Load Flow (Studi Aliran Daya)

Load Flow

Studi aliran beban, u/ mendapatkan :

P

Q

|V|

pada bus yang belum diketahui nilainya

First step in solving the power flow is to create what is known as the bus admittance matrix, often call the Ybus.

Second step is to define the type of the bus

Three Types of Power Flow Buses

There are three main types of power flow buses

Load (PQ) at which P/Q are fixed; iteration solves for voltage magnitude (|V|) and angle ().

Slack at which the voltage magnitude and angle are fixed; iteration solves for P/Q injections

Generator (PV) at which P and |V| are fixed; iteration solves for voltage angle () and Q injection

y = admitansi saluran

Gauss Power Flow

*

* * *i

1 1

* * * *

1 1

*

*1 1,

*

*1,

We first need to put the equation in the correct form

S

S

S

S1

i i

i

i

n n

i i i ik k i ik kk k

n n

i i i ik k ik kk k

n ni

ik k ii i ik kk k k i

ni

i ik kii k k i

V I V Y V V Y V

V I V Y V V Y V

Y V Y V Y VV

V Y VY V

Y = elemen pada matriks Ybus

Modeling Shunts in the Ybus

from other lines

2 2

Since ( )2

2

1 1Note

kcij i j k i

kcii ii k

k k k kk

k k k k k k k

YI V V Y V

YY Y Y

R jX R jXY

Z R jX R jX R X

Two Bus System Example

1 21 1

1 1

2 2

( ) 112 16

2 0.03 0.04

12 15.9 12 16

12 16 12 15.9

cYV VI V jZ j

I Vj j

I Vj j

1

Y1/Z

Gauss Two Bus Power Flow Example

A 100 MW, 50 Mvar load is connected to a generator through a line with z = 0.02 + j0.06 p.u. and line charging of 5 Mvaron each end (100 MVA base). Also, there is a 25 Mvar capacitor at bus 2. If the generator voltage is 1.0 p.u., what is V2?

SLoad = 1.0 + j0.5 p.u.

2

Yc/2=j0.05

V1 = 10o

Gauss Two Bus Example, contd

2

2 bus

bus

22

The unknown is the complex load voltage, V .

To determine V we need to know the .

15 15

0.02 0.06

5 14.95 5 15Hence

5 15 5 14.70

( Note - 15 0.05 0.25)

jj

j j

j j

B j j j

Y

Y

Gauss Two Bus Example, contd*2

2 *22 1,2

2 *2

(0)2

( ) ( )2 2

1 S

1 -1 0.5( 5 15)(1.0 0)

5 14.70

Guess 1.0 0 (this is known as a flat start)

0 1.000 0.000 3 0.9622 0.0556

1 0.9671 0.0568 4 0.9622 0.0556

2 0

n

ik kk k i

v v

V Y VY V

jV j

j V

V

v V v V

j j

j j

.9624 0.0553j

Gauss Two Bus Example, contd

2

* *1 1 11 1 12 2

1

0.9622 0.0556 0.9638 3.3

Once the voltages are known all other values can

be determined, such as the generator powers and the

line flows

S ( ) 1.023 0.239

In actual units P 102.3 MW

V j

V Y V Y V j

12

2

, Q 23.9 Mvar

The capacitor is supplying V 25 23.2 Mvar

Gauss with Many Bus Systems

*( )( 1)

( )*1,

( ) ( ) ( )1 2

( 1)

With multiple bus systems we could calculate

new V ' as follows:

S1

( , ,..., )

But after we've determined we have a better

estimate of

i

i

nvv i

i ik kvii k k i

v v vi n

vi

s

V Y VY V

h V V V

V

its voltage , so it makes sense to use this

new value. This approach is known as the

Gauss-Seidel iteration.

Gauss-Seidel Iteration

( 1) ( ) ( ) ( )2 12 2 3

( 1) ( 1) ( ) ( )2 13 2 3

( 1) ( 1) ( 1) ( ) ( )2 14 2 3 4

( 1) ( 1) ( 1)( 1) ( )2 1 2 3 4

Immediately use the new voltage estimates:

( , , , , )

( , , , , )

( , , , , )

( , , , ,

v v v vn

v v v vn

v v v v vn

v v vv vn n

V h V V V V

V h V V V V

V h V V V V V

V h V V V V V

)

The Gauss-Seidel works better than the Gauss, and

is actually easier to implement. It is used instead

of Gauss.

Inclusion of PV Buses in G-S

i

i

* *

1

( )( ) ( )*

1

( ) ( )

To solve for V at a PV bus we must first make a

guess of Q :

Hence Im

In the iteration we use

k

n

i i ik k i ik

nvv v

i i ikk

v vi i i

S V Y V P jQ

Q V Y V

S P jQ

Inclusion of PV Buses, cont'd

( 1)

( )*( )( 1)

( )*1,

( 1)i i

Tentatively solve for

S1

But since V is specified, replace by V

i

vi

v nvv i

i ik kvii k k i

vi

V

V Y VY V

V

Two Bus PV Example

Bus 1

(slack bus)

Bus 2V1 = 1.0 V2 = 1.05

P2 = 0 MW

z = 0.02 + j 0.06

Consider the same two bus system from the previous

example, except the load is replaced by a generator 3

Two Bus PV Example, cont'd*2

2 21 1*22 2

* *2 21 1 2 22 2 2

2

( ) ( 1) ( 1)2 2 2

1

Im[ ]

Guess V 1.05 0

0 0 0.457 1.045 0.83 1.050 0.83

1 0 0.535 1.049 0.93 1.050 0.93

2 0 0.545 1.050 0.96 1.050 0.96

v v v

SV Y V

Y V

Q Y VV Y V V

v S V V

j

j

j

Generator Reactive Power Limits

The reactive power output of generators varies to maintain the terminal voltage; on a real generator this is done by the exciter

To maintain higher voltages requires more reactive power

Generators have reactive power limits, which are dependent upon the generator's MW output

These limits must be considered during the power flow solution.

Generator Reactive Limits, cont'd

During power flow once a solution is obtained check to make generator reactive power output is within its limits

If the reactive power is outside of the limits, fix Q at the max or min value, and resolve treating the generator as a PQ bus this is know as "type-switching" also need to check if a PQ generator can again

regulate

Rule of thumb: to raise system voltage we need to supply more vars