basic concepts-lecture note

8/12/2019 Basic Concepts-lecture Note

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Module 2

BASIC CONCEPTS

FOR ELECTRIC POWER SYSTEM ANALYSIS

The intention of this module 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 system. This module

can be considered as simply as review of topics that will be utilized throughout this

teaching module. We start by introducing the principal electrical quantities that we

will deal with.

2.1. POWER IN SINGLE PHASE AC CIRCUITS

The electric power systems specialist is almost always more concerned with

describing the rate of change of energy with respect to time (which is the definition

of power) rather than voltage and current. As the power into an element is basically

the product of voltage across and current through it, it seems reasonable to swap the

current for power without losing any information in describing the phenomenon. In

treating sinusoidal steady-state behavior of circuits, some further definitions are

necessary. To illustrate the concepts, we will use a cosine representation of the

waveform.

If the terminals of the load are designated a and n, and if the voltage and

current are expressed by

van= Vmaxcos t and ian= Imaxcos (t - )

the instantaneous power isp = van .ian= VmaxImaxcos t cos (t -) (2.1)

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

for leading current. A positive value ofpexpressed the rate at which energy is being

absorbed by the part of the system between the points a and n. The instantaneous

power is obviously positive when both van and ian are positive but will become

negative when van and ian are opposite in sign. Figure 2.1 illustrates this point.

Positive power calculated as vanianresults when current is flowing in the direction of


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a voltage rise and mean energy is being transferred to the load. Conversely, negative

power calculated as van ian results when current is flowing in the direction of a

voltage rise and means energy is being transferred from the load into the system to

which the load is connected. If van and ian are in phase, as they are in a purely

resistive load, the instantaneous power will never becomes negative. If the current

and voltage are out of phase by 900, as in purely inductive or purely capacitive ideal

circuit element, the instantaneous power will have equal positive and negative half

cycles and its average value will be zero.

Figure 2.1 Current, voltage, and power plotted versus time.

By using trigonometric identities the expression of Eq.(2.1) is reduced to

tIV

tIV

p 2sinsin2

2cos1cos2

maxmaxmaxmax (2.2)

where VmaxImax/2may be replaced by the product of the rms voltage and current |Van|

. | Ian|or |V| . | I|.

For tVvan cosmax ,

tIi

Ri

R coscos

max

max

Figure 2.2 shows Raniv plotted versus t.

Similarly,

tIV

iv

ttIViv

xan

xan

2sinsin2

cossinsin

maxmax

maxmax


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Figure 2.2 Voltage, current in phase with the voltage, and the resulting power

plotted versus time.

which the instantaneous power in the inductance and is the second term in Eq.(2.2).

Figure 2.3 shows ,, xan iv and their product plotted versus t.

Figure 2.3 Voltage, current lagging the voltage by 900, and the resulting power

plotted versus time

Examination of Eq.(2.2) shows that the first term, the term which contains

cos , is always positive and has an average value of

cos2

maxmaxIVp

or, when rms value of voltage and current are substituted,

cosIVp (2.3)

P is the quantity to which the word power refers. P, the average power, is also called

the real power. The fundamental unit for both instantaneous and average power is the

watt, but a watt is such a small unit in relation to power quantities that P is usually

measured in kilowatts or megawatts.


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The cosine of the phase angle between the voltage and the current is called

the power factor. An inductive circuit is said to have a lagging power factor, and a

capacitive circuit is said to have a leading power factor. In other words, the term

lagging power factor and leading power factor indicate, respectively, whether the

current is lagging or leading the applied voltage.

The second term of Eq.(2.2), the term containing sin , is alternately positive

and negative and has an average value of zero. This component of the instantaneous

powerpis called the instantaneous reactive powerand expresses the flow of energy

alternately toward the load and away from the load. The maximum value of this

pulsating power, designated Q, is called reactive power or reactive volt-amperes and

is very useful in describing the operation of a power system, as will become

increasingly evident in further discussion. The reactive power is

sin2

maxmaxIVQ

or (2.4)

sinIVQ

The square root of the sum of the squares of Pand Qis equal to the product

of |V| and | I |, for

IVIVIVQP 222 sincos

Of course P and Q have the same dimensional units, but it is usual to designate the

unit for Q as vars (for volt-amperes reactive). The more practical unit for Q are

kilovars or megavars.

We define a quantity called the complex or apparent power, designated S,

which P and Q are components. By definition,

sincos

sincos

jVIS

jVIVIS

jQPS

(2.5)

Using Eulers identity, we thus have

jeVIS

or


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S = VI /

If we introduce the conjugate current defined by the asterisk(*)

I*= | I | /

It becomes immediately obvious that an equivalent definition of complex or apparent

power is

S = VI* (2.6)

The use of concepts of complex power may prove advantageous in solving

problems of power system analysis.

Consider the situation with an inductive circuit in which the current lags the

voltage by the angle . The conjugate of the current will be in the first quadrant in

the complex plane as shown in figure 2.4(a). Multiplying the phasors by V, we obtain

the complex power diagram shown in figure 2.4(b). Inspection of the diagram as well

as the previous development leads to a relation for the power factor of the circuit:

S

Pcos (2.7)

Example 2.1Consider the circuit composed of a series R-L branch in parallel with

capacitance with the following parameters:

R = 0.5 ohms

XL= 0.8 ohms

Bc= 0.6 siemens

Figure 2-4. Complex power diagrams.

-

I*

V

I

S = V I*

Q = VI sin

P = VI cos

(a) (b)


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Assume

V = 100 /0 V

Calculate the input current and the active, reactive, and apparent power into

the circuit.

Solution

The current into the R-L branch is given by

8.05.0

100

jIZ

= 106.00 /-57.99

0A

The power factor (PF) of the R-L branch is

53.0

99.57coscos 0

Z

ZZ

PF

PF

which is a laggingpower factor.

The current into the capacitance is

AjIc090/601006.0

The input current Itis

0

00

01.28/64.63

90/6099.57/00.106

t

t

Zct

I

I

III

The power factor (PF) of the overall circuit is

88.001.28coscos 0

ttPF lagging

Note that the magnitude of It is less than that of IZ, and that cos is higher

than cos z. This is the effect of the capacitor, and its action is called power

factor correction in power system terminology.

The apparent power into the circuit is

VAS

S

VIS

t

t

tt

0

0

*

01.28/00.6364

01.28/64.630/100

In rectangular coordinates we get


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76.298898.5617 jSt

Thus the active and reactive powers are:

var76.2988

98.5617

t

t

Q

WP

2.2 DIRECTION OF POWER FLOW

For an ac system figure 2.5 shown an ideal voltage source (constant

magnitude, constant frequency, zero impedance) with polarity marks which, as

usually, indicate the terminal which is positive during half cycle is positive

instantaneous voltage. Of course, the positively marked terminal is actually the

negative terminal during the negative half cycle of the instantaneous voltage.

Similarly the arrow indicates the direction of current during the positive half cycle of

current.

Figure 2.5 An ac cicuit representation of an emf and current to illustrate

polarity marks.

In figure 2.5a a generator is expected since the current is positive when

flowing away from the positively marked terminal. However, the positively marked

terminal may be negative when the current is flowing away from it. The approach to

understanding the problem is to resolve phasor Iinto a component along the axis of

the phasor Eand a component 900out of phase with E. The product of |E| and the

magnitude of the component of I which is 900 out of phase with E is Q. If the

component of I along the axis of E is in phase with E, the power is generatedpower

which is being delivered to the system, for this component of current is always

flowing away from the positively marked terminal when that terminal is actually

positive (and toward that terminal when the terminal is negative).P, the real part of

EI*, is positive.

+ -I

E

(a)

+ -I

E

(b)


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If the component of current along the axis of E is negative (180 0out of phase

with E), power is being absorbed and the situation is that of a motor.P, the real part

ofEI*, would be negative.

The voltage and current relationship might be as shown in figure 2.5b, and a

motor would be expected. However, an average power absorbedwould occur only if

the component of the phasor I along the axis of the phasor E was found to be in

phase rather than 1800out of phase with E, so that this component of current would

be always in the direction of the drop in potential. In this caseP, the real part ofEI*

would be positive. Negative P here would indicated generated power.

To consider the sign of Q, figure 2.6 is helpful. In figure 2.6a positivereactive power equal to |I|2X is supplied to the inductance since inductance draws

Figure 2.6 Alternating emf applied (a) to apurely inductive element

and (b) to a purely capasitive element.

positive Q. Than I lags E by 900, and Q, the imaginary part of EI*, is positive. In

figure 2.6b negative Q must be supplied to the capacitance of the circuit, or the

source with the emf E is receiving positive Q from the capacitor, I leads E by 900.

2.3. BALANCED THREE-PHASE SYSTEM

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

voltage having the same magnitude and frequency but time displaced from one

another by 1200. Figure 2.7(a) shows a schematic representation where the three

single phase voltage source appear in a Y connection; a configuration is also

possible. A phasor diagram showing each of the phase voltage is also given in figure

2.7(b). As the phasors revolve at the angular frequency with respect to the

reference line in the counterclockwise (positive) direction, the positive maximum


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value first occurs for phase a and then in succession for phases band c. Stated in a

different way, to an observer in the phasor space, the voltage of phase aarrives first

followed by that of band then of c. For this reason the three phase voltage of figure

2.7 is said to have the phase sequence abc. This is important for certain applications.

For example, in three phase induction motors, the phase sequence determines

whether the motor turn clockwise or counterclockwise.

(a) (b)

Figure 2-7.A Y-connected three-phase system and the corresponding phasor diagram.

Current and Voltage Relation

Balanced three phase system can be studied using technique developed for

single phase circuits. The arrangement of the three single phase voltage into a Y or

configuration requires some modification in dealing with the overall system.

Y Connection

With reference to figure 2.7 the common terminal n is called the neutralor

star(Y)point. 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 line 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 thephase voltages. Analysis of phasor diagrams provides the required

relationships.

~

~

~ a

b

120o

120o

120o

cVcn

Vbn

Van

Reference Line


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Figure 2-8. Illustrating relation between phase and line voltages in a Y connection.

The effective values of the phase voltages are shown in figure 2.8 as V an, Vbn,

and Vcn. Each has the same magnitude, and each is displaced 1200from the other two

phasors. To obtain the magnitude and phase angle of the line voltage from a to b (i.e.,

Vab), we apply Kirchhoffs law:

nbanab VVV (2.8)

This equation states that the voltage existing from a to b is equal to the voltage from

a to n (i.e., Van) plus the voltage from n to b. Thus equation 2.8 can be rewritten as

bnanab VVV (2.8 a)

Since for a balanced system, each phase voltage has the same magnitude, let us set

pcnbnan VVVV (2.9)

where Vpdenotes the effective magnitude of the phase voltage. Accordingly we may

write0

pan 0/VV (2.10)

0pbn 120/VV (2.11)

0p

0pcn

0p

0pcn 120/V240/VV120/V240/VV (2.12)

Substituting equations 2.10 and 2.11 in equation 2.8a yield

012011 /VV pab

0303 /VV pab (2.13)

opcn 120VV

Vab

30o

opan 120VV

opan 0VV


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Similarly we obtain

0903 /VV pbc (2.14)

01503 /VV pca (2.15)

The expressions obtained above for the line voltages show that they constitute

a balanced three phase voltage system whose magnitude are 3 times the phase

voltages. Thus we write

pL VV 3 (2.16)

A current flowing of out line terminal a (or b or c) is the same as that flowing

through the phase source voltage appearing between terminal n and a (or n and b, or

n and c). We can thus conclude that for a Y-connection three phase source, the line

current equals the phase current. Thus

pL II (2.17)

In the above equation, ILdenotes the effective value of the line current and Ipdenotes

the effective value for the phase current.

Connection

We consider now the case when the three single phase sources are rearranged

to form a three phase connection as shown in figure 2.9. It is clear from inspection

of the circuit shown that the line and phase voltage have the same magnitude:

pL VV (2.18)

The phase and line currents, however, are not identical, and the relationship between

them can be obtained using Kirchhoffs current law at one of the line terminals.


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Figure 2-9. -connected three-phase source.

In a manner similar to that adopted for the Y connection source, let us

consider the phasor diagram shown in figure 2.10. Assume the phase currents to be

0

0

120/

120/

0/

pca

pbc

pab

II

II

II

Gambar 2-10. Illustrating relation between phase and line currents in a connection.

The current that follows in the line joining a to a is denoted Iaaand is given

by

abcaaa III

~

~

~

a

bc

Iaa

Ibb

Icc

a

b

c

Ica

Iab

Iaa

Iba

Ica

Vca

Vab

Ibb

IccVbc

30o


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As a result, we have

0/1120/1 0 paa II

which simplifies to

0150/3 paa II

similarly,

0

0

90/3

30/3

pcc

pbb

II

II

Note that a set of balanced three phase currents yields a corresponding set of

balanced line currents that are 3 times the phase values:

pL II 3 (2.19)

where ILdenotes the effective value of any of the three line currents.

Power Relationships

Assume that the three phase generator is supplying a balanced load with the

three sinusoidal phase voltages:

0

0

120sin2

120sin2

sin2

tVtv

tVtv

tVtv

pc

pb

pa

With the currents given by

0

0

120sin2

120sin2

sin2

tIti

tIti

tIti

pc

pb

pa

where is the phase angle between the current and voltage in each phase. The total

power in the load is

titvtitvtitvtp ccbbaa 3

This turns out to be


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120sinsin

120sin120sin

sinsin23

tt

tt

ttIVtp pp

Using a trigonometric identity, we get

2402cos2402cos2coscos33 tttIVtp pp

Note that the last three terms in the above equation are reactive power terms and they

add up to zero. Thus we obtained

cos33 ppIVtp (2.20)

When referring to the voltage level of a three phase system, one invariablyunderstands the line voltages. From the above discussion the relationship between

the line and phase voltages in a Y connection system is

pL VV 3

The power equation thus reads in term of line quantities:

cos33 LL IVp (2.21)

We note the total instantaneous power is constant, having a magnitude of

three times the real power per phase. We may be tempted to assume that the reactive

power is of no importance in a three-phase system since the Q cancel out. However,

the situation is analogous to the summation of balanced three-phase currents and

voltages that also cancel out. Although the sum cancels out this quantities are still

very much in evidence in each phase. We thus extend the concept of complex or

apparent power (S) to three phase system by defining

*

3 3 ppIVS (2.22)

as

cosIVp pp33 (2.23)

sinIVQ pp33 (2.24)

In terms of line values, we can assert that

*

3 3 LLIVS (2.25)

and


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cosIVp LL33 (2.26)

sinIVQ LL33 (2.27)

Example 2.2

A Y connected, balanced three phase load consisting of three

impedances of 10 /300 ohms each shown in figure 2.11 is supplied with

balanced line to neutral voltages:

VV

VV

VV

cn

bn

an

0

0

120/220

240/220

0/220

a. Calculate the phasor currents in each line

b. Calculate the line to line phasor voltages

c. Calculate the total active and reactive power supplied to the load.

Figure 2-11. Load connection for example 2.2.

Solution

a. The phase currents are obtained as

2200

a

b

c

Z = 1030oohms

n


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AI

AI

AI

cn

bn

an

0

0

0

90/2230/10

120/220

210/2230/10

240/220

30/2230/10

220

b. The line voltages are obtained as

0

0

0

0

210/3220

90/322012030/3220

30/3220

240/2200/220

ca

bc

ab

ab

bnanab

V

V

V

V

VVV

c. The apparent power into phase ais given by

VAS

S

IVS

a

a

anana

0

0

*

30/4840

30/22220

The total apparent power is three times the phase value:

00.726069.12574

30/00.1452030/34840 00

jS

VAS

t

t

Thus

var00.7260

69.12574

t

t

Q

WP

Example 2.3

Repeat example 2.2 as if the same three impedance were connected in

a connection.

Solution

From example 2.2 we have


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0

0

0

210/3220

90/3220

30/3220

ca

bc

ab

V

V

V

The currents in each of the impedances are

0

0

00

120/322

120/322

0/32230/10

30/3220

ca

bc

ab

I

I

I

The line currents are obtained with reference to figure 2.12 as

0

0

0

0

210/66

90/66

30/66

120/3220/322

c

bccac

b

abbcb

a

a

caaba

I

III

I

III

I

I

III

The apparent power in the impedance between a and b is

0

0

*

30/14520

0/32230/3220

ab

ab

ababab

S

S

IVS

The total three phase power is then

00.2178004.37724

30/43560 0

jS

S

t

t

As a result,

var00.21780

04.37724

t

t

Q

WP


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Figure 2-12. Load connection for example 2.3.

2.4PER PHASE ANALYSIS

The stage has now been set to introduce the powerful method of per phase

analysis. The justification for the method follows directly from the following

theorem.

Balanced Three-Phase Theorem. Assume that we are given a

1. Balanced three-phase (connected) system with

2. All loads and sources wye connected, and

3. In the circuit model there is no mutual inductance between phases.

Then

(a)

All the neutrals are the same potential,

(b)The phases are completely decoupled, and

(c)All corresponding network variables occur in balanced sets of the same sequence

as the sources.

Outline of Proof. The fact that the neutrals are at the same potential and the

network phases are decoupled follows that; using superposition, the number of

sources can be arbitrary. Without mutual inductance between the phases they are

decoupled and with balanced source it is clear that the responses in phases b and c

lag the corresponding responses occur in balanced sets.

a

c

Z

Ic

Ib

b

Z

Z

Ica

Ibc

Iab

Ia


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Method : per Phase analysis. We next turn to the method of per phase

analysis. Given a balanced three-phase network with no mutual inductance between

phases :

1. Convert all delta-connected sources and loads into equivalent wye connections.

2.

Solve for the desired phase a variables using the phase a circuit with all

neutrals connected.

3. The phase b and phase c variables can then be determined by inspection ;

subtract 1200and 2400, respectively, from the phase angles found in step 2, in the

usual case of abc-sequence. Add 1200 and 2400 in the case of acb-sequence

sources.4.

If necessary, go back to the original circuit to find line-line variables or variables

internal to delta connection.

Note: In using per phase analysis we always pick the neutral as datum. In this

case it is simpler to use a single-subscript notation for phase voltage. We will use Va

rather than Van, etc.

Examples 2.4

Given the balanced three-phase system shown in fig. 2.13 (a), find v1(t) and i2(t)

Figure 2. 13 (a)

SolutionReplace delta by equivalent Wye, Za = -j2/3. Using per phase analysis, we

consider the per phase (phase a) circuit shown in fig. 2.13 (b). Note that while v1

j0.1

j1.0

n

+

-

+

-v1

b

1 + j0.01

cbc c b

a

-j2n

a

oa 45

2

350E

a

i2


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20

appears in the phase circuit, i2 has been suppressed in the process of the -Y

conversion. Carrying out the calculation to find v1, we note that the equivalent

parallel impedance from a to n isj2. Using the voltage-divider law.

Figure 2. 13 (b)

aa EEjj

jv 05,1.

1.02

21

= 045.

2

368

The phasor v1 represents the sinusoid v1(t) = 368 cos ( 45t0). From the

original circuit we see that to find i2(t), we need to find Vab. Thus,

Vab= Va,n-Vbn=''6/3 nVaje

Then using the impedance of the capacitor, we find that

0

'' 165.2

319baI

and it follows that the corresponding sinusoidal is

02 165cos319 tti

Example 2.5

In the circuit shown in figure 2.14, the source phasor voltage is V = 30 /15 0.

Determine the phasor currents I2 and I3 and the impedance Z2. Assume that I1 is

equal to 5 A.

Solution

The voltage V2is given by

+

-

oa 45

2

350E

j0.1

+

-v1 j1.0

32j

a

nn

a


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21

0

0

2

12

94.17/20.25

0/515/30

1

V

IVV

Figure 2-14. Circuit for Problem 2-1.

The current I3is thus

023 94.17/52.2

10

VI

The current I2is obtained using KCL. Thus

0

312

61.16/72.2

III

Finally we have

ohmsj

I

VZ

26.564.7

56.34/28.9 0

2

2

2

Example 2.6

For the circuit of problem 2.5, calculate the apparent power produced by the

source and individual apparent power consumed by the 1-ohm resistor, the

impedance Z2, and the resistance R3. Show that conservation of power hold true.

Solution

The apparent power produced by the source is

~

b

n

-

+Z2I2V I3 R3= 10

I1

a

V2-

+


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22

VAj

VA

VISs

82.3889.144

15/150

0/515/30

0

0

*

1

The apparent power taken by the 1-ohm resistor is

WI

II

IVS abt

25

1

2

1

*

11

*

1

The apparent power taken by the impedance Z2is

VAj

IVS

82.3837.56

56.34/44.68

61.16/72.294.17/20.25

0

0

*

222

The apparent power taken by the resistor R3is

W

IVS

52.63

94.17/52.294.17/20.25

*

323

The total power consumed is

VAj

SSSSt

83.3889.144

321

This is equal to the source apparent power, which proves the principle ofconservation of power.

Example 2.7

A three phase transmission link is rated 100 kVA at 2300 V. when operating

at rated load, the total resistive and reactive voltage drop in the link are, respectively,

2.4 and 3.6 percent of the rated voltage. Determine the rated power and power factor

when the link delivers 60 kW at 0.8 PF lagging at 2300 V.


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Solution :

The active voltage drop per phase is

3

2300024.0 IRVr

The reactive drop is

3

2300036.0 IXVx

But the rated current is

AI 1.2532300

10100

3

As a result,

ohmsX

ohmsR

90.1

27.1

For a load of 60 kW at 0.8 PF lagging, the phase current is

AIl 83.18

8.032300

1060 3

The active and reactive power consumed by the link are thus

var32.20203

43.13503

2

2

XIQ

WRIP

llk

llk

As a result, the apparent power consumed by the link is

32.202043.1350 jSlk

The apparent load power is

VAj

St

000,45000,60

8.0cos/8.0

1060 13

Thus the total apparent power is now obtained by

047.37/74.77296

32.4702043.61350

jSt


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24

As a result

kWP

lagging

t

t

35043.61

79.0

47.37coscos 0

basic concepts-lecture note

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