Mrinal K Pal. Power system stability
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FUNDAMENTALS OF POWER FLOW AND POWER LIMITS
1-11
When R is not equal to zero, the maximum power can be found by taking the partial derivative of
P
R
with respect to
δ
, and equating to zero. For example, from equation (1.34),
0)cos( =+=
∂
∂
αδ
δ
YVV
P
SR
R
which yields
o
90=+
αδ
or
αδ
−=
o
90
and
GVYVVYVYVVP
RSRRSRR
22
max
sin −=−=
α
(1.41)
Problems
1. Show that the maximum power at the sending end occurs at
αδ
+=
o
90 .
2.
Assuming
SR
VV = , find the value of the reactance (X) for a given value of the resistance
(R) for maximum receiving end power. (The solution is
RX 3= .)
Circle Diagram
Equations (1.34) and (1.35) can be rewritten as
)sin(sin
2
αδα
+=+ YVVYVP
SRRR
(1.42)
)cos(cos
2
αδα
+=+ YVVYVQ
SRRR
(1.43)
Squaring (1.42) and 1.43), and adding
()
(
)
()
2
2
2
2
2
cossin YVVYVQYVP
SRRRRR
=+++
αα
(1.44)
Equation (1.44) describes a circle with centers located at
α
sin
2
YV
R
− ,
α
cos
2
YV
R
− , if P
R
and Q
R
are used as the axes of coordinates. The radius of the circle is V
R
V
S
Y. The circle diagram is
shown in Figure 1.7.
Problems
1. Draw the sending-end circle diagram for the system shown in Figure 1.6
2.
Using the circle diagram calculate
δ
, given P
S
= 1.0, V
S
= V
R
= 1.0, R = 0.05, X = 0.3.
It should be noted that the maximum power as given by equation (1.41), or as obtained from the
circle diagram cannot be realized in practice, since this would require an uneconomically large
amount of reactive power to be supplied at the receiving end in order to hold the receiving-end
voltage constant. Also, above a certain level of transmitted power, the reactive support must have
some sort of continuous control, such as provided by synchronous condensers or static var
systems (SVS), in order to avoid voltage instability and collapse. The subject of voltage
instability will be dealt with in more detail in Chapter 10.
FUNDAMENTALS OF POWER FLOW AND POWER LIMITS
1-12
Fig. 1.7 Receiving-end circle diagram for fixed V
S
and V
R
.
If there are synchronous machines at both ends of a transmission line, the power limit is set by
the ability of the machines to stay in synchronism with each other. This limit is called the
(synchronous) stability limit. (2) The steady-state stability limit may be defined as the maximum
power that can be delivered without loss of synchronism when the load is increased gradually
and the machine terminal voltages are adjusted by manual control of excitation. Similarly, the
transient stability limit may be defined as the maximum power that can be delivered without loss
of synchronism following a large disturbance such as a system fault and its subsequent clearing.
Basic Power Flow Calculation
Referring to Figure 1.6, P
R
, Q
R
, and the magnitude of V
S
are given. It is desired to solve for
,,,
SR
PV
δ
and Q
S
. The solution can be obtained in several ways, two of which will be illustrated
here.
Direct solution
From (1.31)
()
jXR
V
jQP
VZIVV
V
jQP
II
R
RR
RRS
R
RR
SR
+
−
+=+=∠
−
==
δ
R
RR
R
RR
R
V
RQXP
j
V
XQRP
V
−
+
+
+=
RR
R
V
K
j
V
K
V
21
+
+=
(1.45)
(2) In this chapter we will follow the classical definitions of power system stability [4]. A more rigorous definition
of stability is given in Chapter 2, which will be followed in subsequent chapters.
FUNDAMENTALS OF POWER FLOW AND POWER LIMITS
1-13
where
RQXPKXQRPK
RRRR
−
=
+=
21
,
From (1.45)
2
2
2
1
2
+
+=
RR
RS
V
K
V
K
VV
which can be rearranged as
() ( )
(
)
02
2
2
2
1
22
1
2
2
=++−+ KKVVKV
RSR
From the above we obtain
(
)
(
)
2
422
2
2
2
1
2
1
2
1
2
2
KKKVKV
V
SS
R
+−−±−
=
V
R
is obtained by taking the appropriate (positive) sign before the radical. Once V
R
is known,
δ
can be calculated from equation (1.45).
Iterative procedure (Newton's method)
Rewriting equations (1.36) and (1.37)
(
)
(
)
δδδ
,sincos
1
2
RSRRR
VfBGVVGVP =−+−=
(1.46)
(
)
(
)
δδδ
,cossin
2
2
RSRRR
VfBGVVBVQ =−+= 1.47)
From (1.46) and (1.47)
() ()
RR
R
RR
VVf
V
VfP ∆
∂
∂
+∆
∂
∂
=∆
δδδ
δ
,,
11
(1.48)
() ()
RR
R
RR
VVf
V
VfQ ∆
∂
∂
+∆
∂
∂
=∆
δδδ
δ
,,
22
(1.49)
At the start of the iterative process
(
)
oo1
,
δ
RRR
VfPP
−
=∆ (1.50)
and
(
)
oo2
,
δ
RRR
VfQQ
−
=∆ (1.51)
where
oo
,
δ
R
V are the starting values of V
R
,
δ
.
The partial derivatives can be easily computed from (1.46) and (1.48). The changes in
δ
and V
R
required to improve upon the values obtained from the previous iteration are obtained by solving
equations (1.48) and (1.49). For example, at the start of the iteration process
RR
QP ∆∆ ,
and the
partial derivatives at
V
Ro
and
δ
o
are computed and ∆
δ
1
and ∆V
R1
are obtained from equations
(1.48) and (1.49). At the end of the first iteration
1o
1
δδδ
∆+= (1.52)
FUNDAMENTALS OF POWER FLOW AND POWER LIMITS
1-14
1o
1
RRR
VVV ∆+= (1.53)
These updated values of
δ
and V
R
are then used in the second iteration. The iteration is continued
until
1
ε
<∆
R
P and
2
ε
<∆
R
Q , where
ε
1
and
ε
2
are the specified tolerances. The above iterative
procedure can be extended to systems of any size.
Selection of starting values for
δ
and V
R
A good starting value for
δ
and V
R
can be obtained by using the approximations
,1cos,0
≈
≈
δ
G and
δ
δ
=
sin
since
G and
δ
are usually small.
To obtain the starting value of
δ, a further approximation, 0.1
≈
≈
SR
VV , can be made since the
voltages are usually fairly close to unity. Using these approximations, the starting value of
δ is
obtained from equation (1.46) as
XPBP
RR
=
−
=
/
o
δ
(1.54)
In general,
V
R
will be close to V
S
, and therefore V
Ro
= V
S
will provide a good starting value for
V
R
. If however, Q
R
is appreciable, a better starting value for V
R
can be obtained from equation
(1.47) using the above approximation, as
XQVBQVV
RSRSR
−
=
+
= /
o
(1.55)
The above method of estimating the starting values can be extended to systems of any size. This
method is advisable in order to avoid divergence in the power-flow solution.
Problem
1. Referring to Figure 1.6, the following quantities are given:
P
R
= 0.85 , Q
R
= 0.35, V
S
= 1.15
Calculate
δ
, V
R
, P
S
and Q
S
.
2. Referring to Figure 1.6, the following quantities are given:
P
R
= 0.8 , Q
R
= 0.4, V
R
= 1.0
Calculate
δ
, V
S
, P
S
and Q
S
.
Steady State Stability
Consider a cylindrical rotor synchronous generator connected to a large power system (infinite
bus) operating at constant speed with constant field current and no saturation, as shown in Figure
1.8.
Fig. 1.8 A cylindrical rotor synchronous generator connected to infinite bus and phasor diagram
FUNDAMENTALS OF POWER FLOW AND POWER LIMITS
1-15
r
a
and x
d
are the armature resistance and synchronous reactance, respectively, and E is the
internal machine voltage or the voltage corresponding to field excitation.
Since
E is produced by the field flux, it is associated with the rotor position. The power output of
the machine is given by
)sin(
sin
2222
2
αδ
α
−
+
+
+
=
dada
xr
VE
xr
E
P (1.56)
where
)/(tan
1
da
xr
−
=
α
Since r
a
is usually small, it can be neglected. Then
δ
sin
d
x
VE
P =
(1.57)
The maximum power output for fixed E and V is
d
x
VE
P =
max
(1.58)
This is termed the pull-out power of a synchronous generator against an infinite bus at constant
field excitation, since any attempt to increase the power output further will result in a loss of
synchronism.
Equation (1.56) represents the power angle characteristic of a cylindrical rotor synchronous
machine connected directly to an infinite bus.
Problem
Calculate the power angle characteristic of a cylindrical rotor generator connected through an
external reactance to an infinite bus of voltage 1.0 pu, as shown in Figure 1.9. E is such that 1.0
pu voltage is obtained at the terminal at P = 1.0 pu.
Fig. 1.9 A cylindrical rotor synchronous generator connected to an infinite bus through an
external reactance.
Steady state stability is the stability of the system under conditions of gradual or relatively slow
changes in load. The load is assumed to be applied at a rate which is slow when compared either
with the natural frequency of oscillation of the major parts of the system or with the rate of
change of field flux in the machine in response to the change in loading.
We have already discussed the maximum power at constant field current against an infinite bus.
Steady state stability limit
The maximum power which a generator can deliver depends upon the field excitation of the
generator. Within limit, the excitation depends upon the system operating condition. Generators
FUNDAMENTALS OF POWER FLOW AND POWER LIMITS
1-16
will usually have their excitations adjusted to hold their terminal voltages at some predetermined
values. The terminal voltage is held either manually or with an automatic voltage regulator.
If the load changes occur slowly or gradually so that steady state conditions may be assumed to
exist, the field excitation will change slowly in order to adjust the terminal or system voltage to
the prescribed value. The steady state stability limit of a generator or system can be defined as
the maximum power that can be transmitted for a slow change in load, the load change occurring
slowly enough to allow for a similar change in excitation to bring the terminal voltage back to
normal after each successive load change. It is important to note that it has been assumed that the
control of excitation is such as to correct the voltage change after each small load change has
occurred. Therefore, this is a stability limit for an infinitesimal change in load with constant field
current. If the change in excitation is assumed to be initiated immediately following the change
in load (as happens with automatic voltage regulator action), the stability limit under such
conditions may be termed dynamic stability limit. The dynamic limit will, in general, be higher
than the steady state stability limit as defined above. However, the dynamic stability limit is
dependent on automatic voltage regulator operation.
Example
Consider the system and its phasor diagram shown in Figure 1.10. Operating conditions are such
that 1.0 pu voltage is maintained at both the infinite bus and the generator terminals. Saturation is
neglected. At zero initial load, the generator excitation is also unity. Therefore
δδδ
sin5.0sin
0.2
0.10.1
sin =
×
=
+
=
ed
xx
VE
P
Fig.1.10 A synchronous machine-infinite bus system and phasor diagram, x
d
= x
e
= 1.0.
As the initial power increases, the initial excitation also increases. For example, at initial power
of 0.5, the excitation voltage E can be calculated, that satisfies the given condition of 1.0 pu
voltage at the generator terminal and the infinite bus, from the phasor diagram. E is found to be,
E = 1.24.
Problem: Verify the above value of E for P = 0.5.
Therefore, the power angle relation is
δδ
sin62.0sin
0.2
24.10.1
=
×
=P
Problem
Calculate the power angle relations for initial power levels of 0.70, 0.75, 0.80, 0.85 and 0.90.
It will be found that at P = 0.85, the initial power corresponds to an initial angular displacement
of 90
o
, as shown in Figure 1.11.
FUNDAMENTALS OF POWER FLOW AND POWER LIMITS
1-17
.
Fig. 1.11 Power angle characteristics at constant field excitation, illustrating determination of
steady state stability limit.
The steady state stability limit corresponds to the point at which, if the receiving system requires
a small additional load ∆P, the generator becomes just unable to deliver it without a change in
the field excitation. Therefore, at the limit dP/d
δ
= 0. This is the condition at P = 0.85 and
δ
=
90
o
, where dP/d
δ
is measured on curve c of Figure 1.11. If a higher load is taken, the initial angle
is greater than 90
o
, and dP/d
δ
becomes negative, when measured along the corresponding power
angle curve for constant field excitation.
Steady state stability criterion
For the simple case of a single generator connected to an infinite bus, the condition for steady
state stability is simply that dP/d
δ
be positive at the given operating point as shown above. The
steady state power limit corresponds to the maximum power that can be delivered up to the
critical point where dP/d
δ
changes from a positive to a negative value. dP/d
δ
is called the steady
state synchronizing power coefficient since it indicates the rate at which the steady or sustained
power changes with changes in electrical angular displacement.
It should be noted that the synchronizing power coefficient can increase considerably by the
action of automatic voltage regulators. In a simplified analysis, the effect of automatic voltage
regulators can be approximately accounted for by suitably adjusting the equivalent machine
reactance. However, as will be seen later, the usual form of instability in the steady state, under
automatic voltage control, is due to lack of damping rather than synchronizing power.
Nevertheless, the concept of steady state stability limit as discussed above is useful in developing
a proper understanding of the subject of power system stability.
Steady state stability of a two-machine system
Many steady state stability problems resolve themselves into a two-machine problem. For
example, a large remotely located generating plant delivering power to the main system can
often be treated as a two-machine problem with the system represented as an equivalent machine
-- in effect as a large equivalent synchronous motor. A value of reactance corresponding to the
total three-phase short circuit current contributed by the receiving system may be used as the
FUNDAMENTALS OF POWER FLOW AND POWER LIMITS
1-18
equivalent reactance of the receiving-end system. A graphical method may be utilized to
determine the stability limit.
For the simplified two-machine case shown in Figure 1.12, the steady-state stability limit occurs
when maximum power is obtained at the receiving end, i.e., when the total angle between the
equivalent synchronous machine groups corresponds to the total impedance angle.
Fig. 1.12 Equivalent two-machine system
At the limit
)/(tan
12
1
12
rx
−
=
δ
, where
mg
xxxx
+
+
=
12
(Note that dP/d
δ
becomes negative first at the receiving end, which determines the steady state
stability limit.)
A further condition is imposed by the operating conditions of the system, i.e., the magnitudes of
the terminal voltages of the equivalent synchronous generator and motor at the sending and
receiving ends, respectively. Using the graphical method, the stability limit can be determined
directly without resorting to a trial and error procedure. The method is as follows:
Select a reference direction for current I. Draw phasors jIx
m
, I(r + jx), and jIx
g
to any convenient
scale as shown in Figure 1.13. In this figure, 1 and 2 refer to the points in the generator and
motor at which voltages are maintained when the final increment of load is added; a and b are the
terminals of the generator and the motor, respectively.
Fig.1.13 Illustration of the graphical method for determination of stability limit.
To locate the origin of the phasor diagram: Connect points 1 and 2 by a straight line and draw a
perpendicular at the mid-point M of this line. Extend the line 2b until it intersects this
perpendicular at point P. With P as center draw a circle passing through points 1 and 2 (i.e., with
FUNDAMENTALS OF POWER FLOW AND POWER LIMITS
1-19
a radius P2). Any point O on the larger of the two arcs (as shown on Fig.1.13) satisfies the
condition that )/(tan
12
1
12
rx
−
==
θδ
(verify this from the geometry of Fig. 1.13).
In order to satisfy the voltage condition, the point O must be so located on the arc that the
distances Oa and Ob are in the ratio of the terminal voltages V
a
and V
b
. This may be done by
determining a series of points with a and b as centers and radii in the ratio V
a
/V
b
. The intersection
of the locus of such points with the arc will completely define the location of the origin O of the
phasor diagram. The scale of the diagram will be determined by the magnitudes of the specified
voltages V
a
and V
b
at the generator and motor terminals.
If a π circuit is used to represent the transmission line as shown in Figure 1.14, the system can be
reduced to the form of Fig. 1.12 by using the Thevenin equivalent shown in Figure 1.15.
Fig. 1.14 Equivalent two-machine system including the effect of shunt capacitance.
Fig. 1.15 Thevenin equivalent of the system of Fig. 1.14.
The same procedure as that used for the system of Figure 1.12 can therefore be used.
Problem
Verify the values of
g
xEE
′′′
,,
21
and
m
x
′
.
If the line resistance is neglected and equal sending- and receiving-end machine terminal
voltages are assumed, Figure 1.13 simplifies to Figure 1.16. At the steady state stability limit
δ
12
will be equal to 90
o
.
Fig. 1.16 Simplification of Fig. 1.13 neglecting r.
FUNDAMENTALS OF POWER FLOW AND POWER LIMITS
1-20
Problem
Show that the maximum power for the case shown in Figure 1.16 is given by
()
(
)
()
()
()
2/2/2/
2/2/
2
max
xxxxx
xxxx
P
mg
mg
+++
++
= (1.59)
(Hint: Use the relations OM
2
= 1M × 2M, and P
max
= E
m
I )
General Network Equations
An electrical network can be analyzed by expressing the electrical characteristics of the
individual network elements and their interconnection relationship in terms of mathematical
equations. The network performance equations can be written in nodal, loop or branch form. In
large power system analyses it is both convenient and customary to use the nodal analysis
approach. In the nodal approach, the bus admittance matrix formulation is generally
computationally more expedient.
Nodal formulation in terms of bus admittance matrix
The general expression for the current toward node k of a network having n independent nodes
(i.e., n buses other than the neutral which is taken as a reference) is
∑
=
=
n
n
ikik
VYI
1
(1.60)
For example, for a four bus system the node equations are
4443432421414
4343332321313
4243232221212
4143132121111
VYVYVYVYI
VYVYVYVYI
VYVYVYVYI
VYVYVYVYI
+++=
+++=
+++=
+
+
+=
(1.61)
which can be written in matrix form as
I = YV (1.62)
where
Y is the network bus admittance matrix, and I and V are the current and voltage vectors,
respectively.
The admittances
,,
2211
YY etc. are called the self or driving-point admittances at the nodes, and
each equals the sum of all the admittances terminating on the node identified by the repeated
subscripts. The other admittances are the mutual or transfer admittances, and each equals the
negative of the admittance connected between the nodes identified by the double subscripts. For
example, if bus 1 is connected to the other buses in the network as shown in Figure 1.17, then
111114121011
jBGyyyY
+
=
+
+
=
where
,
1
,
1
1212
12
1010
10
jxr
y
jxr
y
+
=
+
= etc.