Power System Basics

Power System Basics_r16

Section 1 - Power System Basics

Power System Protection for Engineers PROT 401

1

Copyright SEL 2005

Power System Protection for Engineers

Power System BasicsOverview

The purpose of this introductory section is to provide a review of the main concepts of power systems, especially those concepts most frequently used in the power system protection field.

It is difficult to attempt to study all aspects of power systems in a short period of time. This section serves not only as a brief overview but also as an introduction to conventions used throughout the course.

The concepts presented here can also be found in any textbook on ac circuit analysis or power system analysis.




2

Power System BasicsObjectives

z Explain phasor concepts and their importance in power systems

z Discuss the voltage, current and power relationships in three-phase electric circuits

z Review the power system components and per-unit quantities




3

Sinusoidal Function

2 =period

mY

( )ty

t

t

2

2

mY

( ) ( ) += tYty m cos

The current and voltage of alternating current electrical systems in steady state are normally represented by perfect sinusoidal functions. The figure shows an example of a sinusoidal signal (or function) called y(t) that could be a voltage or a current. The signal is periodic with a period of T seconds. The signal frequency f, in Hertz, is the reciprocal of the period: f = 1/T

The analytical expression of a sinusoidal function of period T is the following:

( ) ( )

locationon depending Hz, 60or 50ff2

anglePhase:frequency(angular)Radian :

AmplitudePeak :Y

tcosYty

m

m

= =

+=




4

Complex Number

Real Axis

Imaginary Axis

Yr

)YIm(b

)YRe(a

jbaY

==

+=

a

b

The analysis of ac linear systems using the time representation of voltages and currents may result in tedious, complicated, and time-consuming computations. For steady state conditions, the complication is reduced through the use of complex numbers. A complex number is composed of two real numbers. One of the real numbers is the real part and the other real number is the imaginary part. The real part of the complex number Y shown in the figure is a, and the imaginary part is b. The imaginary part is always multiplied by j, which is equal to the square root of negative one.

A complex number can be graphically represented as shown in the figure. Two axes are used to represent the real and imaginary parts of the number. The number can be represented as a vector with two components. The following three expressions can be used to represent a complex number:

= =+=

+==

+=

sinYbcosYa

:are partsimaginary and real theand,baYis Y of magnitude theSince

angle) and magnitude using formar (Rectangul sinjYcosYYform) coordinate(Polar YY

form) coordinatear (RectanguljbaY

22

rrrr




5

Phasor Representation

( ) [ ]tj

jjm

eYRe2ty

:Then

Yee2

YY

:asY Phasor the Define

=

==

r

r

r

A phasor is a complex number used to represent an ac voltage or current. The relationship between the phasor and the original signal is given by the following analytical development:

( ) ( ) ( )

( ) ( ) ( )[ ]( ) ( )[ ] [ ]

1j

sinjcose

ee2

YRe2)t(y

eeYReeYRety

tsinjYtcosYRety

tcos2

Y2tcosYty

j

tjjm

tjjm

tjm

mm

mm

=

+=

===

+++=

+=+=

+

:Note

:IdentitysEuler' (*)

:(*)identity sEuler' Using

:is time of function a as signal original The




6


( )tyofvalueRmsYYWherejYYY

YYeY

m

j

:2

sincos

=

+===

r

r

Note that the phasors magnitude Y, is the rms value of the original sinusoidal signal y(t). Note also that the phasor real and imaginary parts, as well as thephasors magnitude and angle are CONSTANT. In other words, with this representation the variable t (time) does not appear in the calculations.




7

Notation

( )

+=

== ++

:ThereforeA"Phasor of Angle" Means

/ )(

A

A

AAeA j

r

r

r




8


Real Axis

Imaginary Axis

= YYrsinY

cosY

As any complex number, a phasor can be graphically represented in the complex plane.




9

Phasor Operations

2*

)(/*

)(/

:tionMultiplica/

/

:Given

AAAABBA

ABBA

BB

AA

==

+=

==

rrrrrr

rr

Phasor operations are the same as for complex numbers. Multiplication of phasors is easiest when they are in polar form.




10

Phasor Operations

nj

nn

jnnnjn

j

eA

eAAeABAe

BA

BA

=====

A

)()( :tionExponentia

)(/ :Division )(

Other operations, like division and exponentiation, of complex number are relatively simple to perform. This would not be so simple if the time representation of the signals was used.




11

Representation of AC Linear CircuitsUsing Phasors and Impedances

V

I

IV Z

IMPEDANCE

( )

REACTANCEXRESISTANCER

jZZjXRZ

ZI

VI

VIVZ IV

I

V

==

+=+===

==

sincos

/rrr

Linear elements of passive ac circuits are represented by their impedances. The impedance of a given circuit element is defined as the complex number resulting from the division of the applied voltage phasor by the resulting current phasor. As with any complex number, the impedance has a real and an imaginary part. The real part is called RESISTANCE (R) and the imaginary part is called REACTANCE (X).Examples of impedances are:

Z = 2 + j6 Ohms Z = 2 - j30 OhmsZ = j10 Ohms (pure reactance)Z = 2.5 Ohms (pure resistance)Z = 0.231 + j0.685 per unit (using a given base impedance)Z = 23.1 + j68.5 % (per unit times 100)

The inverse of impedance is called ADMITTANCE and denoted by Y. For some applications in power systems, the admittance is used instead of the impedance to represent passive elements. The admittance also has real and imaginary parts. The real part is called CONDUCTANCE (G) and the imaginary part is called SUSCEPTANCE (B). In other words:

jBGjXR

1Z1Y +=+== r

r




12

Passive Circuit Element Impedances

I

+ V -

Resistor

R I

+ V -

LI

+ V -

C

Inductor Capacitor

VI

I I

VV

= 0 = 90 = - 90

Even though pure elements do not exist, it is common to model some elements as pure resistors, inductors, and capacitors. The phasor diagrams for these three elements are shown in the figure.

The following expressions are used to calculate the impedances of the three main components of linear ac circuits:

cc

LL

jXC

1jZ,IjXIC

1jV:Capacitor

jXLjZ,IjXILjV:Inductor

RIVZ,IRV:sistorRe

========

===

rrrrrrr

rrrrr




13

I

V

VL VCVR

R L C

VRI

VC

VRVL

VC

V

Series RLC Circuit Phasor Diagram

The following equations serve to calculate the total series impedance of the series R-L-C circuit:

( ) == C1LjXXjX CLC1jLjRjXjXRZ

CL +=+=

jXRIVZ +==




14

Parallel RLC Circuit Phasor Diagram

V

IR

IICIL

IC

I

R L CIR IL IC

IR

V

The parallel R-L-C circuit is the dual of the series R-L-C circuit. It is more convenient in this case to use admittances.




15

Impedance Representation in the Complex Plane

R

X

Z

R

X

r

-r

r-r

ArearZ || r

22 || XRZZ

jXRZ

+==+=

rr

As with any complex number, an impedance can be represented in the complex plane. In the power system protection field, the complex plane used to represent impedances is called the R-X plane. The resulting diagram is sometimes called the R-X diagram.

The R-X diagram is used to study and analyze not only a simple impedance (which is actually a simple point on the complex plane) but also a set of impedances or a variation of impedances. For example, the area inside the circle of radius r and with center at the origin is represented by:

rZ || r




16

Instantaneous Power

)2sin()sin()2cos()cos()cos(

)2cos()cos(

)cos()cos(2

)cos(2

)cos(2

tVItVIVIp

tVIVIp

ttVIivp

tIi

tVv

++=+=

===

= Zi

v

The single-phase ac circuit of the figure serves to provide a review of the concept of instantaneous power for steady state conditions. Note that, for this particular case, it is assumed that the current lags the voltage by an angle of degrees. Note also that the magnitude of each signal is presented as the rms value times the square root of two.

The instantaneous power is obtained by direct multiplication of the two sinusoidal functions representing the voltage and the current. The result is a function with three terms:

1) The first term is constant (does not depend on t), and is equal to VIcos()2) The second term is a perfect sinusoid but at a frequency equal to 2. The magnitude of this term is proportional to the cosine of angle .3) The third term is a perfect sinusoid also at a frequency equal to 2. The magnitude of this term is proportional to the sine of angle .

Note that, if the impedance of the circuit were a perfect resistor, the current and the voltage would be in phase. In other words, the angle would be zero. Note also that the mathematical expression of the instantaneous power can be re-written as:

)t2sin(sinVI))t2cos(1(cosVIp)t2sin(sinVI)t2cos(cosVIcosVIp

++=++=




17

Instantaneous Power

( ) += tVIVIp 2coscosp = vi

ivP

)cos( 10

VIdtpT

PT

Average ==The slide shows how the mathematical expression of the instantaneous power can be manipulated to make its form more evident. The final expression has a constant term (VIcos()) and a double-frequency sinusoidal term. The instantaneous power is shown in the figure as the offset sinusoidal function with an offset equal to the constant term VIcos(), which is shown as the dashed line in the figure. Intuitively, it can be seen that the constant offset is the average power delivered to the impedance.

The figure also shows the voltage and current signals.

The average value can be found by integrating the expression and the result is, as expected, VIcos().




18

Average Power of Elements

I

+ V -

Resistor

R I

+ V -

LI

+ V -

C

Inductor Capacitor

= 0 = 90 = -90tVIVIp 2cos+= tVIp 2sin= tVIp 2sin=

Average = VI= I 2 R

Average = 0 Average = 0

The average power delivered to pure inductors and capacitors is zero. It is said that resistors are the only elements that consume real power (active power).




19

Complex PowerActive and Reactive

=

=

IVReIVReP **

=

=

IVImIVImQ **

jQPIVS * +== rrrI

VZ

Complex power is defined as the product of the voltage phasor and the complex conjugate of the current phasor. The unit of measure for complex power is volt-amperes, or VA.

The average power is also known as the active power, because it is the part of the instantaneous power that actually produces work, or heat. The unit of measure for active power is watts. Another way to determine the active power consists of taking the real part of the multiplication of the voltage and current phasors.

The imaginary part of the complex power, Q, is known as the reactive power. The unit of measure for reactive power is Volt-Amperes-Reactive, or VAR.




20

Complex PowerActive and Reactive

=

= cosVIIVReP

=

= sinVIIVImQ

IV

Z

+=+= sinjZcosZjXRZr

=+== VIjQPIVS *rr

If is the angle of the impedance, and therefore the angle the current lags the voltage, then the active and reactive power can be calculated as functions of .Note that, according to convention, the reactive power of an inductor is positive and the reactive power of a capacitor is negative.




21

Complex PowerApparent Power and Power Factor

( ))FactorPower(cos

SPPF

PowerApparentZIQPSVIS

)PowerComplex(IVjQPSeS

222

*j

==

=+===

=+== r

rrr

IV Z

Also by definition, the magnitude of the complex power (the product of the magnitudes of the voltage and current, S = VI) is called the apparent power.

The ratio of the active power to the apparent power is defined as the power factor. Mathematically, the power factor is the cosine of angle , or p.f. = cos().




22

Power Triangle

P

SQ

It is possible to graphically represent the complex power in the complex plane. The real part is the active power and the imaginary part is the reactive power. This is known as the power triangle.




23

Balanced Three-Phase Systems

z For a balanced three-phase system, the sinusoidal voltages are the same amplitude, displaced in phase by 120.

VA

VB

VC

VA

VC

VB

A-B-C Sequence A-C-B Sequence




24

Y- Connected Loads

EBnECn

EAnn

ZG

ZG

VCn ZG VBn

Ib

Vbn

Zp

n

Zp

Vcn

Zp

VanIa

ZLVAn

ZL

Ic

ZL

ooo 1201200 ===

PCNPBNPAN VVVVVV

The figure shows a circuit diagram of a simple three-phase system. If the ideal three-phase voltage source is perfectly balanced, and all the impedances on each phase of the system are equal, then all voltages and currents in the system will be perfectly balanced.

A wye-connected load consisting of passive impedances is shown for thepurposes of reviewing the equations.

Each impedance of the load receives a line-to-neutral voltage.




25

Y- Connected Loads

o303 ==

PBNANABVVVV

o903 ==

PCNBNBCVVVV

o1503 ==

PANCNCAVVVV

o30V3V PLL =

30

VCNVCA VAB

VAN

VBC

VBN

Each of the impedances of a load is called a phase impedance. The magnitude of the voltage applied to each impedance is the line-to-neutral voltage, or the phase voltage (VP).

The relationship between the line-to-line voltages and the the line-to-neutral voltages can be obtained analytically or graphically. There is a factor equal to the square root of three between the phase voltage and the line-to-line voltage.




26

Y- Connected Loads

==

/

P

P

ANA I

Z

VI

==

o120/P

P

BNB I

Z

VI

==

o120/P

P

CNC I

Z

VI

PL II =

The magnitude of the current passing through each of the impedances is the line current (IL). For a Y-connected load, the line current and the phase currents are equal.




27

- Connected Loads

PLL VV =

Ica

IabZp

Ibc

Iaa

Ibb

IccZp

Zp

For a delta-connected load, the load voltage is equal to the line-to-line voltage.




28

- Connected Loadsoo 120II,120II,0II PCAPBCPAB ===

o1503 ==

PABBCB IIII

o903 ==

PBCCAC IIII

o303 =

PLII

30

IC

ICA

IAB

IAIBCIB

o303 ==

PCAABAIIII

For delta-connected loads, there is a factor equal to the square root of three between the phase and the line currents.




29

Conclusion for Passive Loads

LPLPII,3VV ==

3II,VVLPLP

==

z Y- Connected Loads

z - Connected Loads

This is the conclusion.




30

Power in Balanced Three-Phase Systems

== cosIV3cosIV3PLLPP

== sinIV3sinIV3QLLPP

22 QPIV3IV3SLLPP

+===QjPeSS j +==

== cosS/PPF

Three-phase power is obtained by adding up the power on each phase. The result is that the three-phase power for a balanced three-phase system is three times the power of one of the phases. When the line-to-line voltages and the line currents are used, the factor becomes the square root of three.

Recall that for wye-connected loads, VP = VL/ 3 and IP = IL. When the substitution is made, the equation has 3 divided by root 3, resulting in root 3 times the line voltage and line current.




31

Conclusion on Balanced Power Systems

EBnECn

EAn

n

ZG

ZG

VCn ZG VBn

Ib

Vbn

Zp

n

Zp

Vcn

Zp

Van

Ia

ZLVAn

ZL

Ic

ZL

If the system is perfectly balanced, only one phase needs analysis. The other phases will have the same magnitude with a 120 phase shift.

Ean

ZG Zp

ZL

Ia

+-

One important characteristic of the analysis of three-phase power systems is that it is not necessary to analyze each phase when the system is perfectly balanced. For a balanced system, only one phase needs to be analyzed. The behavior of the other phases is similar. The only difference is that the currents and voltages are shifted 120 with respect to the analyzed phase.




32

One-Line Representation of Power Systems

EBnECn

EAnn

ZG

ZG

VCn ZG VBn

Ib

Vbn

Zp

n

Zp

Vcn

Zp

VanIa

ZLVAn

ZL

Ic

ZL

SystemRepresentation

One-LineRepresentation

The one-line representation saves significant amounts of space. This is more evident for large power systems.




33

Some Power System ComponentsOne-Line Symbols

Generator

Transformer

Transmission Line

General Load Shunt Impedance

These are the symbols used in one-line diagrams for the most common elements of a power system. Other elements are capacitor banks, reactors, phase shifters, motors, power electronics controls, circuit breakers, etc.




34

Power System ExampleOne-Line Diagram




35

In the Substation Yard

Bus

SwitchCircuit

Breaker CurrentTransformer

VoltageTransformer

Switch




36

Mutual CouplingAn Important Concept

MagneticCoupling

InducedVoltages

I1 I2 I1 I2Zm I2 Zm I1

Magnetic coupling between two energized circuits is called mutual coupling. The current in one of the circuits induces a voltage in the other circuit and vice-versa.

Mutual coupling is present in all the components of the three-phase power system. This makes the study, analysis, and computation more complicated.




37

Power System ComponentsPower Generator

ZSIa

Ib

Ic

g

n

Bus

Stator

ZS

ZS

ZmZm

Zm

Ze

Va+

-Vb+

-Vc+

-

VrIr

Ires

Rotor (Field) Circuit

Ia

Ib

Ic

NeutralGroundingDevice

The synchronous generator is a complex machine. Windings in the rotor inject direct current, but the windings move. The movement of the windings is what produces the induced ac voltages in the stator circuits. The stator windings and the rotor windings are magnetically coupled and the equivalent mutual inductances are variable because the machine is in motion. The mathematical model of a typical synchronous generator consists of at least 10 differential equations, including the control. There are simplified models used in practical calculations that provide enough accuracy for many applications.

The generators neutral can be grounded with different methods. In the model shown in the figure, the grounding impedance can take on different characteristics depending upon the method used.




38

Simplified Generator Model

ZS

+Ea

+Eb

+Ec

Ia

Ib

Ic

g

n-

-

-

Bus

ZS

ZS

ZmZm

Zm

Ze

Va+

-Vb+

-Vc+

-

gnccSbmamc

gnbcmbSamb

gnacmbmaSa

VVIZIZIZE

VVIZIZIZE

VVIZIZIZE

++++=++++=++++=

The most common, and simple, method to represent a generator in steady state is with an ideal three-phase source with series and symmetrically coupled impedances.The internal ac sources represent the induced voltages, that is the rotor effect. The equation considers the self impedance, Zs, of each winding of the stator windings and the mutual impedances, Zm, equal for all cases (perfect symmetry). When the system works under perfectly balanced conditions, the ground-to-neutral current and voltage Vgn are both zero.

Although simple and inaccurate, the described model is useful for practical studies and is especially useful for creating equivalent three-phase generators to represent the power system elements behind a given bus. This is the three-phase version of a Thvenin equivalent.




39

Power System ComponentsPower Lines and Cables

(Not to scale)

Overhead lines and cables are among the most important elements of the power system. They transport and distribute the energy along vast geographical regions. There is a large variety of line configurations (conductor arrangement, tower design, etc.). However, the lines can be modeled with an acceptable degree of accuracy with relatively simple models.




40

Magnetic Coupling

a b c

MagneticCoupling

Long lines and cables have mutual coupling among their own phases and among other adjacent lines and cables. This makes the analysis of three-phase power systems a complicated task, specifically for unbalanced conditions. Some special techniques, such as the symmetrical components method, are used to address this complication.




41

Transmission Line Circuit Model

IaIb

Ic

a

b

c

Magnetic Coupling

One of the simplest models for transmission lines consists of the circuit shown in the figure. Both the inductive series impedance and the shunt capacitance are actually distributed along the entire line. These parameters are concentrated in this model to simplify the study of the system in steady state conditions.




42

Transmission Line Circuit ModelWithout Considering Capacitances

IaIb

Ic

a

b

c

Magnetic Coupling

aV bV cV aV bV cV

When studying power system short circuits using phasors, the line capacitance effect can be ignored, resulting in an even simpler line model. Note the mutual coupling representation.




43

Line Equations Without Considering Mutual Coupling Among Phases

ccSc

bbSb

aaSa

VIZVVIZVVIZV

+=+=+=

Ib

Ic

a

b

c

aV bV cV aV bV cV

If the mutual effect is neglected, the line model is too inaccurate. This model is sometimes used for extremely short lines.




44

Line Equations Considering Mutual Coupling Among Phases

ccSbcbacac

bcbcbSabab

acacbabaSa

VIZIZIZVVIZIZIZVVIZIZIZV

+++=+++=+++=

IaIb

Ic

a

b

c

Magnetic Coupling

aV bV cV aV bV cV

The steady state equations of the line, neglecting the capacitance and considering the mutual coupling, are the ones shown in the figure. In this case, the effect of the grounding wire is not shown. This effect can be easily considered by adding a similar coupled equation for each ground wire on the line.

The equations presented in the slide take into account that, in real lines, the self impedances (Zs) are similar for all phases, but the mutual impedances are not all equal. There is some symmetry. For example, Zac=Zca, Zbc=Zcb, etc. These equations are known as the equations for a non-transposed line.




45

Line Equations for a Symmetrical (Transposed) Line

ccSbmamc

bcmbSamb

acmbmaSa

VIZIZIZVVIZIZIZVVIZIZIZV

+++=+++=+++=

IaIb

Ic

a

b

c

Magnetic Coupling

aV bV cV aV bV cV

If the line is considered perfectly symmetrical, the mutual impedances are all equal. This model is known as the transposed line model, and is widely used because the inaccuracy of the model is evident only for very specific applications.

The self impedance has two parts, resistance and reactance. The resistance depends on the conductor used (cross section, material, etc.). The reactance depends on the conductor type and the geometric position of the conductors. The mutual impedance is purely reactive and depends almost exclusively on conductor arrangement and characteristic. The soil characteristics and the grounding wires also have a remarkable influence on the line impedances.

Examples of line impedances for a 13 kV line, with 4/0 ACSR conductor:

Zs = 0.3272 + j1.07 Ohm/km = 0.524 + j1.721 Ohm/mile

Zm = j0.636 Ohm/km = j1.018 Ohm/mile

These values were obtained assuming that the line is perfectly symmetrical.




46

Power System ComponentsTransformer

z Single Phasez Three Phase




47

Single-Phase Transformer

Real Transformer: N1I1 = N2I2+ ERROR

N1 N2

I1 I2

ZLV1

The figure shows the most common representation of a single-phase transformer. The flux confined to the iron core produces the coupling between the two windings. The leakage flux, together with the resistance of the windings and the core non-linearity, produce an error that is negligible in many applications.




48

Single Phase Transformer Symbol

N1 N2

I1V2

+

-

V1

+

-

I2

2

1

2

1

NN

VVTR

N

N =Transformer Ratio:

The transformer ratio is the ratio between the rated voltages for the two terminals of the transformer.

In a well-designed transformer, the ratio between the rated voltages is very close to the turns ratio.




49

Ideal Single Phase Transformer

N1 N2

I1V2

+

-

V1

+

-

I2

221121

2211

2

1

2

1

IVIVSSININ

NN

VV

=====

Perfect relationships can only be applied in the so-called ideal transformer. Even though this device does not exist in reality, the model is used as an auxiliary for more complex models in power systems.




50

Impedance Reflection

2

2

2

1

2

2

1

2

2

1

22

2

12

1

11

2

22

ZNN

NN

IV

NNI

NNV

IVZ

IVZ

=

===

=

N1 N2

I1V2

+

-V1+

-

I2Z2

I1V1+

-(N1/N2)2 Z2

The ideal relationships described in the previous slide lead to the fact that an impedance connected to one side of the ideal transformer is reflected to the other side as the same impedance times the square of the transformer turns ratio.




51

Single Phase TransformerEquivalent Circuit

I2

N1 N2

IE

Rs jXs

Non-Linear

Ideal

Rp jXp

I1

V1 V2+

-

+

-

Usually Not Considered

The figure shows the most popular model for a single-phase transformer.

The equivalent circuit of a single-phase transformer includes the following elements:

An ideal transformer, to represent the effect of the main flux

Leakage reactances, to represent the transformer leakage flux and other effects.

The series resistances, to represent the winding ohmic losses.

The shunt magnetization, a non-linear impedance, is used to represent the behavior of the magnetic circuit (iron core). This branch is usually not considered, especially for short-circuit studies.

There are more accurate models, but this one provides enough accuracy for most practical applications.




52

Single Phase TransformerSimplified Equivalent Circuit

Rs jXs

Ideal

Rp jXp

2

1

2

1

NN

VV

x

x =

I2

N1 N2

I1V1 V2

+

-

+

-

V1x

+

-

V2x

+

-

This is how the equivalent circuit looks after disregarding the magnetization branch. Note that for the two internal (and imaginary voltages V1x and V2y) the ideal transformer relationship still holds. Then, the winding impedances can be reflected to one side or the other, depending on the specific application.




53

Per Unit Representation

z Quantity in Per Unit

z Actual Quantity & Base Value Scalar or complex value of power,

voltage, current, or impedance

z Quantity in Percent Quantity in Per Unit x 100

Actual QuantityBase Value of Quantity




54

Selection of Base Quantities for Single-Phase System

bV

bS

Choose Voltage Base :

Choose Power Base :

The Current Base and Impedance Base are Derived from the Chosen Values:

b

b

b

bb

b

bb S

VIVZ

VSI

2

; ===




55

Effect of Per Unit Calculations on a Transformers Equivalent Circuit

z Step 1: ChooseVoltage Base Vb1Power Base Sb

Rs jXs

Ideal

Rp jXp

V2I2I1

V1+

-

+

-

V1x+

-

V2x+

-

2

1

2

1

NN

VV

x

x =z Step 2: Choose

Voltage Base Vb2 = (N2/N1)Vb1

When using Per Unit values, the power base must remain constant. Therefore, the power base is selected once and then used for all further calculations.




56


I2(pu)

Rs(pu) jXs(pu)Rp(pu) jXp(pu)

V1(pu) V2(pu)+

-

+

-V1x(pu)

+

-

V2x(pu)+

-

I1(pu)

disappears rtransforme ideal The=

====

1)pu(V)pu(V

NN

VVand

NN

VV

;VV)pu(V;VV)pu(V

x2

x1

2

1

b2

b1

2

1

x2

x1

2bx2x21bx1x1




57


I2(pu)

Rs(pu) jXs(pu)Rp(pu) jXp(pu)

V1(pu) V2(pu)+

-

+

-

V1x(pu)=V2x(pu)+

-

I1(pu)

I(pu)V1(pu) V2(pu)+

-

+

-

ttt jXRZ +=

Conclusion: If calculations are performed using per unit values, a proper choice of the base causes the transformer to disappear. This is the main advantage of the per unit method.

The impedance of power transformers normally comes on the nameplate of a transformer and is given as a percentage (Z (pu) x 100) of the rated impedance base for the transformer.

As can be seen from the per unit equivalent circuit, the impedance of a transformer can be measured by making a short circuit test of the transformer. This is because of the following:

Zt = V1(Per Unit)/I1(Per Unit) with V2 = 0 (short circuit)

This is why the transformer impedance is usually referred to as the short circuit impedance.




58

Example

z Single-Phase Distribution Transformer TR = 7200/120 V/V SN = 100 kVA Zt = 3%

Manufacturer provides the transformers series impedance in percent (100 x Z pu) using the transformers rating as the power base.




59

Equivalent Circuit in Per Unit

I(pu)V1(pu) V2(pu)+

-

+

-

tt jXZ =

I(pu)V1(pu) V2(pu)+

-

+

-

pu03.0jZt =

This is the equivalent circuit for the transformer of the previous slide.




60

Three-Phase Banks




61

Y-Y Transformer

N1 N2 Ia

Ib

Ic

Ia(N2/N1)

Ib(N2/N1)

Ic(N2/N1)

a

b

c

2

1

2

1

NN

KVKVTR =




62

-Y Transformer

N1 N2 Ia

Ib

Ic

(Ia-Ib)(N2/N1)

(Ib-Ic)(N2/N1)

(Ic-Ia)(N2/N1)

a

b

c

2

1

2

1

3NN

KVKVTR =




63

Per-Unit System Advantages

z Gives Clear Idea of Relative Values of Similar Quantities

z Per-Unit Impedance of Equipment Based on Rating Fall in Narrow Range




64

Per-Unit System Advantages

z Per-Unit Transformer Values Same Regardless of Transformer Connection or Reference to Primary or Secondary Side

z Per-Unit System Ideal For Computer-Based Power System Analysis and Simulation

Note: The phase shift is considered in per-unit representation. This is done in posterior analysis. The magnitudes, however, are the same in per-unit.




65

Three-Phase Base Quantitiesz Typically Select

Sbase (Three-Phase Value) Vbase (Line-to-Line Value)

base

basebase

base

basebase kV3

kVAI

V3SI ==

( )base

2base

base

basebase S

VI

3VZ ==

( ) ( )base

2base

base

2base

base MVAkV

kVAkV1000Z ==




66

Change of Base

( )2oldbase

old

baseold

base

old

pu

VSZZ

ZZ ==

( )2newbasenewbase

newbase

newpu

VSZZ

ZZ ==

2

newbase

oldbase

oldbase

newbaseold

punewpu V

VSSZZ

=

It is often necessary to convert an impedance to a different base. For example, if a transformer impedance is given on a 20 MVA base, it may need to be converted to a 100 MVA base to match the base of a system model. If the voltage base is not being changed, that portion of the equation becomes equal to 1 and can be ignored.




67

Example

Generator Transformer

z Find Equivalent Reactance on 100 MVA, 115 kV Base Generator: 50 MVA, 13.2 kV, XG = 15% Transformer: 50 MVA, 13.8/115 kV,

XT = 8%




68

Example

pu4345.0XXX

pu16.050

10008.0X

pu2745.08.132.13

5010015.0X

TG

T

2

G

=+==

=

=

=




69

Example of Power Flow Through a 69kV Transmission Line

V1 V2

jQPSI

+= LLL jQPS +=

z Sending-End Voltage (Line-to-Neutral)and Three-Phase Power:

z Load Voltage (Line-to-Neutral)and Three-Phase Power:

)MVAR,MW(45j60S;kV06.37V L2 +==

)MVAR,MW(26.58j65.62S;kV05.689.42V1 +==

12.5 miles

This is an example of steady state functioning of a power line. The voltages are phase-to-neutral voltages, and the powers are three-phase. It is more common to present the voltages in line-to-line volts (for normal load) or in per-unit (or percent) of a given base.




70

Example of Power Flow Througha 69 kV Transmission Line in Per Unit

jQPSI

+= LLL jQPS +=

pu45.0j60.0S;pu09438.0V L2 +==

pu583.0j627.0S;pu05.6077.1V1 +==

V1 V2Base:69 kV100 MVA

12.5 miles

z Sending-End Voltage (Line-to-Neutral)and Three-Phase Power:

z Load Voltage (Line-to-Neutral)and Three-Phase Power:

Power System Basics

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