chapter 1 part2
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powerTRANSCRIPT
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17 March 2015 1
POWER SYSTEM ANALYSIS
Chapter 1:
Power Flow Analysis
(Part 2)
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What are the bus voltages if something happened to the network?. Will the network still operate safely?
How to know whats going on to the network tomorrow?/ next month?. Next year?. With load that keep changing from time and unpredictable? E.g not able to supply enough load, Tripping, shortage of fuel supply, outage due to maintenance schedule etc etc
If you have to analyze all the thousands of s/s, many km;sof lines, how r u going to do it efficiently simulate without need to go to each s/s to get the data? : the answer is
THROUGH COMPUTATION/MODELLING/SIMULATION !!
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Introduction
To determine the steady state analysis of an interconnected power system during normal operation
This system is assumed to be operating under balanced condition and is represented by a single-phase network
The network contains hundreds of nodes and branches with impedances specified in per-unit on a common MVA base
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Load-flow studies are performed to determine the steady-state operation ofan electric power system. It calculates the voltage drop on each feeder, thevoltage at each bus, and the power flow in all branch and feeder circuits.
It also determine if system voltages remain within specified limits undervarious contingency (unforeseen event) conditions, and whether theequipment such as transformers and conductors are overloaded.
Load-flow studies are often used to identify the need for additionalgeneration, capacitive, or inductive VAR support, or the placement ofcapacitors and/or reactors to maintain system voltages within specifiedlimits.
Losses in each branch and total system power losses are also calculated.
Necessary for planning, economic scheduling, and control of an existingsystem as well as planning its future expansion
Pulse of the system
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Back bone of power system
Planning
Operation
Economic scheduling
Exchange of power between utilities
Transient stability
Contingency studies
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Generator supplies the demand & cater for the losses Bus voltages magnitudes close to rated values Generators operates within specified real & reactive
power limit Transmission lines & transformers are not
overloaded
Therefore, the utilities need some kind ofcomputation tool or program in order to successfullyobtained these information for operation purposes-its called power flow or load flow program
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We model the power system using the nodal voltage method.
We present a careful formulation of the basic power flow problem
We investigate its solution by the: Gauss Seidel Method Newton Raphson Method Fast Decoupled Methodand analyze the solution characteristics
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Previously, we have learned how to model the components of power system (Gen, Transformer,T-line) from single line diagram.(recall your earlier power engineering subjects)
Now, were going to learn the steady-state analysis of an interconnected power system during normal operation which have hundreds/thousands of nodes and branches impedances is already converted into per units on a common MVA
base.
The way to analyze these are using network equations such as: If analyze Voltages & Current only and formulate the admittance
matrix Node-Voltage Method (if currents known and want to solve for voltages only , or vice versa)
If analyze Power Flow Iteration Method (if power are known, and will become non-linear) thus, is called power flow/load flow equations. E.g. Gauss Seidel, NR,FD
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The most common way to represent a power systemnetwork is to use the node-voltage method:
Given the voltages of generators at all generatornodes, and knowing all impedances of machines andloads, one can solve for all the currents in the typicalnode voltage analysis methods using Kirchoff's currentlaw (KCL).
First the generators are replaced by equivalent currentsources and the node equations are written in the form:
[I]=[Y][V]I = injected current vector/matrix,
Y = is the admittance vector/matrix
V = is the node voltage vector/matrices.
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Analysis using Gauss
Seidel/NR/FD to solve for P,
Q & V.
1) single line diagram2) reactance/impedance diagram
4) bus admittance
matrices3) Admittance diagram
NOTE : 2,3 & 4 are in PER UNIT!
Using Nodal-
voltage method
(I=YV)Analysis
using Nodal Voltage again to solve V & I
OR
Using conversion
Y=1/Z
Using per unit
system taught
earlier
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Definition : a per-unit system is the expressionof system quantities (power, voltage, current &impedance) as fractions of a defined base unitquantity (instead of VA, volt, ampere, ohms) bythe equation:
Main advantage : Calculations are simplifiedbecause quantities expressed as per-unit are thesame regardless of the voltage level.
actual quantityquantity in per unit
base value of quantity
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The main reasons are : Similar apparatus (generators, transformers, lines) will
have similar per-unit impedances and losses expressedon their own rating, regardless of their absolute size. )
Per-unit quantities are the same on either side of atransformer, independent of voltage level (i.e. It wouldbe very difficult to continually have to refer impedancesto the different sides of the transformers. )
By normalizing quantities to a common base,calculations are simplified.
Overall, the per unit system was developed tomake manual analysis of power systems easier.
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Generally base values of power (Sbase) and voltage(Vbase)are chosen. The base power (Sbase, Pbase, Qbase) may be the rating of a single piece of apparatus such as a motor or generator.
The base voltage (Vbase) is chosen as the nominal rated voltage of the system. All other base quantities (Ibase, Zbase) are derived from these two base quantities
Once the base power and the base voltage are chosen, the base current and the base impedance are determined by the natural laws of electrical circuits. (refer formulas in chapter 2, Glover)
REMEMBER : BASE QUANTITIES HAVE NO ANGLE, ONLY HAVE MAGNITUDE
actual quantityquantity in per unit
base value of quantity
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Normally, we just pick the rated voltage & power
as base value:
Then compute base values for currents and
impedances:
ratedb VV
ratedratedratedb QPSS
b
bb
V
SI
b
2
b
b
bb
S
V
I
VZ
e.g. if Srated=10VA, then, we can select the base power to be:
Sbase = 10VAPbase = 10WattQbase = 10Var
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And the per-unit system is:
b
..V
actualup
VV
b
..I
actualup
II
b
..S
actualup
SS
b
..Z
actualup
ZZ
%100% .. upZZ Percent of base Z
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Per Unit System for single phase
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One-phase circuits
LN,1b1 VV
LN1b IVSS where
neutraltolineLN VV
currentline II
LN,2b2 VV
b1
bb1
V
SI
b2
bb2
V
SI
Note :
1= winding 1
2= winding 2
1V 2V
Zone of
base 1
Zone of
base 2
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b
2
b1
b1
b1b1
S
)(V
I
VZ
b
2
b2
b2
b2b2
S
)(V
I
VZ
*
bSpupu
actualpu IV
SS
cosSb
pupuactual
pu IVP
P
sinSb
pupuactual
pu IVQ
Q
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Per Unit System for three phase
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Three-phase circuits
LbLNbbb IVSS ,,1,3, 33 where
3/bLLbLN VV
)connection wye(assume Lcurrentline III
LLLb IVS 33,
2,2,1,1,3, 33 bLbLLbLbLLb IVIVS
For low voltage winding3/17/2015
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3,
2
1,
3,
1,1,
1,
1,
1
)(3
3 b
bLL
b
bLLbLL
L
LN
bS
V
S
VV
I
VZ
3,
2
2,
2
)(
b
bLL
bS
VZ
*
2
2,3
**
3,
3
1,33
3
S
VZ
SIVIV
IV
S
SS pupupu
bb
LL
b
pu
1,
3,
13 bLL
b
bV
SI
2,
3,
23 bLL
b
bV
SI
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THE PER-UNIT SYSTEM
Change of Base
If the voltage base
are the same:
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G M
T1 T2
T3 T4
1 2 3 4
5 6Load
Line 1
Line 2
220kV
110kV
EXAMPLE:
The one line diagram of a three-phase power system is
shown in figure above. Select a common base of 100MVA
and 22kV on the generator side. Draw an impedance
diagram with all impedances including the load impedance
marked in per-unit. The manufacturers data for each
device is given as follow:
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G : 90MVA 22kV X=18%
T1 : 50MVA 22/220kV X=10%
T2 : 40MVA 220/11kV X=6%
T3 : 40MVA 22/110kV X=6.4%
T4 : 40MVA 110/11kV X=8%
M : 66.5MVA 10.45kV X=18.5%
LINE 1: 48.4
LINE 2: 65.43
3-PHASE LOAD : 57MVA, 0.6 PF LAG AT 10.45kV
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MG
j0.2 j0.1 j0.15
j0.16 j0.54 j0.2
j0.25
j1.2667
0.95
j0.2
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IN EXAMPLE ABOVE BY CHANGING THE BASE FROMGENERATOR TO LINE 1. WHICH THE BASE ARE200kV AND 100MVA. DRAW THE IMPEDANCEDIAGRAM.
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Bus admittance matrix is just a matrix of all interconnected admittances between busses.
Step 1: Number all the nodes of the system from 0 to n. Node 0 is the reference node (or ground node).
Step 2: Replace all generators by equivalent current sources in parallel with an admittance.
Step 3: Replace all lines, transformers, loads to equivalent admittances whenever possible. The rule for this is simple:
where and y, z are generally complex numbers.
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Step 4: The bus admittance matrix Y is then formed by inspection
as follows :
The current vector/matrix is next found from the sources connected
to nodes 0 to n . If no source is connected, the injected current
would be 0.
The equations which result are called the node-voltage equations
and are given the "bus" subscript in power studies thus:
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VYI or
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In order to obtain the
node voltage equations :
Impedances are
expressed in per-unit
on a common MVA
base
Impedances are
converted to
admittance
Nodal solutions is
based upon kirchoffs
current law
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)(0
)()()(0
)()(
)()(
3434
433413132323
322312122202
311321121101
VVy
VVyVVyVVy
VVyVVyVyI
VVyVVyVyI
434334
4343342313223113
323223201122
31321211312101
0
)(0
)(
)(
12
VyVy
VyVyyyVyVy
VyVyyyVyI
VyVyVyyyI
y
Rearrange,
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344334
233223
3113
122112
3444
34231333
232022
13121011
13
12
yYY
yYY
yYY
yYY
yY
yyyY
yyyY
yyyY
y
y
Node equation reduces to:
Y14 = Y41 = 0 , Y24 = Y42 = 0
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Ibus = Ybus Vbus
Bus Admittance Matrix
Vector of
the injected
bus currents
Vector of
bus
voltages
measured
from the
reference
node
Diagonal element self-admittance / driving point admittance
Off-diagonal element mutual admittance / transfer admittance
equal to the negative of the admittance between the node
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Ibus = Ybus Vbus
Bus admittance matrix for the network in previous figure is:
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Figure above shows the one line diagram of a simple four-bus system. Table below gives the line impedances identified by the buses on which these terminate. The shunt admittance at all the buses is assumed negligible.(a) Find the Ybus assuming that the line shown dotted is not connected.
(b) What modifications need to be carried out in Ybus if the line shown dotted is connected.
1
3 4
2
Line, R, pu X, pu
Bus to bus
1-2 0.05 0.15
1-3 0.10 0.30
2-3 0.15 0.45
2-4 0.10 0.30
3-4 0.05 0.15
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(a) Ybus without dotted line.
Line, G, pu B, pu
Bus to bus
1-2 2.0 - 6.0
1-3 1.0 - 3.0
2-3 0.666 - 2.0
2-4 1.0 - 3.0
3-4 2.0 - 6.0
Ybus = 1 j3 0 -1 + j3 00 1.666 j5 -0.666 + j2 -1 + j3
-1 + j3 -0.666 + j2 3.666 j11 -2 + j6
0 -1 + j3 -2 + j6 3 j9
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(b) Ybus with dotted line.
Line, G, pu B, pu
Bus to bus
1-2 2.0 - 6.0
1-3 1.0 - 3.0
2-3 0.666 - 2.0
2-4 1.0 - 3.0
3-4 2.0 - 6.0
Ybus = 3 j9 -2 + j6 -1 + j3 0-2 + j6 3.666 j11 -0.666 + j2 -1 + j3
-1 + j3 -0.666 + j2 3.666 j11 -2 + j6
0 -1 + j3 -2 + j6 3 j9
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