Download - Module 6 Introduction to Power Flow Studies
-
8/3/2019 Module 6 Introduction to Power Flow Studies
1/49
BEE 3243Electric Power Systems
BEE 3243BEE 3243Electric Power SystemsElectric Power Systems
Module 5
Introduction to Power Flow Studies
Module 5Module 5
Introduction to Power Flow StudiesIntroduction to Power Flow Studies
-
8/3/2019 Module 6 Introduction to Power Flow Studies
2/49
BEE 3243 Electric Power Systems Module 5 22
OutlinesOutlines
IntroductionIntroduction
Basic Techniques for Power Flow StudiesBasic Techniques for Power Flow Studies
The Bus Admittance MatrixThe Bus Admittance Matrix Power Flow EquationsPower Flow Equations GaussGauss--Seidel MethodSeidel Method GaussGauss--SeidalSeidal Power Flow SolutionPower Flow Solution
-
8/3/2019 Module 6 Introduction to Power Flow Studies
3/49
BEE 3243 Electric Power Systems Module 5 33
IntroductionIntroduction
Power flow study is also known asPower flow study is also known as load flowload flow study.study.
It is an analysis duringIt is an analysis during steadysteady--statestate
conditions.conditions.
It is used forIt is used forplanningplanning andand controllingcontrolling a system.a system. Assumptions:Assumptions: balancedbalanced conditions andconditions and singlesinglephasephase
analysis.analysis.
Problems:Problems:
determine thedetermine the voltage magnitudevoltage magnitude
andand phase anglephase angle
atat
each bus.each bus.
determine the active and reactive (determine the active and reactive (P & QP & Q) power flow in) power flow ineach lineeach line
each bus haseach bus has fourfour
state variables:state variables: voltage magnitude,voltage magnitude,
voltage phase angle, real power injection, and reactivevoltage phase angle, real power injection, and reactivepower injectionpower injection
-
8/3/2019 Module 6 Introduction to Power Flow Studies
4/49
BEE 3243 Electric Power Systems Module 5 44
IntroductionIntroduction
Each bus hasEach bus has twotwo of the four state variablesof the four state variablesdefined or given.defined or given. Text book:Text book: HadiHadi SaadatSaadat Power System AnalysisPower System Analysis((Chapter 6Chapter 6).).
-
8/3/2019 Module 6 Introduction to Power Flow Studies
5/49
BEE 3243 Electric Power Systems Module 5 55
Types of Buses in Power SystemsTypes of Buses in Power Systems
Types of network buses:Types of network buses: Load BusLoad Bus
ororPQ BusPQ Bus
known real (P) and reactive (Q) power injections.known real (P) and reactive (Q) power injections.
No generator attach to load bus.No generator attach to load bus.
Generator BusGenerator Bus
ororPV BusPV Bus
known real (P) power injection and the voltageknown real (P) power injection and the voltage
magnitude (V).magnitude (V).
Slack BusSlack Bus
ororSwing BusSwing Bus
known voltage magnitude (V) and voltage angle (known voltage magnitude (V) and voltage angle (),),
often it is taken to beoften it is taken to be 1100
p.up.u..
must have one generator as the slack bus.must have one generator as the slack bus. takes up the power slack due to losses in the network.takes up the power slack due to losses in the network.
-
8/3/2019 Module 6 Introduction to Power Flow Studies
6/49BEE 3243 Electric Power Systems Module 566
Basic Techniques for Power Flow StudiesBasic Techniques for Power Flow Studies
Power flow analysis is anPower flow analysis is an iterativeiterative problem.problem.StepsSteps
to be taken in power flow analysis:to be taken in power flow analysis:
1)1) One line diagram orOne line diagram orload flow dataload flow data of a power systemof a power system2)2)
Construct Bus Admittance Matrix (Construct Bus Admittance Matrix (YbusYbus))
3)3)
Calculate theCalculate the power flowpower flow
analysis equationsanalysis equations
Power flow is aPower flow is a nonlinearnonlinear problem and it isproblem and it iscommonly solved by the iterative solution ofcommonly solved by the iterative solution ofnonlinear algebraic equationsnonlinear algebraic equations::
GaussGauss--SeidalSeidal
NewtonNewton--RaphsonRaphson
Fast DecoupledFast Decoupled
-
8/3/2019 Module 6 Introduction to Power Flow Studies
7/49BEE 3243 Electric Power Systems Module 577
Basic Techniques for Power Flow StudiesBasic Techniques for Power Flow Studies
Example ofExample ofload flow input dataload flow input data:: Bus dataBus data
Line dataLine data
Bus no Bus
code
Voltage Load Generator
Magnitude
(p.u.)
Angle
(degree)
P
(MW)
Q
(Mvar)
P
(MW)
Q
(Mvar)
1 1 1.06 0 0.0 0.0 0.0 0.0
2 2 1.043 0 21.70 12.7 40.0 0.03 0 1.0 0 2.4 1.2 0.0 0.0
Line bus no
(From)
Line bus no
(To)
Lineresistance
R (p.u.)
Linereactance
X (p.u.)
linesusceptance
B (p.u.)
Transfor-mer tapsetting
1 2 0.0192 0.0575 0.02640 0.978
-
8/3/2019 Module 6 Introduction to Power Flow Studies
8/49BEE 3243 Electric Power Systems Module 588
The Bus Admittance MatrixThe Bus Admittance Matrix
The matrix equation for relating theThe matrix equation for relating the nodal voltagesnodal voltages to theto the currentscurrents
that flow into and out of a networkthat flow into and out of a network
using theusing the admittanceadmittance values of circuit branches.values of circuit branches.V1
Vi
Ii
yi1
yi2
yin
yi0
V2
Vn
Iinj = YbusVnodeIinj = YbusVnode
ijijij
ijjxrz
y
11
nnnnn
n
n
n V
V
V
YYY
YYY
YYY
I
I
I
2
1
21
22221
11211
2
1
-
8/3/2019 Module 6 Introduction to Power Flow Studies
9/49BEE 3243 Electric Power Systems Module 5 99
The Bus Admittance MatrixThe Bus Admittance Matrix
One line diagram of a power system
-
8/3/2019 Module 6 Introduction to Power Flow Studies
10/49BEE 3243 Electric Power Systems Module 5 1010
The Bus Admittance MatrixThe Bus Admittance Matrix
Impedance Diagram
-
8/3/2019 Module 6 Introduction to Power Flow Studies
11/49BEE 3243 Electric Power Systems Module 5 1111
The Bus Admittance MatrixThe Bus Admittance Matrix
Admittance Diagram
-
8/3/2019 Module 6 Introduction to Power Flow Studies
12/49BEE 3243 Electric Power Systems Module 5 1212
The Bus Admittance MatrixThe Bus Admittance Matrix
Kirchhoffs current law:
-
8/3/2019 Module 6 Introduction to Power Flow Studies
13/49BEE 3243 Electric Power Systems Module 5 1313
The Bus Admittance MatrixThe Bus Admittance Matrix
Rearranging the KCL Equations:Rearranging the KCL Equations:
Matrix Formation of the EquationsMatrix Formation of the Equations::
-
8/3/2019 Module 6 Introduction to Power Flow Studies
14/49
BEE 3243 Electric Power Systems Module 5 1414
The Bus Admittance MatrixThe Bus Admittance Matrix
Completed Matrix Equation:Completed Matrix Equation:
nnnnn
nn
n V
V
V
YYY
YYY
YYY
I
I
I
21
21
2222111211
21
-
8/3/2019 Module 6 Introduction to Power Flow Studies
15/49
BEE 3243 Electric Power Systems Module 5 1515
YY--Bus Matrix Building RulesBus Matrix Building Rules
Square matrixSquare matrix with dimensions equal to thewith dimensions equal to thenumber of buses.number of buses. Convert all network impedances intoConvert all network impedances into admittancesadmittances.. DiagonalDiagonal elements:elements:
OffOff--diagonaldiagonal
elements:elements:
Matrix isMatrix is symmetricalsymmetrical along the leading diagonal.along the leading diagonal.
-
8/3/2019 Module 6 Introduction to Power Flow Studies
16/49
BEE 3243 Electric Power Systems Module 5 1616
Power Flow EquationsPower Flow Equations
V1Vi
Ii
yi1
yi2
yin
yi0
V2
V3
KCL Equations:KCL Equations:
-
8/3/2019 Module 6 Introduction to Power Flow Studies
17/49
BEE 3243 Electric Power Systems Module 5 1717
Power Flow EquationsPower Flow Equations
Power flow equation:Power flow equation:
-
8/3/2019 Module 6 Introduction to Power Flow Studies
18/49
BEE 3243 Electric Power Systems Module 5 1818
GaussGauss--Seidel MethodSeidel Method
GaussGauss--SeidalSeidal is ais a nonlinearnonlinear algebraic equationalgebraic equationsolver. It is a method of successive displacements.solver. It is a method of successive displacements. ItsIts iterative stepsiterative steps::
take a function and rearrange it into the formtake a function and rearrange it into the form x =x = g(xg(x))
make an initial estimate of the variable x:make an initial estimate of the variable x: xx[0][0]
= initial value= initial value
find an iterative improvement offind an iterative improvement ofxx[k[k]]
, that is:, that is: xx[k+1][k+1]
== g(xg(x[k[k]]
))
a solution is reached when the difference between twoa solution is reached when the difference between two
iterations is less than a specified accuracy:iterations is less than a specified accuracy: |x|x[k+1][k+1]
xx[k[k]]||
AccelerationAcceleration factors (factors ())::
can improve the rate of convergence:can improve the rate of convergence:
> 1> 1
modified step: the improvement is found asmodified step: the improvement is found as
xx[k+1][k+1]
= x= x[k][k]
++
[[g( xg( x[k][k]
))
xx[k][k]]]
-
8/3/2019 Module 6 Introduction to Power Flow Studies
19/49
BEE 3243 Electric Power Systems Module 5 1919
GaussGauss--Seidel MethodSeidel Method
Example of the GaussExample of the Gauss--Seidel method:Seidel method:Find x of the equation:Find x of the equation: f(xf(x) = x) = x33--6x6x22+9x+9x--4 = 0.4 = 0.
9x =9x = --xx33+6x+6x22+4+4
Start withStart with initialinitial
guess xguess x[0][0]
= 2,= 2,
)(9
4
9
6
9
123 xgxxx
5173.29
4)2222.2(
9
6)2222.2(
9
1)2222.2(
2222.29
4)2(
9
6)2(
9
1)2(
23]1[]2[
23]0[]1[
xgx
xgx
-
8/3/2019 Module 6 Introduction to Power Flow Studies
20/49
BEE 3243 Electric Power Systems Module 5 2020
GaussGauss--Seidel MethodSeidel Method
0000.4
9988.3
9568.39
4)7398.3(
9
6)7398.3(
9
1)7398.3(
7398.39
4)3376.3(
9
6)3376.3(
9
1)3376.3(
3376.394)8966.2(
96)8966.2(
91)8966.2(
8966.29
4)5173.2(
9
6)5173.2(
9
1)5173.2(
]8[
]7[
23]5[]6[
23]4[]5[
23]3[]4[
23]2[]3[
x
x
xgx
xgx
xgx
xgx
-
8/3/2019 Module 6 Introduction to Power Flow Studies
21/49
BEE 3243 Electric Power Systems Module 5 2121
GaussGauss--Seidel MethodSeidel Method
MatlabMatlab Results of all iterations:Results of all iterations:Iter g dx x1 2.2222 0.2222 2.2222
2 2.5173 0.2951 2.5173
3 2.8966 0.3793 2.8966
4 3.3376 0.4410 3.3376
5 3.7398 0.4022 3.7398
6 3.9568 0.2170 3.9568
7 3.9988 0.0420 3.9988
8 4.0000 0.0012 4.0000
9 4.0000 0.0000 4.0000
-
8/3/2019 Module 6 Introduction to Power Flow Studies
22/49
BEE 3243 Electric Power Systems Module 5 2222
GaussGauss--Seidel MethodSeidel Method
Graphical illustration:Graphical illustration:
Iterations
x = g(x)
Initial value
Solution points
-
8/3/2019 Module 6 Introduction to Power Flow Studies
23/49
BEE 3243 Electric Power Systems Module 5 2323
GaussGauss--Seidel Method withSeidel Method with
Find the root of the equation:Find the root of the equation: f(xf(x) = x) = x33
-- 6x6x
22
+ 9x+ 9x -- 44= 0 with an= 0 with an acceleration factoracceleration factor
of 1.25.of 1.25.
Starting with anStarting with an initialinitial guess of xguess of x[0][0] = 2.= 2.
-
8/3/2019 Module 6 Introduction to Power Flow Studies
24/49
BEE 3243 Electric Power Systems Module 5 2424
GaussGauss--Seidel Method withSeidel Method with
MatlabMatlab results of all iterations:results of all iterations:Iter g dx x
1 2.2222 0.2222 2.2778
2 2.5902 0.3124 2.6683
3 3.0801 0.4118 3.1831
4 3.6157 0.4326 3.7238
5 3.9515 0.2277 4.0084
6 4.0000 -0.0085 3.9978
7 4.0000 0.0022 4.0005
8 4.0000 -0.0005 3.9999
-
8/3/2019 Module 6 Introduction to Power Flow Studies
25/49
BEE 3243 Electric Power Systems Module 5 2525
GaussGauss--Seidel Method withSeidel Method with
Graphical illustration:Graphical illustration:
-
8/3/2019 Module 6 Introduction to Power Flow Studies
26/49
BEE 3243 Electric Power Systems Module 5 2626
GaussGauss--Seidel Method withSeidel Method with
Do not use aDo not use a very largevery large number ofnumber of as the largeras the largerstep sizestep size
may result in an overshoot.may result in an overshoot.
If we take the previous example withIf we take the previous example with = 1.8= 1.8, we, wewill needwill need more iterationsmore iterations to obtain the answer:to obtain the answer:Iter g dx x1 2.2222 0.2222 2.4000
2 2.7484 0.3484 3.0272
3 3.4714 0.4442 3.8268
4 3.9806 0.1538 4.1036
5 3.9927 -0.1109 3.9040
Iter g dx x6 3.9940 0.0900 4.0659
7 3.9971 -0.0688 3.9420
8 3.9978 0.0558 4.0424
9 3.9988 -0.0436 3.9639
10 3.9991 0.0352 4.0273Overshoot
-
8/3/2019 Module 6 Introduction to Power Flow Studies
27/49
BEE 3243 Electric Power Systems Module 5 2727
GaussGauss--Seidel Method withSeidel Method with
Iter g dx x11 3.9995 -0.0278 3.9772
12 3.9997 0.0224 4.0176
13 3.9998 -0.0178 3.9856
14 3.9999 0.0143 4.0113
15 3.9999 -0.0114 3.9908
16 3.9999 0.0091 4.0073
17 4.0000 -0.0073 3.9941
18 4.0000 0.0058 4.0047
Iter g dx x19 4.0000 -0.0047 3.9963
20 4.0000 0.0037 4.0030
21 4.0000 -0.0030 3.9976
22 4.0000 0.0024 4.0019
23 4.0000 -0.0019 3.9985
24 4.0000 0.0015 4.0012
25 4.0000 -0.0012 3.9990
26 4.0000 0.0010 4.0008
-
8/3/2019 Module 6 Introduction to Power Flow Studies
28/49
BEE 3243 Electric Power Systems Module 5 2828
GaussGauss--Seidel Method withSeidel Method with
Graphical illustration:Graphical illustration:
OVERSHOOT
-
8/3/2019 Module 6 Introduction to Power Flow Studies
29/49
BEE 3243 Electric Power Systems Module 5 2929
GaussGauss--SeidalSeidal for a System of n Equationsfor a System of n Equations Consider a system ofConsider a system ofn equationsn equations::
Rearrange each equation for each of the variables:Rearrange each equation for each of the variables:
-
8/3/2019 Module 6 Introduction to Power Flow Studies
30/49
BEE 3243 Electric Power Systems Module 5 3030
GaussGauss--SeidalSeidal for a System of n Equationsfor a System of n Equations Steps:Steps:Assume anAssume an approximate solutionapproximate solution
for the independentfor the independent
variables,variables,
Find the results in aFind the results in a new approximatenew approximate solutionsolution
In theIn the
GaussGauss--SeidelSeidel
method, the updated values of themethod, the updated values of the
variables calculated in the preceding equations are usedvariables calculated in the preceding equations are used
immediately in the solution of the subsequent equations.immediately in the solution of the subsequent equations.
The rate ofThe rate ofconvergenceconvergence can be increased by suitablecan be increased by suitable ..
G S id l P Fl S l i
-
8/3/2019 Module 6 Introduction to Power Flow Studies
31/49
BEE 3243 Electric Power Systems Module 5 3131
GaussGauss--SeidalSeidal Power Flow SolutionPower Flow Solution Previously derivedPreviously derived power flow equationpower flow equation,,
GaussGauss--SeidalSeidal form,form,
G S id l P Fl S l ti
-
8/3/2019 Module 6 Introduction to Power Flow Studies
32/49
BEE 3243 Electric Power Systems Module 5 3232
GaussGauss--SeidalSeidal Power Flow SolutionPower Flow Solution Rewriting the power equation to findRewriting the power equation to find P and QP and Q
thethe real and reactive powersreal and reactive powers
are scheduled for theare scheduled for the loadload
busesbuses
that is, they remain fixedthat is, they remain fixed
the currents and powers are expressed as going into thethe currents and powers are expressed as going into thebusbus
forforgenerationgeneration the powers arethe powers are positivepositive
forforloadsloads
the powers arethe powers are negativenegative
thethe scheduled powerscheduled power is theis the sumsum of the generation and loadof the generation and loadpowerspowers
GG S id lS id l P Fl S l iP Fl S l ti
-
8/3/2019 Module 6 Introduction to Power Flow Studies
33/49
BEE 3243 Electric Power Systems Module 5 3333
GaussGauss--SeidalSeidal Power Flow SolutionPower Flow Solution TheThe complete setcomplete set of equations become:of equations become:
GG S id lS id l P Fl S l tiP Fl S l ti
-
8/3/2019 Module 6 Introduction to Power Flow Studies
34/49
BEE 3243 Electric Power Systems Module 5 3434
GaussGauss--SeidalSeidal Power Flow SolutionPower Flow Solution Since theSince the offoff--diagonaldiagonal elements ofelements ofYYbusbus
,, YYijij
== --yyijij
and theand the diagonaldiagonal
elements,elements, YYiiii
== yyijij
,,
ii
n
ijj
kjijk
i
schi
schi
ki
Y
VYV
jQP
V
,1
][
][*]1[
n
ijj
kjijii
ki
ki
ki
n
ijj
kjijii
ki
ki
ki
VYYVVQ
VYYVVP
,1
][][][*]1[
,1
][][][*]1[
GG S id lS id l P Fl S l tiP Fl S l ti
-
8/3/2019 Module 6 Introduction to Power Flow Studies
35/49
BEE 3243 Electric Power Systems Module 5 3535
GaussGauss--SeidalSeidal Power Flow SolutionPower Flow Solution System characteristics:System characteristics:Since both components (V &Since both components (V & ) are specified for the) are specified for the
slack busslack bus, there are 2(n, there are 2(n -- 1) equations which must be1) equations which must besolved iterativelysolved iterativelyFor theFor the load busesload buses, the real and reactive powers are, the real and reactive powers are
known/ scheduled:known/ scheduled:
the voltage magnitude and angle must be estimatedthe voltage magnitude and angle must be estimated
in per unit, the nominal voltage magnitude is 1in per unit, the nominal voltage magnitude is 1 pupu
the angles are generally close together, so an initial value ofthe angles are generally close together, so an initial value of00
degrees is appropriatedegrees is appropriate
GG S id lS id l P Fl S l tiP Fl S l ti
-
8/3/2019 Module 6 Introduction to Power Flow Studies
36/49
BEE 3243 Electric Power Systems Module 5 3636
GaussGauss--SeidalSeidal Power Flow SolutionPower Flow SolutionFor theFor the generator busesgenerator buses, the real power and voltage, the real power and voltage
magnitude are known,magnitude are known,
the real power is scheduledthe real power is scheduled
the reactive power is computed based on the estimated voltagethe reactive power is computed based on the estimated voltage
valuesvalues
the voltage is computed by Gaussthe voltage is computed by Gauss--Seidel, only the imaginarySeidel, only the imaginary
part is keptpart is kept
the complex voltage is found from the magnitude and thethe complex voltage is found from the magnitude and the
iterative imaginary partiterative imaginary part
Li Fl d LLi Fl d L
-
8/3/2019 Module 6 Introduction to Power Flow Studies
37/49
BEE 3243 Electric Power Systems Module 5 3737
Line Flows and LossesLine Flows and Losses
After solving for bus voltages and angles,After solving for bus voltages and angles, powerpowerflowsflows
andand losseslosses
on the network branches areon the network branches are
calculated.calculated.Transmission lines and transformers are networkTransmission lines and transformers are network
branchesbranches
The direction of positive current flow are defined asThe direction of positive current flow are defined asfollows for a branch element (demonstrated on afollows for a branch element (demonstrated on a
medium length line)medium length line)
Power flow is defined for each end of the branchPower flow is defined for each end of the branch
Li Fl d LLi Fl d L
-
8/3/2019 Module 6 Introduction to Power Flow Studies
38/49
BEE 3243 Electric Power Systems Module 5 3838
Line Flows and LossesLine Flows and Losses
ExampleExample: the power leaving: the power leaving bus ibus i and flowing toand flowing tobus jbus j::
Li Fl d LLi Fl d L
-
8/3/2019 Module 6 Introduction to Power Flow Studies
39/49
BEE 3243 Electric Power Systems Module 5 3939
Line Flows and LossesLine Flows and Losses
Current and power flowsCurrent and power flows::
Power lossesPower losses
E lE ample
-
8/3/2019 Module 6 Introduction to Power Flow Studies
40/49
BEE 3243 Electric Power Systems Module 5 4040
ExampleExample
a)a) Using theUsing the GaussGauss--Seidel methodSeidel method, determine the, determine the phasorphasor values of thevalues of the voltagevoltage at the load buses 2 and 3, accurateat the load buses 2 and 3, accurate
to 4 decimal places.to 4 decimal places.
b)b) Find the slack busFind the slack bus PP andand QQ..c)c)
Determine theDetermine the line flowsline flows
andand line lossesline losses. Construct a. Construct a
power flow diagrampower flow diagram
showing the direction of the line flow.showing the direction of the line flow.
1
2
3
E lExample
-
8/3/2019 Module 6 Introduction to Power Flow Studies
41/49
BEE 3243 Electric Power Systems Module 5 4141
ExampleExample
Line admittances:Line admittances:
AtAt PP--Q busesQ buses, the complex loads in, the complex loads in p.up.u.:.:3216
025.00125.0
1
301003.001.0
1
2010
04.002.0
1
23
13
12
jj
y
jj
y
j
j
y
p.u.452.0386.1
p.u.102.1566.2
3
2
jS
jS
sch
sch
a)
-
8/3/2019 Module 6 Introduction to Power Flow Studies
42/49
ExampleExample
-
8/3/2019 Module 6 Introduction to Power Flow Studies
43/49
BEE 3243 Electric Power Systems Module 5 4343
ExampleExample
For 2For 2ndnd
iteration,iteration,
0353.00011.1
)6226(
)031.09825.0)(3216()005.1)(3010(00.1
452.0386.1
2313
)1(223113)0(*
3
33
)1(3
j
j
jjjjjj
yy
VyVyV
jQP
V
schsch
052.09816.0
)5226(
)0353.00011.1)(3216()005.1)(2010(031.09825.0
102.1566.2
)2(2
j
j
jjjjj
j
V
ExampleExample
-
8/3/2019 Module 6 Introduction to Power Flow Studies
44/49
BEE 3243 Electric Power Systems Module 5 4444
ExampleExample
The remaining iterations until the solution isThe remaining iterations until the solution isconverged with an accuracy of 5 x 10converged with an accuracy of 5 x 10--55
p.up.u.:.:
0459.00008.1
)6226(
)052.09816.0)(3216()005.1)(3010(0353.00011.1452.0386.1
)1(3
j
j
jjjjjj
V
0500.00000.10600.09800.0
0500.00000.10599.09801.0
0499.00001.10598.09801.0
0497.00002.10594.09803.0
0488.00004.10578.09808.0
)7(3
)7(2
)6(3
)6(2
)5(3
)5(2
)4(3
)4(2
)3(3
)3(2
jVjV
jVjV
jVjV
jVjV
jVjV
Final solution
ExampleExample
-
8/3/2019 Module 6 Introduction to Power Flow Studies
45/49
BEE 3243 Electric Power Systems Module 5 4545
ExampleExample
The slack bus power:The slack bus power:b)
Mvar189MW5.409
p.u.890.1095.4
)]05.00.1)(3010()06.098.0)(2010()5020(05.1[05.1
)]()([ 31321213121*
111
j
j
jjjjj
VyVyyyVVjQP
ExampleExample
-
8/3/2019 Module 6 Introduction to Power Flow Studies
46/49
BEE 3243 Electric Power Systems Module 5 4646
ExampleExample
Line currents:Line currents:c)
48.064.0
48.064.0)]05.00.1()06.098.0)[(3216()(
0.10.2
0.10.2)]05.00.1()005.1)[(3010()(8.09.1
8.09.1)]06.098.0()005.1)[(2010()(
2332
322323
1331
311313
1221
211212
jII
jjjjVVyI
jII
jjjjVVyIjII
jjjjVVyI
ExampleExample
-
8/3/2019 Module 6 Introduction to Power Flow Studies
47/49
BEE 3243 Electric Power Systems Module 5 4747
ExampleExample
The line flows are:The line flows are:
Mvar8.44MW4.66
p.u.448.0664.0)48.064.0)(05.00.1(
Mvar243MW6.65
p.u.432.0656.0)48.0656.0)(06.098.0(
Mvar0.90MW0.205
p.u.90.005.2)0.10.2)(05.00.1(
Mvar0.105MW0.210
p.u.05.11.2)0.10.2)(0.005.1(
Mvar0.67MW0.191p.u.67.091.1)8.09.1)(06.098.0(
Mvar0.84MW5.199
p.u.84.0995.1)8.09.1)(0.005.1(
*
32332
*23223
*31331
*13113
*
21221
*12112
j
jjjIVS
.j
jjjIVS
j
jjjIVS
j
jjjIVS
jjjjIVS
j
jjjIVS
ExampleExample
-
8/3/2019 Module 6 Introduction to Power Flow Studies
48/49
BEE 3243 Electric Power Systems Module 5 4848
ExampleExample
Line losses are:Line losses are:
Mvar60.1MW8.0Mvar0.15MW0.5
Mvar0.17MW5.8
322323
311313
211212
jSSS
jSSS
jSSS
L
L
L
ExampleExample
-
8/3/2019 Module 6 Introduction to Power Flow Studies
49/49
ExampleExample
409.5
191
65.6
66.4
44.8
67.0
43.2
205
90.0
199.5
84.0
210.0
105.0
189
256.6
138.6
110.2
45.2
(8.5)
(17.0)
(5)
(15)
(0.8)(1.6)
The power flow diagram:The power flow diagram: