power-flow studies.pptx
TRANSCRIPT
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Power-Flow studies
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Introduction
We should be able to analyze the performance of power systems
both in normal operating conditions and under fault (short-circuit)
condition. The analysis in normal steady-state operation is called a
power-flow study (load-flow study).
Objectives
!t targets on determining the voltages" currents" and real and
reactive power flows in a system under a given load conditions.
The purpose of power flow studies is to plan ahead and account for
various hypothetical situations.
#or instance" what if a transmission line within the power system
properly supplying loads must be ta$en off line for maintenance.
%an the remaining lines in the system handle the re&uired loads
without e'ceeding their rated parameters
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Basic techniques for power-ow studies.
power-flow study (load-flow study) is an analysis of the voltages" currents"
and power flows in a power system under steady-state conditions. !n such astudy" we ma$e an assumption about either a voltage at a bus or the power being
supplied to the bus for each bus in the power system and then determine the
magnitude and phase angles of the bus voltages" line currents" etc. that would
result from the assumed combination of voltages and power flows.
The simplest way to perform power-flow calculations is by iteration*. %reate a bus admittance matri' +bus for the power system,
. a$e an initial estimate for the voltages at each bus in the system,
/. 0pdate the voltage estimate for each bus (one at a time)" based on the
estimates for the voltages and power flows at every other bus and the values
of the bus admittance matri' since the voltage at a given bus depends on the
voltages at all of the other busses in the system (which are just estimates)" theupdated voltage will not be correct. 1owever" it will usually be closer to the
answer than the original guess.
2. 3epeat this process to ma$e the voltages at each bus approaching the correct
answers closer and closer4
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Basic techniques for power-ow studies.
The e&uations used to update the estimates differ for different types of busses.
5ach bus in a power system can be classified to one of three types
*. 6oad bus (78 bus) 9 a buss at which the real and reactive power are specified"
and for which the bus voltage will be calculated. 3eal and reactive powers supplied
to a power system are defined to be positive" while the powers consumed from the
system are defined to be negative. ll busses having no generators are loadbusses.
. :enerator bus (7; bus) 9 a bus at which the magnitude of the voltage is $ept
constant by adjusting the field current of a synchronous generator on the bus (as
we learned" increasing the field current of the generator increases both the
reactive power supplied by the generator and the terminal voltage of the system).We assume that the field current is adjusted to maintain a constant terminal
voltage ;T. We also $now that increasing the prime mover
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Basic techniques for power-ow studies.
/. =lac$ bus (swing bus) 9 a special generator bus serving as the reference bus
for the power system. !ts voltage is assumed to be fi'ed in both magnitude and
phase (for instance" *∠>? pu). The real and reactive powers are uncontrolled thebus supplies whatever real or reactive power is necessary to ma$e the power flows
in the system balance.
!n practice" a voltage on a load bus may change with changing loads. Therefore"
load busses have specified values of 7 and 8" while ; varies with load conditions.
3eal generators wor$ most efficiently when running at full load. Therefore" it is
desirable to $eep all but one (or a few) generators running at *>>@ capacity" while
allowing the remaining (swing) generator to handle increases and decreases in
load demand. ost busses with generators will supply a fi'ed amount of powerand the magnitude of their voltages will be maintained constant by field circuits of
generators. These busses have specific values of 7 and A;iA.
The controls on the swing generator will be set up to maintain a constant voltage
and fre&uency" allowing 7 and 8 to increase or decrease as loads change.
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Constructing Y bus for power-ow
analysis
The most common approach to power-flow analysis is based on the bus admittancematri' Y bus. 1owever" this matri' is slightly different from the one studied previously
since the internal impedances of generators and loads connected to the system are
not included in +bus. !nstead" they are accounted for as specified real and reactive
powers input and output from the busses.5'ample **.* a simple power
system has 2 busses" B transmission
lines" * generator" and / loads.
=eries per-unit impedances are
line
Busto
!us
"eries# $pu%
"eries & $pu%
1 1-2 '.1()'.4
'.5**2- )2.352+
2 2-3 '.1()'.5
'.3*46- )1.+231
3 2-4 '.1()'
.4
'.5**2-
)2.352+
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Constructing Y bus for power-ow analysis
The shunt admittances of the transmission lines are ignored. !n this case" the Y ii terms of the bus admittance matri' can be constructed by summing the admittances
of all transmission lines connected to each bus" and the Y ij (i ≠ j ) terms are just the
negative of the line admittances stretching between busses i and j . Therefore" for
instance" the term Y 11 will be the sum of the admittances of all transmission lines
connected to bus *" which are the lines * and B" so Y 11 C *.DE2D 9 jD.>BFF pu.!f the shunt admittances of the transmission lines are not ignored" the self admittanceY ii at each bus would also include half of the shunt admittance of each transmission
line connected to the bus.
The term Y 12 will be the negative of all the admittances stretching between bus * and
bus " which will be the negative of the admittance of transmission line *" so Y 12 C->.BFF G j./BH.
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Constructing Y bus for power-ow
analysis
The complete bus admittance matri' can be obtained by repeating these calculationsfor every term in the matri'
1.7647 7.0588 0.5882 2.3529 0 1.1765 4.7059
0.5882 2.3529 1.5611 6.6290 0.3846 1.9231 0.5882 2.3529
0 0.3846 1.9231 1.5611 6.6290 1.1765 4.7059
1.1765 4.7059 0.5882 2.3529 1.1765 4.7059 2
bus
j j j
j j j jY
j j j
j j j
− − + − +
− + − − + − += − + − − +
− + − + − + .9412 11.7647 j
−
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Power-ow analsis equations
busY V I =
11 12 13 14 1 1
21 22 23 24 2 2
31 32 33 34 3 3
41 42 43 44 4 4
Y Y Y Y V I
Y Y Y Y V I
Y Y Y Y V I
Y Y Y Y V I
=
21 1 22 2 23 3 24 4 2Y V Y V Y V Y V I + + + =
The basic e&uation for power-flow analysis is derived from the nodal analysise&uations for the power system
#or the four-bus power system shown above" (**.H.*) becomes
where Y ij are the elements of the bus admittance matri'" V i are the bus voltages" and
I i are the currents injected at each node. #or bus in this system" this e&uation
reduces to
(**.H.*)
(**.H.)
(**.H./)
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Power-ow analsis equations
1owever" real loads are specified in terms of real and reactive powers" not ascurrents. The relationship between per-unit real and reactive power supplied to the
system at a bus and the per-unit current injected into the system at that bus is
*S VI P jQ= = +
where V is the per-unit voltage at the bus, I* - comple' conjugate of the per-unitcurrent injected at the bus, 7 and 8 are per-unit real and reactive powers. Therefore"
for instance" the current injected at bus can be found as
* * 2 2 2 2
2 2 2 2 2 2 *
2 2
P jQ P jQV I P jQ I I
V V
+ += + ⇒ = ⇒ =
(**.*>.*)
(**.*>.)
=ubstituting (**.*>.) into (**.H./)" we obtain
2 2
21 1 22 2 23 3 24 4 *
2
P jQY V Y V Y V Y V
V
++ + + = (**.*>./)
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Power-ow analsis equations
=olving the last e&uation for ;" yields
( )2 22 21 1 23 3 24 4*22 2
1 P jQV Y V Y V Y V
Y V
−= − + +
(**.**.*)
=imilar e&uations can be created for each load bus in the power system.
(**.**.*) gives updated estimate for ; based on the specified values of real and
reactive powers and the current estimates of all the bus voltages in the system. Iote
that the updated estimate for ; will not be the same as the original estimate of ;J
used in (**.**.*) to derive it. We can repeatedly update the estimate wile substituting
current estimate for ; bac$ to the e&uation. The values of ; will converge, however"
this would IOT be the correct bus voltage since voltages at the other nodes are alsoneeded to be updated. Therefore" all voltages need to be updated during each
iterationK
The iterations are repeated until voltage values no longer change much between
iterations.
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Power-ow analsis equations
This method is $nown as the :auss-=iedel iterative method. !ts basic procedure is
*. %alculate the bus admittance matri' Y bus including the admittances of all
transmission lines" transformers" etc." between busses but e'clude the
admittances of the loads or generators themselves.
. =elect a slac$ bus one of the busses in the power system" whose voltage will
arbitrarily be assumed as *.>∠>?.
/. =elect initial estimates for all bus voltages usually" the voltage at every load bus
assumed as *.>∠>? (flat start) lead to good convergence.
2. Write voltage e&uations for every other bus in the system. The generic form is
*1
1 N i ii ik k
k ii ik i
P jQV Y V
Y V =≠
− ÷= − ÷ ÷
∑ (**.*.*)
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Power-ow analsis equations
B. %alculate an updated estimate of the voltage at each load bus in succession using(**.*.*) e'cept for the slac$ bus.
E. %ompare the differences between the old and new voltage estimates if the
differences are less than some specified tolerance for all busses" stop.
Otherwise" repeat step B.
D. %onfirm that the resulting solution is reasonable a valid solution typically has busvoltages" whose phases range in less than 2B?.
5'ample **. in a -bus power system" a generator
attached to bus * and loads attached to bus . the series
impedance of a single transmission line connecting them
is >.*Gj>.B pu. The shunt admittance of the line may beneglected. ssume that bus * is the slac$ bus and that it
has a voltage V 1 C *.>∠>? pu. 3eal and reactive powerssupplied to the loads from the system at bus are P 2 C
>./ pu" Q2 C >. pu (powers supplied to the system at
each busses is negative of the above values). Letermine
voltages at each bus for the specified load conditions.
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Power-ow analsis equations
*. We start from calculating the bus admittance matri' Y bus. The Y ii terms can beconstructed by summing the admittances of all transmission lines connected to each
bus" and the Y ij terms are the negative of the admittances of the line stretching
between busses i and j . #or instance" the term Y 11 is the sum of the admittances of
all transmission lines connected to bus * (a single line in our case). The series
admittance of line * is1
1
111 1 0.3846 1.9231
0.1 0.5line
line
Y j j
Y Z
= = = − =+
pplying similar calculations to other terms" we complete the admittance matri' as
0.3846 1.9231 0.3846 1.9231
0.3846 1.9231 0.3846 1.9231bus
j jY j j
− − + = − + −
. Ie't" we select bus * as the slac$ bus since it is the only bus in the system
connected to the generator. The voltage at bus * will be assumed *.>∠>?.
(**.*2.*)
(**.*2.)
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Power-ow analsis equations
/. We select initial estimates for all bus voltages. a$ing a flat start" the initialvoltage estimates at every bus are *.>∠>?.
2. Ie't" we write voltage e&uations for every other bus in the system. #or bus
2 2
2 21 1*22 2,
1
old
P jQV Y V
Y V
−
= − =ince the real and reactive powers supplied to the system at bus are P 2 ->./ pu
and Q2 ->. pu and since Y s and V 1 are $nown" we may reduce the last e&uation
( )( )
( ) ( )
2 1*
2,
*
2,
1 0.3 0.2
0.3846 1.92310.3846 1.9231
1 0.3603 146.31.9612 101.3 1 0
1.9612 78.8
old
old
j
V j V j V
V
− −= − − + −
∠ − °= − ∠ ° ∠ °
∠ − °
(**.*B.*)
(**.*B.)
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Power-ow analsis equations
B. Ie't" we calculate an updated estimate of the voltages at each load bus insuccession. !n this problem we only need to calculate updated voltages for bus "
since the voltage at the slac$ bus (bus *) is assumed constant. We repeat this
calculation until the voltage converges to a constant value.
The initial estimate for the voltage is V 2!" *∠>?. The ne't estimate for the voltage is
( ) ( )
( )
2,1 *
2,
1 0.3603 146.31.9612 101.3 1 0
1.9612 78.8
1 0.3603 146.31.9612 101.3
1.9612 78.8 1 0
0.8797 8.499
old
V V
∠ − °= − ∠ ° ∠ °
∠ − ° ∠ − ° = − ∠ ° ∠ − ° ∠ °
= ∠ − °
This new estimate for V substituted bac$ to the e&uation will produce the second
estimate
(**.*E.*)
1,
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Power-ow analsis equations
( ) ( )2,2 1 0.3603 146.3 1.9612 101.3 1 01.9612 78.8 0.8797 8.499
0.8412 8.499
V ∠ − ° = − ∠ ° ∠ ° ∠ − ° ∠ − ° = ∠ − °
The third iteration will be
( ) ( )2,31 0.3603 146.3
1.9612 101.3 1 01.9612 78.8 0.8412 8.499
0.8345 8.962
V
∠ − °
= − ∠ ° ∠ ° ∠ − ° ∠ − ° = ∠ − °
The fourth iteration will be
( ) ( )2,41 0.3603 146.3
1.9612 101.3 1 01.9612 78.8 0.8345 8.962
0.8320 8.962
V ∠ − ° = − ∠ ° ∠ ° ∠ − ° ∠ − °
= ∠ − °The fifth iteration will be
( ) ( )2,51 0.3603 146.3
1.9612 101.3 1 01.9612 78.8 0.8320 8.9
0.8315 8.994
62V
∠ − ° = − ∠ ° ∠ ° ∠ − ° ∠ − °
∠ −
= °
(**.*D.*)
(**.*D.)
(**.*D./)
(**.*D.2)
1*
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1*
Power-ow analsis equations
This power system converged to the answer in five iterations. The voltages at each
bus in the power system are
E. We observe that the magnitude of the voltage is barely changing and mayconclude that this value is close to the correct answer and" therefore" stop the
iterations.
1
2
1.0 0
0.8315 8.994
V
V
= ∠ °= ∠ − °
D. #inally" we need to confirm that the resulting solution is reasonable. The results
seem reasonable since the phase angles of the voltages in the system differ by only*>?. The current flowing from bus * to bus is
1 2
1
1
1 0 0.8315 8.9940.4333 42.65
0.1 0.5line
V V I
Z j
− ∠ ° − ∠ − °= = = ∠ − °
+
(**.*F.*)
(**.*F.)
1+
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1+
Power-ow analsis equations
The power supplied by the transmission line to bus is
( ) ( ) **
0.8315 8.994 0.4333 42.65 0.2999 0.1997S VI j= = ∠ − ° ∠ − ° = +
This is the amount of power consumed by the loads, therefore" this solution appears
to be correct.
Iote that this e'ample must be interpreted as follows if the real and reactive power
supplied by bus is >./ G j>. pu and if the voltage on the slac$ bus is * ∠>? pu" thenthe voltage at bus will be V C >.F/*B∠-F.HH2?.
This voltage is correct only for the assumed conditions, another amount of power
supplied by bus will result in a different voltage V .
Therefore" we usually postulate some reasonable combination of powers supplied to
loads" and determine the resulting voltages at all the busses in the power system.
Once the voltages are $nown" currents through each line can be calculated.
The relationship between voltage and current at a load bus as given by (**.*.*) is
fundamentally nonlinearK Therefore" solution greatly depends on the initial guess.
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ddin/ /enerator !usses to power-ow studies
t a generator bus" the real power P i and the magnitude of the bus voltage AV i A arespecified. =ince the reactive power for that bus is usually un$nown" we need to
estimate it before applying (**.*.*) to get updated voltage estimates. The value of
reactive power at the generator bus can be estimated by solving (**.*.*) for Qi
**
1 1
1
N N
i ii ik k i i i ii i ik k
k k ii ik i k i
P jQV Y V P jQ V Y V Y V Y V = =
≠ ≠
− ÷ ÷= − ⇔ − = − ÷ ÷ ÷ ÷
∑ ∑
Mringing the case # I into summation" we obtain
*
1
*
1Im
N N
i i ik k i i i ik k
k k P jQ V Y Q V V Y V
==− = ⇒
= − ∑∑Once the reactive power at the bus is estimated" we can update the bus voltage at a
generator bus using P i and Qi as we would at a load bus. 1owever" the magnitude of
the generator bus voltage is also forced to remain constant. Therefore" we must
multiply the new voltage estimate by the ratio of magnitudes of old to new estimates.
(**.>.*)
(**.>.)
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ddin/ /enerator !usses to power-ow studies
Therefore" the steps re&uired to update the voltage at a generator bus are
*. 5stimate the reactive power Qi according to (**.>.),
. 0pdate the estimated voltage at the bus according to (**.*.*) as if the
bus was a load bus,
/. #orce the magnitude of the estimated voltage to be constant by
multiplying the new voltage estimate by the ratio of the magnitude of the
original estimate to the magnitude of the new estimate. This has the
effect of updating the voltage phase estimate without changing the
voltage amplitude.
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ddin/ /enerator !usses to power-ow studies
5'ample **./ a 2-bus power systemwith B transmission lines" generators"
and loads. =ince the system has
generators connected to busses" it
will have one slac$ bus" one generator
bus" and two load busses. ssume that
bus * is the slac$ bus and that it has a
voltage V 1 C *.>∠>? pu. Mus / is agenerator bus. The generator is
supplying a real power P $ C >./ pu to
the system with a voltage magnitude *
pu. The per-unit real and reactivepower loads at busses and 2 are P 2 C>./ pu" Q2 C >. pu" P % C >. pu" Q% C >.*B pu (powers supplied to the system at each
busses are negative of the above values). The series impedances of each bus were
evaluated in 5'ample **.*. Letermine voltages at each bus for the specified load
conditions.
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ddin/ /enerator !usses to power-ow studies
The bus admittance matri' was calculated earlier as
1.7647 7.0588 0.5882 2.3529 0 1.1765 4.7059
0.5882 2.3529 1.5611 6.6290 0.3846 1.9231 0.5882 2.3529
0 0.3846 1.9231 1.5611 6.6290 1.1765 4.7059
1.1765 4.7059 0.5882 2.3529 1.1765 4.7059 2
bus
j j j
j j j jY
j j j
j j j
− − + − +− + − − + − +
=− + − − +
− + − + − + .9412 11.7647 j
−
=ince the bus / is a generator bus" we will have to estimate the reactive power at
that bus before calculating the bus voltages" and then force the magnitude of the
voltage to remain constant after computing the bus voltage. We will ma$e a flat start
assuming the initial voltage estimates at every bus to be *.>∠>?.Therefore" the se&uence of voltage (and reactive power) e&uations for all busses is
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ddin/ /enerator !usses to power-ow studies
(**.2.*)
(**.2.)
(**.2./)
( )
( )
( )
2 2
2 21 1 23 3 24 4*
22 2,
*
3 3
1
3 3
3 31 1 32 2 34 4*
33 3,
3,3 3
3
4 4
4 41 1 42 2 43 3*
44 4,
1
Im
1
1
old
N
ik k
k
old
old
old
P jQV Y V Y V Y V
Y V
Q V Y V
P jQV Y V Y V Y V
Y V
V V V V
P jQV Y V Y V Y V
Y V
=
−= − + +
= −
−= − + +
=
−= − + +
∑
(**.2.2)
(**.2.2)
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ddin/ /enerator !usses to power-ow studies
1
2
3
4
1.0 0
0.964 0.97
1.0 1.84
0.98 0.27
V pu
V pu
V pu
V pu
= ∠ °
= ∠ − °= ∠ °= ∠ − °
The voltages and the reactive power should be updated iteratively" for instance"
using atlab.
%omputations converge to the following solution
(**.B.*)
The solution loo$s reasonable since the bus voltage phase angles is less than 2B?.
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0he inforation deried fro power-ow studies
fter the bus voltages are calculated at all busses in a power system" a power-flowprogram can be set up to provide alerts if the voltage at any given bus e'ceeds" for
instance" ±B@ of the nominal value. This is important since the power needs to besupplied at a constant voltage level, therefore" such voltage variations may indicate
problems4
dditionally" it is possible to determine the net real and reactive power either suppliedor removed from the each bus by generators or loads connected to it. To calculate
the real and reactive power at a bus" we first calculate the net current injected at the
bus" which is the sum of all the currents leaving the bus through transmission lines.
The current leaving the bus on each transmission line can be found as
( )1
N
i ik i k
k k i
I Y V V =≠
= −∑ (**.E.*)
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2,
0he inforation deried fro power-ow studies
The resulting real and reactive powers injected at the bus can be found from
*
i i i i iS V I P jQ= − = +where the minus sign indicate that current is assumed to be injected instead of
leaving the node.
=imilarly" the power-flow study can show the real and reactive power flowing in every
transmission line in the system. The current flow out of a node along a particular
transmission line between bus i and bus j can be calculated as
(**.D.*)
( )ij ij i j I Y V V = −where Y ij is the admittance of the transmission line between those two busses. The
resulting real and reactive power can be calculated as
*
ij i ij ij ijS V I P jQ= − = +
2*
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2*
0he inforation deried fro power-ow studies
lso" comparing the real and reactive power flows at either end of the transmissionline" we can determine the real and reactive power losses on each line.
!n modern power-flow programs" this information is displayed graphically. %olors are
used to highlight the areas where the power system is overloaded" which aids Nhot
spot localization.
7ower-flow studies are usually started from analysis of the power system in itsnormal operating conditions" called the base case. Then" various (increased) load
conditions may be projected to localize possible problem spots (overloads). My
adding transmission lines to the system" a new configuration resolving the problem
may be found. This estimated models can be used for planning.
nother reason for power-flow studies is modeling possible failures of particular linesand generators to see whether the remaining components can handle the loads.
#inally" it is possible to determine more efficient power utilization by redistributing
generation from one locations to other. This variety of power-flow studies is called
economic dispatch.