Download - Analisis Aliran Daya 1
-
7/25/2019 Analisis Aliran Daya 1
1/38
ANALISIS ALIRAN DAYA
-
7/25/2019 Analisis Aliran Daya 1
2/38
Introduction
Perencanaan, desain dan operasi sistem tenagamemerlukan beberapa perhitungan untuk menganalisaperformansi kondisi tunak sistem tenaga pada berbagaikondisi operasi dan untuk menganalisa efek perubahankonfigurasi peralatan.
Pertanyaan dasar pada studi aliran daya adalah:
Diberikan konsumsi daya beban pada pada semua buspada suatu konfigurasi sistem tenaga yang telahdiketahui dan produksi daya peda setiap generator .
Temukan aliran daya pada setiap saluran dan trafo pada
jaringan serta magnitude dan susut phase teganganpada setiap bus.
Analisa solusi permasalahan aliran daya pada berbagaikondisi membantu memastikan bahwa sistem tanagatelah didasain memenuhi kriteria performansi yangdiharapkan.
-
7/25/2019 Analisis Aliran Daya 1
3/38
Manfaat studi aliran daya
Dengan penggunaan studi aliran daya, kitadapat menentukan:
Pembebanan komponen atau rangkaian
Tegangan bus kondisi tunakAliran daya reaktif
Seting tap trafo
Rugi-rugi sistem
Set poin regulator tegangan / exiter generator Unjuk kerja sistem pada kondisi darurat.
-
7/25/2019 Analisis Aliran Daya 1
4/38
Load flow problem
Perhitungan load flow menentukan kondisi sistem tenagapada daya beban dan distribusi pembangkitan yangdiberikan. Dalam kenyataannya aliran daya pada salurandan tegangan bus adalah berfluktuasi sacara konstanpada kisaran kecil akibat adanya perubahan beban. Nilai
fluktuasi yang kecil ini diabaikan pada perhitungan efeksteady state pada sistem.
Oleh karena distribusi beban bervariasi selama periodewaktu yang berbada. Maka diperlukan mendapatkansolusi load flow yang merepresentasikan kondisi sistemyang berbeda seperti pada beban puncak, beban rata-
rata atau beban ringan.
-
7/25/2019 Analisis Aliran Daya 1
5/38
Solusi pada kondisi pembebanan yang berbeda ini dapatdigunakan untuk menentukan:
Mode operasi optimum untuk kondisi normal, misalmenentukan seting piranti kontrol tegangan yangtepat.
Bagaimana sistem akan merespon kondisi tidaknormal misal ketika terjadi outage (pemadaman)pada saluran atau trafo.
Dasar untuk menentukan kondisi kapan diperlukanpenambahan peralatan baru.
Dasar untuk menentukan efektifitas alternatif-alternatifbaru
Dasar untuk menentukan kebutuhan sistem di masayang akan datang.
-
7/25/2019 Analisis Aliran Daya 1
6/38
System Representation (1)
A simplified visual means of representing the complete
system is essential to understanding the operation of the
system under its various possible operating modes =>
single-line diagram
The single-line diagram consists of a drawing identifying
buses and interconnecting lines. Loads, generators,
transformers, reactors, capacitors, etc., are all shown in
their respective places in the system
-
7/25/2019 Analisis Aliran Daya 1
7/38
System Representation (2)
System Data: Most load flow programs perform their
calculations using a per unit representation of the
system, while still some programs work with volts,
amperes, and ohms.
Bus Data: The data includes: bus number, bus name,
bus type, load, shunt, per unit voltage and angle, and bus
base kV. There three bus types, they are: Load bus (P-Q
bus), Generator bus (P-V bus), and Swing bus (Slack
bus)
-
7/25/2019 Analisis Aliran Daya 1
8/38
System Representation (2)
Generator Data:
Real power output in MW.
Maximum reactive power output in MVAR, that is the
machine maximum reactive limit. Minimum reactive power output in MVAR, that is the
machine minimum reactive limit.
Scheduled voltage in per unit
Generator in-service/out-of-service code, or
generator operated continuously, operatedintermittently, or operated as spare.
-
7/25/2019 Analisis Aliran Daya 1
9/38
System Representation (3)
Line Data: Resistance
Reactance
Charging susceptance (shunt capacitance)
Line ratings
Line in-service/out-of-service code
Line-connected shunts
-
7/25/2019 Analisis Aliran Daya 1
10/38
System Representation (4)
Transformer Data:
Impedansi trafo
Tap setting in per unit
Tap angle in degrees Maximum tap position
Minimum tap position
Scheduled voltage range with tap step size
-
7/25/2019 Analisis Aliran Daya 1
11/38
Three Types of Power Flow Buses
There are three main types of power flowbuses
Load (PQ) at which P/Q are fixed;
iteration solves for voltage magnitudeand angle.
Slack at which the voltage magnitudeand angle are fixed; iteration solves for
P/Q injections Generator (PV) at which P and |V| are
fixed; iteration solves for voltage angleand Q injection
-
7/25/2019 Analisis Aliran Daya 1
12/38
Load Flow Bus Specifications
Bus Type P Q V Comments
Load Usual Load Representation
whenQ
-
-
7/25/2019 Analisis Aliran Daya 1
13/38
Power Flow Analysis
When analyzing power systems we knowneither the complex bus voltages nor thecomplex current injections
Rather, we know the complex power beingconsumed by the load, and the power beinginjected by the generators plus their voltagemagnitudes
Therefore we can not directly use the Ybusequations, but rather must use the powerbalance equations
-
7/25/2019 Analisis Aliran Daya 1
14/38
Power Balance Equations
1
bus
1
From KCL we know at each bus i in an n bus system
the current injection, , must be equal to the current
that flows into the network
Since = we also know
i
n
i Gi Di ik k
n
i Gi Di ik k k
I
I I I I
I I I Y V
I Y V
*iThe network power injection is then S i iV I
-
7/25/2019 Analisis Aliran Daya 1
15/38
Power Balance Equations, contd*
* * *i
1 1
S
This is an equation with complex numbers.
Sometimes we would like an equivalent set of real
power equations. These can be derived by defining
n n
i i i ik k i ik k k k
ik ik ik
i
V I V Y V V Y V
Y G jB
V
jRecall e cos sin
iji i i
ik i k
V e V
j
-
7/25/2019 Analisis Aliran Daya 1
16/38
Real Power Balance Equations
* *i
1 1
1
i 1
i1
S ( )
(cos sin )( )
Resolving into the real and imaginary parts
P ( cos sin )
Q ( sin cos
ikn n
ji i i ik k i k ik ik
k k
n
i k ik ik ik ik k
n
i k ik ik ik ik Gi Dik
n
i k ik ik ik ik
P jQ V Y V V V e G jB
V V j G jB
V V G B P P
V V G B
)k Gi DiQ Q
-
7/25/2019 Analisis Aliran Daya 1
17/38
Load Flow Solution Methods
Problem simulations VYI
*
*
i
iii
V
jQPI
VYV
jQP
*
-
7/25/2019 Analisis Aliran Daya 1
18/38
Power Flow Requires Iterative Solution
i
bus
*
* * *i
1 1
In the power flow we assume we know S and the
. We would like to solve for the V's. The problem
is the below equation has no closed form solution:
S
Rath
n n
i i i ik k i ik k
k k
V I V Y V V Y V
Y
er, we must pursue an iterative approach.
-
7/25/2019 Analisis Aliran Daya 1
19/38
Iterative Solution Algorithms
The primary parameters are: Active power into the transmission network
Reactive power into the transmission network
Magnitude of bus voltage (voltage to ground)
Angle of bus voltage referred to a common reference
-
7/25/2019 Analisis Aliran Daya 1
20/38
Gauss Iteration
There are a number of different iterative methods
we can use. We'll consider two: Gauss and Newton.
With the Gauss method we need to rewrite our
equation in an implicit form: x = h(x)
To iterate we fir (0)
( +1) ( )
st make an initial guess of x, x ,
and then iteratively solve x ( ) until we
find a "fixed point", x, such that x (x).
v vh x
h
-
7/25/2019 Analisis Aliran Daya 1
21/38
Gauss Iteration Example
( 1) ( )
(0)
( ) ( )
Example: Solve - 1 0
1
Let k = 0 and arbitrarily guess x 1 and solve
0 1 5 2.61185
1 2 6 2.616122 2.41421 7 2.61744
3 2.55538 8 2.61785
4 2.59805 9 2.61798
v v
v v
x x
x x
k x k x
-
7/25/2019 Analisis Aliran Daya 1
22/38
Stopping Criteria
( ) ( ) ( 1) ( )
A key problem to address is when to stop the
iteration. With the Guass iteration we stop when
with
If x is a scalar this is clear, but if x is a vector we
need to generalize t
v v v vx x x x
( )
2i
2 1
he absolute value by using a norm
Two common norms are the Euclidean & infinity
max x
v
j
n
i ii
x
x
x x
-
7/25/2019 Analisis Aliran Daya 1
23/38
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 k
k k
n n
i i i ik k ik k
k k
n ni
ik k ii i ik k
k k k i
ni
i ik k
ii k k i
V I V Y V V Y V
V I V Y V V Y V
Y V Y V Y V V
V Y VY V
-
7/25/2019 Analisis Aliran Daya 1
24/38
Gauss Two Bus Power Flow Example
A 100 MW, 50 Mvar load is connected to agenerator through a line with z = 0.02 + j0.06p.u. and line charging of 5 Mvar on each end (100MVA base). Also, there is a 25 Mvar capacitor atbus 2. If the generator voltage is 1.0 p.u., what
is V2?
SLoad = 1.0 + j0.5 p.u.
-
7/25/2019 Analisis Aliran Daya 1
25/38
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 .
1 5 150.02 0.06
5 14.95 5 15Hence
5 15 5 14.70(Note - 15 0.05 0.25)
jj
j j
j jB j j j
Y
Y
-
7/25/2019 Analisis Aliran Daya 1
26/38
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 jj V
V
v V v V
j j
j j
.9624 0.0553j
-
7/25/2019 Analisis Aliran Daya 1
27/38
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
1
22
, Q 23.9 Mvar
The capacitor is supplying V 25 23.2 Mvar
-
7/25/2019 Analisis Aliran Daya 1
28/38
Slack Bus
In previous example we specified S2 and V1and then solved for S1 and V2.
We can not arbitrarily specify S at all buses
because total generation must equal totalload + total losses
We also need an angle reference bus.
To solve these problems we define one bus
as the "slack" bus. This bus has a fixedvoltage magnitude and angle, and a
varying real/reactive power injection.
-
7/25/2019 Analisis Aliran Daya 1
29/38
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
n
vv ii ik kv
ii k k i
v v v
i n
v
i
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.
-
7/25/2019 Analisis Aliran Daya 1
30/38
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 v
n 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.
-
7/25/2019 Analisis Aliran Daya 1
31/38
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
-
7/25/2019 Analisis Aliran Daya 1
32/38
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 V V Y V V
v S V V
j
j
j
-
7/25/2019 Analisis Aliran Daya 1
33/38
Generator Reactive Power Limits
The reactive power output of generatorsvaries to maintain the terminal voltage; on areal generator this is done by the exciter
To maintain higher voltages requires morereactive power
Generators have reactive power limits, whichare dependent upon the generator's MW
output These limits must be considered during the
power flow solution.
-
7/25/2019 Analisis Aliran Daya 1
34/38
Generator Reactive Limits, cont'd
During power flow once a solution isobtained check to make generator reactivepower output is within its limits
If the reactive power is outside of thelimits, fix Q at the max or min value, andresolve treating the generator as a PQ bus
this is know as "type-switching"
also need to check if a PQ generator canagain regulate
Rule of thumb: to raise system voltage weneed to supply more vars
-
7/25/2019 Analisis Aliran Daya 1
35/38
Accelerated G-S Convergence
( 1) ( )
( 1) ( ) ( ) ( )
(
Previously in the Gauss-Seidel method we were
calculating each value x as
( )
To accelerate convergence we can rewrite this as
( )
Now introduce acceleration parameter
v v
v v v v
v
x h x
x x h x x
x
1) ( ) ( ) ( )( ( ) )
With = 1 this is identical to standard gauss-seidel.
Larger values of may result in faster convergence.
v v vx h x x
-
7/25/2019 Analisis Aliran Daya 1
36/38
Accelerated Convergence, contd
( 1) ( ) ( ) ( )
Consider the previous example: - 1 0
(1 )
Comparison of results with different values of
1 1.2 1.5 2
0 1 1 1 1
1 2 2.20 2.5 3
2 2.4142 2.5399 2.6217 2.464
3 2.5554 2.6045 2.6179 2.675
4 2.59
v v v v
x x
x x x x
k
81 2.6157 2.6180 2.596
5 2.6118 2.6176 2.6180 2.626
-
7/25/2019 Analisis Aliran Daya 1
37/38
Gauss-Seidel Advantages
Each iteration is relatively fast(computational order is proportionalto number of branches + number ofbuses in the system
Relatively easy to program
-
7/25/2019 Analisis Aliran Daya 1
38/38
Gauss-Seidel Disadvantages
Tends to converge relatively slowly,although this can be improved withacceleration
Has tendency to miss solutions,particularly on large systems
Tends to diverge on cases with negativebranch reactances (common with
compensated lines)
Need to program using complexnumbers