1. POWER (LOAD)
FLOW STUDY
INTRODUCTION
BASIC TECHNIQUES
TYPE OF BUSES
Y BUS MATRIX
POWER SYSTEM COMPONENTS
BUS ADMITTANCE MATRIX
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2. INTRODUCTION
 Power (Load) flow study is the analysis of a
power system in normal steady-state
operation
 This study will determine:
 Voltages
 Currents In a power system under a
given set of load conditions
 Real power
 Reactive power
Why we need load flow study?
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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.
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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.
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5. Terminology
 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.”
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6. Power system components
 Generator
 Transmission Lines
 Load
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Figure 1

7. Generator
 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.
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8. Figure 2
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9.  Figure 2 illustrates several important elements of the
power flow problem.
 First, identify each bus depending on whether
generation and/or load is connected to it.
 A bus may have
 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”)
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10. Basic Technique for Load Flow
Studies
 In a load flow study, assumptions are made
about:
 Voltage at a bus or For each bus
in the system
 Power being supplied to the bus
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11. Types of Buses
 For each bus, there are four possible
variables that characterize the buses
electrical condition.
 The four variables are
 real and reactive power injection, Pi and Qi,
 voltage magnitude and angle, |Vi| and δi ,
respectively
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12. Types of Buses (cont..)
 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.
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13.  Slack or Swing Bus
 Known as reference bus
 Voltage magnitude |Vi| and phase angle δi 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.”
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14. Bus types Quantities Unknown
specified values
Generator Bus |Vi| , Pi Q i , δi
Load Bus Pi , Qi |Vi| , δi
Slack Bus |Vi| , δi Pi , Qi
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15. Bus injection
 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.
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16. Bus injection (cont..)
 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.
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17. Loads
 Loads normally Pk=100
Qk=30
Pk= - 40
Qk= -20
have negative
real and reactive
power injections. (a) (b)
 Figure 3: Illustration of (a)
Pk=100+(-40)=60
positive injection, (b) Qk=30+(-20)=10
negative injection, and (c)
net injection
(c)
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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.
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19. Power Flow solution
 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
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20.  The currents and powers are expressed as
going into the bus
 for generation the powers are positive
 for loads the powers are negative
 the scheduled power is the sum of the generation
and load powers
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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 :-
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22. Forming the Admittance Matrix
1 y13 3 4
I1 y34
2 I4
y12 y23
I2 I3
y1 y4
y2 y3
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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
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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)
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25. Admittance Matrix
 ( y1 + y12 + y13 ) - y12 - y13 0 
 - y 21 ( y 2 + y 21 + y 23 ) - y 23 0 
 
 - y31 - y32 ( y3 + y31 + y32 + y34 ) - y34 
 
 0 0 - y 43 ( y 4 + y 43 )
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26. Matrix Equation
 I1   ( y1 + y12 + y13 ) - y12 - y13 0   V1 
I   - y 21 ( y 2 + y 21 + y 23 ) - y 23 0   2 V 
 2 = 
 I3   - y31 - y32 ( y3 + y31 + y32 + y34 ) - y34   V3 
    
 I4   0 0 - y 43 ( y 4 + y 43 )  V4 
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27. Y-Bus Matrix Building Rules
 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
Yii = y i + ∑y
k =1, k ≠ i
ik
 The off-diagonal elements are the negative of the
admittances connecting buses i and j, i.e., Yij=-yji =
mutual admittance.
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28.  E.g. for a 4 bus system
Y11 Y12 Y13 Y14 
Y Y22 Y23 
Y24 
Y = 21
Y31 Y32 Y33 Y34 
 
Y41 Y42 Y43 Y44 
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29. The power flow equations
 The net complex power injection into a bus
Sk=Sgk-Sdk
Sk=VkIk*
Vk=| Vk|∠δk
Ik =Σ | Ykj|∠θkj | Vj|∠δj
Ik =Σ | Ykj|| Vj|∠ (θkj + δj)
Ik* =Σ | Ykj|| Vj|∠ -(θkj + δj)
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30. Sk=VkIk*
Sk= | Vk|∠δk x Σ | Ykj|| Vj|∠ -(θkj + δj)
Sk= Σ | Ykj| | Vk|| Vj|∠ (δk - δj - θkj )
Pk= Σ | Ykj| | Vk|| Vj|cos (δk - δj - θkj )
Qk= Σ | Ykj| | Vk|| Vj|sin (δk - δj - θkj )
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