EE 369
POWER SYSTEM ANALYSIS
Lecture 12
Power Flow
Tom Overbye and Ross Baldick
1
Announcements
• Homework 9 is 3.20, 3.23, 3.25, 3.27, 3.28,
3.29, 3.35, 3.38, 3.39, 3.41, 3.44, 3.47; due
11/5.
• Midterm 2, Thursday, November 12, covering
up to and including material in HW9
• Homework 10 is: 3.49, 3.55, 3.57, 6.2, 6.9,
6.13, 6.14, 6.18, 6.19, 6.20; due 11/19. (Use
infinity norm and epsilon = 0.01 for any
problems where norm or stopping criterion
not specified.)
2
Transmission System Planning
Source: Federal Energy Regulatory Commission
3
ERCOT
Wind generation capacity in Texas (GW, end of year)
14
12
10
8
6
4
2
0
1999
2000
2001
2002
2003
2004
2005
2006
2007
2008
Source: US Energy Information Administration
2009
2010
2011
2012
•4
ERCOT
• Has considerable wind and expecting
considerable more!
• “Competitive Renewable Energy Zones” study
identified most promising wind sites,
• ERCOT ISO planned approximately $5 billion
(original estimate, now closer to $7 billion) of
new transmission to support an additional 11
GW of wind:
– Used tools such as power flow to identify whether
plan could accommodate wind generation.
• Built by transmission companies.
• Mostly completed by 2014.
5
CREZ Transmission Lines
6
NR Application to Power Flow
We first need to rewrite complex power equations
as equations with real coefficients (we've seen this earlier):
*
n


* *
Si 
 Vi   YikVk   Vi  YikVk
 k 1

k 1
These can be derived by defining
n
*
Vi I i
Yik
Gik  jBik
ji
Vi
Vi e
ik
i   k
Recall e
j
 Vi i
 cos   j sin 
7
Real Power Balance Equations
n
Si  Pi  jQi  Vi  Yik*Vk* 
k 1

n
 Vi Vk
k 1
n
jik
V
V
e
(Gik  jBik )
 i k
k 1
(cosik  j sin ik )(Gik  jBik )
Resolving into the real and imaginary parts:
Pi  PGi  PDi 
Qi  QGi  QDi 
n
 Vi Vk (Gik cosik  Bik sin ik )
k 1
n
 Vi Vk (Gik sin ik  Bik cosik )
k 1
8
Newton-Raphson Power Flow
In the Newton-Raphson power flow we use Newton's
method to determine the voltage magnitude and angle at
each bus in the power system that satisfies power balance.
We need to solve the power balance equations:
n
 Vi Vk (Gik cosik  Bik sin ik )  PGi  PDi  0
k 1
n
 Vi Vk (Gik sin ik  Bik cosik )  QGi  QDi  0
k 1
9
Power Balance Equations
For convenience, write:
Pi ( x ) 
Qi ( x ) 
n
 Vi Vk (Gik cosik  Bik sin ik )
k 1
n
 Vi Vk (Gik sin ik  Bik cosik )
k 1
The power balance equations are then:
Pi ( x )  PGi  PDi  0
Qi ( x )  QGi  QDi  0
•10
Power Balance Equations
• Note that Pi( ) and Qi( ) mean the functions
that expresses flow from bus i into the system
in terms of voltage magnitudes and angles,
• While PGi, PDi, QGi, QDi mean the generations
and demand at the bus.
• For a system with a slack bus and the rest PQ
buses, the power flow problem is to use the
power balance equations to solve for the
unknown voltage magnitudes and angles in
terms of the given bus generations and
demands, and solve for the real and reactive
•11
injection at the slack bus.
Power Flow Variables
Assume the slack bus is the first bus (with a fixed
voltage angle/magnitude). We then need to determine
the voltage angle/magnitude at the other buses.
We must solve f ( x )  0, where:
 2 




 n 
x  
V2




V 
 n 
 P2 ( x )  PG 2  PD 2 




 Pn ( x )  PGn  PDn 
f (x)  
Q2 ( x )  QG 2  QD 2 




Q (x)  Q  Q 
 n
Gn
Dn 
12
N-R Power Flow Solution
The power flow is solved using the same procedure
discussed previously for general equations:
For v  0; make an initial guess of x, x (0)
While f (x ( v ) )   Do
x ( v 1)  x ( v )  [ J ( x ( v ) )]1 f ( x ( v ) )
v
 v 1
End
13
Power Flow Jacobian Matrix
The most difficult part of the algorithm is determining
and factorizing the Jacobian matrix, J (x)
f1
 f1 ( x )
(x)
 x1
x2

f 2
 f 2 ( x )
(x)
x2
J (x )   x1


 f 2 n 2
f 2 n 2
(x)
(x)

x2
 x1
f1
(x) 

x2 n 2

f 2
(x) 

x2 n 2



f 2 n 2
(x)
x2 n 2

14
Power Flow Jacobian Matrix, cont’d
Jacobian elements are calculated by differentiating
each function, fi ( x), with respect to each variable.
For example, if fi ( x) is the bus i real power equation
fi ( x) 
n
 Vi Vk (Gik cosik  Bik sinik )  PGi  PDi
k 1
fi
( x)
 i

fi
( x)
 j
 Vi V j (Gij sin  ij  Bij cos ij ) ( j  i )
n
 Vi Vk (Gik sinik  Bik cosik )
k 1
k i
15
Two Bus Newton-Raphson Example
For the two bus power system shown below, use the
Newton-Raphson power flow to determine the
voltage magnitude and angle at bus two. Assume
that bus one is the slack and SBase = 100 MVA.
Line Z = 0.1j
One
1.000 pu
Two
0 MW
0 MVR
 2 
Unkown: x    ,
 V2 
1.000 pu
200 MW
100 MVR
Also, Ybus
  j10 j10 
 

j
10

j
10


16
Two Bus Example, cont’d
General power balance equations:
n
 Vi Vk (Gik cosik  Bik sin ik )  PGi  PDi  0
k 1
n
 Vi Vk (Gik sin ik  Bik cosik )  QGi  QDi  0
k 1
For bus two, the power balance equations are
(load real power is 2.0 per unit,
while reactive power is 1.0 per unit):
V2 V1 (10sin  2 )  2.0  0
V2 V1 ( 10cos  2 )  V2 (10)  1.0  0
2
17
Two Bus Example, cont’d
P2 ( x )  2.0 
V2 (10sin  2 )  2.0
Q2 ( x ) 1.0 
V2 ( 10cos  2 )  V2 (10)  1.0
2
Now calculate the power flow Jacobian
 P2 ( x ) P2 ( x ) 
  2

 V2

J(x)  
 Q2 ( x ) Q2 ( x ) 
  2

 V2
10sin  2
10 V2 cos 2

 

10
V
sin


10cos


20
V
2
2
2
2

18
Two Bus Example, First Iteration
For v  0, guess x (0)
  2(0)  0
  (0)     . Calculate:
 V2  1


V2(0) (10sin  2(0) )  2.0

f (x (0) )  
 V (0) ( 10cos  (0) )  V (0) 2 (10)  1.0
2
2
 2

10 V2(0) cos 2(0)
J (x (0) )  
10 V2(0) sin  2(0)

 10 0


(0)
(0)   0 10 

10cos  2  20 V2  
10sin  2(0)
1
Solve x
(1)
 2.0
 
1.0 
0 10 0   2.0
  
 1.0 
1
0
10
  
  
 0.2 
 

0.9


19
Two Bus Example, Next Iterations
0.9(10sin( 0.2))  2.0

 0.212 
f (x )  
  0.279 
2

0.9( 10cos( 0.2))  0.9  10  1.0  
 8.82 1.986
(1)
J (x )  


1.788
8.199


(1)
1
 0.2   8.82 1.986  0.212 
 0.233
(2)
x 

 





0.9

1.788
8.199
0.279
0.8586

 
 



 0.0145
 0.236
(2)
(3)
f (x )  
x
 


0.0190
0.8554




f (x
(3)
0.0000906
)
Close enough! V2  0.8554  13.52

 0.0001175
20
Two Bus Solved Values
Once the voltage angle and magnitude at bus 2 are
known we can calculate all the other system values,
such as the line flows and the generator real and
reactive power output
200.0 MW
168.3 MVR
One
-200.0 MW
-100.0 MVR
Line Z = 0.1j
1.000 pu
200.0 MW
168.3 MVR
Two
0.855 pu -13.522 Deg
200 MW
100 MVR
21
Two Bus Case Low Voltage Solution
This case actually has two solutions! The second
"low voltage" is found by using a low initial guess.
Set v  0, guess x
(0)
 0 

. Calculate:

0.25


V2(0) (10sin  2(0) )  2.0
 2 
(0)


f (x ) 

2
 V (0) ( 10cos (0) )  V (0 ) (10)  1.0  0.875
2
2
 2

10 V2(0) cos 2(0)
J (x (0) )  
10 V2(0) sin  2(0)

 2.5 0



(0)
(0)   0

5


10cos  2  20 V2 
10sin  2(0)
22
Low Voltage Solution, cont'd
1
 0   2.5 0   2 
 0.8 
Solve x  

 





0.25
0

5

0.875
0.075

 
 



1.462  (2)  1.42 
 0.921
(2)
(3)
f (x )  
x 
x 



0.534
0.2336
0.220






(1)
Low voltage solution
200.0 MW
831.7 MVR
One
-200.0 MW
-100.0 MVR
Line Z = 0.1j
1.000 pu
200.0 MW
831.7 MVR
Two
0.261 pu -49.914 Deg
200 MW
100 MVR
23
Two Bus Region of Convergence
Graph shows the region of convergence for different initial
guesses of bus 2 angle (horizontal axis) and magnitude
(vertical axis).
Red region
converges
to the high
voltage
solution,
while the
yellow region
converges
to the low
voltage
solution
Maximum
of 15
iterations 24
PV Buses
Since the voltage magnitude at PV buses is
fixed there is no need to explicitly include
these voltages in x nor write the reactive
power balance equations:
– the reactive power output of the generator
varies to maintain the fixed terminal voltage
(within limits), so we can just set the reactive
power product to whatever is needed.
– An alternative is these variations/equations can
be included by just writing the explicit voltage
constraint for the generator bus:
25
|Vi | – Vi setpoint = 0
Three Bus PV Case Example
For this three bus case we have
 2 
x   3 
 
 V2 
 P2 ( x )  PD 2 
f ( x )   P3 ( x )  PG 3   0


Q2 ( x )  QD 2 
Line Z = 0.1j
0.941 pu
One
170.0 MW
68.2 MVR
1.000 pu
Line Z = 0.1j
Three
Two
Line Z = 0.1j
-7.469 Deg
200 MW
100 MVR
1.000 pu
30 MW
63 MVR
26
PV Buses
• With Newton-Raphson, PV buses means that
there are less unknown variables we need to
calculate explicitly and less equations we need
to satisfy explicitly.
• Reactive power balance is satisfied implicitly by
choosing reactive power production to be
whatever is needed, once we have a solved case
(like real and reactive power at the slack bus).
• Contrast to Gauss iterations where PV buses
complicated the algorithm.
27
Modeling Voltage Dependent Load
So far we've assumed that the load is independent of
the bus voltage (i.e., constant power). However, the
power flow can be easily extended to include voltage
dependence with both the real and reactive load. This
is done by making PDi and QDi a function of Vi :
n
 Vi Vk (Gik cosik  Bik sin ik )  PGi  PDi ( Vi )  0
k 1
n
 Vi Vk (Gik sin ik  Bik cosik )  QGi  QDi ( Vi )  0
k 1
28
Voltage Dependent Load Example
In previous two bus example now assume the load is
constant impedance, with corresponding per unit
admittance of 2.0  j1.0 :
P2 ( x )  2.0 V2  V2 (10sin  2 )  2.0 V2
2
2
 0
Q2 ( x ) 1.0 V2  V2 ( 10cos  2 )  V2 (10)  1.0 V2  0
2
2
2
Now calculate the power flow Jacobian
10 V2 cos 2
J(x)  
10 V2 sin  2
10sin  2  4.0 V2

10cos  2  20 V2  2.0 V2 
29
Voltage Dependent Load, cont'd
Again for v  0, guess x
(0)
  2(0)  0
  (0)     . Calculate:
 V2  1 
(0)
(0)
(0) 2


V2 (10sin  2 )  2.0 V2
 2.0
(0)


f (x ) 
 
2
2
 (0)
(0)
(0)
(0)  1.0 
(10)  1.0 V2 
 V2 ( 10cos  2 )  V2
10 4 
(0)
J (x )  

0
12


1
Solve x
(1)
0 10 4   2.0
  
 1.0 
1
0
12
  
  
 0.1667 
 

0.9167


30
Voltage Dependent Load, cont'd
With constant impedance load the MW/MVAr load at
bus 2 varies with the square of the bus 2 voltage
magnitude. This if the voltage level is less than 1.0,
the load is lower than 200/100 MW/MVAr.
160.0 MW
120.0 MVR
One
-160.0 MW
-80.0 MVR
Line Z = 0.1j
1.000 pu
160.0 MW
120.0 MVR
Two
0.894 pu
-10.304 Deg
160 MW
80 MVR
In practice, load is the sum of constant power,
constant impedance, and, in some cases,
constant current load terms: “ZIP” load.
31
Solving Large Power Systems
Most difficult computational task is inverting the
Jacobian matrix (or solving the update equation):
– factorizing a full matrix is an order n3 operation, meaning
the amount of computation increases with the cube of
the size of the problem.
– this amount of computation can be decreased
substantially by recognizing that since Ybus is a sparse
matrix, the Jacobian is also a sparse matrix.
– using sparse matrix methods results in a computational
order of about n1.5.
– this is a substantial savings when solving systems with
32
tens of thousands of buses.
Newton-Raphson Power Flow
 Advantages
– fast convergence as long as initial guess is close to
solution
– large region of convergence
 Disadvantages
– each iteration takes much longer than a Gauss-Seidel
iteration
– more complicated to code, particularly when
implementing sparse matrix algorithms
 Newton-Raphson algorithm is very common in
power flow analysis.
33