Electric Circuits, 9th Edition P43 potx

Thus for a balanced three-phase system, we can focus on determining the voltage or current in one phase, because once we know one phase quantity, we know the others.. Figure 11.4 shows b

396 Sinusoidal Steady-State Power Calculations

Figure P10.61

255/0! rl

V (rms)

40 ft ;200 f) 720 ft

j 1500 ft N 2

rVi

Ideal

6800 ft

10.65 T h e ideal transformer c o n n e c t e d to t h e 10 ft load

in Problem 10.64 is replaced with an ideal

trans-former that h a s a t u r n s ratio of a:\

a) W h a t value of a results in m a x i m u m average

p o w e r being delivered to t h e 10 ft resistor? b) W h a t is t h e m a x i m u m average p o w e r ?

10.62 T h e variable load resistor R L in the circuit shown in

PSPICE Fig PI0.62 is adjusted for m a x i m u m average p o w e r

MULTISIM transfer to i? L

a) Find the m a x i m u m average power

b) What percentage of the average power developed

by the ideal voltage source is delivered to RL

when RL is absorbing maximum average power?

c) Test your solution by showing that t h e p o w e r

d e v e l o p e d by the ideal voltage source equals t h e

p o w e r dissipated in t h e circuit

Figure P10.62

1 1:2 i

500/0° f +

10.63 R e p e a t P r o b l e m 10.62 for the circuit shown in

PSPICE Fig P10.63

MULTISIM

Figure P10.63

40/0! f +

V(vms)\-10.64 Find t h e average p o w e r delivered t o t h e 10 ft

resis-tor in t h e circuit of Fig P10.64

Figure P10.64

2.5 ft

al : 2 0 1 - 7 | 3 0 : 1 1

( i i

\ ) \

Ideal I • [ideal]

10ft

Sections 10.1-10.6 10.66 T h e hair dryer in t h e Practical Perspective uses a

piwrecnvi 6 0 H z s m u s o i d a l voltage of 120 V (rms) T h e h e a t e r

e l e m e n t m u s t dissipate 250 W at t h e LOW setting,

500 W at t h e M E D I U M setting, a n d 1000 W at t h e

H I G H setting

a) Find t h e value for resistor R2 using t h e

specifica-tion for t h e MEDIUM setting, using Fig 10.31

b) Find the value for resistor Ri using the

specifica-tion for t h e LOW setting, using t h e results from part (a) a n d Fig 10.30

c) Is t h e specification for t h e HIGH setting satisfied?

10.67 A s seen in P r o b l e m 10.66, only two i n d e p e n d e n t

PRACTICAL p o w e r specifications can b e m a d e when t w o

resis-PERSPECTIVE * r

PSPICE tors m a k e up t h e healing e l e m e n t of the hair dryer a) Show that t h e expression for t h e H I G H p o w e r

rating (PH ) is

PH = Pi

where PM = t h e MEDIUM p o w e r rating a n d

P L = t h e LOW p o w e r rating

b) If PL = 250 W a n d P M = 750 W , what must the

HIGH p o w e r rating b e ?

10.68 Specify t h e values of R\ a n d R 2 in t h e hair dryer

cir-pRAcncAt c u i t in Fig 10.29 if t h e low p o w e r rating is 240 W

PSPICE a n d t h e high p o w e r rating is 1000 W A s s u m e t h e

MULTISIM supply voltage is 120 V (rms) (Hint: W o r k

P r o b l e m 10.67 first.)

10.69 If a third resistor is a d d e d t o t h e hair dryer circuit in

PRACTICAL Fig 10.29, it is possible t o design to three i n d e p e n d

-PSPICE e n t p o w e r specifications If t h e resistor R$ is a d d e d

MULTISIM j n s e r i e s w i t h t h e Thermal fuse, then t h e correspon-ding LOW, MEDIUM, a n d H I G H p o w e r circuit dia-grams a r e as shown in Fig P10.69 If t h e t h r e e

Problems 397

power settings are 600 W, 900 W, and 1200 W,

respectively, when connected to a 120 V (rms)

sup-ply, what resistor values should be used?

Figure P10.69

LOW MEDIUM HIGH

10.70 You have been given the job of redesigning the hair

PRACTICAL dryer described in Problem 10.66 for use in

'ERSPECTIVE J

DESIGN England The standard supply voltage in England is

PROBLEM ± M -mf i~J f

PSPICE 220 V (rms) What resistor values will you use in

MULTISIM your design to meet the same power specifications? 10.71 Repeat Problem 10.68 using single resistor values from Appendix H Calculate the resulting low, medium, and high power ratings

10.72 Repeat Problem 10.70 using single resistor values from Appendix H Calculate the resulting low, medium, and high power ratings

^ J I

CHAPTER CONTENTS

11.1 Balanced Three-Phase Voltages p 400

11.2 Three-Phase Voltage Sources p 401

11.3 Analysis of the Wye-Wye Circuit p 402

11.4 Analysis of the Wye-Delta Circuit p 407

11.5 Power Calculations in Balanced Three-Phase

Circuits p 410

11.6 Measuring Average Power in Three-Phase

Circuits p 415

/ C H A P T E R O B J E C T I V E S

1 Know how to analyze a balanced, three-phase

wye-wye connected circuit

2 Know how to analyze a balanced, three-phase

wye-delta connected circuit

3 Be able to calculate power (average, reactive,

and complex) in any three-phase circuit

398

Balanced Three-Phase Circuits

Generating, transmitting, distributing, and using large blocks

of electric power is accomplished with three-phase circuits The comprehensive analysis of such systems is a field of study in its own right; we cannot hope to cover it in a single chapter Fortunately, an understanding of only the steady-state sinusoidal behavior of balanced three-phase circuits is sufficient for engi-neers who do not specialize in power systems We define what we mean by a balanced circuit later in the discussion The same cir-cuit analysis techniques discussed in earlier chapters can be applied to either unbalanced or balanced three-phase circuits Here we use these familiar techniques to develop several short-cuts to the analysis of balanced three-phase circuits

For economic reasons, three-phase systems are usually designed to operate in the balanced state Thus, in this introduc-tory treatment, we can justify considering only balanced circuits The analysis of unbalanced three-phase circuits, which you will encounter if you study electric power in later courses, relies heav-ily on an understanding of balanced circuits

The basic structure of a three-phase system consists of volt-age sources connected to loads by means of transformers and transmission lines To analyze such a circuit, we can reduce it to a voltage source connected to a load via a line The omission of the transformer simplifies the discussion without jeopardizing a basic understanding of the calculations involved Figure 11.1 on page 400 shows a basic circuit A defining characteristic of a bal-anced three-phase circuit is that it contains a set of balbal-anced three-phase voltages at its source We begin by considering these voltages, and then we move to the voltage and current relation-ships for the Y-Y and Y-A circuits After considering voltage and current in such circuits, we conclude with sections on power and power measurement

Practical Perspective

Transmission and Distribution of Electric Power

In this chapter we introduce circuits that are designed to

handle large blocks of electric power These are the circuits

that are used to transport electric power from the generating

plants to both industrial and residential customers We

intro-duced the typical residential customer circuit as used in the

United States as the design perspective in Chapter 9 Now we

introduce the type of circuit used to deliver electric power to

an entire residential subdivision

One of the constraints imposed on the design and

oper-ation of an electric utility is the requirement that the utility

maintain the rms voltage level at the customer's premises

Whether lightly loaded, as at 3:00 am, or heavily loaded, as

at midaftemoon on a hot, humid day, the utility is obligated

to supply the same rms voltage Recall from Chapter 10 that

a capacitor can be thought of as a source of magnetizing

vars Therefore, one technique for maintaining voltage levels

on a utility system is to place capacitors at strategic

loca-tions in the distribution network The idea behind this

tech-nique is to use the capacitors to supply magnetizing vars

close to the loads requiring them, as opposed to sending them over the lines from the generator We shall illustrate this concept after we have introduced the analysis of bal-anced three-phase circuits

399

400 Balanced Three-Phase Circuits

Three-phase

Three-phase

voltage •

source

line

/ \

\

Three-phase load

Figure 11.1 • A basic three-phase circuit

11.1 Balanced Three-Phase Voltages

A set of balanced three-phase voltages consists of three sinusoidal volt-ages that have identical amplitudes and frequencies but are out of phase with each other by exactly 120° Standard practice is to refer to the three phases as a, b, and c, and to use the a-phase as the reference phase The three voltages are referred to as the a-phase voltage, the b-phase voltage, and the c-phase voltage

Only two possible phase relationships can exist between the a-phase voltage and the b- and c-phase voltages One possibility is for the b-phase voltage to lag the a-phase voltage by 120°, in which case the c-phase volt-age must lead the a-phase voltvolt-age by 120° This phase relationship is known as the abc (or positive) phase sequence The only other possibility

is for the b-phase voltage to lead the a-phase voltage by 120°, in which case the c-phase voltage must lag the a-phase voltage by 120° This phase relationship is known as the acb (or negative) phase sequence In phasor notation, the two possible sets of balanced phase voltages are

Vh = VI M/ - 1 2 0 '

and

Vh = 1/,,,/ + 120°,

Vc = ^ , , , / - 1 2 0 ° (11.2)

Equations 11.1 are for the abc, or positive, sequence Equations 11.2 are for the acb, or negative, sequence Figure 11.2 shows the phasor dia-grams of the voltage sets in Eqs 11.1 and 11.2 The phase sequence is the clockwise order of the subscripts around the diagram from Va The fact that a three-phase circuit can have one of two phase sequences must be taken into account whenever two such circuits operate in par-allel The circuits can operate in parallel only if they have the same phase sequence

Another important characteristic of a set of balanced three-phase voltages is that the sum of the voltages is zero Thus, from either Eqs 11.1

or Eqs 11.2,

Figure 11.2 A Phasor diagrams of a balanced set of

three-phase voltages, (a) The abc (positive) sequence,

(b) The acb (negative) sequence

Vfl + V„ + Vr = 0 (11.3)

Because the sum of the phasor voltages is zero, the sum of the instanta-neous voltages also is zero; that is,

v & + v h + v c = 0 (11.4)

Now that we know the nature of a balanced set of three-phase volt-ages, we can state the first of the analytical shortcuts alluded to in the introduction to this chapter: If we know the phase sequence and

one voltage in the set, we know the entire set Thus for a balanced

three-phase system, we can focus on determining the voltage (or current) in one

phase, because once we know one phase quantity, we know the others

NOTE: Assess your understanding of three-phase voltages by trying

Chapter Problems 11.2 and 11.3

11.2 Three-Phase Voltage Sources 4 0 1

11.2 Three-Phase Voltage Sources

A three-phase voltage source is a generator with three separate

wind-ings distributed around the periphery of the stator Each winding

com-prises one phase of the generator The rotor of the generator is an

electromagnet driven at synchronous speed by a prime mover, such as a

steam or gas turbine Rotation of the electromagnet induces a sinusoidal

voltage in each winding The phase windings are designed so that the

sinusoidal voltages induced in them are equal in amplitude and out of

phase with each other by 120° The phase windings are stationary with

respect to the rotating electromagnet, so the frequency of the voltage

induced in each winding is the same Figure 11.3 shows a sketch of a

two-pole three-phase source

There are two ways of interconnecting the separate phase windings to

form a three-phase source: in either a wye (Y) or a delta (A)

configura-tion Figure 11.4 shows both, with ideal voltage sources used to model the

phase windings of the three-phase generator The common terminal in the

Y-connected source, labeled n in Fig 11.4(a), is called the neutral terminal

of the source The neutral terminal may or may not be available for

exter-nal connections

Sometimes, the impedance of each phase winding is so small

(com-pared with other impedances in the circuit) that we need not account for it

in modeling the generator; the model consists solely of ideal voltage

sources, as in Fig 11.4 However, if the impedance of each phase winding is

not negligible, we place the winding impedance in series with an ideal

sinusoidal voltage source All windings on the machine are of the same

construction, so we assume the winding impedances to be identical The

winding impedance of a three-phase generator is inductive Figure 11.5

shows a model of such a machine, in which is the winding resistance, and

X w is the inductive reactance of the winding

Because three-phase sources and loads can be either Y-connected

or A-connected, the basic circuit in Fig 11.1 represents four different

configurations:

Source

Y

Y

A

A

Load

Y

A

Y

A

We begin by analyzing the Y-Y circuit The remaining three arrangements

can be reduced to a Y-Y equivalent circuit, so analysis of the Y-Y circuit is

the key to solving all balanced three-phase arrangements We then

illus-trate the reduction of the Y-A arrangement and leave the analysis of the

A-Y and A-A arrangements to you in the Problems

Axis of a-phase winding

Axis of c-phasc windine

\ Axis of b-phase winding

Stator

Figure 11.3 A A sketch of a three-phase voltage source

Figure 11.4 A The two basic connections of an ideal three-phase source, (a) A Y-connected source, (b) A A-connected source

402 Balanced Three-Phase Circuits

R,

R,r

JXu

Figure 11.5 • A model of a three-phase source with winding impedance: (a) a Y-connected source; and

(b) a A-connected source

11.3 Analysis of the Wye-Wye Circuit

Figure 11.6 illustrates a general Y-Y circuit, in which we included a fourth conductor that connects the source neutral to the load neutral A fourth conductor is possible only in the Y-Y arrangement (More about this later.) For convenience, we transformed the Y connections into "tipped-over tees." In Fig 11.6, Zg a, Zg b, and Zg c represent the internal impedance associated with each phase winding of the voltage generator; Zl a, Zl b, and

Zl c represent the impedance of the lines connecting a phase of the source

to a phase of the load; Z0 is the impedance of the neutral conductor con-necting the source neutral to the load neutral; and Z A , ZB, and ZQ

repre-sent the impedance of each phase of the load

We can describe this circuit with a single node-voltage equation Using the source neutral as the reference node and letting VN denote the node voltage between the nodes N and n, we find that the node-voltage equation is

N- Va

Z A + Zi a + -¾

b'n

Zn + Z I I ,lb + Z +

'gb

VN ~ Ve'n

Zc + Zl c + ZfiC = 0

(11.5)

v • ' a n

V • * c n

v

1

t

C

T

7

^6-) V b - n

Zgb

a

I

n

b

c

Zia

z„

Zib

Zlc

A

laA

N

it; c

Z B

Z A

Z c

Figure 11.6 • A three-phase Y-Y system

11.3 Analysis of the Wye-Wye Circuit 403

This is the general equation for any circuit of the Y-Y configuration

depicted in Fig 11.6 But we can simplify Eq 11.5 significantly if we now

consider the formal definition of a balanced three-phase circuit Such a

circuit satisfies the following criteria:

1 The voltage sources form a set of balanced three-phase voltages

In Fig 11,6, this means that Va-n, Vb<n, and Vc<n are a set of

bal-anced three-phase voltages

2 The impedance of each phase of the voltage source is the same In

Fig 11.6, this means that Zg a = Zg b = Zgc

3 The impedance of each line (or phase) conductor is the same In

Fig 11.6, this means that Z ia = Z ^ = Z\ c

4 The impedance of each phase of the load is the same In Fig 11.6,

this means that Z A = Z B = ZQ

There is no restriction on the impedance of a neutral conductor; its

value has no effect on whether the system is balanced

If the circuit in Fig 11.6 is balanced, we may rewrite Eq 11.5 as

< Conditions for a balanced three-phase

circuit

1 ^ +

A)

(11.6)

where

Z& - ZA + Zl a + Zga - Z B + Z ]b + Zgb — Zc + Zl c + Z

gc-The right-hand side of Eq 11.6 is zero, because by hypothesis the

numera-tor is a set of balanced three-phase voltages and Z^ is not zero The only

value of VN that satisfies Eq 11.6 is zero Therefore, for a balanced

three-phase circuit,

Equation 11.7 is extremely important If VN is zero, there is no

differ-ence in potential between the source neutral, n, and the load neutral, N;

consequently, the current in the neutral conductor is zero Hence we may

either remove the neutral conductor from a balanced Y-Y configuration

(It, = 0) or replace it with a perfect short circuit between the nodes n and

N (VN = 0) Both equivalents are convenient to use when modeling

bal-anced three-phase circuits

We now turn to the effect that balanced conditions have on the three

line currents With reference to Fig 11.6, when the system is balanced, the

three line currents are

f

laA

h\i

I

V a ' n "

ZA + Zl a

v„„

-ZB + Z\b

vc-n

-vN

+ z g a

VN

+ z g b

VN

»cC

Z<i>

\w b'n

V c < n

%C + Z\c + Zee Z„

(11.8)

(11.9)

(11.10)

We see that the three line currents form a balanced set of three-phase

cur-rents; that is, the current in each line is equal in amplitude and frequency

and is 120° out of phase with the other two line currents Thus, if we

calcu-late the current Ia A and we know the phase sequence, we have a shortcut

Balanced Three-Phase Circuits

a'

T

X

: )

I

z

•^ga

a Z

ia

c

A

Z A

—o

v 0 „i

Figure 11.7 A A single-phase equivalent circuit

N

for finding IbB and IcC This procedure parallels the shortcut used to find the b- and c-phase source voltages from the a-phase source voltage

We can use Eq 11.8 to construct an equivalent circuit for the a-phase

of the balanced Y-Y circuit From this equation, the current in the a-phase conductor line is simply the voltage generated in the a-phase winding of the generator divided by the total impedance in the a-phase of the circuit Thus Eq 11.8 describes the simple circuit shown in Fig 11.7, in which the neutral conductor has been replaced by a perfect short circuit The circuit

in Fig 11.7 is referred to as the single-phase equivalent circuit of a

bal-anced three-phase circuit Because of the established relationships between phases, once we solve this circuit, we can easily write down the voltages and currents in the other two phases Thus, drawing a single-phase equivalent circuit is an important first step in analyzing a three-phase circuit

A word of caution here The current in the neutral conductor in Fig 11.7 is Ia A, which is not the same as the current in the neutral conduc-tor of the balanced three-phase circuit, which is

I, L Ala A + IhR + I, bB IcC- (11.11)

+ f

-VAB

' - - ? +

T

+

V B C

- +

V AN

VCN

Z B

ZA

-— i N

Zc

Figure 11.8 A Line-to-line and line-to-neutral voltages

Thus the circuit shown in Fig 11.7 gives the correct value of the line cur-rent but only the a-phase component of the neutral curcur-rent Whenever this single-phase equivalent circuit is applicable, the line currents form

a balanced three-phase set, and the right-hand side of Eq 11.11 sums

to zero

Once we know the line current in Fig 11.7, calculating any voltages of interest is relatively simple Of particular interest is the relationship between the line-to-line voltages and the line-to-neutral voltages We establish this relationship at the load terminals, but our observations also apply at the source terminals The line-to-line voltages at the load termi-nals can be seen in Fig 11.8 They are VA B, VBC, and VCA> where the dou-ble subscript notation indicates a voltage drop from the first-named node

to the second (Because we are interested in the balanced state, we have omitted the neutral conductor from Fig 11.8.)

The line-to-neutral voltages are VAN, VBN, and VCN We can now describe the line-to-line voltages in terms of the line-to-neutral voltages, using Kirchhoff s voltage law:

BC V R N VC N , (11.13)

To show the relationship between the line-to-line voltages and the line-to-neutral voltages, we assume a positive, or abc, sequence Using the line-to-neutral voltage of the a-phase as the reference,

VAN = V * / 0 1 ,

BN V A / - 1 2 0 - ,

VCN = v , / + i 2 o ;

(11.15)

(11.16)

(11.17)

11.3 Analysis of the Wye-Wye Circuit 405

where V^ represents the magnitude of the line-to-neutral voltage

Substituting Eqs 11.15-11.17 into Eqs 11.12-11.14, respectively, yields

AB

BC

Vs / 0 ° - ^ / - 1 2 0 ° = V3V, h / 3 0 ° , (11.18)

V* / - 1 2 0 ° - V* /120° = s/Wj, / - 9 0 ° , (11.19)

YC A = J^/12GT - ^ / 0 £ = V3>< 6/150° (11.20)

Equations 11.18-11.20 reveal that

1 The magnitude of the line-to-line voltage is V 3 times the

magni-tude of the line-to-neutral voltage

2 The line-to-line voltages form a balanced three-phase set of voltages

3 The set of line-to-line voltages leads the set of line-to-neutral

volt-ages by 30°

We leave to you the demonstration that for a negative sequence, the only

change is that the set of line-to-line voltages lags the set of line-to-neutral

voltages by 30° The phasor diagrams shown in Fig 11.9 summarize these

observations Here, again, is a shortcut in the analysis of a balanced

sys-tem: If you know the line-to-neutral voltage at some point in the circuit,

you can easily determine the line-to-line voltage at the same point and

vice versa

We now pause to elaborate on terminology Line voltage refers to the

voltage across any pair of lines; phase voltage refers to the voltage across

a single phase Line current refers to the current in a single line; phase

current refers to current in a single phase Observe that in a A

connec-tion, line voltage and phase voltage are identical, and in a Y connecconnec-tion,

line current and phase current are identical

Because three-phase systems are designed to handle large blocks of

electric power, all voltage and current specifications are given as rms

val-ues When voltage ratings are given, they refer specifically to the rating of

the line voltage Thus when a three-phase transmission line is rated at

345 kV, the nominal value of the rms line-to-line voltage is 345,000 V In

this chapter we express all voltages and currents as rms values

Finally, the Greek letter phi (<f>) is widely used in the literature to

denote a per-phase quantity Thus V^„ I,/(, Z^, P^ and Q^ are interpreted

as voltage/phase, current/phase, impedance/phase, power/phase, and

reactive power/phase, respectively

Example 11.1 shows how to use the observations made so far to solve

a balanced three-phase Y-Y circuit

Figure 11.9 A Phasor diagrams showing the relation-ship between line-to-tine and line-to-neutral voltages in

a balanced system, (a) The abc sequence, (b) The acb sequence

Example 11.1 Analyzing a Wye-Wye Circuit

A balanced three-phase Y-connected generator

with positive sequence has an impedance of

0.2 + y'0.5 il/4 and an internal voltage of 120 V/<f>

The generator feeds a balanced three-phase

Y-connected load having an impedance of

39 + /28 fl/4> The impedance of the line

connect-ing the generator to the load is 0.8 + /1,5 (l/4> The

a-phase internal voltage of the generator is

speci-fied as the reference phasor

a) Construct the a-phase equivalent circuit of the system

b) Calculate the three line currents Ia A, Ib B, and IcC c) Calculate the three phase voltages at the load

VAN, VBN< and V CN

d) Calculate the line voltages VAB, V1}C> and \ C A at

the terminals of the load