ECE 421/599
Electric Energy Systems
6 – Power Flow Analysis
Instructor:
Kai Sun
Fall 2014
1
Introduction
•Power flow (load flow) analysis
– Steady-state analysis of an interconnected power system
– To solve the power flow equations for Vi (i.e. |Vi | and δi) and Si (i.e.
Pi and Qi)
– Basis of power system analysis and design
•Assumptions for a power system to be studied
– Balanced conditions
– Represented by a single-phase network
– Modeled by nodes (buses) and branches
– All impedances/admittances are in per unit on a common MVA base
(typically 100MVA)
2
Bus Admittance Matrix
• Branch admittance
E1
1
1
=
zij rij + jxij
y=
ij
• Apply KCL to nodes (buses)1-4
E2
z10=
I01
z20=
z12=
V1
I1 = y10V1 + y12 (V1 − V2 ) + y13 (V1 − V3 )
I 2= y20V2 + y12 (V2 − V1 ) + y23 (V2 − V3 )
z23 =
z13=
=
0 y23 (V3 − V2 ) + y13 (V3 − V1 ) + y34 (V3 − V4 )
=
0 y34 (V4 − V3 )
z34=
Ii : current injected by an equivalent current source
I1 = ( y10 + y12 + y13 )V1 − y12V2 − y13V3
∆
I 01 = ( E1 − V1 ) y10 = E1 y10 + (0 − V1 ) y10 =I1 + (0 − V1 ) y10
− y12V1 + ( y20 + y12 + y23 )V2 − y23V3
I2 =
− y13V1 − y23V2 + ( y13 + y23 + y34 )V3 − y34V4
0=
− y34V3 + y34V4
0
n
Ii
Vi
n
∑ y − ∑ yV
0, j ≠ i
j=
ij
1, j ≠ i
j=
ij
j
Y11 = y10 + y12 + y13
Y12 = Y21 = − y12
Y=
Y=
0
14
41
Y22 = y20 + y12 + y23
Y13 = Y31 = − y13
Y33 = y13 + y23 + y34
Y23 = Y32 = − y23
Y=
Y=
0
42
24
Y44 = y34
Y34 = Y43 = − y34
3
I1 = Y11V1 + Y12V2 + Y13V3 + Y14V4
I 2 = Y21V1 + Y22V2 + Y23V3 + Y24V4
I 3 = Y31V1 + Y32V2 + Y33V3 + Y34V4
I 4 = Y41V1 + Y42V2 + Y43V3 + Y44V4
n
n
0, j ≠ i
j=
1, j ≠ i
j=
 I1  Y11
 I  Y
 2   21
:  :
 =
 I i   Yi1
:   :
  
Yn1
In 
 

Y12 ...
Y22 ...
Y1i ...
Y2 i ...
:
:
Yi 2 ...
Yii ...
:
:
Yn 2 ...
Yin ...
Y1n 
Y2 n 

: 

Yin 
: 

Ynn 

V1 
V 
 2
: 
 
Vi 
: 
 
Vn 


n
I i = Vi ∑ yij − ∑ yijV j = ∑ YijV j
1
j=
I bus = Ybus Vbus
• Ibus: currents injected by external current sources
• Vbus : bus voltages relative to a reference (not included), usually the ground
• Ybus : bus admittance matrix (symmetric and sparse)
– Diagonal elements: called self-admittances or driving point admittances
Yii =
n
∑ y
=j 0, j ≠ i
ij
– Off-diagonal elements: called mutual admittances or transfer admittances
Yij = Y ji = − yij
j≠i
• Zbus=Y-1bus : bus impedance matrix
-1
Vbus = Ybus
I bus = Zbus I bus
4
0 
j 5.00
 − j8.50 j 2.50
 j 2.50 − j8.75

5.00
0
j

Ybus = 
 j 5.00
j 5.00 − j 22.50 j12.50 


0
0
12.50
12.50
j
j
−


How many non-zero elements for a system with N buses
(not including the ground) and M branches? 2M+N
 j 0.50
 j 0.40
−1

Z=
Y=
bus
bus
 j 0.45

 j 0.45
j 0.40
j 0.48
j 0.45
j 0.44
j 0.44
j 0.44
j 0.545
j 0.545
E1
E2
z10=
z20=
z12=
z23=
z13=
j 0.45 
j 0.44 
j 0.545

j 0.625
z34=
• If it is known that E1=1.1∠0o pu and
E2=1.0∠0o pu
I1 =
E1 1.1
=
= − j1.1 pu
z10 j1.0
I2 =
E2 1.0
=
= − j1.25 pu
z20 j 0.8
V1 
 − j1.1  1.050 
V 
 − j1.25 1.040 
2
 =Z 
=

bus
V3 
 0  1.045
 

 

 0  1.045 Why are they
V4 
same?
5
A more general power flow study
z10=
z12=
|V1|=1
z20=
P2=0.5pu
|V2|=1.05pu
z13=
z23=j0.2
z34=
P4+jQ4=1+j0.2 pu
•How to solve all bus voltages and line real and
reactive power flows?
6
Power Flow Equation
• Consider a typical bus of an n-bus system
– All lines represented by equivalent π models
– Admittances are in pu on a common MVA base
• Apply KCL
I=i yi 0Vi + yi1 (Vi − V1 ) + yi 2 (Vi − V2 ) + ... + yij (Vi − V j ) + ... + yin (Vi − Vn )
= ( yi 0 + yi1 + yi 2 + ... yin )Vi − yi1V1 − yi 2V2 − ... − yijV j − ... − yinVn
n
n
Ii =
Vi ∑ yij − ∑ yijV j
j≠i
=j 0=j 1
Si =Pi + jQi =V I
*
i i
Ii =
Pi − jQi
Vi *
j≠i
n
n
Pi − jQi
=
Vi ∑ yij − ∑ yijV j
Vi * =j 0=j 1
j≠i
n
n
Pi − jQi
| Vi | ∠δ i ∑ yij − ∑ yij | V j | ∠δ j
=
| Vi | ∠ − δ i
=j 0=j 1
j≠i
• Solve |Vi|, δi, Pi , Qi and then calculate Pij , Qij
− n complex nonlinear algebraic equations (2×n real equations) with 4×n real quantities
− can be solved by iterative techniques
7
Power Flow Solution
• Determining:
– |Vi| and δ i (magnitude and phase angle of each bus voltage)
– Pij and Qij (real and reactive power flows in each line)
• The system is assumed to be operating under balanced conditions and a
single-phase model is used
• 4 quantities, i.e. |Vi|, δi , Pi and Qi, are associated with each bus
• System buses are usually classified into three types
Slack bus
•
(swing bus or V-δ bus) •
Selected as the reference having |Vi| and δi fixed
Pi and Qi are usually unlimited and can take any values to
make up the gap between system generation and load
Load buses
(P-Q buses)
• Pi and Qi are specified
Regulated buses
(generator buses or
P-V buses)
• Pi and Vi are specified.
• Limits of Qi are also specified
8
More Thinking on Types of Buses
• Need to know two of |Vi|, δ i, Pi and Qi (either constant or observed)
• Relax the other two within upper and lower limits
• May assume more types of buses for a variety of natures of buses
V-δ
|Vi|
δi
X
X
P-Q
P-V
X
Q-V
X
Pi
Qi
X
X
X
X
P-δ
X
Q-δ
X
X
X
9
A more general power flow study
z10=
|V1|=1
z12=
z20=
P2=0.5pu
|V2|=1.05pu
δ 1=0
(slack bus)
z13=
z23=j0.2
z34=
P4+jQ4=1+j0.2 pu
10
Solution of Nonlinear Algebraic Equations
•Gauss-Seidel Method
•Newton-Raphson Method
11
Gauss-Seidel Method: Example 6.2
f ( x) = x3 − 6 x2 + 9 x − 4 = 0
3 roots: x1,2=1 and x3=4
1
6
4
x=
− x3 + x2 + =
g ( x)
9
9
9
x (0) = 2
x
(1)
1 3 6 2 4
(0)
g
(
x
)
=
−
(2) + (2) + =
2.2222
=
9
9
9
|x(1)-x(0)|=0.2222
x(2)= g(x(1))
= 2.5173
|x(2)-x(1)| =0.2951
x(3)= g(x(2))
= 2.8966
|x(3)-x(2)| = 0.3793
x(4)= g(x(3))
= 3.3376
|x(4)-x(3)| = 0.4410
x(5)= g(x(4))
= 3.7398
|x(5)-x(4)| = 0.4022
x(6)= g(x(5))
= 3.9568
|x(6)-x(5)| = 0.2170
x(7)= g(x(6))
= 3.9988
|x(7)-x(6)| = 0.0420
x(8)= g(x(7))
= 4.0000
|x(8)-x(7)| = 0.0012
x(9)= g(x(8))
= 4.0000
|x(9)-x(8)| = <0.0001
12
Gauss-Seidel Method
• To solve nonlinear equation f(x)=0
• Re-write x=g(x)
• Start iteration from an initial estimate x(0)
x(1)=g(x(0))
x(2)=g(x(1))
…
x(k+1)=g(x(k))
• Stop when |x(k+1)-x(k)|≤ε. Solution: x=x(k+1)
13
x=g(x)
1
6
4
y=
g ( x) =
− x3 + x 2 +
9
9
9
(1)
g(x )
g(x(0))
x(0) x(1)
y=x
x(5)
•“Zigzag” graphical illustration
– Can the iteration be faster?
– How to find all roots?
– Does the iteration always converge to a root?
14
Faster iteration?
x ( k +1) =
g ( x(k ) ) =
x ( k ) + [ g ( x ( k ) ) − x ( k ) ] Adjustment on x
• Using an acceleration factor when updating x(k)
x ( k +1) =
x(k ) + α [ g ( x(k ) ) − x(k ) ]
α=1.25
1
6
4
y=
g ( x) =
− x3 + x 2 +
9
9
9
g(x(1))
y=x
g(x(0))
x(0)
x(1)
x(4)
15
All roots? Always convergent?
10
2
g(x) =-1/9x3+6/9x +4/9
8
x
6
4
2
convergent
0
convergent
-2
-4
-6
divergent
-8
-10
-10
Initial value x(0) is important!
-8
-6
-4
-2
0
x
2
4
6
8
10
16
f ( x) = x3 − 6 x 2 + 9 x − 4 = 0
1
6
4
x=
− x3 + x 2 + =
g ( x)
9
9
9
x =− x3 + 6 x 2 − 8 x + 4 =h( x)
10
3
2
h(x) =-x +6x -8x+4
8
h(x) is harder
to converge
6
2
g(x) =-1/9x3+6/9x +4/9
4
2
0
-2
-4
-6
-8
-10
-10
-5
0
x
5
10
17
A system of n equations in n variables
•Using an acceleration factor when updating xi(k)
xi( k +1) =
xi( k ) + α ( xi( kcal+1) − xi( k ) )
xi( kcal+1) =
gi ( x1( k ) , , xn( k ) )
18
G-S Power Flow Solution
n
n
Pi − jQi
=
Vi ∑ yij − ∑ yijV j
*
Vi
=j 0
=j 1, j ≠ i
Yij =
− yij j ≠ i
n
Yii = ∑ yij
j =0
n
= ViYii + ∑ YijV j
=j 1, j ≠ i
n
= ∑ YijV j
j =1
n
Pi ( k ) − jQi( k )
(k )
−
Y
V
∑
ij j
*( k )
V
=j 1, j ≠ i
i
Vi ( k +1) =
Yii
n
n
Pi − jQi =
Vi ∑ YijV j
Qi( k +1) = − Im[Vi*( k ) ∑ YijV j( k ) ]
j =1
*
j =1
Pi
( k +1)
= Re[Vi
n
*( k )
(k )
Y
V
∑ ij j ]
j =1
19
PQ Buses
• |Vi| and δi are unknown
• Pi and Qi are scheduled (generation or load), denoted by Pisch and
Qisch
x(k+1)=g(x(k))
n
Pi sch − jQisch
(k )
Y
V
−
∑
ij j
*( k )
V
=j 1, j ≠ i
i
Vi ( k +1) =
Yii
•Under normal operating conditions:
– Slack bus: |V0|∠δ0 (typically 1∠0o)
– Other buses: |Vi| is close to 1pu or |V0|. For most of cases, there are:
• Generator buses: |Vi|>|V0|, δi > δ0,
• Load buses: |Vi|<|V0|, δi < δ0
•Initial guess could be Vi(0) =1∠0o without a better estimation
20
PV Buses
• Pi=Pisch and |Vi| are specified
• Starting from an initial estimate of δi(0) → Vi(0)=|Vi|∠ δ i(0)
( k +1)
i
Q
= − Im[Vi
n
*( k )
∑Y V
j =1
ij
(k )
j
]
n
Pi sch − jQi( k +1)
(k )
−
Y
V
∑ ij j
Vi *( k )
=j 1, j ≠ i
( k +1)
=
Vci
Yii
• Since |Vi| is specified, only VI,i(k+1)=Im[Vci(k+1)] is retained
( k +1)
=
V
R ,i
| Vi |2 −(VI(,ki +1) ) 2
• Continue the iterations until
Update Vi(k+1)=VR,i(k+1)+j⋅ VI,i(k+1)
| VR(,ki +1) − VR(,ki ) |≤ ε
| VI(,ki +1) − VI(,ki ) |≤ ε
or, the power mismatch, i.e. the largest element in ∆P and ∆Q <ε
• Using acceleration factor α=1.3~1.7
n
Pi sch − jQi( k +1)
− ∑ YijV j( k )
*( k )
Vi
=j 1, j ≠ i
( k +1)
(k )
Vic =
Vi + α (
− Vi ( k ) )
Yii
21
Slack Bus
n
Pi − jQi =
Vi * ∑ YijV j
j =1
Line Flows and Losses
yij
•At bus i:
I ij =I l + I i 0 = yij (Vi − V j ) + yi 0Vi
Sij = Vi I ij∗
•At bus j:
I ji =− I l + I j 0 =yij (V j − Vi ) + y j 0V j
S ji = V j I ∗ji
•Power loss in line i – j:
S Lij= Sij + S ji
22
Tap Changing Transformers
• a is the per unit off-nominal tap position (usually, |a| = 0.9~1.1)
– Complex number for phase shifting transformers
ST=VxIi*= -Vj Ij*
1
Vx = V j
a
yt
Vj
a
y
y
=
− *t Vi + t 2 V j
a
|a |
=
I i yt (Vi − Vx=
) ytVi −
1
I j = − * Ii
a

y
 Ii   t
I  =  y
 j  − t
 a *
Ii =
− a* ⋅ I j
Equivalent circuit if a is real
(ignoring phase shifting)
Non-tap side
Tap side
yt 
a  Vi 

yt  V j 
| a |2 
−
Ybus is not symmetrical with a phase shifting
transformer
23
Example 6.7
(V-δ and P-Q buses)
Using the G-S method to find the power flow solution:
(a) Determine the voltage phasors at P-Q buses 2 and 3
accurate to 4 decimal places
P1, Q1
y23=10-j20
y13=10-j30
y23=16-j32
|V3|, δ3
|V2|, δ2
(b) Find the slack bus real and reactive power
(c) Determine the line flows and losses. Show line flow
directions in a power-flow diagram
(Solve P1, Q1, |V2|, δ2, |V3|, δ3, Sij and Slij)
Step 1. Check what are given
n
Pi sch − jQisch
−
YijV j( k )
∑
*( k )
Vi
=j 1, j ≠ i
Vi ( k +1) =
Yii
Step 2. Set initial estimates and iterate
24
Step 3. Calculate P and Q of the slack bus
n
Pi − jQi =
Vi ∑ YijV j
*
j =1
25
Step 4. Calculate line flows and losses
26
Example 6.8
(V-δ, P-Q and P-V buses)
P1, Q1
y23=10-j20
y13=10-j30
Line charging susceptances are neglected.
Obtain the power flow solution by the G-S
method including line flows and line losses
(Solve P1, Q1, |V2|, δ2, Q3, δ 3, Sij and Slij)
y23=16-j32
|V2|, δ2
Q3, δ 3
Step 1. Check what are given
Step 2. Set initial estimates and iterate
Note: |Vc3(1)|=1.0378 ≠ 1.04=|V3|
VR(1),3
27
Bus 2 (P-Q): Solve |V2|, δ2
n
Pi sch − jQisch
YijV j( k )
−
∑
*( k )
Vi
=j 1, j ≠ i
Vi ( k +1) =
Yii
Bus 3 (P-V): Solve Q3, δ3
( k +1)
i
Q
= − Im[Vi
n
Pi sch − jQi( k +1)
−
YijV j( k )
∑
*( k )
YijV ]
∑
Vi
=j 1, j ≠ i
j =1
Vc(i k +1) =
Yii
n
*( k )
(k )
j
( k +1)
Re[V=
]
3
1.042 − {Im[Vc(3k +1) ]}2
1.03954 − j 0.00833
=
Vc(3)
3
1.03978 − j 0.00873
=
Vc(4)
3
1.03989 − j 0.00893
Vc(5)
=
3
1.03993 − j 0.00900
=
Vc(6)
3
1.03995 − j 0.00903
=
Vc(7)
3
28
Newton-Raphson Method
•Based on Taylor’s series expansion at an initial estimate
of the solution
f ( x) = c
f ( x (0) + ∆x (0) ) =c
df (0) (0) 1 d 2 f (0)
f ( x ) + ( ) ∆x + ( 2 ) ( ∆x (0) ) 2 +  =
c
dx
2! dx
(0)
•Ignore all terms with orders ≥2
Comparison: G-S method ignores
all differential terms (orders ≥1)
∆
df (0) (0)
(0)
( ) ∆x  c − f ( x )＝ ∆c (0)
dx
∆x
(0)
∆c (0)
=
df
( )(0)
dx
(1)
x=
x (0) + ∆x (0)
29
•Iteration 1: (0)→(1)
(0)
(0)
c
c
f
(
x
)
∆
−
x (1) ＝ x (0)＋∆x (0) = x (0) +
＝x (0) +
df
df
( )(0)
( )(0)
dx
dx
∆
•Iteration k+1: (k) →(k+1)
(k )
(k )
∆
−
c
c
f
x
(
)
＝x ( k ) +
x ( k +1) ＝ x ( k )＋∆x ( k ) = x ( k ) +
df ( k )
df ( k )
( )
( )
dx
dx
∆
•Until
∆x
(k )
∆c ( k )
=
df ( k )
( )
dx
| x ( k +1) − x ( k ) |≤ ε
• c=f(x) is actually approximated by the tangent line on the
curve at x(k).
=
x x
(k )
c − f ( x(k ) )
+
df ( k )
( )
dx
It is straight line function
c=kx+b
30
Example 6.4
x
( k +1)
＝x
Let x(0)=6
(k )
df ( x )
= 3x 2 − 12 x + 9
dx
c − f ( x(k ) )
+
df
( )( k )
dx
(
df (0)
) = 3(6) 2 − 12(6) +=
9 45
dx
∆c(0) =−
c f ( x (0) ) =−
0 [(6)3 − 6(6)2 + 9(6) − 4] =
−50
∆x
(0)
∆c (0)
−50
=
= =
−1.1111
df (0)
45
( )
dx
x (1) = x (0) + ∆x (0) = 6 − 1.1111= 4.8889
x (2)
= x (1) + ∆x (1)
= 4.8889 −
13.4431
= 4.2789
22.037
(3)
x=
x (2) + ∆x (2)
= 4.2789 −
2.9981
= 4.0405
12.5797
x (4)
= x (3) + ∆x (3)
= 4.0405 −
0.3748
= 4.0011
9.4914
∆c (0)
∆x (0)
x(2)
x(1)
x(0)
(5)
x=
x (4) + ∆x (4)
= 4.0011 −
0.0095
= 4.0000
9.0126
31
N-dimensional System
f ( x) = c
−1
 df

x ( k +1) = x ( k )＋∆x ( k ) = x ( k ) + ( )( k )  ∆c ( k )
 dx 
∆c ( k ) =−
c f ( x(k ) )
f1 ( x1 , x2 , , xn ) = c1
f 2 ( x1 , x2 , , xn ) = c2

f n ( x1 , x2 , , xn ) = cn
 ∂f1 ( k )
( ∂x )
 1
 ∂f 2 ( k )
( )
J ( k ) =  ∂x1

 
 ∂f
( n )( k )
 ∂x1
Jacobian Matrix:
( k +1)
X = X
(k )
( k +1)
+ ∆X = X
∂f1 ( k )
∂f

)
 ( 1 )( k ) 
∂x2
∂xn

∂f
∂f

( 2 )( k )  ( 2 )( k ) 
∂x2
∂xn



 
∂f n ( k )
∂f n ( k ) 
 (
(
)
) 
∂x2
∂xn

(
(k )
−1
+  J ( k )  ∆C ( k )
 ∆x1( k ) 
 (k ) 
 ∆x2 
∆X ( k ) =


 (k ) 
 ∆xn 
 c1 − ( f1 )( k ) 


c2 − ( f 2 )( k ) 
(k )

∆C =




 cn − ( f n )( k ) 
32
Example 6.5
• Use the N-R method to find the intersections of the curves
x12 + x22 =
4
e x1 + x2 =
1
 2 x1 2 x2 
J =  x1

e
1


• J tells the fastest direction (gradient)
toward a solution
If x1(0)=2, x2(0)= -2:
k
∆C
J
∆x
x
1 -4.0000 4.0000 -4.0000 -0.6424 1.3576
-4.3891 7.3891 1.0000 0.3576 -1.6424
2 -0.5406 2.7152 -3.2848 -0.2989 1.0587
-1.2445 3.8869 1.0000 -0.0825 -1.7249
3 -0.0962 2.1173 -3.4499 -0.0530 1.0056
-0.1576 2.8825 1.0000 -0.0047 -1.7296
4 -0.0028 2.0112 -3.4592 -0.0014 1.0042
-0.0040 2.7336 1.0000 -0.0000 -1.7296
5 -0.0000 2.0083 -3.4593 -0.0000 1.0042
-0.0000 2.7296 1.0000 -0.0000 -1.7296
33
Compared to the Gauss-Seidel Method
•Since higher-order terms are ignored, the N-R method also
needs the initial estimation to be sufficiently close to the
actual solution
•The N-R method converges much faster
– N-R method: quadratic convergence (ignoring the 2nd and
higher orders)
– G-S method: linear convergence (ignoring the 1st and
higher orders)
•The N-R method has some computational issues:
– Needs [J(k)]-1 during each iteration, which is
computationally intense
34
Dealing with [J(k)]-1
∆X
=  J
( k +1)
( k ) −1
 ∆C ( k )
•Try not to update J(k) so often (at least not in every iteration)
•Apply LU decomposition (triangular factorization)
J ( k ) ∆X ( k +1) =
∆C ( k )
L( k )U ( k ) ∆X ( k +1) =
∆C ( k )
In MATLAB, the solution of
J∆X= ∆C can be obtained by
∆X= J \ ∆C
35
Newton-Raphson Power Flow Solution
Pi+jQi
G-S method
n
Pi ( k ) − jQi( k )
(k )
−
Y
V
∑
ij j
*( k )
V
1
,
i
j
j
=
≠
i
Vi ( k +1) =
Yii
X=G(X)
Pi
n
Pi − jQi
=
Ii
= VY
i ii + ∑ YijV j
*
Vi
=j 1, j ≠i
N-R method
n
F(X)=C
= ∑ YijV j
( k +1)
Use polar forms:
| Vi | ∠δ i Yij =
| Yij | ∠θij
Vi =
n
Pi − jQi =
Vi ∑ YijV j
*
j =1
− jQ
n
=
Vi ∑ YijV j( k )
*( k )
j =1
n
Vi * ∑ YijV j= Pi − jQi
j =1
J×∆X=∆C
j =1
( k +1)
i
 J1
J
 3
J 2  Δδ  ΔP 
= 



J 4  Δ | V | ΔQ 
n
= | Vi | ∠ − δ i ∑ | Yij || V j | ∠θij + δ j
j =1
36
Pi = fi (Δ | V |, Δδ)
eq.(1)
n
∑ | V || V | | Y | cos(θ − δ + δ )
j =1
j
i
ij
ij
i
Qi = gi (Δ | V |, Δδ)
j
eq.(2)
n
=
−∑ | V j || Vi | | Yij | sin(θij − δ i + δ j )
j =1
ΔP   J 1
ΔQ  =  J
   3
J 2  Δδ 
J 4  Δ | V |
Algorithm:
• For the slack bus (say bus 1) :
– No need to include bus 1 in J
– calculate P1 and Q1 by (1) and (2) at the end
• For m voltage-controlled (P-V) buses:
– solve δi by (1)
– calculate Qi by (2)
• For the n-1-m load (P-Q) buses left:
– solve |Vi| and δi by (1) and (2)
There are 2n-2-m independent (1)’s and (2)’s
P-V bus
 ∂P2 ( k )

 ∂δ 2
 ∆P2( k )   

 
(k )
∂
P


n


(
k
)
 ∆Pn   ∂δ 2
 (k )  = 
(k )
 ∆Q2   ∂Q2
    ∂δ
 (k )   2
 ∆Qn   

 ∂Qn ( k )

 ∂δ 2
∂P2

∂δ n

(k )

∂P2
∂ | V2 |
(k )

∂Pn

∂δ n
(k )
∂Pn
∂ | V2 |
(k )
∂Q2

∂δ n
(k )
∂Q2
∂ | V2 |
(k )


∂Qn

∂δ n

(k )
∂Qn
∂ | V2 |
(k )
(k )

∂P2


∂ | Vn | 
  ∆δ ( k ) 


2


(k ) 
∂Pn
    J
J 2,( n −1)×( n −1− m )  Δδ 

1,( n −1)×( n −1)

(
k
)
∂ | Vn |  ∆δ n  = 
 


J
J
Δ
|
V
|
3
n
m
n
4
n
m
n
m
,(
−
1
−
)
×
(
−
1)
,(
−
1
−
)
×
(
−
1
−
)





(
k
)
(k )
∂Q2   ∆ | V2 | 


∂ | Vn |    


J: (2n-2-m)×(2n-2-m)
  ∆ | Vn( k ) | 



(k )
∂Qn 


37
∂ | Vn | 
ΔP   J 1
ΔQ  =  J
   3
Jacobian Matrix
J 2  Δδ 
J 4  Δ | V |
• Diagonal and off-diagonal elements of J1~J2 (i ≠ slack bus)
J1
∂Pi
∂δ i
i ≠ slack bus
(n-1)×(n-1)
∑ | V || V || Y | sin(θ − δ + δ )
j ≠i
i
j
ij
ij
i
j
∂Pi
=
− | Vi || V j || Yij | sin(θij − δ i + δ j ) j ≠ i
∂δ j
J3
∂Qi
∂δ i
i ∉ PV and slack buses
(n-1-m)×(n-1)
∑ | V || V || Y | cos(θ − δ + δ )
j ≠i
i
j
ij
ij
i
j
∂Qi
=
− | Vi || V j || Yij | cos(θij − δ i + δ j ) j ≠ i
∂δ j
J2
(n-1)×(n-1-m)
i ∉ PV and slack buses
∂Pi
= 2 | Vi || Yii | cos θii
∂ | Vi |
+ ∑ | V j || Yij | cos(θij − δ i + δ j )
j ≠i
∂Pi
| Vi || Yij | cos(θij − δ i + δ j )
=
∂ |V j |
J4
(n-1-m)×(n-1-m)
j≠i
i ∉ PV and slack buses
∂Qi
= −2 | Vi || Yii | sin θii
∂ | Vi |
− ∑ | V j || Yij | sin(θij − δ i + δ j )
j ≠i
∂Qi
=
− | Vi || Yij | sin(θij − δ i + δ j )
∂ |V j |
j≠i
38
Procedure for Power Flow Solution by N-R Method
1.
Initial values
Load buses: Pisch and Qisch specified, |Vi(0)|∠δi(0)=1∠0 or equal to the slack bus
Voltage-regulated buses: |Vi| and Pisch specified, δi(0)=0 or the slack bus angle
–
–
Calculate Pi(k), ∆Pi(k), Qi(k) and ∆Qi(k)
2.
Load buses: calculate Pi(k) and Qi(k) by eq. (1) and (2) and then
–
∆Pi ( k ) = Pi sch − Pi ( k )
Voltage-controlled buses: calculate Pi(k) by eq. (1) and ∆Pi
–
3.
∆Qi( k ) = Qisch − Qi( k )
(k )
= Pi sch − Pi ( k )
Calculate J1, J2, J3 and J4 and solve ∆|Vi(k)| and ∆δi(k) from
ΔP   J 1
ΔQ  =  J
   3
( k +1)
J 2  Δδ 
J 4  Δ | V |
4.
Calculate | Vi=| | Vi
5.
Iterate Steps 2~5 until
(k )
| +∆ | Vi ( k ) |
| ∆Pi ( k ) |≤ ε
(applying triangular factorization and
Gaussian elimination)
+1)
δ i( k=
δ i( k ) + ∆δ i( k )
| ∆Qi( k ) |≤ ε
39
How to accelerate the N-R method?
Original N-R method
– J(k) is computed at each iteration
– Computing [J(k)]-1 is expensive
x(2) x(1)
x(0)
Ideas: approximate J
– Constant matrix
– Block-diagonal matrix (ignoring
small off-diagonal elements)
40
Fast Decoupled Power Flow Solution
ΔP   J 1
ΔQ  =  J
   3
J 2  Δδ 
J 4  Δ | V |
• Some elements of J may be close to 0
• Transmission lines usually have a high X/R ratio (close to lossless lines),
Pij ≈
Vi V j
X ij
sin(δ i − δ j )
Qij ≈
Vi
[ Vi − V j cos(δ i − δ j )]
X ij
– ∆P is less sensitive to ∆|V| than it is to ∆δ → J2 =∂P/∂|V| ≈0
– ∆Q is less sensitive to ∆δ than it is to ∆|V| → J3 = ∂Q/∂δ ≈0
• F-D method:
ΔP   J 1
ΔQ  ≈  0
  
0  Δδ 
J 4  Δ | V |
=
ΔP J=
[
1Δδ
∂P
]Δδ
∂δ
 ∂Q 
ΔQ = J 4Δ | V |= 
Δ|V|

 ∂ | V |
41
 ΔP   J 1
 ΔQ  =  0
  
Jacobian Matrix
i ≠ slack bus
(n-1)×(n-1)
J1
∂Pi
∂δ i
∑ | V || V || Y | sin(θ − δ + δ )
j ≠i
i
j
ij
ij
i
n
2
j =1
i
j
ij
ij
i
j
i
ii
=−Qi − | Vi |2 Bii
≈ − | Vi | Bii
∂Pi
∂δ j
(Qi << Bii , | Vi | ≈| Vi |≈ 1)
− | Vi || V j || Yij | sin(θij − δ i + δ j )
≈ − | Vi || V j || Yij | sin θij
≈ − | Vi | Bij
2
j≠i
Vi =
| Vi | ∠δ i Yij =
| Yij | ∠θij
J4 (n-1-m)×(n-1-m)
i ∉ PV and slack buses
∂Qi
= −2 | Vi || Yii | sin θii − ∑ | V j || Yij | sin(θij − δ i + δ j )
∂ | Vi |
j ≠i
j
∑ | V || V | | Y | sin(θ − δ + δ )− | V | | Y | sin θ
0   Δδ 
J 4   Δ | V |
ii
≈ −2 | Vi | Bii + | Vi | Bii
= − | Vi | Bii
∂Qi
=
− | Vi || Yij | sin(θij − δ i + δ j )
∂ | Vj |
j≠i
≈ − | Vi | Bij
ΔQ
''
B’ ~(n-1)×(n-1) about
ΔP
'
=
−
B
Δ|V|
= −B Δδ P-Q & P-V buses
|V|
|V|
'' -1 ΔQ
' -1 ΔP
Dot
division
(
./)
Δ
|
V
|
=
−
[B
]
Δδ = −[B ]
|V|
|V|
B”~ (n-1-m)×(n-1-m)
about P-Q buses
42
Qi << Bii
Proof:
Vi
[ Vi − V j cos(δ i − δ j )]
Qij ≈
X ij
Pi+jQi
Qi ≈ Vi ∑ yij [ Vi − V j cos(δ i − δ j )]
j
≈ ∑ yij ⋅ [1 − cos(δ i − δ j )]
j
<< ∑ yij ⋅1 ≈Bii
j
In Example 6.7,
Qi ≈ 0.02 Bii
43
Compared to the N-R Method
 B'−1
 Δδ 
 Δ | V | = −  0



 ΔP 
0  | V |


−1 
B"   ΔQ 
| V |


B’ ~(n-1)×(n-1) about P-Q & P-V buses
B”~ (n-1-m)×(n-1-m) about P-Q buses
•F-D method deals with constant Jacobian matrix: no need to
update in every iteration
•It requires more iterations than the N-R method, but each
iteration requires considerably less time
•Overall, the F-D method is much faster
•The F-D method is very useful in fast contingency screening
44
Examples 6.10 & 6.12
P1, Q1
y23=10-j20
• Obtain the powerflow solution (V2, δ2, δ 3,
P1, Q1 and Q3 ) by the N-R method and the
F-D method
y23=16-j32
|V2|, δ2
Q3, δ 3
(400 + j 250)
S2sch =
−
=
−4.0 − j 2.5 pu
100
sch
P=
3
y13=10-j30
200
= 2.0 pu
100
Slack
P-Q
P-V
31.62278∠1.8925 
 20 − j50 −10 + j 20 −10 + j 30 53.85165∠ − 1.9029 22.36068∠2.0344


=
+ j 32   22.36068∠2.0344 58.13777∠ − 1.1071 35.77709∠2.0344 
Ybus = −10 + j 20 26 − j52 −16


35.77709∠2.0344 67.23095∠ − 1.1737 
 −10 + j 30 −16 + j 32 26 − j 62   31.62278∠1.8925
For F-D Method:
 −52 32 
B′ = 

 32 −62 
B′′ =
 −0.028182
[ B′] = 
 −0.014545
−1
[ −52]
−0.014545
−0.023636
45
N-R Method
•2n-2-m=3 independent equations:
P2 | V2 || V1 || Y21 | cos(θ 21 − δ 2 + δ1 ) + | V22 || Y22 | cos θ 22 + | V2 || V3 || Y23 | cos(θ 23 − δ 2 + δ 3 )
P3 | V3 || V1 || Y31 | cos(θ 31 − δ 3 + δ1 ) + | V3 || V2 || Y32 | cos(θ 32 − δ 3 + δ 2 ) + | V32 || Y33 | cos(θ 33 )
Eq (1)
Q2 =
− | V2 || V1 || Y21 | sin(θ 21 − δ 2 + δ1 ) − | V22 || Y22 | sin θ 22 − | V2 || V3 || Y33 | sin(θ 23 − δ 2 + δ 3 )
Eq (2)
•J has 3x3 elements
 ∂P2
 ∂δ
 ∆P2   2
 ∆P  =  ∂P3
 3   ∂δ 2
 ∆Q2  
 ∂Q2

 ∂δ 2
∂P2
∂δ 3
∂P3
∂δ 3
∂Q2
∂δ 3
∂P2
| V2 || V1 || Y21 | sin(θ 21 − δ 2 + δ1 ) + | V2 || V3 || Y23 | sin(θ 23 − δ 2 + δ 3 )
=
∂δ 2
∂P2
=
− | V2 || V3 || Y23 | sin(θ 23 − δ 2 + δ 3 )
∂δ 3
∂P2 
∂P
= | V || Y | cos(θ − δ + δ ) + 2 | V || Y | cos θ + | V || Y | cos(θ − δ + δ )

∂ |V |
∂ | V2 | 
∂P
 ∆δ 2 
=
− | V || V || Y | sin(θ − δ + δ )
∂δ
∂P3  

δ
∆
2 
∂P
| V || V || Y | sin(θ − δ + δ ) + | V || V || Y | sin(θ − δ + δ )
∂ | V2 |   =
∂δ
 ∆ | V2 |
∂Q2 
∂P
= | V || Y | cos(θ − δ + δ )

∂ |V |
∂ | V2 | 
2
1
21
21
2
1
2
22
22
3
2
32
23
23
2
3
2
3
3
2
23
32
3
2
2
3
3
1
31
31
3
1
3
32
3
2
3
3
3
32
32
3
2
2
∂Q2
=
| V2 || V1 || Y21 | cos(θ 21 − δ 2 + δ1 ) + | V2 || V3 || Y23 | cos(θ 23 − δ 2 + δ 3 )
∂δ 2
∂Q2
=
− | V2 || V3 || Y23 | cos(θ 23 − δ 2 + δ 3 )
∂δ 3
∂Q2
=
− | V1 || Y21 | sin(θ 21 − δ 2 + δ1 ) − 2 | V2 || Y22 | sin θ 22 − | V3 || Y23 | sin(θ 23 − δ 2 + δ 3 )
46
∂ | V2 |
Procedure of the N-R Method
1.
Initial values:
=
V1 1.05∠0 pu
2.
V3 = 1.04 pu
| V2(0) |= 1.0
δ 2(0) = 0.0
δ 3(0) = 0.0
Calculate P2(k), P3(k) and Q2(k) by eq. (1) and (2) and then ∆P2(k), ∆P3(k), and ∆Q2(k) ,
e.g.:
∆P2(0) =P2sch − P2(0) =−4.0 − (−1.14) =−2.8600
∆P3(0) = P3sch − P3(0) = 2.0 − (0.5616) =1.4384
∆Q2(0) =Q2sch − Q2(0) =−2.5 − (−2.28) =−0.2200
3.
Calculate J, solve ∆|V2(k)|, ∆δ2(k) and ∆δ3(k) and then |V2(k+1)|, δ2(k+1) and δ3(k+1) , e.g.:
(0)

 −2.8600   54.28000 −33.28000 24.86000   ∆δ 2


1.4384  =


(0)


  −33.28000 66.04000 −16.64000   ∆δ 3


(0)
 −0.2200   −27.1400 16.64000 49.72000   ∆ | V2 |
∆δ 2(0) =−0.045263 δ 2(1) =(0) + (−0.045263) =−0.0452653
∆δ 3(0) =−0.007718 δ 3(1) =(0) + (−0.007718) =−0.007718
∆ | V2(0) |=−0.026548
| V2(1) |=1 + (−0.026548) =0.97345
(1)
 −0.099218   51.724675 −31.765618 21.302567   ∆δ 2 
0.021715  =
 −32.981642 65.656383 −15.379086   ∆δ (1) 

 
 3 
 −0.050914   −28.538577 17.402838 48.103589   ∆ | V2(1) |


∆δ 2(1) =−0.001795 δ 2(2) =−0.045263 + (−0.001795) =−0.04706
∆δ 3(1) =−0.000985 δ 3(2) =0.007718 + (−0.000985) =−0.00870
∆ | V2(1) |=−0.001767 | V2(2) |=0.973451 + (−0.001767) =0.971684
4.
Stop after 3 iterations: =
V2 0.97168∠ − 2.696°
5.
Calculate:
=
V3 1.04∠ − 0.4988°
=
ε 2.5 ×10−4
Q3 =
− | V3 || V1 || Y31 | sin(θ31 − δ 3 + δ1 )− | V3 || V2 || Y32 | sin(θ32 − δ 3 + δ 2 )− | V3 |2 | Y33 | sin θ33 = 1.4617 pu
=
P1 | V1 |2 | Y11 | cos θ11 + | V1 || V2 || Y12 | cos(θ12 − δ1 + δ 2 )+ | V1 || V3 || Y13 | cos(θ13 − δ1 + δ 3 )
Q1 =
− | V1 |2 | Y11 | sin θ11 − | V1 || V2 || Y12 | sin(θ12 − δ1 + δ 2 )− | V1 || V3 || Y13 | sin(θ13 − δ1 + δ 3 )
= 2.1842 pu
= 1.4085 pu
47
Procedure of the F-D Method
1.
Initial values:
=
V1 1.05∠0 pu
2.
V3 = 1.04 pu
| V2(0) |= 1.0
δ 2(0) = 0.0
δ 3(0) = 0.0
Calculate P2(k), P3(k) and Q2(k) by eq. (1) and (2) and then ∆P2(k), ∆P3(k), and ∆Q2(k) , e.g.:
∆P2(0) =P2sch − P2(0) =−4.0 − (−1.14) =−2.8600
∆P3(0) = P3sch − P3(0) = 2.0 − (0.5616) =1.4384
∆Q2(0) =Q2sch − Q2(0) =−2.5 − (−2.28) =−0.2200
3.
Use B’ and B” to solve ∆|V2(k)|, ∆δ2(k) and ∆δ3(k) and then |V2(k+1)|, δ2(k+1) and δ3(k+1) ,
e.g.:
 −2.8600 
2(1)  0  (0.060483)  0.060483
 ∆δ 
 −0.028182 −0.014545  1.0   −0.060483 
−
 (0)  =
 1.4384  =


−
−
0.014545
0.023636
∆
δ


3(1)  0  (0.008989)  0.008989
  −0.008909
 3 
 1.04 
(1)
 1  0.22 
V
 1  (0.0042308)  0.995769
(0)
2
  0.0042308
 | V2 |    
 52   1.0 
(0)
2
4.
Stop after 14 iterations:
=
V2 0.97168∠ − 2.696°
5.
Calculate:
=
V3 1.04∠ − 0.4988°
Q3 = 1.4617 pu
P1 = 2.1842 pu
Q1 = 1.4085 pu
48