Power-flow studies.pptx

8/15/2019 Power-flow studies.pptx

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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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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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/. =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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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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=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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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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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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*. 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.)

15


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/. 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.)

16


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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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( ) ( )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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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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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.

2'


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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.

(**.>.*)

(**.>.)

21


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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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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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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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(**.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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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?.

26


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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.*)

2,


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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*


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.

Documents

Power-flow studies.pptx