Post on 07-Jul-2018
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AC Power Flow
Dr. Shashidhara M Kotian,
IC, EEE F312 Power Systems
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• AC power flow analysis is basically a steady-state analysis of the AC
transmission and distribution grid.
• Essentially, AC power flow method computes the steady state
values of bus voltages and line power flows from the knowledge of
electric loads and generations at different buses of the system under
study.
• We will look into the power flow solution of the AC transmission grid
only (the solution methodology of AC distribution grid will not be
covered).
• Only a balanced system is considered in which the transmission
lines and loads are balanced (the impedances are equal in all the
three phases) and the generator produces balanced three phase
voltages (magnitudes are equal in all the phases while the angular
difference between any two phases is 120 degree).
Load Flow Analysis
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Basic power flow equation
Where,
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Therefore, for a ‘n’ bus system
Or,
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Complex power injected at bus ‘i’ is given by,
Now,
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• It can be seen that for any ith bus, there are two
equations (load-flow equations).
• Therefore, for a ‘n’-bus power system, there are
altogether ‘2n’ load-flow equations.
• There are four variables (Vi, θi, Pi and Qi) associated
with the ith bus.
• Thus for the ‘n’-bus system, there are a total of ‘4n’
variables.
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• As there are only ‘2n’ equations available, out of these
‘4n’ variables, ‘2n’ quantities need to be specified and
remaining ‘2n’ quantities are solved from the ‘2n’ load-
flow equations.
• As ‘2n’ variables are to be specified in a ‘n’ bus system,
for each bus, two quantities need to be specified.
• For this purpose, the buses in a system are classified
into three categories and in each category, two different
quantities are specified.
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Classification of buses (‘n’ bus,‘m’ generator power system)
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Therefore, in a ‘n’ bus, ‘m’ generator system,
the unknown quantities are: Vi (total ‘n-m’ of them) and θi(total ‘n-1’ of them). Therefore, total number of unknown
quantities is ‘2n-m-1’.
the specified quantities are: Pi (total ‘n-1’ of them) and Qi(total ‘n-m’ of them). Hence total number of specified
quantities is also ‘2n-m-1’.
As the number of unknown quantities is equal to the number of
specified quantities, the load-flow problem is well-posed.
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• The load-flow equations represent a set of simultaneous,non-linear, algebraic equations.
• As the set of equations is non-linear, no closed form,
analytical solution for these equations exist.
• Hence, these equations can only be solved by using
suitable numerical iterative techniques.
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Gauss Seidel Load Flow(GSLF) technique
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Now, initially to understand the basic GSLF procedure,
let us assume that m = 1, i.e., there is only one generator
(which is also the slack bus) and the rest ‘(n-1)’ buses
are all load buses.
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Initial Guess:
As any power system is generally expected to operate at
the normal steady-state operating condition (with the bus
voltage magnitudes maintained between 0.95 - 1.05
p.u.), all the unknown bus voltage are initialized to
p.u.
This process of initializing all bus voltage to
is called flat start.
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GSLF without PV bus
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For a system having multiple generators, the bus voltageinitialisation is carried out in a two step procedure:
i) the load buses are initialised with flat start.
ii) the magnitudes of the voltages of the PV buses are
initialised with the corresponding specified voltage
magnitudes while initialising all these voltage angles to
0o
where V jsp is thespecified bus voltage magnitude of the jth generator)
GSLF with PV bus
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Assume that the ‘m’ generators are connected to the first‘m’ buses (bus ‘1’ being the slack bus) and remaining
‘(n-m)’ buses are load buses.
Complete GSLF algorithm
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An Example on GSLF
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Bus Data
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YBUS matrix
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Use of acceleration factor
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Using an accelerating factor of 1.6
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Bus 4 is PV bus
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A4(1)
=
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Thank You!