Competency Training and Certification Program in Electric Power Distribution System Engineering
Certificate in
Power System Modeling and Analysis Training Course in
Fundamental Principles and Methods in Power System Analysis
U. P. NATIONAL ENGINEERING CENTER NATIONAL ELECTRIFICATION ADMINISTRATION
Training Course in Fundamental Principles and Methods in Power System Analysis
Course Outline 1. Circuit Conventions and Notations 2. Power System Representation 3. Per Unit Quantities 4. Symmetrical Components 5. Network Equations and Methods of Solution
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
2
Training Course in Fundamental Principles and Methods in Power System Analysis
Circuit Conventions & Notations
Voltage and Current Directions
Double Subscript Notation
Voltage, Current and Phasor Notation
Complex Impedance and Phasor Notation
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
3
Training Course in Fundamental Principles and Methods in Power System Analysis
4
Circuit Conventions & Notations Voltage and Current Directions Polarity Marking of Voltage Source Terminals: Plus sign (+) for the terminal where positive current comes out
+
I
+ Specification of Load Terminals: Plus sign (+) for the terminal whereVs positive current enters Specification of Current Direction: Arrows for the positive current (i.e., from the source towards the load)
U. P. National Engineering Center National Electrification Administration
ZA
I
a
VA I
b
-
I +
VB
ZB -
o
n
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
Circuit Conventions & Notations Double Subscript Notation The letter subscripts on a voltage indicate the nodes of the circuit between which the voltage exists. The first subscript denotes the voltage of that node with respect to the node identified by the second subscript.
ZA
a
+ +
VS = Vao
VA I = Iab
-
b
-
+
Zb
Vb = Vbn -
o U. P. National Engineering Center National Electrification Administration
n
The current direction is from first subscript to the second subscript .
Vao - IabZA - Vbn = 0 Vao - Vbn Iab = ZA
Competency Training & Certification Program in Electric Power Distribution System Engineering
5
Training Course in Fundamental Principles and Methods in Power System Analysis
6
Circuit Conventions & Notations Voltage, Current and Phasor Notation 1
V=
v 0.5
i 0 -4
π
-2
0
π
2
4
π
π
Vm ∠0o volts 2
Im I= ∠ − θ o amperes 2
-0.5
-1
v = Vmε i = I mε
jω t j (ω t − θ )
V
θ
ε
j ωt
j =
= cos ωt + j sin ωt
I
−1
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
7
Circuit Conventions & Notations Complex Impedance and Phasor Notation + +
VS = Vmε
VL
i(t)
jω t
R (Resistance)
-
L (Inductance)
-
Applying Kirchoff’s voltage law,
di ( t ) = Vmε Ri ( t ) + L dt
jω t
The first order linear differential equation has a particular jω t i ( t ) = K ε solution of the form . Hence, RK ε
jω t
+ j ω LK ε
U. P. National Engineering Center National Electrification Administration
jω t
= Vmε
jω t
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
8
Circuit Conventions & Notations Complex Impedance and Phasor Notation Solving for the current
Vm ε i( t ) = R + jω L
jω t
Dividing voltage by current to get the impedance,
v( t ) Z = = i( t )
V m ε jω t Vm ε R + jω L
jω t
= R + jω L
Therefore, the impedance Z is a complex quantity with real part R and an imaginary (j) part ωL U. P. National Engineering Center National Electrification Administration
ωL
Z
θ R
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
Circuit Conventions & Notations Complex Impedance and Phasor Notation + +
-
-
For Capacitive Circuit,
1 Z = R − j( ) ωC Z= |Z|ejφφ
VL
or
.1 ωC
C (Capacitance) R
φ
VS = Vmε
i(t)
jω t
R (Resistance)
Z
Z = |Z|(cosφ φ + jsinφ φ) or Z = |Z|∠φ ∠φ
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
9
Training Course in Fundamental Principles and Methods in Power System Analysis
10
Circuit Conventions & Notations Complex Impedance and Phasor Notation Z= |Z|ejφφ
or
Z = |Z|(cosφ φ + jsinφ φ) or Z = |Z|∠φ ∠φ
v = 141.4 cos(ω ωt + 30°) volts i = 7.07 cos(ω ωt) amperes Vmax = 141.4 |V| = 100 V = 100∠ ∠30 Imax = 7.07
|I| = 5
100 ∠ 30 Z = = 20 ∠ 30 5 ∠0
I = 5∠ ∠0
Z = 20(cos 30 + j sin 30) = 17.32 + j10 U. P. National Engineering Center National Electrification Administration
10
20 30°
17.32
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
Power System Representation
Electrical Symbols
Three-Line and Single-Line Diagram
Equivalent Circuit of Power System Components
Impedance and Admittance Diagram
Bus Admittance Matrix
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
11
Training Course in Fundamental Principles and Methods in Power System Analysis
Power System representation Electrical Symbols G
Generator
Switch
Circuit Breaker or
Transformer Fuse Transmission or Distribution Line Bus U. P. National Engineering Center National Electrification Administration
Node
Competency Training & Certification Program in Electric Power Distribution System Engineering
12
Training Course in Fundamental Principles and Methods in Power System Analysis
Power System representation 3-phase wye neutral ungrounded 3-phase delta connection 3-phase wye neutral grounded Current Transformer Potential Transformer
U. P. National Engineering Center National Electrification Administration
V
Voltmeter
A
Ammeter
R
Protective Relay
Competency Training & Certification Program in Electric Power Distribution System Engineering
13
Training Course in Fundamental Principles and Methods in Power System Analysis
Power System Representation Three Line Diagram
A
B
C
Circuit Recloser
Main Bus
Circuit Breaker
The three-line diagram is used to represent each phase of a threeTransformer phase power system. CTs
R R
Distribution Lines
Relays
R
R U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
14
Training Course in Fundamental Principles and Methods in Power System Analysis
Power System Representation Single Line Diagram The three-line diagram becomes rather cluttered for large power systems. A shorthand version of the three-line diagram is referred to as the Single Line Diagram.
Bus CB
Transformer Distribution Line Recloser
R CT and Relay
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
15
Training Course in Fundamental Principles and Methods in Power System Analysis
16
Power System Representation Equivalent Circuit of Power System Components: Generator a Ia
Za
R a + jX
s
Ea
+
Eb Zb
Ec
Zc
Ib
Ia
Eg b
Ic
-
+
Va -
c
Three-Phase Equivalent U. P. National Engineering Center National Electrification Administration
Single-Phase Equivalent Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
17
Power System Representation Equivalent Circuit of Power System Components: Transformer Primary A B C
Y11 Y12 Y13 Y14 Y15 Y16
Secondary a
Y21 Y22 Y23 Y24 Y25 Y26 Y31 Y32 Y33 Y34 Y35 Y36
b
Y41 Y42 Y43 Y44 Y45 Y46
c
Y51 Y52 Y53 Y54 Y55 Y56 Y61 Y62 Y63 Y64 Y65 Y66
6x6 Admittance Matrix Core Loss
Three-Phase Equivalent U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
18
Power System Representation Equivalent Circuit of Power System Components: Transformer R H + jX H
v Iex
+
r VH
Rc
a 2 R X + ja 2 X X
jX m
r IH
+
r aV X
RT = R H + a 2 R X 2 XT = X H + a X X ZT
-
-
Single-Phase Equivalent
U. P. National Engineering Center National Electrification Administration
+
+
-
-
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
Power System Representation Equivalent Circuit of Power System Components: Transmission & Distribution Lines T&D Lines can be represented by an infinite series of resistance and inductance and shunt capacitance.
Δl
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
19
Training Course in Fundamental Principles and Methods in Power System Analysis
20
Power System Representation Equivalent Circuit of Power System Components: Distribution Lines
Equivalent π-Network
A B C
1/2
Yaa
Yab
Yac
Yba
Ybb
Ybc
Yca
Ycb
Ycc
Capture Unbalanced Characteristics U. P. National Engineering Center National Electrification Administration
Zaa
Zab
Zac
Zba
Zbb
Zbc
Zca
Zcb
Zcc
1/2
a b c Yaa
Yab
Yac
Yba
Ybb
Ybc
Yca
Ycb
Ycc
Three-Phase Equivalent Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
21
Power System Representation Equivalent Circuit of Power System Components: Long Transmission Lines Length = Longer than 240 km. (150 mi.)
Z ' = Z c sinh γ l
+• Vs
γl Y' 1 = tanh 2 2 Zc
IS
-•
Characteristic Impedance
ZC =
z y
U. P. National Engineering Center National Electrification Administration
• + Y' 2
IR
VR
• -
Propagation Constant
γ=
zy
Single-Phase Equivalent Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
22
Power System Representation Equivalent Circuit of Power System Components: Medium-Length Transmission Lines Length = 80 – 240 km. (50 - 150 mi.)
Z = (r + jx L )l
+• Vs
-•
IS
Y 1 / ωc = 2 2
• + Y 2
IR
VR
• -
Single-Phase Equivalent U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
Power System Representation Equivalent Circuit of Power System Components: Short Transmission Lines Length = up to 80 km. (50 mi.)
+•
Z = (r + jx L )l
Vs
-•
Is = IR
• + VR
• Single-Phase Equivalent
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
23
Training Course in Fundamental Principles and Methods in Power System Analysis
Power System Representation Single Phase Equivalent of Balanced Three-Phase System c
Ic
c
Eco = |E| ∠120° V o
a
Ia
ZR a
n
Eao = |E|∠0° V
ZR
Ebo= |E| ∠240° V b
ZR
b
Ib U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
24
Training Course in Fundamental Principles and Methods in Power System Analysis
Power System Representation a
o
Ia
a
Eao = |E|∠ 0° V
E∠0 Ia = = I∠ − θ Z R ∠θ
n
ZR
o
n
ZR
b c
Ebo= |E| ∠240° V Eco = |E| ∠120° V
E∠240 Ib = = I∠( 240 − θ ) Z R ∠θ
ZR
Ib
b c
Ic ZR
o
E∠120 Ic = = I∠( 120 − θ ) Z R ∠θ
n
ZR U. P. National Engineering Center National Electrification Administration
Note: Currents are Balanced
Competency Training & Certification Program in Electric Power Distribution System Engineering
25
Training Course in Fundamental Principles and Methods in Power System Analysis
26
Power System Representation Single Phase c Representation of a Balanced Three-Phase System b
Ic Eco = |E| ∠120° V
c
ZR
Ia a
o
a
n
Eao = |E|∠0° V
ZR
Ebo= |E| ∠240° V Ia
a
Ia
Eao = |E|∠ 0° V o U. P. National Engineering Center National Electrification Administration
ZR b
a
ZR
n Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
27
Power System Representation Impedance and Admittance Diagrams a Bus
Gen
Line
1 2 3 4
a b c
1-3 2-3 1-4 2-4 3-4
c
1
b
3
2
4 Single Line Diagram U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
28
Power System Representation Impedance and Admittance Diagrams 0
Ea
Ea
za 1
za
Eb
zc
Generator
1
zb 3
zd
0
zf
z13 1
Ec
2
ze zh
zg
3 Line 4 0
U. P. National Engineering Center National Electrification Administration
Impedance Diagram Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
Power System Representation Impedance and Admittance Diagrams
Zg Eg
VL
VL
+
-
IL
Is
Zp
IL
Eg = ISZP Zg = Zp
The two sources will be equivalent if VL and IL are the same for both circuits. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
29
Training Course in Fundamental Principles and Methods in Power System Analysis
30
Impedance and Admittance Diagrams a Bus
Gen
Line
1 2 3 4
a b c
1-3 2-3 1-4 2-4 3-4
c
1
b
3
2
4 Single Line Diagram U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
31
Impedance and Admittance Diagrams 0
Ea
Ea
za 1
za
Eb
zc
Generator
1
zb 3
zd
0
zf
z13 1
Ec
2
ze zh
zg
3 Line 4 0
U. P. National Engineering Center National Electrification Administration
Impedance Diagram Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
Impedance and Admittance Diagrams Equivalent Sources
Zg Eg
VL
VL
+
-
IL
Is
Zp
IL
Eg = ISZP Zg = Zp
The two sources will be equivalent if VL and IL are the same for both circuits. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
32
Training Course in Fundamental Principles and Methods in Power System Analysis
33
Impedance and Admittance Diagrams 0
I1 = Ea/za y01 = 1/za I2 = Eb/zb y02 = 1/zb
I1
y01 1
y13
I3 = Ec/zc y03 = 1/zc
I3
y03 3
y14 = 1/zf
y23 = 1/ze
y24 = 1/zg y34 = 1/zh
U. P. National Engineering Center National Electrification Administration
y02 2
y23
y14
y13 = 1/zd
I2
y24 4
Admittance Diagram Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
Nodal Voltage Equations Applying Kirchoff’s Current Law
at node 1: I 1 = V 1 y 01 + (V 1 − V 3 ) y 13 + (V 1 − V 4 ) y 14
at node 2: I 2 = V 2 y 02 + (V 2 − V 3 ) y 23 + (V 2 − V 4 ) y 24 at node 3:
I 3 = V 3 Y03 + (V 3 − V 2 ) y 23 + (V 3 − V 4 ) y 34 + (V 3 − V 1 ) y 13
at node 4: 0 = (V 4 − V1 ) y 14 + (V 4 − V 2 ) y 24 + (V 4 − V 3 ) y 34 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
34
Training Course in Fundamental Principles and Methods in Power System Analysis
Nodal Voltage Equations Rearranging the equations,
I 1 = V1 ( y 01 + y13 + y14 ) − V 3 y13 − V 4 y14 I 2 = V 2 ( y 02 + y 23 + y 24 ) − V 3 y 23 − V 4 y 24 I 3 = V 3 ( y 03 + y 23 + y 34 + y 13
0 = V 4 ( y 14 + y 24 + y 34
)−
) − V 2 y 23
− V 4 y 34 − V 1 y 13
V 1 y 14 − V 2 y 24 − V 3 y 34
In matrix form,
0 - y13 - y14 ⎡I1 ⎤ ⎡y01 + y13 + y14 ⎤⎡V1 ⎤ ⎥ ⎢ ⎢I ⎥ ⎢ ⎥ 0 y02 + y23 + y24 - y23 - y24 ⎢ 2⎥ = ⎢ ⎥⎢V2 ⎥ ⎢I3 ⎥ ⎢ ⎥⎢V3 ⎥ - y13 - y23 y03 + y23 + y34 + y13 - y34 ⎢ ⎥ ⎢ ⎥⎢ ⎥ y y y y y y − − − + + 14 24 34 14 24 34 ⎦⎣V4 ⎦ ⎣0 ⎦ ⎣ U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
35
Training Course in Fundamental Principles and Methods in Power System Analysis
Nodal Voltage Equations ⎡I1 ⎤ ⎢ ⎥ ⎢I2 ⎥ ⎢I3 ⎥ = ⎢ ⎥ ⎢M ⎥ ⎢I ⎥ ⎣ n⎦
⎡ Y 11 ⎢ ⎢ Y 21 ⎢ ⎢ Y 31 ⎢ ⎢M ⎢ ⎢Y n 1 ⎣
Y 12 Y 22 Y 32 M Yn2
Y 13 L Y 1 n ⎤ ⎡V 1 ⎤ ⎥⎢ ⎥ ⎥ V2 ⎥ Y 23 L Y 2 n ⎥⎢ Y 33 L Y 3 n ⎥ ⎢V 3 ⎥ ⎥⎢ ⎥ M M ⎥⎢ M ⎥ ⎥ Y n 3 L Y nn ⎥ ⎢V ⎥ ⎦⎣ n⎦
[ I ] = [Ybus][V] Yii = self-admittance, the sum of all admittances terminating on the node (diagonal elements) Yij = mutual admittance, the negative of the admittances connected directly between the nodes identifed by the double subscripts U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
36
Training Course in Fundamental Principles and Methods in Power System Analysis
Power System Representation
[YBUS] =
⎡ Y 11 ⎢ ⎢ Y 21 ⎢ ⎢ Y 31 ⎢ ⎢M ⎢ ⎢Y n 1 ⎣
Y 12 Y 22 Y 32 M Yn2
Y 13 L Y 1 n ⎤ ⎥ Y 23 L Y 2 n ⎥ ⎥ Y 33 L Y 3 n ⎥ ⎥ M M ⎥ ⎥ Y n 3 L Y nn ⎥ ⎦
Ybus is also called Bus Admittance Matrix
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
37
Training Course in Fundamental Principles and Methods in Power System Analysis
Per Unit Quantities
The Per Unit System
Per Unit Impedance
Changing Per Unit Values
Consistent Per Unit Quantities of Power System
Advantages of Per Unit Quantities
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
38
Training Course in Fundamental Principles and Methods in Power System Analysis
The Per Unit System Per Unit Value Per Unit Value
=
Actual Value Base Value
Per-unit Value is a dimensionless quantity Per-unit value is expressed as decimal
Percent U. P. National Engineering Center National Electrification Administration
=
Actual Value 100
Competency Training & Certification Program in Electric Power Distribution System Engineering
39
Training Course in Fundamental Principles and Methods in Power System Analysis
The Per Unit System Per Unit Value Per Unit Power
Per Unit Voltage
Per Unit Current
Per Unit Impedance
=
Actual Value of Power
=
Actual Value of Voltage
=
Actual Value of Current
=
Actual Value of Impedance
U. P. National Engineering Center National Electrification Administration
Base Power
Base Voltage
Base Current
Base Impedance Competency Training & Certification Program in Electric Power Distribution System Engineering
40
Training Course in Fundamental Principles and Methods in Power System Analysis
The Per Unit System Per Unit Calculations PU Voltage PU Current
= PU Impedance
PU Power = PU Voltage x PU Current
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
41
Training Course in Fundamental Principles and Methods in Power System Analysis
The Per Unit System Per Unit Calculations I
Example: +
+ Zline = 1.4 ∠75° Ω Zload = 20 ∠30° Ω
Vs = ?
-
2540 ∠0° V
Determine Vs
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
42
Training Course in Fundamental Principles and Methods in Power System Analysis
The Per Unit System Per Unit Calculations Choose: Base Impedance = 20 ohms (single phase) Base Voltage = 2540 volts (single phase) PU Impedance of the load = 20∠30°/20 = ______ p.u. PU Impedance of the line = 1.4∠75°/20 = ______ p.u. PU Voltage at the load = 2540∠0° /2540 = ______ p.u. Line Current in PU = PU voltage / PU impedance of the load = ______ / ______ = ______ p.u. PU Voltage at the Substation = Vload(pu) + IpuZLine(pu) = ________ + _______ x _______ = _______ p.u. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
43
Training Course in Fundamental Principles and Methods in Power System Analysis
The Per Unit System Per Unit Calculations 1.0 ∠-30° p.u.
+
0.07 ∠75° p.u.
1.05∠2.70°
+
1.0∠30° p.u.
-
1.0∠0° p.u.
-
The magnitude of the voltage at the substation is 1.05 p.u. x 2540 Volts = _______ Volts U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
44
Training Course in Fundamental Principles and Methods in Power System Analysis
The Per Unit System Establishing Base Values 1. Base values must satisfy fundamental electrical laws (Ohm’s Law and Kirchoff’s Laws) 2. Choose any two electrical parameters • Normally, Base Power and Base Voltage are chosen 3. Calculate the other parameters • Base Impedance and Base Current
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
45
Training Course in Fundamental Principles and Methods in Power System Analysis
The Per Unit System Establishing Base Values For a Given Base Power and Base Voltage, Base Power Base Current = Base Voltage (Base Voltage)2
Base Voltage Base Impedance =
= Base Current
U. P. National Engineering Center National Electrification Administration
Base Power
Competency Training & Certification Program in Electric Power Distribution System Engineering
46
Training Course in Fundamental Principles and Methods in Power System Analysis
The Per Unit System Establishing Base Values For Single Phase System,
For Three Phase System,
Ibase =
Pbase(1φ) -----------Vbase(1φ)
Ibase =
Zbase =
Vbase(1φ) -----------Ibase(1φ)
Vbase(LN) Zbase = -----------Ibase(L)
[Vbase(1φ)]² = -----------Pbase(1φ) U. P. National Engineering Center National Electrification Administration
Pbase(3φ) -----------√3Vbase(LL)
[Vbase(LL)]² = -----------Pbase(3φ)
Competency Training & Certification Program in Electric Power Distribution System Engineering
47
Training Course in Fundamental Principles and Methods in Power System Analysis
The Per Unit System Establishing Base Values • Base Values can be established from Single Phase or Three Phase Quantities
Base MVA 1φ Base kV 1φ =
1 = Base MVA 3φ 3 Base kV LL 3
• Base MVA is the same base value for Apparent, Active and Reactive Power • Base Z is the same base value for Impedance, Resistance and Reactance U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
48
Training Course in Fundamental Principles and Methods in Power System Analysis
The Per Unit System Establishing Base Values Example: Base kVA3φφ
= 30,000 kVA = 30 MVA
Base kVA1φφ
= 10,000 kVA = 10 MVA
Base kVLL
= 120 kV
Base kVLN
= 69.282 kV
Base Z
= (69.282)2/10 = 480 ohms
= (120)2/30 = 480 ohms 30 x1000 Base Current = 3 ( 120 )
Base Z
= 144.34 Amps U. P. National Engineering Center National Electrification Administration
10 x1000 Base Current = 69.282 = 144.34 Amps Competency Training & Certification Program in Electric Power Distribution System Engineering
49
Training Course in Fundamental Principles and Methods in Power System Analysis
Per Unit Impedance Generators
Manufacturers provide the following impedance in per unit: 1.
Armature Resistance, Ra
2.
Direct-axis Reactances, Xd”, Xd’ and Xd
3.
Quadrature-axis Reactances, Xq”, Xq’ and Xq
4.
Negative Sequence Reactance, X2
5.
Zero Sequence Reactance, X0
}
Positive Sequence Impedances
The Base Values used by manufacturers are: 1.
Rated Capacity (MVA, KVA or VA)
2.
Rated Voltage (kV or V)
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
50
Training Course in Fundamental Principles and Methods in Power System Analysis
Per Unit Impedance Transmission and Distribution Lines
R( X
pu )
L ( pu )
= =
X C ( pu ) = U. P. National Engineering Center National Electrification Administration
R(立
)
Z
base
X
L( 立 )
Z base
X C( 立 ) Z base
Competency Training & Certification Program in Electric Power Distribution System Engineering
51
Training Course in Fundamental Principles and Methods in Power System Analysis
Per Unit Impedance Transformers
The ohmic values of resistance and leakage reactance of a transformer depends on whether they are measured on the high- or low-tension side of the transformer
The impedance of the transformer is in percent or per unit with the Rated Capacity and Rated Voltages taken as base Power and Base Voltages, respectively
The per unit impedance of the transformer is the same regardless of whether it is referred to the high-voltage or lowvoltage side
The per unit impedance of the three-phase transformer is the same regardless of the connection U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
52
Training Course in Fundamental Principles and Methods in Power System Analysis
Per Unit Impedance Example A single-phase transformer is rated 110/440 V, 2.5 kVA. The impedance of the transformer measured from the lowvoltage side is 0.06 ohms. Determine the impedance in per unit (a) when referred to low-voltage side and (b) when referred to high-voltage side
Solution 0.110 2 Low-voltage Zbase = 2.5 / 1000 PU Impedance, Zpu = U. P. National Engineering Center National Electrification Administration
= ______ ohms = ______ p.u.
Competency Training & Certification Program in Electric Power Distribution System Engineering
53
Training Course in Fundamental Principles and Methods in Power System Analysis
Per Unit Impedance If impedance had been measured on the high-voltage side, the ohmic value would be 2
⎛ 440 ⎞ Z = 0 . 06 ⎜ ⎟ = _______ ⎝ 110 ⎠
ohms
High Voltage, Zbase =
= _______ ohms
PU Impedance, Zpu =
= _______ p.u.
Note: PU value of impedance referred to any side of the transformer is the same U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
54
Training Course in Fundamental Principles and Methods in Power System Analysis
Changing Per-Unit Values Example: Consider a three-phase transformer rated 20 MVA, 67 kV/13.2 kV voltage ratio and a reactance of 7%. The resistance is negligible. a) What is the equivalent reactance in ohms referred to the high voltage side? b) What is the equivalent reactance in ohms referred to the low voltage side? c) Calculate the per unit values both in the high voltage and low voltage side at 100 MVA. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
55
Training Course in Fundamental Principles and Methods in Power System Analysis
Changing Per-Unit Values SOLUTION: a) Pbase = 20 MVA Vbase = 67 kV (high voltage) ( kV)² Zbase = ( MVA)
= ________ ohms
Xhigh = Xp.u. x Zbase = _______ x _______= _______ ohms
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
56
Training Course in Fundamental Principles and Methods in Power System Analysis
Changing Per-Unit Values SOLUTION: b) Pbase = 20 MVA Vbase = 13.2 kV (low voltage) Zbase =
(
kV)²
(
MVA)
= ________ ohms
Xlow = Xp.u. x Zbase = _______ x _______= _______ ohms
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
57
Training Course in Fundamental Principles and Methods in Power System Analysis
Changing Per-Unit Values c) Pbase = 100 MVA Vbase,H = 67 kV (67)² Zbase,H = = ________ ohms 100 ohms Xp.u.,H = = ______ p.u. ohms Vbase,L = 13.2 kV (13.2)² Zbase,L = = _______ ohms 100 ohms Xp.u.,L = = ______ p.u. ohms
Note that the per unit quantities are the same regardless of the voltage level. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
58
Training Course in Fundamental Principles and Methods in Power System Analysis
59
Changing Per-Unit Values Three parts of an electric system are designated A, B and C and are connected to each other through transformers, as shown in the figure. The transformer are rated as follows: A-B B-C SOURCE
10 MVA, 3φ, 13.8/138 kV, leakage reactance 10% 10 MVA, 3φ, 138/69 kV, leakage reactance 8% A
B A-B
C B-C
LOAD 300 Ω/ φ PF=100 %
Determine the voltage regulation if the voltage at the load is 66 kV. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
60
Changing Per-Unit Values Solve using actual quantities XAB=10%
XBC=8%
Vc
VA A-B 13.8/138 kV
U. P. National Engineering Center National Electrification Administration
300 Ί/ φ PF=100%
B-C 138/69 kV
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
Changing Per-Unit Values SOLUTION USING PER UNIT METHOD: Pbase = 10 MVA VA,base = 13.8 kV VB,base = 138 kV VC,base = 69 kV ZLOAD,p.u. =
(69)² ZC,base = -------- = _____ ohms 10
---------- = ______ + j _____ p.u.
VC,p.u. = -------------- = _______ p.u. Ip.u. = ---------------- = _______ p.u. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
61
Training Course in Fundamental Principles and Methods in Power System Analysis
Changing Per-Unit Values VA = _______ + ( ________ ) x ( ________ + ________ ) = ________ + j ________ p.u. = _________ p.u. VNL - VL V.R. = ---------------- x 100% VNL V.R. = ------------------------ x 100% =
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
62
Training Course in Fundamental Principles and Methods in Power System Analysis
Changing Per-Unit Values Consider the previous example, What if transformer A-B is 20 MVA instead of 10 MVA. The transformer nameplate impedances are specified in percent or per-unit using a base values equal to the transformer nameplate rating. The PU impedance of the 20 MVA transformer cannot be added to the PU impedance of the 10 MVA transformer because they have different base values The per unit impedance of the 20 MVA can be referred to 10 MVA base power U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
63
Training Course in Fundamental Principles and Methods in Power System Analysis
Changing Per-Unit Values Convert per unit value of 20 MVA transformer, Pbase = 20 MVA (Power Rating) Vbase,H = 138 kV (Voltage Rating) (138)² Zbase,H = ---------- = _______ ohms 20 Xactual,H = 0.10 p.u. x _______ ohms = _______ ohms At Pbase = 10 MVA (new base) (138)² Zbase,H = ---------- = 1904.4 ohms 10 95.22 Xp.u.(new) = ---------- = 0.05 p.u. 1904.4 U. P. National Engineering Center National Electrification Administration
The per unit impedance of the 20MVA and 10 MVA transformer can now be added.
Competency Training & Certification Program in Electric Power Distribution System Engineering
64
Training Course in Fundamental Principles and Methods in Power System Analysis
Changing Per-Unit Values Note that the transformer can have different per unit impedance for different base values (i.e., the actual ohmic impedances of the equipment is independent of the selected base values), then
Zactual = Zpu1 • Zbase1
Zactual = Zpu2 • Zbase2
Z pu 1 ⋅ Z base 1 = Z pu 2 ⋅ Z base 2 Z base Z pu 2 = Z pu 1 ⋅ Z base U. P. National Engineering Center National Electrification Administration
1 2
Competency Training & Certification Program in Electric Power Distribution System Engineering
65
Training Course in Fundamental Principles and Methods in Power System Analysis
66
Changing Per-Unit Values Recall:
Z base
( base =
voltage
base Power
(kV
Then,
Z
pu 2
= Z
LL , base 1
MVA
pu 1
(kV
Z pu 2
3 φ , base 1 2 LL , base 2
⎛ kV LL ,base 1 = Z pu 1 ⎜⎜ ⎝ kV LL ,base 2
U. P. National Engineering Center National Electrification Administration
)
2
MVA or,
)
2
)
3 φ , base 2
⎞ ⎟ ⎟ ⎠
2
⎛ MVA 3 φ ,base 2 ⎜ ⎜ MVA 3 φ ,base 1 ⎝
⎞ ⎟ ⎟ ⎠
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
Changing Per-Unit Values
Z p .u .( new ) = Z p .u .( given ) x
[ Vbase( given ) ] [ Vbase( new ) ]
2
2
x
Pbase( new ) Pbase( given )
Example A three-phase transformer is rated 400 MVA, 220Y/22 Δ kV. The impedance measured on the low-voltage side of the transformer is 0.121 ohms (approx. equal to the leakage reactance). Determine the per-unit reactance of the transformer for 100 MVA, 230 kV base values at the high voltage side of the transformer. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
67
Training Course in Fundamental Principles and Methods in Power System Analysis
Changing Per-Unit Values Solution On its own base the transformer reactance is (
X=
) (
)2
(
)
= ________ pu
On the chosen base the reactance becomes X=(
)x
(
)2
(
)2
x
U. P. National Engineering Center National Electrification Administration
(
= )________ pu
Competency Training & Certification Program in Electric Power Distribution System Engineering
68
Training Course in Fundamental Principles and Methods in Power System Analysis
Consistent Per Unit Quantities of Power System Procedure: a)
Establish Base Power and Base Voltages • Declare Base Power for the whole Power System • Declare Base Voltage for any one of the Power System components • Compute the Base Voltages for the rest of the Power System Components using the voltage ratio of the transformers
Note: Define each subsystem with unique Base Voltage based on separation due to magnetic coupling U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
69
Training Course in Fundamental Principles and Methods in Power System Analysis
Consistent Per Unit Quantities of Power System b) Compute Base Impedance and Base Current • Using the Declared Base Power and Base Voltages, compute the Base Impedances and Base Currents for each Subsystem c) Compute Per Unit Impedance • Using the declared and computed Base Values, compute the Per Unit values of the impedance by: Dividing Actual Values by Base Values Changing Per Unit Impedance with change in Base Values U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
70
Training Course in Fundamental Principles and Methods in Power System Analysis
Consistent Per Unit Quantities of Power System Use Base Power = 100 MVA T1
T2
G2
L1
G1
G3 Generator 1 (G1):
300 MVA; 20 kV; 3φ; Xd” = 20 %
Transmission Line(L1):
64 km; XL = 0.5 Ω / km
Transformer 1 (T1): 3φ; 350 MVA; 230 / 20 kV; XT1 = 10 % Transformer 2 (T2): 3-1φ; 100 MVA; 127 / 13.2 kV; XT2 = 10 % Generator 2 (G2): 200 MVA; 13.8kV, Xd” = 20 % Generator 3 (G3): 100 MVA; 13.8kV, Xd” = 20 % U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
71
Training Course in Fundamental Principles and Methods in Power System Analysis
72
Consistent Per Unit Quantities of Power System T1
T2
G2
Transmission Line
G1 G3 1
XT1 2
XLINE
3
XT2 4
XG1 E1
U. P. National Engineering Center National Electrification Administration
XG2 E2
XG3 E3
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
Consistent Per Unit Quantities of Power System a) & b) Establish Base Power, Base Voltages, Base Impedance, and Base Current Base Power: SubSystem
Vbase (kV)
U. P. National Engineering Center National Electrification Administration
Zbase (Ohm)
Ibase (Amp)
Competency Training & Certification Program in Electric Power Distribution System Engineering
73
Training Course in Fundamental Principles and Methods in Power System Analysis
Consistent Per Unit Quantities of Power System c) Compute Per Unit Impedance G1: G2: G3: L1: T1: T2: U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
74
Training Course in Fundamental Principles and Methods in Power System Analysis
Advantages of Per-Unit Quantities The computation for electric systems in per-unit simplifies the work greatly. The advantages of Per Unit Quantities are: 1. Manufacturers usually specify the impedances of equipments in percent or per-unit on the base of the nameplate rating. 2. The per-unit impedances of machines of the same type and widely different rating usually lie within a narrow range. When the impedance is not known definitely, it is generally possible to select from tabulated average values. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
75
Training Course in Fundamental Principles and Methods in Power System Analysis
Advantages of Per-Unit Quantities 3. When working in the per-unit system, base voltages can be selected such that the per-unit turns ratio of most transformers in the system is equal to 1:1. 4. The way in which transformers are connected in threephase circuits does not affect the per-unit impedances of the equivalent circuit, although the transformer connection does determine the relation between the voltage bases on the two sides of the transformer. 5. Per unit representation yields more meaningful and easily correlated data. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
76
Training Course in Fundamental Principles and Methods in Power System Analysis
Advantages of Per-Unit Quantities 6. Network calculations are done in a much more handier fashion with less chance of mix-up • between phase and line voltages • between single-phase and three-phase powers, and • between primary and secondary voltages.
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
77
Training Course in Fundamental Principles and Methods in Power System Analysis
Symmetrical Components
Sequence Components of Unbalanced Phasor
Sequence Impedance of Power System Components
Practical Implications of Sequence Components of Electric Currents
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
78
Training Course in Fundamental Principles and Methods in Power System Analysis
Sequence Components of Unbalanced Phasor In a balanced Power System, Generator Voltages are three-phase balanced Line and transformer impedances are balanced Loads are three-phased balanced Single-Phase Representation and Analysis can be used for the Balanced Three-Phase Power System
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
79
Training Course in Fundamental Principles and Methods in Power System Analysis
Sequence Components of Unbalanced Phasor In a practical Power Systems, Lines are not transposed. Single-phase transformers used to form three-phase banks are not identical. Loads are not balanced. Presence of vee-phase and single phase lines. Faults Single-phase Representation and Analysis cannot be use for an unbalanced three-phase power system. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
80
Training Course in Fundamental Principles and Methods in Power System Analysis
Sequence Components of Unbalanced Phasor Any unbalanced three-phase system of phasors may be resolved into three balanced systems of phasors which are referred to as the symmetrical components of the original unbalanced phasors, namely:
a) POSITIVE-SEQUENCE PHASOR b) NEGATIVE-SEQUENCE PHASOR c) ZERO-SEQUENCE PHASOR U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
81
Training Course in Fundamental Principles and Methods in Power System Analysis
Sequence Components of Unbalanced Phasor REFERENCE PHASE SEQUENCE: abc Phase c
120째 120째
Phase a
120째
Phase b
Positive Sequence Phasors are three-phase, balanced and have the phase sequence as the original set of unbalanced phasors.
Negative Sequence Phasors are three-phase, balanced but with a phase sequence opposite to that original set of unbalnced phasors.
Zero Sequence Phasors are single-phase, equal in magnitude and in the same direction.
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
82
Training Course in Fundamental Principles and Methods in Power System Analysis
Sequence Components of Unbalanced Phasor Each of the original unbalanced phasor is the sum of it’s sequence components. Thus,
Va = Va1 + Va2 + Va0 Vb = Vb1 + Vb2 + Vb0 Vc = Vc1 + Vc2 + Vc0 Where, Va1 – Positive Sequence component of Voltage Va Va2 – Negative Sequence component of Voltage Va Va0 – Zero Sequence component of Voltage Va U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
83
Training Course in Fundamental Principles and Methods in Power System Analysis
Sequence Components of Unbalanced Phasor OPERATOR “a” An operator “a” causes a rotation of 120° in the counter clockwise direction of any phasor.
a = 1 ∠ 120° a² = 1 ∠ 240° a³ = 1 ∠ 0°
U. P. National Engineering Center National Electrification Administration
aV
120°
V
Operating V by a
Competency Training & Certification Program in Electric Power Distribution System Engineering
84
Training Course in Fundamental Principles and Methods in Power System Analysis
Sequence Components of Unbalanced Phasor The original Phasor and Positive Sequence components in terms of phase a Vc = aVa
Vb in terms of Va
120°
Vb = a² Va Vb1 = a² Va1
120°
Va
120°
Vb = a2Va
U. P. National Engineering Center National Electrification Administration
Vc in terms of Va Vc = a Va Vc1 = a Va1 Competency Training & Certification Program in Electric Power Distribution System Engineering
85
Training Course in Fundamental Principles and Methods in Power System Analysis
Sequence Components of Unbalanced Phasor The Negative Sequence components in terms of phase a
Vb in terms of Va
Vb2 = aVa2
Vb2 = aVa2 120° 120°
Va2
120°
Vc2 = a2Va2
U. P. National Engineering Center National Electrification Administration
Vc in terms of Va Vc2 = a²Va2 Competency Training & Certification Program in Electric Power Distribution System Engineering
86
Training Course in Fundamental Principles and Methods in Power System Analysis
Sequence Components of Unbalanced Phasor The Zero Sequence components in terms of phase a
Vb in terms of Va Vb0 = Va0 Va0= Vb0 = Vc0
Vc in terms of Va Vc0 = Va0
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
87
Training Course in Fundamental Principles and Methods in Power System Analysis
88
Sequence Components of Unbalanced Phasor Writing again the phasors in terms of phasor Va and operator “a”,
Va = Va0 + Va1 + Va2 Vb = Va0 + a²Va1 + aVa2 Vc = Va0 + aVa1 + a²Va2 Computing for Va0, Va1 & Va2
V a0
[
1 1 = [V a + V b + V c ] V a 1 = V a + aV b + a 2 V c 3 3 1 V a2 = V a + a 2 V b + aV c 3
[
U. P. National Engineering Center National Electrification Administration
]
]
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
Sequence Components of Unbalanced Phasor Vc = 8 ∠143.1
EXAMPLE: Determine the symmetrical components of the following unbalanced voltages.
Va = 4 ∠0
Vb = 3 ∠-90 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
89
Training Course in Fundamental Principles and Methods in Power System Analysis
Sequence Components of Unbalanced Phasor For Phasor Va: 1 V a0 = ( V a + V b + V c ) 3 1 = ( 4 ∠ 0 + 3 ∠ - 90 + 8 ∠ 143.1) 3 = 1 ∠ 143.05 1 V a 1 = ( V a + aV b + a 2V c ) 3 1 = [4 ∠ 0 + (1 ∠ 120)(3 ∠ - 90) + (1 ∠ 240)(8 ∠ 143.1) ] 3 = 4.9 ∠ 18.38 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
90
Training Course in Fundamental Principles and Methods in Power System Analysis
Sequence Components of Unbalanced Phasor For Phasor Va:
Va 2
1 = ( V a + a 2Vb + aV c ) 3 1 = [4 ∠0 + (1 ∠ 240)(3 ∠ - 90) + (1 ∠120)(8 ∠143.1) ] 3 = 2.15 ∠ − 86.08
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
91
Training Course in Fundamental Principles and Methods in Power System Analysis
Sequence Components of Unbalanced Phasor Components of Vb can be obtained by operating the sequence components of phasor Va.
V b0 = V a0 = 1 ∠ 143.05
= 1 ∠ 143.05
V b1 = a 2V a 1
V b2
= (1 ∠ 240)(4.9 ∠ 18.38) = 4.9 ∠ 258.38 = 4.9 ∠ - 101.62 = aV a2 = (1 ∠ 120)(2.15 ∠ - 86.08) = 2.15 ∠ 33.92
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
92
Training Course in Fundamental Principles and Methods in Power System Analysis
Sequence Components of Unbalanced Phasor Similarly, components of phasor Vc can be obtained by operating Va.
V c0 = V a0 V c1
= 1 ∠ 143.05 = aV a 1
= 1 ∠ 143.05
= (1 ∠ 1 2 0)(4.9 ∠ 18.38) = 4.9 ∠ 1 3 8.38 V c2 = a 2 V a2 = (1 ∠ 2 4 0)(2.15 ∠ - 86.08) = 2.15 ∠ 1 5 3.92 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
93
Training Course in Fundamental Principles and Methods in Power System Analysis
94
Sequence Components of Unbalanced Phasor Vc1
Va0 = Vb0 = Vc0 143.05 째
Va1 18.38 째
Zero Sequence Components Vc2
Vb2 86.08 째
Vb1 Positive Sequence Components U. P. National Engineering Center National Electrification Administration
Va2 Negative Sequence Components Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
Sequence Components of Unbalanced Phasor Add Sequence Components Graphically
Va1 Va0
Vc2
Va = 4 ∠0
Components of Va
Va2
Vb0 Vb1
Vb = 3 ∠-90
Vc1 Vc = 8 ∠143.1
Vb2 Vc0
Components of Vb
Components of Vc U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
95
Training Course in Fundamental Principles and Methods in Power System Analysis
Sequence Components of Unbalanced Phasor The results can be checked either mathematically or graphically.
V a = V a0 + V a1 + V a2 = 1 ∠ 143 . 05 + 4.9 ∠ 18.38 + 2.15 ∠ - 86.08 = 4 ∠0 V b = V b0 + V b1 + V b2 = 1 ∠ 143.05 + 4.9 ∠ - 101.62 + 2.15 ∠ 33.92 = 3 ∠ - 90 V c = V c0 + V c1 + V c2 = 1 ∠ 143 . 05 + 4.9 ∠ 138.38 + 2.15 ∠ 153.92 = 8 ∠ 143.1 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
96
Training Course in Fundamental Principles and Methods in Power System Analysis
Sequence Components of Unbalanced Phasor Ia a
Ia1 + Ia2 + Ia0
a b
b
c
c Ib
Ib1 + Ib2 + Ib0
Ic Ic1 + Ic2 + Ic0
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
97
Training Course in Fundamental Principles and Methods in Power System Analysis
98
Sequence Impedance of Power System Components Sequence Networks +
+
+
Ia1 Va1
Ia2
Z1
Va2
Ia0
Z2
Va0
Z0
+ Vf -
-
Va1 = V f – I a1 Z 1
Positive Sequence
-
V =-I Z a2 a2 2
Negative Sequence
U. P. National Engineering Center National Electrification Administration
V =-I Z ao ao o
Zero Sequence
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
Sequence Impedance of Power System Components
In general,
Z1 ≠ Z2 ≠ Z0
for generators
Z1 = Z2 = Z0
for transformers
Z1 = Z2 ≠ Z0
for lines
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
99
Training Course in Fundamental Principles and Methods in Power System Analysis
100
Practical Implications of Sequence Components of Electric Currents ZERO-SEQUENCE CURRENTS: IA0
Ia0 A
IC0 C
a B
3I0
IB0
Ic0
b c
3Io
Ib0
The neutral return (ground) carries the in-phase zerosequence currents. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
101
Practical Implications of Sequence Components of Electric Currents ZERO-SEQUENCE CURRENTS: I0 = 0
IC0
Ia0 a
C A I0 = 0 I0 = 0
B
IA0
b c
IB0
In-phase zero-sequence currents circulates in the delta-connected transformer windings. There is “balancing ampere-turns� for the zero-sequence currents. U. P. National Engineering Center National Electrification Administration
Ic0 3Io
Ib0
The neutral return (ground) carries the in-phase zerosequence currents.
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
102
Practical Implications of Sequence Components of Electric Currents NEGATIVE-SEQUENCE CURRENTS: A three-phase unbalanced load produces a reaction field which rotates synchronously with the rotor-field system of generators. Any unbalanced condition will have negative sequence components. This negative sequence currents rotates counter to the synchronously revolving field of the generator. The flux produced by sequence currents cuts the rotor field at twice the rotational velocity, thereby inducing double frequency currents in the field system and in the rotor body. The resulting eddy-currents are very large and cause severe heating of the rotor.
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
103
Network Equations and Methods of Solution
Network Equations
Matrix Representation of System of Equations
Type of Matrices
Matrix Operations
Direct Solutions of System of Equations
Iterative Solutions of System of Equations
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
104
Network Equations The standard form of n independent equations: ⎡ Y 11 Y 12 Y 13 L Y 1 n ⎤ ⎡V 1 ⎤ ⎡I1 ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ V Y 21 Y 22 Y 23 L Y 2 n 2⎥ ⎢I2 ⎥ ⎢ ⎥⎢ ⎢ I 3 ⎥ = ⎢ Y 31 Y 32 Y 33 L Y 3 n ⎥ ⎢V 3 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ M M M ⎥⎢ M ⎥ ⎢M ⎢M ⎥ ⎢ ⎥ ⎢ ⎥ ⎢I ⎥ Y Y Y L Y ⎢ n1 n2 n3 nn ⎥ V n n ⎣ ⎦ ⎣ ⎦ ⎣
⎦
[ I ] = [Ybus][V] Ypp = self-admittance, the sum of all admittances terminating on the node (diagonal elements) Ypq = mutual admittance, the negative of the admittances connected directly between the nodes identifed by the double subscripts U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
105
Network Equations a Bus
Gen
Line
1 2 3 4
a b c
1-3 2-3 1-4 2-4 3-4
c
1
b
3
2
4 Single Line Diagram U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
106
Network Equations 0
Ea
Ea
za 1
za
Eb
zc
Generator
1
zb 3
zd
0
zf
z13 1
Ec
2
ze zh
zg
3 Line 4 0
U. P. National Engineering Center National Electrification Administration
Impedance Diagram Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
107
Network Equations 0
I1 = Ea/za y01 = 1/za I2 = Eb/zb y02 = 1/zb
I1
y01 1
y13
I3 = Ec/zc y03 = 1/zc
I3
y03 3
y14 = 1/zf
y23 = 1/ze
y24 = 1/zg y34 = 1/zh
U. P. National Engineering Center National Electrification Administration
y02 2
y23
y14
y13 = 1/zd
I2
y24 4
Admittance Diagram Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
108
Network Equations I 1 = V1 ( y 01 + y13 + y14 ) − V 3 y13 − V 4 y14 I 2 = V 2 ( y 02 + y 23 + y 24 ) − V 3 y 23 − V 4 y 24
I 3 = V 3 ( y 03 + y 23 + y 34 + y 13 ) − V 2 y 23 − V 4 y 34 − V 1 y 13 0 = V 4 ( y 14 + y 24 + y 34 ) − V 1 y 14 − V 2 y 24 − V 3 y 34
In matrix form,
0 - y13 - y14 ⎡I1 ⎤ ⎡y01 + y13 + y14 ⎤⎡V1 ⎤ ⎢I ⎥ ⎢ ⎥⎢V ⎥ 0 y y y y y + + 02 23 24 23 24 ⎢ 2⎥ = ⎢ ⎥⎢ 2 ⎥ ⎢I3 ⎥ ⎢ ⎥⎢V3 ⎥ - y13 - y23 y03 + y23 + y34 + y13 - y34 ⎢ ⎥ ⎢ ⎥⎢ ⎥ y14 + y24 + y34⎦⎣V4 ⎦ − y24 − y34 ⎣0 ⎦ ⎣ − y14 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
109
Matrix Representations of System of Equations
System of n Linear Equations
In the following system of equations:
a11 x1 + a12 x 2 + a13 x 3 = y 1 a 21 x1 + a 22 x 2 + a 23 x 3 = y 2 a 31 x1 + a 32 x 2 + a 33 x 3 = y 3 x1, x 2, and x3, are unknown variables, a11, a12,‌ ‌, a33 are the coefficient of these variables and y1, y2, and y3 are known parameters. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
110
Matrix Representations of System of Equations The coefficients form an array
⎡ a11 ⎢ A = a21 ⎢ ⎢⎣a31
a12 a22 a32
a13 ⎤ ⎥ a23 ⎥ a33 ⎥⎦
which is called the Coefficient Matrix of the system of equations. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
111
Matrix Representations of System of Equations Similarly, the variables and parameters can be written in matrix form as.
⎡ x1 ⎤ ⎥ ⎢ X = x2 ⎢ ⎥ ⎢⎣ x3 ⎥⎦
and
U. P. National Engineering Center National Electrification Administration
⎡ y1 ⎤ ⎥ ⎢ Y = y2 ⎢ ⎥ ⎢⎣ y3 ⎥⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
112
Matrix Representations of System of Equations
System of Equations in Matrix Form
The system of equations in matrix notation is
⎡ a11 ⎢a ⎢ 21 ⎢⎣ a31
a13 ⎤ ⎡ x1 ⎤ ⎡ y1 ⎤ ⎥ ⎢ ⎥ ⎢ ⎥ a23 x2 = y 2 ⎥⎢ ⎥ ⎢ ⎥ a33 ⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣ y 3 ⎥⎦
a12 a 22 a32
AX = Y U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
113
Matrix Representations of System of Equations Sequence Components of Unbalanced Phasor
Va = Va0 + Va1 + Va2 Vb = Va0 + a²Va1 + aVa2 Vc = Va0 + aVa1 + a²Va2 Rearranging and writing in matrix form
⎡Va ⎤ ⎡1 1 ⎢V ⎥ = ⎢1 a 2 ⎢ b⎥ ⎢ ⎢⎣Vc ⎥⎦ ⎢⎣1 a U. P. National Engineering Center National Electrification Administration
1 ⎤ ⎡Va 0 ⎤ ⎥ ⎢ ⎥ a ⎥ ⎢Va1 ⎥ a 2 ⎥⎦ ⎢⎣Va 2 ⎥⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
114
Matrix Representations of System of Equations V a0
1 = [V a + V b + V c ] 3 V a2
Va1
[
[
1 = V a + aV b + a 2 V c 3
1 = V a + a 2 V b + aV 3
c
]
]
In matrix form
⎡1 1 ⎡Va 0 ⎤ ⎢V ⎥ = 1 ⎢1 a ⎢ a1 ⎥ 3 ⎢ ⎢⎣1 a 2 ⎢⎣Va 2 ⎥⎦ U. P. National Engineering Center National Electrification Administration
1 ⎤ ⎡Va ⎤ a 2 ⎥⎥ ⎢⎢Vb ⎥⎥ a ⎥⎦ ⎢⎣Vc ⎥⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
115
Definition of a MATRIX A matrix consists of a rectangular array of elements represented by a single symbol. [A] is a shorthand notation for the matrix and aij designates an individual element of the matrix. A horizontal set of elements is called a row and a vertical set is called a column. The first subscript i always designates the number of the row in which the element lies. The second subscript j designates the column. For example, element a23 is in row 2 and column 3.
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
116
Definition of a MATRIX ⎡ a11 ⎢a ⎢ 21 [A] = [a ij ] = ⎢ a31 ⎢ ⎢ M ⎢⎣am1
a12 a22 a32 M am2
a13 a23 a33
K K K
am3
K
a 1n ⎤ a2n ⎥ ⎥ a3n ⎥ ⎥ M ⎥ amn ⎥⎦
The matrix has m rows and n columns and is said to have a dimension of m by n (or m x n).
[aij]mxn U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
117
Definition of a Vector A vector X is defined as an ordered set of elements. The components x1, X2…, Xn may be real or complex numbers or functions of some dependent variable.
⎡ x1 ⎤ ⎢x ⎥ 2 ⎢ ⎥ X = ⎢M ⎥ ⎢ ⎥ ⎣ xn ⎦ “n” defines the dimensionality or size of the vector. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
118
Matrices with only one row (n = 1) are called Row Vectors while those with one column (m=1) are called Column Vectors. The elements of a vectors are denoted by single subscripts as the following:
R = [r1
r2
L rn ]
⎡ c1 ⎢c 2 ⎢ C = ⎢ M ⎢ ⎣cm
⎤ ⎥ ⎥ ⎥ ⎥ ⎦
Thus, R is a row vector of dimension n while C is a column vector of dimension m. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
119
Type of Matrices Square Matrix Upper Triangular Matrix Lower Triangular Matrix Diagonal Matrix Identity or Unit Matrix Null Matrix Symmetric Matrix Skew-symmetric Matrix
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
120
Type of Matrices A square matrix is a matrix in which m = n. For a square, the main or principal diagonal consists of the elements of the form aii; e.g., for the 3 x 3 matrix shown
⎡a11 ⎢ A = a21 ⎢ ⎢⎣a31
a12 a22 a32
a13 ⎤ ⎥ a23 ⎥ a33 ⎥⎦
the elements a11, a22, and a33 constitute the principal diagonal. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
121
Type of Matrices An upper triangular matrix is one where all the elements below the main diagonal are zero.
⎡u11 ⎢ U= 0 ⎢ ⎢⎣ 0 U. P. National Engineering Center National Electrification Administration
u12 u22 0
u13 ⎤ ⎥ u23 ⎥ u33 ⎥⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
122
Type of Matrices A lower triangular matrix is one where all elements above the main diagonal are zero.
⎡l11 ⎢ L = l21 ⎢ ⎢⎣l31 U. P. National Engineering Center National Electrification Administration
0 l22 l32
0⎤ ⎥ 0 ⎥ l33 ⎥⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
123
Type of Matrices A diagonal matrix is a square matrix where all elements off the diagonal are equal to zero. Note that where large blocks of elements are zero, they are left blank.
⎡d11 ⎢ D= 0 ⎢ ⎢⎣ 0 U. P. National Engineering Center National Electrification Administration
0 d22 0
0⎤ ⎥ 0 ⎥ d33 ⎥⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
124
Type of Matrices An identity or unit matrix is a diagonal matrix where all elements on the main diagonal are equal to one.
⎡1 ⎢ I = ⎢0 ⎢⎣ 0 U. P. National Engineering Center National Electrification Administration
0 1 0
0⎤ ⎥ 0⎥ 1 ⎥⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
125
Type of Matrices The null matrix is matrix whose elements are equal to zero.
⎡0 ⎢ N= 0 ⎢ ⎢⎣0 U. P. National Engineering Center National Electrification Administration
0 0 0
0⎤ ⎥ 0 ⎥ 0 ⎥⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
126
Type of Matrices A symmetric matrix is one where aij = aji for all i’s and j’s.
⎡5 ⎢ S= 1 ⎢ ⎢⎣2
1 3 7
2⎤ ⎥ 7 ⎥ 8 ⎥⎦
U. P. National Engineering Center National Electrification Administration
a12 = a21 = 1 a13 = a31 = 2 a23 = a32 = 7 Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
127
Type of Matrices A skew-symmetric matrix is a matrix which has the property aij = -aji for all i and j; this implies aii = 0
⎡0 ⎢ K= 5 ⎢ ⎢⎣− 3
−5 0 −6
U. P. National Engineering Center National Electrification Administration
3⎤ ⎥ 6 ⎥ 0⎥⎦
a12 = −5 a 21 = +5
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
128
Matrix Operations Addition of Matrices Product of a Matrix with a Scalar Multiplication of Matrices Transpose of a Matrix Kron Reduction Method Determinant of a Matrix Minors and Cofactors of a Matrix Inverse of a Matrix
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
129
Addition of Matrices Two matrices A = [aij] and B = [bij] can be added together if they are of the same order (mxn). The sum C = A + B is obtained by adding the corresponding elements.
C = [cij] = [aij + bij]
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
130
Addition of Matrices Example:
⎡1 A = ⎢ ⎣2
4 7
0⎤ 3 ⎥⎦
⎡5 B = ⎢ ⎣0
6⎤ 1 ⎥⎦
2 1
then,
⎡(1 + 5) A+ B= ⎢ ⎣(2 + 0) ⎡(1 − 5) A− B = ⎢ ⎣(2 − 0)
(4 + 2) (7 + 1) (4 − 2) (7 − 1)
U. P. National Engineering Center National Electrification Administration
(0 + 6)⎤ ⎡6 =⎢ ⎥ (3 + 1)⎦ ⎣2 (0 − 6)⎤ ⎡− 4 =⎢ ⎥ (3 − 1)⎦ ⎣ 2
6 8 2 6
6⎤ ⎥ 4⎦ − 6⎤ ⎥ 2⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
131
Addition of Matrices Example:
⎡1+ j2 A = ⎢4 + j1 ⎢ ⎢⎣6 + j3
4 − j1 5 + j3 1 − j1
6 − j3⎤ ⎡3 + j2 1 + j1⎥ B = ⎢2 + j1 ⎢ ⎥ ⎢⎣7 − j5 8 + j9⎥⎦
2 − j1 4 + j6 5 − j4
7 + j5⎤ 5 + j4⎥ ⎥ 6 + j5⎥⎦
then,
⎡(1+ j2) +(3+ j2) A+ B= ⎢(4+ j1) +(2+ j1) ⎢ ⎢⎣(6 + j3) +(7 − j5) U. P. National Engineering Center National Electrification Administration
(4− j1) +(2− j1) (6 − j3) +(7 + j5)⎤ (5+ j3) +(4+ j6) (1+ j1) +(5+ j4) ⎥⎥ (1− j1) +(5− j4) (8+ j9) +(6 + j5)⎥⎦ Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
132
Addition of Matrices ⎡ 4 + j4 A + B = ⎢ 6 + j2 ⎢ ⎢⎣13− j2 ⎡− 2 + j0 ⎢ A− B = 2 + j0 ⎢ ⎢⎣−1+ j8 U. P. National Engineering Center National Electrification Administration
6 − j2
13+ j2 ⎤ 9 + j9 6 + j5 ⎥ ⎥ 6 − j5 14+ j14⎥⎦ 2 + j0 −1− j8⎤ ⎥ 1− j3 − 4 − j3 ⎥ − 4 + j3 2 + j4 ⎥⎦ Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
133
Product of a Matrix with a Scalar A matrix is multiplied by a scalar k all elements mn by k , that is,
⎡ ka11 ⎢ ka 21 ⎢ kA = Ak = ⎢ M ⎢ ⎣kam1 U. P. National Engineering Center National Electrification Administration
ka12 ka22 M kam 2
by multiplying
L ka1n ⎤ ⎥ L ka2 n ⎥ M M ⎥ ⎥ L kamn ⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
134
Product of a Matrix with a Scalar Example:
⎡4 ⎢ A= 5 ⎢ ⎢⎣6
⎡4 ⎢ B = kA = 3 5 ⎢ ⎢⎣6
3⎤ ⎥ 2 ⎥ 1⎥⎦
3⎤ ⎥ 2 ⎥ 1⎥⎦
U. P. National Engineering Center National Electrification Administration
and
k =3
⎡ 12 ⎢ B = 15 ⎢ ⎢⎣18
9⎤ ⎥ 6 ⎥ 3 ⎥⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
135
Product of a Matrix with a Scalar Example:
⎡4 + j1 ⎢ A = 5 - j2 ⎢ ⎢⎣6 + j3
3 − j6⎤ ⎥ 2 + j5 and k = 3 ⎥ 1 − j4⎥⎦
⎡12+ j3 ⎡4 + j1 3- j6⎤ ⎢ ⎢ ⎥ B = kA=3 5 - j2 2 + j5 B = 15− j6 ⎢ ⎢ ⎥ ⎢⎣18+ j9 ⎢⎣6 + j3 1- j4⎥⎦ U. P. National Engineering Center National Electrification Administration
9 − j18⎤ ⎥ 6 + j15 ⎥ 3 − j12⎥⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
136
Multiplication of Matrices Two matrices A = [aij] and B = [bij] can be multiplied in the order AB if and only if the number of columns of A is equal to the number of rows of B . That is, if A is of order of (m x l), then B should be of order (l x n). If the product matrix is denoted by C = A B, then C is of order (m x n).
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
137
Multiplication of Matrices An easy way to check whether two matrices can be multiplied.
[A ]m x l [B ]l x n
= [C ]m x n
Interior dimensions are equal multiplication is possible Exterior dimensions define the dimensions of the result U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
138
Multiplication of Matrices If the product matrix is denoted by C = A B, then C is of order (m x n). The elements cij are given by l
cij = ∑ aik bkj k =1
U. P. National Engineering Center National Electrification Administration
for all i and j.
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
139
Multiplication of Matrices Example:
then
⎡ a11 ⎢ A = ⎢a21 ⎢⎣a31
a12 ⎤ ⎥ a22 ⎥ and a32 ⎥⎦ 3 x 2
⎡ a11 ⎢ C = A B = a 21 ⎢ ⎢⎣ a31 2
cij = ∑ aik bkj k =1
⎡b11 b12 ⎤ B=⎢ ⎥ ⎣b21 b22 ⎦ 2 x 2
a12 ⎤ ⎡b11 ⎥ a22 ⎢ ⎥ ⎣b21 a32 ⎥⎦
b12 ⎤ ⎥ b22 ⎦
c 11 = a 11 b 11 + a 12 b 21
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
140
Multiplication of Matrices 2
cij = ∑ aik bkj k =1
⎡ (a11b11 + a12b21 ) (a11b12 + a12b22 ) ⎤ ⎢ ⎥ C = AB = ⎢( a21b11 + a22b21 ) ( a21b12 + a22b22 )⎥ ⎢⎣ (a31b11 + a32b21 ) ( a31b12 + a32b22 )⎥⎦
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
141
Multiplication of Matrices Example:
then
⎡1 4⎤ ⎢ ⎥ A = ⎢2 5⎥ ⎢⎣3 6⎥⎦ 3 x 2
and
⎡1 ⎢ C = AB = ⎢ 2 ⎢⎣ 3
U. P. National Engineering Center National Electrification Administration
⎡7 8 ⎤ B=⎢ ⎥ ⎣9 0 ⎦ 2 x 2
4⎤ ⎡7 ⎥ 5⎥⎢ 9 ⎣ 6 ⎥⎦
8⎤ ⎥ 0⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
142
Multiplication of Matrices ⎡ (1x7 + 4 x9) (1x8 + 4 x0) ⎤ ⎥ ⎢ C = AB = ⎢(2 x7 + 5x9) (2 x8 + 5x0)⎥ ⎢⎣(3x7 + 6 x9) (3x8 + 6 x0)⎥⎦3 x 2 ⎡ 43 C = ⎢⎢ 59 ⎢⎣ 75 U. P. National Engineering Center National Electrification Administration
8 ⎤ 16 ⎥⎥ 24 ⎥⎦ 3 x 2
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
143
Multiplication of Matrices Example:
⎡1+ j2 4− j1 6 − j3⎤ ⎡3+ j2 2− j1 7 + j5⎤ A= ⎢4+ j1 5+ j3 1+ j1⎥ B= ⎢2+ j1 4+ j6 5+ j4⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣6 + j3 1− j1 8 + j9⎥⎦ ⎢⎣7 − j5 5− j4 6 + j5⎥⎦ c 11 = (1 + j 2 )(3 + j 2 ) + (4 − j 1 )(2 + j 1 ) + (6 − j 3 )(7 − j 5 ) = 35 − j 41
c 12 = (1 + j 2 )(2 − j 1 ) + (4 − j 1 )(4 + j 6 ) + (6 − j 3 )(5 − j 4 ) = 44 − j 16
c 13 = (1 + j 2 )(7 + j 5 ) + (4 − j 1)(5 + j 4 ) + (6 − j 3 )(6 + j 5 ) = 72 + j 42
c21 = (4 + j1)(3 + j 2 ) + (5 + j 3 )(2 + j1) + (1 + j1)(7 − j 5 ) = 29 + j 24
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
144
Multiplication of Matrices c 21 = (4 + j 1 )(3 + j 2 ) + (5 + j 3 )(2 + j 1 ) + (1 + j 1 )(7 − j 5 ) = 29 + j 24 c 22 = (4 + j 1 )(2 − j 1 ) + (5 + j 3 )(4 + j 6 ) + (1 + j 1 )(5 − j 4 ) = 20 + j 41 c 23 = (4 + j 1 )(7 + j 5 ) + (5 + j 3 )(5 + j 4 ) + (1 + j 1 )(6 + j 5 ) = 37 + j73
c 31 = (6 + j 3 )(3 + j 2 ) + (1 − j 1)(2 + j 1) + (8 + j 9 )(7 − j 5 ) = 116 + j 43 c 32 = (6 + j 3 )(2 − j 1) + (1 − j 1)(4 + j 6 ) + (8 + j 9 )(5 − j 4 ) = 101 + j 15 c 33 = (6 + j 3 )(7 + j 5 ) + (1 − j 1)(5 + j 4 ) + (8 + j 9 )(6 + j 5 ) = 39 + j 144
⎡ 35 − j 41 [A] x[B] = ⎢⎢ 29 + j 24 ⎢⎣116 + j 43 U. P. National Engineering Center National Electrification Administration
44 − j16 20 + j 41 101 + j15
72 + j 42 ⎤ 37 + j73 ⎥ ⎥ 39 + j144⎥⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
145
Transpose of a Matrix If the rows and columns of an m x n matrix are interchanged, the resultant n x m matrix is the transpose of the matrix and is designated by AT.
For the matrix
The transpose is
⎡ a11 A=⎢ ⎣a12 ⎡ a11 AT = ⎢a21 ⎢ ⎢⎣ a31
U. P. National Engineering Center National Electrification Administration
a21 a22
a31 ⎤ a32 ⎥⎦ 2 x 3
a12 ⎤ a22 ⎥ ⎥ a32 ⎥⎦ 3 x 2
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
146
Transpose of a Matrix Example:
then,
⎡1 A=⎢ ⎣2 ⎡1 T ⎢ A = 3 ⎢ ⎢⎣5
U. P. National Engineering Center National Electrification Administration
3 4
5⎤ 6 ⎥⎦ 2 x 3 2⎤ ⎥ 4 ⎥ 6 ⎥⎦ 3 x 2
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
147
Transpose of a Matrix Example:
⎡1 + j 4 A=⎢ ⎣2 − j 5
3 − j6 4 + j1
5 + j 2⎤ ⎥ 6 − j 3⎦
then,
⎡1 + j 4 T ⎢ A = 3 − j6 ⎢ ⎢⎣5 + j 2 U. P. National Engineering Center National Electrification Administration
2 − j 5⎤ ⎥ 4 + j1 ⎥ 6 − j 3⎥⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
148
Determinant of a Matrix Determinant of a 2 x 2 Matrix Two simultaneous equations:
a11 x1 + a12 x2 = y1 a21 x1 + a22 x2 = y2
(1) (2)
In Matrix Form
⎡ a11 ⎢a ⎣ 21
a12 ⎤ ⎡ x1 ⎤ ⎡ y1 ⎤ =⎢ ⎥ ⎥ ⎢ ⎥ a22 ⎦ ⎣ x2 ⎦ ⎣ y 2 ⎦
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
149
Determinant of a Matrix The solutions of two simultaneous equations can be obtained by eliminating the variables one at a time. Solving for x2 in terms of x1 from the second equation and substituting this expression for x2 in the first equation, the following is obtained:
a22x2 = y2 − a21x1
y2 a21 x2 = − x1 a22 a22 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
150
Determinant of a Matrix substituting x2 and solving for x1
y 2 a21 a11 x1 + a12 ( − x1 ) = y1 a22 a22 a11 a22 x1 + a12 y 2 − a12 a21 x1 = a22 y1 ( a11 a22 − a12 a21 ) x1 = a22 y1 − a12 y2 a22 y1 − a12 y2 x1 = a11 a22 − a12 a21 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
151
Determinant of a Matrix Then, substituting x1 in either equation (1) or (2), x2 is obtained
a11 y2 − a21 y1 x2 = a11a22 − a12a21
The expression (a11a22 – a12a21) is the value of the determinant of the coefficient matrix A, denoted by |A|.
a11 a12 | A| = a21 a22 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
152
Determinant of a Matrix The determinant of A is can be determined by reducing the size of the original matrix by eliminating rows. For example:
⎡ 1 [ A ] = ⎢⎢ − 1 ⎢⎣ − 6 1 | A |= − 1 −6
4
2⎤ 1 ⎥⎥ 2 ⎥⎦
1
2
− 2 4
1 2
1 − 2
U. P. National Engineering Center National Electrification Administration
Eliminate row 1 by striking out the row1 and jth column Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
153
Determinant of a Matrix The determinant of A is 1 1 2
| A |= − 1 −6
| A |= ( −1 ) 1 1+1
− 2 4
1 2
−2
1
4
2
+ ( −1 ) 1 1+2
−1
1
−6
2
+ ( −1 ) 2 1+3
−1
−2
−6
4
| A |= ( +1)(1)[( −2)( 2) − (1)( 4) ] + ( −1)(1)[( −1)( 2) − (1)( −6) ] + ( +1)( 2)[( −1)( 4) − ( −2)( −6) ] = (1)[− 4 − 4 ] − (1)[− 2 + 6 ] + ( 2)[− 4 − 12 ] | A |= −44 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
154
Determinant of a Matrix Minors and Cofactors of a Matrix The determinant obtained by striking out the ith row and jth column is called the minor element aij. Example:
a11 a 21 a31
a12 a22 a32
a13 a12 a 23 = a32 a33
a13 a33
The minor of a 21 = ( a 12 a 33 − a 32 a 13 ) U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
155
Determinant of a Matrix Minors and Cofactors of a Matrix The cofactor of an element aij designated by Aij is
Aij = ( − 1 ) Example:
i+ j
(minor
of a ij )
A 21 = ( − 1) 2 +1 (the min or of a 21 ) = (-1) 3 (the min or of a 21 )
A 21 = - 1 (the minor of a 21 )
Since the minor of a 21 = ( a 12 a 23 − a 32 a 13 )
∴ the cofactor of A21 = − 1( a 12 a 33 − a 32 a 13 ) U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
156
Determinant of a Matrix Minors and Cofactors of a Matrix
⎡1 ⎢- 1 A = Example: ⎢ ⎢⎣- 6
1 -2 4 1
A11 = ( −1 )1+1 − 1
1
2⎤ 1⎥ ⎥ 2⎥⎦
−2
−6
4
1 A12 = ( −1 )1+2 − 1 −6
1 −2 4
U. P. National Engineering Center National Electrification Administration
2 1 = 1[( −2 )( 2 ) − ( 1 )( 4 )] = −8 2 2 1 = −1[( −1 )( 2 ) − ( 1 )( −6 )] = −4 2 Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
157
Determinant of a Matrix Minors and Cofactors of a Matrix 1 1 2 A13 = ( −1 )1+3 − 1 − 2 1 = 1[( −1 )( 4 ) − ( −2 )( −6 )] = −16 −6 4 2
A21 = ( −1 )2+1
A22 = ( −1 )2+2
1 −1
1 −2
−6
4
1 −1
1 −2
−6
4
U. P. National Engineering Center National Electrification Administration
2 1 = −1[( 1 )( 2 ) − ( 4 )( 2 )] = 6 2
2 1 = 1[( 1 )( 2 ) − ( 2 )( −6 )] = 14 2 Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
158
Determinant of a Matrix Minors and Cofactors of a Matrix
1 A23 = ( −1 )2+3 − 1 −6 1 A31 = ( − 1 )3 + 1 − 1 −6 1 A32 = ( −1 )3+ 2 − 1 −6
1 −2 4
2 1 = −1[( 1 )( 4 ) − ( 1 )( −6 )] = −10 2
1 −2 4 1 −2 4
U. P. National Engineering Center National Electrification Administration
2 1 = 1[( 1 )( 1 ) − ( −2 )( 2 )] = 5 2 2 1 = −1[( 1 )( 1 ) − ( −1 )( 2 )] = −3 2 Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
159
Determinant of a Matrix Minors and Cofactors of a Matrix
1 A33 = ( −1)3+3 − 1 −6
1 −2 4
2 1 = 1[(1 )( −2 ) − ( −1 )(1 )] = −1 2
Therefore the cofactors of matrix A are:
A11 = −8
A21 = 6
A31 = 5
A12 = −4
A22 = 14
A32 = −3
A13 = −16
A23 = −10
A33 = −1
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
160
Determinant of a Matrix Minors and Cofactors of a Matrix and in matrix form:
⎡ A11 Cofactors of A = ⎢⎢ A21 ⎢⎣ A31
A12 A22 A32
A13 ⎤ A23 ⎥⎥ A33 ⎥⎦
⎡− 8 Cofactors of A = ⎢⎢ 6 ⎢⎣ 5
−4 14 −3
− 16 ⎤ − 10 ⎥⎥ − 1 ⎥⎦
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
161
Inverse of a Matrix Division does not exist in matrix algebra except in the case of division of a matrix by a scalar. However, for a given set of equations.
a 11 x 1 + a 12 x 2 + a 13 x 3 = y 1 a 21 x 1 + a 22 x 2 + a 23 x 3 = y 2 a 31 x 1 + a 32 x 2 + a 33 x 3 = y 3 or in matrix form [AX] = [Y]. It is desirable to express x1, x2, and x3 a function of y1, y2, and y3, i.e.. [X] = [BY], where B is the inverse of A designated by A-1. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
162
Inverse of a Matrix If the determinant of A is not zero, the equations can be solved for x ’s as follows;
A 11 A 21 A 31 x1 = y1 + y2 + y3 | A| | A| | A|
x2
A 12 A 22 A 32 = y1 + y2 + y3 | A| | A| | A|
A 13 A 23 A 33 x3 = y1 + y2 + y3 | A| | A| | A| where A11, A12, ‌, A33 are cofactors of a11, a12,,a33 and |A| is the determinant of A. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
163
Inverse of a Matrix Thus,
B = A -1
⎡ A 11 ⎢| A | ⎢ A 12 ⎢ = ⎢| A | ⎢ A 13 ⎢ ⎣| A |
A 21 | A | A 22 | A | A 23 | A |
A 31 ⎤ | A |⎥ ⎥ A 32 ⎥ or | A |⎥ A 33 ⎥ ⎥ | A |⎦
+ A -1 A = | A|
A+ is called the adjoint of A. It should be noted that the elements of adjoint A+ are the cofactors of the elements of A, but are placed in transposed position. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
164
Inverse of a Matrix Example: Get the inverse of A
⎡1 A = ⎢- 1 ⎢ ⎢⎣- 6
1 -2 4
2⎤ 1⎥ ⎥ 2⎥⎦
A
-1
A+ = | A|
the Adjoint of A is
⎡A11 A+ = ⎢A12 ⎢ ⎢⎣A13
A21 A22 A23
U. P. National Engineering Center National Electrification Administration
A31⎤ ⎡ − 8 A32 ⎥ = ⎢ − 4 ⎥ ⎢ A33 ⎥⎦ ⎢⎣−16
6 14 −10
5⎤ − 3⎥ ⎥ −1⎥⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
165
Inverse of a Matrix Hence, the inverse of matrix A is
6 5⎤ ⎡ −8 ⎥ ⎢ −4 14 3 − ⎥ ⎢ + ⎢⎣− 16 − 10 − 1⎥⎦ A −1 A = = A − 44
A
−1
⎡ −8 1 ⎢ = − ⎢ −4 44 ⎢⎣− 16
U. P. National Engineering Center National Electrification Administration
⎡ −−448 ⎢ = ⎢ −−444 ⎢⎣ −−16 44
6 −44 14 −44 −10 −44
5 −44 −3 −44 −1 −44
⎤ ⎥ ⎥ ⎥⎦
5 ⎤ ⎥ 14 − 3⎥ − 10 − 1⎥⎦ 6
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
166
Kron Reduction Method ⎡ y1 ⎤ ⎡ a11 ⎢ y ⎥ ⎢a ⎢ 2 ⎥ = ⎢ 21 ⎢ y3 ⎥ ⎢a31 ⎢ ⎥ ⎢ ⎣ 0 ⎦ ⎣a41
⎡ y1 ⎤ ⎡a11 ⎢ y ⎥ ⎢a ⎢ 2 ⎥ = ⎢ 21 ⎢ y3 ⎥ ⎢a31 ⎢ ⎥ ⎢ ⎣ 0 ⎦ ⎣a41
a12 a22
a13 a23
a 32 a42
a33 a43
a12
a13
a22 a32 a42
a23 a33 a43
a14 ⎤ ⎡ x1 ⎤ a24 ⎥ ⎢ x2 ⎥ ⎥⎢ ⎥ a34 ⎥ ⎢ x3 ⎥ ⎥⎢ ⎥ a44 ⎦ ⎣ x4 ⎦
a14 ⎤⎡ x1 ⎤ ⎥ ⎢ ⎥ a24 x2 ⎥⎢ ⎥ a34 ⎥⎢x3 ⎥ ⎥⎢ ⎥ a44 ⎦⎣x4 ⎦
The four equations can be reduced to three equations by Kron Reduction Method since the independent variable of the fourth equation is zero. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
167
Kron Reduction Method ⎡ y1 ⎤ [Y1 ] = ⎢⎢ y2 ⎥⎥ ⎢⎣ y3 ⎥⎦
⎡a11 [ A1 ] = ⎢⎢a21 ⎢⎣a31
a12 a22 a32
a13 ⎤ a23 ⎥ ⎥ a33 ⎥⎦
⎡ a14 ⎤ [A2 ] = ⎢⎢a24 ⎥⎥ ⎢⎣a34 ⎥⎦
⎡ x1 ⎤ [X1] = ⎢⎢x2 ⎥⎥ ⎢⎣ x3 ⎥⎦
[Y 2 ] = [0 ]
[A ] = [a
a42
a43 ]
[A4 ] = [a44 ]
[X 2 ] = [x4 ]
3
41
⎡Y1 ⎤ ⎡ A1 ⎢ 0 ⎥ = ⎢A ⎣ ⎦ ⎣ 3 U. P. National Engineering Center National Electrification Administration
A2 ⎤⎡ X 1 ⎤ ⎥ ⎢ ⎥ A4 ⎦⎣ X 2 ⎦ Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
168
Kron Reduction Method
⎡Y1 ⎤ ⎡ A1 ⎢ 0 ⎥ = ⎢A ⎣ ⎦ ⎣ 3
A2 ⎤⎡ X1 ⎤ ⎥ ⎢ ⎥ A4 ⎦⎣X2 ⎦
[Y1 ] = [ A1 ][ X 1 ] + [ A 2 ][X 2 ] [0 ] = [A3 ][X 1 ] + [A 4 ][X 2 ]
(1) (2)
From (2)
[ A 4 ][X 2 ] = − [A3 ][X 1 ] [X 2 ] = − [A4 ]− 1 [A3 ][X 1 ] U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
169
Kron Reduction Method Substitute [X2] to equation (1)
[Y1 ] = [A1 ][X 1 ] + [A2 ](− [A4 ]−1 [A3 ][X 1 ]) Thus,
[Y1 ] = {[A1 ] − [A2 ][A4 ] [A3 ]}[X 1 ] −1
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
170
Kron Reduction Method Example:
⎡ y1 ⎤ ⎡1 ⎢ y ⎥ ⎢8 ⎢ 2⎥ = ⎢ ⎢ y3 ⎥ ⎢3 ⎢ ⎥ ⎢ ⎣ 0 ⎦ ⎣2
2 7 4 4
3 6 5 7
4 ⎤ ⎡ x1 ⎤ 5 ⎥ ⎢x2 ⎥ ⎥⎢ ⎥ 6 ⎥ ⎢ x3 ⎥ ⎥⎢ ⎥ 8 ⎦ ⎣ x4 ⎦
⎞ ⎡ x1 ⎤ ⎡ y1 ⎤ ⎛ ⎡1 2 3⎤ ⎡4⎤ ⎢ y ⎥ = ⎜ ⎢8 7 6⎥ − ⎢5⎥[8]−1 [2 4 7]⎟ ⎢ x ⎥ ⎟⎢ 2 ⎥ ⎢ 2⎥ ⎜⎢ ⎥ ⎢ ⎥ ⎟⎢ x ⎥ ⎢⎣ y3 ⎥⎦ ⎜⎝ ⎢⎣3 4 5⎥⎦ ⎢⎣6⎥⎦ ⎠⎣ 3 ⎦ U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
171
Kron Reduction Method ⎞ ⎡ x1 ⎤ ⎡ y1 ⎤ ⎛ ⎡1 2 3⎤ ⎡4⎤ ⎜ ⎢ y ⎥ = ⎢8 7 6⎥ − ⎢5⎥[0.125][2 4 7]⎟ ⎢ x ⎥ ⎟⎢ 2 ⎥ ⎢ 2⎥ ⎜⎢ ⎥ ⎢ ⎥ ⎟⎢ x ⎥ ⎢⎣ y3 ⎥⎦ ⎜⎝ ⎢⎣3 4 5⎥⎦ ⎢⎣6⎥⎦ ⎠⎣ 3 ⎦
⎞ ⎡ x1 ⎤ ⎡ y1 ⎤ ⎛ ⎡1 2 3⎤ ⎡ 0.5 ⎤ ⎜ ⎢ y ⎥ = ⎢8 7 6⎥ − ⎢0.625⎥[2 4 7 ]⎟ ⎢ x ⎥ ⎟⎢ 2 ⎥ ⎥ ⎥ ⎢ ⎢ 2⎥ ⎜⎢ ⎟⎢ x ⎥ ⎢⎣ y3 ⎥⎦ ⎜⎝ ⎢⎣3 4 5⎥⎦ ⎢⎣ 0.75 ⎥⎦ ⎠⎣ 3 ⎦ 2 3.5 ⎤ ⎞ ⎡ x1 ⎤ ⎡ y1 ⎤ ⎛ ⎡1 2 3⎤ ⎡ 1 ⎜ ⎢ y ⎥ = ⎢8 7 6⎥ − ⎢1.25 2.5 4.375⎥ ⎟ ⎢ x ⎥ ⎢ 2⎥ ⎜⎢ ⎥ ⎢ ⎥ ⎟⎢ 2 ⎥ ⎢⎣ y3 ⎥⎦ ⎜⎝ ⎢⎣3 4 5⎥⎦ ⎢⎣ 1.5 3 5.25 ⎥⎦ ⎟⎠ ⎢⎣ x3 ⎥⎦ U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
172
Kron Reduction Method The 4x4 matrix was reduced to a 3x3matrix.
⎡ y1 ⎤ ⎡1 ⎢ y ⎥ ⎢8 ⎢ 2⎥ = ⎢ ⎢ y3 ⎥ ⎢3 ⎢ ⎥ ⎢ ⎣ 0 ⎦ ⎣2
2 7
3 6
4 4
5 7
4 ⎤ ⎡ x1 ⎤ 5 ⎥ ⎢x2 ⎥ ⎥⎢ ⎥ 6 ⎥ ⎢ x3 ⎥ ⎥⎢ ⎥ 8 ⎦ ⎣ x4 ⎦
0 − 0.50⎤ ⎡ x1 ⎤ ⎡ y1 ⎤ ⎡ 0 ⎢ y ⎥ = ⎢6.75 4.5 1.625 ⎥ ⎢ x ⎥ ⎢ 2⎥ ⎢ ⎥⎢ 2 ⎥ ⎢⎣ y3 ⎥⎦ ⎢⎣1.50 1.0 − 0.25⎥⎦ ⎢⎣ x3 ⎥⎦ U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
173
Direct Solutions of System of Equations
Cramer’s Rule of Determinants
Matrix Inversion Method
Gaussian Elimination Method
Gauss-Jordan Method
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
174
Solutions of System of Equations by Cramer’ s Rule The system of three linear equations in three unknowns x1, x2, x3:
a11 x1 + a12 x 2 + a13 x3 = y1 a 21 x1 + a 22 x 2 + a 23 x3 = y 2 a31 x1 + a 32 x 2 + a 33 x3 = y 3
written in matrix form as :
⎡ a11 ⎢a ⎢ 21 ⎢⎣ a 31
a12 a 22 a 32
a13 ⎤ ⎡ x1 ⎤ ⎡ y 1 ⎤ a 23 ⎥ ⎢ x 2 ⎥ = ⎢ y 2 ⎥ ⎥⎢ ⎥ ⎢ ⎥ a 33 ⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣ y 3 ⎥⎦
U. P. National Engineering Center National Electrification Administration
or
AX = Y
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
175
can be solved by Cramer’s Rule of determinants. The determinant of coefficient matrix A is
a 11
a 12
a 13
| A |= a 21
a 22
a 23
a 31
a 32
a 33
x1 can be obtained by :
x1 =
y1
a 12
a 13
y2 y3
a 22 a 32 | A|
a 23 a 33
Note that values in the numerator are the values of the determinant of A with the first column were replaced by the Y vector elements. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
176
Similarly, x2 and x3 can be obtained by:
x2 =
a 11 a 21
y1 y2
a 13 a 23
a 31
y3 | A|
a 33
a 11 a 21
a 12 a 22
y1 y2
a 31
a 32 | A|
y3
and
x3 =
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
177
Solutions of System of Equations by Cramer’s Rule Example:
x 1 + x 2 + 2x 3 = 3 - x 1 − 2x 2 + x 3 = 7 - 6x 1 + 4x 2 + 2x 3 = 14
⎡ 1 ⎢- 1 ⎢ ⎢⎣ - 6
1 - 2 4
U. P. National Engineering Center National Electrification Administration
2 ⎤ ⎡ x1 ⎤ ⎡ 3 ⎤ ⎥ ⎢ ⎥ ⎢ ⎥ 1 x2 = 7 ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣ 14 ⎥⎦ 2 ⎥⎦ ⎢⎣ x 3 ⎥⎦ Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
178
Solutions of System of Equations by Cramer’s Rule ⎡1 A = ⎢- 1 ⎢ ⎢⎣- 6
x1 =
1 -2 4
2⎤ 1⎥ ⎥ 2⎥⎦
1
1
| A|= -1 −6
2
−2 4
1 = − 44 2
y1
a 12
a 13
3
1
2
y2 y3
a 22 a 32 | A|
a 23 a 33
7 14
−2 4 - 44
1 2
U. P. National Engineering Center National Electrification Administration
x1 =
88 = = −2 − 44
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
179
Solutions of System of Equations by Cramer’s Rule
x2 =
x3 =
a 11 a 21
y1 y2
a 13 a 23
a 31
y3 | A|
a 33
a 11 a 21 a 31
a 12 a 22 a 32 | A|
Therefore,
y1 y2 y3
1
x2 =
−1 −6
⎡ 1 ⎢− 1 ⎢ ⎢⎣ − 6 x3 =
x1 = - 2
U. P. National Engineering Center National Electrification Administration
3
2
7 14 - 44 1
1 2
−2 4 - 44
x
2
3⎤ 7 ⎥ ⎥ 14 ⎥⎦
= -1
44 = = −1 − 44
− 132 =3 = − 44
x
3
=3
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
180
Solutions of System of Equations by Matrix Inversion The system of equations in matrix form can be manipulated as follows:
AX = Y A AX = A Y -1
-1
IX = A Y -1
X =A Y -1
Hence, the solution X can be obtained by multiplying The inverse of the coefficient matrix by the constant matrix Y. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
181
Solutions of System of Equations by Matrix Inversion Example:
x 1 + x 2 + 2x 3 = 3 - x 1 − 2x 2 + x 3 = 7 - 6x 1 + 4x 2 + 2x 3 = 14 ⎡ 1 A = ⎢- 1 ⎢ ⎢⎣ - 6
U. P. National Engineering Center National Electrification Administration
1 -2 4
2⎤ 1⎥ ⎥ 2 ⎥⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
182
Solutions of System of Equations by Matrix Inversion ⎡ −8 1 ⎢ -1 −4 A =− ⎢ 44 ⎢⎣− 16
6 14 − 10
⎡ −8 1 ⎢ X = A -1Y = − −4 ⎢ 44 ⎢⎣ − 16 ⎡ − 8( 3 ) + 1 ⎢ -1 X =A Y =− − 4( 3 ) + ⎢ 44 ⎢⎣− 16( 3 ) + U. P. National Engineering Center National Electrification Administration
6 14 − 10 6( 7 )
5⎤ − 3⎥ ⎥ − 1⎥⎦ 5 ⎤⎡ 3 ⎤ − 3⎥ ⎢ 7 ⎥ ⎥⎢ ⎥ − 1⎥⎦ ⎢⎣14 ⎥⎦ +
14( 7 ) + − 10( 7 ) +
5( 14 ) ⎤ − 3(14) ⎥ ⎥ − 1( 14 )⎥⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
183
Solutions of System of Equations by Matrix Inversion ⎡ 88 ⎤ ⎡ − 2 ⎤ 1 ⎢ -1 X =A Y =− 44 ⎥ = ⎢ − 1 ⎥ ⎥ ⎢ ⎥ 44 ⎢ ⎢⎣ − 132 ⎥⎦ ⎢⎣ 3 ⎥⎦ Therefore:
x1 = - 2 x2 = - 1 x3 = 3 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
184
Gaussian Elimination Method The following are the rules in matrix manipulation: (1) Interchange rows (2) Multiply row by constant (3) Add rows
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
185
Gaussian Elimination Method Example:
⎡1 ⎢−1 ⎢ ⎢⎣−6
x1 + x2 + 2x3 = 3 - x1 − 2x2 + x3 = 7 - 6x1 + 4x2 + 2x3 = 14
1 −2 4
2 1 2
M M M
3⎤ 7⎥ ⎥ 14⎥⎦
Add row 1 to row 2 to get row 2. Add 6 times row 1 to row 3 to get row 3.
⎡1 ⎢0 ⎢ ⎢⎣0
1
2
M
−1 10
3 14
M M
U. P. National Engineering Center National Electrification Administration
3⎤ ⎥ 10 ⎥ 32⎥⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
186
Gaussian Elimination Method ⎡1 ⎢0 ⎢ ⎢⎣0
M
1
2
−1 10
3 M 14 M
3⎤ 10⎥ ⎥ 32⎥⎦
Multiply row 2 by 10 then add to row 3 to obtained row 3.
⎡1 ⎢0 ⎢ ⎢⎣0
1
2
M
−1 0
3 44
M M
U. P. National Engineering Center National Electrification Administration
3⎤ 10 ⎥ ⎥ 132⎥⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
187
Gaussian Elimination Method By Back Substitution:
0x1 + 0 x2 + 44 x3 = 132 0 x1 − x2 + 3( 3 ) = 10 x1 + ( −1 ) + 2( 3 ) = 3 Therefore:
x3 = 3 U. P. National Engineering Center National Electrification Administration
x2 = - 1
x1 = -2
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
Gaussian Elimination Method The two phases of Gauss Elimination: forward elimination & back substitution. The primes indicate the number of times that the coefficients and constants have been modified.
⎡a11 ⎢a ⎢ 21 ⎢⎣a31
a12 a22 a32
a13 M c1 ⎤ a23 M c2 ⎥ ⎥ a33 M c3 ⎥⎦ ⇓
⎡a11 ⎢ ⎢ ⎢⎣
a12 a'22
188
a13 M c1 ⎤ a'23 M c2' ⎥ ⎥ a"33 M c3" ⎥⎦
Forward elimination
⇓ x3 = c3" / a"33 x2 =( c2 −a23x3 ) / a22 '
'
'
Back substitution
x1 =( c1 −a12x2 −a13x3 ) / a11
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
189
Gauss-Jordan Method From Gauss Elimination Method
⎡1 ⎢0 ⎢ ⎢⎣0
1
2
M
−1 0
3 44
M M
1
2
M
1 0
−3 44
M M
3⎤ 10 ⎥ ⎥ 132⎥⎦
Multiply row 2 by -1.
⎡1 ⎢0 ⎢ ⎢⎣0 U. P. National Engineering Center National Electrification Administration
3 ⎤ ⎥ − 10 ⎥ 132⎥⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
190
Gauss-Jordan Method Divide row 3 by 44.
⎡1 ⎢0 ⎢ ⎢⎣0
1
2
M
1 0
−3 1
M M
3 ⎤ − 10⎥ ⎥ 3 ⎥⎦
Multiply row 2 by -1 then add to row 1 to get row 1.
⎡1 ⎢0 ⎢ ⎢⎣0
0 1 0
U. P. National Engineering Center National Electrification Administration
5 −3 1
M M M
13 ⎤ − 10⎥ ⎥ 3 ⎥⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
191
Gauss-Jordan Method Multiply row 3 by -5 then add to row 1 to get row 1. Multiply row 3 by 3 then add to row 2 to get row 2.
⎡1 ⎢0 ⎢ ⎢⎣0
0
0
M
1 0
0 1
M M
U. P. National Engineering Center National Electrification Administration
− 2⎤ − 1⎥ ⎥ 3 ⎥⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
192
Gauss-Jordan Method Therefore:
Then,
⎡1 ⎢0 ⎢ ⎢⎣0
0 1 0
0 ⎤ ⎡ x1 ⎤ ⎡ − 2 ⎤ 0 ⎥ ⎢ x2 ⎥ = ⎢ − 1⎥ ⎥⎢ ⎥ ⎢ ⎥ 1 ⎥⎦ ⎢⎣ x 3 ⎥⎦ ⎢⎣ 3 ⎥⎦
x1 = − 2 x2 = −1 x3 = 3
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
193
Gauss-Jordan Method
The Gauss-Jordan method is a variation of Gauss Elimination. The major differences is that when an unknown is eliminated in the GJM, it is eliminated from all other equations rather than just the subsequent ones.
In addition, all rows are normalized by dividing them by their pivot elements. Thus, the elimination steps results in an identity matrix rather than a triangular matrix.
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
Gauss-Jordan Method Graphical depiction of the Gauss-Jordan Method. The superscript (n) means that the elements of the right-hand-side vector have been modified n times (for this case, n=3).
⎡ a 11 ⎢a ⎢ 21 ⎢⎣ a 31
a 12
a 13
M
a 22 a 32
a 23 a 33
M M
c1 ⎤ c2 ⎥ ⎥ c 3 ⎥⎦
↓ ⎡1 ⎢0 ⎢ ⎢⎣0
0 1
0 0
M M
0
1
M
c 1(n) ⎤ c 2(n) ⎥⎥ c 3(n) ⎥⎦
↓ x1 x2
U. P. National Engineering Center National Electrification Administration
194
=
c 1(n)
=
c 2(n)
x3 =
c 3(n)
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
195
Iterative Solutions of System of Equations
Gauss Iterative Method
Gauss-Seidel Method
Newton-Raphson Method
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
196
Iterative Solutions of System of Equations An iterative method is a repetitive process for obtaining the solution of an equation or a system of equation. It is applicable to system of equations where the main-diagonal elements of the coefficient matrix are larger in magnitude in comparison to the off-diagonal elements. The Gauss and Gauss-Seidel iterative techniques are for solving linear algebraic solutions and the NewtonRaphson method applied to the solution of non-linear equations. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
197
Iterative Solutions of System of Equations The solutions starts from an arbitrarily chosen initial estimates of the unknown variables from which a new set of estimates is determined. Convergence is achieved when the absolute mismatch between the current and previous estimates is less than some pre-specified precision index for all the variables.
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
198
Gauss Iterative Method Example:
4x 1 − x 2 + x 3 = 4 x 1 + 4x 2 + x 3 = 6 x 1 + x 2 + 3x 3 = 5
Assume a convergence index of Îľ = 0.001 and the following initial estimates:
a) x1 = x 2 = x3 = 0.0 0
0
0
b) x1 = x 2 = x 3 = 0.5 0
U. P. National Engineering Center National Electrification Administration
0
0
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
199
Gauss Iterative Method Solution: a) The system of equation must be expressed in standard form.
x
k +1 1
x
k +1 2
x 3k + 1
1 k k = ( 4 + x2 - x3 ) 4 1 = ( 6 - x 1k - x 3k ) 4 1 = ( 5 - x 1k - x 2k ) 3
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
200
Gauss Iterative Method a) with x1 0 = x 2 0 = x3 0 = 0 Iteration 1 (k = 0):
1 x1 = ( 4 + 0 - 0 ) = 1.0 4 1 1 x2 = ( 6 - 0 - 0 ) = 1.5 4 1 x31 = ( 5 - 0 - 0 ) = 1.6667 3 Δ x 10 = 1 − 0 = 1 1
Δ x 20 = 1 . 5 − 0 = 1 . 5 Δ x 30 = 1 . 6667 − 0 = 1 . 6667 max
Δ x 30
U. P. National Engineering Center National Electrification Administration
= 1.6667
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
201
Gauss Iterative Method Iteration 2 (k = 1):
1 x1 = ( 4 + 1.5 - 1.6667 ) = 0.958333 4 1 2 x2 = ( 6 - 1.0 - 1.6667 ) = 0.833333 4 1 2 x3 = ( 5 - 1.0 - 1.5 ) = 0.833333 3 Δx11 = 0.958325 − 1 = 0.041667 2
Δx21 = 0.833333 − 1.5 = 0.66667 Δx31 = 0.833333 − 1.6667 = −0.83334 max Δx31 = 0.83334 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
202
Gauss Iterative Method Iteration 3 (k = 2):
1 ( 4 + 0.8333 - 0.8333 ) = 1.0 4 1 3 x 2 = ( 6 - 0.9583 - 0.8333 ) = 1.0521 4 1 3 x 3 = ( 5 - 0.9583 - 0.8333 ) = 1.0695 3 Δ x 12 = 1 − 0 . 958325 = 0 . 041667 x1 3 =
Δ x 22 = 1 . 0521 − 0 . 833325 = 0 . 21877 Δ x 32 = 1 . 0695 − 0 . 8333 = 0 . 23617 max Δ x 32 = 0.23617 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
203
Gauss Iterative Method Iteration 4 (k = 3):
1 x1 = ( 4 + 1.0521 - 1.0695 ) = 0.9956 4 1 x2 4 = ( 6 - 1.0 - 1.0695 ) = 0.9826 4 1 x3 4 = ( 5 - 1.0 - 1.0521 ) = 0.9826 3 Δx13 = 0.9956 − 1 = −0.0044 4
Δx23 = 0.9826 − 1.0521 = −0.0695 Δx33 = 0.9826 − 1.0695 = −0.0869 max Δx33 = 0.0869 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
204
Gauss Iterative Method Iteration 5 (k = 4):
1 x 1 = ( 4 + 0.9826 - 0.9826 ) = 1.0 4 1 5 x 2 = ( 6 - 0.9956 - 0.9826 ) = 1.0054 4 1 5 x 3 = ( 5 - 0.9956 - 0.9826) = 1.0073 3 Δ x 14 = 1 − 0 . 9956 = 0 . 0044 5
Δ x 24 = 1 . 0054 − 0 . 9826 = − 0 . 0228 Δ x 34 = 1 . 0073 − 0 . 9826 = 0 . 0247 max Δ x 34 = 0.0247 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
205
Gauss Iterative Method Iteration 6 (k = 5):
1 x1 = ( 4 + 1.0054 - 1.0073 ) = 0.9995 4 1 6 x 2 = ( 6 - 1.0 - 1.0071 ) = 0.9982 4 1 6 x 3 = ( 5 - 1.0 - 1.0054 ) = 0.9982 3 Δx15 = 0.9995 − 1 = −0.0005 6
Δx 25 = 0.9982 − 1.0054 = −0.0072 Δx 35 = o .9982 − 1.0073 = −0.0091 max Δx35 = 0.0091 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
206
Gauss Iterative Method Iteration 7 (k = 6):
1 x 1 = ( 4 + 0.9982 - 0.9982 ) = 1.0 4 1 7 x 2 = ( 6 - 0.9995 - 0.9982 ) = 1.0006 4 1 7 x 3 = ( 5 - 0.9995 - 0.9982) = 1.0008 3 Δ x 16 = 1 − 0 . 9995 = 0 . 0005 7
Δ x 62 = 1 . 0006 − 0 . 9982 = 0 . 0024 Δ x 36 = 1 . 0008 − 0 . 9982 = 0 . 0026 max Δ x 36 = 0.0026 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
207
Gauss Iterative Method Iteration 8 (k = 7):
1 x1 = ( 4 + 1.0006 - 1.0008 ) = 0.9995 4 1 8 x2 = ( 6 - 1.0 - 1.0008 ) = 0.9998 4 1 8 x3 = ( 5 - 1.0 - 1.0008 ) = 0.9998 3 Δx17 = 0.9995 − 1 = −0.0005 8
Δx27 = 0.9998 − 1.0006 = −0.0008 Δx37 = 0.9998 − 1.0008 = −0.0010 max Δx37 = 0.0010 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
208
Gauss Iterative Method The Gauss iterative method has converged at iteration 7. The method yields the following solution.
x 1 = 0.9995 x 2 = 0.9998 x 3 = 0.9998
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
209
Gauss Iterative Method b) with x 1 = x 2 = x 3 = 0.5 0
0
0
Iteration 1 (k = 0):
x1 1 = x21 = x31 =
Δ x 10 = Δ x 20 = Δ x 30 = max Δ x = U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
210
Gauss Iterative Method Iteration 2 (k = 1):
x1 = 2
x2 2 = x3 2 =
Δx11 = Δx21 = Δx31 = max Δx 1 = U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
211
Gauss Iterative Method Iteration 3 (k = 2):
x1 3 = x2 = 3
x3 3 =
Δx12 = Δx22 = Δx32 = max Δx 2 = U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
212
Gauss Iterative Method Iteration 4 (k = 3):
x1 = 4
x2 4 = x3 4 =
Δx13 = Δx23 = Δx33 = max Δx 3 = U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
213
Gauss Iterative Method Iteration 5 (k = 4):
x1 5 = x2 = 5
x3 5 =
Δx14 = Δx24 = Δx34 = max Δx = 4
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
214
Gauss Iterative Method Iteration 6 (k = 5):
x1 = 6
x2 6 = x3 6 =
Δx15 = Δx25 = Δx35 = max Δx 5 = U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
215
Gauss Iterative Method Iteration 7 (k = 6):
x1 7 = x2 7 = x3 7 =
Δx16 = Δx26 = Δx36 = max Δx 6 = U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
216
Gauss Iterative Method Iteration 8 (k = 7):
x18 = x2 8 = x3 8 =
Δx17 = Δx27 = Δx37 = max Δx7 = U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
217
Gauss Iterative Method =
x
1
x
2
=
x
3
=
Note: Number of iterations to achieve convergence is also dependent on initial estimates U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
218
Gauss Iterative Method Given the system of algebraic equations,
a 11 x 1 + a 12 x 2 + L + a 1n x n = y 1 a 21 x 1 + a 22 x 2 + L + a 2n x n = y 2 ↓
↓
↓
↓
a 31 x 1 + a 32 x 2 + L + a nn x n = y 3 In the above equation, the x’s are unknown. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
219
Gauss Iterative Method From the first equation,
a 11 x 1 = y 1 − a 12 x 2 K − a 1n x n 1 x1 = ( y 1 − a 12 x 2 K − a 1n x n ) a 11
Similarly, x2, x3…xn of the 2nd to the nth equations can be obtained.
1 x2 = (b 2 − a 21 x 1 − a 23 x 3 − K a 2n x n ) a 22 ↓
↓
↓
↓
1 ( bn − a n1 x 1 − a n2 x 2 − K − a n,n -1 x n -1 ) xn = a nn U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
220
Gauss Iterative Method In general, the jth equation may be written as
1 n xj = ( b j − ∑ i = 1 a ji x i ) a jj
equation “a”
i≠ j
j = 1, 2,K n
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
221
Gauss Iterative Method In general, the Gauss iterative estimates are:
x1 x2
xn
k +1
k +1
k +1
a 13 a 1n y1 a 12 k k k = − x2 − x 3 − ... − xn a 11 a 11 a 11 a 11 a 2n y2 a 21 a 23 k k k = − x1 − x 3 − ... − xn a 22 a 22 a 22 a 22
a n, n - 1 yn a n1 k a n2 k k = − x1 − x 2 − ... − x n -1 a nn a nn a nn a nn
where k is the iteration count U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
222
Gauss Iterative Method From an initial estimate of the unknowns (x10, x20,…xn0), updated values of the unknown variables are computed using equation “a”. This completes one iteration. The new estimates replace the original estimates. Mathematically, at the kth iteration,
k +1
xj
1 n = ( b j − ∑ i=1 a a jj i≠ j
k
ji
x )
equation “b”
i
j = 1, 2,K n U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
223
Gauss Iterative Method A convergence check is conducted after each iteration. The latest values are compared with their values respectively.
Δx = x k
k +1 j
− x
k j
equation “c”
j = 1, 2,K n The iteration process is terminated when
max | Δ x | < ε k j
k = itermax U. P. National Engineering Center National Electrification Administration
(convergen t) (non - convergent )
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
225
Gauss-Seidel Method Solution: a) The system of equation must be expressed in standard form.
x
k +1 1
x
k +1 2
x
k +1 3
1 k k ( 4 + x2 - x3 ) = 4 1 k +1 k = ( 6 - x1 - x3 ) 4 1 k +1 k +1 = ( 5 - x1 - x2 ) 3
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
224
Gauss-Seidel Method Example: Solve the system of equations using the Gauss-Seidel method. Used a convergence index of Îľ = 0.001
4x
1
â&#x2C6;&#x2019; x2 + x3 = 4
x 1 + 4x
+ x3
2
x 1 + x 2 + 3x
=6 3
= 5
x 1 = x 2 = x 3 = 0.5 0
U. P. National Engineering Center National Electrification Administration
0
0
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
226
Gauss-Seidel Method with x 1 0 = x 2 0 = x 3 0 = 0.5 Iteration 1 (k =0):
1 ( 4 + 0.5 - 0.5 ) = 1.0 4 1 1 ( 6 - 1.0 - 0.5 ) = 1.125 x2 = 4 1 x31 = ( 5 - 1.0 - 1.125 ) = 0.9583 3 Δ x 10 = 1 − 0 . 5 = 0 . 50 x11 =
Δ x 20 = 1 . 125 − 0 . 50 = 0 . 625 Δ x 30 = 0 . 9583 − 0 . 50 = 0 . 4583 max | Δ x 20 | = 0.625 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
227
Gauss-Seidel Method Iteration 2 (k = 1):
1 x1 2 = ( 4 + 1.125 - 0.9583 ) = 1.0417 4 1 2 x2 = ( 6 - 1.0417 - 0.9583 ) = 1.0 4 1 x3 2 = ( 5 - 1.0417 - 1.0 ) = 0.9861 3 Δx11 = 1.0417 − 1 = 0.0417
Δx21 = 1 − 1.125 = −0.125 Δx31 = 0.9861 − 0.9583 = 0.0323 max | Δx21 | = 0.125 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
228
Gauss-Seidel Method Iteration 3 (k = 2):
1 x 1 = ( 4 + 1.0 - 0.9861 ) = 1.0035 4 1 3 x 2 = ( 6 - 1.0035 - 0.9861 ) = 1.0026 4 1 x 3 3 = ( 5 - 1.0035 - 1.0026 ) = 0.9980 3 Δ x 12 = 1 . 0035 − 1 . 0417 = − 0 . 0382 3
Δ x 22 = 1 . 0026 − 1 = 0 . 0026 Δ x 32 = 0 . 9980 − 0 . 9861 = 0 . 0119 max | Δ x 32 | = 0.0119 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
229
Gauss-Seidel Method Iteration 4 (k = 3):
1 x1 4 = ( 4 + 1.0026 - 0.9980 ) = 1.0012 4 1 4 x2 = ( 6 - 1.0012 - 0.9980) = 1.0002 4 1 4 x3 = ( 5 - 1.0 - 1.0012 - 1.0002) = 0.9995 3 Δx13 = 1.0012 − 1.0035 = 0.0023
Δx23 = 1.0002 − 1.0026 = −0.0024 Δx33 = 0.9995 − 0.9980 = 0.0015 max | Δx23 | = 0.0024 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
230
Gauss-Seidel Method Iteration 5 (k = 4):
1 ( 4 + 1.0002 - 0.9995 ) = 1.0002 4 1 x 2 5 = ( 6 - 1.0002 - 0.9995) = 1.0001 4 1 5 x 3 = ( 5 - 1.0002 - 1.0001) = 0.9999 3 Δ x14 = 1 .0002 − 1 .0012 = − 0 .001 x1 5 =
Δ x 24 = 1 .0001 − 1 .0002 = − 0 .0001 Δ x 34 = 0 .9999 − 0 .9995 = 0 .0004 max | Δ x 4 | = 0.001 < ε U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
231
Gauss-Seidel Method The Gauss-Seidel Method has converged after 4 iterations only with the following solutions:
x 1 = 1.0002 x 2 = 1.0001 x 3 = 0.9999
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
232
Gauss-Seidel Method The Gauss-Seidel method is an improvement over the Gauss iterative method. As presented in the previous section, the standard form of the jth equation may be written as follows.
1 xj = (bj − a jj
∑
n i=1 i≠ j
a ji x i )
j = 1, 2, K n
From an initial estimates (x10, x20,…xn0), an updated value is computed for x1 using the above equation with j set to 1.This new value replaces x10 and is then used together with the remaining initial estimates to compute a new value for x2. The process is repeated until a new estimate is obtained for xn. This completes one iteration. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
233
Gauss-Seidel Method Note that within an iteration, the latest computed values are used in computing for the remaining unknowns. In general, at iteration k,
x
k +1 j
1 = (bj − a jj
∑
n i=1
a
i≠ j
α
ji
xi )
j = 1, 2, K n where α = k
if i > j
= k + 1 if i < j After each iteration, a convergence check is conducted. The convergence criterion applied is the same with Gauss Iterative Method. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
234
Gauss-Seidel Method An improvement to the Gauss Iterative Method k +1
x1
k +1
x2
y1 a 12 a 1n k k = − x 2 − ... − xn a 11 a 11 a 11 a 2n k y2 a 21 k + 1 = − x 1 − ... − xn a 22 a 22 a 22
ai,i-1 k+1 ai,i+1 k+1 ain k+1 yi aij k+1 xi = − xi −...− xi-1 − xi+1 − xn aii aii aii aii aii k+1
k +1
xn
a n, n - 1 yn a n1 k + 1 k +1 = − x 1 − ... − x n -1 a nn a nn a nn
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
235
Newton-Raphson Method Solve the non-linear equations
x12 â&#x2C6;&#x2019; 4x2 = 4 2x1 â&#x2C6;&#x2019; x2 = 2 The Newton-Raphson method is applied when the system of equations is non-linear. Consider a set of 2 non-linear equations in 2 unknowns.
y 1 = f 1 ( x1 , x 2 )
y 2 = f 2 ( x1 , x 2 ) U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
236
Newton-Raphson Method The system of non-linear equations can be linearized using the first order Taylor’s Series y1 = f 1 ( x 0 ) +
∂f 1 0 ∂f 0 0 ( x ) Δ x1 + 1 ( x 0 ) Δ x 2 ∂ x1 ∂x2
y2 = f 2 ( x 0 ) +
∂f 2 0 ∂f 0 0 ( x ) Δ x1 + 2 ( x 0 ) Δ x2 ∂x1 ∂x2
Where: x0 = (x10, x20) are set of initial estimates fi(x0) = the function fi (x1,x2) evaluated using the set of initial estimates. 0 ∂f i ( x ) = the partial derivatives of the function fi(x1,x2) ∂x j evaluated using the set of original estimates. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
237
Newton-Raphson Method The equation may be written in matrix form as follows:
⎡ y1 − f1 ( x )⎤ ⎡ = ⎢ ∂f ( x 0 ) ⎢ 0 ⎥ 2 ( ) y f x − ⎢ 2 ⎣ 2 ⎦ ⎣ ∂x1 0
∂f 1 ( x o ) ∂x 1
⎤ ⎡ Δx 0 ⎤ 1 ⎥ 0⎥ ∂f 2 ( x 0 ) ⎢ ⎥ Δx 2 ⎦ ∂x 2 ⎦ ⎣ ∂f 1 (x 0 ) ∂x 2
The matrix of partial derivatives is known as the Jacobian. The linearized system of equations may be solved for ∆x’s which are then used to update the initial estimates.
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
238
Newton-Raphson Method At the kth iteration: k +1
= x1 + Δx1
k +1
= x2 + Δx2
x1
x2
k
k
k
k
Convergence is achieved when
y1 − f1(xk ) ≤ ε y2 − f2 (xk ) ≤ ε Where ε is pre-set precision indices. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
239
Newton-Raphson Method Solve the non-linear equation
x12 − 4x2 = 4 2x1 − x2 = 2 use: x1 = 1, x2 = −1 0
Solution:
0
First, form the Jacobian
f1 = x − 4 x2 2 1
f 2 = 2 x1 − x 2
∂f1 = 2 x1 ∂ x1 ∂f2 = 2 ∂ x1
U. P. National Engineering Center National Electrification Administration
∂f1 = −4 ∂x2 ∂f 2 = −1 ∂x2
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
240
Newton-Raphson Method In Matrix form
⎡ ∂f 1 ( x ) 0 ⎢ − y f ( x ) ⎡ 1 ⎤ ∂x1 1 ⎢ y − f ( x )⎥ = ⎢ ∂f ( x 0 ) ⎢ 2 ⎣ 2 2 0 ⎦ ⎢⎣ ∂x1 0
⎡4 − ( x1 − 4 x2 )⎤ ⎡ 2 x1 ⎢ 2 − ( 2 x − x )⎥ = ⎢ 2 ⎣ ⎣ 1 2 ⎦ 2
U. P. National Engineering Center National Electrification Administration
∂f 1 ( x ) ⎤ ∂x2 ⎥ ⎡ Δx1 ⎤ 0 ⎥⎢ ∂f 2 ( x ) ⎥ ⎣Δx2 ⎥⎦ ∂x2 ⎥⎦ 0
- 4 ⎤ ⎡ Δx1 ⎤ ⎢ ⎥ ⎥ − 1⎦ ⎣Δx2 ⎦
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
241
Newton-Raphson Method Iteration 0:
f 1 ( x 0 ) = 12 − 4( −1 ) = 5 , y1 = 4 ∂f 1 = 2(1) = 2 ∂x1 ∂f 1 = −4 ∂x2
f 2 ( x 0 ) = 2( 1 ) − ( − 1 ) = 3 , y 2 = 2 ∂f 2 =2 ∂x1 ∂f 2 = −1 ∂x 2
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
242
Newton-Raphson Method The equations are:
In matrix form:
4 − 5 = ( 2 )Δx + ( −4 )Δx2 0 1
2 − 3 = ( 2 )Δx1 + ( −1 )Δx2 0
0
0
⎡ − 1⎤ ⎡ 2 ⎢ − 1⎥ = ⎢ 2 ⎣ ⎦ ⎣
− 4 ⎤ ⎡ Δ x10 ⎤ ⎢ ⎥ 0⎥ − 1 ⎦ ⎣Δx2 ⎦
Solving,
Δ x1
0
= − 0 .5
Δ x
0
= 0
2
Thus,
x1 = 1 + ( −0.5 ) = 0.5 1
x2 = −1 + 0 = −1 1
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
243
Newton-Raphson Method Repeating the process with the new estimates, Iteration 1:
f 1 ( x1 ) = ( 0.5 )2 − 4( −1 ) = 4.25 , y1 = 4 ∂f 1 ( x1 ) = 2( 0.5 ) = 1.0 ∂x1 ∂f 1 ( x ) = −4 ∂x2 1
f 2 ( x 1 ) = 2 ( 0 .5 ) − ( − 1 ) = 2
, y2 = 2
∂f 2 ( x 1 ) = 2 ∂x1 ∂f 2 ( x 1 ) = −1 ∂x2
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
244
Newton-Raphson Method The equations are:
In matrix form:
4 − 4.25 = Δx1 − 4 Δ x 2
1
2 − 2 = 2 Δ x1 − Δ x 2
1
1
1
⎡ − 0.25 ⎤ ⎡ 1 ⎢ 0 ⎥ = ⎢2 ⎣ ⎦ ⎣
− 4 ⎤ ⎡ Δ x1 1 ⎤ ⎢ ⎥ − 1 ⎥⎦ ⎣ Δx 2 1 ⎦
Solving,
Δ x 1 = 0 . 03571 1
Δ x 2 = 0 . 07143 1
Thus,
x 1 = 0 .5 + 0 .03571 = 0 .53571 2
x 2 = − 1 + 0 .07143 = − 0 .92857 2
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
245
Newton-Raphson Method Repeating the process with the new estimates, Iteration 2:
f 1 ( x 2 ) = ( 0.53571 )2 − 4( −0.92857 ) = 4.001265 , y1 = 4 ∂f 1 ( x 2 ) = 2( 0.53571 ) = 1.07142 ∂x1 ∂f 1 ( x 1 ) = −4 ∂x2
f 2 ( x 2 ) = 2( 0.53571 ) − ( −0.92857 ) = 1.99999 ≅ 2 ∂f 2 ( x 2 ) =2 ∂x 1
y2 = 2
∂f 2 ( x 2 ) = −1 ∂x 2 U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
246
Newton-Raphson Method The equations are:
4 − 4 . 001265 = 1 . 07142 Δ x 1 − 4 Δ x 2 2
2 − 2 .0 = 2 Δ x1 − Δ x 2 2
Solving,
2
2
In matrix form:
⎡ − 0 .001265 ⎤ ⎡1 .07142 =⎢ ⎢ ⎥ 0 2 ⎣ ⎦ ⎣
− 4 ⎤ ⎡ Δ x1 2 ⎤ ⎢ 2 ⎥ ⎥ − 1 ⎦ ⎣Δx2 ⎦
Δ x 1 = − 0 . 00018 2
Δ x 2 = 0 . 00036 2
Thus,
x 1 = 0 . 53571 − 0 . 00018 = 0 . 53553 3
x 2 = − 0 . 92857 − 0 . 00035 = − 0 . 92893 3
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
247
Newton-Raphson Method Substituting to the original equation:
f 1 ( x1 ) = ( 0.53553 )2 − 4( −0.92893 ) = 4.0025124 , y1 = 4 3 f 2 ( x1 ) = 2( 0.53553 ) − ( −0.92893 ) = 1.99928 , y 2 = 2 3
y 1 − f 1 = − 0 . 0025 y 2 − f 2 = 0 . 00072
Therefore,
x 1 = 0 .53553 x 2 = − 0 .92893 Note the rapid convergence of the Newton-Raphson Method. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
248
Newton-Raphson Method The Newton-Raphson method is applied when the system of equations is non-linear. Consider a set of n non-linear equations in n unknowns.
y 1 = f 1 ( x 1 , x 2 ,K , x n ) y 2 = f 2 ( x 1 , x 2 ,K , x n ) M y n = f n (x 1 , x 2 , K , x n ) U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
249
Newton-Raphson Method The system of non-linear equations can be linearized using Taylor’s Series
∂f1 0 ∂f1 0 ∂f1 0 0 0 0 y1 = f1 ( x ) + ( x )Δx1 + ( x )Δx2 + K + ( x )Δxn ∂x1 ∂x2 ∂xn 0
∂f 2 0 ∂f 2 0 ∂f 2 0 0 0 0 y2 = f 2 ( x ) + ( x )Δx1 + ( x )Δx2 + K+ ( x )Δxn ∂xn ∂x2 ∂x1 0
M
M
M
M
∂f n 0 ∂f1 0 ∂f1 0 0 0 0 yn = f n ( x ) + ( x )Δx1 + ( x )Δx2 + K+ ( x )Δxn ∂x1 ∂x2 ∂xn 0
U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
250
Newton-Raphson Method Where: X0 = (x10,x20, …, xn0) = set of initial estimates fi(x0) = the function fi (x1,x2, …, xn) evaluated using the set of initial estimates.
∂f i ( x 0 ) = the partial derivatives of the ∂x j function fi(x1,x2,…,xn) evaluated using the set of original estimates. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
251
Newton-Raphson Method The equation may be written in matrix form as follows:
⎡ y1 − f 1 ( x ) ⎤ ⎡ ⎢ ⎥ ⎢ 0 ∂f 2 ( x 0 ) ⎢ y 2 − f 2 ( x )⎥ = ⎢ ∂x1 ⎢ ⎥ ⎢ M M ⎢ ⎥ ⎢ 0 ∂f n ( x 0 ) ⎢⎣ y n − f n ( x )⎥⎦ ⎢⎣ ∂x1 0
∂f 1 ( x o ) ∂x1
∂f 1 (x 0 ) ∂x 2 ∂f 2 ( x 0 ) ∂x 2
M ∂f n ( x 0 ) ∂x 2
0 ⎤ ⎡ Δx1 ⎤ K 0 ⎥⎢ 0⎥ ∂f 2 ( x ) Δx 2 ⎥ K ∂x n ⎥ ⎢ ⎥⎢ M ⎥ M ⎥ ⎢ 0⎥ ∂f n ( x 0 ) ⎥ K ⎣Δxn ⎥⎦ ∂x n ⎦ ⎢ ∂f 1 (x 0 ) ∂x n
The matrix of partial derivatives is known as the Jacobian. The linearized system of equations may be solved for ∆x’s which are then used to update the initial estimates. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
252
Newton-Raphson Method At the kth iteration:
x j = x j + Δx j k =1
k
k
j = 1, 2,...,n
Convergence is achieved when
y j − f j ( x ) ≤ ε1 or
k
j = 1,2,...,n
Δx j ≤ ε 2
j = 1, 2, ..., n
k
Where ε1 and ε2 are pre-set precision indices. U. P. National Engineering Center National Electrification Administration
Competency Training & Certification Program in Electric Power Distribution System Engineering
Training Course in Fundamental Principles and Methods in Power System Analysis
U. P. National Engineering Center National Electrification Administration
253
Competency Training & Certification Program in Electric Power Distribution System Engineering