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POWER SYSTEM 2 - EPO 643 1
POWER (LOAD) FLOW STUDY
INTRODUCTION
BASIC TECHNIQUES
TYPE OF BUSES
Y BUS MATRIX
POWER SYSTEM COMPONENTS
BUS ADMITTANCE MATRIX
POWER SYSTEM 2 - EPO 643 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 powerWhy we need load flow study?
In a power system under a given set of load conditions
POWER SYSTEM 2 - EPO 643 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 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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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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Power system components
Generator Transmission Lines Load
Figure 1
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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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Figure 2
POWER SYSTEM 2 - EPO 643 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”)
POWER SYSTEM 2 - EPO 643 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 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 i , respectively
POWER SYSTEM 2 - EPO 643 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.
POWER SYSTEM 2 - EPO 643 13
Slack or Swing Bus 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.”
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Bus types Quantities specified
Unknown values
Generator Bus |Vi| , Pi Qi , i
Load Bus Pi , Qi |Vi| , i
Slack Bus |Vi| , i Pi , Qi
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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.
POWER SYSTEM 2 - EPO 643 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.
POWER SYSTEM 2 - EPO 643 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
(c)
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.
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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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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
POWER SYSTEM 2 - EPO 643 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 22
Forming the Admittance Matrix
y4 y3
y1 y2
I4
I3 I2
I1
4 3
2
1
y34
y23
y13
y12
POWER SYSTEM 2 - EPO 643 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 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)
POWER SYSTEM 2 - EPO 643 25
Admittance Matrix
)y y (y-00
y-)y yy y (y-y-
0y- )y y y (y-
0y-y- )y y y (
43 443
34343231 33231
232321 221
131213121
POWER SYSTEM 2 - EPO 643 26
Matrix Equation
4
3
2
1
43 443
34343231 33231
232321 221
131213121
4
3
2
1
)y y (y-00
y-)y yy y (y-y-
0y- )y y y (y-
0y-y- )y y y (
V
V
V
V
I
I
I
I
POWER SYSTEM 2 - EPO 643 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
The off-diagonal elements are the negative of the admittances connecting buses i and j, i.e., Yij=-yji = mutual admittance.
N
ikkikiii yyY
,1
POWER SYSTEM 2 - EPO 643 28
E.g. for a 4 bus system
44434241
34333231
24232221
14131211
YYYY
YYYY
YYYY
YYYY
Y
POWER SYSTEM 2 - EPO 643 29
The power flow equations
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)
POWER SYSTEM 2 - EPO 643 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 )