1Chapter 7

Alternators & Three-Phase Circuits 7.1 INTRODUCTION

Large amounts of electric power invariably are generated, transmitted and used by polyphase devices.Three-phase systems, which have three source voltages of equal magnitude but out-of phase with each otherby 120(, are the most common. Three-phase transmission systems, consisting of one or two groups of threelarge conductors suspended from large steel towers, are a frequent sight on a country drive or a drive near theoutskirts of any large city. The advantage of polyphase systems are numerous. Polyphase motors andgenerators make more efficient use of materials hence are not as expensive as single phase equipment of thesame rating. Most motors having a rating above one horsepower are three-phase. Not only is the purchase costof a three-phase motor less than a single-phase motor, but its operating cost is less, it is easier to start, and ithas less vibration. The latter is due to the fact that the power (and torque) of a single-phase motor is pulsatingwith time while the power (and torque) of polyphase motors is steady. In situations where it is desirable tochange the number of phases from three to six or, twelve etc., this is easily and efficiently done by usingtransformer banks. An example of this is in the conversion of ac to dc electronically. The ripple componentof the rectified voltage is much less with a 6-phase or 12-phase voltage than with three-phase or single-phasevoltages, thus reducing or eliminating filtering requirements.

7.2 ALTERNATORS

Types of Alternators

The elementary AC generator has already been discussed in Chapter 5. The reader should go backand review pages 5-3 through 5-6. The term alternator refers to the alternating nature of the voltage andcurrent. There are two types of alternators: the revolving-armature type and the revolving-field type. We havediscussed the revolving-armature type alternator in Chapter 5. Its characteristic feature is that the armaturerotates through a stationary magnetic field, and the generated AC is brought to the load by means of slip ringsand brushes. The revolving-armature alternator is found only in alternators of small power rating and is notgenerally used. This is because a rotating armature requires slip rings and brushes to conduct the current fromthe armature to the load. The result for higher voltage alternators is arc-overs and short circuits and henceinefficiency and damage at the slip ring/brush interface at high generated voltages. Figure 7-1 shows thephysical differences between the two types of alternators.

The revolving-field type alternator has a stationary armature and a rotating magnetic field. Despitethis major difference, the physics and the operation of the alternator is essentially the same as discussed forthe elementary generator (a conductor in a magnetic field with relative motion between them). The advantageof having a stationary armature is that the generated voltage can be connected directly to the load withouthaving to pass across the slip rings and brushes. Therefore, high-voltage alternators are usually of the rotatingfield type. The voltage applied to generate the rotating field is a small DC voltage (called a field excitation

2Figure 7-1: Types of Alternators

voltage) and, therefore, the problem of arc-over at the slip rings is not encountered.The maximum current that can be supplied by an alternator depends upon the maximum heating loss

that can be sustained in the armature. This heating loss (which is an I2R power loss) acts to heat the conductorsand, if excessive, to destroy the wire insulation. Therefore, alternators are rated in terms of this maximumcurrent and in terms of the voltage output. The rating is expressed in a combined unit called a volt-ampere(VA), or in more practical units, a kilovolt-ampere (KVA).

Alternato rConstruction

Alternators having high kilovolt-ampere ratings are of the turbine-driven, high-speed type. The primemover for this type of alternator is a high-speed steam turbine which is driven by steam under high pressure.Due to the high speed of rotation, the rotor field winding of the turbine-driven alternator is cylindrical andsmall in diameter with windings firmly imbedded in slots in the rotor face as shown in Figure 7-2. Thewindings are arranged to form two or four distinct magnetic poles. Only with this type of construction can therotor withstand the terrific centrifugal force developed at high speeds without flying apart.

Slower speed alternators, which are driven by engines, water power, geared turbines or electricmotors, have a salient-pole rotor. In this type of rotor a number of separately wound pole pieces are bolted tothe frame of the rotor. The field windings are connected in series, with the ends of the windings connect toslip rings mounted on the rotor shaft. The windings on the rotor are supplied DC current for magnetic fieldexcitation by a DC generator called an "exciter.

3Figure 7-2: Types of Alternator Rotors

The armature windings are wound on the stationary part of the alternator called a stator. The ACvoltage generated in the armature is applied directly to the load. The stators of all alternators are essentiallythe same. The stator consists of a laminated iron core with the armature windings embedded in this core. Thestator is laminated and the laminations insulated from one another by a non-conducting coating to preventinduced eddy currents which would needlessly heat the stator and lead to the waste of energy. The core issecured to the stator frame (the outer case of the machine).

Single-Phase Alternator

A single-phase alternator has all the armature conductors connected in series; essentially one windingacross which an output voltage is generated. The schematic diagram of Figure 7-3 illustrates a two-pole,single-phase alternator. The stator is two pole because the winding is wound in two distinct pole groups, bothpoles being wound in the same direction around the stator frame. Observe that the rotor also consists of twopole groups, adjacent poles being of opposite polarity. As the rotor (field) turns, its poles will induce ACvoltages in the stator (armature) windings. Since one rotor pole is in the same position relative to a stator pole

4Figure 7-3: Single Phase Alternator

as any other rotor pole, both the stator poles are cut by equal amounts of magnetic lines of force at any time.As a result, the voltages induced in the two poles of the stator winding have the same amplitude or value at anygiven instant. The two poles of the stator winding are connected to each other so that the AC voltages are inphase, so they add. Assume that rotor pole 1, a south pole, induces a voltage with the polarity as shown instator pole 1. Since rotor pole 2 is a north pole, it will induce the opposite voltage polarity in stator pole 2, inrelation to the polarity of the voltage induced in stator pole 1. In order that the voltages in the two poles beadditive, poles 1 and 2 are connected as shown. Observe that the two stator poles are connected in series sothat the voltages induced in each pole add to give a total voltage that is twice the voltage in any one pole.

The rotor shown in Figure 7-3 has the form of a permanent bar magnet which rotates within the stator.A more practical rotor, which uses a DC electromagnet, is shown in cross section in Figure 7-4 (a) below. Thefield excitation winding is shown wound in slots in this cylindrical rotor. The field winding produces the DCmagnetic field when supplied with DC current from an external source connected through brushes and sliprings.

The rotor can have more than two poles but they must come in pairs of north and south. An exampleof a four pole rotor in cross section is shown in Figure 7-4 (b) below. This rotor is a salient pole rotor meaningthat the pole faces protrude out from the rotor shaft. Rotors with the field winding imbedded in slots in thesurface, such as the one shown in Figure 7-4 (a) are capable of turning at higher speeds because they are lesssusceptible to centrifugal forces. Salient pole rotors are designed to spin at lower rpm. If a salient pole rotoris spun too fast it will fly apart.

5 (a) Cross Section of a 2 Pole Rotor (b) Cross Section of a Salient 4 Pole RotorFigure 7-4

A picture of an actual alternator rotor is shown in Figure 7-5 below. Note the slip rings for connectionof the rotor winding to an external DC source through carbon brushes riding on the surfaces of the slip rings.Also note the large number of poles for a rotor meant to rotate at a relatively slow speed.

Figure 7-5: Multiple Pole Rotor

6Three-Phase Alternator

The three-phase alternator has three single-phase windings spaced so that the voltage induced in anyone is phase-displaced by 120 degrees from the other two. A schematic diagram of a three-phase statorshowing all the coils becomes complex, and it is therefore difficult to see what is actually happening. Asimplified schematic shows all the windings of a single-phase lumped together as one winding, as illustratedin Figure 7-6. The rotor is omitted for simplicity. The voltage waveforms generated across each phase aredrawn on a graph phase-displaced 120 degrees from each other. The three-phase alternator is essentially threesingle-phase alternators whose generated voltages are out of phase by 120 degrees. The three phases areindependent of each other.

.

Figure 7-6: Three Phase Alternator

An alternative depiction of a three winding stator along with the rotor is shown in Figure 7-7. Thealternator of Figure 7-7 is shown schematically and in cross section. The wires in each of the three windingsrun in and out of the page along the axis of the cylindrical rotor.

Figure 7-7: Three Phase Alternator Winding in Cross Section

Rather than have six leads come out of the three-phase alternator, one lead from each phase can beconnected together to form a "wye connection. The point of connection is called the neutral, and the voltagefrom this point to any one of the line leads will be the phase voltage. The line voltage across any two line leadsis the vector sum of the individual phase voltages. The line voltage is 1.73, ( ), times the phase voltage.3Since the windings form only one path for current flow between phases, the line and phase currents are equal.

7A three-phase stator can also be connected so that the phases form a delta connection. In the deltaconnection the line voltages are equal to the phase voltages, but the line currents will be equal to the vector sumof the phase currents. Since the phases are 120 degrees out of phase, the line current will be 1.73, ( ), times3the phase current. Both "wye" and the "delta" connections are used in alternators. On board U. S. Navy ships,most alternators are delta connected for ease of maintenance and better battle damage resistance. Figure 7-8shows schematically both types of three phase stator connections.

Figure 7-8: Three Phase Stator Connections

Frequency and Voltage Regulation

The frequency of the AC generated by an alternator depends upon the number of poles and the speedof the rotor (Figure 7-9). When a rotor has rotated through an angle so that two adjacent rotor poles (a northand a south) have passed one winding, the voltage induced in that one winding will have varied through acomplete cycle of 360 electrical degrees. The more poles there are, the lower the speed of rotation must be fora given frequency. A two pole machine must rotate at twice the speed of a four-pole machine to generate thesame frequency. The magnitude of the voltage generated by an alternator can be varied by adjusting thecurrent on the rotor which changes the strength of the magnetic field. Remember that the voltage induced ona wire moving through a magnetic field is proportional to the strength of the field and the velocity of the wire.This means that the magnitude of the voltage could also be increased by increasing the speed of the rotor butthis would have the undesirable effect of also changing the frequency of the AC produced. Generally, the linefrequency is kept constant at 60 Hz. The line frequency depends upon the number of pairs of poles and thespeed of rotation. For example, a two pole alternator produces one electrical cycle for each completemechanical rotation. A four pole alternator will produce two electrical cycles for each mechanical rotationbecause two north and two south poles move by each winding on the stator for one complete revolution of therotor. In equation form this relationship becomes: f = (nRotor)(p/2)/60 = (nRotorp)/120 where nRotor is thespeed of the rotor in revolutions per minute, p is the number of poles and f is the electrical line frequencyproduced by the alternator. The speed of the rotor must be divided by 60 to change from revolutions perminute to revolutions per second.

In an alternator the output voltage varies with the load. In addition to the IR drop, there is another voltage drop in the windings called the IXL drop. The IXL drop is due to the inductive reactance of thearmature windings. Both the IR drop and the IXL drop decrease the output voltage as the load increases. Thechange in voltage from no-load to full-load is called the voltage regulation of an alternator. A constantvoltage output from an alternator is maintained by varying the field strength as required by changes in load.

8Figure 7-9: Speed/Frequency Relationship for an Alternator

7.3 GENERATION OF THREE-PHASE VOLTAGES

Faradays law governs the calculation of generated voltages. In most rotating machinery the principleused is that a voltage is induced in a coil where there is relative motion of a magnetic field with respect to acoil of wire. In d-c machines the magnetic field is established by the stationary part of the machine while thevoltage is induced and energy is converted in the rotating member. This voltage is alternating and a switchconsisting of the commutator segments and brushes is used to obtain a d-c output voltage.

As with d-c machines, the a-c generator has a field structure to produce the required magnetic fieldand an armature winding in which the voltage is induced and energy is converted.

The a-c generator could use the configuration of the d-c machine just described, only connect the rotorterminals to slip rings where the alternating voltage could be picked off by brushes and made available externalto the machine. However, in a-c generators having ratings in excess of a few kVA the functions of the rotorand stator are reversed. The rotor establishes the magnetic field and the armature winding is placed in slotson the stationary part of the machine.

Figure 7.10(a) shows a cross-section of an elementary a-c generator. The rotor is mechanicallyconnected to a prime mover such as a turbine which turns the rotor at a constant speed. The winding on therotor is connected to a d-c source via slip rings and brushes. The mmf developed by this winding establishesa magnetic field, two typical lines of which are shown. In this elementary machine the armature windingconsists of a single turn placed in slots in the stator located diametrically opposite each other. The letters a

9and a are used to identify the terminals and sides of the coil. The arch-shaped connection between coilsides at the rear of the machine gives mechanical clearance so the rotor can turn freely, or be removed forrepairs.

With the rotor in the position shown in Figure 7.10 there is no flux linking coil a-a. When the primemover has turned the rotor 90( CCW, the maximum possible flux links the coil. After another 90( CCWrotation, the flux linking the coil has returned to zero. Rotation of the rotor 270( CCW from the originalposition again gives maximum flux linkage, but opposite in sense to the 90( position. When the rotor has beenturned through 360(, the voltage induced in the coil will have gone through one cycle. Assume for the momentthat the rotor is held stationary and that as one moves around the inner circumference of the stator the flux isfound to vary sinusoidally. Then when the rotor turns, the flux linking the coil is

(7-1) ( )t wtm m= =sin sin2 ft webers

Since the induced voltage goes through one cycle for each revolution, the frequency of the generatedvoltage of this two-pole generator is the same as the number of revolutions per second or f = nrps= nrpm/60.If the coil nas N series turns rather than one, then from the laws of Faraday and Lenz the voltage induced incoil A is

(7-2)v N ddt

fN N V t V taa m m m m' = = = = 2 cos2 ft = cos cos volts,

where and the rms value of the induced voltage is = =2 2f rad / s and V fNm m(7-3)V V fNm m= =/ .2 4 44

Figure 7.10. Elementary a-c generator. (a) Cross-sectional view.(b) Typical stator coil with arch-shaped back end connection.

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Figure 7.11. Cross-section of the stator of an elementary three-phase generator.

The voltage vaa= vaa = -Vm cos 7t. The generator shown in cross-section in Figure 7.11 is an elementary three-phase generator with the stator having two additional pairs of slots. The coils placed in slots b-b and slots c-care shaped like the coil in Figure 7.10(b). The waveform of the voltage induced in coils B and C will be thesame as that induced in coil A. However, due to the fact that side b is displaced from side a by 120( in thedirection of the rotation, the instantaneous voltage vbb will lag vaa by 120( or vbb = Vmcos(7t-120(). By thesame reasoning because side C is displaced 240( vcc will lag vcc by 240( because side C is displaced 240(from side a in the direction of rotation. Thus vcc = Vm Vmcos(7t-240(). Each of these windings is anindependent voltage source and is referred to as one phase of a three-phase generator.

The elementary generator makes inefficient use of materials and a practical three-phase generator willhave many more series turns per phase distributed in additional slots. However the relative location of thephases with respect to each other will remain as described above and each phase will have the same numberof turns so that the maximum voltage of each phase will be the same. The A-phase winding, for example,might occupy slots in that part of the circumference from slot a to slot c and from slot a to c. Conductors ofthe B-phase would lie in slots from b to a and b to c. Conductors of the B-phase would lie in slots from bto a and b to a, while the C-phase would be located in slots in the remaining two sixths of the circumference.Symbolically the three phases could be represented by three independent sources (Figure 7.12(a) or three coils(Figure 7.12(b)).

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Figure 7.12 Symbolic representation of the three sources of a three-phase generator.

Tabulated below are the expression for the instantaneous values of the voltage of each phase andthe corresponding phasor. . A plot of the instantaneous voltage variations appears in FigureV Vm= / 27.13(a) while the corresponding phasors are shown in Figure 7.13(b).

(7-4)v V taa m' = cos V Vaa ' /= 0(7-5)v V tbb m' )= cos( 120 V Vbb' = 120

(7-6a)v V tcc m' )= cos( 240 V Vcc ' = 240

(7-6b)= Vm cos ( t +120 ) = + V 120We define phase sequence (also phase rotation) as the sequence or order in which the phase voltages

reach their positive maximum values. As discussed earlier for any given event in the A phase (for example,the positive maximum) the same event occurs 120( later in the B phase and 240(later in the C phase. Thusthe phase sequence of this generator is ABC. The phase sequence can be changed to ACB by reversing thedirection of rotation of the poles. If phase sequence ABC is called the positive phase sequence, then ACB isthe negative sequence.

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Figure 7.13. (a) Instantaneous variations of the voltages of a three-phase generator.(b) Corresponding phasors.

7.4 THREE-PHASE CONNECTIONS; VOLTAGE AND CURRENT RELATIONS

Since each of the phases of the three-phase generator are independent, each phase could be connectedto individual single-phase loads. To do this, however, would invalidate all the advantages cited earlier for thethree-phase system. In connecting the three independent sources to make a bona-fide three-phase generatoran important objective is to have a system that is balanced and symmetrical, i.e., corresponding voltages shouldall have the same magnitude and they should be 120( out-of-phase with each other. The two ways of makingconnections with the properties mentioned are the delta connection and the wye (or star). (The individualphases of a three-phase load would also be connected in one of these two configurations.

Y-Connection

The parts of Figure 7.14 show several of the ways that a wye-connection of the sources of Figure 7.12 couldbe drawn. The important thing about the Y-connection is that three similar terminals are connected together.In this example the primed ends of the three coils are connected together at a single point called the neutral(N). Four leads (a,b,c,N) would be brought out of the stator of the wye-connected generator.

One consequence of connecting the three phases in wye is that two magnitudes of voltage areavailable. One of these is the phase voltage available between any of the terminals a, b or c and the neutral,while the other voltage would be the line voltage available between any pair of lines such as a to b, b to c, orc to a. (In the future when you read the phrase line voltage you may find it helpful to translate this mentallyas line-to-line voltage. The latter phrase is more accurate but common parlance uses the term line voltage.)

In drawing the wye-connection as in Figure 7.14 there need be no relation between the magnetic axesof a coil in the actual generator and the orientation of the symbol that represents it in Figure 7.14. Nor is thereany relation between the orientation of the terminals a, b, and c in Figure 7.14 and the phase sequence.

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Figure 7.14. Methods of schematically showing the connection of a wye-connected three phase generator.

Figure 7.15 A wye-connected generator with meters connected to measure currents and voltages.

Our next objective is to determine the relation between line current and phase current in the wyeconnection and the relation between line voltage and phase voltage. Subscript p will be used to designatephase quantities and subscript L will denote line quantities. Figure 7.14 (c) has been reproduced in Figure7.15. A typical phase current and phase voltage have been labeled as well as a typical line current and linevoltage. Meters are shown connected so as to read these four quantities.

If one starts at the neutral N and proceeds through the source in phase a, through the ammeter in serieswith that phase, out of a terminal a, through the ammeter in series with the line to terminal A, it is obvious thatthis is a series circuit. Since the current everywhere in a series circuit is the same, we conclude that in the wye-connection

(7-7)I Ip L=

The two ammeters will have identical readings.

14

To determine the other relation we write KVL around the closed path AaNbBA. (The ammeters areideal and there is no voltage drop across them.) Thus . NowV V V V VaN bN AB AB aN bN = = 0 or V

and and these voltage are described in Equation (7-4) and (7-5). Using thoseV VaN aa= " V VbN bb= 'relations

V V V V j V j Vj V V

AB = = + = + =

0 120 0 0 5 0 86615 0 866 1732 30

( ) ( . . )( . . ) . volts.

The magnitude of the line voltage is 1.732V or larger than the magnitude of the phase voltage. Thus in3the wye connection

(7-8)V VL p= 3 '

and the voltmeter reading line reads times the reading of the meter connected across the phase terminals.3In a balanced-symmetrical system (the only kind we will assume) if phasor line current ,I IA =

then phasor line current will have the same magnitude. The neutral current N can be found by writing KCLI Bat node N. Thus . Phasors are shown on the phasor diagram ofI I I IA B C N+ + + = 0 I I IA B C, and Figure 7.16(a). These three phasors make angles of 120 with each other. When they are added, as shown inFigure 7.16(b) the phasors form the sides of an equilateral triangle and their sum must be zero. Therefore

in a balanced wye-connection. In some installations the neutral wire is omitted. If the load is notI N = 0balanced, then the neutral is needed to provide a return path for the resultant current.

Figure 7.16 (a) Phasor diagram of the line currents of a balanced system;(b) phasor sum of these currents is zero.

Delta-Connection

Figure 7.17 shows several ways of schematically showing a delta connection. In this connection the threephases form a closed series path. Another way of describing the connection is that the individual phases areconnected between pairs of lines. The connection must be made so that the net voltage around the series loop

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is zero. Figure 7.17(a) is redrawn in Figure 7.18. The result of writing KVL around the closed path of thedelta is . Assume these voltages are given by Equation (7-4), (7-5) and (7-6)V V Vaa bb cc' ' '+ + = 0respectively, and the phasor diagram shown in Figure 7.13(b). Upon addition, these phasors will form anequilateral triangle and their sum will be zero. Thus the net voltage around the delta connected windings iszero.

Figure 7.18 shows an ammeter inserted in series with phase A to measure phase current and anammeter in series with line A to measure line current. The voltmeters are connected to measure the voltageof a typical phase and a typical line.

Figure 7.17 Methods of schematically showing the connection of a delta-connected three-phase generator.

Figure 7.18. A delta-connected generator with meters connected to measure currents and voltages.

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Our next objective is to determine the relation between phase and line quantities for the deltaconnection. From inspection of Figure 7.18 it is clear that voltmeters Vp and VK are connected between thesame pair of nodes, therefore they are in parallel and thus

Vp =VL (7-9)

To show the relation between line and phase current in the delta connection we write KCL at nodea-b to obtain or . Assume a balanced symmetrical system so that ifI I Ia a b b a' ' = 0 I I IA a a b b= ' '

is given by , then and . Using toI a a' I Ia a' = 0 I b b' = 120 I Ic c' = + 120 I Ia a b b' ' and solve for gives . This relation is to hold for any value of . For convenienceI A I I IA = 120we choose to evaluate it when =0. Thus

This result says that in theI A I I I j I j I I j= = + = + 0 120 0 0 5 866 15 8661 1 732 30( ) ( . . . .balanced delta the relation between the magnitude of line and phase current is

(7-10)I L p= 31

The discussion up to this point has been confined solely to three-phase generators. Much of thisarticle applies to three-phase loads as well. Three-phase loads may be connected in either wye or delta. Forthe load to be balanced the magnitude and power factor (phase angle) of each of the three phase impedancesmust be identical. Then relations (7-7) through (7-10) hold for the loads also.

EXAMPLE 7-1

A generator having phase voltage of 150 volts is connected in delta and feeds a three-phase wye-connected load consisting of impedance per phase. Compute all phase and line voltages andZ = 4 30currents.

Solution

Figure 7.19.a. VL = line voltage - generator phase voltage =VPg = 150 volts.b. Load phase voltage =V Vpl L= =/ .3 86 6 volts.c. Load line current = load phase current = V Z Apl / . / . .= =86 6 4 21 65

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d. IL= Generator line current = load line current = 21.65A.e. I I Apg Lg= / . .3 12 5

Summary

Generator (delta) Load (wye)Phase voltage 150 86.6 volts

Line Voltage 150 150 volts

Phase Current 12.5 21.65 amperes

Line Current 21.65 21.65 amperes

7.5 THREE-PHASE POWER

If the impedance of each phase of a balanced load is , and the rms value of phaseZ Zp p= /voltage and current are Vp and Ip, then the instantaneous power to phase A is given by

(7-11)P v i V t ta a a p p p= = 2 21cos cos

= +2V I t t tp pcos cos sinp p cos sin = +2 2V I wt t tp p pcos cos cos sin sin

= +

+

2 12

12

2 12

2V I t tp p p pcos cos sin sin

(7-12)= + +V I V I t V I tp p p p p p p p pcos cos cos sin sin 2 2

The average value of pa is Instantaneously, however pa (t) varies about this averageP V Ip p p p= cos .value. The instantaneous power to phases b and c would be given by

(7-13)P v i V t I tb b b p p p= = 2 120 2 120cos cos

(7-14)P v i V t V tc c c p p p= = + + +2 120 2 120cos cos

If (7-13 and (7-14 are expanded and added to (7-12) the sum is

Total instantaneous power watts. (7-15)= = =P V I Pp p p p3 3cos

Several items worth noting can be inferred from Equation (7-15). First, the total instantaneous power is aconstant it is independent of time. This is significantly different from the result for any one of the threephases. As far as motors are concerned this means that power (and hence torque) of the three-phase motor will

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be constant whereas the power (torque) of a single-phase motor will pulsate and vary. As a consequence thepolyphase motor will operate with much less vibration than the single phase motors. (This is somewhat likecomparing the power and torque variations of a turbine with a reciprocating engine.) Secondly, because theinstantaneous power is constant, it is also the average power and the total average power is three times theaverage power of one phase. If the load is wye-connected, then . Using theseV V I Ip L p p= =/ 3 and relations (7-15) can be put in the form

. (7-16)P V I wattstot L L p3 cos

It can be shown that this relation is also valid for a delta connected system. Similar relations hold for totalreactive and apparent power. Thus

, (7-17)P V I V IX L L p L L ptot = =3 3sin sin var s

(7-18)P VI I V IA P P L Ltot = =3 3 volt amperes

The terms power factor and reactive factor have meaning in polyphase systems only if the load is balanced.In that event the power factor (reactive factor) of one phase.

EXAMPLE 7-2

For the system described in EXAMPLE 7-1, calculate total power delivered to the load apparentpower and also the power factor.

SolutionP V IL L= 3 cos = 3 150 21 65 30x x x. cosP = 4871 wattsVA V IL L= =3 5625 volt - amperesPower Factor = cos 30( = .866 laggingAlso, Power Factor = Real Power/Apparent Power

= 4371/5625 = .866

EXAMPLE 7-3

A three-phase generator is rated 1500 kVA, 460V, 60 Hz

a. Determine VL, VP, IL, IP if the generator is wye-connected.

b. Repeat part (a) for a delta connection.

c. What are the power and reactive power output of this machine if it delivers rated kVA at a power factorof 0.6 lagging? Draw a power triangle.

Solution

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a. By convention, the voltage given in rating information is line voltage. Thus VL = 460 V.#V V Vp L= = =/ / . .#3 460 3 265 6

To fine the phase current, first find the apparent power per phase, i.e. Pap.

Pap = 1,500,000/3 = 500,000 volt-amperes

Since Pap = VpIp, the current per phase can be found as

Ip = Pap/Vp, = 500,000/265.6 = 1882 A. #.

In the Y, IL = Ip = 1882 A.#.

b. In the delta Vp = VL=460 V#

The current per phase can be found from Ip = Pap/Vp or

Ip = 500,000/460 - 1087 A.#

In the delta I I I x AL p L= = =3 3 1087 1882 or

c. Draw the power triangle first. The hypotenuse is the apparent power, while the base and altituderespectively are the power (real or average) and reactive power. Since power factor = cos,

= = cos . .1 0 6 53 2

From trigonometry andP P x kWA= = cos . # 1500 0 6 900

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P P x kVARX A= = sin . .# 1500 0 8 1200

EXAMPLE 7-4

A three-phase induction motor is rated 250 horsepower, 460 V, and a full-load has an efficiency of85% and a power factor of 0.8 lagging. When operating under rated conditions determine:

a. the line currentb. the apparent power

c. the apparent equivalent impedance per phase if the motor is wye-connected.

d. At the instant of starting the current to the motor is five times rated current. What apparent power issupplied the motor at the instant of starting?

Solution

a. Find the total power input then use (7-16)

P PEfficiency Hpx

WHp

x Win out= = =250 7461

0 85219400

.

From (7-16) I PV x x

ALL

= = =

32194003 460 0 8

344cos .

.#

b. Draw a power triangle. The power input has just been found. The power factor was given. Thus

P P VA KVAA in= = =/ . , . .#0 8 274 200 274 2(As an aside P P x KVARx A= = =sin . . . .36 8 274 2 0 6 164 5

c. Phase values must be used. Since the motor is wye-connected,Z V Ieq = / .

, and .V V Vp L= = =/ / .3 460 3 265 6 I I Ap L= = 344

Thus Z ohms phasep = =265 6 344 0 772. / . / .#

(The complex description of this impedance is 0.772/36.8( Ohms.)

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d. Since the motor takes 5 times rated current at starting, and apparent power varies directly with current (see(7-18) the apparent power at starting is

P xP x kVAA Astart rated= = =5 5 274 2 1371. #

EXAMPLE 7-5

Three identical impedances each 1.0 +j1.732 ohms are connected in Y to a 460 volts, three phase balancedpower system. Determine

a. the line current

b. the power factor of the load

c. draw the power triangle for the load

Solution

a. First write Zp in polar form.Z jp = + = 1 1732 2 60. .

Next find phase current.

I V Z x Ap p p/ .= =460

312

132 8

In the wye connection I I Ap p= = 132 8. #

b. Load power factor = power factor of any phase = cos 60(=0.5 lagging.

c. Sketch the shape of the power triangle, then determine magnitudes. The power triangle and the impedtriangles are similar, hence = 60(.

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The power per phase can be easily found from

P I R x Wp p p= = =2 2132 8 1 17 630( . ) ,

The total power = 3 3 17 630 52 900 52 9P x W kWp = = =, , . .#

From trigonometry P P kVAA = =/ cos . #60 1058

and P P kVARX A= =sin . #60 916

EXAMPLE 7-6

The loads of EXAMPLE 7-4 and 7-5 are connected in parallel to a 460 V three-phase system.

a. Determine the current supplied the combined load.

b. A synchronous motor rated 100 kVA is also connected in parallel with the other two loads. This motoroperates at a power factor of 0.5 leading. Draw a power triangle for the combined load and determine thetotal current supplied the combination.

Solution

a. Power and reactive power must be conserved. First determine the apparent power to the combined loadsby adding the power triangles of the two loads. The result is shown in dashed lines on the figure. Thecomplex apparent power is given by

P jA = + = +( . . ) ( . . )219 4 52 9 164 5 916= + = 272 3 256 373 7 43 2. . .j kVA

Using (7-18) the line current is

Ix

AL373 700

3 460469, .#=

b. Due to the leading power factor, the power triangle of the synchronous motor will be in the fourth quadrant.The complex apparent power of this load will be

P j kVAAsyn = = = 100 0 5 100 60 50 86 61cos . .Combining this apparent power with the complex apparent power obtained for the other two loads in part(a) gives

.P j j kVAA = + + = = = ( . ) ( . ) . . . .272 3 256 50 86 6 322 3 169 4 364 1 27 7

The accompanying figure shows the power triangles and how they are combined. The apparent power tothe three loads is 364.1 kVA. Note that even though a third load was added, the resultant apparent powerrequired is less than with the original two loads. This result occurred because the third load operated at alow, leading power factor. Using (7-18) to find the new line current gives

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I AL = =364 100

3 460457,( ) .#

7.6 THREE-PHASE CONNECTIONS OF TRANSFORMERS

Three-phase transformer connections are commonly encountered. A cluster of three transformers mounted ona utility pole is a frequent sight in a distribution system serving a business or light industrial area. There maybe a separate transformer for each phase in which case the group is sometimes called a bank of three-phasetransformers. However, the transformer may be a bona-fide three-phase device as shown in Figure 7.20. Thereare four variations in connecting the primary/secondary windings. These are summarized in Table 7.1.

Figure 7.20. Core and windings of a three-phase transformer. The individual primary and secondaries not shown.

Table 7.1 WAYS OF CONNECTING THREE-PHASE TRANSFORMERSPrimary Secondary Symbol

delta delta

wye wye YY

delta wyeY

wye delta Y

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The circuit analysis of a three-phase transformer connection is no different from the circuit analysis of anyother three-phase device. However, we do need to show the details of a typical three-phase transformerconnection such as the one in Figure 7.21. There the primaries are connected in delta, while the secondariesare connected to a 4-wire system in which the neutral is brought out. The same care used in connecting thewindings of a three-phase generator must be observed here. In the secondary connection, for example, threesimilar terminals (all the X2's) are connected together to form the neutral. The other secondary terminals areconnected to the respective lines A, B and C. For the delta-connected primary, the windings are connected inseries, so that upon exiting at the H2 terminal of one phase, one enters the H1 terminal of the next phase thenexits at terminal H2 of the second phase, etc. The three nodes which result are connected to the three lines.

Figure 7.21. Connection diagram of a delta-wye transformer bank.

The delta-delta connection has the advantage that if one phase becomes disabled, it can be removedand three-phase voltages and power are still available. The remaining two phases are said to be operating inopen-delta. It is for this reason that transformer banks using three identical transformers connected delta-deltaare commonly used in the naval service. To illustrate, assume that Figure 7.22 represents a normal delta-deltatransformer bank. If and the phase rotation is ABC, then V VAB p= 0 V VBC P= 120

Figure 7.22. Open-delta connection.

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and . A typical phase current IP and a typical line current IL are shown. TheirV VCA P= + 120magnitudes are related by . In the delta connection VL = Vp and the apparent power of the bankI IL p= 3

is . Now imagine that phase AB becomes disabled and is removed as suggested by the dashed line3V IL Lof Figure 7.13. The phases producing are intact as would be those voltages. What is V BC CA and V V ABafter the transformer is removed? This voltage is given by which can be rewritten asV V VAB AC CB= +

. Thus isV V V V V V VAB CA BC p p p p= = + = = 120 120 180 0 V ABunchanged and a three-phase load connected to lines A, B and C would be unaffected. It can be seen, however,that in the open delta the current in the two remaining phases is now the old line current. Unless the load isreduced, the remaining two phases will overheat and be damaged. The line current must be reduced to

of its old value to prevent damage. Thus the kVA rating of an open-delta connection is1 3 0 577/ .=57.7% the rating of the full delta-delta connection.

A popular type of starter for a-c motors uses an open-delta auto-transformer as shown in Figure 7.23.The device delivers a three-phase voltage which is a prime consideration. The turns ratio is on the order of 4:3or 2:1 which is suitable ratio for an autotransformer. Use of an autotransformer also permits economy ofmaterial. The use of the open delta dispenses with one phase which gives further economy. This type starterwould be used infrequently and then only for say 5 to 10 seconds. The long time thermal capacity of thewindings is thus not a problem.

Figure 7.23. A-C motor starter consisting of an open-delta autotransformer.

7.6 SUMMARY

Each load or generator in a three-phase system consists of three phases.

The three phases may be connected in wye delta.

In a balanced three-phase generator the voltages generated have the same magnitude but are 120( out ofphase with each other.

In a balanced three-phase load, the three phase impedances have the same magnitude and phase angle.

For balanced systems:

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Delta: ; V Vline phase= I Iline phase= 3

Wye: ; V VL phase= 3 I Iline phase=

Total Power = 3 3V I V Ip p p L L pcos cos =Total reactive power = 3 3V I V Ip p p L L psin sin =

Total apparent power = 3 3V I V Ip p L L= Power triangle methods can be extended to solving three-phase circuit problems.

A useful method of solving three-phase circuit problems in balanced systems is to determine per phasevalues of current, voltage, impedance, power, etc., then determine total values (of P, PX, or PA) or linevalues (of V and I).

The open-delta transformer bank can supply three-phase power but must be derated to 57.7% of a delta-delta bank.

7.7 QUESTIONSQ7.1 What is the primary advantage of the rotating field alternator as compared to the rotating armature

alternator?Q7.2 How does the construction of high and low speed rotors differ?Q7.3 What is a salient pole rotor?Q7.4 Is the source of the field current to a rotating field alternator DC or AC? Why?Q7.5 What hardware is involved in the conduction if current to and from the rotor of an alternator?Q7.6 What is the limiting factor in the maximum output power of a rotating field alternator? Why?Q7.7 Where is the armature located in a rotating field alternator?Q7.8 List some advantages that polyphase systems have over single phase systems.Q7.9 What is the speed of rotation for a 6-pole 50 Hz alternator? 8-pole 60 Hz?Q7.10 What is the line frequency for a 10-pole alternator spinning at 800 rpm? 4-pole at 900 rpm?Q7.11 What might be the advantages of a slower speed of alternator rotation?Q7.12 What is true of the value of the instantaneous power provided to a balanced three phase load and how

does this characteristic provide an advantage when running rotating machinery on three phase power?Q7.13 Describe an important advantage that a delta-delta connected transformer has in comparison with a

system in which at least one side is wye connected.

7.8 PROBLEMS

P7.1 Calculate the rpm for 60 Hz operation of a three-phase, 8-pole, 220V, 500 kW generator. For 50 Hz operation.

P7.2 What portion of a cycle will the volage of a four pole generator pass as the shaft rotates 60 degrees?

P7.3 A generator having phase voltage of 200 Volts is connected in delta and feeds a three-phase wye-connectedload consisting of a phase impedance equal to 3 +j4 Ohms per phase. Compute all phase and line voltages andcurrents.

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P7.4 For the system described in problem P7.1, calculate the total power delivered to the load, the total apparentpower and the power factor.

P7.5 A three-phase generator is rated 2000 kVA and 400 Volts at 60 Hz.a. Determine the line and phase voltages and the line and phase currents at the generator and at rated output

if the generator is wye-connected.

b. Repeat part (a) for a delta-connection.

c. What are the real and reactive powers output by this machine if it delivers rated kVA at a power factor of0.7 lagging? Draw a power triangle for this case.

P7.6 A three-phase induction motor is rated 300 horsepower and 400 volts and at full-load has an efficiency of 90%and a power factor of 0.7 lagging. When operating under rated conditions determine:

a. the line current,

b. the apparent power,

c. and the apparent equivalent impedance per phase if the motor is wye-connected.

d. If at the instant of starting the motor, the current is five times rated current, what apparent power issupplied the motor at the instant of starting?

P7.7 Three identical impedances each 1.732 +j1 Ohms are connected in wye to a 400 Volt, three-phase balancedpower system. Determine:

a. the line current,

b. and the power factor of the load.

c. Draw the power triangle for the load.

P7.8 The loads of problems P7.4 and P7.5 are connected in parallel to a 460 Volt, three-phase system.

a. Determine the current supplied to the combined load.

b. If a three phase synchronous motor rated at 80 kVA and operating at a power factor of 0.5 leading is nowadded in parallel with the other two loads, determine the total current supplied to the combined load anddraw a power triangle for this case.

P7.9 A wye-connected generator having 240 Volts per phase is delivering power to a delta-connected motor whichhas an impedance of 16 +j20 Ohms per phase.

a. Calculate line voltage and current magnitudes.

b. Calculate the power factor and active, reactive and apparent powers.

P7.10 A balanced wye-connected load composed of three equal impedance of Ohms is connected to a 12010 30 Volt three-phase source.

a. Calculate the line current magnitude.

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b. Calculate the power factor and active, reactive, and apparent powers.

c. What would the per phase impedance need to be for a delta connected load drawing the same current atthe same power factor when attached to the same source?

d. Calculate the per-phase capacitance needed in a balanced, wye-connected capacitive load to correct thepower factor to unity when connected in parallel with the above load. Repeat for a delta-connectedcapacitive load.

P7.11 A balanced delta load having a 7 Ohm resistance in series with a 14 Ohm inductive reactance in each leg isconnected to a three-phase, three-wire, wye-connected generator having a line voltage of 208 Volts. Calculatethe magnitudes of the

a. phase voltage of the generator,

b. phase voltage of the load,

c. phase current of the load,

d. and the line current.

e. Calculate the value of capacitance that must be used in each leg of a balanced delta-connected capacitiveload which will correct the power factor to a 0.9 lagging when connected in parallel with the above load.Repeat for a wye-connected, balanced capacitive load.

P7.12 Calculate the total active, reactive, and apparent powers taken by a balanced wye-connected load consistingof 30+j10 Ohms per phase when connected to a 440 Volt three-phase source.

P7.13 Calculate the minimum kVA rating of a three-phase ships service generator in order to carry the followingloads. At what power factor will it operate?

Kwatts KVARS (lagging)Electronic Sensors 100 30

Weapons Control and Drive 90 28

Propulsion Auxiliaries 50 50

Hotel and Convenience 50 50

Heat, Light, and Ventilation 100 20

P7.14 Design a three-phase, 15 kW heater for operation on a 208 Volt (line-to-line) system. Specify the resistanceand power rating of each resistance assuming:

a. a wye-connection, and then

b. a Delta-connection..

P7.15 Calculate the real and reactive power delivered to a balanced, wye-connected three-phase load if the per-phase

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impedance is 4 + j3 Ohms and the line voltage is 50 Volts. At what power factor is this power delivered?

P7.16 The line current delivered to a balanced, three phase, delta-connected load is 22 Amps and the per-phaseimpedance is 4 + j5 Ohms.

a. Determine the line voltage.

b. Determine the total complex power delivered to the load and the power factor at which this power isdelivered.

P7.17 A balanced, three-phase, wye-connected load operates on a line voltage equal to 280 Volts and consumes atotal of 1200 Watts at a power factor of 0.6 lagging. Determine the equivalent per-phase impedance in bothrectangular and polar forms.