Component of power system

7/30/2019 Component of power system

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1

1. COMPONENTS OF ELECTRIC POWER SYSTEMS

1.1 Introduction

The intention of this chapter is to lay the groundwork for the study of electric power

systems. This is done by developing some basic tools involving concepts, definitions, and some

procedures fundamental to electric power systems. The chapter can be considered as a simple

review of topics that will be utilized throughout this work. We start by introducing the principal

electrical quantities that we will deal with in subsequent chapters.

1.2 Power Concepts

1.2.1 Single-Phase Systems

The electric power systems specialist is in many instances more concerned with electric

power in the circuit rather than the currents. To study steady-state behavior of circuits, some

further definitions are necessary. Consider a sinusoidal voltage, v(t) in given by

( ) ( )tVtv m cos= (1.1)

( ) ( ) = tIti m cos (1.2)Note that in this case, the current lags the voltage by an angle . The instantaneous power isdefined as

( ) ( ) ( )

( ) ( ) ( ) ==

tItVtp

titvtp

mm coscos(1.3)

P

V

I

V I cos

t

Figure 1.1. Current, Voltage, and Power Plotted Versus Time.

The angle in these equations is positive for current lagging the voltage and negative for current

leading the voltage. Using the trigonometric identity


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2

( ) ( )[ ] ++= coscoscoscos21 ,

the instantaneous power can be written as:

( ) ( )[ ] += tIV

tp mm 2coscos2

(1.4)

A more useful quantity is the average power that is being delivered. This can be obtained by

averaging the instantaneous power over a specified time-period, typically for one cycle. Since

the average of ( ) t2cos is zero, through one complete cycle, the average power, P,becomes

cos2

mm IVP = (1.5)

It is more convenient to use the effective (rms) values of voltage and current than the maximumvalues. Substituting rmsm VV 2= and rmsm II 2= , we get

cosrmsrms IVP = (1.6)where, cos() is called the power factor (PF).

Lagging power factor means the current lags the voltage by an angle , see Figure 1.2.

Leading power factor means the current leads the voltage by an angle , see Figure 1.3.

I

Isin

V

IcosI

Isin

V

Icos

Figure 1.2. Phasor Diagram for Figure 1.3. Phasor Diagram for

Lagging Power Factor. Leading Power Factor.

COMPLEX POWER

If the phasor expressions for voltage and current are known, the calculation of real and

reactive power is accomplished conveniently in complex form. For a certain load or part of a

circuit, the rms values of the voltage drop and the current flow are expressed as:

== IIVV and0 It is common convention in the electric power industry to set the voltage angle as the angular

reference. The complex power or the apparent power S is defined as the product of voltage

times the conjugate of current, or

sincos

*

IVjIVS

IVIVS

+=

==(1.7)

where is the phase angle between the voltage and current. Equation (1.7) can be written as:

( )VAQjPS += (1.8)where


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( ) ( )

( ) ( )VArIVQ

WIVP

sin

cos

=

=(1.9)

The power factor is therefore:

S

P

QP

P

PQ =

+=

=

22arctancoscos (1.10)

POWER TRIANGLE

Equation 1.8 suggests a graphical method of obtaining the overall P, Q, and phase angle,

, for several loads in parallel. A power triangle can be drawn for an inductive load as shown inFigure 1.4. The signs ofP and Q are important in knowing the direction of the power flow, that

is, whether power is being generated or absorbed when a voltage and current are specified.

S = V I*

Q = V Isin

P = V Icos

Figure 1.4. Power Triangle for an Inductive Load.

EXAMPLE 1.1

Consider the circuit shown in Figure 1.5 with the following parameters:

R = 0.5 L = 2.122 mH

C= 1600 F V= 1000 V

R

L

C

+

-

+

-

V I2 I1

IS

Figure 1.5. Circuit of Example 1.1.

Find the following: (a) the source current, (b) the active, reactive, and apparent power

into the circuit, (c) the power factor of the circuit.

a) source current

( )( )( ) ( )( )( )

A8.275.63A903.60A0.580.106

A903.60F106.1s/r377V100

A0.580.106H10122.2s/r3775.0

V0100

21

3

2

31

=+=+====

=+

=

+=

III

jCjVI

jLjR

VI

S


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4

b) power flows

( )( )

Var2962

W5617

VA29625617

VA8.276350A8.275.63V0100*

==

+====

Q

P

jS

IVS

c) power factor

( ) ( ) lagging88.08.27coscos

8.27

====

PF

1.2.2 Three-Phase SystemsThe major assumption of all the electric power presently used is generated, transmitted,

and distributed using balanced three-phase voltage systems. Three-phase operation is preferable

to single-phase because a three-phase system is more efficient than a single-phase system, and

the flow of power is constant.

A balanced three-phase voltage system is composed of three-single phase voltage sources

having the same magnitude and frequency but time-displaced from one another by 120 of acycle as shown in the phasor diagrams of Figure 1.6.

120

120

120

VA

VB

VC

IA

IB

IC

120

120

120

Figure 1.6. Phasor Diagrams of a Balanced Three-Phase System.

There are two possible connections of loads and sources in three-phase systems: wye-connection and delta-connection. Figure 1.7 shows the two types of connections for three-

identical impedances.

Because of the connections, new voltages and current quantities can be defined. Starting

with the voltage, the line voltage or line-to-line voltage is that quantity between two supply lines

or load terminals. For voltage variables, line-to-line quantities have subscripts with two phases

(i.e., ab, bc, or ca). The line-to-neutral or line-to-ground voltage is that quantity between a

supply line and the neutral or ground node of the circuit. The voltage variables for these have

subscripts of only one phase or one phase and n for neutral or g for ground. For currents, a

similar convention is used. A current flowing through an impedance or source between two lines

is given a double subscript notation denoting the two phases. A current flowing though an

impedance or source between a line and the neutral point may be either a single subscript or onefollowed by an n. A line current flowing from a source to a load may be a single or a double


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subscript notation denoting the particular phase or the symbols representing the nodes at each

end of the line (i.e., a or Aa).

Z Z

Z

Z

Z

Z

n

Ia

Ic

Ib

Ia

Ic

IabIca

Ibc

Ib

Vca

Vca

Van

c

b

c b

a a

Figure 1.7. (a) Wye-Connected Load (b) Delta-Connected Load

CURRENT AND VOLTAGE RELATIONS IN THE WYE CONNECTION

The voltage appearing between any two of the line terminals a, b, and c have different

relationships in magnitude and phase to the voltages appearing between any one terminal and the

neutral point n. The set of voltages Vab, Vbc, and Vca are called the line voltages, and the set of

voltages Van, Vbn, and Vcn are referred to as the phase voltages. Analysis of a phasor diagram

provides the required relationships.

30Van = Vp 0

Vbn = Vp -120

Vcn = Vp 120 Vab = 3 Vp 30

Vca = 3 Vp 150

Vbc = 3 Vp -90

Figure 1.8. Phasor Diagram of the Phase and Line Voltages of a Wye-Connection.

The effective values of the phase voltage are shown in Figure 1.8 as V an, Vbn, and Vcn.

Each has the same magnitude, and each is displaced 120 from the other two phasors. The

relation between the line voltage and the phase voltage at the terminal a and b can be written as:


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==

=

303

1200

p

pp

bnanab

V

VV

VVV

(1.11)

Similarly,

=

=

1503

903

pca

pbc

VV

VV

Thus the relation between line-to-line voltage, VL and phase voltage Vp for a balanced wye-

connected, three-phase voltage system is

+= 303 pL VV (1.12)

The current flowing out of a line terminal is the same current that is flowing through the

phase terminal. Thus the relation between the line current IL and phase current Ip for a wye-

connected, three-phase system is

pL II = (1.13)

CURRENT AND VOLTAGE RELATIONS IN THE DELTA CONNECTION

Consider now the case when three single-phase sources are rearranged to form a three-

phase delta connection as shown in Figure 1.9. It is clear from the circuit shown that the line and

phase voltages are the same. Thus:

pLVV = (1.14)

Ia

Ib

Iab

Ica Ibc

Ic

c b

a

Vab

Figure 1.9. Delta-Connected, Three-Phase Source.

The phase and line currents, however, are not identical and the relationships between them can

be obtained as:


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==

=

120

120

0

pca

pbc

pab

II

II

II

Also, from Figure 1.9, the relation between the line and phase currents can be obtained as:

=

==

1503

0120

p

ppabcaa

I

IIIII

Similarly,

=

=

903

303

pc

pb

II

II

The phasor diagram in Figure 1.10 illustrates these relations. Thus the relation between line and

phase currents for a balanced delta-connected system is:

+= 303 pL II (1.15)

Note that in the equations above, VL, Vp, IL, and Ip are the rms or effective values of

voltages and currents.

Iab = Ip 0

Ibc = Ip -120

Ica = Ip 120 Ib = 3 Ip 30

Iba = - Iab

Ia = 3 Ip150

Ic = 3 Ip -90

Figure 1.10. Relation between Phase and Line Currents in a Delta Connection.

POWER RELATIONSHIPS

Assume that a balanced three-phase voltage source is supplying a balanced load. The

three sinusoidal phase voltages can be written as:

( )

( )

( )+=

=

=

120sin2)(

120sin2)(

sin2)(

tVtV

tVtV

tVtV

pc

pb

pa

with the currents given by


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( )

( )

( )

+=

=

=

120sin2)(

120sin2)(

sin2)(

tItI

tItI

tItI

pc

pb

pa

where is the phase angle between the current and voltage in each phase or the power factorangle.

The three-phase power can be defined as:

coscoscos3 ++= ccbbaa IVIVIVP Using a trigonometric identity, we get the following:

( ) ( ) ( )[ ]{ } +++= 2402cos2402cos2coscos33 tttIVP pp The summation of the last three terms, in the above equation, is zero. Thus the three-phase

power can be obtained as:

( ) cos33 ppIVP = (1.16)

In wye-connected systems, Lp II = and 3Lp VV = , and in delta-connected systems,

3Lp II = and Lp VV = . Thus, the power equation, Equation 1.16, reads in terms of linequantities:

( ) cos33 LLIVP = (1.17)Note that Equations 1.16 and 1.17 apply for both wye- and delta-connected systems.

COMPLEX POWER

The above analysis can be extended to include the reactive power, Q, or to get the

apparent power, S, for a three-phase system.

=

=

LL

pp

IV

IVS

3

3 *3(1.18)

If = 0pp VV and = pp II , then

=pp

IVS 33

or in complex notation:

( )

33

3 sincos3

jQP

jIVS pp

+=

+=

where

cos3cos33 LLpp IVIVP == (1.19)

sin3sin33 LLpp IVIVQ == (1.20)


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EXAMPLE 1.2

A wye-connected, balanced three-phase load consisting of three impedances of 1030each as shown in Figure 1.11, is supplied with a balanced set of line-to-neutral voltages:

==

=

120220

240220

0220

VV

VV

VV

cn

bn

an

a) Calculate the phasor currents in each line, b) calculate the line-to-line phasor voltages and

show the corresponding phasor diagram, and c) calcualte the apparent power, active power, and

reactive power supplied to the load.

Z = 1030

Z

Z

n

Ia

Ic

Ib

Van = 220V 0

c

b

a

Figure 1.11 Load Connection for Example 1.2.

a) The phase currents of the loads are obtained as:

= =

=

=

=

=

90223010120220

210223010

240220

30223010

0220

AVI

AV

I

AV

I

cn

bn

an

b) The line voltages are obtained as

=

==

303220

2402200220

V

VV

VVV baab

or


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( )

=

+=

+=

303220

3003220

303

V

V

VV ananab

Similarly,

( )

=

+=

+=

903220

302403220

303

V

V

VV bnbnbc

( )

=

+=

+=

1503220

301203220

303

V

V

VV cncnca

30Van = 220V 0

Vca Vab = 220 3 V 30Vcn

Vbn

Vbc

Figure 1.13. Relation between Phase and Line Voltages in a Wye-Connection.

c) The powers are given by:

( )( )

VArQ

WP

VAj

AV

IVIVSananpp

0.7260

69.12574

0.726069.12574

300.14520302202203

33

3

3

3

=

=+=

==

==


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EXAMPLE 1.3

A delta-connected, balanced three-phase load consisting of three impedances of

1030 each as shown in Figure 1.13, is supplied with a balanced set of line-to-line voltages:

V1502203

V902203

V302203

=

=

=

ca

bc

ab

V

V

V

a) Calculate the phasor voltage across each phase load, b) calculate the phase and line currents

and show the corresponding phasor diagram, and c) calculate the apparent power, active power,

and reactive power supplied to the load.

Z Z

Z

Ia

Ic

IabIca

Ibc

Ib

Vca

cb

a

Vab

Vca

Figure 1.13 Load connection for example 1.3.

a) For a delta connected load,

V1502203

V902203

V302203

=

=

=

=

ca

bc

ab

LLphs

V

V

V

VV

b) The phase currents in each of the impedances are:

A120223

A240223

A02233010

V302203

=

=

=

=

ca

bc

ab

I

I

I

The line currents can be obtained as:

A30661202230223 == =caaba

III


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12

or

( ) ====

30663003223303

303

AAII

II

aba

pL

Similarly,

==

==

9066303

21066303

AII

AII

cac

bcb

-30

Iab = 3 220 A

IbIa = 66-30 A

Ibc

Ica

Ic

Figure 1.14 Relation between Phase and Line Currents in a Delta Connection.

c) The apparent, active, and reactive powers are computed.

( )( )

VAR0.21780

W04.37724

VA0.2178004.37724

VA304356002233022033

33

3

3

3

=

=+=

==

==

Q

P

j

IVIVS ababpp

g

1.3 Power System Representation

A major portion of the modern power system utilizes three-phase as circuits and devices.

A balanced three-phase system is solved as a single-phase circuit made of one line and the

neutral return. Standard symbols are used to indicate the various components. The simplified

one-line diagram is called the single-line diagram. From the one-line diagram the impedance, or

reactance, diagram can be conveniently developed, as shown in the following section. A further

advantage of the one-line diagram is in the power flow studies. The one-line diagram rather

becomes second nature to power system engineers as they attempt to visualize a widespread

complex network.


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generator or motor

transformer

transmission line

oil circuit breaker

air circuit breaker

static load

delta connection

ungrounded wye connection

grounded wye connection

Figure 1.15. Symbolic Representation of Elements of a Power System.

Using the symbols in Figure 1.15, a section of a one-line diagram of a power system is

shown in Figure 1.16.

G1

G2

M

Figure 1.16. A One-Line Diagram of a Portion of a Power System.

1.3.1 Equivalent Circuit and Reactance Diagram

We note from Figure 1.16 that the power system components are: generators,

transformers, transmission lines, and loads. Equivalent circuits of these components may then be

interconnected to obtain a circuit representation for the entire system. In other words, the one-

line diagram may be replaced by an impedance diagram or a reactance diagram (if resistances are

neglected).

Thus, corresponding to Figure 1.16, the impedance and reactance diagrams are shown in

Figures 1.17(a) and 1.17(b), respectively, on a per phase basis. In the equivalent circuit of the

components in Figure 1.17(a) is based on the following assumptions:

1. A generator can be represented by a voltage source in series with an inductive

reactance. The internal resistance of the generator is negligible compared to the

reactance.

2. The motor load is inductive.

3. The static load has a lagging power factor.4. A transformer is represented by a series impedance on a per phase basis.


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5. The transmission line is of medium length and can be represented by a T section.

The reactance diagram, shown in Figure 1.17(b), is drawn by neglecting all theresistances, static loads, and capacitances of the transmission line. Reactance diagrams are

generally used for short-circuit calculation, whereas the impedance diagram is used for power-

flow studies.

Transmission LineTransformer Transformer Motor &

Static Load

Generators

(a) Impedance Diagram

Transmission LineTransformer Transformer Motor &

Static Load

Generators

(b) Corresponding Reactance Diagram

Figure 1.17. Electrical Diagrams of the System Illustrated in Figure 1.16.

1.4 Per-Unit Representation

The per-unit (pu) system is used extensively in power system calculations. The

representation simplifies the vast scaling of sizes from super-large generation and transmissionnetworks to the industrial distribution system or a residential load. The definition of the per-unit

value of any quantity is given as:

dimensionsametheofvaluebase

valueactualvaluepu = (1.21)

In any electrical network, a minimum of four base quantities is required to define completely a

per-unit system: voltage, current, power, and impedance (or admittance).

powerbase

poweractualpowerunitper = (1.22)


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voltagebase

voltageactualvoltageunitper = (1.23)

currentbase

currentactualcurrentunitper = (1.24)

impedancebase

impedanceactualimpedanceunitper = (1.25)

If any two of these quantities are chosen arbitrarily, the other two become fixed. For example,

selecting base values for voltage and power fixes the base values for current and impedance.

Therefore, on a per phase basis the following relationships hold:

voltagebasepowerbasecurrentbase = (1.26)

currentbase

voltagebaseimpedancebase = (1.27)

EXAMPLE 1.4

Calculate the base impedance and base current for a single-phase system if the base

voltage is 7.2 kV and the base apparent power is 10 MVA.

( ) ( )

A9.1388

V7200

VA000,000,10

184.5

VA000,000,10

V720022

=

==

=

==

Base

Base

Base

Base

Base

Base

BaseBase

I

V

SI

Z

S

VZ

g

For three-phase systems, we use the total or three-phase power and the line-to-line

voltage as the base. For currents and impedance values, it is common practice to convert the

system to a wye-connected network, and use the phase current and phase impedance as the bases.Hence, the base impedance and the base current can be computed directly from the three-phase

values of the base apparent power and the line-to-line base voltage. Delta-connections for the

moment are converted to wye-connected equivalents.

Recall for wye-connections,pLL

VV = 3 , and from Equations 1.18 and 1.27 we find:

BaseLL

Base

BaseLL

Base

Basep

Base

BaseV

S

V

S

V

SIcurrentbase

===

33

3,

331 (1.26)

Base

BaseLL

Base

Basep

Base

I

V

I

VZimpedanceBase

3,

== (1.27)


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and

( ) ( ) ( )Base

BaseLL

Base

BaseLL

Base

Basep

BaseS

V

S

V

S

VZ

=== 3

2

3

2

1

2

3

3(1.28)

EXAMPLE 1.5

A three-phase system delivers 18,000 kW to a pure resistive wye-connected load. The

line-to-line voltage at the load terminals is 108 kV. Assuming the three-phase power base is

30,000 kVA and the voltage base is 120 kV, find the following per unit quantities for the load:

a) the per unit voltage,

b) the per unit power,

c) the per unit current, and

d) the per unit impedance.

a) The line-to-line base voltage is:

kVVBaseLL 120=

and the phase base voltage is:

kVkV

VV BaseLL

Basep

2.693

120

3

==

=

The actual line-to-neutral voltage is:

kVkV

VV LL

p

4.623

108

3

==

=

The per unit voltage is:

pukV

kV

kV

kV

V

V

V

VVBaseLL

LL

Basep

p

pu

900.0120

108

2.69

4.62===

==

b) The three-phase base power is:

kVAS 000,303 =

and the single-phase base power is:


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17

kVAkVA

SS

000,103

000,30

331

1

==

=

The per unit power is:

puVA

W

VA

W

S

P

S

PP

BaseBase

pu

600.0000,30

000,18

000,10

6000

3

3

1

1

===

==

c) The per unit current is:

pupu

puj

V

SI

pu

pu

pu

667.0900.0

0.0600.0=

+=

=

The base current is:

( )A

kV

kVA

V

SI

BaseLL

Base

Base

3.1441203

000,30

3

3

=

=

=

To verify the results, first calculate the actual current, which is:

ApuA

IIIpuBasep

2.96)667.0)(3.144( ==

=

Calculate the load current from actual voltage and power:

AII

A

kV

kVAI

pp

p

2.96

2.96

4.62

000,6

==

==

d) The per unit impedance is:

pujpu

pu

I

VZ

pu

pu

pu

0.0350.1667.0

900.0+==

=

The base impedance is:


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18

( )

( )==

==

48030

1202

3

2

31

MVA

kV

S

V

I

VZ

Base

Base

Base

Base

Base

To verify the results, first calculate the actual impedance, which is:

==

=

648)350.1)(480( pu

ZZZpuBasep

Calculate the load resistance from actual voltage and current:

==

==

648

6482.96

4.62

pp

p

ZZ

A

kVZ

1.4.2 Changing the Base of Per-Unit Quantities

Often the per-unit impedance of a component of a system is expressed on a base other

than the one selected as base for the part of the system in which the component is located. Since

all impedances in any one part of 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 base to another. From Equations (1.25) and (1.27) the per unit impedance can be given as:

( )23

BaseLL

BaseActualp

Base

Actualp

puV

SZ

Z

ZZ

== (1,29)

Equation (1.29) shows that per-unit impedance is directly proportional to base power and

inversely proportional to the square of the base voltage. Therefore, to change from per-unit

impedance on a given base to per-unit impedance on a new base, the following equation applies:

2

3

3

=

newBaseLL

oldBaseLL

oldBase

newBase

puoldpunewV

V

S

SZZ

(1.30)

Equation (1.30) is important in changing the per-unit impedance given on a particular

base to a new base. In some problems, when transformers are involved we must choose morethan one voltage base for each primary and secondary side of the transformers and one power

base for the entire system. The following examples will illustrate the procedure.

EXAMPLE 1.6

A three-phase 13.0 kV transmission line delivers 8 MVA at 13.6 kV to a resistive load.

The per phase impedance of the line is (0.01 + j0.05) p.u. on a 13.0 kV, 8 MVA base. What is

the voltage drop across the line in per unit and in volts?

a) Choose the base voltage to be 13 kV and the base power equal to 8 MVA.

The base current and the load current are:


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19

=

=

=

=

=

=

0956.03.355

06.339

06.3396.133

0.8

3.355

0.133

0.8

puA

AI

AkV

MWI

A

kV

MVAI

puLoad

Load

Base

The voltage drop is calculate as:

=+=+=

=

7.780487.00478.000956.0)05.001.0)(0956.0( jj

ZIV linelinedrop

b) The actual values for the line-to-line voltage drop and the phase voltage drop are:

==

==

7.783663

)0.13()7.780487.0(

7.78633)0.13)(7.780487.0(

VkV

V

VkVV

pdrop

LLdrop

EXAMPLE 1.7

The per phase reactance of a three-phase, 220 kV, 6.25 kVA transmission line is 8.4 .Find the reactance value in per unit, based on the rated values of the line. Convert the per unit

reactance value to a 230 kV, 7.5 kVA base.

a)

( )

( )==

=

744.7000,250,6

000,2202

3

2

VA

V

S

VZ

Base

BaseLL

Base

pu

Z

XX

Base

pu

085.1744.7

4.8

==

=

b)

pukV

kV

kVA

kVApu

V

V

S

SXX

newBase

oldBase

oldBase

newBase

puoldpunew

19.1230

220

25.6

5.7085.1

2

2

=

=

=


20/21

20

EXAMPLE 1.8

Consider the system in Figure 1.18. Find the new per-unit values for each element of thesystem above based on a 2.0 MVA system base. Draw the impedance diagrams of the system.

1 MVA

11 kV

j0.1 pu

0.5 MVA

11 kV

j0.15 pu

2 MVA

11/33 kV

j0.15 pu

3 MVA

33/11 kV

j0.1 pu

2 MVA

12 kV

j0.05 pu

10 + j20

Figure 1.18. Small Power System of Example 1.7.

The per-unit values for each element of the three-phase system shown above are as follows:

Machine 1 1.00 MVA, 11 kV, Z = j0.1 pu



Transmission Line Z = 10 + j20

Transformer 1 2.00 MVA, 11 / 33 kV, Z = j0.15 pu

Transformer 2 3.00 MVA, 33 / 11 kV, Z = j0.10 pu

a) The power base for the entire system is 2.00 MVA. The base voltages are chosen for the

following areas as:

Zone 1 kVVB 111 =

Zone 2 kVVB 333311

112 ==

Zone 3 kVVB 111133

333 ==

1 MVA

11 kV

j0.1 pu

0.5 MVA

11 kV

j0.15 pu

2 MVA

11/33 kV

j0.15 pu

3 MVA

33/11 kV

j0.1 pu

2 MVA

12 kV

j0.05 pu

10 + j20

Zone 1

11 kV

Zone 2

33 kV

Zone 3

11 kV


21/21

Figure 1.19. Three Voltage Zones of Example 1.7.

The per-unit values for each element of the above system can be obtained by using Equation 1.30as follow:

Machine 1: pujkV

kV

MW

MWjZ newpu 20.0

11

11

0.1

0.210.0

2

=

=

Machine 2: pujkV

kV

MW

MWjZ

newpu 60.011

11

5.0

0.215.0

2

=

=

Transformer 1: pujZ newpu 15.0=

Transmission Line: ( ) == 5.5440.2

332

MVAkVZ

Base

pujj

Zpu 037.0

5.544

20=

=

Transformer 2: pujkV

kV

MW

MWjZ

newpu 067.033

33

0.3

0.210.0

2

=

=

Machine 3: pujkV

kV

MW

MWjZ

newpu06.0

11

12

0.2

0.205.0

2

=

=

b) The reactance diagram is shown in Figure 1.20.

L1T1

G1

j0.2 j0.6

j0.15

G2

j0.037 j0.067

T2

G3

j0.6

Figure 1.20. Reactance Diagram of Example 1.7.

Date post:	04-Apr-2018
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