Power flow analysis
Subject lecturer: Dr. XU Zhao
Department of Electrical Engineering
Hong Kong Polytechnic University
Email: eezhaoxu@polyu.edu.hk
eezhaoxu@polyu edu hk
Room: CF632
Tel: 27666160
Lecture Overview
•
•
•
•
•
•
2
Load Flow Objective
Basic Line & Bus Equations
Newton--Raphson Method
Newton
Decoupled & Fast Decoupled Method
Application: Design and Operation
Conclusions
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Load flow: Objectives and applications
• Busbar (Node) voltages
• Flow of real and reactive powers in the branches of
the network
• Power system augmentation studies to plan
expansion
p
to the network to meet future
requirements
• Effect of temporary loss of generation and
transmission circuits on system loading
3
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Load Flow: Objectives and applications
• Optimum
O ti
system
t
running
i conditions
diti
and
d load
l d
distribution
• Generator scheduling and reactive scheduling to
minimizes losses
• Optimum rating and tap-range of transformers
• Optional placement of reactive compensation
• Improvements from change of conductor size and
system voltage
• Starting point of other studies such as fault analysis
and stability analysis
4
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
A Simple 3 Bus Power System
2
1
I1
AC
I2
AC
y12= y21
y12
y12'
2
'
y13 = y31
2
3
I3
y23=y32
y13'
y13'
y23'
y23'
2
2
2
2
Network elements given in
admittance format
5
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Nodal analysis
• Using KCL (nodal voltage method) to write out nodal
current balance equations
I1=(V1-V3)Y13+V1(Y12’/2+Y13’/2)-(V2-V1)Y12
I2=(V2-V1)Y12+V2(Y12’/2+Y23’/2)-(V3-V2)Y23
I3=(V
( 3-V2)Y
) 23+V3(Y
( 13’/2+Y
’/
’/ ) ( 1-V3)Y
) 13
23’/2)-(V
6
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Network Performance Equation
• Y11 = (y
( 12+ y13+ y12’/2+
/2 y13’/2) ---------self
lf admittance
d itt
• (- y12)= Y12 , (-y13 ) = Y13 ---mutual admittance
⎡ I 1 ⎤ ⎡Y 11 Y 12 Y 13 ⎤ ⎡V 1 ⎤
⎢ I 2 ⎥ = ⎢Y 21 Y 22 Y 23⎥ ⎢V 2 ⎥
⎢ ⎥ ⎢
⎥⎢ ⎥
⎢⎣ I 3 ⎥⎦ ⎢⎣Y 31 Y 32 Y 33⎥⎦ ⎢⎣V 3 ⎥⎦
7
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Solving for Bus Currents
For example,
p , ina two
pprevious
nodecase
system
y assume
with Ybus
⎡ 1.0 ⎤
V=⎢
⎥
0
0.8
8
−
j
0.2
0
2
⎣
⎦
Bus
1
Bus
2
Then
⎡12 − j15.9 −12 + j16 ⎤ ⎡ 1.0 ⎤ ⎡ 5.60 − j 0.70 ⎤
=⎢
⎢ −12 + j16 12 − j15
⎥
⎢
⎥
⎥
15.9
9
0
0.8
8
−
j
0
0.2
2
−
5
5.58
58
+
j
0.88
0
88
⎣
⎦⎣
⎦ ⎣
⎦
Therefore the power injected at bus 1 is
S1 = V1I1* = 1.0 × (5.60 + j 0.70) = 5.60 + j 0.70
S2 = V2 I 2* = (0.8
(0 8 − j 0.2)
0 2) × (−5.58
5 58 − j 0.88)
0 88) = −4.64
4 64 + j 0.41
0 41
8
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Solving for Bus Voltages
For example,
example in previous case assume
⎡ 5.0 ⎤
I=⎢
⎥
−
4
4.8
8
⎣
⎦
Then
−1
⎡12 − j15.9 −12 + j16 ⎤ ⎡ 5.0 ⎤ ⎡ 0.0738 − j 0.902 ⎤
=⎢
⎢ −12 + j16 12 − j15
⎥
⎢
⎥
⎥
15.9
9
−
4
4.8
8
−
0
0.0738
0738
−
j
1.098
1
098
⎣
⎦ ⎣
⎦ ⎣
⎦
Therefore the power injected is
S1 = V1I1* = (0.0738 − j 0.902) × 5 = 0.37 − j 4.51
S 2 = V2 I 2* = ( −00.0738
0738 − j11.098)
098) × ( −44.8)
8) = 00.35
35 + j 55.27
27
9
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Effect of Shunt Elements
• System
S t
power flow
fl
can be
b controlled
t ll d by
b switching
it hi
of shunt capacitor banks, shunt reactors, and static
VAR systems.
systems If there is a shunt capacitor of Y
Y’ at
bus 1, the self-admittance at bus 1 changes to Y11 =
(y12+ y13+ y12’/2+ y13’/2) + Y’ and no other elements
will be affected.
• Both tap-changing and regulating transformers are
modelled by a transformer with off nominal turns
ratio a and equivalent series admittance Yeq (as
shown in Figure next page).
page)
Then the
corresponding elements in the YBus (Yii, Yij and Yjj)
g
need to be changed.
10
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Tap-changing and Regulating Transformers
• Equivalent π Circuit
i
a Yeq
(1 a)Yeq
(1-a)
j
(a 2-a) Yeq
G
Ground
d
11
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Line Flow Equations
• The
Th reall and
d reactive
ti power flow
fl
from
f
bus
b i to
t b
bus j
is given by: Pij -jQij = Vi* [yij(Vi-Vj)+ Vi (yij’/2)]
• Similarly at bus j the power flow from j to i will be
given by:Pji -jQji = Vj* [yij(Vj-Vi)+ Vj (yij’/2)]
• The algebraic sum of above two equations gives the
line loss over line i-j
i
y'ij/2
yij
j
y'ij/2
Ground
12
Electrical Engineering, HKPU
Noteconjugate!
EE3741 Ass. Prof Zhao Xu
Power Balance Equations
From KCL we know at each bus i in an n bus system
the current injection, I i , must be equal to the current
that flows into the network
I i = I Gi − I Di =
n
∑ Iik
k =1
Since I = Ybus V wee also kno
know
I i = I Gi − I Di =
n
∑ YikVk
k =1
The network power injection is then Si = Vi I i*
13
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Power Balance Equations, cont’d
*
n
⎛
⎞
Si = Vi I i* = Vi ⎜ ∑ YikVk ⎟ = Vi ∑ Yik*Vk*
⎝ k =1
⎠
k =1
This is an equation with complex numbers.
q
set of real
Sometimes we would like an equivalent
power equations. These can be derived by defining
n
Yik = Gik + jBik
=
jθ i
=
Vi Vi e = Vi ∠θ i
θ ik = θ i − θ k
Recall e jθ = cosθ + j sin θ
14
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Real Power Balancen Equations
n
Si = Pi + jQi = Vi ∑ Yik*Vk* = ∑ Vi Vk e jθik (Gik − jBik )
k =1
=
n
∑ Vi Vk
k =1
k =1
(cosθ ik + j sin θ ik )(Gik − jBik )
Resolving into the real and imaginary parts
Pi =
Qi =
15
n
∑ Vi Vk (Gik cosθ ik + Bik sinθ ik ) = PGi − PDi
k =1
n
∑ Vi Vk (Gik sinθ ik − Bik cosθ ik ) = QGi − QDi
k =1
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Power Flow Requires
q
Iterative Solution
I the
In
th power flow
fl we assume we know
k
Si and
d the
th
Ybus . We would like to solve for the V's. The problem
is the below equation has no closed form solution:
*
n
⎛
⎞
Si = Vi I i* = Vi ⎜ ∑ YikVk ⎟ = Vi ∑ Yik*Vk*
⎝ k =1
⎠
k =1
Rath
er, wealgebraic
must pursue
an iterative
Nonlinear
equation
set, notapproach.
exactly
solvable but can be solved through iterative
solvable,
process
n
V—magnitude and angle !
16
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Gauss Iteration
There are a number of different iterative methods
we can use. We'll consider two: Gauss and Newton.
With the Gauss method we need to rewrite our
equation in an implicit form: x = h(x)
To iterate we first make an initial guess of x, x (0) ,
and then iteratively solve x (v +1) = h( x ( v ) ) until we
find a "fixed point",
point" x,
x? such that x = h(x)
(x).
?
17
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Gauss Iteration Example
Example: Solve x - x − 1 = 0
x
( v +1)
=1+ x
(v)
Let k = 0 and arbitrarily guess x
18
(0)
= 1 and solve
k
x(v )
k
x(v)
0
1
5
2.61185
1
2
6
2.61612
2
2 41421
2.41421
7
2.61744
2 61744
3
2.55538
8
2.61785
4
2.59805
9
2.61798
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Stopping Criteria
A keyy problem
p
to address is when to stop
p the
iteration. With the Guass iteration we stop when
with
i h Δx ( v ) = x ( v +1) − x ( v )
Δx ( v ) < ε
If x is a scalar this is clear, but if x is a vector we
need to generalize the absolute value by using a norm
Δx ( v )
j
<ε
T common norms are the
Two
th Euclidean
E lid
& infinity
i fi it
Δx 2 =
19
Electrical Engineering, HKPU
n
2
Δ
x
∑ i
i =1
Δx ∞ = max i Δx i
EE3741 Ass. Prof Zhao Xu
Gauss Power Flow
We first need to put the equation in the correct form
Si =
*
n
⎛
⎞
* *
= Vi ⎜ ∑ YikkVk ⎟ = Vi ∑ YikkVk
⎝ k =1
⎠
k =1
n
*
Vi I i
n
n
k =1
k =1
S*i = Vi* I i = Vi* ∑ YikVk = Vi* ∑ YikVk
S*i
=
*
Vi
20
n
∑ YikVk
k =1
= YiiVi +
n
∑
k =1,
1 k ≠i
n
⎞
1 ⎛ S*i
Vi =
⎜ * − ∑ YikkVk ⎟
⎟
Yii ⎜⎝ V
k =1,k ≠i
⎠
i
Electrical Engineering, HKPU
YikVk
EE3741 Ass. Prof Zhao Xu
Gauss Two Bus Power Flow Example
p
•A
A 100 MW, 50 Mvar load is connected to a generator
•through a line with z = 0.02 + j0.06 p.u. and line charging of 5 Mvar on
each end (100 MVA base). Also, there is a 25 Mvar capacitor at bus 2. If the
generator voltage is 1.0 p.u., what is V2?
21
Electrical Engineering, HKPU
SLoad = 1.0 + j0.5 p.u.
EE3741 Ass. Prof Zhao Xu
Gauss Two Bus Example, cont’d
The unknown is the complex load voltage
voltage, V2 .
To determine V2 we need to know the Ybus .
1
= 5 − j15
0.02 + j 0.06
⎡5 − j14.95 −5 + j15 ⎤
Hence Ybus = ⎢
⎥
−
5
+
j
15
5
−
j
14
14.70
70
⎣
⎦
( Note B22 = - j15 + j 0.05 + j 0.25)
22
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Gauss Two Bus Example, cont’d
n
⎞
1 ⎛ S*2
V2 =
⎜ * − ∑ YikVk ⎟
Y22 ⎝ V2 k =1,k ≠i
⎠
⎛ -1 + j 0.5
⎞
1
− (−5 + j15)(1.0∠0) ⎟
V2 =
⎜
*
5 − j14.70 ⎝ V2
⎠
Guess V2(0) = 1.0∠0 (this is known as a flat start)
v
0
1
2
23
V2( v )
1.000
1
000 + j 00.000
000
0.9671 − j 0.0568
0.9624 − j 0.0553
Electrical Engineering, HKPU
v
3
4
V2( v )
00.9622
9622 − j 0.0556
0 0556
0.9622 − j 0.0556
EE3741 Ass. Prof Zhao Xu
Gauss Two Bus Example, cont’d
V2 = 0.9622
0 9622 − j 00.0556
0556 = 00.9638
9638∠ − 33.3
3°
Once the voltages are known all other values can
be determined, such as the generator powers and the
line flows
S1* = V1* (Y11V1 + Y12V2 ) = 1.023 − j 0.239
In actual units P1 = 102.3 MW, Q1 = 23.9 Mvar
2
The capacitor is supplying V2 25 = 23
23.2
2 Mvar
24
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Slack Bus
zIn previous example we specified S2 and V1 and then
solved for S1 and V2.
zWe can not arbitrarily
y specify
p
y S at all buses because total
generation must equal total load + total losses
zWe also need an angle reference bus.
zTo solve these problems we define one bus as the "slack"
bus. This bus has a fixed voltage magnitude and angle,
and a varying real/reactive power injection.
25
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Gauss with Many Bus Systems
With multiple
p bus systems
y
we could calculate
new Vi ' s as follows:
Vi(v +1)
⎞
n
1 ⎛ S*i
⎜ ( v )* − ∑ YikVk( v ) ⎟
=
⎟
Yii ⎜ V
k
=
1
1,
k
≠
i
⎝ i
⎠
=
But after
26
(v)
(v)
(v )
hi (V1 ,V2 ,...,Vn )
we've determined Vi(v +1)
we have a better
estimate
i
off its
i voltage
l
, so it
i makes
k sense to use this
hi
new value. This approach is known as the
Gauss-Seidel iteration.
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Gauss-Seidel Iteration
Immediatelyy use the new voltage
g estimates:
V2( v +1) = h2 (V1 ,V2( v ) ,V3( v ) ,…,Vn( v ) )
( v +1))
V3
V4( v +1))
=
=
( v +1))
(v)
(v )
h2 (V1 ,V2 ,V3 ,…,Vn )
h2 (V1 ,V2( v +1)) ,V3( v +1)) ,V4( v ) …,Vn(v ) )
#
Vn( v +1) = h2 (V1 ,V2( v +1) ,V3( v +1) ,V4( v +1) …,Vn( v ) )
The Gauss-Seidel
Gauss Seidel works better than the Gauss
Gauss, and
is actually easier to implement. It is used instead
of Gauss.
27
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Three Types of Power Flow Buses
z There are three main types of power flow buses
– Load (PQ) at which P/Q are fixed; iteration solves for voltage
magnitude and angle.
– Slack at which the voltage magnitude and angle are fixed; iteration
solves for P/Q injections
– Generator (PV) at which P and |V| are fixed; iteration solves for
voltage
g angle
g and Q injection
j
zspecial coding is needed to include PV
buses in the Gauss-Seidel iteration
28
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
BUS TYPES
Bus Type No of Bus
Slack i=1 1
Voltage Ng
Controlled
Load Bus N
N- Ng -1
1
Total
29
N
Electrical Engineering, HKPU
Quantities No. of available No of state
Specified Equations
variables
δi and Vi
0
0
δ1 V1
Pi |V i |
Ng
Pi Q I
Ng (only P
Equations)
2(N- Ng -1)
2(N
1)
2N
2N- Ng -2
2N- Ng -2
2(N- Ng -1)
2(N
1)
EE3741 Ass. Prof Zhao Xu
Inclusion of PV Buses in G-S
To solve for Vi at a PV bus we must first make a
guess of Qi :
n
Si* = Vi* ∑ YikVk = Pi + jQi
k =1
Hence
(v )
Qi
⎡ ( v )* n
(v ) ⎤
= − Im ⎢Vi ∑ YikV ⎥
k
⎣
⎦
k =1
In the iteration we use
30
Electrical Engineering, HKPU
(v)
Si
= Pi +
(v )
jQi
EE3741 Ass. Prof Zhao Xu
Inclusion of PV Buses, cont'd
T t ti l solve
Tentatively
l for
f Vi( v +1)
( v +1)
Vi
⎞
n
1 ⎛ Si( v )*
(v)
⎜ (v )* − ∑ YikVk ⎟
=
⎟
Yii ⎜ V
1,
k
k
i
=
≠
⎝ i
⎠
But since Vi is specified, replace Vi( v +1) by Vi
Magnitude is known
31
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Two Bus PV Example
Consider
C
id th
the same two
t
bus
b
system
t
from
f
the
th previous
i
example, except the load is replaced by a generator
z = 0.02 + j 0.06
Bus 1
(slack bus)
V1 = 1.0
10
Bus 2
V2 = 1.05
1 05
P2 = 0 MW
32
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Two Bus PV Example, cont'd
⎞
1 ⎛ S2*
V2 =
⎜ * − Y21V1 ⎟
Y22 ⎝ V2
⎠
Q2 = − Im[Y21V1V2* + Y22V2V2* ]
Guess V2 = 11.05
05∠0°
v
S2( v )
V2( v +1)
V2( v +1)
0 0 + j 0.457 1.045∠ − 0.83° 1.050∠ − 0.83°
1 0 + j 0.535
0 535 11.049
049∠ − 00.93
93° 11.050
050∠ − 00.93
93°
2 0 + j 0.545 1.050∠ − 0.96° 1.050∠ − 0.96°
33
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Generator Reactive Power Limits
zThe reactive power output of generators varies to maintain the
terminal voltage; on a real generator this is done by the exciter
zTo maintain higher voltages requires more reactive power
zGenerators have reactive power limits,
limits which are dependent upon
the generator's MW output
zThese limits must be considered during the power flow solution.
34
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Generator Reactive Limits, cont'd
z During power flow once a solution is obtained check to make generator
reactive power output is within its limits
z If the reactive power is outside of the limits, fix Q at the max or min
value, and resolve treating the generator as a PQ bus
– this is know as "type-switching"
– also need to check if a PQ generator can again regulate
z Rule of thumb: to raise system voltage we need to supply more vars
35
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Accelerated G-S Convergence
Previously in the Gauss
Gauss-Seidel
Seidel method we were
calculating each value x as
x ( v +1) = h( x ( v ) )
g
we can rewrite this as
To accelerate convergence
x
( v +1)
= x
(v)
+ h( x
(v )
)−x
(v)
Now introduce acceleration parameter α
x ( v +1) = x ( v ) + α (h( x ( v ) ) − x ( v ) )
With α = 1 this is identical to standard gauss-seidel.
Larger values
l
off α may result
l in
i faster
f
convergence.
36
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Accelerated Convergence, cont’d
Consider the pprevious example:
p
x - x −1 = 0
x ( v +1) = x ( v ) + α (1 + x ( v ) − x ( v ) )
Comparison of results with different values of α
37
k
α =1
α = 1.2
α = 1.5 α = 2
0
1
1
1
1
1
2
2 20
2.20
22.5
5
3
2
2.4142
2.5399
2.6217
2.464
3
2.5554
2.6045
2.6179
2.675
4
2 5981
2.5981
22.6157
6157
2.6180
2 6180
2.596
2 596
5
2.6118
2.6176
2.6180
Electrical Engineering, HKPU
2.626
EE3741 Ass. Prof Zhao Xu
Gauss-Seidel Advantages
zEach iteration is relatively fast (computational order is
proportional to number of branches + number of buses in the
system
zR l ti l easy tto program
zRelatively
Gauss-Seidel Disadvantages
g
zTends to converge relatively slowly, although this can be
improved with acceleration
zHas tendency to miss solutions,
solutions particularly on large systems
zTends to diverge on cases with negative branch reactance
(common with compensated lines)
zNeed
N d tto program using
i
complex
l
numbers
b
38
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Page 337 Example 9.2
9 2 Power system
analyses
39
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
40
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
41
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Newton-Raphson Algorithm
z The second major power flow solution method is the Newton
Newton-Raphson
Raphson
algorithm
z Key idea behind Newton-Raphson is to use sequential linearization
General form of problem: Find an x such that
f ( xˆ ) = 0
42
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Newton-Raphson Method
1 For each guess of xˆ, x(v) , define
1.
Δx(v) = xˆ - x(v)
2. Represent f ( xˆ ) by a Taylor series about f ( x)
(v)
df
(
x
) (v)
(v)
f ( xˆ ) = f ( x ) +
Δx +
dx
1 d 2 f ( x(v) ) (v) 2
+
Δx
+ higher order terms
2
2 dx
(
43
Electrical Engineering, HKPU
)
EE3741 Ass. Prof Zhao Xu
Newton-Raphson Method
44
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Newton-Raphson Method, cont’d
3. Approximate
pp
f ( xˆ ) byy neglecting
g
g all terms
except the first two
f ( xˆ ) = 0 ≈ f ( x
(v)
(v)
df ( x ) ( v )
)+
Δx
dx
4. Use this linear approximation to solve for Δx ( v )
−1
⎡ df ( x ) ⎤
(v)
f
(
x
)
Δx
= −⎢
⎥
⎣ dx ⎦
5. Solve for a new estimate of x̂
(v )
(v )
x ( v +1) = x ( v ) + Δx ( v )
45
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Newton-Raphson Method, cont’d
46
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
NR method general formulation
Use Newton
Newton-Raphson
Raphson to solve f ( x) = x 2 - 2 = 0
The equation we must iteratively solve is
Δx
(v )
Δx
(v )
−1
⎡ df ( x ) ⎤
(v )
= −⎢
f
(
x
)
⎥
d
⎣ dx
⎦
1 ⎤ (v ) 2
⎡
= − ⎢ ( v ) ⎥ (( x ) - 2)
⎣2x ⎦
(v )
x (v +1)) = x ( v ) + Δx ( v )
x
47
( v +1)
= x
Electrical Engineering, HKPU
(v)
1 ⎤ (v) 2
⎡
− ⎢ ( v ) ⎥ (( x ) - 2)
⎣2x ⎦
EE3741 Ass. Prof Zhao Xu
Jacobian matrix
(v+1)
x
1 ⎤ (v) 2
⎡
= x − ⎢ (v) ⎥ ((x ) -2)
⎣2x ⎦
(v)
Guess x(0) =1. Iteratively solving we get
48
v
x(v)
f (x(v) )
Δx(v)
0
1
−1
0.5
1
1.5
0.25
−0.08333
2
1.41667
141667
6.953
6953×10−3
−2.454
2454×10−3
3
1.41422
6.024×10−6
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Jacobian matrix
(v+1)
x
1 ⎤ (v) 2
⎡
= x − ⎢ (v) ⎥ ((x ) -2)
⎣2x ⎦
(v)
Guess x(0) =1. Iteratively solving we get
49
v
x(v)
f (x(v) )
Δx(v)
0
1
−1
0.5
1
1.5
0.25
−0.08333
−3
2
1.41667
141667
6.953
6953×10
3
1.41422
6.024×10−6
Electrical Engineering, HKPU
−3
−2.454
2454×10
EE3741 Ass. Prof Zhao Xu
Bus type and Jacobian formulation
50
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
Iteration process
51
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
NR steps
52
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
NR steps
53
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
NR example
54
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
NR example
55
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
NR example
56
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
NR example
57
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
NR example
58
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
NR example
59
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
NR example
60
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
NR example
61
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
NR example
62
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
NR example
63
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu
NR example
64
Electrical Engineering, HKPU
EE3741 Ass. Prof Zhao Xu