Different methods of load flow solutions

1. Load Flow Analysis
Part- II
Prepared by – A. L. Jamadar
1

2. NR Application to Power Flow
*
* * *
1 1
We first need to rewrite complex power equations
as equations with real coefficients (we've seen this earlier):
These can be derived by defining
n n
i i i i ik k i ik k
k k
ik ik ik
i
S V I V Y V V Y V
Y G jB
V
 
 
   
 

 
@
Recall e cos sin
ij
i i i
ik i k
j
V e V
j



  
 
 

 
@
@
2

3. Real Power Balance Equations
* *
1 1
1
1
1
( )
(cos sin )( )
Resolving into the real and imaginary parts:
( cos sin )
( sin
ik
n n
j
i i i i ik k i k ik ik
k k
n
i k ik ik ik ik
k
n
i Gi Di i k ik ik ik ik
k
n
i Gi Di i k ik ik
k
S P jQ V Y V V V e G jB
V V j G jB
P P P V V G B
Q Q Q V V G

 
 

 



    
  
   
  
 


 cos )ik ikB 
3

4. 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 equ
1
1
ations:
( cos sin ) 0
( sin cos ) 0
n
i k ik ik ik ik Gi Di
k
n
i k ik ik ik ik Gi Di
k
V V G B P P
V V G B Q Q
 
 


   
   


4

5. Power Balance Equations
•5
1
1
For convenience, write:
( ) ( cos sin )
( ) ( sin cos )
The power balance equations are then:
( ) 0
( ) 0
n
i i k ik ik ik ik
k
n
i i k ik ik ik ik
k
i Gi Di
i Gi Di
P V V G B
Q V V G B
P P P
Q Q Q
 
 


 
 
  
  


x
x
x
x

6. 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, 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 then
use solution to calculate the real and reactive
injection at the slack bus. •6

7. Power Flow Variables
2
n
2
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 ( ) , where:
n
V
V









f x 0
x
M
M
2 2 2
2 2 2
( )
( )
( )
( )
( )
G D
n Gn Dn
G D
n Gn Dn
P P P
P P P
Q Q Q
Q Q Q
   
  
  
   
    
   
   
     
x
x
f x
x
x
M
M
7

8. N-R Power Flow Solution
(0)
( )
( 1) ( ) ( ) 1 ( )
The power flow is solved using the same procedure
discussed previously for general equations:
For 0; make an initial guess of ,
While ( ) Do
[ ( )] ( )
1
End
v
v v v v
v
v v

 


 
 
x x
f x
x x J x f x
8

9. Power Flow Jacobian Matrix
1 1 1
1 2 2 2
2 2 2
1 2 2 2
2 2 2 2 2 2
1 2
The most difficult part of the algorithm is determining
and factorizing the Jacobian matrix, ( )
( ) ( ) ( )
( ) ( ) ( )
( )
( ) ( )
n
n
n n n
f f f
x x x
f f f
x x x
f f f
x x x


  
  
  
  
  
  
  
J x
x x x
x x x
J x
x x
L
L
M O O M
L
2 2
( )
n
 
 
 
 
 
 
 
 
 
 
x
9

10. Power Flow Jacobian Matrix, cont’d
1
Jacobian elements are calculated by differentiating
each function, ( ), with respect to each variable.
For example, if ( ) is the bus real power equation
( ) ( cos sin )
i
i
n
i i k ik ik ik ik Gi
k
f x
f x i
f x V V G B P P 

   
1
( ) ( sin cos )
( ) ( sin cos ) ( )
Di
n
i
i k ik ik ik ik
i k
k i
i
i j ij ij ij ij
j
f
x V V G B
f
x V V G B j i
 

 




  


  


10

11. Two Bus Newton-Raphson Example
Line Z = 0.1j
One Two1.000 pu 1.000 pu
200 MW
100 MVR
0 MW
0 MVR
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.
2
2
10 10
Unkown: , Also,
10 10
bus
j j
V j j
    
      
x Y
11

12. Two Bus Example, cont’d
1
1
General power balance equations:
( cos sin ) 0
( sin cos ) 0
For bus two, the power balance equations are
(load real power is 2.0 per unit,
while react
n
i k ik ik ik ik Gi Di
k
n
i k ik ik ik ik Gi Di
k
V V G B P P
V V G B Q Q
 
 


   
   


2 1 2
2
2 1 2 2
ive power is 1.0 per unit):
(10sin ) 2.0 0
( 10cos ) (10) 1.0 0
V V
V V V


 
    12

13. Two Bus Example, cont’d
2 2 2
2
2 2 2 2
2 2
2 2
2 2
2 2
2 2 2
2 2 2 2
( ) 2.0 (10sin ) 2.0
( ) 1.0 ( 10cos ) (10) 1.0
Now calculate the power flow Jacobian
( ) ( )
( )
( ) ( )
10 cos 10sin
10 sin 10cos 20
P V
Q V V
P P
x x
V
Q Q
x x
V
V
V V




 
 
  
    
  
  
 
  
   
 
    
x
x
J x
13

14. Two Bus Example, First Iteration
(0)
2(0)
(0)
2
(0) (0)
2 2
(0)
2(0) (0) (0)
2 2 2
(0) (0) (0)
2 2 2(0)
(0) (0)
2 2
0
For 0, guess . Calculate:
1
(10sin ) 2.0 2.0
( )
1.0( 10cos ) (10) 1.0
10 cos 10sin
( )
10 sin 10cos
v
V
V
V V
V
V



 

   
     
   
 
           


x
f x
J x
(0) (0)
2 2
1
(1)
10 0
0 1020
0 10 0 2.0 0.2
Solve
1 0 10 1.0 0.9
V

 
         
       
         
       
x
14

15. Two Bus Example, Next Iterations
(1)
2
(1)
1
(2)
0.9(10sin( 0.2)) 2.0 0.212
( )
0.2790.9( 10cos( 0.2)) 0.9 10 1.0
8.82 1.986
( )
1.788 8.199
0.2 8.82 1.986 0.212 0.233
0.9 1.788 8.199 0.279 0.8586
(

    
           
 
   
         
                
f x
J x
x
f (2) (3)
(3)
2
0.0145 0.236
)
0.0190 0.8554
0.0000906
( ) Close enough! 0.8554 13.52
0.0001175
V
   
    
   
 
     
 
x x
f x
15

16. Two Bus Solved Values
Line Z = 0.1j
One Two1.000 pu 0.855 pu
200 MW
100 MVR
200.0 MW
168.3 MVR
-13.522 Deg
200.0 MW
168.3 MVR
-200.0 MW
-100.0 MVR
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
16

17. Two Bus Case Low Voltage Solution
(0)
(0) (0)
2 2
(0)
(0) (0) (0
2 2 2
This case actually has two solutions! The second
"low voltage" is found by using a low initial guess.
0
Set 0, guess . Calculate:
0.25
(10sin ) 2.0
( )
( 10cos )
v
V
V V


 
   
 


 
x
f x 2)
(0) (0) (0)
2 2 2(0)
(0) (0) (0) (0)
2 2 2 2
2
0.875(10) 1.0
10 cos 10sin 2.5 0
( )
0 510 sin 10cos 20
V
V V
 
 
 
         
 
          
J x
17

18. Low Voltage Solution, cont'd
1
(1)
(2) (2) (3)
0 2.5 0 2 0.8
Solve
0.25 0 5 0.875 0.075
1.462 1.42 0.921
( )
0.534 0.2336 0.220

       
                 
      
       
     
x
f x x x
Line Z = 0.1j
One Two1.000 pu 0.261 pu
200 MW
100 MVR
200.0 MW
831.7 MVR
-49.914 Deg
200.0 MW
831.7 MVR
-200.0 MW
-100.0 MVR
Low voltage solution
18

19. 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
iterations19

20. PV Buses
Since the voltage magnitude at PV buses is fixed
there is no need to explicitly include these
voltages in x nor explicitly include the reactive
power balance equations at the PV buses:
– the reactive power output of the generator varies to
maintain the fixed terminal voltage (within limits), so
we can just use the solved voltages and angles to
calculate the reactive power production to be
whatever is needed to satisfy reactive power balance.
– An alternative is to keep the reactive power balance
equation explicit but also write an explicit voltage
constraint for the generator bus:
|Vi | – Vi setpoint = 0 20

21. Three Bus PV Case Example
Line Z = 0.1j
Line Z = 0.1j Line Z = 0.1j
One Two1.000 pu
0.941 pu
200 MW
100 MVR
170.0 MW
68.2 MVR
-7.469 Deg
Three 1.000 pu
30 MW
63 MVR
2 2 2
3 3 3
2 2 2
For this three bus case we have
( )
( ) ( ) 0
( )
D
G
D
P P
P P
V Q Q


   
      
   
     
x
x f x x
x
21

22. 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. 22

23. 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
1
1
load. This
is done by making and a function of :
( cos sin ) ( ) 0
( sin cos ) ( ) 0
Di Di i
n
i k ik ik ik ik Gi Di i
k
n
i k ik ik ik ik Gi Di i
k
P Q V
V V G B P P V
V V G B Q Q V
 
 


   
   


23

24. Voltage Dependent Load Example
2 2
2 2 2 2 2
2 2 2
2 2 2 2 2 2
In previous two bus example now assume the load is
constant impedance, with corresponding per unit
admittance of 2.0 1.0:
( ) 2.0 (10sin ) 2.0 0
( ) 1.0 ( 10cos ) (10) 1.0 0
Now
j
P V V V
Q V V V V



   
     
x
x
2 2 2 2
2 2 2 2 2
calculate the power flow Jacobian
10 cos 10sin 4.0
( )
10 sin 10cos 20 2.0
V V
V V V
 
 
 
     
J x
24

25. Voltage Dependent Load, cont'd
(0)
2(0)
(0)
2
2(0) (0) (0)
2 2 2(0)
2 2(0) (0) (0) (0)
2 2 2 2
(0)
(1)
0
Again for 0, guess . Calculate:
1
(10sin ) 2.0 2.0
( )
1.0
( 10cos ) (10) 1.0
10 4
( )
0 12
0
Solve
1
v
V
V V
V V V



   
     
   
              
 
  
 

 

x
f x
J x
x
1
10 4 2.0 0.1667
0 12 1.0 0.9167

      
       
      
25

26. Voltage Dependent Load, cont'd
Line Z = 0.1j
One Two1.000 pu
0.894 pu
160 MW
80 MVR
160.0 MW
120.0 MVR
-10.304 Deg
160.0 MW
120.0 MVR
-160.0 MW
-80.0 MVR
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.
26
More generally, load can be modeled as the sum of:
constant power, constant impedance, and, in some cases,
constant current load terms: “ZIP” load.

27. 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
tens of thousands of buses. 27

28. 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.
28