Summary of Papers
1. P. Sauer and M. Pai, “Power System Steady-State Stability and the Load Flow Jacobian,” IEEE
Transactions on Power Systems, Vol. 5, No. 4, Nov. 1990
2. V. Ajjarapu and C. Christy, “The Continuation Power Flow: A Tool for Steady-State Voltage
Stability Analysis,” IEEE Transactions on Power Systems, Vol. 7, No. 1, Feb., 1992.
3. S. Greene, I. Dobson, and F. Alvarado, “Sensitivity of the Loading Margin to Voltage Collapse
with Respect to Arbitrary Parameters,” IEEE Transactions on Power Systems, Vol. 12, No. 1,
Feb. 1997, pp. 232-240.
4. S. Greene, I. Dobson, and F. Alvarado, “Contingency Ranking for Voltage Collapse via
Sensitivities from a Single Nose Curve,” IEEE Transactions on Power Systems, Vol. 14, No. 1,
Feb. 1999, pp. 262-272.
1
Voltage Security
Voltage security is the ability of the system to maintain
adequate and controllable voltage levels at all system load buses.
The main concern is that voltage levels outside of a specified
range can affect the operation of the customer’s loads.
Voltage security may be divided into two main problems:
1. Low voltage: voltage level is outside of pre-defined range.
2. Voltage instability: an uncontrolled voltage decline.
You should know that
• low voltage does not necessarily imply voltage instability
• no low voltage does not necessarily imply voltage stability
• voltage instability does necessarily imply low voltage
2
Resources
There have been several individuals that have significantly
progressed the field of voltage security. These include:
• Ajjarapu from ISU
• Van Cutsem: See the book by Van Cutsem and Vournas.
• Alvarado, Dobson, Canizares, & Greene:
There are a couple other texts that provide good treatments of
the subject:
• Carson Taylor: “Power System Voltage Stability”
• Prabha Kundur: “Power System Stability & Control”
3
Our treatment of voltage security will proceed as follows:
• Voltage instability in a simple system
• Voltage instability in a large system
• Brief treatment of bifurcation analysis
• Continuation power flow (path following) methods
• Sensitivity methods
4
Voltage instability in a simple system
Consider the per-phase equivalent of a very simple three
phase power system given below:
V1
V2
Z=R+jX
Node 1
+
I
Node 2
+
V2
V1
_
_
S12
SD=-S12
5
Z  R  jX  Y  G  jB
Note B>0
S12  P12  jQ12
P12 | V1 |2 G  | V1 || V2 | G cos(1   2 ) | V1 || V2 | B sin( 1   2 )
Q12 | V1 |2 B  | V1 || V2 | B cos(1   2 ) | V1 || V2 | G sin( 1   2 )
Let G=0. Then….
P12 | V1 || V2 | B sin( 1   2 )
Q12 | V1 |2 B  | V1 || V2 | B cos(1   2 )
6
Now we can get SD=PD+jQD=-(P21+jQ21) by
•- exchanging the 1 and 2 subscripts in the previous equations.
•- negating
PD   P21   | V1 || V2 | B sin(  2  1 )
| V1 || V2 | B sin( 1   2 )
QD  Q21   | V2 |2 B  | V1 || V2 | B cos( 2  1 )
  | V2 |2 B  | V1 || V2 | B cos(1   2 )
Define 12 =1- 2
PD | V1 || V2 | B sin 12
QD   | V2 |2 B  | V1 || V2 | B cos12
7
Define:  is the power factor angle of the load, i.e.,
  V2  I
Then we can* also express SjD as:
S D  V2 I | V2 || I | e
| V2 || I | (cos   j sin  )
sin 
| V2 || I | cos  (1  j
)
cos 
 PD (1  j tan  )
Define β=tan. Then
S D  PD  jQD  PD (1  j )
Note that phi, and
therefore beta, is
positive for lagging,
negative for leading.
8
So we have developed the following equations….
PD | V1 || V2 | B sin 12
QD   | V2 |2 B  | V1 || V2 | B cos12
S D  PD  jQD  PD (1  j )
Equating the expressions for PD and for QD, we have:
QD  PD    | V2 |2 B  | V1 || V2 | B cos 12
PD | V1 || V2 | B sin 12
PD   | V2 |2 B | V1 || V2 | B cos 12
Square both equations and add them to get…..
9
PD 2  ( PD   | V2 |2 B) 2 | V1 |2 | V2 |2 B 2 (sin 2 12  cos 2 12 )
 PD  ( PD   | V2 |2 B) 2 | V1 |2 | V2 |2 B 2
2
Manipulation yields:
| V | 
2 2
2


PD
 2 PD 
2
2

 | V1 |  | V2 |  2 1   2  0
B
 B

2
Note that this is a quadratic in |V2|2. As such, it has the solution:
1/ 2
| V1 |
PD  | V1 | PD  PD
2 
| V2 | 



  | V1 | 

2
B
B  B

 4
2
4
2
10
Let’s assume that the sending end voltage is |V1|=1.0 pu
and B=2 pu. Then our previous equation becomes:
2 1  PD  1  PD ( PD  2  ) 
| V2 | 
2
1/ 2
You can make
the P-V plot using
the following
matlab code.
% pf = 0.97 lagging
beta=0.25
pdn=[0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.78];
v2n=sqrt((1-beta.*pdn - sqrt(1-pdn.*(pdn+2*beta)))/2);
pdp=[0.78 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0];
v2p=sqrt((1-beta.*pdp + sqrt(1-pdp.*(pdp+2*beta)))/2);
pd1=[pdn pdp];
v21=[v2n v2p];
% pf = 1.0
beta=0
pdn=[0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.99];
v2n=sqrt((1-beta.*pdn - sqrt(1-pdn.*(pdn+2*beta)))/2);
pdp=[0.99 0.9 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0];
v2p=sqrt((1-beta.*pdp + sqrt(1-pdp.*(pdp+2*beta)))/2);
pd2=[pdn pdp];
v22=[v2n v2p];
% pf = .97 leading
beta=-0.25
pdn=[0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3];
v2n=sqrt((1-beta.*pdn - sqrt(1-pdn.*(pdn+2*beta)))/2);
pdp=[1.3 1.2 1.1 1.0 0.9 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0];
v2p=sqrt((1-beta.*pdp + sqrt(1-pdp.*(pdp+2*beta)))/2);
pd3=[pdn pdp];
v23=[v2n v2p];
plot(pd1,v21,pd2,v22,pd3,v23)
11
Plots of the previous equation for different power factors
|V2|
Real power loading, PD
12
Some comments regarding the PV curves:
1. Each curve has a maximum load. This value is typically called the
maximum system load or the system loadability.
2. If the load is increased beyond the loadability, the voltages will
decline uncontrollably.
3. For a value of load below the loadability, there are two
voltage solutions. The upper one corresponds to one that can be
reached in practice. The lower one is correct mathematically, but I
do not know of a way to reach these points in practice.
4. In the lagging or unity power factor condition, it is clear that the
voltage decreases as the load power increases until the loadability.
In this case, the voltage instability phenomena is detectable, i.e.,
operator will be aware that voltages are declining before the
loadability is exceeded.
•5. In the leading case, one observes that the voltage is flat, or perhaps
even increasing a little, until just before the loadability. Thus, in
the leading condition, voltage instability is not very detectable.
The leading condition occurs during high transfer conditions when the
load is light or when the load is highly compensated.
13
QV Curves
We consider our simple (lossless) system again, with the equations
PD | V1 || V2 | B sin 12
QD   | V2 |2 B  | V1 || V2 | B cos12
Now, again assume that V1=1.0, and for a given value of PD
and V2, compute 12 from the first equation, and then Q from the
second equation. Repeat for various values of V2 to obtain a QV
curve for the specified real load PD.
You can make the P-V plot using the following matlab code.
v1=1.0;
b=1.0;
pd1=0.1
v2=[1.1,1.05,1.0,.95,.90,.85,.80,.75,.70,.65,.60,.55,.50,.45,.40,.35,.30,.25,.20,.15];
sintheta=pd1./(b*v1.*v2);
theta=asin(sintheta);
qd1=-v2.^2*b+v1*b*v2.*cos(theta);
plot(qd1,v2);
The curve on the next page illustrates….
14
Q-V Curve
|V2|
QD
15
Homework
1. Draw the PV-curve for the following cases, and for each, determine the loadability.
a. B=2, |V1|=1.0, pf=0.97 lagging
b. B=2, |V1|=1.0, pf=0.95 lagging
c. B=2, |V1|=1.06, pf=0.97 lagging
d. B=10, |V1|=1.0, pf=0.97 lagging
Identify the effect on loadability of power factor, sending-end voltage, and line reactance.
2. Draw the QV-curves for the following cases, and for each, determine the maximum QD.
a. B=1, |V1|=1.0, PD=0.1
b. B=1, |V1|=1.0, PD=0.2
c. B=1, |V1|=1.06, PD=0.1
d. B=2, |V1|=1.0, PD=0.1
Identify the effect on maximum QD of real power demand, sending-end voltage, and line
reactance.
16
Some comments regarding the QV Curves
• In practice, these curves may be drawn with a power flow program
by
1. modeling at the target bus a synchronous condenser (a
generator with P=0) having very wide reactive limits
2. Setting |V| to a desired value
3. Solving the power flow.
4. Reading the Q of the generator.
5. Repeat 2-4 for a range of voltages.
• QV curves have one advantage over PV curves:
They are easier to obtain if you only have a power flow (standard
power flows will not solve near or below the “nose” of PV curves
but they will solve completely around the “nose” of QV curves.)
17
Voltage instability in a large system:
Influential factors:
• Load modeling
• Reactive power limits on generators
• Loss of a circuit
• Availability of switchable shunt devices
Two important ideas on which understanding of the above
influences rest:
1. Voltage instability occurs when the reactive power supply
cannot meet the reactive power demand of the network.
•
•
•
•
Transmission line loading is too high
Reactive sources (generators) are too far from load centers
Generator terminal voltages are too low.
Insufficient load reactive compensation
2. Reactive power cannot be moved very far in a network
(“vars do not travel”), since I2X is large.
Implication: The SYSTEM can have a var surplus but experience
voltage instability if a local area has a var deficiency.
18
Load modeling
In analyzing voltage instability, it is necessary to consider the network
under various voltage profiles.
Voltage stability depends on the level of current drawn by the loads.
The level of current drawn by the loads can depend on the voltage seen
by the loads.
Therefore, voltage instability analysis requires a model of how the
load responds to load variations.
Thus, load modeling is very influential in voltage instability analysis.
19
Exponential load model
A typical load model for a load at a bus is the exponential model:

V 
P  P0  
 V0 
V 
Q  Q0  
 V0 

where the subscript 0 indicates the initial operating conditions.
The exponents  and  are specific to the type of load, e.g.,


Incandescent lamps
1.54
Room air conditioner
0.50
2.5
Furnace fan
0.08
1.6
Battery charger
2.59
4.06
Electronic compact florescent 1.0
0.40
Conventional florescent
2.07
3.21
20
Polynomial load model
The ZIP or polynomial model is a special case of the more general
exponential model, given by a sum of 3 exponential models with
specified subscripts:
  V 2

V
P  P0  p1    p2  p3 
V0
  V0 

p1  p2  p3  1.0
  V 2

V
Q  Q0 q1    q2  q3 
V0
  V0 

q1  q2  q3  1.0
where again the subscript 0 indicates the initial operating conditions.
Usually, values p2 and q2 are the largest.
So this model is composed of three components:
• constant impedance component (p1, q1) - lighting
• constant current component (p2, q2) – motor/lighting
• constant power component (p3,, q3) – loads served by LTCs
21
Effect of Load modeling
Understanding the effect of each component on voltage instability
depends on understanding two ideas:
1. Voltage instability is alleviated when the demand reduces. This
is because I reduces and I2X reactive losses in the circuits reduce.
2. Since voltage instability causes voltage decline, alleviation of
voltage instability results if demand reduces with voltage decline.
This gives the key to understanding the effect of load modeling.
• constant impedance load (p1) is GOOD since demand
reduces with square of voltage.
• constant current load (p2) is OK since demand reduces
with voltage.
• Constant power load (p3) is BAD since demand does
not change as voltage declines.
22
Some considerations in load modeling
The effects of voltage variation on loads, and thus of loads on
voltage instability, cannot be fully captured using exponential or
polynomial load models because of the following three aspects.
• Thermostatic load recovery
• Induction motor stalling/tripping
• Load tap changers
23
Thermostatic load recovery
Heating load is the most common type of thermostatic load, and it
is one for which we are all quite familiar. Although much heating is
done with natural gas as the primary fuel, some heating is done
electrically, and even gas heating systems always contain some
electric components as well, e.g., the fans.
Other thermostatic loads include space heaters/coolers, water
heaters, and refrigerators.
When voltage drops, thermostatic loads initially decrease in power
consumption. But after voltages remain low for a few minutes, the
load regulation devices (thermostats) will start the loads or will
maintain them for longer periods so that more of them are on at
the same time. This is referred to as thermostatic load recovery,
and it tends to exacerbate voltage problems at the high voltage
level.
24
Induction motor stalling/tripping
Three phase induction motors comprise a significant portion of
the total load and so its response to voltage variation is important,
especially since it has a rather unique response.
Consider the steady-state induction motor per-phase equivalent
model.
Za=R1+jX1
V1
Zb=
Rc//jXm
X’2
I’2
R’2+R’2(1-s)/s
=R’2 / s
25
Induction motor stalling/tripping
The (referred to stator)
I '2 
rotor current is given by:
Zb
Vth  V1
where
Z a  Zb
Vth
Z th  ( R'2 / s)  jX '2
and
Z a Zb
Z th  Z a // Z b 
Z a  Zb
Under normal conditions, the slip s is typically very small, less than 0.05
(5%). In this case, R’2/s >> R’2, and I’2 is small.
But as voltage V1 decreases, the electromagnetic torque developed
decreases as well, the motor slows down. Ultimately, the motor may stall.
In this case, s=1, causing R’2/s = R’2. Thus, one sees that the current I’2 is
much larger for stalled conditions than for normal conditions. Because of X1
and X’2 of the induction motor, the large “stall” current represents a large
reactive load.
Large motors have undervoltage tripping to guard against this, but smaller26
motors (refrigerators/air conditioners) may not.
Tap changers:
Load tap changers (LTC, OLTC, ULTC, TCUL) are transformers that
connect the transmission or subtransmission systems to the distribution
systems. They are typically equipped with regulation capability that
allow them to control the voltage on the low side so that voltage
deviation on the high side is not seen on the low side.
t:1
V1 and t are
given in pu.
HV side
V1
V1/t
LV side
In per unit, we say that the tap is t:1, where
• t may range from 0.85-1.15 pu
• a single step may be about 0.005 pu (5/8%=0.00625 is very common)
• a change of one step typically requires about 5 seconds.
• there is a deadband of 2-3 times the tap step to prevent excessive tap change
Under low voltage conditions at the high side, the LTC will decrease t
in order to try and increase V1/t.
27
Tap changers:
Thus, as long as the LTC is regulating (not at a limit), a voltage
decline on the high side does not result in voltage decline at the
load, in the steady-state, so that even if the load is constant Z,
it appears to the high side as if it is constant power. So a simple
load model for voltage instability analysis, for systems using LTC,
is constant power!
There are 2 qualifications to using such a simple model (constant power):
1. “Fast” voltage dips are seen at the low side (since LTC
action typically requires minutes), and if the dip is low enough,
induction motors may trip, resulting in an immediate decrease in
load power.
2. Once the LTC hits its limit (minimum t), then the low side
voltage begins to decline, and it becomes necessary to model
the load voltage sensitivity.
28
Generator capability curve:
Field current limit due to field heating,
enforced by overexcitation limiter on If.
Q
Qmax
Typical
approximation
used in power
flow programs.
Armature current limit due to
armature heating, enforced by
operator control of P and If.
P
Qmin
Limit due to steady-state instability (small
internal voltage E gives small |E||V|Bsin),
and due to stator end-region heating from
induced eddy currents, enforced by
underexcitation limiter (UEL).
29
Effect of generator reactive power limits:
1. Voltage instability is typically preceded by generators hitting their upper reactive
limit, so modeling Qmax is very important to analysis of voltage instability.
2. Most power flow programs represent generator Qmax as fixed. However, this
is an approximation, and one that should be recognized. In reality, Qmax is not fixed.
The reactive capability diagram shows quite clearly that Qmax is a function of P and
becomes more restrictive as P increases. A first-order improvement to fixed Qmax
is to model Qmax as a function of P.
3. Qmax is set according to the Over-eXcitation Limiter (OXL). The field circuit has a
rated steady-state field current If-max, set by field circuit heating limitations. Since
2
heating is proportional to  I f dt, we see that smaller overloads can be tolerated
overload
for longer times.
time
Therefore, most modern OXLs are set with a time-inverse characteristic:
4. As soon as the OXL acts to limit If, then no
further increase in reactive power is possible.
2.0
OXL characteristic
When drawing PV or QV curves, the
If
action of a generator hitting Qmax, will
Irated
manifest itself as a sharp discontinuity
120
1.0
in the curve.
10
30
Overload time (sec) 
Effect of OXL action on PV curve:
One generator hits reactive limit
|V|
o
No reactive
limits modeled
P
(demand)
Note: Georgia Power Co. models its loadability
limit at point x, not point o.
31
Loss of a circuit
Compare reactive losses with and without second circuit
Assume both circuits have reactance of X.
I/2
I
X
I/2
X
Qloss=(I/2)2X+ (I/2)2X=I2X/2
P
P
Qloss=I2X
Implication: Loss of a circuit will always increase reactive losses
in the network. This effect is compounded by the fact
that losing a circuit also means losing its line charging
32
capacitance.
Kundur, on pp. 979-990, has an excellent example which illustrates
many of the aforementioned effects. The illustration was done using
a long-term time domain simulation program (Eurostag).
33
Influence of switched shunt capacitors
I
I
P
P
|V|
Without
capacitor
With capacitor
P
(demand)
34
But, shunt compensation has some drawbacks:
• It produces reactive power in proportion to the square of the
the voltage, therefore when voltages drop, so does the reactive
power supplied by the capacitor.
• It has a maximum compensation level beyond which stable
operation is not possible (See pg. 972 of Kundur, and next slide).
(A synchronous condenser and an SVC do not have these 2 drawbacks)
• It results in a flatter PV curve and therefore makes voltage
instability less detectable. Therefore, as the load grows in areas
lacking generation, more and more shunt compensation is used to
keep voltages in normal operating ranges. By so doing, normal
operating points progressively approach loadability.
35
V1=1.0
Each QV curve/Capacitor characteristic
intersection shows the operating point. Note
that for the first three operating points, a
small increase in Q-comp (indicated by
arrows) results in voltage increase, but for
the last operating point (950), more Q-comp
(say 960) results in a voltage decrease.
V2
PL
QL=0
S=|V2|2B*Sbase
with |V2|=1.0
675 Mvar 450 Mvar
|V2|
300 Mvar
1.2
950 Mvar
1.0
0.8
QV-curves drawn
using synchronous
condensor approach.
1600
1400
0.6
1200
1000
800
600
Capacitive Mvars
400
200
36
Bifurcation analysis (ref: A. Gaponov-Grekhov, “Nonlinearities in action” and
also Van Cutsem & Vournas, “Voltage stability of electric power systems.”)
A bifurcation, for a dynamic system, is an acquisition of a new
quality by the motion the dynamic system, caused by small changes
in its parameters. A power system that has experienced a bifurcation
will generally have corresponding motion that is undesirable.
Consider representing the dynamics of the power system as:
x  F ( x, y, p )
0  G ( x, y , p )
Eqts. 1
A differential-algebraic system (DAS):
Here x represents state variables of the system (e.g., rotor angles, rotor
speed, etc), y represents the algebraic variables (bus voltage magnitudes
& voltage angles), and p represents the real and reactive power injections
at each bus. The function F represents the differential equations for the
generators, and the function G represents the power flow equations. 37
Types of bifurcations
There are at least two types of bifurcation:
• Hopf: two eigenvalues become purely imaginary:
a birth of oscillatory or periodic motion.
• Saddle node: a disappearance of an equilibrium state.
The stable operating equilibrium coalesces with an unstable
equilibrium and disappears. The dynamic consequence of a
generic saddle node bifurcation is:
a monotonic decline in system variables.
So we think it is the saddle node bifurcation that causes
voltage instability.
38
The unreduced Jacobian:
The Jacobian matrix of eqts. 1 is
F X
J 
G X
FY 

GY 
and it is referred to as the unreduced Jacobian of the DAS, where
 x 
x 
 0   J y 
 
 
Eqt. 2
39
The reduced Jacobian:
We may reduce eq. 2 by eliminating the variable y
x   F X
 0   G
   X
F Y   x 
y 

GY   
This means we need to force the top right hand submatrix to 0, which we can
do by multiplying the bottom row by -FYGY-1 and then adding to the top row.
1


x 
F X  F Y G Y G X 0  x 
 y 
0
GY   
  G X
This results in:
1
x  F X  F Y G Y G X x


So that the reduced Jacobian matrix is a Schur’s complement:
1
A  F X  F Y GY G X
40
Stability:
Fact 1 : The conditions for a saddle node bifurcation are
x  F ( x, y, p )
1. Equilibrium:
F X
J 
0  G ( x, y , p )
2. Singularity of the unreduced Jacobian
G X
 det(J)=0 (a 0 eigenvalue, J noninvertible) .
FY 
GY 
Implication 1: The stability of an equilibrium point of the DAS depends on
the eigenvalues of the unreduced Jacobian J. The system will experience a
SNB as parameter p increases when J has a zero eigenvalue.
Fact 2: The determinant of a Schur’s complement times the determinant of
GY gives the determinant of the original matrix: det(J)=det(A)*det(GY)
if GY is nonsingular.
Implications 2:
1. If GY is nonsingular, then singularity of A implies singularity of J so
that we may analyze eigenvalues of A to ascertain stability.
2. The fact that GY may be nonsingular, yet A singular, means that
load flow convergence is not a sufficient condition for voltage 41
stability.
Singularity of load flow Jacobian:
Implications 2:
1. If GY is nonsingular, then singularity of A implies singularity of J so
that we may analyze eigenvalues of A to ascertain stability.
2. The fact that GY may be nonsingular, yet A singular, means that
load flow convergence is not a sufficient condition for voltage
stability.
GY
Singular
Singular
Nonsingular
Nonsingular
A
J
Singular
(unstable)
Nonsingular
(stable)
42
Singularity of load flow Jacobian:
So voltage instability analysis using only a load flow Jacobian may yield
optimistic results when compared to results from analysis of A,
that is, stable points (based on Gy) may not be really stable.
=> However, I believe it is true that points identified as unstable using
the load flow Jacobian will be really unstable (Schur’s complement
does not support that singularity of GY implies singularity of J,
however, because it is only valid if GY is nonsingular).
GY
Singular
(unstable)
Nonsingular
(stable)
A
Singular
(unstable)
Nonsingular
(stable)
J
Singular
(unstable)
Nonsingular
(stable)
Note: Sauer and Pai, 1990, provide an in-depth analysis of the relation
between singularity of GY and singularity of J, and show some special
43
cases for which singularity of GY implies singularity of J.
Singularity of load flow Jacobian:
So, we assume that load flow Jacobian analysis provides an upper
bound on stability.
Fact: The bifurcation (zero eigenvalue of GY) of the load flow
Jacobian corresponds to the “turn-around point” (i.e., the “nose”
point) of a P-V or Q-V curve drawn using a power flow program.
This can be proven using an optimization approach.
See pp. 218-220 of the text by Van Cutsem and Vournas.
We have previously denoted the power flow equations as G(x,y,p)=0,
but now we denote them as G(y,p)=0, without the dependence on the
state variables x (which relate to the machine modeling and include,
minimally,  and  of each machine).
44
So we turn our effort to identifying the saddle node bifurcation
(SNB) for the power flow Jacobian matrix.
The Jacobian can reach a SNB in many ways. For example,
• increase the impedance in a key tie line
• increase the generation level at a generator with weak transmission, while
decreasing generation at all other generators.
|V|
• increase the load at a single bus
• increase the load at all buses.
In all cases, we are looking for the “nose” point of the
V- curve, where  is the parameter that is being increased.)


Most applications focus on the last method (increase load at all buses).
Key questions here are:
• “direction” of increase: are bus loads increased proportionally, or in some other way?
• dispatch policy: how do the generators pick up the load increase ?
We will assume proportional load increase with “governor” load flow
(generators pick up in proportion to their rating)
45
Define: critical point - the operating conditions, characterized
by a certain value of , beyond which operation is not
acceptable.
Question 1:
What can cause the critical point to differ
from the SNB point ?
|V|


Question 2:
How can knowledge of the critical point provide a security
measure?
Question 3:
Does the P-V curve provide a forecast of the system trajectory ?
46
Solution approaches to finding *, the value of  corresponding to SNB.
Approach 1: Search for * using some iterative search procedure.
1. i=1
2. Using (i), solve power flow using Newton-Raphson.
Here, we iteratively solve G(y,p)=0. At each step,
we must solve for y in the eqt: GY y = p
3. If solved,
(i+1)= (i)+ .
i=i+1
go to 2
else if not solved,
*= (i+1)
endif
4. End
But big problem: as  gets close to *, GY becomes ill-conditioned
(close to singular). This means that at some point before the critical
point, step 2 will no longer be feasible.
47
Approach 2: Use the continuation power flow (CPF).
Predictor step
Corrector step
No.
Pass * ?
Select
continuation
parameter
Yes.
Stop
48
The predictor step:
The power flow equations are functions of the bus voltages and
bus angles and the bus injections:
0  G ( y, p)
Augment the power flow equations so that they are functions of 
(dependence on p is carried through the dependence on ).
pp0 
0  G ( y,  )
Now recognize that y   
 
V 
so that
0  G( ,V ,  )
If we want to compute the change in the power flow equations dG
due to small changes in the variables , V, and ,
• that move us closer to the loadability point
• as we move from one solution i to another “close” solution i+1, then
dG= G((i),V(i),(i))- G((i+1),V(i+1),(i+1)) = 0 – 0 = 0
49
dG
dG
dG
dG 
d 
dV 
d
d
dV
d
Here, each set of partial derivatives are evaluated at the operating conditions
corresponding to the old solution. If the power flow equations are linear with the 3
sets of variables in the region between the old solution and the (close) new one, the
following is satisfied:
dG
dG
dG
dG 
d 
dV 
d  0
d
dV
d
Eq. 3
 G
GV
 d 


G  dV   0
 d 
BUT, we have added one unknown, , to the power flow problem without adding a
corresponding equation, i.e., in G(,V,)=0, there are are N equations but N+1
variables, so that in eq. 3, the matrix [G GV, G], has N rows (the number of eqts
being differentiated) and N+1 columns (the number of variables for which each eqt
is differentiated). So we need another equation in order to solve this. What to do ?
50
The answer to this can be found by identifying how we will be using using the
solution to eqt. 3. Note the solution corresponding to the “new” point is:
 (i 1, p )   (i )   d ' 
 (i 1, p )   (i )   
V
  V   dV '
 (i 1, p )   (i )   d ' 

    
Here the “p” indicates
that this is the
“predicted” point.
If we define  to be the “step size,” then we can rewrite this as
 (i 1, p )   (i ) 
 d 
 (i 1, p )   (i ) 
where  d  ' 


 d 
V

V


d
V

  
 
dV '   dV 
(
i

1
p
)
(
i
)

  
 d 
 
 

  
 d ' 
 d 
51
We call the update vector (with the differentials) the
“tangent” vector, denoted by t.
 d 
t  dV 
 d 
This vector provides the direction to move in order
to find a new solution (i+1,p) from the old one (i).
We can think of this in terms of the following picture…..
52
Tangent vector
|V|

53
Note: In specifying a direction using an n-dimensional vector, only n-1 of the
elements are constrained - one element can be chosen to be any value we
like.
For example, consider a 2-dimensional vector….
x2=x1tan(30) so:
x2
Direction
= 30o
x1
- the direction is specified by
selecting x1=1, x2=0.5774,
- the direction is specified by
selecting x1=0.5, x2=0.2246.
So we can set one of the tangent vector elements to
any value we like, then compute the other elements.
This provides us with our other equation….
54
Suppose that we set the k-th parameter in the tangent vector to be
1.0. Then our equation given as eq. 3 can be augmented to become:
G


GV
ek
 d 
G     0 
 dV   1 
  d   
 
where
ek = [ 0
0
...
0
1
0
...
0]
k
To select , we would have:
ek = [ 0
0
...
0
0
0
...
1]
Which would
force d=1.
55
The parameter for which we select k is called the continuation
parameter, and it can be any load level (or group of load levels),
or it can be a voltage magnitude. Initially, when the solution is
far from the nose, the continuation parameter is typically .
 (i 1, p )   (i ) 
 d 
 (i 1, p )   (i ) 
dV 
V

V



 

 
 (i 1, p )   (i ) 
 d 

 

y
( i 1, p )
y
(i )
 t
The parameter  is called the step size, and it can be selected
using various techniques. The simplest of these is to just
set it to a constant. Let’s try this on our simple problem
formulated at the beginning of these slides.
56
HOMEWORK #2, Due Monday, Jan 26.
1. Using the equations at the bottom of slide 7, with the left-hand side (PD
and QD) and also V1 given by the problem statement, we know everything
except V2 and theta.
2. Now, just bring the right hand side of these 2 equations over to the
left-hand side, and you have the 2 equations that correspond to G(y,p)=0.
3. Solve these equations to get the corresponding power flow solution (but
you do not need Newton-Raphson to do this – you can just use the equation
at the bottom of slide 10).
4. Now you need to replace the value specified in the equations for PD
(which is 0.4 according to the problem statement) with 0.4*lambda. This
gives you the equations in the form of slide 49: 0=G(theta,V,lambda).
Note, however, that G is really two equations: G1 and G2.
5. Now you need to formulate the equations on the slide 55. This is a matter
of taking derivatives and then evaluating those derivatives at the
solution that you obtained above. Note, however, the each element in the
matrix of slide 55 actually represents 2 elements. That is:
PD | V1 || V2 | B sin 12
QD   | V2 |2 B  | V1 || V2 | B cos12
| dG1/dtheta
dG1/dV
dG1/dlambda|
| dG2/dtheta
dG2/dV
dG2/dlambda|
#9 and #10 will be
| 0
0
1
|
6. Evaluate each of the above matrix elements at the solution obtained in
explained in next
step 3.
few slides.
7. Then solve these equations for the tangent vector. You can do this by
inverting the above matrix (use matlab or a calculator to do this) and
then multiply the right-hand-side by this inverted matrix.
8. Then take a “step” using an appropriately chosen step size per the
equation on slide 56.
9. Beginning from your predicted point that you identified in step 8 of #2a, develop
equations for approach a, solve them, and identify the resulting corrected point in terms of voltage and power.
10. Repeat #9 except implement approach b.
57
Corrector step
Note, however, that the predicted point will satisfy the
power flow equations only if the power flow equations are
linear, which they are not.
So our point needs correction. This leads to the corrector step.
There are two different approaches for performing the
corrector step.
Approach a: Perpendicular intersection method.
Approach b: Parameterization method
58
Approach a: perpendicular intersection
Here, we find the intersection between the power flow
equations (the PV curve) and a plane that is perpendicular to
the tangent vector.
|V|
Solve simultaneously,
for y(i+1)
0  G( y
y
( i 1)
y
( i 1)
( i 1, p )
, )
y(i)
t
y(i+1,p)
y(i+1)
 t  0
The last equation says the inner
(dot) product of 2  vectors is zero.

Use Newton-Raphson to solve the above (requires only 1-3 iterations since we have
good starting point). If no convergence, cut step size () by half and repeat.
59
Approach b: Parameterization
The corrector step is performed by
•identifying a continuation parameter (see slide 62) – can be λ
• fixing it at the value found in the predictor step;
• then solving the power flow equations.
|V|
Solve simultaneously,
for y(i+1)
G( y ,  )

0
( i 1)
 yk
  
( i 1)
yk(i+1)
y(i)
t
y(i+1,p)
Vertical corrections
correspond to a fixed
load-continuation
parameter, horizontal
corrections to a fixed
voltage-continuation
parameter.
y(i+1)
yk(i+1)

Here,
is the continuation parameter; it is the variable
that corresponds to the k-th
element dyk(i+1) in the tangent vector and is usually λ at first but often becomes something else
as the nose point is neared. The parameter  is the value to which yk is set, which would be the
value found in the predictor step. As in approach a, we can solve this using Newton-Raphson.
60
If no convergence, cut step size () by half and repeat.
Detection of critical point:
We will know that we have surpassed the critical point
when the sign of d in the tangent vector becomes
negative, because it is at this point where the loading
reaches a maximum point and begins to decrease.
|V|
 increasing
x
 decreasing

61
Selection of continuation parameter:
The continuation parameter is selected from among 
The one changing and the state variables in y according to the one that is
the most with λ is
most sensitive andchanging the most with . This will be the parameter that
represents a
has the largest element in the tangent vector.
variable that we
want to be careful • relatively unstressed conditions (far from nose): generally 
with as we look for
another solution, • relatively stressed conditions (close to nose): generally the
so it makes sense to
voltage magnitude of the weakest bus, as it changes a great
keep it constant.
deal as  is changed, when we are close to *.
Typically, yk
is going to be
one of these.
 (i 1, p )   (i ) 
 d 
 (i 1, p )   (i ) 


V
  V    dV 
 (i 1, p )   (i ) 
 d 

 

62
Selection of continuation parameter (unstressed condition):
The continuation parameter is selected from among 
and the state variables in y according to the one that is
changing the most with . This will be the parameter that
has the largest element in the tangent vector.
• relatively unstressed conditions (far from nose): generally .
=> This looks like below.
|V|
y(i)
y(i+1,p)
y(i+1)
Here,  is fixed.

63
Selection of continuation parameter (stressed condition):
• relatively stressed conditions (close to nose): generally the
voltage magnitude of the weakest bus. Here, the voltage being
plotted is chosen as the continuation parameter.
|V|
y(i)
y(i+1,p)
y(i+1)
Here, |V| is fixed.

“Essentially, a variable is fixed as a parameter (the voltage), and
the parameter () is treated as a variable. This process of selecting
a variable to fix is sometimes called the parameterization step.”
-Scott Greene, Ph.D. dissertation, 1998.
64
A central question:
How does the continuation technique alleviate the illconditioning problem experienced by a regular power flow ?
Refer to the solutions procedures for the two corrector approaches.
Perpendicular interesection
Solve simultaneously,
for y(i+1)
0  G( y
y
( i 1)
y
( i 1)
( i 1, p )
, )
 t  0
Parameterization
Solve simultaneously,
for y(i+1)
G ( y (i 1)  )

0
 yk (i 1)   
In both cases, we use Newton-Raphson to solve, so we need to obtain the
Jacobian. But the Jacobian is slightly different than in normal power flow.
65
The Jacobian of the power flow equations is just Gy, but the
Jacobian of the equations in the two corrector approaches
will have an extra row and column.
G y

C y
G xk 

C xk 
Here, C is the additional equation, and xk is the selected
continuation parameter.
This addition of a row and column to the Jacobian has the
effect of improving the conditioning so that the previously
singular points can in fact be obtained. In other words, the
additional row and column provides that this Jacobian is
nonsingular at * where the standard Jacobian is singular.
66
Known codes for continuation methods:
1. Claudio Canizarres at University of Waterloo: C-code
See http://www.power.uwaterloo.ca/~claudio/claudio.html
UWPFLOW is a research tool that has been designed to calculate local bifurcations related to system
limits or singularities in the system Jacobian. The program also generates a series of output files that
allow further analyses, such as tangent vectors, left and right eigenvectors at a singular bifurcation
point, Jacobians, power flow solutions at different loading levels, voltage stability indices, etc
2. I have Matlab code that does it – from Scott Greene.
3. Venkataramana Ajjarapu (ISU): Fortran code
4. Powertech has a program
67
Calculation of sensitivities for voltage instability analysis
What is a sensitivity ?
It is the derivative of an equation with respect to a variable.
It shows how parameter 1 changes with parameter 2.
It is: exact when parameter 2 depends linearly on parameter 1.
It is approximate when parameter 2 depends nonlinearly on parameter 1,
but it is quite accurate if it is only
used close to where it is calculated.
68
G
y
Consider the system characterized by G(y). Then
is the sensitivity of the equation G with respect to y,
evaluated at y*.
G(y)
y*
Slope is G/y
evaluated at y*.
y
y y*
y
It’s usefulness is that once it is calculated, it can be used to
QUICKLY evaluate f(y) from G(y)G(y*)+ (G/y|y*)y,
BUT ONLY AS LONG AS y IS CLOSE TO y*.
69
Consider parameter p: we desire to obtain the sensitivity of
G(y,p) to p. Typical parameters p would be a bus load, a bus
power factor, or a generation level.
Very important to distinguish between
• voltage sensitivities
• voltage instability sensitivities
What is the difference between them in terms of
• what they mean ?
• how to compute them ?
70
Sensitivities for bus voltage
These we compute at the current operating condition.
|V|
For a given continuation parameter, they can be obtained
from the first predictor step in the continuation power flow.
Recall that this provides us with  d 
dV 
t

the tangent vector, given by:
 
 d 

Current
operating point
The tangent vector is the vector of
sensitivities with respect to a small
change in , so the portion of the vector
designated as dV is exactly the voltage
sensitivities.
71
Sensitivities for voltage instability
Here, it is important to realize that the measure of voltage instability,
the loading margin, depends on an operating condition
different from the present operating condition.
The implication is that we must look at sensitivities
of the loading margin, not of the voltage.
|V|
Loading margin

Current
operating point
So we want the sensitivities
evaluated at this point, i.e.,
the SNB point.
72
Derivation of loading margin sensitivities at SNB point.
Let S be the vector of real and reactive load powers,
and k be the direction of load increase.
S  S0   k
Also, define L as the loading margin (a scalar), so that
the load powers resulting in the SNB point are given by:
S  S 0  Lk
We desire to find the sensitivity of the loading margin L to a
change in the parameter p. We denote this sensitivity by Lp.
73
Consider the system characterized by
G(y,S, p) = 0
We want the sensitivity of
the loading margin to p.
Assumption: the system has a SNB at (y*,S*, p*), i.e., :
1. G(y*,S*, p*) = 0 (an equilibrium point)
2. Gy(y*,S*, p*) is singular (zero eigenvalue), and
w is a left eigenvector of Gy(y*,S*, p*), corresponding
to the zero eigenvalue so that (by definition of the left
eigenvector)
wT Gy(y*,S*, p*) =0 wT=0
Note that Gy(y*,S*, p*), being singular, cannot be inverted, but
we can compute it (that is, Gy (y*,S*, p*)), and its eigenvectors.
3. wT GS(y*,S*, p*)  0
74
The points (y,S, p) satisfying numbers 1 and 2 correspond to SNB
points,
and we can obtain a curve of such points by
varying p about its nominal value p*.
Linearization of this curve about the SNB point results in
G y y  G S * S  G p p  0
*
*
where the notation |* indicates the derivatives are evaluated at the SNB point.
Pre-multiplication by the left eigenvector w results in:
w G y y  w G S * S  w G p p  0
T
T
*
T
*
By #2 on the previous slide, the first term in the above is zero. So...75
w G S * S  w G p p  0
T
T
Eqt. *
*
Now recall the relation of the load powers to the loading margin….
S  S 0  Lk
 S  L k
Substituting this expression for the load powers into eqt. *,
w G  * Lk  w G p p  0  Lw G  * k  w G p p
T
T
T
T
*
*
And the loading margin sensitivity to parameter p is:
L  w G p *
 Lp 
 T
*
p w G S k
T
*
So p may be, for example,
real power load at a bus (to
detect the most effective load
shedding) or reactive power
at a bus (to determine where
76
to site a shunt cap).
w Gp
T
Some comments about computing Lp
Lp 
*
*
T
w GS k
*
• The left eigenvector w must be computed for the
Jacobian Gy evaluated at the SNB point.
• You only need to compute w and GS once, independent of
how many sensitivities you need. Methods to compute the left eigenvector
w include QR or inverse iteration.
• The vector of derivatives with respect to the parameter p, which is Gp, is
typically sparse. For example, if you want to compute the sensitivity to a
bus power, then there would be only 1 non-zero entry in Gp.
• The matrix of derivatives with respect to the load powers, GS, using constant
power load models, is a diagonal matrix with ones in the rows corresponding
to load buses. This is because a particular load variable would ONLY occur
in the equation corresponding to the bus where it is located, and for these
equations, these variables appear linearly with 1 as coefficient.
77
Some comments about extensions
• Multiple sensitivities may be computed using Gp (a matrix) instead of Gp (a vector).
In this case, the result is a vector.
T
Lp 
*
w Gp
*
T
w GS k
*
• Getting multiple sensitivities can be especially attractive when we want to find
the sensitivity to several simultaneous changes. One good example is to find the
sensitivity to changes in multiple loads.
• A special case of this is to find the sensitivity to changes at ALL loads, which is
very typical, given a particular loading direction k . Then
Lall loads *   ki L pi
i
• A sensitivity to a line outage may be obtained by letting p contain elements
corresponding to the outaged line parameters.
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Some comments about extensions
• A sensitivity to a line outage may be obtained by letting p contain elements
corresponding to the outaged line parameters: R (series conductance), X (series
reactance), and B (line charging). Then use the multiple parameter approach.
w Gp
Zpq=R + jX
T
Lp 
*
*
T
w GS k
*
L  L p * p
p
q
jB
jB
• Here, p = [R X B]T.
• Note that p is NOT SMALL ! Therefore L may have considerable error.
For that reason, this one needs to be careful about using this approach to
compute the actual loading margins following contingencies.
• However, it certainly can be used for RANKING contingencies. One might
consider having a “quick approximation” and a “long exact” risk calculation. 79
Some comments about alternatives
• Greene, et al., also propose a quadratic sensitivity which
requires calculation of a second order term Lpp . This is used
together with the linear sensitivity according to
1
L  Lp p  Lpp (p) 2
*
2
*
It requires significantly more computation but can provide greater
accuracy over a larger range of p.
• Invariant Subspace Parametic Sensitivity (ISPS) by Ajjarapu.
Advantages:
– based on differential-algebraic model
– provides sensitivities at ANY point on the P-V curve
80