so a decelerating area exists.

Difference between the three types

Factors affecting transient stability

Transient stability is the ability of a system to remain in synchronism after a large sudden disturbance. From the swing equation $\frac{2H}{\omega_s}\frac{d^2\delta}{dt^2} = P_m - P_{max}\sin\delta$ and the equal area criterion, it depends on how much the rotor accelerates during the fault and how much decelerating power is available afterwards. The main factors are:

1. Pre-fault loading: heavier loading means a larger initial angle $\delta_0$ and less margin.
2. Type of fault: 3-phase faults are most severe, then LLG, LL and LG.
3. Location of fault: a fault near the generator reduces power transfer to almost zero.
4. Fault clearing time: the longer the fault stays, the larger the accelerating area; clearing must be faster than the critical clearing time.
5. Network reactance during and after the fault: higher post-fault reactance (e.g. one of two lines lost) lowers $P_{max}$.
6. Inertia constant $H$: a larger $H$ slows the rotor swing.
7. Generator internal voltage and excitation response: higher $E$ and fast excitation increase power transfer.
8. Reclosing scheme: fast and single-pole reclosing restores transfer capability quickly.

Stability of a power system is its ability to remain in synchronism (all machines running at the same electrical speed) under normal operation and to return to a stable operating state after a disturbance.

Steady state stability

Steady state stability is the ability of the system to remain in synchronism when subjected to small and gradual changes in load. For a generator connected to an infinite bus through reactance $X$:

$$P_e = \frac{EV}{X}\sin\delta = P_{max}\sin\delta$$

 Pe
  ^          Pmax
  |        .-''-.
  |  Pm  .'  |   '.
  |----a'----|----b'.--
  |   /|     |      \
  |  / |stable  unstable
  +-+--+-----+--------+--> d
    0  d0   90       180

• At point $a$ ($\delta_0 < 90^\circ$): if load increases slightly, $\delta$ increases and $P_e$ also increases, matching the load. The operation is stable.
• At point $b$ ($\delta > 90^\circ$): an increase in $\delta$ reduces $P_e$, the rotor accelerates further and synchronism is lost.
• If $P_m$ is raised slowly, $\delta$ increases until $\delta = 90^\circ$, where $P_e = P_{max}$. This maximum is the steady state stability limit:

$$P_{ss} = \frac{EV}{X}$$

Ways to raise it: increase $E$ (excitation) or $V$, reduce $X$ (parallel lines, series compensation).

Synchronizing power coefficient

Synchronizing power coefficient $P_s$ is the rate of change of electrical power output with rotor angle at the operating point:

$$P_s = \left.\frac{\partial P_e}{\partial\delta}\right|_{\delta_0} = \frac{EV}{X}\cos\delta_0 = P_{max}\cos\delta_0\ \ \text{(pu power/rad)}$$

It gives the change in power that tends to pull the rotor back to synchronism for a small change in rotor angle: $\Delta P_e = P_s\,\Delta\delta$. It is also called the stiffness of the coupling between the machine and the system.

Importance:

• Stability criterion: steady state stability requires $P_s > 0$, i.e. $0 < \delta_0 < 90^\circ$. At $\delta_0 = 90^\circ$, $P_s = 0$ (limit).
• Stability margin: a larger $P_s$ means a stiffer system and a larger margin.
• Natural frequency of oscillation: $f_n = \frac{1}{2\pi}\sqrt{P_s\pi f/H}$, used to study rotor oscillations and damping.
• Machines operated at light load (small $\delta_0$) have large $P_s$ and are more stable.

Synchronizing power coefficient $P_s$ is the rate of change of electrical power output with rotor angle at the operating point:

$$P_s = \left.\frac{\partial P_e}{\partial\delta}\right|_{\delta_0} = \frac{EV}{X}\cos\delta_0 = P_{max}\cos\delta_0\ \ \text{(pu power/rad)}$$

It gives the change in power that tends to pull the rotor back to synchronism for a small change in rotor angle: $\Delta P_e = P_s\,\Delta\delta$. It is also called the stiffness of the coupling between the machine and the system.

Derivation

For a machine connected to an infinite bus, $P_e = P_{max}\sin\delta$. If the rotor angle increases from $\delta_0$ by $\Delta\delta$:

$$\Delta P_e = P_{max}\sin(\delta_0 + \Delta\delta) - P_{max}\sin\delta_0 \approx (P_{max}\cos\delta_0)\Delta\delta = P_s\Delta\delta$$

With $P_m$ constant, an increase of $\Delta\delta$ makes $P_e > P_m$ when $P_s > 0$; the rotor decelerates and returns to $\delta_0$. So $P_s$ acts like the spring constant of the electrical coupling.

 Ps = Pmax cos(d0)
  ^
 Pmax.
  |    '.
  |      '.
  +--------'.------> d0
  0          90 deg (Ps = 0, limit)

Applications

1. Steady state stability check: the operating point is stable only if $P_s > 0$, i.e. $\delta_0 < 90^\circ$; the limit is $P_s = 0$.
2. Measure of stability margin/stiffness: larger $P_s$ (lower loading, lower reactance, higher excitation) means a more stable machine.
3. Natural frequency of rotor oscillation: from the linearised swing equation $M\frac{d^2\Delta\delta}{dt^2} + P_s\Delta\delta = 0$,

$$f_n = \frac{1}{2\pi}\sqrt{\frac{P_s}{M}}, \quad M = \frac{H}{\pi f}$$

used for small-signal (dynamic) stability studies and design of damper windings and power system stabilisers. 4. Parallel operation and synchronising of alternators: $P_s$ gives the synchronising power that keeps paralleled machines in step and shares load changes. 5. Hunting analysis: used to find the period of oscillation of a machine after a small disturbance.

Example: $P_{max} = 2$ pu, $P_m = 1$ pu → $\delta_0 = 30^\circ$, $P_s = 2\cos30^\circ = 1.732$ pu/rad.

Synchronizing power coefficient $P_s$ is the rate of change of electrical power output with rotor angle at the operating point:

$$P_s = \left.\frac{\partial P_e}{\partial\delta}\right|_{\delta_0} = \frac{EV}{X}\cos\delta_0 = P_{max}\cos\delta_0\ \ \text{(pu power/rad)}$$

It gives the change in power that tends to pull the rotor back to synchronism for a small change in rotor angle: $\Delta P_e = P_s\,\Delta\delta$. It is also called the stiffness of the coupling between the machine and the system.

Small-signal analysis

Let the machine operate at $\delta_0$ with $P_m = P_{max}\sin\delta_0$, and let the angle change by a small amount: $\delta = \delta_0 + \Delta\delta$. Then

$$P_e = P_{max}\sin(\delta_0+\Delta\delta) \approx P_{max}\sin\delta_0 + (P_{max}\cos\delta_0)\Delta\delta = P_m + P_s\Delta\delta$$

Substituting in the swing equation $M\dfrac{d^2\delta}{dt^2} = P_m - P_e$ (with $M = H/\pi f$):

$$M\frac{d^2\Delta\delta}{dt^2} + P_s\,\Delta\delta = 0$$

The characteristic equation is $Ms^2 + P_s = 0$, so

$$s = \pm\sqrt{-\frac{P_s}{M}}$$

• $P_s > 0$: roots are imaginary, $s = \pm j\omega_n$ with $\omega_n = \sqrt{P_s/M}$. $\Delta\delta$ oscillates with constant amplitude (and decays once damping is included). The rotor is pulled back: the operating point is stable.
• $P_s < 0$: one root is real and positive, $\Delta\delta$ grows exponentially; the machine falls out of step: unstable.
• $P_s = 0$ ($\delta_0 = 90^\circ$): the steady state stability limit.

The frequency of natural oscillation is

$$f_n = \frac{1}{2\pi}\sqrt{\frac{P_s}{M}} = \frac{1}{2\pi}\sqrt{\frac{\pi f P_s}{H}}\ \text{Hz}$$

Physical reasoning

 Pe
  ^         .-''-.
  |  Pm   .' Ps<0 '.
  |-----a'---------b.---
  |  Ps>0           \
  +-+---+-----+------+--> d
    0   d0    90    180

At point $a$ ($\delta_0 < 90^\circ$, $P_s > 0$) a small increase in $\delta$ raises $P_e$ above $P_m$, decelerating the rotor back. At point $b$ ($\delta > 90^\circ$, $P_s < 0$) the same increase lowers $P_e$, the rotor accelerates further and synchronism is lost. Hence a positive synchronizing power coefficient is necessary for a stable equilibrium.

The swing equation describes the relative motion of the rotor (load angle $\delta$) of a synchronous machine with respect to the synchronously rotating air-gap field when the mechanical and electrical powers are not balanced.

Derivation

Consider the rotor of a synchronous generator. Let

• $J$ = moment of inertia of rotor and turbine (kg·m²)
• $\theta_m$ = angular position of the rotor with respect to a stationary axis (mech. rad)
• $T_m$ = mechanical (shaft) torque, $T_e$ = electromagnetic torque (N·m)

Step 1: Newton's law of rotation. Accelerating torque = moment of inertia × angular acceleration:

$$J\frac{d^2\theta_m}{dt^2} = T_a = T_m - T_e$$

Step 2: Measure angle from a synchronously rotating frame. The rotor angle $\delta_m$ is the angle of the rotor relative to a reference axis rotating at synchronous speed $\omega_{sm}$:

$$\theta_m = \omega_{sm}t + \delta_m \;\Rightarrow\; \frac{d\theta_m}{dt} = \omega_{sm} + \frac{d\delta_m}{dt}, \quad \frac{d^2\theta_m}{dt^2} = \frac{d^2\delta_m}{dt^2}$$

so

$$J\frac{d^2\delta_m}{dt^2} = T_m - T_e$$

Step 3: Convert torque to power. Multiply by the rotor speed $\omega_m$ (power = torque × speed):

$$J\omega_m\frac{d^2\delta_m}{dt^2} = P_m - P_e$$

$M = J\omega_m$ is the angular momentum (inertia constant). Since the speed changes very little during a swing, $\omega_m \approx \omega_{sm}$ and $M$ is taken as constant:

$$M\frac{d^2\delta_m}{dt^2} = P_m - P_e$$

Step 4: Introduce the inertia constant $H$. $H$ is the stored kinetic energy at synchronous speed per MVA rating $S$:

$$H = \frac{\tfrac12 J\omega_{sm}^2}{S}\ \ \text{(MJ/MVA or s)} \;\Rightarrow\; M = J\omega_{sm} = \frac{2HS}{\omega_{sm}}$$

Substituting and dividing by $S$:

$$\frac{2H}{\omega_{sm}}\frac{d^2\delta_m}{dt^2} = \frac{P_m - P_e}{S} = P_m - P_e\ \ \text{(pu)}$$

Step 5: Use electrical angle. For a machine with $p$ poles, $\delta = \frac{p}{2}\delta_m$ and $\omega_s = \frac{p}{2}\omega_{sm}$, so $\delta/\omega_s = \delta_m/\omega_{sm}$:

$$\boxed{\frac{2H}{\omega_s}\frac{d^2\delta}{dt^2} = P_m - P_e} \qquad \text{or} \qquad \frac{H}{\pi f}\frac{d^2\delta}{dt^2} = P_m - P_e$$

with $\delta$ in electrical radians and $\omega_s = 2\pi f$. If $\delta$ is in electrical degrees, $\dfrac{H}{180f}\dfrac{d^2\delta}{dt^2} = P_m - P_e$.

For a machine connected to an infinite bus through reactance $X$, $P_e = \dfrac{EV}{X}\sin\delta = P_{max}\sin\delta$, so

$$\frac{H}{\pi f}\frac{d^2\delta}{dt^2} = P_m - P_{max}\sin\delta$$

This is the swing equation. It is a second-order non-linear differential equation; its solution $\delta(t)$ describes how the rotor swings after a disturbance. If damping is included, a term $D\frac{d\delta}{dt}$ is added on the left.

The swing equation describes how the rotor angle $\delta$ of a synchronous machine changes when the mechanical input and electrical output are not balanced.

Derivation

Consider the rotor of a synchronous generator. Let

• $J$ = moment of inertia of rotor and turbine (kg·m²)
• $\theta_m$ = angular position of the rotor with respect to a stationary axis (mech. rad)
• $T_m$ = mechanical (shaft) torque, $T_e$ = electromagnetic torque (N·m)

Step 1: Newton's law of rotation. Accelerating torque = moment of inertia × angular acceleration:

$$J\frac{d^2\theta_m}{dt^2} = T_a = T_m - T_e$$

Step 2: Measure angle from a synchronously rotating frame. The rotor angle $\delta_m$ is the angle of the rotor relative to a reference axis rotating at synchronous speed $\omega_{sm}$:

$$\theta_m = \omega_{sm}t + \delta_m \;\Rightarrow\; \frac{d\theta_m}{dt} = \omega_{sm} + \frac{d\delta_m}{dt}, \quad \frac{d^2\theta_m}{dt^2} = \frac{d^2\delta_m}{dt^2}$$

so

$$J\frac{d^2\delta_m}{dt^2} = T_m - T_e$$

Step 3: Convert torque to power. Multiply by the rotor speed $\omega_m$ (power = torque × speed):

$$J\omega_m\frac{d^2\delta_m}{dt^2} = P_m - P_e$$

$M = J\omega_m$ is the angular momentum (inertia constant). Since the speed changes very little during a swing, $\omega_m \approx \omega_{sm}$ and $M$ is taken as constant:

$$M\frac{d^2\delta_m}{dt^2} = P_m - P_e$$

Step 4: Introduce the inertia constant $H$. $H$ is the stored kinetic energy at synchronous speed per MVA rating $S$:

$$H = \frac{\tfrac12 J\omega_{sm}^2}{S}\ \ \text{(MJ/MVA or s)} \;\Rightarrow\; M = J\omega_{sm} = \frac{2HS}{\omega_{sm}}$$

Substituting and dividing by $S$:

$$\frac{2H}{\omega_{sm}}\frac{d^2\delta_m}{dt^2} = \frac{P_m - P_e}{S} = P_m - P_e\ \ \text{(pu)}$$

Step 5: Use electrical angle. For a machine with $p$ poles, $\delta = \frac{p}{2}\delta_m$ and $\omega_s = \frac{p}{2}\omega_{sm}$, so $\delta/\omega_s = \delta_m/\omega_{sm}$:

$$\boxed{\frac{2H}{\omega_s}\frac{d^2\delta}{dt^2} = P_m - P_e} \qquad \text{or} \qquad \frac{H}{\pi f}\frac{d^2\delta}{dt^2} = P_m - P_e$$

with $\delta$ in electrical radians and $\omega_s = 2\pi f$. If $\delta$ is in electrical degrees, $\dfrac{H}{180f}\dfrac{d^2\delta}{dt^2} = P_m - P_e$.

For a machine connected to an infinite bus through reactance $X$, $P_e = \dfrac{EV}{X}\sin\delta = P_{max}\sin\delta$, so

$$\frac{H}{\pi f}\frac{d^2\delta}{dt^2} = P_m - P_{max}\sin\delta$$

This is the swing equation. It is a second-order non-linear differential equation; its solution $\delta(t)$ describes how the rotor swings after a disturbance. If damping is included, a term $D\frac{d\delta}{dt}$ is added on the left.

Importance of the inertia constant $H$

$$H = \frac{\text{stored kinetic energy at synchronous speed (MJ)}}{\text{machine rating (MVA)}}$$

1. Rate of rotor swing: from $\frac{d^2\delta}{dt^2} = \frac{\pi f}{H}(P_m - P_e)$, a larger $H$ gives a smaller acceleration for the same power imbalance, so $\delta$ rises more slowly after a fault.
2. Critical clearing time: for a terminal fault $t_{cr} = \sqrt{\frac{4H(\delta_{cr}-\delta_0)}{\omega_s P_m}}$, so $t_{cr} \propto \sqrt H$. Higher $H$ allows more time for breakers to clear the fault.
3. Narrow range: $H$ lies in a narrow range for each machine type (about 2-9 s), e.g. turbo-alternators 4-9 s, hydro generators 2-4 s, so typical values can be assumed when data are missing.
4. Equivalent machine: machines swinging together can be combined: $H_{eq} = \sum H_i S_i / S_{base}$.
5. Frequency response: $H$ also decides how fast the system frequency falls after loss of generation.

Classification of power system stability

             Power system stability
       +----------------+-----------------+
  Rotor angle       Frequency          Voltage
  stability         stability          stability
   +------+                         +--------+
 Small-  Transient            Large-dist. Small-dist.
 signal  (large-dist.)
 (steady state / dynamic)

1. Rotor angle stability: ability of synchronous machines to remain in synchronism.
   • Steady state (small-signal) stability: small, gradual changes; limit $EV/X$.
   • Dynamic stability: small disturbances with controllers (AVR, governor, PSS) acting; concerns damping of oscillations over several seconds.
   • Transient stability: large sudden disturbances such as faults; first swing, about 1 s.
2. Voltage stability: ability to keep acceptable voltages at all buses; lost through lack of reactive power, leading to voltage collapse.
3. Frequency stability: ability to maintain frequency after a large imbalance between generation and load.

Derivation

Consider the rotor of a synchronous generator. Let

• $J$ = moment of inertia of rotor and turbine (kg·m²)
• $\theta_m$ = angular position of the rotor with respect to a stationary axis (mech. rad)
• $T_m$ = mechanical (shaft) torque, $T_e$ = electromagnetic torque (N·m)

Step 1: Newton's law of rotation. Accelerating torque = moment of inertia × angular acceleration:

$$J\frac{d^2\theta_m}{dt^2} = T_a = T_m - T_e$$

Step 2: Measure angle from a synchronously rotating frame. The rotor angle $\delta_m$ is the angle of the rotor relative to a reference axis rotating at synchronous speed $\omega_{sm}$:

$$\theta_m = \omega_{sm}t + \delta_m \;\Rightarrow\; \frac{d\theta_m}{dt} = \omega_{sm} + \frac{d\delta_m}{dt}, \quad \frac{d^2\theta_m}{dt^2} = \frac{d^2\delta_m}{dt^2}$$

so

$$J\frac{d^2\delta_m}{dt^2} = T_m - T_e$$

Step 3: Convert torque to power. Multiply by the rotor speed $\omega_m$ (power = torque × speed):

$$J\omega_m\frac{d^2\delta_m}{dt^2} = P_m - P_e$$

$M = J\omega_m$ is the angular momentum (inertia constant). Since the speed changes very little during a swing, $\omega_m \approx \omega_{sm}$ and $M$ is taken as constant:

$$M\frac{d^2\delta_m}{dt^2} = P_m - P_e$$

Step 4: Introduce the inertia constant $H$. $H$ is the stored kinetic energy at synchronous speed per MVA rating $S$:

$$H = \frac{\tfrac12 J\omega_{sm}^2}{S}\ \ \text{(MJ/MVA or s)} \;\Rightarrow\; M = J\omega_{sm} = \frac{2HS}{\omega_{sm}}$$

Substituting and dividing by $S$:

$$\frac{2H}{\omega_{sm}}\frac{d^2\delta_m}{dt^2} = \frac{P_m - P_e}{S} = P_m - P_e\ \ \text{(pu)}$$

Step 5: Use electrical angle. For a machine with $p$ poles, $\delta = \frac{p}{2}\delta_m$ and $\omega_s = \frac{p}{2}\omega_{sm}$, so $\delta/\omega_s = \delta_m/\omega_{sm}$:

$$\boxed{\frac{2H}{\omega_s}\frac{d^2\delta}{dt^2} = P_m - P_e} \qquad \text{or} \qquad \frac{H}{\pi f}\frac{d^2\delta}{dt^2} = P_m - P_e$$

with $\delta$ in electrical radians and $\omega_s = 2\pi f$. If $\delta$ is in electrical degrees, $\dfrac{H}{180f}\dfrac{d^2\delta}{dt^2} = P_m - P_e$.

For a machine connected to an infinite bus through reactance $X$, $P_e = \dfrac{EV}{X}\sin\delta = P_{max}\sin\delta$, so

$$\frac{H}{\pi f}\frac{d^2\delta}{dt^2} = P_m - P_{max}\sin\delta$$

This is the swing equation. It is a second-order non-linear differential equation; its solution $\delta(t)$ describes how the rotor swings after a disturbance. If damping is included, a term $D\frac{d\delta}{dt}$ is added on the left.