chapter 1 part2

17 March 2015 1

POWER SYSTEM ANALYSIS

Chapter 1:

Power Flow Analysis

(Part 2)

17 March 2015 2

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 !!

2

17 March 2015 3UNIVERSITI TEKNIKAL MALAYSIA MELAKA

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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

UNIVERSITI TEKNIKAL MALAYSIA

MELAKA

17 March 2015 9

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


MELAKA

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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.


MELAKA

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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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MELAKA

17 March 2015 2020

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

3/17/2015


MELAKA

17 March 2015 2121

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

3/17/2015


MELAKA

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Per Unit System for three phase

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MELAKA

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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


MELAKA

17 March 2015 2424

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

3/17/2015


MELAKA

17 March 2015 25

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.


MELAKA

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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:


MELAKA

VYI or

17 March 2015 33

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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chapter 1 part2

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