2. Dynamic State
• Dynamic phenomena have different physical
origins and they occur in different time scales
• A system is in a dynamic state if the time
derivative of any system variable is not zero.
• To describe a dynamic system mathematically
differential equations must be employed
Introduction to Power System Stability
3. Introduction
Exists different dynamic phenomena with
different characteristics that can be initiated
by:
– Local phenomena: minor part or single component
involve interactions between different parts of the system
– causing fault with subsequent change in network topology
» they can cause system instabilities
» can lead to black outs in large parts of the system
‒ Controller actions or switching of lines or other
components by system operators.
‒ Normal- not endanger the stability of the system
4. Classification of power system dynamics
Based on their character dynamic phenomena
in power systems are usually classified as:
• Wave dynamics
• Electromagnetic dynamics
• Electromechanical dynamics
• Wave dynamics (from microseconds to
milliseconds): the propagation of
electromagnetic waves caused by lightning
strikes or switching operations (The fastest
dynamics).
5. Classification…
Electromagnetic dynamics:
Takes place in the machine windings following a
disturbance, operation of the protection system
or the interaction between the electrical
machines and the network (milliseconds to a
second)
Electromechanical dynamics:
This is due to the oscillation of the rotating
masses of the generators and motors that occur
following a disturbance, operation of the
protection system and voltage and prime mover
control (from seconds to several seconds).
6. Time frame of the basic power system
dynamic phenomena
7. • Wave dynamics occur in the network and
basically do not propagate beyond the
transformer windings.
• The electromagnetic phenomena mainly
involve the generator armature and damper
windings and partly the network.
• electromechanical phenomena, namely the
rotor oscillations and accompanying network
power swings, mainly involve the rotor field
and damper windings and the rotor inertia.
Occurrences of power system dynamics
8. Power System Stability Definition
IEEE/CIGRE Joint Task Force on Stability Terms and Definitions
Prabha Kundur, John Paserba, Venkat Ajjarapu, et.al.
• Power system stability is the ability of an
electric power system, for a given initial
operating condition, to regain a state of
operating equilibrium after being subjected to
a physical disturbance, with most system
variables bounded so that practically the entire
system remains intact.
9. Elaborating remarks on the definition
• Remark 1. It is not necessary that the system
regains the same steady state operating
equilibrium as prior to the disturbance.
• Remark2. It is important that the final steady
state operating equilibrium after the fault is
an acceptable steady state.
10. Classification of power system
stability
To achieve a better overview and structure
of stability analyses of power systems.
The classification here is based on the
physical mechanism responsible for the
instability.
• Angle stability
• Voltage stability
• Frequency stability
11.
12. Voltage stability
• Voltage stability is the ability of a power
system to maintain steady acceptable voltages
at all buses in the system under normal
operating conditions and after being
subjected to a disturbance.
• Large disturbance voltage stability
• Small disturbance voltage stability
13. Frequency stability
• It refers to the ability of a power system to
maintain steady frequency following a severe
disturbance between generation and load.
• If the total power fed into the system by the
prime movers is less than what is consumed b
the loads, including losses, this imbalance wil
influence the frequency of the whole system.
– the imbalance is not local but global.
15. The swing equation
The equation governing the motion of the
rotor of a synchronous machine
Based on the fundamental principle in
dynamics the accelerating torque is the
product of the moment of inertia and angular
acceleration.
𝐽
𝑑2
𝜃 𝑚
𝑑𝑡2
= 𝑇𝑎
16. Derivation of the swing equation
𝐽
𝑑2
𝜃 𝑚
𝑑𝑡2
= 𝑇𝑎 = 𝑇 𝑚 − 𝑇𝑒
Where:
J = The total moment of inertia of the synchronous machine (kgm2)
𝜽 𝒎= is the mechanical angle of the rotor field axis with respect to the
stator reference or fixed reference frame (rad) .
17. Derivation of ….cont’d
𝑻 𝒎= Mechanical torque from the turbine (Nm). Positive Tm
corresponds to mechanical torque fed into the
machine, i.e. normal operation as a generator in steady state.
𝑻 𝒆= Electrical torque on the rotor (in Nm). Positive Te in normal
operation as generator.
• As the rotor is continuously rotating at synchronous speed in steady
state 𝜽 𝒎 will also be continuously varying with respect to time.
• To make the angle 𝜽 𝒎 constant in steady state we can measure this
angle with respect to a synchronously rotating reference instead of a
stationary reference.
𝜃 𝑚 = 𝛿 𝑚 + 𝜔 𝑚𝑠t
18. Where, 𝛿 𝑚 is the angle between the rotor field axis and the reference axis
rotating synchronously at rps, 𝜔 𝑚𝑠. If we differentiate the above equation
with respect to time we get
𝑑𝜃 𝑚
𝑑𝑡
=
𝑑𝛿 𝑚
𝑑𝑡
+ 𝜔 𝑚𝑠
𝑑2
𝜃 𝑚
𝑑𝑡2
=
𝑑2
𝛿 𝑚
𝑑𝑡2
And
But, the rate of change of the rotor mechanical angle 𝜃 𝑚 with respect to
time is the speed of the rotor. Hence,
𝜔 𝑚 =
𝑑𝜃 𝑚
𝑑𝑡
By substitution:
𝑑𝛿 𝑚
𝑑𝑡
= 𝜔 𝑚 − 𝜔 𝑚𝑠
Derivation of ….cont’d
𝜃 𝑚 = 𝛿 𝑚 + 𝜔 𝑚𝑠t
19. 𝐽
𝑑2 𝛿 𝑚
𝑑𝑡2
= 𝑇 𝑚 − 𝑇𝑒
Multiplying both sides of this equation by the mechanical angular velocity
m, we have
If the angular acceleration is expressed in terms of electrical angles:
Therefore, again by substitution:
𝜔 𝑚 𝐽
𝑑2
𝜃 𝑚
𝑑𝑡2
= 𝑃𝑚 − 𝑃𝑒
2
𝑝
𝜔 𝑚 𝐽
𝑑2
𝜃𝑒
𝑑𝑡2
= 𝑃𝑚 − 𝑃𝑒
4
𝑝𝜔 𝑚
(
1
2
𝜔 𝑚
2 𝐽)
𝑑2
𝜃𝑒
𝑑𝑡2
= 𝑃𝑚 − 𝑃𝑒
20. Dividing both sides by the rating of the machine 𝑆 𝑟𝑎𝑡𝑒𝑑
Derivation ...cont’d
4
𝑝𝜔 𝑚
(
1
2
𝜔 𝑚
2
𝐽)
𝑆𝑟𝑎𝑡𝑒𝑑
𝑑2
𝜃𝑒
𝑑𝑡2
=
𝑃𝑚 − 𝑃𝑒
𝑆𝑟𝑎𝑡𝑒𝑑
2
𝜔 𝑒
(
1
2
𝜔 𝑚
2 𝐽)
𝑆 𝑟𝑎𝑡𝑒𝑑
𝑑2 𝜃𝑒
𝑑𝑡2
=
𝑃𝑚 − 𝑃𝑒
𝑆𝑟𝑎𝑡𝑒𝑑
2𝐻
𝜔0𝑒
𝑑2
𝜃𝑒
𝑑𝑡2
= 𝑃𝑚
𝑝𝑢
− 𝑃𝑒
𝑝𝑢
where superscript pu indicates that the mechanical and
electrical powers are expressed in p.u. on the rating of
the machine.
21. (
1
2
𝜔 𝑚
2 𝐽)
𝑆𝑟𝑎𝑡𝑒𝑑
= 𝐻
2𝐻
𝜔0𝑒
𝑑2
𝜃𝑒
𝑑𝑡2
= 𝑃𝑚
𝑝𝑢
− 𝑃𝑒
𝑝𝑢
Where, H is inertia constant
of the synchronous machine
22. During disturbances the angular velocity of the rotor
will not deviate significantly from the nominal values
→ 𝜔0𝑒 ≈ 𝜔0
2𝐻
𝜔0
𝑑2 𝜃
𝑑𝑡2
= 𝑃𝑚 − 𝑃𝑒
The above equation for a system with an
electrical frequency f,Hz can be written as:
In electrical radians
𝐻
𝜋𝑓
𝑑2 𝜃
𝑑𝑡2
= 𝑃𝑚 − 𝑃𝑒
𝐻
180𝑓
𝑑2
𝜃
𝑑𝑡2
= 𝑃𝑚 − 𝑃𝑒 In mechanical degrees
23. The Swing Equation for the Synchronous Machine
(Nonlinear - Second order diff equation)
2𝐻
𝜔0
𝑑2
𝜃
𝑑𝑡2
= 𝑃𝑚 − 𝑃𝑒
This equation will give an insight into the fundamental
relations governing the dynamics during rotor
oscillations.
Physical meaning:
The difference between mechanical power fed into the
machine and the electrical output power will cause a
motion of the rotor relative to a rotation with constant
angular velocity 𝜔0.
24. Swing Curve
• When the swing equation is solved we obtain
the expression for 𝜃as a function of time, 𝜃(t).
• A graph of the solution is called the swing
curve of the machine.
– inspection of the swing curves of all the machines
of the system will show whether the machines
remain in synchronism after a disturbance
(Examples in a separate slide)
25. Analysis of the swing equation
• The parameters included in this equation and their
influence will give an insight into the fundamental relations
governing the dynamics during rotor oscillations.
• The difference between mechanical power fed into the
machine and the electrical output power will cause a
motion of the rotor relative to a rotation with constant
angular velocity 𝜔0.
2𝐻
𝜔0
𝑑2
𝜃
𝑑𝑡2
= 𝑃𝑚 − 𝑃𝑒
26. With fairly good accuracy the electrical power can thus be
written as:
with Ps being the synchronizing power and Pd being the
damping power.
The swing equation given above is an ordinary differential equation (ODE) of
the second order.
ODE of higher order can be written as a system of first order ODE, which is
more practical.
By introducing
𝜔 =
𝑑𝜃
𝑑𝑡
; In most cases it is not possible to solve the
equation analytically and one has to use
numerical integration techniques for solution.
Numerical integration requires usually that the
ODE is in first order.
2𝐻
𝜔0
𝑑2
𝜃
𝑑𝑡2
= 𝑃𝑚 − 𝑃𝑒
27. Can be written as
• The angular velocity introduced in swing equation
above denotes the frequency with which the rotor
oscillates relative to a system rotating with the
synchronous and constant angular velocity 0.
𝜃
𝜔
=
𝜔
𝜔 𝑂
2𝐻
(𝑃𝑚 − 𝑃𝑒
28. Power swings in a simple system
In large and complicated systems it is often hard
to distinguish the fundamental phenomena from
the more irrelevant ones. It is therefore of
importance to study simple systems to get an
insight into and understanding of the basics, that
can be used in the analysis of more complex
systems.
29. • Now the swing equation can be written as:
ttanconsPP mm 0
This is the swing equation in final form for the
simplified system including the assumptions made.
32. • The following conclusions can be deduced
from the figure with regard to the equilibrium
points:
• If Pm0< Pe,max there are two equilibrium points,
i.e. 𝜃0and π − 𝜃0 for 0 ≤ θ ≤ π.
• If Pm0 = Pe,max there is exactly one equilibrium
point 𝜃0 = π/2 for 0 ≤ θ ≤ π.
• If Pm0> Pe,max there is no equilibrium point.
33. • We have seen that swing equation is a non linear
function of the power angle.
• There is no formal solution possible since swing
equation is nonlinear dif. eq. and therefore
numerical techniques must be developed to solve
the SW eq.
34. Solution of Swing Equation
Point-by-point method
(Reading?)
The point-by-point solution of
swing equation consists of two
processes which are carried out
alternatively
E.W. Kimbark, Power system stability, V1, John Wiley & Sons, 1995
35. 1. The first process is the computation
of the angular positions and angular
speeds at the end of each interval
from the knowledge of the angular
positions and speeds at the
beginning of the time interval and
the accelerating power assumed for
the interval.
36. 2.The second process is the computation of
accelerating power of each machine from
the angular positions of all the machines
of the system . The second process
requires the knowledge of network
solution.
37. In the first method it is assumed
that the accelerating power is
constant through out the time
interval ∆t and has the value
computed for the beginning of the
interval