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POWER SYSTEM STABILITY
LESSON SUMMARY-1:-
- Introduction
- Classification of Power System Stability
- Dynamic Equation of Synchronous Machine
Power system stability involves the study of the dynamics of the power system under disturbances. Power system stability implies that its ability to return to normal or stable operation after having been subjected to some form of disturbances.
From the classical point of view power system instability can be seen as loss of synchronism (i.e., some synchronous machines going out of step) when the system is subjected to a particular disturbance. Three type of stability are of concern: Steady state, transient and dynamic stability.
Steady-state Stability : -
Steady-state stability relates to the response of synchronous machine to a gradually increasing load. It is basically concerned with the determination of the upper limit of machine loading without losing synchronism, provided the loading is increased gradually.
Dynamic Stability : -
Dynamic stability involves the response to small disturbances that occur on the system, producing oscillations. The system is said to be dynamically stable if theses oscillations do not acquire more than certain amplitude and die out quickly. If these oscillations continuously grow in amplitude, the system is dynamically unstable. The source of this type of instability is usually an interconnection between control systems.
Transient Stability : -
Transient stability involves the response to large disturbances, which may cause rather large changes in rotor speeds, power angles and power transfers. Transient stability is a fast phenomenon usually evident within a few second.
Power system stability mainly concerned with rotor stability analysis. For this various assumptions needed such as:
For stability analysis balanced three phase system and balanced disturbances are considered. Deviations of machine frequencies from synchronous frequency are small. During short circuit in generator, dc offset and high frequency current are present. But for analysis of stability, theses are neglected. Network and impedance loads are at steady state. Hence voltages, currents and powers can be computed from power flow equation.
Dynamics of a Synchronous Machine :-
The kinetic energy of the rotor in synchronous machine is given as:
KE=1/2Jws^2 x10 -6^ MJoule……………………...………...… (1)
Where J= rotor moment of inertia in kg-m^2
ws = synchronous speed in mechanical radian/sec.
Speed in electrical radian is
wse = (P/2) ws = rotor speed in electrical radian/sec…….....….(2)
Where P = no. of machine poles
From equation (1) and (2) we get
KE= MJ………………..……….. (3)
or KE= MJ
Where M = = moment of inertia in
MJ.sec/elect. radian……..... (4)
We shall define the inertia constant H, such that
GH = KE = MJ…………………….…….…………. (5)
Where G = three-phase MVA rating (base) of machine
H = inertia constant in MJ/MVA or MW.sec/MVA
Te = Ti ……………………………………….. (10)
Here we have neglected any retarding torque due to rotational losses. Therefore we have
Te ws= Ti ws……………………………………….. (11)
And Te ws - Ti ws = Pi - Pe = 0……………...…………………… (12)
When a change in load or a fault occurs, then input power Pi is not equal to Pe. Therefore left side of equation is not zero and an accelerating torque comes into play. If Pa is the accelerating (or decelerating) power, then
Pi- Pe = …………………...………… (13)
Where D = damping coefficient
θe = electrical angular position of the rotor
It is more convenient to measure the angular position of the rotor with respect to a synchronously rotating frame of reference. Let
δ = θe -ws.t ……………………………..……………… (14)
So …………………………...……………… (15)
Where δ is power angle of synchronous machine.
δ
θe
Ws
Rotor Field
Reference rotatingaxis
Reference axis
(Fig.2 Angular Position of rotor with respect to reference axis)
Neglecting damping (i.e., D = 0) and substituting equation (15) in equation (13) we get
MW……………………………….. (16)
Using equation (6) and (16), we get
MW……………………………… (17)
Dividing throughout by G, the MVA rating of the machine,
pu………………………….... (18)
Where …………………………………………. (19)
or pu………………….…………… (20)
Equation (20) is called Swing Equation. It describes the rotor dynamics for a synchronous machine. Damping must be considered in dynamic stability study.
Multi Machine System:-
In a multi machine system a common base must be selected. Let
Gmachine = machine rating (base)
Gsystem = system base
Equation (20) can be written as:
………………… (21)
So pu on system base…………...…… (22)
Where ………………………….(23)
= machine inertia constant in system base
Machines Swinging in Unison (Coherently) :-
Let us consider the swing equations of two machines on a common system base, i.e.,
………………………...……. (24)
Now
= 10
= = 108 elect.deg./sec^2 So, α = acceleration = 108 elect.deg./sec^2 (c) 12 cycles = 12/60 = 0.2sec.
Change in δ = ½ α.(Δt)^2 = ½.108.(0.2)^2 =2.16 elect.deg
Now α = 108 elect.deg./sec^2 = 60 x (108/360ᵒ) rpm/sec = 18 rpm/sec
Hence rotor speed at the end of 12 cycles
=
= ( ) rpm
= 1803.6 rpm.
(d) Heq = =32MJ/MVA
LESSON SUMMARY-3:-
- Power flow under steady state
- Steady-state Stability
- Examples
Power Flow under Steady State:-
Consider a short transmission line with negligible resistance.
VS = per phase sending end voltage VR = per phase receiving end voltage Vs leads VR by an angle δ x = reactance of per transmission line
I + (^) jx +
- -
Vs VR
(Fig.3-A short transmission line)
On the per phase basis power on sending end,
SS = PS + j QS =VSI*…………………….………….. (29)
From Fig.3 I is given as
I =
or I* = ………….………………………. (30)
From equation (29) and (30), we get
SS = …………………………………. (31)
Now VR =|VR| so, VR = VR*^ = |VR|
VS =|VS| =|VS|
Equation (31) becomes
SS = PS + jQS = δ δ
So PS = δ……………………...…………….. (32)
and Qs= ………………...………...…..………(33)
Similarly, at the receiving end we have
SR = PR + j QR = VRI*………………………………. (34)
Proceeding as above we finally obtain
QR = .................................................. (35)
PR = ……………………………..…….. (36)
Therefore for lossless transmission line,
PS = PR = ………………………...…….. (37)
In a similar manner, the equation for steady-state power delivered by a lossless synchronous machine is given by
The system stability to small changes is determined from the characteristic equation
Mp^2 + = 0…………………………….. (45)
Where two roots are p = …………………………….. (46)
As long as is positive, the roots are purely imaginary and conjugate
and system behavior is oscillatory about. Line resistance and damper windings of machine cause the system oscillations to decay. The system is therefore stable
for a small increment in power so long as (^) δ
When (^) δ is negative, the roots are real, one positive and the other
negative but of equal magnitude. The torque angle therefore increases without bound upon occurrence of a small power increment and the synchronism is soon
lost. The system is therefore unstable for (^) δ
is known as synchronizing coefficient. This is also called stiffness of
synchronous machine. It is denoted as Sp. This coefficient is given by
δ ………………………… (47)
If we include damping term in swing equation then equation (43) becomes
δ
or δ
or δ
or δ …………………….. (48)
Where and …………………………… (49)
So damped frequency of oscillation, ……………..……….. (50)
And Time Constant, T = = …………………………….. (51)
Example2:-
Find the maximum steady-state power capability of a system consisting of a generator equivalent reactance of 0.4pu connected to an infinite bus through a series reactance of 1.0 p.u. The terminal voltage of the generator is held at1.10 p.u. and the voltage of the infinite bus is 1.0 p.u.
Solution:-
Equivalent circuit of the system is shown in Fig.4.
(Fig.4 Equivalent circuit of example2)
δ ……………………………….. (i)
I = …………………………….. (ii)
Using equation (i) and (ii)
δ θ
δ θ θ θ
δ θ θ …………….. (iii)
As shown in Fig.5, in an unstable system, δ increases indefinitely with time and machine looses synchronism. In a stable system, δ undergoes oscillations, which eventually die out due to damping. From Fig.4, it is clear that, for a system to be
stable, it must be that = 0 at some instant. This criterion ( = 0) can simply be
obtained from equation (52).
(Fig. 5 A plot of δ (t))
Multiplying equation (52) by , we have
This upon integration with respect to time gives
dδ ………………………...………. (54)
Where = accelerating power and δ is the initial power angle
before the rotor begins to swing because of a disturbance. The stability ( = 0)
criterion implies that
dδ ………………………………..……… (55)
For stability, the area under the graph of accelerating power versus δ must be zero for some value of δ; i.e., the positive (accelerating) area under the graph must be equal to the negative (decelerating) area. This criterion is therefore know as the equal area criterion for stability and is shown in Fig. 6.
(Fig.6 Power angle characteristic)
Application to sudden change in power input:-
In Fig. 6 point ‘a’ corresponding to the δ is the initial steady-state operating point. At this point, the input power to the machine, , where is the developed power. When a sudden increase in shaft input power occurs to , the accelerating power , becomes positive and the rotor moves toward point ‘b’
We have assumed that the machine is connected to a large power system so that |Vt| does not change and also xd does not change and that a constant field current maintains |Eg|. Consequently, the rotor accelerates and power angle begins to
increase. At point and δ =δ 1. But is still positive and δ overshoots ‘b’,
the final steady-state operating point. Now is negative and δ ultimately reaches a maximum value δ 2 or point ‘c’ and swing back towards point ‘b’. Therefore the rotor settles back to point ‘b’, which is ultimate steady-state operating point.
In accordance with equation (55) for stability, equal area criterion requires
Area A 1 = Area A 2
or δ dδ δ dδ ………….… (56)
or
…………………………. (57)
But
Which when substituted in equation (57), we get
9 MW per phase = 3x313.42 = 940.3 MW (3-φ) of input shaft power.
LESSON SUMMARY-5:-
- Critical clearing angle and critical clearing time
- Application of equal area criterion a) Sudden loss of one parallel line
Critical Clearing Angle and Critical Clearing Time:-
If a fault occurs in a system, δ begins to increase under the influence of positive accelerating power, and the system will become unstable if δ becomes very large. There is a critical angle within which the fault must be cleared if the system is to remain stable and the equal area criterion is to be satisfied. This angle is known as the critical clearing angle.
(Fig. 8 Single machine infinite bus system)
Consider a system as shown in Fig. 8 operating with mechanical input at steady angle. as shown by point ‘a’ on the power angle diagram as shown in Fig. 9. Now if three phase short circuit occur at point F of the outgoing radial line , the terminal voltage goes to zero and hence electrical power output of the generator instantly reduces to zero i.e., and the state point drops to ‘b’. The acceleration area A1 starts to increase while the state point moves along b-
c. At time tc corresponding clearing angle δc, the fault is cleared by the opening of
the line circuit breaker. tc is called clearing time and δc is called clearing angle.
After the fault i s cleared, the system again becomes healthy and transmits power Pe = Pmax sinδ, i.e., the state point shifts to‘d’ on the power angle
curve. The rotor now decelerates and the decelerating area A2 begins to increase
while the state point moves along d-e. For stability, the clearing a ngle, δc, must
be such that area A1 = area A 2.
(Fig. 9 δ characteristics)
Expressing area A1 =Area A2 mathematically we have,
δ δ dδ
δ δ δ dδ δ δ
δ δ δ δ δ δ
δ δ δ δ …………....……. (60)
Also …………………………………. (61)
Using equation (60) and (61) we get,
δ δ δ δ δ δ δ δ …………………… (62) Where δ = clearing angle, δ = initial power angle, and δ = power angle to
which the rotor advances (or overshoots) beyond δ
For a three phase fault with Pe =0, ………………………………………….. (63)
Integrating equation (63) twice and utilizing the fact that and t = 0 yields
δ t δ …………………..……… (64)
Application of the Equal Area Criterion:-
(1) Sudden Loss of One of parallel Lines:-
Pi (^) Infinite Bus |V|∟ 0 ᵒ
(a)
Pi |V|∟ 0 ᵒ
Switched Off
Eg∟ δ Xd
X
X
(b) (Fig. 11 Single machine tied to infinite bus through two parallel lines) Consider a single machine tied to infinite bus through parallel lines as shown in
Fig. 11(a). The circuit model of the system is given in Fig. 11(b).
Let us study the transient stability of the system when one of the lines is
suddenly switched off with the system operating at a steady load. Before switching
off, power angle curve is given by
δ δ
Immediately on switching of line 2, power angle curve is given by
δ δ
In Fig. 12, wherein as. The system is
operating initially with a steady state power transfer at a torque angle δ
on curve I.
δ (^0) δ 1 δ (^2) π δ
(Fig. 12 Equal area criterion applied to the opening of one of the two lines in
parallel)
On switching off line2, the electrical operating point shifts to curve II (point b).
Accelerating energy corresponding to area A 1 is put into rotor followed by
decelerating energy for δ > δ1. Assuming that an area A 2 corresponding to
decelerating energy (energy out of rotor) can be found such that A 1 = A 2 , the
system will be stable and will finally operate at c corresponding to a new rotor
angle is needed to transfer the same steady power.
If the steady load is increased (line Pi is shifted upwards) a limit is finally
reached beyond which decelerating area equal to A 1 cannot be found and therefore,
the system behaves as an unstable one. For the limiting case, δ 1 has a maximum
value given by
δ δ δ
LESSON SUMMARY-6:-
- Sudden short circuit on one of parallel lines a) Short circuit at one end of line b) Short circuit at the middle of a line
- Example
Sudden Short Circuit on One of Parallel Lines:-
(1) Short circuit at one end of line:- Let us a temporary three phase bolted fault occurs at the sending end of one of
the line.
