Load Flow GS
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Transcript of Load Flow GS
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Load Flow (Studi Aliran Daya)
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Load Flow
Studi aliran beban, u/ mendapatkan :
P
Q
|V|
pada bus yang belum diketahui nilainya
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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
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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
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y = admitansi saluran
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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
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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
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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
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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
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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
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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
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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
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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.
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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.
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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
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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
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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
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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
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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.
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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