The load flow problem

LOAD FLOW
A. J. Conejo
Univ. Castilla – La Mancha
2011
22 July 2011
1
The load flow problem
0. References
1. Introduction
2. Problem formulation
Two-bus case
Matrix YBUS
General equations
Bus classification
Variable types and
limits
22 July 2011
2
The load flow problem
3. The Gauss-Seidel solution technique
Introduction
Algorithm initialization
PQ Buses
PV Buses
Stopping criterion
22 July 2011
3
The load flow problem
4. The Newton-Raphson solution technique
Introduction
General fomulation
Load flow case
Jacobian matrix
Solution outline
22 July 2011
4
The load flow problem
5.
6.
7.
8.
Fast decoupled AC load flow
Adjustment of bounds
DC load flow
Comparison of load flow solution methods
22 July 2011
5
The load flow problem
References
1. A. Gómez Expósito, A. J. Conejo, C.
Cañizares. “Electric Energy Systems.
Analysis and Operation”. CRC Press, Boca
Raton, Florida, 2008.
2. A. R. Bergen, V. Vittal. “Power Systems
Analysis”. Second Edition. Prentice Hall,
Upper Saddle River, New Jersey, 1999.
22 July 2011
6
1. Introduction
22 July 2011
7
Introduction
• A snapshot of the system
• Most used tool in steady state power system
analysis
• Knowning the demand and/or generation of
power in each bus, find out:
– buses voltages
– load flow in lines and transformers
22 July 2011
8
Introduction
• The problem is described throught a nonlineal system of equations
• Need of iterative solution techniques
• Solution technique: accuracy vs. computing
time
22 July 2011
9
Introduction
• Applications:
1. On-line analyses
• State estimation
• Security
• Economic analyses
22 July 2011
10
Introduction
2. Off-line analyses
• Operation analyses
• Plannig analyses
Network expansion planning
Power exchange planning
Security and adecuacy analyses
- Faults
- Stability
22 July 2011
11
2. Problem formulation
22 July 2011
12
Problem formulation
Two-bus case
We want to find out the relationship between
j
Si  Pi  jQi and Vi  Vie i in all buses of the power
system
I1
V1
22 July 2011
I2
YS
YG
YG
Transmission line π model
V2
13
Problem formulation
Two-bus case
Using Kirchhoff laws:
I1  V1 YG  ( V1  V2 ) Y S
I2  V2 YG  ( V2  V1) Y S
Matrix notation:
 Y S   V1   Y11 Y12   V1 
 I1   Y G  Y S
  V   
  V 
I   
Y G  Y S   2   Y 21 Y 22   2 
 2    YS
IBUS   YBUS  VBUS 
22 July 2011
14
Problem formulation
Two-bus case
• Complex power injected in each bus:
S1  SG1  SD1  P1  jQ1  V1  I1*
S2  SG2  SD2  P2  jQ2  V2  I2*
*
*
S1  V1  I1*  V1  ( Y11
V1*  Y12
V2* )
*
*
S2  V2  I2*  V2  ( Y21
V1*  Y22
V2* )
22 July 2011
15
Problem formulation
Two-bus case continuation
Notation:
Yik  Yik e j ik
Vi  Vi e j i
Replacing:
2
P1  jQ1  V1  Y1k Vk e j( 1 k 1k )
k 1
2
P2  jQ2  V2  Y2k Vk e j( 2 k 2k )
k 1
22 July 2011
16
Problem formulation
Two-bus case continuation
Therefore, the no lineal equations for the 2
buses network are:
2
Pi  Vi   Yik Vk cos (i  k  ik )
k 1
2
Qi  Vi   Yik Vk sin (i  k  ik )
k 1
i  1, 2
22 July 2011
17
Problem formulation
Matrix Ybus
•
Two bus case
YBUS
22 July 2011
 Y11 Y12   YG  YS



 Y21 Y22    YS
 YS 
YG  YS 
18
General building rules
Matrix Ybus
1. Self admittance of node i, Yii ,equals the
algebraic sum of all the admittances
connected to node i
2. Mutual admittance between nodes i and
k, Yik ,equals the negative of the sum of
all admittances connecting nodes i and k
3. Yik  Yki
22 July 2011
19
Problem formulation
Matrix Ybus
•
Caracteristics of YBUS
1. YBUS
is symmetric
is very sparse
(>90% for more than 100 buses)
2. YBUS
22 July 2011
20
Ybus example
Shunt element YG  j 0.01 (two per line)
Series element ZS  j 0.1
YBUS
22 July 2011
j10
j10 
 j19.98
  j10
 j19.98
j10 


j10
 j19.98 
 j10
21
Problem formulation
General equations
• 2n equations (static load flow equations)
n
Pi  PGi  PDi  Vi   Yik Vk cos (i  k  ik )
1
k 1
n
Qi  QGi  QDi  Vi   Yik Vk sin (i  k  ik )
2
k 1
i  1,..., n
n  bus system
22 July 2011
22
Problem formulation
General equations
Polar representation for voltages and rectangular
for admittances
n
Pi  Vi   Vk (Gik cos ik  Bik senik )
[1]
Qi  Vi   Vk (Gik senik  Bik cos ik )
[2]
k 1
n
k 1
where
Y ik  Gik  j Bik
ik  i  k
22 July 2011
23
Problem formulation
General equations
• 4n variables
Vi, i,Pi, Qi
;
i  1,...,n
• If 2n variables are specified, the other 2n are
determined by equations [1] and [2]
22 July 2011
24
Problem formulation
BUS Classification
1. PQ buses
Pi known ( PDi known, PGi zero)
Qi known (QDi known, QGi zero)
Vi unknown
i unknown
22 July 2011
25
Problem formulation
BUS Classification
2. PV buses
Pi known
Qi unknown
Pi known ( PGi specified, PDi known)
Vi known (specified)
Qi unknown ( QGi unknown, QDi known)
i unknown
22 July 2011
26
Problem formulation
BUS Classification
3. Slack bus, generator with large capacity.
V1 known (specified)
1 known (specified, typically 1  0 
reference)
P1 unknown ( PD1 known, PG1 unknown)
Q1 unknown ( QD1 known, QG1 unknown)
22 July 2011
27
Problem formulation
Variable types and limits
• Power balance
n
n
i1
i1
 PGi   PDi  PLOSS
n
n
i1
i1
 QGi   QDi  QLOSS
22 July 2011
28
Problem formulation
Variable types and limits
• Variable types
 Control variables
PGi (excepting slack bus)
QGi or V i
 Non-control variables
PDi QDi
 State variables
Vi i
22 July 2011
29
Problem formulation
Variable types and limits
• Variable limits
 Voltage magnitude Vi min 
 Power angle i  k

Vi

Vi max
i  k max (every existing line)
 Power limits
PGi,min  PGi  PGi,max
QGi,min  QGi  QGi,max
22 July 2011
30
3.The Gauss-Seidel solution
technique
22 July 2011
31
Gauss-Seidel solution technique
No lineal system:
f (x)  0

x  F( x )
Iteration
x (r 1)  F( x (r ) )
Stoping rule
22 July 2011
x (r 1)  x (r )  
32
Gauss-Seidel solution technique
 Example
f ( x)  x 2  5x  4  0
x  0 . 2 x 2  0 .8
x (r 1)  0.2( x (r ) )2  0.8 ; x ( 0 )  2
r  0,
x (1)  0.2  22  0.8  1.6
r  1,
x ( 2 )  0.2  1.6 2  0.8  1.312
r  2,
x ( 3 )  0.2  1.312 2  0.8  1.1442
...
r  5,
x ( 6 )  1.0103
r  6,
x ( 7 )  1.0042
...
r  10,
22 July 2011
x (11)  1.0001
r  11,
x
r  12,
x (13 )  1.0000
(12 )
 1.0000
Many iterations!
33
Gauss-Seidel solution technique
Algorithm beginning
1) Known
Pi
i  2,...,n
Qi
i  m  1,...,n
Vi
i  2,...,m
V1
slack bus
(PV & PQ Buses)
(PQ Buses)
(PV Buses)
V i min , V i max i  m  1,...,n (PQ Buses)
QGi,min , QGi,max
22 July 2011
i  2,...,m (PV Buses)
34
Gauss-Seidel solution technique
Algorithm beginning
2) Build YBUS
3) Initialize voltages
Vi  Vi
i  i
22 July 2011
0
0
i  m  1,...,n
i  2,...,n
35
Gauss-Seidel solution technique
PQ buses
4)
*
Si
PQ buses
 Pi  jQi 
*
V i Ii

*
Vi
Y ii V i 


n
1 Pi  jQi

Vi 

Y
V
ik
k


Y ii  V i*
k 1


k i
P  jQi
Ai  i
Y ii
Y ik
Bik 
Y ii
22 July 2011
* n
V i Y ik
k 1
k i

Vk
i  m  1,...,n
i  m  1,...,n ; k  1,...,n ; k  i
36
Gauss-Seidel solution technique
PQ buses
At iteration (r+1) and bus i, the available values of
voltages at previous buses are used:
Vi
(r 1)
22 July 2011

Ai
(r ) *
(Vi )
i1
  Bik V k
k 1
(r 1)

n
 Bik V k
(r )
k i1
37
Gauss-Seidel solution technique
PV buses
5) PV buses
*
Si
 Pi  jQi 
*
V i Ii

* n
V i Y ik V k
k 1

 * n

Qi   V i  Y ik V k 
 k 1

22 July 2011
38
Gauss-Seidel solution technique
PV buses
At iteration (r+1):
Q(ir 1)
 (r ) * i1
(r 1)
(r ) * n
(r ) 
 ( V i )  Y ik V k  ( V i )  Y ik V k 


k 1
k i
i  angleV i 
(ir 1)
22 July 2011
(r 1)
Ai
Pi  jQ(ir 1)

Y ii
n
 A (r 1) i1
(r 1)
(r ) 

i
 angle (r )   Bik V k   Bik V k 
k i1
( V i )* k 1

39
Gauss-Seidel solution technique
PV buses
Beware of limits!
Q(ir 1)  Qi,min

Q(ir 1)  Qi,min
Q(ir 1)  Qi,max

Q(ir 1)  Qi,max
&
&
i becomes PQ
i becomes PQ
(more on this below)
22 July 2011
40
Gauss-Seidel solution technique
PV buses
6) Stop criterion
Vi(r 1)  Vi(r )   ; i  2,...,n
6.1) Slack bus power (after convergence)
P1  jQ1 
22 July 2011
* n
V1 Y1k V k
k 1

41
Gauss-Seidel solution technique
Algorith final calculations
6.2) Compute line currents (after convergence)
ILik  ( V i  V k ) YLik ;
Iik 0  V i Y ik 0
ILki  ( V k  V i ) YLik ;
Iki0  V k Y ki 0
Vi
i
Vk
ILik
YLik
ILki
Sik Iik 0
Iki0 Sik
Y ik 0
22 July 2011
k
Y ki0
42
Gauss-Seidel solution technique
Algorith final calculations
6.2) Compute line complex power (after convergence)
*
*
*
*
*
Ski  V k ( V k  V i )YLik  V k V k Y ki 0 ;
*
*
*
*
*
Sik  V i ( V i  V k )YLik  V i V i Y ik 0 ;
Vi
i
Vk
ILik
YLik
ILki
Sik Iik 0
Iki0 Sik
Y ik 0
22 July 2011
k
Y ki0
43
Gauss-Seidel solution technique
Algorith final calculations
6.3) Compute losses (after convergence)
Sloss , ik  Sik  Ski ;  k, i
Sloss 
 Sloss , ik
 k,i
7)
If no convergence, go to step 4.
22 July 2011
44
Gauss-Seidel solution technique
Algorithm improvement
• Acceleration factor (in order to decrease
the number of iterations):

~
(r )
(r 1)
(r )
Vi(r 1)  V i   V i
 Vi

  1.6 (generally recommended)
22 July 2011
45
Gauss-Seidel
Matlab code
function [Vfinal,angfinal,nite,P,Q,errorplot,tiempo]=Gaussgen(m,n,Ybus,Vmodini,Angini,P,Q,tol,Vmax,Vmin,Qmax,Qmin)
%-------------------------------------------------------------------------------------------------------------------------------------------------%-function [Vmod,ang,nite,P,Qerrorplot,tiempo]=Gaussgen(m,n,Ybus,Vmodini,Angini,P,Q,tol,Vmax,Vmin,Qmax,Qmin)
%-Resuelve de forma general problema de carga por el m´etodo de Gauss-Seidel
%-donde:
%-2...m nudos PV; (m=1 cuando no hay nudos PV)
%-n nudos totales
%-Ybus matriz de admitancias
%-Vmodini tensiones iniciales modulo
%-Angini angulos iniciales RADIANES
%-P potencia activa inicial
%-Q potencia reactiva inicial
%-tol tolerancia para error en tension y potencia reactiva
%-Vmax Vmin valores limites aceptables para las tensiones
%-Qmax Qmin valores limites aceptables para las potencias reactivas
%-nite n´umero de iteraciones
%-Vfinal vector con todas las potencias para cada iteraci´on
%-angfinal igual pero con los ´angulos
%-tiempo, tiempo invertido en hacer las operaciones
%--------------------------------------------------------------------------------------------------------------------------------------------------%calculo de la matriz B
B=zeros(n,n);
for t=1:n
for k=1:n
B(t,k)=Ybus(t,k)/Ybus(t,t);
end
end
22 July 2011
46
Gauss-Seidel
Matlab code 2
%calculos los valores en coordenadas cuadrangulares de la tension
V=zeros(n,1);
for a=1:n
V(a)=Vmodini(a)*exp(i*Angini(a));
end
ang=Angini;
%empieza el bucle:
error=1; %valores iniciales para poder entrar en el bucle
errorQ=1;
nite=0;
tic;
while max(abs(error))>tol | max(abs(errorQ))>tol
nite=nite+1;
Vmod=abs(V);
Vini=V;
ang=angle(V);
Qini=Q;
%calculo las reactivas para los nudos PV
if m>1
for l=2:m
AQ=0;
for k=1:n
AQ=AQ+Ybus(l,k)*V(k);
end
Q(l)=-imag((V(l)')*AQ);
end
22 July 2011
47
%calculo las A para todos los nudos:
for a=1:n
A(a)=(P(a)-i*Q(a))/Ybus(a,a);
end
%calculo los angulos nudos PV:
22 July 2011
for l=2:m
Aang=0;
for k=1:n
Aang=Aang+B(l,k)*V(k);
end
ang(l)=angle(A(l)/((V(2))')-Aang+B(l,l)*V(l));
end
end
for a=1:n
A(a)=(P(a)-i*Q(a))/Ybus(a,a);
end
%ahora actualizo los voltajes otra vez:
for a=1:n
V(a)=Vmod(a)*exp(i*ang(a));
end
%ahora calculo los nudos PQ
AV=zeros(n,1);
for p=m+1:n
for k=1:n
AV(p)=AV(p)+B(p,k)*V(k);
end
V(p)=A(p)/((V(p))')-AV(p)+B(p,p)*V(p);
end
error=Vini-V;
errorQ=Qini-Q;
errorplot(1,nite)=norm(abs(error));
Vfinal(:,nite)=abs(V);
angfinal(:,nite)=ang*180/pi; %paso a grados
end
Gauss-Seidel
Matlab code 3
48
Gauss-Seidel
Matlab code 4
%calculos los valores de potencia para el nudo slack:
S1=0;
for t=1:n
S1=S1+(V(1)')*(Ybus(1,t)*V(t));
end
P(1)=real(S1);
Q(1)=-imag(S1);
tiempo=toc;
%alerta por si se sobrepasan valores aceptables:
if max(Vmod)>Vmax | min(Vmod)<Vmin
disp('¡¡SE SOBREPASAN LIMITES TENSIONES!!')
end
if max(Q)>Qmax | min(Q)<Qmin
disp('¡¡SE SOBREPASAN LIMITES REACTIVA!!')
end
ang=(180/pi)*angle(V);
Vmo=abs(V);
%represento los errores por iteraci´on:
plot(1:nite,errorplot,'o');grid on;xlabel('iteraci´on');ylabel('error por iteraci´on');title('evoluci´on error tesi´on');
%------------------------------------------------------------%
%
%
% Realizado por Carlos Ruiz Mora octubre 2006 %
%
%
%------------------------------------------------------------%
22 July 2011
49
Gauss-Seidel example
22 July 2011
50
G-S Example
Solution tolerance is set to 0.1 MVA.
ONE
TWO
THREE
22 July 2011
51
G-S Example
Data below. Base power is SB=100 MVA:
Bus
Voltage (p.u.)
Power
Lin
1
1.02
(slack)
1-2
0.02+0.04j
2
1.02
PG=50 MW
1-3
0.02+0.06j
3
-
PC=100 MW
QC=60 MVAr
2-3
0.02+0.04j
(each)
22 July 2011
Impedance (p.u.)
52
Solution procedure
1.
Data and unknown:
Bus
Type
Data
Unknown
1
Slack
V1=1.02 1=0.0
P1 Q1
2
PV
V2=1.02 P2=0.5
δ 2 Q2
3
PQ
P3=-1.0 Q2=-0.6
δ3 V3
22 July 2011
53
Solution procedure
YBUS
ONE
TWO
YBUS
 15  35 j  10  20 j  5  15 j 
  10  20 j 30  60 j  20  40 j


  5  15 j  20  40 j 25  55 j 
THREE
22 July 2011
54
Solution procedure
3. Voltage magnitude initialization (iteration 0):
V3  V30  1
2  02  0
PQ bus
All buses but the reference one
3  03  0
Vector form:
1.02
V 0  1.02


 1 
22 July 2011
0
0  0
 
0
55
Solution procedure
Per iteration:
- PV buses
(Q, δ)
i=2,...,m
(bus 2)
- PQ buses
(V, δ)
i=m+1,...,n
(bus 3)
- Stopping criterios:
a) convergence  Sslack and power flows;
b) no converge  new iteration
22 July 2011
56
Solution procedure
4. PV buses: iteration (r+1) :
bus 2:

Q(2r 1)   ( V2(r ) )* Y21V1( r 1)  ( V2(r ) )* Y22 V2( r )  ( V2(r ) )* Y23 V3( r )
2  angle [ V2 ]
22 July 2011
A (2r 1)
P2  jQ(2r 1)

Y22
57

Solution procedure
(2r 1)
 A (2r 1)
( r 1)
(r ) 
 angle (r 1) *  B21V1  B23 V3 
( V2 )

where
Bik 
22 July 2011
Yik
is a constant
Yii
58
Solution procedure
5. Buses PQ iteration (r+1):
bus 3:
V3(r 1)
A3
 (r ) *  B31V1(r 1)  B32 V2(r 1)
( V3 )
where
P3  jQ3
A3 
Y33
Are constants for PQ buses
Yik
Bik 
Yii
22 July 2011
59
Solution procedure
6.
Stopping criterion:
  10 3
| Vi(r 1)  Vi(r ) |
&

i  2, 3; j  2
| Q(jr 1)  Q(jr ) |
22 July 2011
60
Solution procedure
If convergence:
6.1) Slack power
*
Sslack
 P1  jQ1  V1* ( Y11V1  Y12 V2  Y13 V3 )
S12 , S21, S13 , S23 , S31, S32
6.2) Power flows 
*
Sik  Vi ( Vi - V *k )YLik
*
22 July 2011
61
Solution procedure
6.3) Line losses:
Slosss ,ik  Sik  Ski
k, i  1,2,3
Sloss  Sloss ,12  Sloss ,13  Sloss ,23
7.
If no coveergence, the procedure continues in Step 4.
22 July 2011
62
Implementation
MATLAB:
22 July 2011
63
Solution
11 iterations needed to attain the solution
Iteration (pu)
1
2
3
...
10
11
P1 ,Q1(slack)
-
-
-
...
-
0.5083
0.0716
P2
0.5
0.5
0.5
...
0.5
0.5
Q2
0.81
0.4084
0.4696
...
0.5493
0.5501
P3
-1.0
-1.0
-1.0
...
-1.0
-1.0
Q3
-0.6
-0.6
-0.6
...
-0.6
-0.6
22 July 2011
64
Solution
Iteration
1
2
3
...
10
11
V1
1 (º )
1.02
1.02
1.02
...
1.02
1.02
0
0
0
...
0
0
V2
2 (º )
1.02
1.02
1.02
...
1.02
1.02
0.0675
-0.1596
-0.2885
...
-0.4667
-0.4685
V3
1.0041
1.0042
1.0043
...
1.0043
1.0043
3 (º )
-0.5746
-0.7336
-0.8278
...
-0.9580
-0.9593
22 July 2011
65
Solution
Errors for |V| & |Q|:
Iteration
1
2
3
...
10
11
Max. Error V
0.0041
9.68·10-5
5.05·10-5
...
1.15·10-6
6.74·10-7
Max. Error Q
0.8160
0.4076
0.0612
...
0.0014
8.11·10-4
22 July 2011
66
Final Solution
Bus
P (MW)
Q (MVAr )
V (pu)
1.02  0º
1
50.83
7.16
2
50.00
55.01
3
-100
-60.00
22 July 2011
1.02
-0.4685º
1.0043  -0.9593º
67
Checking
Checking using Power-World:
ONE
TWO
THREE
22 July 2011
68
Checking
Bus 1
Bus 2
Bus 3
Solution
G-S
PW
G-S
PW
G-S
PW
V (pu)
1.02
1.02
1.02
1.02
1.0043
1.0043
δ(º)
0
0
-0.4685
-0.4710
-0.9593
-0.9612
P (MW)
50.83
50.98
50.00
50.00
-100
-100
Q (MVAr)
7.16
7.10
55.01
55.12
-60
-60
22 July 2011
69
Checking
Variable
V
δ
P
Q
Error Max. (%)
0%
0.53 %
0.29%
0.84 %
22 July 2011
70
4. The Newton-Raphson solution
technique
22 July 2011
71
The Newton-Raphson solution technique
Introduction
• One variable
f ( x )  0,
f ( x ( 0 ) )  0,
f ( x (0 )  x ( 0 ) )  0
(0)
(0)
(0)
( 0 )  df 
f ( x  x )  0  f ( x )  x  
 dx 
x ( 0 )  
f ( x(0) )
 df 
 
 dx 
x (r 1)  x (r ) 
22 July 2011
(0)
 ...
x (1)  x ( 0 )  x ( 0 )
;
(0)
f ( x (r ) )
 df 
 
 dx 
(r )
72
The Newton-Raphson solution technique
Introduction
• Example
f ( x )  x 2  5 x  4  0;
x
( r 1)
x
(r )
df ( x )
 2x  5
dx
( x ( r ) )2  5 x ( r )  4 ( 0 )

; x  0 .5
(r )
2x  5
x
(1)
0 .5 2  5  0 . 5  4
 0 .5 
 0.9375
2  0 .5  5
x
( 2)
0.9375 2  5  0.9375  4
 0.9375 
 0.9988
2  0.9375  5
x
(3)
0.9988 2  5  0.9988  4
 0.9988 
 1.0000
2  0.9988  5
Fast!
22 July 2011
73
The Newton-Raphson solution technique
General formulation
• General case
f ( x )  0 : Rn  Rn
f ( x ( 0 ) )  0,
f ( x ( 0 )  x ( 0 ) )  0
f ( x ( 0 )  x )  f ( x ( 0 ) )  J( 0 )  x ( 0 )  0
x
(0)
 
( 0 ) 1
J
f ( x(0) )
x (1)  x ( 0 )  x ( 0 )
x
(r 1)
22 July 2011
x
(r )
 
(r ) 1
J
f ( x (r ) )
74
The Newton-Raphson solution technique
General formulation
 x1 
x 
x   2
 
x 
 n
22 July 2011
 f1 
f 
f   2

f 
 n
 f1
 x
 1
 f2
J   x1

 fn

 x1
f1
x 2
f2
x 2

fn
x 2




f1 
x n 

f2 
x n 

fn 

x n 
75
The Newton-Raphson solution technique
Load flow case
x 
(r )
 (2r ) 
 ( r ) 
 3 
n
  
(r )  fiP ()  Vi   Vk (Gik cos ik  Bik sinik )
n
k 1
  (r ) 
n
 Vm1 

 fiQ ()  Vi   Vk (Gik sinik  Bik cos ik )
k 1
  
 Vn(r )1 
 (r ) 
 Vn 
Pi ()  fiP, Pi  Pi,sp
Qi ()  fiQ, Qi  Qi,sp
 2 ,..., n ; Vm1,..., Vn
22 July 2011
76
The Newton-Raphson solution technique
Load flow case
Using Taylor:
(r )
(r )
(r )
(r )
(r )
(r )
(r )
(r )
Pi,sp
 f 
 f 
 f 
 f 
 fiP(r )   iP  (2r )  ...   iP  (nr )   iP  Vm(r)1  ...   iP  Vn(r )
  2 
 n 
 Vm1 
 Vn 
Qi,sp
 f 
 f 
 f 
 f 
 fiQ(r )   iQ  (2r )  ...   iQ  (nr )   iQ  Vm(r)1  ...   iQ  Vn(r )
  2 
 n 
 Vm1 
 Vn 
The increments below should be 0:
Pi(r )  Pi,sp  fiP(r ) ;
22 July 2011
Q(ir )  Qi,sp  fiQ(r )
77
The Newton-Raphson solution technique
Load flow case
Matrix notation:
 P2(r )    f  (r )

   2P 
      2 
 P2(r )   ...
 (r )   
Qm1   ... (r )
     fnQ 

 

(r )


 Qn    2 
22 July 2011
(r )





2
 f2P 

   

 Vn   
(r ) 




...
n


(r ) 
...  Vm1 
(r )
 fnQ     

   (r ) 
 n    Vn 
(r )
... ...
... ...
... ...
... ...
78
The Newton-Raphson solution technique
Jacobian matrix
• PQ buses generate 2 Jacobian rows
corresponding to ΔP and ΔQ
• PV buses generate 1 Jacobian row
corresponding to ΔP
22 July 2011
79
The Newton-Raphson solution technique
Jacobian elements
• Jacobian dimension
2  NPQ  NPV
22 July 2011
NPQ
Number of PQ buses
NPV
Number of PV buses
80
The Newton-Raphson solution technique
Jacobian elements
• Jacobian dimension
P(r )  H(r ) N(r )   (r 1) 

 V (r 1)
 (r )    (r )
(r ) 
L  
Q   J
V (r ) 
V (r 1)
(r 1)
instead
of

V
V (r )
for improving computational efficiency
22 July 2011
81
The Newton-Raphson solution technique
Jacobian elements
 m  k
Hkm
Pk

 Vk Vm (Gkm sinkm  Bkm cos km )
m
Nkm
Pk
 Vm
 Vk Vm (Gkm cos km  Bkm sinkm )
Vm
Jkm
Qk

  Vk Vm (Gkm cos km  Bkm sinkm )
m
Lkm
Qk
 Vm
 Vk Vm (Gkm sinkm  Bkm cos km )
Vm
22 July 2011
82
The Newton-Raphson solution technique
Jacobian elements
k  m
Hkk
Pk

 Qk  Bkk Vk2
k
Nkk  Vk
Pk
 Pk  Gkk Vk2
Vk
Jkk
Qk

 Pk  Gkk Vk2
k
Lkk
Qk
 Vk
 Qk  Bkk Vk2
Vk
22 July 2011
 Note :
km  k  m
Y km  Gkm  jBkm
83
The Newton-Raphson solution technique
Solution outline
1. Build YBUS
2. Specify 1  0
Pi
i  2,...,n
Qi
i  m  1,...,n
Vi
i  2,...,m
V1
3.
i , i  2,...,n
Initialize 
 V i , i  m  1,...,n
22 July 2011
84
The Newton-Raphson solution technique
Solution outline
4. Compute Pi(r ) (for PV & PQ Buses), Q(ir ) (for PQ Buses)
5. If Pi(r )  P & Q(ir )  Q
then compute
1) P1  jQ1

2) line flows
3) Stop

else go on
6. Compute submatrices H(r ), N(r ), J(r ), L(r )
22 July 2011
85
The Newton-Raphson solution technique
Solution outline
7. Solve
 (r 1)  H(r ) N(r )  1  P(r ) 
V (r 1)
   (r )
(r )  
(r ) 
L  Q 

V (r )   J
8. Update
(r 1)  (r )  (r 1)
V (r 1)  V (r )  V (r 1)
PV & PQ Buses
PV Buses
9. Go to step 4
22 July 2011
86
The Newton-Raphson, Matlab Code
function [V_modulo, V_fase, P, Q, N_iter, T_calculo, Error_p] = nraphson(V_modulo_ini, V_fase_ini, P_ini, Q_ini, Y, npv)
%
% [V_modulo, V_fase, P, Q, N_iter, T_calculo, Error_p] = nraphson(V_modulo_ini, V_fase_ini, P_ini, Q_ini, Y, npv)
%
% Obtencion de indices de nodos: 1
-> SLACK;
%
2,...,M -> PV;
%
M+1,...,N -> PQ
n = length(V_modulo_ini);
m = npv + 1;
% Parametros del Metodo
Tol = 0.0001;
Error_p = 1;
N_iter = 0;
% Tolerancia del metodo (Perdida de potencia).
% Error inicial (A un valor mayor que Tol).
% Numero de iteraciones.
% Valores iniciales
V_modulo= V_modulo_ini';
V_fase = V_fase_ini';
P = P_ini';
Q = Q_ini';
j=sqrt(-1);
22 July 2011
87
The Newton-Raphson, Matlab Code
tic;
while (Error_p > Tol)
if (N_iter >50)
error('Demasiadas iteraciones');
break
end
V = V_modulo.*exp(j*V_fase);
S = V.*conj(Y*V);
DP = P(2:n)-real(S(2:n));
DQ = Q(m+1:n)-imag(S(m+1:n));
PQ = [DP ; DQ];
Error_p = norm(PQ,2);
% Bucle principal (Se permiten 50 iteraciones como mucho).
% Expresion compleja de la tension
% Expresion compleja de la potencia
% Incremento de potencia activa (nudos PV y PQ)
% Incremento de potencia reactiva (nudos PQ)
% Error en esa iteracion
DS_DA = diag(V)*conj(Y*j*diag(V)) + diag(conj(Y*V))*j*diag(V);
DS_DV = diag(V)*conj(Y*diag(V./V_modulo)) + diag(conj(Y*V))*diag(V./V_modulo);
% Construccion del Jacobiano
J = [real(DS_DA(2:n , 2:n)) real(DS_DV(2:n , m+1:n))
imag(DS_DA(m+1:n , 2:n)) imag(DS_DV(m+1:n , m+1:n))]
dx=J\PQ;
22 July 2011
% indices: 1...n-1 = fases en PV y PQ; n...final = modulos en PQ
88
The Newton-Raphson, Matlab Code
V_fase (2:n) = V_fase(2:n) + dx(1:n-1);
% Actualizamos la fase de las tensiones (nudos PV y PQ)
V_modulo (m+1:n)= V_modulo(m+1:n) + dx(n:end); % Actualizamos el modulo de las tensiones (nudos PQ)
N_iter = N_iter + 1;
disp('Pulse una tecla para continuar')
pause
end
P=real(S);
Q=imag(S);
V_fase=V_fase*180/pi;
T_calculo=toc;
%
%
%
%
%
%
%
% Incremento el numero de iteraciones
% Calculo de la potencia activa
% Calculo de la potencia reactiva
% Paso de Radianes a grados
**** ENTRADAS ****
V_modulo_ini = Modulo de la tension para comenzar a iterar (conocidos en SLACK y PV).
V_fase_ini = Fase de la tension para comenzar a iterar (conocido en SLACK).
P_ini
= Potencia activa en los nodos (conocido en PV y PQ).
Q_ini
= Potencia reactica en los nodos (conocido en PQ).
Y
= Matriz de admitancias nodales.
22 July 2011
89
The Newton-Raphson, Matlab Code
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
**** SALIDAS ****
V_modulo = Modulo de la tension en todos los nodos.
V_fase
= Fase de la tension en todos los nodos.
P
= Potencia activa en todos los nodos.
Q
= Potencia reactiva en todos los nodos.
N_iter
= Numero de iteraciones.
T_calculo = Tiempo de calculo.
Error_p
= Error.
**** OBSERVACIONES ****
Los nodos deben estar ordenador asi:
*1
: SLACK.
* 2...m : PV.
* m+1...n : PQ.
**** FECHA Y AUTOR ****
Laura Laguna - Octubre de 2005
22 July 2011
90
Ejemplo resuelto por el método
de Newton-Raphson
22 July 2011
91
Newton-Raphson Example
• Checking with Matlab and PowerWorld
22 July 2011
92
Newton-Raphson Example
ONE
Bus
Voltage p.u
Power
1
1.02
-
TWO
THREE
2
1.02
PG=50 MW
3
-
PC= 100 MW
QC=60 MVAr
22 July 2011
93
Newton-Raphson Example
Data:
22 July 2011
Line
Impedance p.u.
1-2
0.02+0.04j
1-3
0.02+0.06j
2-3
0.02+0.04j each
94
Newton-Raphson Example
•
3 buses:
• Bus1: Slack
• Bus 2: PV
• Bus 3: PQ
•
Voltage magnitude at bus 3 initialized at 1.02.
•
Angles intitialized to zero.
22 July 2011
95
Newton-Raphson Example
Y bus:
 15.0000 - 35.0000j

Y   - 10.0000 + 20.0000j
 - 5.0000 + 15.0000j

- 10.0000 + 20.0000j
30.0000 - 60.0000j
- 20.0000 + 40.0000j
- 5.0000 + 15.0000j 

- 20.0000 + 40.0000j 
25.0000 - 55.0000j 
Intitialization:
 2  3  0
V3  1.02
22 July 2011
96
Newton-Raphson Example
Residuals:
3
Pi  Piesp - Vi ∑Vj (Gij cos ij  Bijsen(ij )) i  2,3
j1
3
Qi  Qiesp - Vi ∑Vj (Gijsenij - Bij cos(ij )) i  3
j1
22 July 2011
97
Newton-Raphson Example
Checking:
cal
P2cal  0.3624·1014 
 P2   0.5  P2   0.5 
 cal  




 1 
cal 
P

0


P


1

P

3
 3  

 
 3 

cal

15
cal
Q3  7.2475·10 
Q3   0.6  Q3   0.6

 

No tolerance satisfied: the process continues.
22 July 2011
98
Newton-Raphson Example
Jacobian:
 H22 H23 N23 


J   H32 H33 N33 
M

M
L
33
33 
 32
22 July 2011
 62.4240

J    41.6160
 20.8080

 41.6160
57.2220
 26.0100
 20.4000 

25.5000 
56.1000 
99
Newton-Raphson Example
First iteration:


  2  - 0.4557
 3   - 0.9406

V
 
 3  - 0.0154
V3 

22 July 2011
1.0200


V  1.0200 ;
1.0046


0




   - 0.4557
 - 0.9406


100
Ejemplo por Newton-Raphson
Residuals:
P2cal   0.4958 
 P2   0.5  0.4958   0.0042 
 cal  
   P     1  ( 0.9835)    0.0165
P


0
.
9835
 3  

 3 
 

cal
Q3   0.5874
Q3   0.6  ( 0.5874)  0.0126


No convergence.
22 July 2011
101
Newton-Raphson Example
Jacobian for iteration 2:
 61.8860

J    40.8145
 20.8409

22 July 2011
 41.1614  20.0540 

56.0994
24.1371 
 26.2162 54.6707 
102
Newton-Raphson Example
State variables at iteration 2


 2   - 0.0154 
 3    - 0.0206 

V
 
-4
 3  - 3.0024·10 
V3 

22 July 2011
1.0200 


V  1.0200 ;
1.0043 


0




   - 0.4710 
 - 0.9612 


103
Newton-Raphson Example
Residuals:
5
P2cal   0.5000 
 P2   0.5  0.5000   0.0664·10 
 cal  






5 
P3     1.0000   P3     1  ( 1.0000)     0.5454·10 
6 
Qcal
  0.6000






Q

0
.
6

(

0
.
6000
)

4
.
9698
·
10
3
3

 
 



Tolerance OK.
22 July 2011
104
Newton-Raphson Example
Jacobian iteration 3:
 61.8860

J    40.8145
 20.8409

 41.1614  20.0540 

56.0994
24.1371 
 26.2162 54.6707 


-5




0.6458·10
2 

 3   - 0.7534·10-5 

V
 
-7 

3
1.1106·10

 

V
3

22 July 2011
1.0200 


V  1.0200 ;
1.0043 


0




   - 0.4710 
 - 0.96810 


105
Newton-Raphson Example
Final power values:
P1   0.5098 
P    0.5000 
 2 

P3   1.0000
22 July 2011
 Q1   0.0710 
Q    0.5513 
 2 

Q3   0.6000
106
Newton-Raphson Example
convergence
22 July 2011
107
Newton-Raphson Example
convergence
22 July 2011
108
Newton-Raphson Example
convergence
22 July 2011
109
Newton-Raphson Example
convergence
22 July 2011
110
Power World
5 0 ,9 7 7 M W
7 ,0 9 5 M var
5 0 ,0 0 0 M W
5 5 ,1 2 6 M var
UNO
C ERO
1 ,0 2 0 pu
0 ,0 0 0 Deg
1 ,0 2 0 pu
-0 ,4 7 1 Deg
DO S
1 ,0 0 4 3 pu
-0 ,9 6 1 Deg
100 MW
6 0 M var
22 July 2011
111
Newton-Raphson Example
Bus
V(p.u)
δ
P
Q
0 (PW)
1.02
0
50.9763
7.0955
0
1.02
0
50.9755
7.0954
1 (PW)
1.02
-0.4710
50
55.1256
1
1.02
-0.4710
50.0006
55.1228
2 (PW)
1.0043
-0.9612
-100
-60
2
1.0043
-0.9612
-99.9989
-59.9970
22 July 2011
112
Newton-Raphson Example
• Largest error below 0.1 MVA.
• More effective technique than Gauss Seidel.
• Convergence is fast (if adequate initialization).
22 July 2011
113
Including tap-changing transformers
22 July 2011
114
Tap-Changing transformer
• A tap changing transformer makes the
admitance matrix dependent on the
transformer parameter t.
• The Jacobian matrix also depends on t.
22 July 2011
115
Admitance matrix
Equivalent circuit:
t:1
i
Ycc
Ycc
i
j
t 1
22 July 2011
1 t
Ycc 2
t
t
j
t 1
Ycc
t
116
Admitance matrix
•
General building rules
1. Self admittance of node i, Yii , equals the
algebraic sum of all the admittances
connected to node i
2. Mutual admittance between nodes i and
k, Yii , equals the negative of the sum of
all admittances connecting nodes i and k
3. Yik  Yki
22 July 2011
117
Tap-changing
Example
22 July 2011
118
Tap-changing
Example
1
Yshunt 13
Z13
Z 23
3
2
Yshunt 13 Yshunt 23
Yshunt 23
Ycc
The tap-changing
transformer
controls voltage
of bus 4
22 July 2011
1 t
Ycc 2
t
t
4
t 1
Ycc
t
119
Tap-changing
Example
1
Y11 
 Yshunt 13 ; Y12  0 ;...
Z13
... Y33
1
1
1  t Ycc

 Yshunt 13 
 Yshunt 23  Ycc 2 
t
Z13
Z 23
t
Y34  
Y44 
Ycc
Ycc
22 July 2011
t
t 1
 Ycc
 Ycc
t
t
It does not depend
on t!
120
Tap-changing
Example
 Y11 Y12
Y
Y22
21
Y
 Y31 Y32
Y
 41 Y42
22 July 2011
Y14 
Y23
Y24 

Y33 ( t ) Y34 ( t )

Y43 ( t ) Y44 
Y13
121
Tap-changing transformer
Load flow equations:
n
Pi  Vi   Vk Gik cos ik  Bik sin ik 
k 1
n
Qi  Vi   Vk Gik sin ik  Bik cos ik 
k 1
22 July 2011
122
Tap-changing transformer
Taylor Expansion:
(r )
Pi
dato
f
(r )
iP
(r )
(r )
(r )
(r )
 fiP 
 fiP 
 fiP 
(r )
(r )
(r )




...





V
2
n
m 1  ...





  2 
 n 
 Vm1 
(r )
 fiP 
 fiP 
(r )
(r )
...  

V


t
n

 t 

V


 n
(r )
(r )
 f 
 f 
 f 
Qidato  fiQ(r )   iQ  (2r )  ...   iQ  (nr )   iQ  Vm(r)1  ...
  2 
 n 
 Vm1 
(r )
(r )
 fiQ 
 fiQ 
(r )
(r )
...  

V


t
n

 t 

V


 n
22 July 2011
123
Tap-changing transformer
• The admitance matrix depend on t.
• The Jacobian matrix has a new column.
• The new variable (t) replace the voltage value
at the corresponding PQ bus.
22 July 2011
124
Tap-changing transformer
• Increment Δt is divided by t to improve
computational efficiency.
• The Jacobian matrix maintains its # of column
& rows.
• The variable t is considered in the last place.
22 July 2011
125
Tap-changing transformer
The Jacobian matrix blocks are:
H(r ) N(r ) C(r ) 
J   (r )
(r )
(r ) 
L
D 
M
H, N, M & L are the standard blocks.
Submatrix C & D have as many columns as
the # of tap-changing transformers.
22 July 2011
126
Tap-changing transformer
Derivatives with respect to t:
n
fiP
Bik
 Gik

Ci  t 
 t  Vi  ∑Vk 
cos ik 
sinik 
t
t
 t

k 1
n
fiQ
 Gik
Di  t 
 t  Vi  ∑Vk 
sinik
t
 t
k 1
Bik

cos ik 
t

C & D multiplied by t!
22 July 2011
127
Tap-changing transformer
Iterative procedure analogous, but:
• If t hit any of its limit, it is fixed to the limit &
the corresponding bus becomes PQ.
• If the procedure converge, fix t at its closest
integer value & continues the iteration with
that t fixed.
22 July 2011
128
Tap-changing
Example
1.- Slack
2.- PV
3.- PQ
4.- PQ
5.- PQV
22 July 2011
129
Tap-changing
Example
Data:
Line impedances: 0.001+j0.05 p.u.
Shunt admitances: j0.05 p.u. (2 per line).
Transformer: 0.9<t<1.1, steps of 0.005; Zcc=j0.1.
22 July 2011
130
Tap-changing
Example
Bus
Voltage (pu).
Power
1
1.00
Slack
2
1.00
Pg=150MW
3
--
Pc=50MW, Qc=10MVAr
4
--
Pc=0MW,
5
1.00
22 July 2011
Qc=0MVAr
Pc=100MW,Qc=50MVAr
131
Tap-changing
Example
1
Ylínea
Yshunt
Yshunt
2
Ylínea
Yshunt
Ylínea
4
Yshunt
Yshunt
Yshunt
Ycc
1 t
Ycc 2
t
22 July 2011
3
5
t
t 1
Ycc
t
132
Tap-changing
Example
Admitance matrix:
YBUS
 Yl  Ys
 Y
l

 0

 0

 0

22 July 2011
Yl
0
0
2Yl  2Ys
0
Yl
0
Yl  Ys
Yl
Yl
0
0
Yl
Ycc
(1 t )
2Yl  2Ys 
 Ycc 2
t
t
Ycc
t
0 
0 

0 
Ycc 

t 
Ycc 

133
Tap-changing
Example
0
0
0 
 0.3998  j19.942  0.3998  j19.992
 0.3998  j19.992 0.7997  j39.884

0

0
.
3998

j
19
.
992
0


Ybus  
0
0
0.3998  j19.942  0.3998  j19.992 0 


0

0
.
3998

j
19
.
992

0
.
3998

j
19
.
992
0
.
7997

j
49
.
884
j
10



0
0
0
j10
 j10
Variable initial values:
2  3   4  5  0;
V3  V4  1;
t  1;
22 July 2011
134
Tap-changing
Example
Increment calculations:
 P2   1.5  P2cal   1.5 
 P  
  0 .5 
cal 

0
.
5


P
3 
 3 


 P4    P4cal   0 

 


cal 



P
 5    1  P5    1 
 Q3    0.1  Qcal    0.1
3
 

 

cal
  0 
Q 4    Q 4
 Q   0.5  Qcal   0.5


 5 
5 
22 July 2011
135
Tap-changing
Example
Jacobian
0
 39.984

0
19.992

 19.992  19.992
J
0
0


0
 0.3998

0.3998
 0.3998

0
0
22 July 2011
 19.992
0
0
 0.3998
 19.992
0
0.3998
 0.3998
49.984
 10  0.3998
0.7997
 10
10
0
0
0.3998
0
19.892
 19.992
 0.7997
0
 19.992
49.754
0
0
0
 10
0 
0 

0 
0 
0 

 10 
10 
136
Tap-changing
Example
First iteration:
  2 
  
0

 3 
1 
0
  4   0.0993 





  
 1 



0
.
0744
0


 5






 V3    0.1744  V  0.9623 ;    0.0993 ; t  0.918

 V3  
 0.968 
 0.0744 



0
.
0377
 V 






4


 1 
 0.1744 
 0.032 

 V4 
 t    0.082 
 t 
22 July 2011
137
Tap-changing
Example
Admitance matrix:
Ybus
0
0
0
 0.3998  j19.942  0.3998  j19.99

 0.3998  j19.992 0.7997  j39.884

0
 0.3998  j19.992
0



0
0
0.3998  j19.94  0.3998  j19.992
0



0

0
.
3998

j
19
.
992

0
.
399

j
19
.
99
0
.
7997

j
51
.
750
j
10
.
8933



0
0
0
j10.893
 j10 
22 July 2011
138
Tap-changing
Example
Power computation:
0
 P1  

P   1.4521 
 2 

P3    0.4657 
P  

0
.
069
 4 

P5    1.0527 
22 July 2011
 Q1    0.05 
Q   0.5647 
 2 

Q3    0.1407 
Q  

0
.
1027
 4 

Q5    0.492 
139
Tap-changing
Example
Power increments:
 P2   1.5  1.4521   0.0479 
 P   0.5  0.4657   0.0343 
 3 
 

 P4    0.069
   0.069 
 P   
   0.0527 

1

1
.
0527
 5 
 

 Q3    0.1  0.1407   0.0407 

 
 

Q 4    0.1027   0.1027 
 Q5    0.5  0.492    0.008 
22 July 2011
140
Tap-changing
Example
Jacobian matrix:
 39.9193

0

 19.2698
J
0


0

 1.8242

0
22 July 2011
0
 19.3273
0
0
1.0523
18.6075
 18.6075
0
 0.0954
 0.8359
 18.6261
48.3879
 10.492
0.0912
0.8183
0
 10.492
10.492
0
 1.0527
 0.8359
0.8359
0
18.3261
 18.6075
 0.0912
 0.6803
 1.0527
 18.6261
48.5934
0
1.0527
 1.0527
0
 10.492


0

 1.0527 
1.0527 

0

 11.746 
10.492 
0
141
Tap-changing
Example
Second iteration
  2 
  
 0.0002 
 3 
0
  4    0.005 
 1 




  
 1 
  0.0002 

0
.
003


 5







V
 3   0.0021  V  0.9607 ;    0.1043 ; t  0.9144

 V3  
0.9644 
 0.0744 



0
.
0016
 V 






4


 1 
 0.1723 
 0.0037 

 V4 
 t   0.0039 
 t 
22 July 2011
142
Tap-changing
Example
Admitance matrix:
Ybus
0
0
0
 0.3998  j19.942  0.3998  j19.992

 0.3998  j19.992 0.7997  j39.884

0

0
.
3998

j
19
.
992
0




0
0
0.3998  j19.942  0.3998  j19.992
0


0

0
.
3998

j
19
.
992

0
.
3998

j
19
.
992
0
.
7997

j
51
.
8447
j
10
.
9365



0
0
0
j10.9365
 j10 
22 July 2011
143
Tap-changing
Example
Power calculations:
 P1   0.0031 
P   1.4998 
 2 

P3    0.4998 
P  

0
.
000
 4 

P5    1.0000 
22 July 2011
 Q1    0.0501
Q   0.6391 
 2 

Q3    0.1000 
Q  

0
.
0006
 4 

Q5   0.4999 
144
Tap-changing
Example
Power increments:
 P2   1.5  1.4998   0.2065 
 P   0.5  0.4998   0.1969 
 3 
 

0.000
 P4  
  0.0057 
 P   
   0.0258   10 3

1

1
.
0000
 5 
 

 Q3    0.1  0.1000   0.0339 

 
 


Q

0
.
0006

0
.
6375
 4 
 

 Q5   0.5  0.4999   0.1218 
22 July 2011
145
Tap-changing
Example
Tap to the closest feasible value:
t  0.9144  t  0.915
Admitance matrix:
Ybus
0
0
 0.3998  j19.942  0.3998  j19.992
 0.3998  j19.992 0.7997  j 39.884
0
 0.3998  j19.992


0
0
0.3998  j19.942  0.3998  j19.992

0
 0.3998  j19.992  0.3998  j19.992 0.7997  j51.8282


0
0
0
j10.929
22 July 2011

0 
0 

j10.929
 j10 
0
146
Tap-changing
Example
Power calculation:
 P1   0.0031 
P   1.4998 
 2 

P3    0.4998 
P  

0
.
0007
 4 

P5   0.9993 
22 July 2011
 Q1    0.0501
Q   0.6391 
 2 

Q3    0.1000 
Q  


0
.
0075
 4 

Q5   0.4926 
147
Tap-changing
Example
Power increment calculations:
  2 


 P2   1.5  1.4998   0.0002   3 
 P   0.5  0.4998    0.0002    
 3 
 
  4
 P4    0.0007   0.0007   5 
 P   
   0.0007  ;  V3 

1

0
.
9993
 5 
 
 V 
 Q3    0.1  0.1000   0.0000   3 

 
 
  V4 
0.0075
Q 4  
  0.0075   V4 
 Q5   0.5  0.4926   0.0074   V 
5
V 
 5 
22 July 2011
148
Tap-changing
Example
Jacobian:
 39.2449

0

 19.1935
J
0


0

 1.872

0
22 July 2011
0
 19.253
0
0
1.103
18.5072
 18.5072
0
 0.1307
 0.8689
 18.5271
48.2132
 10.4926
0.1282
0.7431
0
 10.4926
10.4926
0
 0.9993
 0.8689
0.8689
0
18.3071
 18.5072
 0.1282
 0.7445
 0.9993
 18.5271
48.1983
0
0.9993
 0.9993
0
 10.4926


0

0.9993 
 0.9993 

0

 10.4926 
9.5074 
0
149
Tap-changing
Example
Final values
  2 
  
 3   0.0007 
  4   0.0289 
0
 1 




  
 1 
  0.0002 

0
.
0152


5









3

V
 3    0.1699  10  V  0.9607 ;    0.1043 

 V3  
0.9644 
 0.0744 



0
.
0552
 V 




4




 0.9991
 0.1725 
 0.0558 

V
 4 
 V5    0.8522 
V 
 5 
22 July 2011
150
Tap-changing
Example
Final power values:
 P1   0.0031 
P   1.5000 
 2 

P3    0.5000 
P  

0
.
0000
 4 

P5    1.0000 
22 July 2011
 Q1    0.0501
Q   0.6402 
 2 

Q3    0.1000 
Q  

0
.
000
 4 

Q5   0.5000 
151
Power World
22 July 2011
152
5. Fast decoupled AC load flow
22 July 2011
153
Fast decoupled AC load flow
Two simplifications:
– Do not build Jacobian at each iteration (small error
introduced, then, the procedure needs more
iterations to reach the solution)
– Decoupling between P-δ and Q-V (not
recommended in system highly loaded and/or with
low voltage levels)
22 July 2011
154
Fast decoupled AC load flow
( r 1)

 P(r )  H(r ) N(r )   
 (r 1)

 (r )    (r )
( r )  V
L  
Q   J
V (r ) 
Assume:
i)
V i  1 .0
ii) Gik  Bik
i
i, k
iii) cos(i  k )  1.0
sin(i  k )  0.0
iv) Qk  Bkk
22 July 2011
155
Fast decoupled AC load flow
We have:
H  B~1
,
L  B~2 , N  0 , J  0
~

B
 ik  Bik 
~ ~
B1 , B2  ~

Bii  Bii 
  
22 July 2011
 Elements of YBUS
156
Fast decoupled AC load flow
~
( r 1) 
 P(r )  B


0


1
 (r )   
~   (r 1) 
Q   0 B 2  V

 
 

~
P(r )  B1 (r 1)
~
Q(r )  B 2 V (r 1)

(1)
( 2)
     
V     
( r 1)
( r 1)
~ 1
 B1 P(r )
~ 1
 B 2 Q(r )
Use Newton-Raphson Iteration
22 July 2011
157
Fast decoupled load flow. Flow diagram
DATA INPUT
Calculate Delta P
Real
power
coverged
?
YES
NO
Reactive
power
coverged
?
YES
NO
Solve (1) and update angles
Calculate Delta Q
NO
Reactive
power
coverged
?
NO
YES
Real
power
coverged
?
YES
NO
Solve (2) and update V
22 July 2011
OUTPUT RESULTS
158
Fast decoupled load flow. Matlab Code
function [V_modulo, V_fase, P, Q, N_iter, T_calculo, Error_p] = desacoplado(V_modulo_ini, V_fase_ini, P_ini, Q_ini, Y,
npv)
%
% [V_modulo, V_fase, P, Q, N_iter, T_calculo, Error_p] = desacoplado(V_modulo_ini, V_fase_ini, P_ini, Q_ini, Y, npv)
%
% Obtencion de indices de nodos: 1
-> SLACK;
%
2,...,M -> PV;
%
M+1,...,N -> PQ
n = length(V_modulo_ini);
m = npv + 1;
% Parametros del Metodo
Tol = 0.0001;
Error_p = 1;
N_iter = 0;
% Valores iniciales
V_modulo= V_modulo_ini';
V_fase = V_fase_ini';
V
= V_modulo.*exp(j*V_fase);
P
= P_ini';
Q
= Q_ini';
% Tolerancia del metodo (Perdida de potencia).
% Error inicial (A un valor mayor que Tol).
% Numero de iteraciones.
% Expresion compleja de la tension
DS_DA = diag(V)*conj(Y*j*diag(V)) + diag(conj(Y*V))*j*diag(V);
DS_DV = diag(V)*conj(Y*diag(V./V_modulo)) + diag(conj(Y*V))*diag(V./V_modulo);
22 July 2011
159
Fast decoupled load flow. Matlab Code
%J
%
H
L
= [real(DS_DA(2:n , 2:n)) real(DS_DV(2:n , m+1:n))
% Construccion del Jacobiano
imag(DS_DA(m+1:n , 2:n)) imag(DS_DV(m+1:n , m+1:n))];
= imag(-Y(2:end , 2:end))
= imag(-Y(m+1:end,m+1:end))
j=sqrt(-1);
tic;
while (Error_p > Tol)
if (N_iter >50)
error('Demasiadas iteraciones');
break
end
V = V_modulo.*exp(j*V_fase);
S = V.*conj(Y*V);
DP = P(2:n)-real(S(2:n));
DQ = Q(m+1:n)-imag(S(m+1:n));
% Bucle principal (Se permiten 50 iteraciones como mucho).
% Expresion compleja de la tension
% Expresion compleja de la potencia
% Incremento de potencia activa (nudos PV y PQ)
% Incremento de potencia reactiva (nudos PQ)
Dfase = H\DP;
Dmodulo = L\DQ;
PQ = [DP ; DQ];
Error_p = norm(PQ,2);
22 July 2011
% Error en esa iteracion
160
Fast decoupled load flow. Matlab Code
V_fase (2:n) = V_fase(2:n) + Dfase;
% Actualizamos la fase de las tensiones (nudos PV y PQ)
V_modulo (m+1:n)= V_modulo(m+1:n) + Dmodulo; % Actualizamos el modulo de las tensiones (nudos PQ)
N_iter = N_iter + 1;
% Incremento el numero de iteraciones
% disp('Pulse una tecla para continuar')
% pause
end
P=real(S);
Q=imag(S);
V_fase=V_fase*180/pi;
T_calculo=toc;
%
%
%
%
%
%
%
% Calculo de la potencia activa
% Calculo de la potencia reactiva
% Paso de Radianes a grados
**** ENTRADAS ****
V_modulo_ini = Modulo de la tension para comenzar a iterar (conocidos en SLACK y PV).
V_fase_ini
= Fase de la tension para comenzar a iterar (conocido en SLACK).
P_ini
= Potencia activa en los nodos (conocido en PV y PQ).
Q_ini
= Potencia reactica en los nodos (conocido en PQ).
Y
= Matriz de admitancias nodales.
22 July 2011
161
Fast decoupled load flow. Matlab Code
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
**** SALIDAS ****
V_modulo
V_fase
P
Q
N_iter
T_calculo
Error_p
= Modulo de la tension en todos los nodos.
= Fase de la tension en todos los nodos.
= Potencia activa en todos los nodos.
= Potencia reactiva en todos los nodos.
= Numero de iteraciones.
= Tiempo de calculo.
= Error.
**** OBSERVACIONES ****
Los nodos deben estar ordenador asi:
*1
: SLACK.
* 2...m : PV.
* m+1...n : PQ.
**** FECHA Y AUTOR ****
Laura Laguna - Nobiembre de 2005
22 July 2011
162
Fast decoupled load flow: Exaple
• Tolerance 0.1 MVA.
ONE
TWO
THREE
22 July 2011
163
Fast decoupled load flow: Exaple
• Power base SB = 100MVA
22 July 2011
Bus
Voltage p.u.
Power
1
1.02
-
2
1.02
PG=50 MW
3
-
PC=100 MW
QC=60 MVAr
Line
Impedance p.u.
1-2
0.02+0.04j
1-3
0.02+0.06j
2-3
0.02+0.04j
(cada una)
164
Fast decoupled load flow: Exaple
• Admitance matrix
 15  35 j  10  20 j  5  15 j 
Ybus   10  20 j 30  60 j  20  40 j


  5  15 j  20  40 j 25  55 j 
22 July 2011
165
Fast decoupled load flow: Exaple
• Data and unknown:
Bus
Type
Data
Unknown
1
Slack and
reference
V1 = 1.02
δ1=0.0
P1, Q1
2
PV
P2 =0.5
V2 = 1.02
δ2, Q2
3
PQ
P3 = -1.0
Q3 = -0.6
δ3, V3
• Inicialization:
δ2 = δ3 = 0.0 ;
22 July 2011
V3 = 1.02 p.u.
166
Fast decoupled load flow: Exaple
• Flow diagram
Vi(0) , δi(0), Piesp, Qiesp
Compute: Delta P
Compute: Pical, Qical
Convergence P?
Yes
Convergence Q?
Yes
NO
Solve subproblem 1 and update angles
NO
Compute: Delta Q
Convergence Q?
Yes
Convergence P?
Yes
NO
Solve subproblem 2 and update voltages
NO
Final results
22 July 2011
167
Fast decoupled load flow: Exaple
Compute
 60  40 
H
; L  55 
  40 55 
Calculate P and Q:
cal
P2cal  0.3624·1014 
 P2   0.5  P2   0.5 
 cal  




 1 
cal 
P

0


P


1

P

3
 3  

 
 3 

cal

15
cal
Q3  7.2475·10 
Q3   0.6  Q3   0.6

 

No convergence
22 July 2011
168
Fast decoupled load flow: Exaple
• First iteration:


  2  - 0.4213
 3    - 1.3481

V
 
 3  - 0.0109
V3 

22 July 2011
1.0200 


V  1.0200 ;
 1.0091 


0




   - 0.4213 
 - 1.3481 


169
Fast decoupled load flow: Exaple
Residuals:
cal
P2cal   0.7385 
 P2   0.5  P2   0.2385
 cal  
   P     1  Pcal    0.3003 
P


1
.
3003
3
 3  
 

 3 

cal 
Qcal






 0.4583

0
.
1417

Q

0
.
6

Q
3 

 3 
 3  
Error = 0.5976 → no convergence
22 July 2011
170
Fast decoupled load flow: Exaple
• System state (second iteration)


  2  - 0.0372
 3    0.2857 

V
 
 3  - 0.0083
V3 

22 July 2011
1.0200 


V  1.0200 ;
1.0008 


0




   - 0.4585 
 - 1.0264 


171
Fast decoupled load flow: Exaple
•Residuals:
cal
P2cal   0.6578 
 P2   0.5  P2   0.1578
 cal  
   P     1  Pcal    0.1936 
P


1
.
1936
3
 3  
 

 3 

cal
cal
Q3   0.7444
Q3   0.6  Q3   0.1444 


•Error = 0.2885 → no convergence
22 July 2011
172
Fast decoupled load flow: Exaple
• System state (third iteration):


  2  - 0.0315
 3    0.1788 

V
 
 3   0.0026 
V3 

1.0200 


V  1.0200 ;
1.0034 


0




   - 0.4900 
 - 0.8836 


• The process continues.
22 July 2011
173
Fast decoupled load flow: Exaple
•After 9 iterations:
cal
3
P2cal   0.5006 
 P2   0.5  P2   0.5954·10 
 cal  
   P     1  Pcal    0.7006·103 
P


1
.
0007
3
 3  
 


 3 
cal 
3 
Qcal








0
.
5992

Q

0
.
6

Q

8
.
3346
·
10
3 

 3 
 3  


•Error = 0.0012 → convergence attained
22 July 2011
174
Fast decoupled load flow: Exaple
• System state at iteration 10:


-3




0.1592·10
2 

 3    0.6141·10-3 

V
 
-5 

3
1.5154·10

 

V
3

22 July 2011
1.0200 


V  1.0200 ;
1.0043 


0




   - 0.4710 
 - 0.9614 


175
Fast decoupled load flow: Exaple
•Residuals:
cal
3
P2cal   0.5003 
 P2   0.5  P2   0.2863·10 
 cal  





cal 
3 
P3     1.0004   P3     1  P3    0.3516·10 
cal 
4 
Qcal








0
.
6003

Q

0
.
6

Q
3
.
3349
·
10
3 

 3 
 3  


•Error = 5.6285·10-4 → convergence attained
22 July 2011
176
Fast decoupled load flow: Exaple
• State at iteration 11:


-3




0.0567·10
2 

 3    0.3251·10-3 

V
 
-6 

3
6.0634·10

 

V
3

 0.5098 


P   0.5003 
  1.0004 


22 July 2011
1.0200 


V  1.0200 ;
1.0043 


0




   - 0.4711
 - 0.9611


 0.0711 


Q   0.5515 
  0.6003 


177
Fast decoupled load flow: Exaple
convergence
22 July 2011
178
Fast decoupled load flow: Exaple
convergence
22 July 2011
179
Fast decoupled load flow: Exaple
convergence
22 July 2011
180
Fast decoupled load flow: Exaple
convergence
22 July 2011
181
6. Variable limits
22 July 2011
182
Variable limits
Physical considerations
1.
Voltage magnitudes out of limits
a PQ BUS does not meet
max
v min

v

v
i
i
i
Action: Warning!
2.
voltage problem
Flow magnitudes out of limits
A line does not meet
- S max ≤ S ≤ S max
ij
ij
ij
Action: Warning!
22 July 2011
overloading of lines problem
183
Variable limits
Physical Considerations
3.
Reactive Power out of limits
A PV BUS does not meet
Qi
min
 Qi 
Qi
max
Action: Wrong Formulation!
Specified voltage cannot be attained
Formulate the problem properly
22 July 2011
184
Variable limits
Computational Considerations
Changing PV to PQ
no
no
Q hits
limit?
QG>Qmax?
Q=Qmax ?
Yes
>Vdato?
No
<Vdato?
No
22 July 2011
Change to P-V
V=Vdato
Change to
P-Q
Q=Qmax
Yes
Change to
P-Q
Q=Qmin
QG<Qmin?
No
Stay as
P-Q
Yes
V
Yes
No
yes
V
yes
Stay as
P-V
Change to P-V
V=Vdato
Stay as
P-Q
185
7. DC load flow
22 July 2011
186
DC load flow
• Approximate solution.
• Two simplifications:
– In network model: do not consider series
resistences and shunt admittances
– Assume Vi=1 at all buses
22 July 2011
187
DC load flow
Approximate analytical solution
V V
i j
P 
 sin
ij
ij
X
ij
2
V V
V
i j
i
Q 
 cos 
ij
ij
X
X
ij
ij
Assume
i)
Vi ≈ 1.0 ∀i
ii) sinij ≈ ij
iii) cos ij ≈ 1.0
22 July 2011
188
DC load flow
Approximate analytical solution
Pij 
ij
Xij
 Bijij  Biji  Bij j


n
n


Pi   Pij   Bij  i   Bij j
j1
j1
j1


j i
j i
 ji 
P  B´
n
where
22 July 2011
Bii'
n
  Bij
j1
ji
Bij'   Bij
189
DC load flow
Continuation
Qij 
Vi  V j
Xij
 Bij ( V i  V j )  Bij V i  Bij V j
n

n


Qi   Qij   Bij V i   Bij V j
 j1 
j1
j1


ji
ji
 ji 
n
Q  B' V 
22 July 2011
190
DC load flow
Continuation
Solution
P  B' 
Q  B' V 
 1 
     ;
n 
PQ & PV
PQ
Buses
Buses
 V1 
 
V      ;
 Vn 
 
P2 
P     ;
Pn 
Qm1 
Q    
 Qn 
n
 '

B

B
 ii  ij 
1
'
j1
B 
;
B


ij
ji
Xij
 '

Bij  Bij 
 
22 July 2011
191
DC Power flow: example
22 July 2011
192
DC Power flow: example
ONE
TWO
THREE
22 July 2011
193
DC Power flow: example
Data
Bus
Voltage p.u.
Power
0
1.02
-
1
1.02
PG=50MW
2
-
PC=100MW, QC=60MVAr
22 July 2011
Line
Impedance p.u.
0-1
0.02+0.04j
0-2
0.02+0.06j
1-2
0.02+0.04j (both)
194
DC Power flow: example
 41.67
B    25

 16.67
 P0   41.67
 0.5     25

 
 1.0  16.67
 Q0   41.67
 Q     25
 1  
 0.6  16.67
22 July 2011
 25  16.67 
75
 50 

 50 66.67 
 25  16.67   0 
75
 50   1 
 
 50 66.67   2 
 25  16.67  1.02 
75
 50  1.02 


 50 66.67   V2 
195
DC Power flow: example
 P0   41.67  25  16.67 

 

75
 50 
 0.5     25
  1.0    16.67  50 66.67 

 

22 July 2011
0
 
 1 
 
 2
196
DC Power flow: example
Po= -25δ1 - 16.67δ2
0.5= +75δ1 - 50δ2
-1.0= -50δ1 + 66.67δ2
Q0= 41.67*1.02 - 25*1.02 - 16.67V2
Q1= -25*1.02 + 75*1.02 - 50V2
-0.6= -16.67*1.02 - 50*1.02 + 66.67V2
22 July 2011
197
DC Power flow: example
Solution:
22 July 2011
P0
0.5p.u.=50MW
Q0
0.15p.u.=15MVAr
Q1
0.45p.u.=45MVAr
δ1
-1.1459º
δ2
-0.3819º
V2
1.011p.u.
198
DC Power flow: example
PowerWorld comparison:
Var
DC
PowerWorld DC
PowerWorld G-S
P0
0.50p.u.
0.50p.u.
0.509p.u.
Q0
0.15p.u.
0.00p.u.
0.07p.u.
Q1
0.45p.u.
0.00p.u.
0.55p.u.
δ1
-0.3819
-0.3820
-0.47
δ2
-1.1459
-1.1459
-0.96
V2
1.0056p.u.
1.0000p.u.
1.0043p.u.
22 July 2011
199
DC Power flow: example
0,00 MW
0,00 Mvar
50,00 MW
0,00 Mvar
CERO
UNO
DOS
TRES
22 July 2011
100,00 MW
60,00 Mvar
200
DC Power flow: example
Data.
Bus
Voltage p.u.
Power
0
1.02
-
1
1.02
PG=50MW
2
-
PC=0MW, QC=0MVAr
3
-
PC=100MW, QC=60MVAr
22 July 2011
Line
Impedance p.u.
0-1
0.02+0.04j
0-2
0.02+0.06j
1-2
0.02+0.04 (both)
2-3
0.1j
201
DC Power flow: example
0 
 41.67  25  16.67
  25
75
 50
0 

B
 16.67  50 76.67  10 
 0

0

10
10


 P0   41.67
 0.5    25


 0.0   16.67
  1 .0   0

 
 Q0   41.67
 Q    25
 1 
 0.0   16.67
  0 .6   0

 
22 July 2011
 25  16.67
0  0 
75
 50
0   1 
 
 50 76.67  10   2 
 
0
 10
10  3 
 25  16.67
0  1.02
75
 50
0  1.02


 50 76.67  10   V2 


0
 10
10   V3 
202
DC Power flow: example
P0 = -25δ1 -16.67δ2-0*δ3
0.5 = 75δ1 - 50δ2-0*δ3
0.0 = -50δ1 + 76.67δ2-10δ3
-1.0 = -0*δ1 - 10δ2+10δ3
Q0 = 41.67*1.02 - 25*1.02 - 16.67V2 + 0*V3
Q1 = -25*1.02 + 75*1.02 - 50V2 + 0*V3
0.0 = -16.67*1.02 - 50*1.02 + 76.67V2-10*V3
-0.6 = 0*1.02-0*1.02 - 10*V2 + 10*V3
22 July 2011
203
DC Power flow: example
PowerWorld comparison:
Var
DC
PowerWorld G-S
PowerWorld DC
P0
0.50p.u.
0.51p.u.
0.50p.u.
Q0
0.00p.u.
0.11p.u.
0.00p.u.
Q1
0.00p.u.
0.67p.u.
0.00p.u.
δ1
-0.3819º
-0.48º
-0.382º
δ2
-1.1459º
-0.91º
-1.1459º
δ3
-6.8755º
-7.05º
-6.8755º
V2
1.011p.u.
1.002p.u.
1.000p.u.
V3
0.951p.u.
0.932p.u.
1.000p.u.
22 July 2011
204
8. Comparison of load flow
solution methods
22 July 2011
205
Comparison of load flow methods
1. Gauss-Seidel (G-S)
 Simple technique
 Iteration time increases linearly with the
number of buses. Lower iteration time than NR. Seven times faster in large systems
 Linear rate of convergence. Many iterations
required for getting close to the solution
 Number of iteration increases with the
number of buses
22 July 2011
206
Comparison of load flow methods
(Continuation)
2. Newton-Raphson (N-R)







22 July 2011
Widely used
Iteration-time increases linearly with the number of
buses
Quadratic rate of convergency. A few iterations for
getting close to the solution
Number of iterations independent of the number of
buses of the system
The Jacobian is a very sparse matrix
Method non-sensitive to slack bus choice and the
presence of series capacitors
Sensitive to initial solution
207
Comparison of load flow methods
(Continuation)
3. AC decoupled
 B has to be computed and factorized only once
It requires more iterations than NewtonRaphson method
Iteration time is 5 times lower than NewtonRaphson´s iteration time
~
Useful for analyzing topology changes because B
can be easily modified
Used in planning and contigency analyses
~
22 July 2011
208
Comparison of load flow methods
(Continuation)
4. DC Decoupled
 Analytical, approximate and non-iterative method
 Good approximation for  , not that good
approximation for V 
 Used in reliability analyses
 Used in optimal pricing calculations
 Good for getting an initial point
22 July 2011
209