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Power Systems Analysis - Lecture 7Contingency Analysis,
Decoupled Analysis, Control of Power Flows
7.1 Contingency Analysis: In general we use computational tools
such as the Matlab scripts in HW5 orPowerWorld to solve power flow
problems. These tools are often used in industry to assess how
vulnerablethe systems steady state operation will be to failures
within the system. Federal regulations place require-ments on what
constitutes reliable service (i.e. good power quality). These
reliability conditions include
There are no transmission line or transformer limit violations.
Transformers and transmission linesall have physical constraints
that limit the voltages or current that these devices can carry.
Violatingthese limits can result in device failure and as these
devices are extremely expensive, it is essentialthat we manage the
system to stay within the device ratings.
Bus voltages must be maintained between 0.95 and 1.05 pu. We
measure power quality in terms of itsdistance from the nominal
voltage and its deviation from the nominal frequency. Voltage
levels mustbe maintained relatively constant as all user devices
expect such a voltage to function properly.
The line frequency stays within 59.5 Hz to 60.5 Hz. This is not
actually dealt with in the power flowanalysis and must be evaluated
using simulations and dynamic models. But as noted before, manyuser
devices expect power to be delivered with a given line
frequency.
In the case of the loss of a single device, these voltage and
frequency standards must be maintained.This is sometimes known as n
1 reliability. n 1 contingency analysis is the procedure used
toverify that the system has this level of n 1 reliability and the
power flow tools discussed above areoften used to check this.
These reliability conditions are legally binding in the sense
that the North American Electric ReliabilityCorporation (NERC) has
legal authority to enforce these reliability standards. So if
utilities fail to maintainadequate power quality, they can be
subject to fines and other legal action.
Figure 1: 37 bus system in PowerWorld
We consider a rather simple example of contin-gency analysis
that examines the impact of line out-ages. This example is worked
out in the textbookusing PowerWorld. Consider the 37 bus
systemwhose one-line diagram is shown in Figure 1.
We generate a single solution to the power flowproblem using the
Newton method and then checkfor bus violations by going to the
PowerWorldMenu
case information Network Buses
We can also check line and transformer loadingthrough the menu
items
case information Network Lines and Transformers
The result from checking the buses is shown on the left side of
Figure 2 in which we see that bus JO345 ismost highly loaded at
103%.
We can force a line overload in this system by opening the line
from bus TIM69 to HANNAH69. We againsolve the power flow problem
and the display on the right side of Figure 2 shows there are two
violations,
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Figure 2: PWorld Contingency analysis for 37-bus system that
shows bus overload (right) and line overloads(left) after one of
the transmission lines has been opened
thereby implying the system does not have n 1 reliability. The
objective would be to use this tool tofix that problem by finding
alternative configurations of the system for which such failures do
not result inviolations of the reliability conditions.
Electric utilities use elaborate computer programs such as
PWorld for power-flow studies. These studiesare aimed at evaluating
the adequacy of a complex interconnected network. Important
information is ob-tained concerning the design and operation of
systems that have not yet been built and the effects of
changingexisting systems through these studies. A power-flow study
for a system operating under actual or nominaloperating conditions
is called a base case. The results from the base case form a
benchmark for comparisonof changes in network flows and voltages
under abnormal or contingency conditions. Transmission
planningengineers can use these studies to discover system
weaknesses such as low voltages, line overload, or load-ing
conditions that seem excessive. These weaknesses can be removed by
making design studies involvingchanges and/or additions to the base
case system. The system model is then subjected to
computer-basedcontingency testing to discover whether weaknesses
arise under contingency conditions involving abnormalgeneration
schedules or load levels.
A typical commercial power flow program can handle systems of
more than 2000 buses, 3000 lines, and500 transformers. Data
supplied to the computer must include the numerical values of the
line and bus dataand an indication of whether a bus is the slack
bus, a regulated bus where the voltage magnitude is heldconstant by
generation of reactive power Q, or a load bus with fixed P and
Q.
The output of the program consists of a number of tables.
Usually the most important information to beconsidered first is the
table that lists each bus number and name, bus-voltage magnitude in
per unit andphase angle, generation and load at each bus in
megawatts and megavars, and megavars of static capacitorsor
reactors on the bus. The totals of system generation and loads are
listed in megatwatts and megavars. Forthe system in example 6.3 in
lecture 6 an example of such an output is shown in the following
two tables
Bus Information
bus namevolts(pu)
angle(deg)
(MW) Gen (Mvar) Gen (MW) Load (Mvar) Load type
1 Birch 1.000 0. 186.81 114.50 50.00 30.99 slack2 Elm 0.982
-0.976 0. 0. 170.0 105.35 PQ3 Pine 0.969 -1.872 0. 0. 200.0 123.94
PQ4 Maple 1.020 1.523 318.00 181.43 80.00 49.58 PV
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Line Flowto Bus Name (MW) (Mvar)
2 Elm 39.69 22.303 Pine 98.12 61.211 Birch -38.46 -31.244 Maple
-131.54 -74.111 Birch -97.09 -63.574 Maple -102.91 -60.372 Elm
133.25 74.923 Pine 104.75 56.93
7.2 Regulating Transformers: Regulating (tap changing)
transformers can be used to control the real andreactive power
flows in a circuit. We now develop the bus admittance equations to
include such transformersin power-flow studies. Figure 3 is a
impedance diagram of a regulating transformer. The admittance Y
inper unit is the reciprocal of the per-unit impedance of the
transformer which has the transformation ratio1 : t as shown. The
admittance Y is shown on the side of the ideal transformer nearest
node j, which isthe tap-changing side. This designation is
important in using the equations which are to be derived. If weare
considering a transformer with off-nominal turns ratio, t may be
real or imaginary such as 1.02 for anapproximate 2% boost in
voltage magnitude or ejpi/60 fo a 3 phase shift.
Vi
+ +tVi
- -
Y
Vj+
-
tY
(1-t)Yt(1-t)Y j+
-Vi
+
-V
Figure 3: Tap-changing transformer: (left) impedance diagram
(right) pi model.
The left side of Figure 3 shows currents Ii and Ij entering the
two buses and the voltages are Vi and Vjreferred to the reference
node. The complex expressions for power into the ideal transformer
from bus i andbus j are, respectively,
Si = ViIi , Sj = tViI
j
Since we are assuming an ideal transformer with no losses, the
power Si into the ideal transformer from busi must equal Sj out of
the ideal transformer on the bus j side, and so we obtain
Ii = tIjThe current can be expressed by
Ij = (Vj tVi)Y = tYVi + YVjMultiplying by t and substituting Ii
for tIj yields,
Ii = ttYVi tYVj
setting tt = |t|2 and rearranging into theYbus matrix form we
get[Yii YijYji Yjj
] [ViVj
]=
[ |t|2Y tYt Y
] [ViVj
]=
[IiIj
](1)
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The equivalent pi circuit corresponding to these values of node
admittances can be found only if t is realbecause then Yij = Yji.
This pi model is shown on the right hand side of Figure 3.
If a particular transmission line in a system is carrying too
small or too large a reactive power, a regulatingtransformer can be
provided at one end of the line to make the line transmit a larger
or smaller reactivepower. Any appreciable drop in voltage on the
primary of a transformer due to a change of load may makeit
desirable to change the tap settings on the tap-changing
transformer in order to maintain proper voltage atthe load. We can
investigate voltage-magnitude adjustment at a bus by means of the
automatic tap-changingfeature in the power flow program. For
instance in the four-bus system considered in example 6.3,
supposewe wish to raise the load voltage at bus 3 by inserting a
magnitude-regulating transformer between the loadand the bus. With
real t, in the equation (1), we set i = 3 and assign the number 5
to bus j from which theload is now to be served. To accommodate the
regulator in the power-flow equations, Ybus of the networkis
expanded by one row and one column for bus 5, and the elements of
buses 3 and 5 in the matrix of theequation are then added to the
previous bus admittance matrix to give
Ybus(new) =
Y11 Y12 Y13 0 0Y21 Y22 0 Y24 0Y31 0 Y33 + t
2Y Y34 tY0 Y42 Y43 Y44 0
0 0 tY 0 Y
The Yij elements correspond to the parameters already in the
network before the regulator was added. Thevector of state
variables depends on how bus 5 is treated within the power flow
analysis. There are twoalternatives
Tap t can be regarded as an independent parameter with a
respecified value before the power-flowsolution begins. Bus 5 can
then be treated as a load bus with angle 5 and voltage magnitude
|V5| tobe determined with the other state variables. In this
case
x =[2 3 4 5, |V2|, |V3|, |V5|
]T Voltage magnitude at bus 5 can be prespecified. Tap t then
replaced |V5| as the state variable to be
determined along with 5 at the voltage controlled bus 5. In this
case
x =[2 3 4 5, |V2|, |V3|, t
]T
V1
+ +
- -
jX=j0.1
+
-
V2tV1
jX=j0.1
0.8
j0.6
I1 I2
I2(a)
I2(b)
I1(a)
I1(b)
Figure 4: Example 7.1 - impedance diagram
Example 7.1: Two transformers are connected inparallel to supply
an impedance to neutral to phaseof 0.8 + j0.6 per unit at a voltage
ofV2 = 1.00per unit. Transformer Ta has a voltage ratio equalto the
ratio of the base voltages on the two sides ofthe transformer. This
transformer has an impedanceof j0.1 per unit on the appropriate
base. The sec-ond transformer Tb also has an impedance of j0.1per
unit on the same base but has a step-up to-ward the load of 1.05
times that of Ta (secondarywindings on 1.05 tap). Figure 4 shows
the equiv-alent circuit with transformer Ta represented by
itsimpedance. Find the complex power transmitted tothe load through
each transformer.
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The admittance Y of each transformer is given by 1/j0.1 = j10
per unit. Hence, the currents intransformer Ta can be determined
from the bus admittance equation.[
I(a)1
I(a)2
]=
[Y YY Y
] [V1V2
]=
[ j10 j10j10 j10
] [V1V2
]and the currents in transformer Tb with t = 1.05 are given by
equation (1),[
I(b)1
I(b)2
]=
[t2Y tYtY Y
] [V1V2
]=
[ j11.025 j10.5j10.5 j10
] [V1V2
]
From the figure, we see that I1 = I(a)1 + I
(b)1 and likewise I2 = I
(a)2 + I
(b)2 , which means that the preceding
two matrix equations can be directed added to obtain[I1I2
]=
[ j21.025 j20.5j20.5 j20
] [V1V2
]The load current I2 is
I2 = V2Zload
= 1.00.8 + j0.6
= 0.8 + j0.6 per unit
So the second row of the equation is
I2 = 0.8 + j0.6 = j20.5V1 j20(1.0)
that we can solve to get the bus 1 voltage phasor,
V1 =0.8 + j20.6
j20.5= 1.0049 + j0.0390 per unit
SinceV1 andV2 are now known, we return to the admittance
equation for transformer Ta to obtain
I(a)2 = j10V1 j10V2 = j10(1.0049 + j0.0390 1.0)
= 0.390 + j0.049 per unit
and similarly for transformer Tb current
I(b)2 = j10.5V1 j10V2 = j10.5(1.0049 + j0.0390) j10
= 0.41 + j0.551 per unit
Hence the complex power outputs of the transformers are
STa = V2[I
(a)2
]= 0.39 + j0.049 per unit
STb = V2[I
(b)2
]= 0.41 + j0.551 per unit
7.3 Decoupled Power Flow Analysis: Inverting the Jacobian is the
most computationally intensive part ofthe Newton-Raphson recursion.
Several approximations of the Jacobian are used to reduce this
complexity.One common group of methods is referred to as decoupled
power flow (DPF).
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The Newton-Raphson recursion in power flow is based on the
following equation
P (x(k)) P (x(k))|V|
Q(x(k))
Q(x(k))|V|
[ (k)|V(k)|
]=
[P (x(k))
Q(x(k))
]= f(x(k))
where
P (x(k)) =
P2(x(k)) + PD2 PG2
...Pn(x
(k)) + PDn PGn
, Q(x(k)) = Q2(x
(k)) +QD2 QG2...
Qn(x(k)) +QDn QGn
In many power systems, the off-diagonal matrices, P (x
(k))|V| and
Q(x(k)) are small, so we can approximate
them as zero and thereby reduce the update equation to
[
P (x(k)) 0
0 Q(x(k))
|V|
] [(k)
|V(k)|]
=
[P (x(k))
Q(x(k))
]Separating these equations yields the following pair of
updates
(k) = [P (x(k))
]1P (x(k))
|V(k)| = [Q(x(k))
|V(k)|
]1Q(x(k))
This approximation occurs in power systems in which |Gij | |Bij
|. In general ij is small so we can letsin ij 0. With this last set
of assumptions we then see that
Pi|Vj | = |Vi|(Gij cos ij +Bij sin ij) 0Qij
= |Vi||Vj |(Gij cos ij +Bij sin ij) 0
thereby providing a justification for the DPF approximation.A
modification of this approach known as fast decoupled power flow
(FDPF) goes one step further and
assumes that |Gij | = 0, sin ij = 0 and cos ij = 1. We assume
the slack bus voltage is |V1| = 1 and sothe update equation for all
other buses becomes
(k) = B1P (x(k))
|V(k)|
|V(k)| + B1 Q(x(k))
|V(k)|The susceptance matrix now becomes a good approximation to
the Jacobian and is essentially independentof voltage and angle.
This means, of course, that we only have to do the matrix inversion
once, at thebeginning of the algorithm, thereby avoiding
recomputation of the Jacobian and its inverse in each
iteration.
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Figure 5: Example 7.2
Example 7.2 FDPF Three Bus Example: This ex-amples one line
diagram is shown in Figure 5. Thebus admittance matrix for this
system is
Ybus = j
34.3 14.3 2014.3 24.3 1020 10 30
Since well treat bus 1 as the slack bus, the reducedsusceptance
matrix is
Br =
[ 24.3 1010 30
]If we use the fast decoupled power flow analysis, then we only
have to invert the approximate Jacobian once,
J1 = B1r =[ 0.0477 0.01590.0159 0.0389
]So for the initial iteration (k = 0) we select a flat start in
which
(0) =
[
(0)2
(0)3
]=
[00
], |V(0)| =
[|V(0)2 ||V(0)3 |
]=
[11
]We also need the initial power mismatch (per unit) for the real
and reactive power on bus 2. We assume
P (0) =
[22
], Q(0) =
[11
]The first recursion k = 1 yields[
(1)2
(3)3
]=
[00
]+
[ 0.0477 0.01590.0159 0.0389
] [22
]=
[ 0.1272.1091
][|V(1)2 ||V(1)3 |
]=
[11
]+
[ 0.0477 0.01590.0159 0.0389
] [11
]=
[0.93640.9455
]
We need to recompute the mismatches P(k)i
|V(k)i |and Q
(k)i
|V(k)i |to do the next recursion
Pi(x(k))
|V(k)i |=
n`=1
|V(k)` |(Gi` cos
(k)i` +Bi` sin i`
)+PDi PGi|V(k)i |
Qi(x(k))
|V(k)i |=
n`=1
|V(k)` |(Gi` sin
(k)i` Bi` cos i`
)+QDi QGi|V(k)i |
The power mismatches computed after this recursion are[
(2)2
(2)3
]=
[ 0.12720.1091
]+
[ 0.0477 0.01590.0159 0.0389
] [0.1510.107
]=
[ 0.13610.1156
][V
(2)2
V(2)3
]=
[0.93640.9455
]+
[ 0.0477 0.01590.0159 0.0389
] [0.1850.1780
]=
[0.92480.9357
]We then compute the mismatch as before and if it is small
enough, then stop. Otherwise continue. Like alllazy bods, we write
a MATLAB script to do the recursion for us
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%fpdf 3 bus exampleB = [-34.3 14.3 20; 14.3 -24.3 10; 20 10
-30];Br = B(2:3,2:3);
Pd2 = 2; Pg2 = 0;Qd2 = 1; Qg2 = 0;Pd3 = 2; Pg3 = 0;Qd3 = 1; Qg3
= 0;V1=1;del1=0;V2=1;del2=0;V3=1;del3=0;
del = [del2 ; del3];V = [V2 ; V3];delp=del;Vp = V;DelPV =
[Pd2;Pd3];DelQV = [Qd2;Qd3];
k=0;data=[ k del V norm([DelPV;DelQV])];
while (norm([DelPV;DelQV])> 1.e-4);
del = delp + inv(Br)*DelPVV = Vp + inv(Br)*DelQV
V2=V(1);V3=V(2);del2=del(1);del3=del(2);DelPV =
[V1*B(2,1)*sin(del2-del1)+V3*B(3,2)*sin(del2-del3)+(Pd2-Pg2)/V2;
V1*B(1,3)*sin(del3-del1)+V2*B(2,3)*sin(del3-del2)+(Pd3-Pg3)/V3];DelQV
=
[-V1*B(2,1)*cos(del2-del1)-V2*B(2,2)-V3*B(3,2)*cos(del2-del3)+(Qd2-Qg2)/V2;
-V1*B(1,3)*cos(del3-del1)-V2*B(2,3)*cos(del3-del2)-V3*B(3,3)+(Qd3-Qg3)/V3];
data = [data ; k del V norm([DelPV; DelQV])];
delp = del; Vp = V; k=k+1;if (k >= 10) break; end;
end;data
So after the final recursion, we get
k 2 3 |V2| |V3| P2/|V2| P3/|V3| Q2/|V2| Q3/|V3| mismatch0 0 0
1.0000 1.0000 2.0000 2.0000 1.0000 1.0000 3.16231 -0.1272 -0.1091
0.9364 0.9455 0.1506 0.1081 0.1850 0.1780 0.31672 -0.1361 -0.1156
0.9248 0.9357 0.0313 0.0192 0.0306 0.0262 0.05453 -0.1379 -0.1169
0.9229 0.9342 0.0053 0.0028 0.0058 0.0047 0.00964 -0.1382 -0.1171
0.9225 0.9339 0.0010 0.0005 0.0010 0.0008 0.00175 -0.1382 -0.1171
0.9225 0.9338 0.0002 0.0001 0.0002 0.0001 0.00036 -0.1382 -0.1171
0.9225 0.9338 0.0000 0.0000 0.0000 0.0000 0.0001
which means that the final answer is
V2 = 0.9224 7.9183, V3 = 0.9338 6.0793
The next most restrictive approximation is called the DC power
flow approximation. In this case, wecompletely ignore the reactive
power and assume that all voltages are always at 1.0 per unit and
we ignore
8
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line conductance. This turns the power flow into a set of linear
equations that can be directly solve as
= B1P
G
T1 T2345 kVLine 1 - 50 mi
2
3
400 MVA 15 kV
Y Y
G
Line
2 -
100
mi
Line
3 -
200
mi
400 MVA 15/345 kV
800 MVA 345/15 kV
280 Mvar 800 MW
800 MVA 15 kV
520 MW
40 Mvar 80 MW
1 4 5
Figure 6: Example 7.3
Example 7.3: Lets consider the five bus systemshown in Figure 6.
With bus 1 as the slack bus, theB matrix and P vector are
B =
30 0 10 20
0 100 100 010 100 150 4020 0 40 110
, P =8.04.400
So the DC power flow solution for the bus angles is
= B1P =
0.3260.0090.0350.072
radians =18.70.5212.004.125
degrees7.4 Control of Power Flows: The following means are used
to control a systems power flows
Control of the generators real power output and voltage level
Switching of shunt capacitor banks and static var compensators to
adjust line voltages (reactive power
support)
Control of tap-changing and regulating transformers to control
voltageThe objective is to find the settings for these controls
that ensure n 1 reliability.
P,Q
I ++
--V = |V|E = |E|
jX
Figure 7: Generator Model
The main method for controlling power flows is to adjustthe
power delivered by the generator. This is usually donethrough
control of the power angle and voltage. To see whyangle and voltage
control power output consider the generatormodel shown in Figure 7.
In this modelV is the terminal volt-age and E is called the
excitation voltage. The power angle is and X is the synchronous
reactance of the machine.
The generator current is
I =|E|ej V
jX
so the complex power delivered by the machine is
S = P + jQ = VI = V( |E|ej V
jX)
=|V||E|(j cos + sin ) j|V|2
X
and so the real and reactive powers are
P = Re(S) =|V||E|X
sin
Q = Im(S) =|V|X
(E cos V)
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The important thing to note about this last equation is that if
the excitation voltage is held relative constantthen we can control
the real power output through control of the power angle, . On the
other hand since ourexpression for Q has a cos in it that would be
close to 1 for small , then we should see that the
excitationvoltage E can be used to control the reactive power
delivered/absorbed by the generator. In practice, onewould use
power-flow analysis in conjunction with an optimization algorithm
to determine the optimalsetting for the generator real and reactive
power output. This problem will be discussed later and is calledthe
optimal power flow or OPF problem.
Another way of controlling voltage in a line is through shunt
capacitor banks. The addition of a shuntcapacitor bank will modify
the generator circuit shown in Figure 8. In this case we have a
generator withexcitation voltage E connected to a transmission line
whose series impedance is R + jX . Before thecapacitor bank is
connected the bus voltage V equals |E|. When the capacitor bank is
connected, wethen get a current flow Ic which essentially causes
voltage seen at the receiving end V to rotate by anamount that is
determine by the size of the capacitor. As discussed earlier,
capacitor banks act as reactivepower loads and they are essentially
used to control bus voltage.
+
-
E = |E| V +
-
I C jX R
C
sw
V
E = |E|
I C
RI C
jXI C
Figure 8: Shunt Capacitor Bank Controls
The last important class of controls we already discussed above.
These are tap-changing transformers thatare used to directly adjust
the voltage over one of the transformer windings. There are other
available controlmechanisms that include the switching on/off of
feeder lines (through breakers). Such controls can be usedto
redirect power flows and thereby avoid overloading specific lines
in the system. We can also, sometimes,directly command changes to
the load. This is sometimes referred to as demand-side management.
It wouldinvolve, for instance, slightly adjusting the setpoint of
an air conditioner unit, so that the user doesnt see theimpact, but
there is a small decrease in actual power consumption by the unit.
While doing this for a singleunit may not be that effective, if one
aggregates such small efficiencies over tens of thousands of
customers,the benefits can be very important to the utility in
terms of helping to reduce the variability in the demandside of the
system.
Determining the operating point of these controls is usually
done in a trial-and-error basis using com-puter tools such as Power
World to test out different scenarios and see which ones satisfy
the n1 reliabilitycondition. In some cases, engineering judgement
and some simple relationships characterizing the sensitiv-ity of
line power to generator control, or bus voltage to reactive power
controls can be used to estimate howmuch reactive or real power
should be adjusted. These estimates would be used as starting
points for asolution that would then be adjusted through the use of
computer tools.
Z=.5j
jQ G2
V2
S = 1 puD1
V = 1 0
S 12-S 21 S = 1 puD2
S G1
Figure 9: Voltage Sag at end of Transmis-sion Line
As an example, lets consider the problem of sizing a
shuntcapacitor to handle the voltage sag at the end of a
transmissionline. For most lines, the resistance is small so that Z
= jX . Inthis case the real power sent equals the active power
received(no losses). But the sent and received reactive power will
bedifferent. In particular, the real power sent and received
be-tween bus 1 and 2 in the system shown in Figure 9 is
P12 = P21 = |V1||V2|X
sin 12
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The reactive power sent and received however, will be different
since
Q12 =|V1|2X |V1||V2|
Xcos 12
Q21 =|V2|2X |V2||V1|
Xcos 12
If the generated reactive power at bus 2, QG2 is zero, then what
is the voltage magnitude at bus 2? WithQG2 = 0 we have
0 =|V2|2|X
|V2||V1|X
cos 12
Since P12 = P12 = PD2 = 1, we have
1 =|V1||V2|
Xsin 12
Simplifying the first equation yields,
|V2| = |V1| cos 12and simplifying the second equation
yields,
|V2| = X|V1| sin 12 .
SinceV1 = 10 we can further simplify to see that |V2| = cos 12.
Inserting this into the second relationyields,X = sin 12 cos 12. In
Figure 9,X = 1/2. So we now end up with two equations and two
unknowns
1
2= sin 12 cos 12, |V2| = cos 12
Solving both equations we get 12 = 45 and so |V2| = 1/2.Note
that the voltage at the receiving end is 1/2 pu, this is a
tremendous loss in voltage and clearly does
not satisfy the requirement that the user voltages stay within 5
percent of 1 pu. So we add some reactivepower support by connecting
a capacitor to the end of the line. So if we require |V2| = 1,
then
P12 =|V1||V2|
Xsin 12 = 2 sin 12 = 1
which means the angle across the line 12 must equals 30. The
reactive power absorbed by the shuntcapacitor is QG2 and
satisfies
QG2 = Q21 =|V2|2|X
|V2||V1|X
cos 12 = 2 2 cos 30 = 0.268
So we must size the capacitor to deliver 0.268 pu of reactive
power support.
Figure 10: Line SensitivityProblem
In many cases, we need to control the amount of power flowing
through atransmission line. To do this, we need to to determine how
a change in gener-ation at bus k affects the power flow on a line
from bus i to bus j (Figure 10).Well refer to this as the line
sensitivity problem. The assumption here is thatthe change in
generation is absorbed by the slack bus.
One way to address this problem is to compare the power flow
before andafter generator output was changed. The following example
for a three bus
11
-
Figure 11: Line Sensitivity Example (left) initial power flows
with line violation (right) adjusted systemwith line violation
removed
system solved using the PowerWorld tool shows how this is done.
Initially our system has a 124% overloadon the line from bus 1 to
2. But by increasing the generation at bus 3 by 95 MW (and also
decreasinggeneration by a similar amount on bus 1), there is a 31.3
MW drop in power flow across line to 1 and 2,which removes the
overload condition as shown in Figure 11.
Computing such control sensitivities through this
trial-and-error process, however, is very tedious and
timeconsuming. An alternative approach would analytically estimate
these values and then use the simulator tocheck or adjust your
answer. To see how this is done, consider the power flow from bus i
to bus j
Pij |Vi||Vj |Xij
sin(i j) i jXij
So Pij ijXij and so we need to compute is the sensitivity of the
power angle across the line as afunction of a change in generation
power, ijPGk .
We can obtain such analytic sensitivities from the fast
decoupled power flow (FDPF) analysis. From FDPFwe know
= B1P (x)
So to get the change in due to a change of generation at bus k,
we set P (x) to zero except at theposition k which we set equal to
1. We then use this test P vector to compute using the
precedingFDPF equation. The resulting vector can then be used to
evaluate the change in line power over all linesin the network.
Lets return to the three bus example again. For that earlier
example, Zline = j0.1 and the bus admittancematrix is
Ybus = j
20 10 1010 20 1010 10 20
Since bus 1 is the slack bus the reduced susceptance matrix
is
Br =
[ 20 1010 20
]Hence for a change of generation at bus 3 we get[
23
]=
[ 20 1010 20
]1 [01
]=
[0.03330.0667
]
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and so P31 = 0.066700.1 = 0.667 pu. Similar computations for the
other lines are P32 = 0.33 puand P21 = 0.33 pu. Now lets compare
this to the earlier PowerWorld result. Using PowerWorld, wefound
that a 95 MW change in generation at bus 3 resulted in a drop of
31.3 MW across line 1 and 2. This iscertainly consistent with the
sensitivities computed above using the FDPF approach to
analytically assessingline sensitivities.
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