ac powerflow part1

8/18/2019 AC PowerFlow Part1

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AC Power Flow

Dr. Shashidhara M Kotian,

IC, EEE F312 Power Systems


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BITS Pilani, K K Birla Goa Campus

• 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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7/34BITS Pilani, K K Birla Goa Campus

• 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.



• 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.



Classification of buses (‘n’ bus,‘m’ generator power system)



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.



• 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.



Gauss Seidel Load Flow(GSLF) technique



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.



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.



GSLF without PV bus



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



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!

ac powerflow part1

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