The document discusses power flow analysis, which determines voltages, currents, and real and reactive power flows in a power system under specified conditions. It provides details on: 1) Creating an admittance matrix and making initial voltage estimates. 2) Iteratively updating the estimates until voltages converge within a tolerance. 3) Equations used depend on whether the bus is a load, generator, or slack bus. 4) Power flow analysis provides important information about the system such as all bus voltages.

1. ECE 333 Green Electric Energy April 2018 1
ECE 333
Green Electric Energy
Dr. Karl Reinhard
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
reinhrd2@illinois.edu
Lecture 18
Introduction to Power Flow Analysis

2. ECE 333 Green Electric Energy April 2018 2
Introduction
We seek to analyze the power system performance under steady
state conditions.
The analysis in normal steady-state operation is called a power-flow study
(load-flow study) and it targets on determining the Voltages, Currents, and
Real and Reactive Power Flows in a system under specified generation and
load conditions.
At each bus, We make an assumption about either
• a Voltage at a bus or
• the Power being supplied to the bus
Then determine
• Bus voltage magnitude and phase angles
• Line currents, etc. that would result

3. ECE 333 Green Electric Energy April 2018 3
Basics for Power-flow Studies.
The way ahead…. to find the power-flow solution via iteration:
1. Create a bus admittance matrix Ybus for the power system;
2. Make an initial estimate for the voltages at each bus in the system;
3. Iterate to find conditions that satisfy the system’s load flow equations.
• Update the voltage estimate for each bus (one at a time), based on the estimates
for the voltages and power flows at every other bus and the values of the bus
admittance matrix.
• Since the voltage at a given bus depends on the voltages at all of the other busses
in the system (which are just estimates), the updated voltage will not be correct.
However, it will usually be closer to the answer than the original guess.
4. Repeat this process to make the voltages at each bus approaching the correct
answers to within a set tolerance level…

4. ECE 333 Green Electric Energy April 2018 4
Basics for power-flow studies
The equations used to update the estimates differ for each of 3 bus types.
1. Load bus (PQ bus) – All buses not having a generator
• Real and reactive power (P and Q)are specified
• Bus voltage magnitude and phase angle (V and q) will be calculated
• Real and reactive powers supplied to a power system are defined to be
positive
• Powers consumed from the system are defined to be negative.
2. Generator bus (PV bus) –
• Voltage and real power supplied are specified
• Bus phase angle (q) will be calculated during iteration
• Reactive power will be calculated after the case’s solution is found

5. ECE 333 Green Electric Energy April 2018 5
Basics for Power-flow Studies.
3. Slack bus (swing bus) –
• Special generator bus serving as the reference bus for the power system.
• Voltage is fixed – both magnitude and phase (for instance, 10˚ pu).
• Real and reactive powers are uncontrolled – supplies whatever real or
reactive power is necessary to make the power flows in the system balance.
Key Points:
• Voltage on a load bus (P-Q bus) changes as the load varies – P and Q are
fixed, while V (magnitude and angle) vary with load conditions.
• Generators (@ P-V buses) work most efficiently when running at full load – P
and V are fixed
• Slack bus generator varies P and Q that it supplies to balance Complex power
– V and Angle reference are fixed.

6. ECE 333 Green Electric Energy April 2018 6
Ybus for Power-flow Analysis
12 32 42 2
I I I I
  
3
1 4
2 2
21 21 23 24 23 24
1 1 1 V
V V
V I
Z Z Z Z Z Z
 
     
 
 
I12
I32
I42
I2
2 3
2 1 2 4
2
21 23 24
V V
V V V V
I
Z Z Z

 
  
1 21 2 22 3 23 4 24 2
VY V Y V Y V Y I
   
The basic equation for power-flow analysis is derived from the nodal
analysis equations for the power system:
V
V I R I
R
  

7. ECE 333 Green Electric Energy April 2018 7
Power-flow Analysis Equations
bus
V I Z Y V I
  
11 12 13 14 1 1
21 22 23 24 2 2
31 32 33 34 3 3
41 42 43 44 4 4
Y Y Y Y V I
Y Y Y Y V I
Y Y Y Y V I
Y Y Y Y V I
     
     
     

     
     
   
 
The basic equation for power-flow analysis is derived from the nodal analysis
equations for the power system:
For the four-bus power system shown above, (1) becomes
where Yij are the elements of the bus admittance matrix, Vi are the bus voltages, and
Ii are the currents injected at each node.
21 1 22 2 23 3 24 4 2
Y V Y V Y V Y V I
   
(1)
(2)
(3)
For bus 2 in this system, this equation
reduces to

8. ECE 333 Green Electric Energy April 2018 8
Ybus for Power-flow Analysis
1.7647 7.0588 0.5882 2.3529 0 1.1765 4.7059
0.5882 2.3529 1.5611 6.6290 0.3846 1.9231 0.5882 2.3529
0 0.3846 1.9231 1.5611 6.6290 1.1765 4.7059
1.1765 4.7059 0.5882 2.3529 1.1765 4.7059 2
bus
j j j
j j j j
Y
j j j
j j j
    
      

    
      .9412 11.7647
j
 
 
 
 
 

 
line
#
Bus to
bus
Series
Z (pu)
Series Y (pu)
1 1-2 0.1+j0.4 0.5882-j2.3529
2 2-3 0.1+j0.5 0.3846-j1.9231
3 2-4 0.1+j0.4 0.5882-j2.3529
4 3-4 0.5+j0.2 1.1765-j4.7059
5 4-1 0.5+j0.2 1.1765-j4.7059
Example: a simple power system has 4 buses, 5 transmission lines, 1 generator,
and 3 loads. Series per-unit impedances are:
Note: Ybus symmetric construction – Off diagonal elements are -Yij = -Yji

9. ECE 333 Green Electric Energy April 2018 9
Ybus For Power-flow Analysis
1.7647 7.0588 0.5882 2.3529 0 1.1765 4.7059
0.5882 2.3529 1.5611 6.6290 0.3846 1.9231 0.5882 2.3529
0 0.3846 1.9231 1.5611 6.6290 1.1765 4.7059
1.1765 4.7059 0.5882 2.3529 1.1765 4.7059 2
bus
j j j
j j j j
Y
j j j
j j j
    
      

    
      .9412 11.7647
j
 
 
 
 
 

 
line
#
Bus to
bus
Series
Z (pu)
Series Y (pu)
1 1-2 0.1+j0.4 0.5882-j2.3529
2 2-3 0.1+j0.5 0.3846-j1.9231
3 2-4 0.1+j0.4 0.5882-j2.3529
4 3-4 0.5+j0.2 1.1765-j4.7059
5 4-1 0.5+j0.2 1.1765-j4.7059
Example: a simple power system has 4 buses, 5 transmission lines, 1 generator,
and 3 loads. Series per-unit impedances are:
Note: Ybus symmetric construction – On diagonal elements:
i j
ii ij
Y Y

 

10. ECE 333 Green Electric Energy April 2018 10
Power-flow Analysis Equations
However, real loads are specified in terms of real and reactive powers, not as
currents. The relationship between per-unit real and reactive power supplied to the
system at a bus and the per-unit current injected into the system at that bus is:
*
S VI P jQ
  
where V is the per-unit voltage at the bus; I* - complex conjugate of the per-unit
current injected at the bus; P and Q are per-unit real and reactive powers. Therefore,
for instance, the current injected at bus 2 can be found as
 
*
2 2
* * 2 2
2 2 2 2 2 2 *
2 2
P jQ
P jQ
V I P jQ I I
V V


     
(4)
(5)
Substituting (5) into (3), we obtain
 
*
2 2
21 1 22 2 23 3 24 4 *
2
P jQ
Y V Y V Y V Y V
V

   
(6)
 
*
* * * *
21 1 2 22 2 2 23 3 2 24 4 2 2 2
Y VV Y V V Y V V Y V V P jQ
    

11. ECE 333 Green Electric Energy April 2018 11
Power-flow Analysis Equations
(6)
 
*
* * * * *
2 2 2 21 1 2 22 2 2 23 3 2 24 4 2
S P jQ Y VV Y V V Y V V Y V V
     
* *
1 1
( )
ik
n n
j
i i i i ik k i k ik ik
k k
S P jQ V Y V V V e G jB
q
 
    
 
1
(cos sin )( )
n
i k ik ik ik ik
k
V V j G jB
q q

  

1
1
Resolving into the real and imaginary parts:
( cos sin )
( sin cos )
n
i i k ik ik ik ik Gi Di
k
n
i i k ik ik ik ik Gi Di
k
P V V G B P P
Q V V G B Q Q
q q
q q


   
   



12. ECE 333 Green Electric Energy April 2018 12
Power-flow Analysis Equations
(6)
1
1
( cos sin )
( sin cos )
n
i i k ik ik ik ik Gi Di
k
n
i i k ik ik ik ik Gi Di
k
P V V G B P P
Q V V G B Q Q
q q
q q


   
   


2 2 2 2
n
2 2 2 2
( )
( )
( )
( )
( )
G D
n Gn Dn
G D
n n Gn Dn
P P P
P P P
V Q Q Q
V Q Q Q
q
q
 
   
   
   
 
   
 
   
 
   
   
   
 
 
 
x
x
x f x
x
x
G
B
Y

13. ECE 333 Green Electric Energy April 2018 13
Non-Linear – Newton Raphson Solution
2
n
2
n
V
V
q
q
 
 
 
 
  
 
 
 
 
x
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

14. ECE 333 Green Electric Energy April 2018 14
Information from power-flow studies
The basic information contained in the load-flow output is:
• All bus voltage magnitudes and phase angles w.r.t the slack bus.
• All bus active and reactive power injections.
• All line sending- and receiving-end complex power flows.
• Individual line losses can be deduced by subtracting receiving-end
complex Power from sending-end complex power.
• Total system losses – deduced by summing complex power at all
loads and generators and subtracting the totals.

15. ECE 333 Green Electric Energy April 2018 15
Information From Power-flow Studies
The most important information obtained from the
load-flow is the system voltage profile.
A power-flow program can be set up to provide
alerts if the voltage at any given bus exceeds, for
instance, 5% of the nominal value
 such voltage variations may
indicate problems…
• If │V│ varies greatly over the system, large reactive flows will result; this, in
turn, will lead to increased real power losses and, in extreme cases, an
increased likelihood of voltage collapse.
• When a particular bus has an unacceptably low voltage, the usual practice is
to install capacitor banks in order to provide reactive compensation to the load.