106.pdf

Balabanian, N. Three-Phase Circuits The Engineering Handbook. Ed. Richard C. Dorf Boca Raton: CRC Press LLC, 2000

1998 by CRC PRESS LLC

106Three-Phase Circuits

106.1 Relationships between Voltages and Currents106.2 Line Voltages106.3 Power Relationship106.4 Balanced Source and Balanced Load106.5 Other Types of Interconnections

Norman BalabanianUniversity of Florida, Gainesville

A very important use of electricity is the driving of industrial equipment, such as electric motors, inthe AC steady state. Suppose that the instantaneous AC voltage and current of such a load is givenby

v(t) =p2jV j cos(!t + )

i(t) =p2jIj cos (!t+ )

(106:1)

Then the power to the load at any instant of time is

p(t) = jV jjIj[cos ( ) + cos (2!t + + )] (106:2)

The instantaneous power has a constant term and a sinusoidal term at twice the frequency. Thequantity in brackets fluctuates between a minimum value of cos ( ) 1 and a maximum valueof cos ( ) + 1 . This fluctuation of power delivered to the load has a great disadvantage whenthe load is an electric motor. A motor operates by receiving electric power and transmittingmechanical (rotational) power at its shaft. If the electric power is delivered to the motor in spurts,the motor is likely to vibrate. For satisfactory operation in such a case, a physically larger motor,with a larger shaft and flywheel, will be needed to provide more inertia for smoothing out thefluctuations than would be the case if the delivered power were constant.

This problem is overcome in practice by the use of what is called a three-phase system. Thischapter will provide a brief discussion of three-phase power systems.

Consider the circuit in Fig. 106.1. This is an interconnection of three AC sources to three loadsconnected in such a way that each source/load combination shares the return connection from O toN. The three sources can be viewed collectively as a single source, and the three loadswhich areassumed to be identicalcan be viewed collectively as a single load. Each of the individual


sources and loads is referred to as one phase of the three-phase system.

Figure 106.1 Flow of power from source to load.


106.1 Relationships between Voltages and CurrentsThe three sources are assumed to have the same frequency; for this reason, they are said to besynchronized. It is also assumed that the three phase voltages have the same rms values and thatthe phase difference between each pair of voltages is 120 (2=3 rad). Thus, the voltages can bewritten:

va =p2jV j cos (!t + 1) $ Va = jV jej0

vb =p2jV j cos (!t + 2) $ Vb = jV jej120

vc =p2jV j cos (!t + 3) $ Vc = jV jej120

(106:3)

The phasors representing the sinusoids have also been shown. For convenience the angle va hasbeen chosen as the reference for angles; vb lags va by 120 and vc leads va by 120 .Observe that the principle value of the angle lying between 180 is used. One could add 360 tothe negative angle and use the value 240 instead.

There are two options for choosing the sequence of the phases. Once the particular phase that isto be the reference for angles is chosen and named "a," there are two possible sequences for theother two: either "abc" or "acb." This fact is hardly earthshaking; all it means is that the leadingand lagging angles can be interchanged. Obviously, nothing fundamental is different in the secondsequence. Hence, the discussion that follows is limited to the abc phase sequence.

Because the loads are identical, the rms values of the three currents shown in Fig. 106.1 will alsobe the same and the phase difference between each pair of them will be 120 . Thus, the currentscan be written as

i1 =p2jIj cos (!t + 1) $ I1 = jIjej1

i2 =p2jIj cos (!t + 2) $ I2 = jIjej(1120 )

i3 =p2jIj cos (!t + 3) $ I3 = jIjej(1+120 )

(106:4)

Perhaps a better form of visualizing the voltages and currents is a graphical one. Phasor diagramsfor the voltages and the currents are shown separately in Fig. 106.2. The value of angle 1 willdepend on the load. Something significant is clear from these diagrams. First, V2 and V3 are eachthe other's conjugate. So if they are added, the imaginary parts cancel and the sum will be real, asillustrated by the construction in the voltage diagram. Furthermore, the construction shows thissum to be negative and equal in magnitude to V1 . Hence, the sum of the three voltages is zero. Thesame is true of the sum of the three currents, as can be established graphically by a similarconstruction. The same results can be confirmed analytically by converting the phasor voltages andcurrents into rectangular form.


By Kirchhoff's current law applied at node N in Fig. 106.1, we find that the current in the returnline is the sum of the three currents in Eq. (106.4). But since this sum was found to be zero, thereturn line carries no current. Hence, it can be removed entirely without affecting the operation ofthe system. The resulting circuit is redrawn in Fig. 106.3. It can be called a three-wire three-phasesystem. Because of its geometrical form, this connection of both the sources and the loads is said tobe a wye (Y) connection, even though it is an upside-down Y.

Figure 106.3 Wye-connected three-phase system.

Notice that the circuit in Fig. 106.3 is planar, with no lines crossing any other lines. Thatsimplicity has been achieved at a price. Notice how the sequence (abc) of sources has been laid out

Figure 106.2 Voltage and current phasor diagrams.


geometrically. Clearly, with the connections shown, the sequence of the loads is not the same asthat of the sources. Having the same sequence would require interchanging the connections of theb and c sources with the bottom two loads. Doing that would result in one branch crossing another.However, nothing fundamental would change with either connection, assuming equal loads.

SYSTEM OF DISTRIBUTION BY ALTERNATING CURRENTSCharles P. SteinmetzPatented January 29, 1895#533,244

By the 1890's, Edison's light bulb had taken over the lighting of big cities, but his high-voltageDC method of supplying power had not. Alternating current (AC) was becoming the standardpower source. It was cheaper and more efficient to transmit AC over long distances because itcould be transformed up to high voltages and low currents (thinner wires, less losses) fortransmission and then transformed back down to lower voltages for household current.

Steinmetz, while at General Electric's laboratory in Schenectady, New York, developed thesystem for 3-phase AC. This permitted polyphase industrial motors and machines to operate on thesame system as single-phase electric light. This system of power distribution is substantially thesame one in use throughout the world today. (1993, DewRay Products, Inc. Used withpermission.)



106.2 Line VoltagesIn the three-wire three-phase system in Fig. 106.3, the neutral point O is not accessible, so phasevoltages cannot be measured. The voltages that are available for measurement are theline-to-line or simply the line voltages: Vab , Vbc , and Vca . By Kirchhoff's voltage law,

Vab = Va Vb = jV j jV jej120 =p3jV jej30

Vbc = Vb Vc = jV jej120 jV jej120 =p3jV jej90 (106:5)

Vca = Vc Va = jV jej120 jV j =p3jV jej150

The interesting result is that all the line-voltage magnitudes are equal at p3 times the

phase-voltage magnitude. Thus, a 220 V line voltage corresponds to a phase voltage of 127 V. Theline-voltage angles have the same mutual relationships as the phase-voltage angles; they areseparated by 120 .

106.3 Power RelationshipThe instantaneous power delivered by each of the sources has the form given in Eq. (106.2),consisting of a constant term representing the average power and a double-frequency sinusoidalterm. The latter, being sinusoidal, can be represented by a phasor also. The only caveat is that adifferent frequency is involved here, so this power phasor should not be mixed with the voltageand current phasors in the same diagram or calculations. Let jSj = jV jjIj be the apparent powerdelivered by each of the three sources, and let the three power phasors be Sa , Sb , and Sc ,respectively. Then,

Sa = jSjej(1+1 ) = jSjej1

Sb = jSjej(2+2 ) = jSjej(120+1120 ) = jSjej(1+120 )

Sc = jSjej(3+3 ) = jSjej(120+1+120 ) = jSjej(1120 )(106:6)

It is evident that the phase relationships between these three phasors are the same as the onesbetween the voltages and the currents. That is, the second leads the first by 120 and the third lagsthe first by 120 . Hence, just as with the voltages and the currents, the sum of these three powerphasors will also be zero. This is a very significant result. It constitutes the motivation for usingthree-phase power over the pulsating power of a single-phase system. Although the instantaneouspower delivered by each load has a constant component and a sinusoidal component, when the


three powers are added, the sinusoidal components add to zero, leaving only the constants. Thus,the total power delivered to the three loads is constant.

To determine the value of this constant power, let's use Eq. (106.2) as a model. The contributionof the kth source to the total (constant) power is jSj cos (k k ) . It can be easily verified thatk k = 1 1 = 1 . The first equality follows from the relationships between the valuesfrom Eq. (106.3) and between the values from Eq. (106.4). The choice of 1 = 0 leads to the lastequality. Hence, each phase contributes an equal amount to the total average power. If P is the totalaverage power, then

P = Pa + Pb + Pc = 3Pa = 3jV jjIj cos (1 1) (106:7)

Although the angle 1 has been set equal to zero, it is shown in this equation for the sake ofgenerality.

A similar result can be obtained for the reactive power. The reactive power of the kth phase isjSj sin(k k ) = jSj sin(1 1) . If Q is the total reactive power, then

Q = 3jSj sin(1 1)

106.4 Balanced Source and Balanced LoadWhat has just been described is a balanced three-phase three-wire power system. The three sourcesin practice are not three independent sources but consist of three different parts of the samegenerator. The same is true of the loads. An AC power generator consists of (a) a rotor that is rotated by a prime mover (say a turbine) andproduces a magnetic field that also rotates, and (b) a stator on which is wound one or more coils ofwire. In three-phase systems the number of coils is three. The rotating magnetic field induces avoltage in each of the coils. The frequency of the induced voltage depends on the number ofmagnetic poles created on the rotor and the speed of rotation. These are fixed so as to"synchronize" with the 60 Hz frequency of the power system. The 120 leading and lagging phaserelationships between these voltages is obtained by distributing the conductors of the coils aroundthe circumference of the stator so that they are separated geometrically by 120 . Thus, the threesources described in the text are in reality a single physical device, a single generator. Similarly,the three loads might be the three windings on a three-phase motor, again a single physical device.Or they might be the windings of a three-phase transformer.What has been described is ideal in a number of ways. First, the circuit can be unbalancedforexample, by the loads being somewhat unequal. Second, since the real devices whose ideal modelis a voltage source are coils of wire, each source should be accompanied by a branch consisting ofthe coil inductance and resistance. Third, since the power station (or the distribution transformer atsome intermediate point) may be at some distance from the load, the parameters of the physicalline carrying the power (the line inductance and resistance) must also be inserted in series betweenthe source and the load.

The analysis of this chapter does not apply to an unbalanced system. An entirely new analytical


technique is required to do full justice to such a system.The technique for analyzing unbalanced circuits utilizes what are called symmetricalcomponents.An understanding of balanced circuits is a prerequisite to tackling the unbalancedcase.

The last two of the conditions that make the circuit less than ideal (winding and line impedances)introduce algebraic complications but change nothing fundamental in the preceding theory. If thesetwo conditions are taken into account, the appropriate circuit takes the form shown in Fig. 106.4.Here the internal impedance of a source (the winding impedance labeled Zw ) and the lineimpedance Zl connecting that source to its load are both connected in series with thecorresponding load. Thus, instead of the impedance in each phase being Z, it is Z + Zw + Zl .Hence, the rms value of each current is

jIj = jV jjZ + Zw + Zl j(106:8)

instead of jV j=jZj . All other previous results remain unchangednamely, that the sum of thephase currents add to zero and that the sum of the phase powers is a constant. The detailedcalculations just become a little more complicated.

Figure 106.4 Three-phase circuit with nonzero winding and line impedances.


106.5 Other Types of InterconnectionsAll of the preceding development was based on both the sources and the loads being connected ina wye connection. Although the upside-down Y structure looks geometrically a little different froman upside-down tee circuit, electrically, the two are exactly the same. The wye is not, however, theonly possible way to connect the phases of a three-phase system. Another possibility, the deltaconnection, so named because it looks like the Greek letter , is shown in Fig. 106.5. (In thisfigure the boxes can represent either sources or impedances.)

Figure 106.5 Wye connection and delta connection.

By proper choice of the branch parameters, a tee can be made equivalent to a pi () at theterminals. We note that the delta is just an upside-down pi. As a pi, the junction between A and Bis usually extended as the common terminal of a two-port.

If the structures in Fig. 106.5 are to be equivalent, the line voltages Vab , Vbc , and Vca should bethe same in both circuits. Similarly, the currents into the terminals should be the same in both. Notethat, in the delta connection, the phase voltages are not evident; the only voltages available are theline voltages. Thus the voltages in the delta are the line voltages given in Eq. (106.3). In the wyethe phase currents are also the currents in the lines. For the delta, however, the line currents are thedifference between two phase currents, as noted in Fig. 106.5. For the line currents, a set ofequations similar to Eq. (106.5) can be written in terms of the phase currents. Since the same 120difference of angle exists between the phase currents as between the phase voltages, we wouldexpect that the result for currents would be similar to the result for voltages in Eq.(106.5)namely, that the line-current magnitudes in a delta connection would be p3 times thephase-current magnitudes.

In a three-phase circuit the sources, the loads, or both, can be replaced by a delta equivalent; fourdifferent three-phase circuits can therefore be imagined: wye-wye, wye-delta, delta-wye, anddelta-delta. There are no fundamental differences in analyzing each of these four circuits.

Example. A balanced, 120 V, three-wire three-phase transmission system in a wye-wyeconnection is represented by the circuit in Fig. 106.4. Assume that the winding impedances are


negligible but that the line impedances are given by Zl = 0:1 + j0:2 - . Each load impedance isZ = 20 + j5 - . The following quantities are to be determined: (a) the line current magnitude; (b)the magnitude of the voltage across each load; (c) the average power, reactive power, and apparentpower delivered to the load by each phase; (d) the average power, reactive power, and apparentpower delivered by each source; and (e) the fraction of the power delivered by the system that islost in the lines.Solution. The solution is completely straightforward. First, the line current is found by dividingthe phase voltage by the sum of the load and line impedances; the load voltage follows from theproduct of the load impedance by the line current. Thus,

jIj = 120p(20 + 0:1)2 + (5 + 0:2)2

= 5:78 A

jVL j = jIjjZ j = 5:78p

202 + 52 = 119:16 V

The power calculations then follow:

jSL j = jVL jjIj = 119:16(5:78) = 688:7 VA orjSL j = jIj2 jZL j = 5:782

p202 + 52 = 688:7 VA

PL = RL jIj2 = 20(5:78)2 = 668:2 WQL = XL jIj2 = 5(5:78)2 = 167:0 VAR

=

qjSL j2 P 2L =

p688:72 668:22 = 166:8 VAR

Perhaps the best way to find the power delivered by the sources is to determine the power lost inthe line and then add this to the load power. Carrying out this approach leads to the followingresult:

Pl = 0:1jIj2 = 3:34 W Ps = 3:34 + 668:2 = 671:5 WQl = 0:2jIj2 = 6:68 VAR Qs = 6:68 + 167:0 = 173:7 VAR

Finally, the fraction of the source power that is lost in the line is 3.34/671.5 = 0.005 or0.5%.

Defining TermsDelta connection: The sources or loads in a three-phase system connected end-to-end, forming a

closed path, like the Greek letter .Phasor: A complex number representing a sinusoid; its magnitude and angle are the rms value

and the phase of the sinusoid, respectively.


Wye connection: The three sources or loads in a three-phasor system connected to have onecommon point, like the letter Y.

Referencesdel Toro, V. 1992. Electric Power Systems. Prentice Hall, Englewood Cliffs, NJ.Gungor, B. R. 1988. Power Systems. Harcourt Brace Jovanovich, San Diego, CA.Peebles, P. Z. and Giuma, T. A. 1991. Principles of Electrical Engineering. McGraw-Hill, New

York.


The Engineering HandbookContentsThree-Phase Circuits106.1 Relationships between Voltages and Currents106.2 Line Voltages106.3 Power Relationship106.4 Balanced Source and Balanced Load106.5 Other Types of InterconnectionsDefining TermsReferences

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