INTRODUCTION BASIC TECHNIQUES TYPE OF BUSES Y BUS MATRIX POWER SYSTEM COMPONENTS BUS ADMITTANCE MATRIX  Power (Load) flow study is the analysis of a power system in normal steady-state operation  This study will determine:

1. POWER (LOAD)
FLOW STUDY
1
INTRODUCTION
BASIC TECHNIQUES
TYPE OF BUSES
Y BUS MATRIX
POWER SYSTEM COMPONENTS
BUS ADMITTANCE MATRIX

2. INTRODUCTION
 Power (Load) flow study is the analysis of a
power system in normal steady-state
operation
 This study will determine:




Voltages
Currents
Real power
Reactive power
Why we need load flow study?
In a power system under a
given set of load conditions
2

3.  The power flow problem was originally
motivated within planning environments
where engineers considered different network
configurations necessary to serve an
expected future load.
 Later, it became an operational problem as
operators and operating engineers were
required to monitor the real-time status of the
network in terms of voltage magnitudes and
circuit flows.
POWER SYSTEM 2 - EPO 643 3

4. 
POWER SYSTEM 2 - EPO 643 4
A power flow solution procedure is a numerical
method that is employed to solve the power flow
problem.
 A power flow program is a computer code that
implements a power flow solution procedure.
The power flow solution contains the voltages and
angles at all buses, and from this information, we
may compute the real and reactive generation and
load levels at all buses and the real and reactive
flows across all circuits.


5. Terminology
POWER SYSTEM 2 - EPO 643 5
 The above terminology is often used with the
word “load” substituted for “power,” i.e., load
flow problem, load flow solution procedure,
load flow program, and load flow solution.
 However, the former terminology is preferred
as one normally does not think of “load” as
something that “flows.”

6. POWER SYSTEM 2 - EPO 643 6
Power system components
 Generator
Transmission Lines
Load


Figure 1

7. Generator
POWER SYSTEM 2 - EPO 643 7

 Generators have maximum and minimum real and
reactive power capabilities.
Maximum reactive power capability:
 maximum reactive power that the generator may produce
when operating with a lagging power factor.
minimum reactive power capability:
 maximum reactive power the generator may absorb when
operating with a leading power factor.
These limitations are a function of the real power
output of the generator,

 as the real power increases, the reactive power limitations
move closer to zero.

8. Figure 2
POWER SYSTEM 2 - EPO 643 8

9. 
POWER SYSTEM 2 - EPO 643 9
Figure 2 illustrates several important elements of the
power flow problem.
 First, identify each bus depending on
generation and/or load is connected to it.
A bus may have
whether



 generation only (buses B1, B2, and B3),
load only (buses B5, B7, and B9),
neither generation or load (buses B4, B6, and B8).
 both generation and load (leads us to define “bus
injection”)

10. Basic Technique for Load Flow
Studies
 In a load flow study, assumptions are made
about:

 Voltage at a bus or
Power being supplied to the bus
For each bus
in the system
POWER SYSTEM 2 - EPO 643 10

11. Types of Buses
POWER SYSTEM 2 - EPO 643 11
 For each bus, there are
variables that characterize
four possible
the buses
electrical condition.
 The four variables are
 real and reactive power injection, Pi and Qi,
 voltage magnitude and angle, |Vi| and i ,
respectively

12. Types of Buses (cont..)
POWER SYSTEM 2 - EPO 643 12
 Generation Bus



Also called the P-V bus or voltage-controlled buses
Voltage magnitude |Vi| and real power Pi are specified
Able to specify (and therefore to know) the voltage
magnitude of this bus.
Most generator buses fall into this category, independent of
whether it also has load

 Load Bus



Also called the P-Q bus
Real power Pi and Qi are specified
All load buses fall into this category, including buses that
have not either load or generation.

13.  Slack or Swing Bus
POWER SYSTEM 2 - EPO 643 13


Known as reference bus
Voltage magnitude |Vi| and phase angle i are
specified
There is only one swing bus, and it can be
designated by the engineer to be any generator
bus in the system.
This generator “swings” to compensate for the
network losses, or, one may say that it “takes up
the slack.”



14. Bus types Quantities
specified
Unknown
values
Generator Bus |Vi| , Pi Qi , i
Load Bus Pi , Qi |Vi| , i
Slack Bus |Vi| , i Pi , Qi
POWER SYSTEM 2 - EPO 643 14

15. Bus injection
POWER SYSTEM 2 - EPO 643 15
 An injection is the power (P or Q), that is being
injected into or withdrawn from a bus by an element
having its other terminal (in the per-phase
equivalent circuit) connected to ground. Such an
element would be either a generator or a load.
Positive injection is defined as one where power is
flowing from the element into the bus.
Negative injection is then when power is flowing
from the bus, into the element.



16. Bus injection (cont..)
POWER SYSTEM 2 - EPO 643 16
 Generators normally have positive real power
injections, although they may also be assigned
negative real power injections when they are
operating as a motor.
 Generators may have either positive or negative
reactive power injections:
 positive if the generator is operating lagging and delivering
reactive power to the bus,
 negative if the generator is operating leading and
absorbing reactive power from the bus, and
 zero if the generator is operating at unity power factor.

17. Loads
 Loads normally
have negative
real and reactive
power injections.
 Figure 3: Illustration of (a)
positive injection, (b)
negative injection, and (c)
net injection
Pk=100
Qk=30
(a)
Pk= - 40
Qk= -20
(b)
Pk=100+(-40)=60
Qk=30+(-20)=10
POWER SYSTEM 2 - EPO 643 17
(c)

18. 
POWER SYSTEM 2 - EPO 643 18
Figure 3 illustrates the net injection as the algebraic
sum when a bus has both load and generation;
 In this case, the net injection for both real and
reactive power is positive (into the bus).
Thus, the net real power injection is Pk=Pgk-Pdk, and
the net reactive power injection is Qk=Qgk-Qdk.
We may also refer to the net complex power
injection as Sk=Sgk-Sdk, where Sk=Pk+jQk.



19. Power Flow solution
POWER SYSTEM 2 - EPO 643 19
 Most common and important tool in power
system analysis



also known as the “Load Flow” solution
used for planning and controlling a system
assumptions: balanced condition and single phase analysis
 The utility wants to know the voltage profile
 the nodal voltages for a given load and generation
schedule
From the load flow solution –
 the voltage magnitude and phase angle at each bus could
be determined and hence the active and reactive power
flow in each line could be calculated

20.  The currents and powers are expressed as
going into the bus
POWER SYSTEM 2 - EPO 643 20


 for generation the powers are positive
for loads the powers are negative
the scheduled power is the sum of the generation
and load powers

21. The Bus Admittance Matrix
 The matrix equation for relating the nodal
voltages to the currents that flow into and out
of a network using the admittance values of
circuit branches is given by :-
POWER SYSTEM 2 - EPO 643 21

22. Forming the Admittance Matrix
y4
y3
y1
y2
I4
I3I2
4
I1
3
2
1
y34
y23
y13
y12
POWER SYSTEM 2 - EPO 643 22

23. From Kirchoff’s Current Law (KCL) –
 the current injections be equal to the sum of
the currents flowing out of the bus and into
the lines connecting the bus to other buses,or
to the ground.
 Therefore, recalling Ohm’s Law, I=V/Z=VY,
the current injected into bus 1 may be written
as:
I1=(V1-V2)y12 + (V1-V3)y13 + V1y1
POWER SYSTEM 2 - EPO 643 23

24. 
POWER SYSTEM 2 - EPO 643 24
I2 = (V2 - V1)(y21) + V2 y2 + (V2 – V3)y23
 I1= V1( y1 + y12 + y13) + V2(-y12) + V3(-y13)
 I2= V1(-y21) + V2( y2 + y21 + y23) + V3(-y23)
 I3= V1(-y31) + V2(-y32) + V3( y3 + y31 + y32+ y34) +
V4(-y34)
 I4= V3(-y43) + V4( y4 + y43)

25. Admittance Matrix
POWER SYSTEM 2 - EPO 643 25
(y  y )





4 43




(y3y31y32y34)
(y y y )
-y
-y32
0
-y34
0
0-y-y
-y31
0
-y13-y12
43
2321 23221
(y1y12y13)


26. Matrix Equation
POWER SYSTEM 2 - EPO 643 26
y )
V

43  4
 2
V3
V
V1

(y3y31y32y34)
 4
 2
(y1y12y13)
I3 
  
I 
I
I1
443
-y34
232 21 23
-y32
0
(y y y )21
-y31
0
-y13-y12
(y-y
0
0-y-y

27. Y-Bus Matrix Building Rules
POWER SYSTEM 2 - EPO 643 27
 The matrix is symmetric, i.e., Yij=Yji.
A diagonal element Yii = Self Admittance
 is obtained as the sum of admittances for all branches
connected to bus i, including the shunt branch
N
 The off-diagonal elements are the negative of the
admittances connecting buses i and j, i.e., Yij=-yji =
mutual admittance.
 yi  yik
k1,ki
Yii

28. 
POWER SYSTEM 2 - EPO 643 28
E.g. for a 4 bus system



21 22 23 24 
Y44 Y41
Y32 Y33 Y34 
Y42 Y43
Y31
Y Y Y
Y11 Y12 Y13 Y14 
Y
Y  

29. The power flow equations
POWER SYSTEM 2 - EPO 643 29
 The net complex power injection into a bus
Sk=Sgk-Sdk
Sk=VkIk*
Vk=| Vk|k
Ik = | Ykj|kj | Vj|j
Ik = | Ykj|| Vj| (kj + j)
Ik
* = | Ykj|| Vj| -(kj + j)

30. Sk=VkIk*
Sk= | Vk|k x  | Ykj|| Vj| -(kj + j)
Sk=  | Ykj| | Vk|| Vj| (k - j - kj )
Pk=  | Ykj| | Vk|| Vj|cos (k - j - kj )
Qk=  | Ykj| | Vk|| Vj|sin (k - j - kj )
POWER SYSTEM 2 - EPO 643 30