PSA_Chapter_02.pdf

DRAFT and INCOMPLETE

Table of Contents

from

A. P. Sakis Meliopoulos

Power System Modeling, Analysis and Control

Chapter 2 _____________________________________________________________ 2

Basic Power Systems Concepts ____________________________________________ 2

2.1 Introduction ____________________________________________________________ 2

2.2 Phasors and Power Concepts ______________________________________________ 2

2.3 Network Analysis - Basic Concepts _________________________________________ 9

2.4 Conservation of Complex Power __________________________________________ 12

2.5 Three Phase Power Systems ______________________________________________ 13

2.6 Symmetrical Components ________________________________________________ 17

2.7 Balanced Operation _____________________________________________________ 20

2.8 Delta-Wye Transformations ______________________________________________ 25

2.9 Introduction to the Power Transmission Problem ____________________________ 33

2.10 Non-sinusoidal Operation _______________________________________________ 39

2.11 Summary ____________________________________________________________ 41

2.12 Problems ____________________________________________________________ 42

Power System Modeling, Analysis and Control: Chapter 2, Meliopoulos

.2 Copyright © A. P. Sakis Meliopoulos – 1990-2013

Chapter 2 Basic Power Systems Concepts

2.1 Introduction

In this chapter we discuss a number of basic power system concepts which appear in

almost all power system analysis problems. We focus on the steady state operation of

power systems. Most of the time, an electric power system operates under sinusoidal

steady state conditions, i.e. the voltages and currents anywhere in the system are

considered to be near perfect sinusoids. We refer to this operating condition as Sinusoidal

Steady State Condition (SSSC). This assumption and the resulting analysis methods are

appropriate to describe the operation of the system for a rather large number of

applications, such as power flow analysis, short circuit analysis, transient stability, etc. It

is also possible that certain components of the system may distort the sinusoidal

waveform. Power electronic based devices connected to the power system may result in

deviations from the pure sinusoidal steady state operation. The voltage and electric

current waveforms may be periodic but they are not pure sinusoidal. We refer to this

operating condition as Periodic Steady State Condition (PSSC). The basic concepts

discussed in this chapter are applicable for the analysis of power systems under

sinusoidal steady state conditions. For completeness, a short discussion of the periodic

steady state conditions is also provided.

2.2 Phasors and Power Concepts

Many power system analysis problems are based on the following assumptions:

(a) the power system operates under steady state conditions

(b) the power system excitation is a pure sinusoid

(c) the power system comprises only linear elements

Accepting above assumptions, the electric current and voltage anywhere in the system

will be a pure sinusoid written in the form:

tIti m cos)(

tVtv m cos)(

where Im is the maximum value of the electric current

Vm is the maximum value of the voltage

is the angular frequency

is the phase of the electric current

is the phase of the voltage


Copyright © A. P. Sakis Meliopoulos – 1990-2013 Page 2.3

A number of interesting properties apply to this condition. First let's examine the root

mean square (rms) value of the electric current and voltage:

2

)(cos1

)(1

0

22

0

2 mT

m

T

rms

VdttV

Tdttv

TV

Similarly,

2

mrms

II

Above relations tell us that under SSSC, the root mean square value of the electric current

or voltage is equal to the maximum value divided by the square root of 2. Thus we can

write:

tIti rms cos2)(

tVtv rms cos2)(

Another interesting property can be obtained by writing the above equations in the

following equivalent form:

)(2Re)( tj

rmseIti

)(2Re)( tj

rmseVtv

Where Re is the real part of the argument. The above relationships are identities,

which can be shown by recalling Euler's identity:

sincos je j

The above relationships can also be written as:

tjj

rms eeIti 2Re)(

tjj

rms eeVtv 2Re)(

Note that the quantities Irmsej and Vrmse

j are complex quantities. Let's define

j

rmseII ~

jrmseVV

~

Then,


Page 2.4 Copyright © A. P. Sakis Meliopoulos – 1990-2013

tjeIti ~2Re)(

tjeVtv ~2Re)( (2.1)

Let's examine a useful geometric interpretation of above equation. Consider Figure 2.1.

The complex quantity tjeI ~2 is a vector in the complex plane. Note that the angle of

this vector with respect to the real axis is t+, i.e. it is time varying. The projection of

this vector on the real axis is the instantaneous value of the electric current. As time

progresses, the vector tjeI ~2 rotates with angular speed equal to . Its projection on

the real axis is always i(t). As a result one needs to know only I~

to reconstruct i(t). The

quantity I~

is called the phasor of the electric current. Based on this definition, it would

be more precise to call it the root mean square phasor of the electric current. For

simplicity we refer to I~

as the phasor current or the complex current. A similar

construction and discussion applies to tjeV ~2 . It is customary to drop the factor 2 and

to plot the complex current or complex voltage on the complex plane at time t=0. This

construction results to what is called the phasor diagram. The phasor diagram for the

current and voltage of Figure 2.1 is illustrated in Figure 2.2.

t

i(t)Real Axis

Imagin

ary

Axis

tjeI ~2

tjeV ~2

Figure 2.1 Geometric Interpretation of Equation (2.1)



Real Axis

Imagin

ary

Axis

V~

I ~

Figure 2.2 Phasor Representation of Equation (2.1)

Now, consider a one-port two-terminal device, illustrated in Figure 2.3a. The voltage and

electric current at the port of the device are v(t) and i(t) respectively. The instantaneous

power flowing into the device is:

)2cos()cos(

)cos()cos(2

)()()(

tIVIV

ttIV

titvtp

rmsrmsrmsrms

rmsrms

Note that the power consists of two terms: one which is independent of time and another

term which is a sinusoidal function of time. The average power flowing into the device is

)cos()(1

0

rmsrms

T

IVdttpT

P

Note that the average power equals the product of the RMS voltage and current, times the

cosine of the angle - . The last term is referred to as the power factor:

power factor cos( - )

The instantaneous power consists of a constant term and a sinusoidal term which has an

angular frequency double of that of the voltage or electric current. This power is

pulsating, i.e. flows “in and out” of the device with zero net flow. This power has been

coined the reactive power many decades ago. In 1932, Fryze provided a theory which

simplified the representation of reactive power. Specifically, Fryze postulated that the

“apparent” power, Sa, of the device is simply the product of its voltage times the current:



rmsrmsa IVS

Then he postulated that the reactive power Q is related to S and P as follows:

)(sin

)(cos1

222

222

222

rmsrms

rmsrms

a

IV

IV

PSQ

or

)sin( rmsrmsIVQ

It is customary to select only the plus sign resulting in the following definition of reactive

power.

)sin( rmsrms IVQ

This theory, in essence, postulates that the real and reactive power are projections of the

apparent power on the real and imaginary axes of the complex plane. Algebraically, this

can be stated as follows:

jQP

jSS

IjVIV

eIV

IVS

aa

rmsrmsrmsrms

j

rmsrms

rmsrms

)sin()cos(

)sin()cos(

~~

)(

*

Above relationship involves four quantities, S, Sa, P, and Q. All of them express power

but they are all different physical quantities. To distinguish them, the following

nomenclature has been adopted many years ago and it is used in power engineering:

Quantity Name Units

S Complex Power VA (Volt Ampere)

Sa Apparent Power VA (Volt Ampere)

P Real Power W (Watt)

Q Reactive Power VAr (Volt Ampere, reactive)



v(t)

+

-

i(t)

SinglePort

Device

(a)

(b)

Figure 2.3 Electric Power into a Single Port Device (a) Single Port Circuit, (b) Input Voltage, Current and Power

Program XfmHms - Page 1 of 1

c:\books\md_psa\examples\psa-ch10-ex10-2c - Aug 31, 2002, 18:09:04.000000 - 30000.0 samples/sec - 3000 Samples

4.040 4.050 4.060 4.070 4.080

-100.0 k

-60.00 k

-20.00 k

20.00 k

60.00 k

100.0 kVoltage (V)

-100.0

-60.00

-20.00

20.00

60.00

100.0 Current (A)

-2.000 M

0.000

2.000 M

4.000 M

6.000 M

8.000 MPower (W)



Now let’s examine the term )sin( aSQ . If the phase of the electric current, , is

less than , then 0)sin( and Q > 0. In this case we say that the electric current

“lags” the voltage because if the voltage and electric current phasors are depicted on the

same complex plane as it is illustrated in Figure 2.4, the voltage phasor appears to be

'ahead' of the electric current phasor. On the other hand, if < , then Q < 0. Again, by

observing the voltage and current phasors we conclude that in this case, the current

phasor “leads” the voltage phasor. In power engineering it is customary to specify the

power by stating the apparent power and the power factor, i.e. the value of the term

)cos( , which appears in the real power expression. Since )cos()cos( ,

the power factor alone can not determine whether > or < . Additional information

must be given. Most of the time we are given the value of )cos( (power factor) and

whether the electric current is lagging or leading the voltage. For example, power factor

0.8 lagging means that 8.0)cos( and the electric current phasor lags the voltage

phasor, i.e. > .

Q

P

S = V I~ ~*

Real Axis

Imag

inary

Axi

sV

olt

s, A

mp

s, o

r V

ars

Volts, Amps, or Watts

Positive

Direction

of Rotation

V~

I~

I~*

Figure 2.4. Phasor Representation of Voltage, Electric Current, Real Power, and Reactive Power

Example E2.1: Consider the single phase transmission line of Figure E2.1. The electric

current and voltage at one terminal of a single phase transmission line are:

AtI o ),10cos(141~

VtV o ),5cos(390~

Compute the rms values of the electric current and voltage, the rms phasors of the electric

current and voltage, the apparent power, the complex power, and the real and reactive

power flowing into the line.



V~

I~

Single Phase

Transmission Line

Figure E2.1: Illustration of a Single Phase Line

Solution: The computed values are:

VArQ

WP

jIVS

VAS

VeV

AeI

VV

AI

rmsrms

a

j

rms

j

rms

rms

rms

117,7

560,26

117,7560,26~~498,27

8.275~

7.99~

8.275

7.99

*

5

10

0

0

2.3 Network Analysis - Basic Concepts

An electric power system consists of the interconnection of many devices. For many

analysis purposes, each device is converted into an equivalent circuit. In this way the

entire power system is converted into the interconnection of the equivalent circuits of all

devices, i.e. an equivalent network. The analysis of this network can be performed with

one of two methods: (a) Loop Analysis, and (b) Nodal Analysis. In this book we prefer to

analyze networks using the nodal method. There is a good reason for this preference:

Recent advances in computer analysis methods indicate that nodal analysis is superior to

other methods from the computational efficiency point of view. Specifically, nodal

analysis requires the manipulation of the admittance matrix while loop analysis requires

the manipulation of the impedance matrix of a system. For electric power systems, the

admittance matrix is sparse while the impedance matrix is full. Advances in sparsity

techniques (see Appendix A) have made the manipulations of the admittance matrix

much more efficient than those of the impedance matrix. The efficiency gains, for large

scale systems such as electric power networks, are dramatic and therefore present

methods of power system analysis are nodal analysis based. A brief description of the

nodal analysis method follows.



Consider a node i of the network and the elements connected to node i, as illustrated in

Figure 2.5. For device k1, the electric current at the terminal connected to node i is a

linear combination of the voltages at the terminals of device k1, i.e.

j

jij

k

i Vyi 1 (2.2)

Device k1

Node iTo

Device

k3

To

Device

k2

3k

ii

2k

ii

1k

ii

Figure 2.5. Illustration of a Node i and Devices k1, k2, and k3 Connected to Node i

At node i, the sum of all currents should be zero by Kirchoff's current law. For the

example illustrated in Figure 2.5:

0321 k

i

k

i

k

i iii (2.3)

Since each electric current in equation 2.3 can be expressed as a linear combination of the

terminal voltages (as in equation 2.2), substitution of the electric currents will result in a

linear equation in terms of node voltages (i.e. the only unknowns are the node voltages).

Observe that one such equation can be written for each node in the system. Each one of

these equations contains, as unknowns, a subset of the nodal voltages. For a system with

N nodes, N such equations are written in terms of the N unknown node voltages. The

result is a consistent system of N equations in N unknowns. Solution of these equations

will yield the node voltages. Subsequently, substitution of the node voltages into

equation (2.2) will yield the electric current at any terminal of a device. The procedure is

illustrated with an example.

Example E2.2: Consider the system of Figure E2.2. A generator connected to node 1

injects an electric current of 100 Amperes. Compute the complex power absorbed by the

load at bus 3 and the complex power delivered by the generator.

Solution: The nodal equations (three nodes) are:

III~~~

1312

0~~

2321 II



0~~~

33231 III

where

)~~

(2.0~

2112 VVjI

)~~

(2.0~

2121 VVjI

)~~

(1.0~

3113 VVjI

)~~

(1.0~

3131 VVjI

)~~

(1.0~

3223 VVjI

)~~

(1.0~

3232 VVjI

33

~001.0

~VI

Upon substitution of the electric currents in terms of the voltages, the following equations

result.

0

0

100

~

~

~

2.0001.01.01.0

1.03.02.0

1.02.03.0

3

2

1

V

V

V

jjj

jjj

jjj

1 2

3

-j0.2S12

~I 21

~I

13

~I

31

~I

23

~I

32

~I

Load

3

~I

AI 100~

-j0.1S -j0.1S

0.001S

Figure E2.2 Example Power Circuit

Solution of these equations yields:



VjV 600000,100~

1

VjV 400000,100~

2

VV 000,100~

3

The complex power absorbed by the load at bus 3 is:

VAVVIVS 000,000,10~~

001.0~~ *

33

*

333

WP 000,000,103

03 Q

The complex power delivered by the generator is

VAjIVS 60000000,000,10~~ *

111

WP 000,000,101

VArQ 600001

2.4 Conservation of Complex Power

Another concept, which is a consequence of Kirchoff’s current law, is the concept of

conservation of complex power at a node. Consider a network node i, and apply

Kirchoff’s current law at this node:

0)(

iKk

k

i

j

ji

where jk

ii is the electric current flowing into node i of device kj. K(i) is the set of devices

connected to node I of the network. Assuming sinusoidal steady state conditions, the

above equation is converted to:

0~

)(

iKk

k

i

j

jI

The above equation is first conjugated and then multiplied by the voltage phasor at node

i, yielding:

0~~ *

)(

iKk

k

ii

j

jIV



Note that jk

ii IV~~

is the complex power jkS flowing into circuit (or device) kj at node i ,

noting that device kj is connected to node i. Thus

0)(

iKk

k

j

jS

The above equation states that the sum of all complex power values flowing into a node,

from all circuits connected to this node, is zero. We will refer to this general principle as

the conservation of complex power.

Example E2.3: A single phase generator supplies two electric loads of 5 MW+j2 MVAr

and 2.3 MW+j1.2 MVAr respectively, as shown in Figure E2.3. The voltage at the

terminals of the generator is 7.2 kV. Compute the electric current supplied by the

generator.

5 MW + j2MVAr

2.3MW + j1.2MVAr

I~

V~

Figure E2.3 A Generator Supplying Two Electric Loads

Solution: Applying the principle of conservation of complex power at the indicated node

and observing that the complex power of the generator is *~~IV :

MVArjMWIV 2.33.7~~ *

kAekAjI j 067.23* 107.14444.00139.1~

Thus, kAeI j 067.23107.1~

2.5 Three Phase Power Systems

The idea of a three phase system came about by trying to maximize the amount of power

that can be transferred or generated with respect to materials used. The majority of

power systems comprise three phase arrangements which consist of the interconnection

of three phase generators, three phase lines, three phase transformers, and other

supporting equipment.

Any three phase element has three phases and possibly a neutral (sometimes a fifth wire

may be present, a safety ground). This is illustrated in Figure 2.6. A number of

definitions pertinent to three phase systems will be introduced next.



ThreePhaseDevice

A

B

C

N

Ground

AI~

BI~

CI~

NI~

ANV~

BNV~

CNV~

Figure 2.6. A General Three Phase Element

In general, a three phase system may be constructed as a single integrated device with

three phases on a common construction, for example, a three phase generator, a three

phase transformer, etc. or it may be constructed with three single phase elements

connected in a three-phase arrangement. Examples are electric loads, transformers, and

motors. The three phases may be connected in a delta or a wye configuration or any

combination of these two. The delta and wye connections are illustrated in Figure 2.7.

For the description of three-phase systems, the following definitions are introduced:

A

B

C

A

B

C

N

(a) (b) Figure 2.7 Three Elements Forming a Three-Phase System

(a) Delta Connection, (b) Wye Connection

Balanced Set of Three-Phase Voltages. A set of three-phase voltages, va(t), vb(t), vc(t),

is called balanced if and only if:



The voltages vary sinusoidally with time.

The amplitudes of the voltages are equal.

There is a 120o phase difference between any two.

As an example, the following set of three-phase voltages is balanced:

)cos(2)( tVtva (2.4a)

)120cos(2)( o

b tVtv (2.4b)

)240cos(2)( o

c tVtv (2.4c)

In equations (2.4), the phase difference between the phase A and the phase B is +120o.

This phase relationship among the three phases will be called the positive phase

sequence. A three-phase generator generates a set of three-phase voltages that are nearly

balanced and of the positive phase sequence. It is expedient to introduce the concept of

an ideal three-phase source, which is illustrated in Figure 2.8.

eab(t) ebc(t)

eca

(t)

B

A C

aV~

bV~

cV~

eb(t) ec(t)

ea(t)

B

A

C

aV~

bV~

cV~

N

(a) (b)

Figure 2.8 Ideal Three-Phase Voltage Source (a) Delta Connected, (b) Wye Connected

An ideal three-phase source generates a set of balanced three-phase voltages. Most three-

phase sources generate a balanced set of voltages of the positive phase sequence. (Note

that it is possible to define other phase sequences, such as the “negative” and the “zero”

sequence. These are introduced in a subsequent section). It is apparent that a set of

balanced three-phase voltages is uniquely determined by the voltage magnitude V, the

angular frequency , the phase angle , and the phase sequence. Alternatively, it can be

uniquely defined by the phasor of the phase A voltage, ja VeV

~, and the phase



sequence. Throughout the text, when the phase sequence is not specified, it will be

assumed to be the positive sequence.

Balanced Set of Three-Phase Currents. The definition of a balanced set of 3-phase

currents is similar to the one for the voltage, stated above. Specifically, a set of three-

phase currents, ia(t), ib(t), ic(t), is called balanced if and only if:

The currents vary sinusoidally with time.

The amplitudes of the electric currents are equal.

There is a 120o phase difference between any two.

As an example, the following set is balanced:

)cos(2)( tIta

i (2.5a)

)120cos(2)( otItb

i (2.5b)

)240cos(2)( otItc

i (2.5c)

The set above, as in the case of voltages, is called a balanced set of three phase electric

currents of positive phase sequence.

A related definition is introduced for a three-phase system, next.

Symmetric Three-Phase System. A three-phase passive system is called symmetric if

and only if the following two statements are true:

It is a linear system.

A balanced set of three-phase currents flows into the system when it is excited

with a balanced set of three-phase voltages.

The definition of a symmetric three-phase system is illustrated in Figure 2.9. Practical

three-phase systems comprise three-phase components that are symmetric or nearly

symmetric. Three-phase transformers are symmetric three-phase devices, three-phase

synchronous generators are nearly symmetric devices, overhead transmission lines are

nearly symmetric, and so on. Traditional power system analysis techniques (i.e. load

flow, fault analysis, transient stability techniques, etc.) have been developed based on the

assumption of symmetric three-phase systems.



ia(t)

ib(t)

ic(t)vc(t)

va(t)

vb(t)Passive

Linear

3-Phase

System

Ideal

3-Phase

Source

)(),(),()(),(),( tititi

rrentsBalancedCu

tVtVtV

ltagesBalancedVo

cbacba

Figure 2.9. Definition of a Symmetric Three-Phase System

Most practical three phase power system elements are not symmetric; but they are near

symmetric. By this we mean that by modeling a specific power element as symmetric,

we commit a small error. In most applications, i.e. power flow, stability analysis, etc.,

this error is acceptable.

2.6 Symmetrical Components

Most of the time, the majority of a three phase power system operates under balanced

conditions. Whenever the balanced operation is disturbed, the analysis of the system can

be performed in two ways:

(a) By analysis of the entire three phase system.

(b) By decomposing the system into three balanced and symmetric systems and then

analyzing each system individually.

The second method is known as the method of symmetrical components. It is introduced

next:

Consider a set of three phase voltages and electric currents (cba VVV

~,

~,

~), and (

cba III~

,~

,~

),

respectively. These sets are transformed via a transformation T into another set of

voltages and electric currents (021

~,

~,

~VVV ), and (

021

~,

~,

~III ) respectively.

The transformation is defined as follows:

120

1 ~~VTVabc

120

1~~ITIabc

where



,

111

1

1

3

1 2

2

aa

aa

T 0120jea ,

1

1

111

2

21

aa

aaT

c

b

a

abc

V

V

V

V~

~

~

~

0

2

1

120~

~

~

~

V

V

V

V

c

b

a

abc

I

I

I

I~

~

~

~

0

2

1

120~

~

~

~

I

I

I

I

The importance of this transformation relies on the fact that it can transform the equations

of a three phase device (generally a set of coupled equations), into three uncoupled

equations. This property of the transformation will be illustrated by an example.

Example E2.4: Consider a simple model of a three phase generator. The generator

consists of three inductors (one for each phase A, B, and C) and the 3 phase voltage

sources, Ea, Eb, and Ec. Each inductor has a self inductance Ls and mutual inductance

with the other two equal to Lm, as illustrated in Figure E2.4a. Express the terminal

voltages in terms of the electric currents and source voltages and then transform these

equations using the symmetrical transformation. Then compute the complex power

delivered by the generator in terms of the symmetrical components.

b

a

c

n

aE~

bE~

cE~

Ls

Lm

Ls

Ls

Lm

Lm

aI~

bI~

cI~

Figure E2.4a A Simplified Three Phase Source



Solution: The voltages at the generator terminals are:

acbmasan EIILjILjV

~)

~~(

~~

bcambsbn EIILjILjV

~)

~~(

~~

cbamcscn EIILjILjV

~)

~~(

~~

In compact matrix notation, the above equations are written as follows:

abcabcabc EZIV

The vectors of voltages, currents and the matrix Z are defined by the equivalence of the

matrix equation to the original set of equations.

Note that above equations are coupled, i.e. each terminal voltage is a function of all three

phase currents. Now let's apply the symmetrical transformation as follows:

120

1 ~~VTVabc

120

1~~ITI abc

120

1 ~~ETEabc

Substitution into the matrix equation yields:

120

1

120

1

120

1 ~~~ETIZTVT

Upon pre-multiplication of above equation by the matrix T:

120120

1

120

~~~EITZTV

It can bee shown by direct evaluation that the product of the three matrices TZT-1

is a

diagonal matrix. Thus, expanding the above matrix equation we obtain:

111

~~)(

~EILLjV ms

222

~~)(

~EILLjV ms

000

~~)2(

~EILLjV ms

Note that above equations are decoupled, i.e. the voltage and current variables of each

equation do not appear in the other two. These equations can be represented by the three

independent circuits illustrated in Figure E2.4b. It should be also noted that if the 3 phase

source is balanced, then 0~~

02 EE . The three networks of Figure E2.4b are referred to

as the positive sequence equivalent circuit, the negative sequence equivalent circuit and

the zero sequence equivalent circuit, respectively.



The complex power delivered by the generator is (the negative sign is due to the fact that

the electric current is in the direction into the generator).

*

00

*

22

*

11

*

120

1

120

1-

*T

abc

***

~~3

~~3

~V~

3~

V~

T-=

~V~~~~~~~

IVIVIIT

IIVIVIVS

T

abcccnbbnaan

2

~E

Ls-Lm

2

~I

2

~V

1

~E

Ls-Lm

1

~I

1

~V

0

~E

Ls+2L

m

0

~I

0

~V

Figure E2.4b The Sequence Networks of the Simplified Three Phase Source of Figure E2.4a

2.7 Balanced Operation

The balanced operation of an electric power system is characterized by the fact that, at

every location of the system, the three phase voltages and currents are balanced. A three

phase electric power system is in balanced operation whenever the following two

conditions are met: (a) the excitation of the system is balanced (balanced three phase

sources), and (b) all system components are symmetric three phase components. In

practical power systems these two conditions are not met. However, most systems come

very close to meeting these conditions. For this reason, most analysis methods are based

on the assumption of balanced operation, i.e. (a) all sources are balanced and (b) all

power system components are symmetric.

The analysis of a balanced three phase system can be drastically simplified by observing

that the voltage and electric current at the three phases are equal in magnitude and the



phase angle between any two is 120o. Consider for example Figure 2.10. Observe that, if

balanced operation is assumed, the electric current in the neutral is equal to zero and the

voltage at any point along the neutral is zero. Therefore all points on the neutral can be

represented with only one electric node. This fact isolates the three phases. The

representation of phase A is illustrated in Figure 2.11. It is apparent now that the phase A

branch can be analyzed by considering Figure 2.11 only. Once the voltages and currents

of the circuit of Figure 2.11 have been determined, the voltages and currents at the

remaining phases (B and C) can be obtained by adding a phase angle of -120o and -240o

to the phase A voltage and current phasor arguments. The circuit of Figure 2.11 is called

the per phase equivalent circuit.

bE~

ZL

aE~

cE~

Zn

Zp

Zp

Zp

ZL

ZL

Figure 2.10. A Symmetric Three Phase System in Balanced Operation

aE~

Zp

ZL

Figure 2.11. Equivalent Circuit of Phase A of System of Figure 2.10

It is important to note that the per phase equivalent circuit can be derived in a rigorous

mathematical way using symmetrical components. For this purpose consider again the

system of Figure 2.10. The circuit equations describing this system are:

)~~~

(~~~

cbanaLapa IIIZIZIZE

)~~~

(~~~

cbanbLbpb IIIZIZIZE

)~~~

(~~~

cbancLcpc IIIZIZIZE

In compact matrix notation:



abcabc IZE~~

where

nLpnn

nnLpn

nnnLp

ZZZZZ

ZZZZZ

ZZZZZ

Z

Application of the symmetrical component transformation yields:

120

1

120

~

~ITTZE

Note that

nLp

Lp

Lp

ZZZ

ZZ

ZZ

TZT

300

00

001

By writing explicitly above equations:

111

~~~IZIZE Lp

222

~~~IZIZE Lp

0000

~3

~~~IZIZIZE nLp

On the other hand:

0

0

~

~~120

a

abc

E

ETE

0

0

~

~~120

a

abc

I

ITI

Thus above equations become:

111

~~~IZIZE Lp

22

~~0 IZIZ Lp

000

~3

~~0 IZIZIZ nLp



Note that the first equation corresponds to the per phase equivalent circuit of the system

of Figure 2.10. The second and third equations yield the trivial results 0~

2 I and

0~

0 I respectively.

In the previous analysis, the per phase equivalent circuit is overly simplistic. Let’s

consider a more complicated but also more realistic case. Again considering the system

of Figure 2.12a, the line has mutual impedance between any two phases, Zm and between

any phase and the neutral, Zmn. In this case, the model equations of this system are:

cbamncbanaLcbamncbmasa IIIZIIIZIZIIIZIIZIZE ~

cbamncbanbLcbamncambsb IIIZIIIZIZIIIZIIZIZE ~

cbamncbancLcbamnabmcsc IIIZIIIZIZIIIZIIZIZE ~

ZL

Zn

Zs

ZL

ZL

ZmnZmn

Zmn

Zm

Zm

Zm

Zs

ZsbE

~

aE~

cE~

Figure 2.12a A Symmetric Three Phase System in Balanced Operation Circuit Diagram of the System

In compact matrix notation

abcabc IZE~~

where

mnnLsnmnmnmnm

nmnmmnnLsnmnm

nmnmnmnmmnnLs

ZZZZZZZZZZ

ZZZZZZZZZZ

ZZZZZZZZZZ

Z

222

222

222

Application of the symmetrical components transformation yields.

120

1

120

~~ITZTE

Note that



mnnmLs

mLs

mLs

ZZZZZ

ZZZ

ZZZ

TZT

63200

00

001

1

~E

Zs - Zm

ZL

1

~I

Zs - Zm

ZL

2

~I

Zs + 2Zm + 3Zn - 6Zmn

ZL

0

~I

Figure 2.12b A Symmetric Three Phase System in Balanced Operation. Positive, Negative and Zero Sequence Networks

Again, the above equations yield:

0

2

11

~)632(= 0

~)(= 0

~)(

~

IZZZZZ

IZZZ

IZZZE

mnnLms

mLs

mLs

The above equations are represented by the equivalent circuits of Figure 2.12b.

Note that

a

a

II

EE~~

~~

1

1

0~

2 I , and

0~

0 I



In this case, the positive sequence network of Figure 2.12b is the phase A equivalent

circuit of the system.

The above example reveals that whenever the three phase system consists of wye

connected elements, the per phase equivalent circuit can be determined, in most cases, by

inspection. However, many power system elements are delta connected. In this case, a

transformation of the delta-connected element into a wye-connected equivalent element

simplifies the problem. The transformation is performed on the basis of identical

voltage-current relationships. The next section introduces these transformations.

2.8 Delta-Wye Transformations

In this section we examine transformations of delta-connected power apparatus into

equivalent wye-connected power apparatus and vice-versa. The two elements are

equivalent in the sense that they obey the same terminal voltage-current equations.

Consider a delta connected voltage source. The general form of this component is

illustrated in Figure 2.13. We will seek the transformation of this system into simple

single phase equivalents and in case of balanced operation into a per phase equivalent.

Zs

Zs

Zs

Zm Zm

Zm

abE~

bcE~

caE~

A

B

C

aI~

bI~

cI~

N

Figure 2.13 A Delta Connected Three Phase Source

Let the self impedance of each leg be sZ and the mutual impedance Zm . The equations

describing the source are:

abcabcmabsab EIIZIZV

~)

~~(

~~



bccaabmbcsbc EIIZIZV

~)

~~(

~~

cabcabmcasca EIIZIZV

~)

~~(

~~

Note also that

caaba III

~~~

abbcb III

~~~

bccac III

~~~

The transformation of above equations using symmetrical components proceeds as

follows. First we write the equation in compact matrix notation.

ca

bc

ab

ca

bc

ab

smm

msm

mms

ca

bc

ab

E

E

E

I

I

I

ZZZ

ZZZ

ZZZ

V

V

V

~

~

~

~

~

~

~

~

~

Upon solution of the above equations for the currents:

ca

bc

ab

ca

bc

ab

smm

msm

mms

ca

bc

ab

E

E

E

V

V

V

YYY

YYY

YYY

I

I

I

~

~

~

where )2/()( 22

mmssmss ZZZZZZY

)2/( 22

mmssmm ZZZZZY

Note that the phase currents and voltages can be expressed as:

ca

bc

ab

ca

bc

ab

c

b

a

abc

I

I

I

D

I

I

I

I

I

I

I~

~

~

~

~

~

110

011

101

~

~

~

~

abc

T

c

b

a

T

c

b

a

ca

bc

ab

VD

V

V

V

D

V

V

V

V

V

V~

~

~

~

~

~

~

101

110

011

~

~

~

abc

T

c

b

a

T

c

b

a

ca

bc

ab

ED

E

E

E

D

E

E

E

E

E

E~

~

~

~

~

~

~

101

110

011

~

~

~



With the above definition, the matrix equation becomes:

abc

T

abc

T

abc EDYDVDYDI~~~

Substituting the phase currents and voltages with their symmetrical components, the

above equation becomes:

120

1

120

1

120

~~~ETTDYDVTTDYDI TT

Note that:

000

0)(30

00)(31

ms

ms

T YY

YY

TTDYD

Upon substitution, the above equations become:

111

~)(3

~)(3

~EYYVYYI msms

222

~)(3

~)(3

~EYYVYYI msms

000

~0

~0

~EVI

These equations are represented by the equivalent circuit of Figure 2.14.

1

~E

3(Ys - Ym) 1

~I

1

~V

0

~V

2

~E

3(Ys - Ym) 2

~I

2

~V

0

~I

Figure 2.14 Equivalent Sequence Network Representation of the Source of Figure 2.13



Now recall that

0

2

1

2

22

2

0

2

1

1

~

~

~

0)1()1(

0

011

~

~

~

~

~

~

E

E

E

aa

aaaa

aa

E

E

E

TD

E

E

ET

ca

bc

ab

Observe that

03031 jea

0302 31 jea

0902 3 jeaa

0902 3 jeaa

The first two equations are

2

30

1

30 ~3

~3

~ 00

EeEeE jj

ab

2

90

1

90 ~3

~3

~ 00

EeEeE jj

bc

Solution for 21

~,

~EE yields

bc

j

ab EeEE~

3

1~

3

1~ 0120

1

bc

j

ab EeEE~

3

1~

3

1~ 060

2

In the above equation, note that E0 is undefined. In the case of balanced source:

EEab

~~

0120~~ j

bc eEE

0240~~ j

ca eEE

Thus:

030

1

~

3

1~ j

abeEE

0~

2 E

In the case of balanced operation, the above transformations are equivalent to a delta-wye

transformation. Specifically, consider the wye-connected system of Figure 2.15. Under

balanced conditions, this system is equivalent to the system of Figure 2.13, as long as the

following relationships hold:



B

A

C

aE~

bE~

cE~

3(Ys-Ym )

aI~

bI~

cI~

aV~

bV~

cV~

3(Ys-Y

m ) 3(Y

s-Y

m )

Figure 2.15 A Wye Connected Source

030~

3

1~ j

aba eEE

0120~~ j

ab eEE

0240~~ j

ac eEE

22 2

)2(3)(3

mmss

msms

ZZZZ

ZZYY

Special Cases: A number of special cases of the transformations discussed earlier are

presented.

Case 1: Consider three impedances connected in a delta arrangement, i.e.

0~~~

cabcab EEE

0mZ

0sZ

The delta-wye transformation for this case is shown in Figure 2.16a, where 3/ sY ZZ .



Z Z

Z

ZY ZY

ZY

Z=3ZY

a

b

c

a

b

c

(a)

0303~~ j

anab eEE

anE~

bnE~

cnE~

abE~

caE~

bcE~

a

b

c

a

b

c

(b)

0303~~ j

anab eEE anE~

bnE~

cnE~

abE~

caE~

bcE~

a

b

c

a

b

c

Z

Z

Z

ZY

ZY ZY

Z=3ZY

(c)

Figure 2.16 Three Special Cases of Delta-Wye Transformations

Case 2: Consider an ideal voltage source connected in a delta arrangement, i.e.

0 ms ZZ

0~

abE

0120~~ j

abbc eEE

0240~~ j

abca eEE



The delta-wye transformation for this case is shown in Figure 2.16b, where 0303

~~ j

anab eEE .

Case 3: Consider a balanced voltage source with internal impedance but without mutual

coupling among phases, i.e.

0mZ

0sZ

0~

abE

0120~~ j

abbc eEE

0240~~ j

abca eEE

The delta-wye transformation for this case is shown in Figure 2.16c, where 3/ sY ZZ

and 0303

~~ j

aab eEE .

The usefulness of above transformations will be illustrated with an example.

Example E2.5: Two three phase synchronous machines, one wye connected and the

other delta connected, are interconnected with a three phase transmission line as it is

shown in Figure E2.5a. Each synchronous machine is modeled with an equivalent

voltage source. The system operates under balanced conditions and

kVeE

kVeE

j

ab

j

An

0

0

12

0

0.25~

4.14~

(1) Determine the per phase equivalent circuit, (2) Compute the electric currents

cba III~

,~

,~

of the three phase line, (3) Compute the three phase complex power flowing

into machine 1, and (4) Compute the three phase complex power flowing into machine 2.

AnE~

BnE~

CnE~

abE~

caE~

bcE~

j3j5

j3

j3

j5

j5

j15

j15

j15

j1.2

j1.2j1.2

a

b c

aI~

cI~

bI~

SynchronousMachine 1

SynchronousMachine 2

Figure E2.5a Two Synchronous Machines Interconnected with a Three Phase Line



Solution: (1) The synchronous machine 1 will be transformed into a wye connected

machine. Using the results illustrated in Figure 2.16c.

kVeeeE jjj

a

000 181230 434.140.253

1~

Ohmsjj

Za 0.13

0.3

The per phase equivalent (or positive sequence) circuit is illustrated in Figure E2.5b:

j10

j1.2

aI~

j1.0

kVeE j

a

018434.14~ kVeE j

An

04.14~

aV~

AV~

Figure E2.5b Per Phase Equivalent Circuit of System in Figure E2.5a

(2) First, from the circuit of Figure E2.5b we compute

kAej

eI j

j

a

0

0

4.17118

369.02.12

4.14434.14~

Next, the electric currents at the other two phases are determined by the fact that the

system is balanced, i.e.

kAeI

kAeI

j

c

j

b

0

0

6.68

4.51

369.0~

369.0~

(3) First, we compute the voltage at the terminals of the synchronous machine 1:

kVeIjEV j

aaa

55.16377.14~

0.1~~

Then, the complex power flowing into machine 1 is:

MVArjMWIVS aa 199.2762.15~~

3 *

1

Since the real power flow into machine 1 is positive, this machine operates as a

motor.



(4) First, we compute the voltage at the terminal of the synchronous machine 2:

kVeIjEV j

aAnA

75.134.14~

2.1~~

Then, the complex power flowing into machine 2 is:

MVArjMWIVS AA 893.1762.15~~

3 *

2

Since the real power flow into machine 2 is negative, this machine operates as a

generator.

2.9 Introduction to the Power Transmission Problem

In this section we examine the basic problem of power transmission. For this purpose,

we consider power transmission through a single transmission line. For generality, we

assume that at both ends of the line there are both generators and loads as illustrated in

Figure 2.17. As we discussed earlier, under the assumption of symmetry and balanced

operating conditions, the system of Figure 2.17a can be replaced by its per phase

equivalent circuit (or the positive sequence network) shown in Figure 2.17b. The same

circuit can also be drawn in single line diagram form, as shown in Figure 2.17c.

Source 1 Source 2

2

~aV

2

~bV

2

~cV

1

~aV

1

~bV

1

~cV

Electric

Load 1

Electric

Load 2

(a)

1

~V

Z

2

~V

(b)



2

~V1

~V

Sd1 Sd2

Z

(c)

Figure 2.17 A Simplified Two Source Electric Power System

Assume that the voltage phasors at the two ends of the line are 1

~V and 2

~V respectively,

or:

1

11

~ jeVV

2

22

~ jeVV

The power flows at the two terminals of the line are:

)(21

2

1

*

21

11221

*

3

*

3~~

~3

je

Z

VV

Z

V

Z

VVVS

)(21

2

2

*

12

22112

*

3

*

3~~

~3

je

Z

VV

Z

V

Z

VVVS

Another popular form of above equations is the so called hybrid from where the voltages

are expressed in polar coordinates and the impedance is converted into admittance

expressed in Cartesian coordinates, i.e.

YjbgjxrZ

0.10.1

, thus

22 xr

rg

, and

22 xr

xb

Using this notation, the complex power S21 becomes:

jQP

bgVVjbVj

bgVVgV

jVVjbgVjbgS

)cos()sin(33

)sin()cos(33

)sin()cos()(3)(3

121221

2

2

121221

2

2

121221

2

221

Thus:

)cos()sin(33

)sin()cos(33

121221

2

2

121221

2

2

bgVVbVQ

bgVVgVP



This equation provides the complex power flowing into the line from the bus 2 terminal,

as a function of the voltage magnitudes at the line terminals and the terminal voltage

phase angle difference. An important special case of the presented power transmission

problem is the case where generation exists only in one line terminal. This case will be

examined next.

Electric Load Supplied via a Transmission Line. This case corresponds to the system of

Figure 2.17 with the source 2 removed. For simplicity, we assume that source 1 is a very

large generator which implies the following: (a) it can supply any amount of power and

(b) it can control the voltage at bus 1 to a constant value independent of the load. We

will examine the operation of the system as the electric load changes. Recall the

conservation of complex power law. The complex power conservation equation applied

to bus 2 is:

02221 dd jQPS

Where:

222 ddd SjQP is the load complex power at bus 2.

Assuming constant power factor, the load reactive power is expressed as Qd2 = Pd2

where is a constant. In this case:

0)1( 212 dPjS

or:

0)1(

)cos()sin(33)sin()cos(33

2

121221

2

2121221

2

2

dPj

bgVVjbVjbgVVgV

Without loss of generality, we can assign 1=0.0. Then:

0)1(cossin33sincos33 22221

2

22221

2

2 dPjbgVVjbVjbgVVgV

The above equation links the voltage magnitude to the electric load Pd2 at bus 2. For a

given Pd2, the equation can be numerically solved for V2. A typical variation of the

voltage magnitude V2 versus the electric load Pd2 is shown in Figure 2.18.



V2

Pd2

Qd2 = Pd2

1.0Power Factor

Leading

< 0

Power Factor

Lagging

> 0

PmaxPmax

0.0

Figure 2.18 Typical Variation of the Voltage Magnitude V2 versus Electric Load

For a special case whereby the line resistance is neglected (g=0), a closed form solution

can be obtained for V2. Specifically in this case the power equations become:

2221

2

22

2212

cos33

sin3

dd

d

PbVVbVQ

bVVP

The above equations can be solved for V2 as a function of Pd2 as follows:

2221 )sin(3 dPVbV (2.6a)

2

22221 3)cos(3 bVPVbV d (2.6b)

Upon squaring and adding above equations:

2

22

4

2

22

2

22

21 6913 VbPVbPVbV dd

Rewrite equation as:

019692

2

22

2

2

1

2

2

4

2

2 dd PVVbbPVb

Solution for V2 yields:



2

2

2

2222

1

2

2

2

1

2

22

218

)1(36)96(96

b

PbVbbPVbbPV

ddd (2.7)

The related problem of transfer capability is defined as the maximum Pd2 which can be

transferred through the line. For this purpose we need to determine the maximum Pd2 for

which the equation for V2 above has a solution, i.e. the quantity inside the square root is

non-negative:

0)1(36)96( 2

2

2222

1

2

2 dd PbVbbP

The above inequality is a quadratic equation in terms of Pd2. When the left side is equal

to zero, the solution for Pd2 provides the maximum power that can be transferred through

the line, Pd max:

09124 4

1

4

max

2

1

32

max

2 VbPVabPb dd

Solution of this equation yields:

22

1max, 12

3

V

bPd

Note that the quadratic equation will have two solutions. Only one is admissible (the one

given above) since the other solution provides a negative power. Upon substitution of the

maximum transferable power into the equation for the voltage magnitude, the following

relationship is obtained:

221

2 112

V

V

In this analysis we have assumed that the source is infinitely large, i.e. it can supply any

amount of real power and it can control the voltage at bus 2 to 1.0 pu. It is expedient to

ask the question: how much reactive power must be generated by the source to sustain the

real power transfer capability. For this purpose, the reactive power output of the source

can be computed as a function of the transferred real power Pd2. The complex power

supplied by the source is:

*

2

*

111

~~)(

~VVjbgVSg

The reactive power supplied by the source is:

)Im( 11 gg SQ



Note that for each value of transferred power Pd2, the voltage magnitude V2 can be

computed from equation (2.7) and the phase angle (1-2) can be computed from equation

(2.6a). Then using above equations, the reactive power of the source can be computed.

The equations are complicated and are left as an exercise for the reader.

The transfer capability of transmission circuits is demonstrated with an example, next.

Example E2.6: A two bus, one generating unit electric power system operates in such a

way that 3/115~1 kVV , and the electric load at bus 2 has a unity power factor. The

positive sequence impedance of the line is Z=j29.75 ohms. The system is illustrated in

Figure E2.6.

(a) Compute the maximum real power P which can be transmitted through the power

line, Pmax.

(b) Compute the magnitude of the voltage ~V2 , i.e. V2, and the reactive power generated

by the source versus P for the range P = [ 0.0 to Pmax].

1 2j29.751

~V jeVV 22

~

S = P + j0.0 Figure E2.6

Solution: (a) For this system

mhosb 03361.075.29

1

kVkV

V 395.663

1151 , and 0.0

The maximum power that can be transferred is:

MW 27.22212

3 22

1max,

V

bPd

At this power the voltage will be

kV 948.462

1

2 V

V

(b) Upon substitution of numerical values into equation (2.6), the voltage magnitude and

source reactive power versus transferred power is obtained.



MWinPkVPV ,34027.9826.268,858,4148.204,2 2

2 kV,

MWinPMVArV

PVQg ,

8269.440.16953.6534.444

2

2

2

21

Using above equations, Table E2.6 is generated which provides the voltage magnitude

and source reactive power for several values of transferred power is the range 0 to 222.27

MWs. Note that at the maximum power transfer, the reactive power generation at the

source equals (numerically) the transferred real power. This means that the power factor

of the source at this point will be 0.707.

Table E2.6 Tabulated Values of Voltage Magnitude and Source Reactive Power vs.

Transferred Power

Transferred Power

(MW) Voltage Magnitude at

Load, V2 (kV) Source Reactive

Power, Qgen (MVAr)

0.00 66.395 0.0

50.00 65.970 5.74

100.00 64.600 23.78

150.00 61.900 58.22

200.00 56.270 125.31

222.27 46.948 222.27

2.10 Non-sinusoidal Operation

Recent developments in power system components resulted in elements which operate by

switching and therefore changing topologies within one period of operation. These

devices distort the sinusoidal waveform of voltages and currents. The distorted

waveform is described in terms of harmonics resulting from a Fourier analysis.

Waveform distortion necessitates a new way of analysis and system description.

Consider a point in an electric power system. Let v(t) and i(t) be the voltage and electric

current at this point. If v(t) and i(t) are not pure sinusoids but periodic, they can be

expanded into a Fourier series, i.e.

i

ii tiatv cos)(

i

ii tibti cos)(

The real power is computed as the average power flow, i.e.



T

t

dttitvT

P0

)()(1

Using above Fourier expansion, the real power is

i

ii

ii baP )cos(

2

Note that the real power can be determined directly from the definition of real power.

However, for reactive power it is conceptually difficult to develop a simple defining

equation as it has been done for the case of pure sinusoidal waveforms. To overcome this

difficulty, the concept of distortion power is introduced. Specifically, the distortion power

D is defined from the equation

2222 DQPS b

where

S is the apparent power defined as the product of the voltage and electric current

rms values, i.e. S = VrmsIrms

P is the real power

Qb is the reactive power of the fundamental component,

i.e. )cos(2

11

11 ba

Qb

D is the distortion power.

The concepts discussed above will be illustrated with an example.

Example E2.7: An electric load is connected to an ideal voltage source. The voltage of

the source is:

Vttv cos1002)(

The electric load distorts the current waveform resulting in an electric current equal to:

Attti 00 213cos10210cos502)(

Compute: (a) The real power absorbed by the electric load, (b) the reactive power at the

fundamental frequency, and (c) the distortion power.

Solution:

(a) The real power is



W

dtttT

dtttT

dttitvT

P

TT

T

924,4W 10cos000,5

)213cos(cos000,2

)10cos(cos000,10

)()(1

0

0

0

0

0

0

(b) The reactive power, of the fundamental frequency, is

VAr 24.86810sin)50()100( bQ

(c) First we compute the rms values of the voltage and current:

VVrms 100

AI rms 9902.501050 22 , and

VAIVS rmsrms 02.099,5

The distortion power is

VAd 19.100024.868492402.5099 222222 bQPSD

Note that we use the convention VAd for the units of distortion power.

2.11 Summary

In this chapter we have provided an overview of basic concepts utilized in the analysis of

power systems. We have introduced the concepts of sinusoidal steady state operation,

phasors, symmetrical components, complex power conservation, and network analysis

methods. The basic concepts were applied to address the problem of power transfer

capability for a simple electric power system. In later chapters, these concepts will be

utilized to formulate and solve a variety of analysis problems. For completeness, we have

also introduced the concepts of reactive power and distortion power under non-sinusoidal

but periodic steady state operation of an electric power system.



2.12 Problems

Problem P2.1. A single phase electric load is connected to a 480 volt source and absorbs

15 kW at a power factor of 0.80, electric current lagging.

(a) Compute the reactive power of the load.

(b) Compute the complex power.

(c) Compute the apparent power.

(d) Assume the line voltage phase angle to be zero degrees and compute the electric

current phasor (magnitude and phase) and the expression of the instantaneous

power, p(t), as a function of time.

Solution: (a) Expressing the real power of the system:

AIkWI 55.221538.0480 0869.368.0cos

VArIQ 248,11sin3

4803

(b) kVArjkWjIVS 248.11156.08.055.223

4803

~~3 *

(c) kVASa 747.18

(d) 0869.3655.22

~ jeI

VAtttttp 00 869.36coscos495,37869.36cos55.222cos3

48023

Problem P2.2. Figure P2.2 illustrates a three phase power system which consists of a

wye connected source, a delta connected synchronous motor, a delta connected load, and

two transmission lines. For simplicity the internal impedance of both source and motor is

assumed to be zero. The system is symmetric and operates under balanced conditions.

The following have been measured:

kVeE j

a

002.7~

, kVeE j

AB

050.12~

a) Compute the electric currents aI

~ and ABI

~ shown in Figure P2.1. Provide both

magnitude and phase.

b) What is the total real power absorbed by the motor?



2 + j10

A

B C

j210

j210j2

10

2 + j10

2 + j10 2 + j10

2 + j10

2 + j10

b c

a

Source Motor

j3

j3 j3

j3

j3 j3

bE~

cE~

ABE~

CAE~

BCE~

aE~

ABI~

aI~

bI~

cI~

Figure P2.2

Solution: (a) The positive sequence network is illustrated in the figure.

70

j7Ia

~2

6.928e-j250

7.2e-j00

j7 IA

~2V

~

Kirchoff’s current law at node 1 yields:

072

928.6~

70

~

72

2.7~ 025

j

eV

j

V

j

V j

Solution of above for the voltage yields:

kVeV j 048.11567.6~

Then:



Amperesej

VI j

a

035.142079.072

2.7~

~

Amperesee

j

eVI j

jj

AB

0

00

4.403025

129.0372

93.6~

~

(b) The power absorbed by the motor is:

MVAej

eVeS j

j0

0

0 4.35

*25

25 635.472

93.6~

93.63

MWSP 778.3Re

Problem P2.3. Determine whether a three phase wye connected, neutral grounded

element has an equivalent delta connected element.

Solution: No. The delta arrangement does not represent the neutral.

Problem P2.4. A three phase electric load consumes a total of 25 MW of real power

with a 0.85 power factor, electric current leading when connected to a 12 kV line to line

system. Compute the line electric current and determine the minimum reactive power

required to correct the power factor to 0.95.

Solution: 079.31)85.0arccos( 1.415,1

3)(

~ jj

LL

line eepfV

PI

MVArMWpfQ 49.1579.31tan25)85.0( 0

MVArMWpfQ 22.819.18tan25)95.0( 0

Needed MVAr is the difference: -7.27 MVAr (inductive).

Problem P2.5. A single phase electric load is connected to a 480 volt source and absorbs

15 kW at a power factor of 0.80, electric current lagging.

(a) Compute the reactive power of the load.

(b) Compute the complex power.

(c) Compute the apparent power.

(d) Assume the line voltage phase angle to be zero degrees and compute the electric

current phasor (magnitude and phase) and the expression of the instantaneous

power, p(t), as a function of time.

Problem P2.6. The illustrated power system in Figure P2.6 is symmetric and operates

under balanced conditions. The voltages of the machines are: kVeE j

a

002.7~

,

kVeEAB

050.12~

.



(a) Draw the per phase equivalent circuit.

(b) Compute the complex power absorbed by the delta connected capacitor

bank.

(c) Compute the complex power (total three phase) flowing into the

synchronous machine 2.

(d) Which synchronous machine operates as a generator and which as a

motor?

j10

A

B C

j10

j10

j5

b c

a

Synchronous

Machine 2

j5

j5

- j580 - j580

- j580

Capacitor

Bank

Synchronous

Machine 1

j4

j4j4 j2j2

j2

bE~

cE~

ABE~

CAE~

BCE~

aE~

ABI~

Figure P2.6

Solution: (a) Build the positive sequence network: Line 1: j6 ohms, capacitor: -j580/3

ohms, Line 2: j3 ohms, Machine 1: kVeE j

a

002.7~

, Machine 2:

kVeEAN

025928.6~

(b) Solution of the circuit yields the following voltage at the capacitor bank:

kVeECN

051.1694.6~ . The capacitor power is:

MVArEjS CN 747.03

5803 2

1

(c) Machine 2 power (going into machine)

MVArjMWEj

EVS AN

ANCN 443.010.7~

3

~~

3

*

2

(d) Since machine 2 absorbs real power, it operates as motor. Therefore, machine 1

operates as generator.

Problem P2.7. The positive sequence equivalent circuit of a three bus system is

illustrated in the Figure P2.7. Generator G1 injects an electric current 1

~I while generator

G2 injects an electric current 2

~I at buses 1 and 2 respectively. (a) Compute the complex

power absorbed at the electric load if 1

~I = 1,000 A and 2

~I = 1,200e j50

A, (b) Compute



the complex power flow in circuit 2-3, and (c) compute the complex power provided by

generating unit G2.

1

~I

2

~I

G1 G2

1 2

3

Electric

Load

0.03 - j0.002

-j1

5-j1

10-j2

Figure P2.7

Solution: The nodal equations are:

0

200,1

000,1

~

~

~

002.203.5151

15315210

121031005

3

2

1

je

V

V

V

jjj

jjj

jjj

VeVjV j 0684.6

1 549,73560,8049,73~

VeVjV j 0646.6

2 477,73505,8983,72~

VeVjV j 0541.6

3 102,73328,8626,72~

phaseperMVArjMWVjVS 688.10317.160~

002.003.0~ *

333

phaseperMVArjMWVVjVS 85.2168.147~~

15~ *

32223

phaseperMVArjMWeVS j

g 532.2172.88200,1~ *5

22

Problem P2.8. Consider a single phase power system which has the equivalent circuit

shown in the Figure P2.8. All pertinent data are given in the Figure.

(a) Compute the voltage phasors (magnitude and phase) at nodes 1, 2, and 3.

(b) Compute the complex power, S = P + jQ, absorbed by the electric load.



1 2

3

Electric

Load

0.01 - j0.001

-j1

-j2

j0.001

AIg 1000~

Figure P2.8

Solution: (a) The nodal equations are:

3

2

1

~

~

~

001.301.021

2999.10

101

0

0

1000

V

V

V

jjj

jj

jjA

Upon solution:

Volts

e

V

V

V j

0.000,100

0.050,100

0.005,100

~

~

~ 057.0

3

2

1

(b)

100000,1001.001.0~~

3 jjVIL

MVArjMWIVS L 0.100.100~~ *

3

Problem P2.9. Consider the simplified electric power system of Figure P2.9 consisting

of a balanced generator, a symmetric transmission line and a symmetric electric load.

Each phase of the symmetric line has a self impedance of j9 ohms. The mutual

impedance between any two phases is j4 ohms with the indicated polarity. Other system

parameters are indicated in the figure. Compute the real power absorbed by the electric

load.



bc

a

b c

a j9

j9

j9

j4

j4

j4

R+jX

R+jXR+jX

R = 35

X = 15

bE~

cE~

aE~

kVeE j

a

092.6~

Figure P2.9

Solution: The positive sequence model is:

6.92 e

j15

35

j5I~

j00

S/3

kAej

I j 0745.291717.02035

92.6~

MVArjMWMVAjIjIVS 327.1095.31717.04510515353~~

322*

Thus:

MWPload 095.3

Problem P2.10. Consider a simplified electric power system consisting of a three phase

source, a symmetric three phase transmission line and a symmetric three phase electric

load. Assume that the source voltage is 230 kV line to line and remains constant. The

positive sequence series impedance of the line is j57.5 ohms. The power factor of the

electric load is 0.90 current lagging and remains constant.

Compute the maximum total power that can be transferred from the source to the load

under the specified conditions.



Compute and graph the voltage at bus 2 and the reactive power of the source versus the

transferred power.

What will the maximum transfer capability if the power factor is 0.9 current leading and

remains constant?

Solution: The maximum power transfer is:

22

1max 12

3 V

bP

where:

4843.0842.25cos

842.25sin0

0

Thus: MWP 33.288max

The graph of the voltage and reactive power as a function of P is given below.

Leading power factor:

4843.0842.25cos

842.25sin0

0

Thus: MWP 88.733max

Problem P2.11. Figure P2.11 illustrates a simplified power system consisting of two

synchronous machines, a phase shifting transformer and a line. Assume that the

synchronous machines are ideal voltage sources and that the phase shifter consists of

ideal transformers. Specifically, all illustrated transformers are ideal with a

transformation ratio of secondary to primary voltage equal to 0.5

5.0

Pr geimaryVolta

oltageSecondaryV. Note that the primary is designated as P and the secondary

as S. The voltages of the synchronous machines are:

000 240

11

120

11

0

1

~~,

~~,66.8

~ j

ac

j

ab

j

a eEEeEEkVeE 000 240

22

120

22

0

2

~~,

~~,0.9

~ j

ac

j

ab

j

a eEEeEEkVeE



Compute the total electric power absorbed or generated by the synchronous machine 2.

bE1

~cE1

~

aE1

~

bE2

~cE2

~

aE2

~0.2-j1.0

0.2-j1.0

0.2-j1.0

S P P

P

P

S

S

S

S

PS

P

Figure P2.11

Solution: Since the phase shifter is an ideal transformer we can compute the voltage on

the right hand side of the phase shifter. Specifically, the phase A voltage on the right

hand side of the transformer is:

kVjEEEV cbaa 75.366.8~~

25.0~~

111

kAejjI j 049.168399.30.975.366.82.0~

MVArjMWIES a 429.2941.99~~

3 *

2 , absorbed power

Problem P2.12. Consider the simplified electric power system of Figure P2.12 consisting

of a balanced generator, a symmetric transmission line and a symmetric electric load.

Each phase of the symmetric line has a self impedance of j9 ohms. The mutual

impedance between any two phases is j4 ohms. Other system parameters are indicated in

the figure. Compute the real power absorbed by the electric load.



Ea

~ +

j9

j9

j9

j4

j4

j4

R+jX R+jX

R+jX

R = 36 X = 15

Ea=6.92 e j0 kV~

Figure P2.12

Solution: The positive sequence model is:

6.92 kV

j5

12

j5

V

I

~

~

kAej

I j 08.39443.01012

92.6~

MVArjMWMVAjIjIVS 944.2065.7443.015365123~~

322*

Thus:

MWPload 065.7

Problem P2.13. Figure P2.13 illustrates a three phase power system which consists of a

wye connected source, a delta-wye connected transformer, a three phase transmission line

and an induction motor. The figure provides the impedances of the various components

of the system. Note that the system is fully symmetric and for each device the impedance

of only one phase is provided. If applicable one mutual impedance is also provided. It

should be understood that the impedances of the other phases are identical and the mutual

impedances between any two phases are also identical. The transformer ratio is 5.16t .

At a certain instant of time, the system operates under balanced conditions and the phase

A current and voltage of the induction motor is:



AmpereseI j

ma

07.210.285~ , kVoltseV j

ma

0095.7~

a) Compute the real and reactive power absorbed by the induction motor – provide the

total power.

b) Compute the power factor of the induction motor.

c) Compute the electric current gaI~

of the generator phase A and the voltage gaV~

of

the generator of the same phase.

b) Compute the real and reactive power delivered at the terminals of the generator –

provide the total power.

j0.02

-j0.02

Iga Vga

~~1:t j0.210.1

j0.06

j0.800.02 ImaVma

~ ~

Generator Transmission Line MotorTransformer

Ea

~

Figure P2.13

Solution: Since the system operates under balanced conditions, the voltages and currents

of phases B and C at any point of the system is same as phase A voltages and currents

with a phase angle difference of 120 and 240 degrees respectively.

(a) MVArjMWMVAeIVS j

mamam 513.2315.6797.6~~

307.21*

(b) 9291.07.21cos 0 pf

(c) The voltage at the right hand side of the transformer, phase A is:

mamamcmbmamamata IjVIIjIjIjVV~

95.012.0~~~

06.0~

21.01.0~

80.002.0~~

VoltseIjVV j

mamata

069.1088.085,8~

95.012.0~~

The voltage and current on the left hand side of the transformer is:

VoltseVt

eVVVV j

ta

j

gagbgagab

00 69.1120 01.490~1

1~~~~



Thus: VoltseVt

eV j

ta

j

ga

00 31.281

120 90.282~1

1~

07.215.702,4

~~ j

mata eItI

00 7.51240 97.144,81~~~~ jj

tatctaga eeIIII

(d) MVArjMWMVAeIVS j

gagag 744.2345.6913.6~~

3039.23*

Problem P2.14. Consider the normalized model of a transmission line in Figure P2.x.

Under the present operating conditions, the voltages at the terminals of the transmission

line are (magnitude in pu, phase in radians):

67.0

1 02.1~ jeV ,

25.0

2 99.0~ jeV , phase angles are in radians

Compute the complex power flow (real power and reactive power) at the left side of the

transmission line.

Figure P2.14

Solution:

PSA_Chapter_02.pdf

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