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Learning Outcomes

Students should be able to

Identify the symbols and draw the one-line

diagram Construct the impedance diagram and find

the per unit values.

Solve the per unit problems of a single-phaseand three-phase circuit.

Identify the advantages of per unit system.

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Power System Representation :

One Line Diagram Represent or portray the interconnection of the power system

components used extensively in power flow studies

Also referred as Single-line Diagram

In a one line diagram, lines on paper (or on a computer screen)represent wires.

Advantage: Simplicity Rather than drawing all three wires in a three-phase

system, it is normal to simplify things by representing allthree phases with one line thus the name one linediagram

Equivalent circuit of the components are replaced by their

standard symbols The completion of the circuit through the neutral isomitted.

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One Line Diagram - Symbols

two-winding

transformercurrent transformer

two-winding

transformer

generator

bus

voltage transformer

capacitor

circuit breaker

transmission line

delta connection

wye connection

circuit breaker

fuse

surge arrestor

static load disconnect

Symbols used in one line diagram (from ANSI and IEEE)

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Exercise

Draw a one line diagram for a simple

power system which composes of the

following equipments :

Generator (1 unit), transformer,

transmission line, transformer and load

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L7

L6

L2

L1

L3

1 2

3

54

L5L4

generator

transformer

bus coding

circuit breaker

bus

load

line coding

tie line connection

with neighboring

system

transmission line

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One-line diagram - symbols

The main component of a one-line (or

single line) diagram are : Buses,

Branches, Loads, Machines, 2 windingTransformers, Switched Shunts,

Reactor and Capacitor Banks.

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One-line diagram - explanation

Buses are represented as a dot, circle or a thick line.

The bus name (EAST500) and number (202) are given, aswell as the voltage measured on the line (510.5kV and1.021V in per unit).

Final characteristic given is the angle(-26.1 degree). The voltage is indicated by the color of the bus. Red

indicates 500kV

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Branches are represented as a thin line.

The real power P, as shown on the branch below, flowsfrom 478MW to -473MW and the reactive power Q, flowsfrom 89.9MVA flow to -229.4MVA.

The power flows from positive number to the negativenumber, and the number on top is the real power whilethe number on the bottom is the reactive power.

The voltage is indicated by the color of the bus. Redindicates 500kV

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Loads are represented as a triangle with the IDnumber located inside the triangle.

The real power, PLOAD, is donated by the numberon top(250MW), and the reactive power QLoad, is

denoted by the number on bottom (100Mvar) The voltage is indicated by the color of the load. In

this example, black indicates 230kV

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Machines are represented as a circle with the ID numberlocated inside the circle.

The real power PGEN, is denoted by the number on top(321.0MW), and the reactive power QGEN, is denoted by

the number on bottom (142.3RMVAR). The R indicatesthis machine is in voltage regulation mode, and it iscontrolling a specific bus to a voltage set point whichrequires it to generate 142.3MVAR.

The voltage is indicated by the color of the machine. In

this example, red indicates 500kV.

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One-line diagram - explanation Reactor bank are represented as an inductor at the end

of a line. The number on the top (or on the left if thereactor is shown vertical) indicates that the real chargeadmittance, GSHUNT of the line (3.6MW).

If you see a SW instead of a number, you are looking ata switched shunt compensator.

The number on bottom indicates the initial reactivecharge admittance, BSHUNT, of the line ( 490.0MVAR )

The voltage is indicated by the color of the reactor. In

this example, red indicates 500kV

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Capacitor bank are represented as a capacitor at the endof a line. The number on the top (or on the left if thecapacitor is shown vertical) indicates that the realcharge admittance, GSHUNT of the line (0.0MW).

If you see a SW instead of a number, you are looking ata switched shunt compensator.

The number on bottom indicates the initial reactivecharge admittance, BSHUNT, of the line (-1080.0MVAR ).

The voltage is indicated by the color of the capacitor.

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Switched shunts are represented as either a capacitor orinductor at the end of a line. The SW shown on top (oron the left if the shunt is shown vertical) indicates thatthis unit is a switched shunt compensator. Permanently

installed reactor or capacitor bank. The number on bottom indicates the initial reactive

charge admittance, BINIT(253.MVAR or 599.6MVAR)

The voltage is indicated by the color of the shunt. In thisexample, red indicates 21.6kV

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T i l di i d i i i d b i i

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Typical medium sized city transmission and sub-transmission system.

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Per Unit System

The per unit system is widely used in

the power system industry to express

values of voltages, currents, powers,and impedances of various power

equipment.

It is mainly used for transformers andAC machines

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One line Diagram and Per unit system

A one-line diagram is a simplified

graphical representation of a three

phase system, used extensively inpower flow studies

Per unit system used extensively in

one-line diagram to further simplifythe process

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Per-Unit System

An interconnected power system typically consistsof many different voltage levels (sizes and nominalvalues) given a system containing severaltransformers and/or rotating machines.

A simple power system

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Base value

Specify the base values of current and

voltage, base impedance,

kilovoltamperes can be determined Quantities and base value selected

voltage, base value in kilovolts, kV

current, base value in ampere, A

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Base values

Generally the following two base values are chosen :

The base power = nominal power of the equipment

The base voltage = nominal voltage of theequipment

The base current and The base impedance are

determined by the natural laws of electrical circuits

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Base values

Usually, the nominal apparent power

(S) and nominal voltage (V) are taken

as the base values for power (Sbase) andvoltage (Vbase).

The base values for the current (Ibase)

and impedance (Zbase) can becalculated based on the first two base

values.

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Base value

For single phase system

A,currentbase

V,voltagebase,impedanceBase

kV,voltagebase

kVAbaseA,currentBase

LN

LN

1

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Base value

For single phase system

11

11

1

2

LN

MVAbaseMW,powerBase

kVAbasekW,powerBase

MVA)KV,voltagebase(,impedanceBase

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Base value - example

Base kVA3 = 30,000 kVA

and Base kVLL = 120 kVA

therefore Base kVA 1 = 30,000 / 3 = 10,000 kVA

and Base kV LN= 120 / 3 = 69.2 kVA

valuebased

quantitytheofvalueactualquantityanyofvalueunitper

For actual line-to-line voltage 108 kV, the line-to-neutral

voltage, VLNis 108/ 3 = 62.3

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Per unit value - example

and

Per-unit voltage = 108/120 (3) OR

= 62.3/69.2 (1)

= 0.9

For three-phase power of 18,000 kW,

Per-unit power = 18,000/30,000 (3 ) OR

= 6,000/10,000(1)

= 0.6

valuebased

quantitytheofvalueactualquantityanyofvalueunitper

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Per unit value

e.g. in a synchronous generator with

13.8 kV as its nominal voltage, instead

of saying the voltage is 12.42 kV, wesay the voltage is 0.9 p.u.

p. u voltage = 12.42/13.8 = 0.9 p.u

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Per Unit System

base

base

base

basebase

base

basebase

basebase

SV

IVZ

V

SI

SV

2

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exercise

A generator has an impedance of 2.65

ohms. What is its impedance in per-

unit, using bases 500MVA and 22kV

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Base value in 3 Circuit

,3

3

B

BB

BBB

V

SI

IVS

Usually, the 3-phase SB or MVAB and line-to-line VB or kVB

are selected

IB and ZB dependent on SB and VB

B

B

B

BB

BBB

S

V

I

VZ

ZIV

23/

3

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base

pu SS

S

basepu II

I

basepu

VV

V

base

puZZ

Z

ZZ

2

base

base

base

puV

S

ZZ

pu

base

2

basepubase Z

S

VZZ Z

Per Unit System

Voltage, current, kilovoltamperes and impedance

are quantities often expressed in per unit value

Conversion of Per Unit Values

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Transformer Voltage Base

V1/V2

Vb1 Vb2

1

1

22 bb V

V

VV

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Per Unit System

The percent impedance

e.g. in a synchronous generator with 13.8 kV

as its nominal voltage, instead of saying thevoltage is 12.42 kV, we say the voltage is 0.9

p.u.

100%Z

base

actual%

Z

Z

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Example : Three-Phase Transformer

Consider, for example, a three-phase two-windingtransformer. The following typical parameters couldbe provided by the manufacturer:

Nominal power = 300 kVA total for three phases

Nominal frequency = 60 HzWinding 1: connected in wye, nominal

voltage = 25 kV RMS line-to-line

resistance 0.01 pu, leakage reactance = 0.02 pu

Winding 2: connected in delta, nominal

voltage = 600 V RMS line-to-line

resistance 0.01 pu, leakage reactance = 0.02 pu

Magnetizing losses at nominal voltage in % ofnominal current:

Resistive 1%, Inductive 1%

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Example : Three-Phase Transformer

Base power 300 kVA/3 = 100e3 VA/phase

Base voltage 25 kV/sqrt(3) = 14434 V RMS

Base current 100e3/14434 = 6.928 A RMS

Base impedance 14434/6.928 = 2083

Base resistance 14434/6.928 = 2083

Base inductance 2083/(2*60)= 5.525 H

The base values for each single-phase transformer are first

calculated: For winding 1:

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p.u. to SI

The value of the winding resistance and leakage

inductances expresses in SI units are

For winding 1: R1=0.01 * 2083 = 20.83 L1 = 0.02 * 5.525 = 0.1105 H

For winding 2: R2 = 0.01 * 3.60 = 0.0360

L1 = 0.02 * 0.009549 = 0.191 mH

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p.u. to SI

For the magnetizing branch, magnetizing losses of 1%

resistive and 1% inductive mean a magnetizing resistance

Rm of 100 pu and a magnetizing inductance Lm of 100 pu.

Therefore, the values expressed in SI units referred towinding 1 are

Rm = 100*2083 = 208.3 k

Lm = 100*5.525 = 552.5 H

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excercise

Region 1

Vbase1 = 13.8 kV

Region 2

Vbase2 = Vbase1(110kV/13.8kV) = 110 kV

Region 3

Vbase3 = Vbase2(14.4kV/120kV) = 13.2 kV

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Excercise

2. Calculate corresponding base impedances for each region

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exercise

Region 1

Region 2

Region 3

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Per Unit in 3 Circuit

Simplified:

Concerns about using phase or line

voltages are removed in the per-unitsystem

Actual values of R, XC and XL for lines,

cables, and other electrical equipment

typically phase values.

It is convenient to work in terms of base

VA (base volt-amperes)

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Change of Base

The impedance of individual generators &

transformer, are generally in terms of

percent/per unit based on their own ratings.

Impedance of transmission line in ohmic

value

When pieces ofequipment with various

different ratings are connected to a system,it is necessary to convert their impedances

to a per unit value expressed on the same

base.

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Change of Base

In other word, since all impedances in any

one part of the a system must be expressed

on the same impedance base when making

computations, it is necessary to have a

means of converting per-unit impedances

from one based to another.

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ZV

S

Z

ZZ

old

B

oldB

old

B

old

pu 2

Z

V

S

Z

ZZ

new

B

new

B

new

B

new

pu 2

then,valueactualisZifand,Vbasevoltage&

Sbasepoweron theimpedanceunitperthebeZ

oldB

oldB

oldpu

new

B

new

B

Vbasevoltagenew&

Sbasepowernewon theimpedanceunitpernewthebenewpuZ

1

2

Change of Base

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Change of base

oldB

2oldBold

puS

VZZ

From 1 and 2

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2

new

B

old

Bold

B

new

Boldpu

newpu

VV

SSZZ

old

B

new

Bold

pu

new

puS

SZZ

Then, the relationship between the old and the

new per unit value is obtained as

If the voltage base are the same,

Change of Base

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Example

The reactance of a generator designated X is given as

0.25 per unit based on the generators nameplate

rating of 18 kV, 500 MVA. The base for calculations is

20kV, 100 MVA. Find X on the new base2

new

B

old

B

old

B

new

Bold

pu

new

puV

V

S

SZZ

25.0Zoldpu kV18VoldB MVA500S

oldB

kV20VnewB MVA100SnewB

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Example

unitper0405.020

18

500

100

25.0"XZ

2newpu

25.0Zoldpu kV18VoldB MVA500SoldB

kV20VnewB MVA100SnewB

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Change of base

4. Calculate the p.u. resistances and

inductances at G1 and T1

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Exercise

RG1,pu = 0.1 per unit

XG1,pu = 0.9 per unit

RT1,pu = 0.01 per unit

XT1,pu = 0.05 per unit

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Excercise

5. Calculate per unit value for resistance and inductance of

transmission line

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Exercise

6. Calculate per unit value for resistance and inductance of

T2 and M2

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Exercise

unitper119.0MA50

MA100

kV2.13

kV4.14)05.0(X

unitper238.0

MA50

MA100

kV2.13

kV4.14)01.0(R

2

pu,2T

2

pu,2T

unitper405.2MA50

MA100

kV2.13

kV8.38)05.0(X

unitper219.0MA50

MA100

kV2.13

kV8.38)01.0(R

2

pu,2M

2

pu,2M

At T2

At M2

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Per-Unit System the per-unit-system which choosing a common set of

base parameter in term of which quantities are defined.The different voltage levels disappear and the overallsystem reduces to a set of impedances.

Per-phase, per unit equivalent circuit of the simple power system

G1s

impedance

T1s

impedance

Transmission

lines impedance

T2s

impedance

Motors

impedance

Writing Node Equations for

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Equivalent Circuit

Once the per-phase, per unit

equivalent circuit of a power system is

created, it may be used to find thevoltages, currents, and powers present

at various points in a power system.

Most common technique used to solvesuch circuit is nodal analysis


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Equivalent Circuit

The simple three-

phase power

system

containing three

busses connectedby three

transmission

lines, generator

to bus 1, a load

to bus 2 and a

motor to bus 3.

Draw the

impedance

diagram and

respective nodes

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Equivalent CircuitThe diagram showing current sources at nodes 1, 2and 3; all other equipments are admittance

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Equivalent Circuit

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Advantages

Transformer equivalent circuit can besimplified by properly specifying basequantities.

Give a clear idea of relative magnitudes ofvarious quantities such as voltage,current, power and impedance.

Avoid possibility of making seriouscalculation error when referring quantitiesfrom one side of transformer to the other.

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Advantages

Per-unit impedances of electricalequipment of similar type usually liewithin a narrow numerical range when

the equipment ratings are used as basevalues. Manufacturers usually specify the

impedances of machines and transformers

in per-unit or percent in nameplaterating.

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Advantages

Why Use the Per Unit System Instead of the Standard SI Units?

Here are the main reasons for using the per unit system:

When values are expressed in pu, the comparison of electrical quantities with their "normal" values isstraightforward.

For example, a transient voltage reaching a maximum of 1.42 pu indicates immediately that this voltageexceeds the nominal value by 42%.

The values of impedances expressed in pu stay fairly constant whatever the power and voltage ratings.

For example, for all transformers in the 3 kVA to 300 kVA power range, the leakage reactance variesapproximately between 0.01 pu and 0.03 pu, whereas the winding resistances vary between 0.01 pu and0.005 pu, whatever the nominal voltage. For transformers in the 300 kVA to 300 MVA range, the leakagereactance varies approximately between 0.03 pu and 0.12 pu, whereas the winding resistances varybetween 0.005 pu and 0.002 pu.

Similarly, for salient pole synchronous machines, the synchronous reactanceXd is generally between 0.60and 1.50 pu, whereas the subtransient reactanceX'd is generally between 0.20 and 0.50 pu.

It means that if you do not know the parameters for a 10 kVA transformer, you are not making a majorerror by assuming an average value of 0.02 pu for leakage reactances and 0.0075 pu for windingresistances.

The calculations using the per unit system are simplified. When all impedances in a multivoltage powersystem are expressed on a common power base and on the nominal voltages of the differentsubnetworks, the total impedance in pu seen at one bus is obtained by simply adding all impedances inpu, without taking into consideration the transformer ratios.

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Review

Study the examples.

Try to solve the problems in your

tutorials.

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Impedance Diagram

G1

Load A

1

G2

Load B

G3

2

G4

T1 T2

G2G1 G3 G4

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