Three-Phase A.C. Circuits

Transcription

1 Chapter 3 Three-Phase A.C. Circuits earning Outcomes This chapter introduces the concepts and principles of the three-ase electrical supply, and the corresponding circuits. On completion you should be able to: 1 Describe the reasons for, and the generation of the three-ase supply. 2 Distinguish between star (3 and 4-wire) and delta connections. 3 State the relative advantages of three-ase systems compared with single-ase-systems. 4 Solve three-ase circuits in terms of ase and line quantities, and the power developed in three-ase balanced loads. 5 Measure power dissipation in both balanced and unbalanced three-ase loads, using the 1, 2 and 3-wattmeter methods, and hence determine load power factor. 6 Calculate the neutral current in a simple unbalanced 4-wire system. 3.1 Generation of a Three-Phase Supply In order to understand the reasons for, and the method of generating a three-ase supply, let us firstly consider the generation of a singlease supply. Alternating voltage is provided by an a.c. generator, more commonly called an alternator. The basic principle was outlined in Fundamental Electrical and Electronic Principles, Chapter 5. It was shown that when a coil of wire, wound on to a rectangular former, is rotated in a magnetic field, an alternating (sinusoidal) voltage is induced into the coil. You should also be aware that for electromagnetic induction to take place, it is the relative movement between conductor and magnetic flux that matters. Thus, it matters not whether the field is static and the conductor moves, or vice versa. For a practical alternator it is found to be more convenient to rotate the magnetic field, and to keep the conductors (coil or winding) stationary. 85

2 86 Further Electrical and Electronic Principles In any rotating a.c. machine, the rotating part is called the rotor, and the stationary part is called the stator. Thus, in an alternator, the field system is contained in the rotor. The winding in which the emf is generated is contained in the stator. The reasons for this are as follows: In this context, the term fi eld refers to the magnetic fi eld. This fi eld is normally produced by passing d.c. current through the rotor winding. Since the winding is rotated, the current is passed to it via copper slip-rings on the shaft. The external d.c. supply is connected to the slip-rings by a pair of carbon brushes. (a) When large voltages are generated, heavy insulation is necessary. If this extra mass has to be rotated, the driving device has to develop extra power. This will then reduce the overall efficiency of the machine. Incorporating the winding in the stator allows the insulation to be as heavy as necessary, without adversely affecting the efficiency. (b) The contact resistance between the brushes and slip-rings is very small. However, if the alternator provided high current output (in hundreds of ampere), the I 2 power loss would be significant. The d.c. current (excitation current) for the field system is normally only a few amps or tens of amps. Thus, supplying the field current via the slip-rings produces minimal power loss. The stator winding is simply connected to terminals on the outside of the stator casing. (c) For very small alternators, the rotor would contain permanent magnets to provide the rotating field system. This then altogether eliminates the need for any slip-rings. This arrangement is referred to as a brushless machine. The basic construction for a single-ase alternator is illustrated in Fig The conductors of the stator winding are placed in slots E G J stator C N A rotor B M S D K H Fig. 3.1 F

3 Three-Phase A.C. Circuits 87 around the inner periery of the stator. The two ends of this winding are then led out to a terminal block on the casing. The rotor winding is also mounted in slots, around the circumference of the rotor. This figure is used to illustrate the principle. A practical machine would have many more conductors and slots. Since the conductors of the stator winding are spread around the whole of the slots, it is known as a distributed winding. As the rotor field sweeps past these conductors an emf is induced in each of them in turn. These individual emfs reach their maximum values only at the instant that the rotating field cuts them at 90. Also, since the slots have an angular displacement between them, then the conductor emfs will be out of ase with each other by this same angle. In Fig. 3.1 there are a total of twelve conductors, so this ase difference must be 30. The total stator winding emf will therefore not be the arithmetic sum of the conductor emfs, but will be the asor sum, as shown in Fig The ratio of the asor sum to the arithmetic sum is called the distribution factor. For the case shown (a fully distributed winding) the distribution factor is E 30 D F G J H B C K A M AB, CD etc. are conductor (coil) emfs. AM is the asor sum Fig. 3.2 Now, if all of the stator conductors could be placed into a single pair of slots, opposite to each other, then the induced emfs would all be in ase. Hence the asor and arithmetic sums would be the same, yielding a distribution factor of unity. This is not a practical solution. However, if the conductors are concentrated so as to occupy only one third of the available stator slots, then the distribution factor becomes In a practical single-ase alternator, the stator winding is distributed over two thirds of the slots. et us return to the option of using only one third of the slots. We will now have the space to put two more identical windings into the stator. Each of the three windings could be kept electrically separate, with their own pairs of terminals. We would then have three separate singlease alternators in the same space as the original. Each of these would

4 88 Further Electrical and Electronic Principles also have a good distribution factor of The three winding emfs will of course be mutually out of ase with each other by 120, since each whole winding will occupy 120 of stator space. What we now have is the basis of a three-ase alternator. The term three-ase alternator is in some ways slightly misleading. What we have, in effect, are three identical single-ase alternators contained in the one machine. The three stator windings are brought out to their own separate pairs of terminals on the stator casing. These stator windings are referred to as ase windings, or ases. They are identified by the colours red, yellow and blue. Thus we have the red, yellow and blue ases. The circuit representation for the stator winding of such a machine is shown in Fig In this figure, the three ase windings are shown connected, each one to its own separate load. This arrangement is known as a three-ase, six-wire system. However, three-ase alternators are rarely connected in this way. OAD 2 OAD 1 OAD 3 Fig. 3.3 Since the three generated voltages are sinewaves of the same frequency, mutually out-of-ase by 120, then they may be represented both on a waveform diagram using the same angular or time axis, and as asors. The corresponding waveform and asor diagrams are shown in Figs. 3.4 and 3.5 respectively. In either case, voltage Y B θ (deg) Fig. 3.4

5 Three-Phase A.C. Circuits 89 B Y Fig. 3.5 we need to select a reference asor. By convention, the reference is always taken to be the red ase voltage. The yellow ase lags the red by 120, and the blue lags the red by 240 (or, if you prefer, leads the red by 120 ). The windings are arranged so that when the rotor is driven in the chosen direction, the ase sequence is red, yellow, blue. If, for any reason, the rotor was driven in the opposite direction, then the ase sequence would be reversed, i.e. red, blue, yellow. We shall assume that the normal sequence of, Y, B applies at all times. It may be seen from the waveform diagram that at any point along the horizontal axis, the sum of the three voltages is zero. This fact becomes even more apparent if the asor diagram is redrawn as in Fig In this diagram, the three asors have been treated as any other vector quantity. The sum of the vectors may be determined by drawing them to scale, as in Fig. 3.6, and the resultant found by measuring the distance and angle from the beginning point of the first vector to the arrowhead of the last one. If, as in Fig. 3.6, the first and last vectors meet in a closed figure, the resultant must be zero. B Y Fig Three-Phase, Four-Wire System It is not necessary to have six wires from the three ase windings to the three loads, provided there is a common return line. Each winding will have a start (S) and a finish (F) end. The common connection mentioned above is achieved by connecting the corresponding ends

6 90 Further Electrical and Electronic Principles of the three ases together. For example, either the three F ends or the three S ends are commoned. This form of connection is shown in Fig. 3.7, and is known as a star or Y connection. With the resulting 4-wire system, the three loads also are connected in star configuration. The three outer wires are called the lines, and the common wire in the centre is called the neutral. S S F F F S Fig. 3.7 If the three loads were identical in every way (same impedance and ase angle), then the currents flowing in the three lines would be identical. If the waveform and/or asor diagrams for these currents were drawn, they would be identical in form to Figs. 3.4 and 3.5. These three currents meet at the star point of the load. The resultant current returning down the neutral wire would therefore be zero. The load in this case is known as a balanced load, and the neutral is not strictly necessary. However it is difficult, in practice, to ensure that each of the three loads are exactly balanced. For this reason the neutral is left in place. Also, since it has to carry only the relatively small out-ofbalance current, it is made half the cross-sectional area of the lines. et us now consider one of the advantages of this system compared with both a single-ase system, and the three-ase 6-wire system. Suppose that three identical loads are to be supplied with 200 A each. The two lines from a single-ase alternator would have to carry the total 600 A required. If a 3-ase, 6-wire system was used, then each line would have to carry only 200 A. Thus, the conductor csa would only need to be 1/3 that for the single-ase system, but of course, being six lines would entail using the same total amount of conductor material. If a 4-wire, 3-ase system is used there will be a saving on conductor costs in the ratio of 3.5:6 (the 0.5 being due to the neutral). If the power has to be sent over long transmission lines, such as the National Grid System, then the 3-ase, 4-wire system yields an enormous saving in cable costs. This is one of the reasons why the power generating companies use three-ase, star-connected generators to supply the grid system.

7 Three-Phase A.C. Circuits elationship between ine and Phase Quantities in a Star-connected System Consider Fig. 3.8, which represents the stator of a 3-ase alternator connected to a 3-ase balanced load. The voltage generated by each of the three ases is developed between the appropriate line and the neutral. These are called the ase voltages, and may be referred to in general terms as, or specifically as N, YN and BN respectively. However, there will also be a difference of potential between any pair of lines. This is called a line voltage, which may be generally referred to as, or specifically as Y, YB arid B respectively. I I IN N I I I B Y Fig. 3.8 A line voltage is the asor difference between the appropriate pair of ase voltages. Thus, Y is the asor difference between N and YN. In terms of a asor diagram, the simplest way to subtract one asor from another is to reverse one of them, and then find the resulting asor sum. This is, mathematically, the same process as saying that a b a ( b). The corresponding asor diagram is shown in Fig BN YN C 30 B 30 0 A N YN Fig. 3.9

8 92 Further Electrical and Electronic Principles Note : If YN is reversed, it is denoted either as YN or as NY. We shall use the first of these. The asor difference between N and YN is simply the asor sum of N ( YN ). Geometrically this is obtained by completing the parallelogram as shown in Fig This parallelogram consists of two isosceles triangles, such as OCA. Another property of a parallelogram is that its diagonals bisect each other at right angles. Thus, triangle OCA consists of two equal right-angled triangles, OAB and ABC. This is illustrated in Fig Since triangle OAB is a 30, 60, 90 triangle, then the ratios of its sides AB:OA:OB will be 1:2: 3 respectively. C B O A Fig Hence, OB 3 OA 2 3. OA so OB 2 but OC 2 OB 3. OA and since OC Y, and OA then 3 Using the same technique, it can be shown that: Y N 3 and 3 YB YN B BN N Thus, in star configuration, 3 (3.1) The complete asor diagram for the line and ase voltages for a star connection is shown in Fig Also, considering the circuit diagram of Fig. 3.8, the line and ase currents must be the same. Hence, in star configuration, I I (3.2) We now have another advantage of a 3-ase system compared with single-ase. The star-connected system provides two alternative voltage outputs from a single machine. For this reason, the stators of all alternators used in electricity power stations are connected in star configuration. These machines normally generate a line voltage

9 Three-Phase A.C. Circuits 93 B BN Y N 30 YN YB Fig of about 25 k. By means of transformers, this voltage is stepped up to 400 k for long distance transmission over the National Grid. For more localised distribution, transformers are used to step down the line voltage to 132 k, 33 k, 11 k, and 415. The last three of these voltages are supplied to various industrial users. The ase voltage derived from the 415 lines is 240, and is used to supply both commercial premises and households. Worked Example 3.1 Q A 415, 50 Hz, 3-ase supply is connected to a star-connected balanced load. Each ase of the load consists of a resistance of 25 Ω and inductance 0.1 H, connected in series. Calculate (a) ase voltage, (b) the line current drawn from the supply, and (c) the power dissipated. A Whenever a three-ase supply is specified, the voltage quoted is always the line voltage. Also, since we are dealing with a balanced load, then it is necessary only to calculate values for one ase of the load. The figures for the other two ases and lines will be identical to these. 415 ; f 5 0 H z ; 2 5 Ω ; 0.1 H I I 25 Ω 0.1 H Fig. 3.12

10 94 Further Electrical and Electronic Principles (a) so Ans (b) Since it is possible to determine the impedance of a ase of the load, and we now know the ase voltage, then the ase current may be calculated: X 2πf ohm 2π hence X Ω Z 2 X2 ohm Z Ω 240 I amp Z so I 598. A 2 In a star-connected circuit, I I Ph therefore I 5.98 A Ans The power in one ase, P P I 2 watt W and since there are three ases, then the total power is: P 3 P watt hence P 268. kw Ans 3.4 Delta or Mesh Connection If the start end of one winding is connected to the finish end of the next, and so on until all three windings are interconnected, the result is the delta or mesh connection. This connection is shown in Fig The delta connection is not reserved for machine windings only, since a 3-ase load may also be connected in this way. I F S I I S F I F S I I Fig. 3.13

11 Three-Phase A.C. Circuits elationship between ine and Phase Quantities in a Delta-connected System It is apparent from Fig that each pair of lines is connected across a ase winding. Thus, for the delta connection: (3.3) It is also apparent that the current along each line is the asor difference of a pair of ase currents. The three ase currents are mutually displaced by 120, and the asor diagram for these is shown in Fig Using exactly the same geometrical technique as that for the ase and line voltages in the star connection, it can be shown that: I 3 I (3.4) I B() I () I Y() Fig The asor diagram for the ase and line currents in delta connection is as in Fig Note that the provision of a neutral wire is not applicable with a delta connection. However, provided that the load I I I I 30 I I Fig. 3.15

12 96 Further Electrical and Electronic Principles is balanced, there is no requirement for one. Under balanced load conditions the three ase currents will be equal, as will be the three line currents. If the load is unbalanced, then these equalities do not exist, and each ase or line current would have to be calculated separately. This technique is beyond the scope of the syllabus you are now studying. Worked Example 3.2 Q A balanced load of ase impedance 120 Ω is connected in delta. When this load is connected to a 600, 50 Hz, 3-ase supply, determine (a) the ase current, and (b) the line current drawn. A Z 120 Ω ; 600 ; f 5 0 H z The circuit diagram is shown in Fig I I 600 Z 120 Ω Fig (a) In delta, I so, I 600 Z 5 A Ans 600 amp 120 (b) in delta, I 3I amp 3 5 therefore I 866. A Ans 3.6 Power Dissipation in Star and Delta-connected oads We have seen in Example 3.1 that the power in a 3-ase balanced load is obtained by multiplying the power in one ase by 3. In many practical situations, it is more convenient to work with line quantities. P I cos φ watt where cos φ is the ase power factor. total power, P 3 P 3 I cos φ watt... [] 1

13 Three-Phase A.C. Circuits 97 Considering a STA-connected load, I 3, and I Substituting for and I in eqn [1]: P 3 I 3 cos φ For a delta-connected load, therefore P 3 I cos φ watt (3.5) and I I 3 and substituting these values into eqn [1] will yield the same result as shown in (3.5) above. Thus, the equation for determining power dissipation, in both star and delta-connected loads is exactly the same. However, the value of power dissipated by a given load when connected in star is not the same as when it is connected in delta. This is demonstrated in the following example. Worked Example 3.3 Q A balanced load of ase impedance 100 Ω and power factor 0.8 is connected (a) in star, and (b) in delta, to a 400, 3-ase supply. Calculate the power dissipation in each case. A Z 10 0 Ω ; cos φ 0.8; (a) the circuit diagram is shown in Fig I I 400 Z 100 Ω Fig. 3.17

14 98 Further Electrical and Electronic Principles so 400 volt I I amp Z 100 and I A P 3 I cos φ watt therefore P kw Ans (b) The circuit diagram is shown in Fig I I 400 Z 100 Ω I so I but I so I Fig Z 4A 400 amp 100 3I A P 3 I cos φ watt therefore P 38. 4kW Ans Comparing the two answers for the power dissipation in the above example, it may be seen that: Power in a delta-connected load is three times that when it is connected in star configuration. (3.6) Worked Example 3.4 Q A balanced star-connected load is fed from a 400, 50 Hz, three-ase supply. The resistance in each ase of the load is 10 Ω and the load draws a total power of 15 kw. Calculate (a) the line current drawn, (b) the load power factor, and (c) the load inductance. A ; f 5 0 H z ; 10 Ω ; P W

15 Three-Phase A.C. Circuits Ω 400 Fig (a) The power in one ase will be one third of the total power, so P but, P P watt 5 kw 3 3 I 2 watt P 5000 so, I I 10 I A Ans (b) P 3 I cos P so, cos φ p.f. 3 I hence, p.f Ans φ watt (c) φ cos X tan φ so, X X ohm 2πf 100π and 822. mh Ans φ Z X Fig. 3.20

16 100 Further Electrical and Electronic Principles Worked Example 3.5 Q A balanced delta-connected load takes a ase current of 15 A at a power factor of 0.7 lagging when connected to a 115, 50 Hz, three-ase supply. Calculate (a) the power drawn from the supply, and (b) the resistance in each ase of the load. A 115 ; f 5 0 H z ; I 15 A; cos φ 0.7 I I 15 A 115 Fig (a) I 3I amp 3 15 I A P 3 I cos φ watt so P W Ans (b) P and P P W 3 3 I 2 watt P so I ohm Ω Ans 3.7 Star/Delta Supplies and oads As explained in Section 3.3, the distribution of 3-ase supplies is normally at a much higher line voltage than that required for many users. Hence, 3-ase transformers are used to step the voltage down to the appropriate value. A three-ase transformer is basically three single-ase transformers interconnected. The three primary windings may be connected either in star or delta, as can the three secondary windings. Similarly, the load connected to the transformer secondary windings may be connected in either configuration. One important point to bear in mind is that the transformation ratio (voltage or turns ratio) refers to the ratio between the primary ase to the secondary ase winding. The method of solution of this type of problem is illustrated in the following worked example.

17 Three-Phase A.C. Circuits 101 Worked Example 3.6 Q Figure 3.22 shows a balanced, star-connected load of ase impedance 25 Ω and power factor 0.75, supplied from the delta-connected secondary of a 3-ase transformer. The turns ratio of the transformer is 20:1, and the star-connected primary is supplied at 11 k. Determine (a) the voltages 2, 3 and 4, (b) the currents I 1, I 2, and I 3, and (c) the power drawn from the supply. I 1 I 20:1 3 I k Z 25Ω cosφ 0.75 Fig A ; N p /N s 20/1; Z 25 Ω ; cos φ 0.75 (a) k Ans N N s p N s so 3 volt Np 20 hence Ans and Ans (b) In order to calculate the currents, we shall have to start with the load, and work back through the circuit to the primary of the transformer: I3 amp Z 25 I A Ans I I2 3 3 I A Ans I1 Ns I2 Np NsI so I1 amp Np 20 hence I 02. 2A Ans 1 1

18 102 Further Electrical and Electronic Principles (c) P 3 I cos φ watt therefore P 302. kw Ans Worked Example 3.7 Q The star-connected stator of a three-ase, 50 Hz alternator supplies a balanced delta-connected load. Each ase of the load consists of a coil of resistance 15 Ω and inductance 36 mh, and the ase voltage generated by the alternator is 231. Calculate (a) the ase and line currents, (b) the load power factor, and (c) the power delivered to the load. A f 50 Hz; 15 Ω ; 36 mh; 231 I 1 I Ω 36 mh Fig (a) For the alternator: I I I 231 For the load: X 2πf ohm 100 π X Ω Z 2 X ohm Z Ω 400 I I2 amp Z I A Ans I I1 3I I A Ans (b) p.f. cos φ Z p.f. 08. lagging Ans

19 Three-Phase A.C. Circuits 103 (c) P 3I cos φ watt P kw Ans Alternatively, P 3 P 3 I watt P kw Ans Measurement of Three-ase Power In an a.c. circuit the true power may only be measured directly by means of a wattmeter. The principle of operation of this instrument is described in Fundamental Electrical and Electronic Principles, Chapter 5. As a brief reminder, the instrument has a fixed coil through which the load current flows, and a moving voltage coil (or pressure coil) connected in parallel with the load. The deflection of the pointer, carried by the moving coil, automatically takes into account the ase angle (or power factor) of the load. Thus the wattmeter reading indicates the true power, P I cos φ watt. If a three-ase load is balanced, then it is necessary only to measure the power taken by one ase. The total power of the load is then obtained by multiplying this figure by three. This technique can be very simply applied to a balanced, star-connected system, where the star point and/or the neutral line are easily accessible. This is illustrated in Fig W 1 N B Y Fig In the situation where the star point is not accessible, then an artificial star point needs to be created. This is illustrated in Fig. 3.25, where the value of the two additional resistors is equal to the resistance of the wattmeter voltage coil. In the case of an unbalanced star-connected load, one or other of the above procedures would have to be repeated for each ase in turn. The total power P P 1 P 2 P 3, where P 1, P 2 and P 3 represent the three separate readings.

20 104 Further Electrical and Electronic Principles W 1 B Y Fig For a delta-connected load, the procedure is not quite so simple. The reason is that the ase current is not the same as the line current. Thus, if possible, one of the ases must be disconnected to allow the connection of the wattmeter current coil. This is shown in Fig Again, if the load was unbalanced, this process would have to be repeated for each ase. W 1 B Y Fig The Two-Wattmeter Method The measurement of three-ase power using the above methods can be very awkward and time-consuming. In practical circuits, the power is usually measured by using two wattmeters simultaneously, as shown in Fig The advantages of this method are: (a) Access to the star point is not required. (b) The power dissipated in both balanced and unbalanced loads is obtained, without any modification to the connections. (c) For balanced loads, the power factor may be determined.

21 Three-Phase A.C. Circuits 105 W 1 I B B I B YB Y W 2 I Y Fig Considering Fig. 3.27, the following statements apply: Instantaneous power for W1, p1 vbi watt and for W2, p2 vybiy watt total instantaneous power p1 p2 v i v i B YB Y [1] Now, any line voltage is the asor difference between the appropriate pair of ase voltages, hence: v v v and v v v B N BN YB YN BN and substituting these into eqn [1] yields: p1 p2 i( vn vbn) iy( vyn vbn) vni vyn iy vbn ( i iy ) but, i i Y i B i.e. the asor sum of three equal currents is zero. therefore p p v i v i v i 1 2 N YN Y BN B The instantaneous sum of the powers measured by the two wattmeters is equal to the sum of instantaneous power in the three ases. Hence, total power, P P1 P2 P PY PB watt (3.7) Consider now the asor diagram for a resistive-inductive balanced load, with the two wattmeters connected as in Fig This asor diagram appears as Fig. 3.28, below.

22 106 Further Electrical and Electronic Principles I B BN B φ φ N φ (30 φ) I I Y YN (30 φ) B YB Fig The power indicated by W 1, P 1 I cos (30 φ ) (3.8) and that for W 2, P 2 I cos (30 φ ) (3.9) From these results, and using Fig. 3.29, the following points should be noted: wattmeter readings (30 φ) Fig If the load p.f. 0.5 (i.e. φ 60 ); both meters will give a positive reading. 2 If the load p.f. 0.5 (i.e. φ 60 ); W 1 indicates the total power, and W 2 indicates zero. 3 If the load p.f. 0.5 (i.e. φ 60 ); W 2 attempts to indicate a negative reading. In this case, the connections to the voltage coil

23 Three-Phase A.C. Circuits 107 of W 2 need to be reversed, and the resulting reading recorded as a negative value. Under these circumstances, the total load power will be P P 1 P 2. 4 The load power factor may be determined from the two wattmeter readings from the equation: φ 3 P P P P tan (3.10) hence, power factor, cos φ can be determined. Worked Example 3.8 Q The power in a 3-ase balanced load was measured, using the two-wattmeter method. The recorded readings were 3.2 kw and 5 kw respectively. Determine the load power and power factor. A P1 32. kw; P2 5kW P P1 P2 watt ( 32. 5) kw therefore, P 82. kw Ans φ tan 1 P2 P 1 3 P P tan hence, φ and p.f. cos φ Ans 2 1 Worked Example 3.9 Q A 3-ase balanced load takes a line current of 24 A at a lagging power factor of 0.42, when connected to a 415, 50 Hz supply. If the power dissipation is measured using the two-wattmeter method, determine the two wattmeter readings, and the value of power dissipated. Comment on the results. A I 24 A; cos φ 0.42; 415 φ cos P1 I cos (30 φ) watt cos ( ) therefore, P kw Ans P2 I cos ( 30 φ) watt cos ( ) therefore, P W Ans P P1 P2 watt 8142 ( ) hence, P kw Ans To obtain the negative reading on W 2, the connections to its voltage coil must have been reversed.

24 108 Further Electrical and Electronic Principles Worked Example 3.10 Q A delta-connected load has a ase impedance of 100 Ω at a ase angle of 55, and is connected to a 415 three-ase supply. The total power consumed is measured using the two-wattmeter method. Determine the readings on the two meters and hence calculate the power consumed. A Z 10 0 Ω ; φ 55 ; 415 W 1 I 415 Z 100 Ω 415 I W 2 I I Fig amp A Z 100 3I A P I 1 cos ( 30 φ) watt cos 25 P kw Ans P I cos ( 30 φ) watt cos 85 2 P 260 W Ans 2 P P1 P2 watt P kw Ans 3.10 Neutral Current in an Unbalanced Three-ase oad We have seen that the neutral current for a balanced load is zero. This is because the asor sum of three equal currents, mutually displaced by 120, is zero. If the load is unbalanced, then the three line (and ase) currents will be unequal. In this case, the neutral has to carry the resulting out-of-balance current. This current is simply obtained by calculating the asor sum of the line currents. The technique is basically the same as that used previously, by resolving the asors into horizontal and vertical components, and applying Pythagoras theorem. The only additional fact to bear in mind is that both horizontal and vertical components can have negative values. This is illustrated by the following example.

25 Three-Phase A.C. Circuits 109 Worked Example 3.11 Q An unbalanced, star-connected load is supplied from a 3-ase, 415 source. The three ase loads are purely resistive. These loads are 25 Ω, 30 Ω and 40 Ω, and are connected in the red, yellow and blue ases respectively. Determine the value of the neutral current, and its ase angle relative to the red ase current. A 415 ; 2 5 Ω ; Y 3 0 Ω ; B 40 Ω The circuit diagram is shown in Fig I 25 Ω Y I Y I N 30 Ω 40 Ω B I B Fig volt I amp; IY amp; I B amp I 96. A I 8A I 6A Y B Y B The corresponding asor diagrams are shown in Fig I B I φ H.C. I Y.C. I N Fig. 3.32

26 110 Further Electrical and Electronic Principles Horizontal components, H.C. I I cos 60 I cos 60 Y B 96. ( ) ( ) so, H.C. 26. A ertical components,.c. I B sin 60 sin 60 ( ) ( ) so.c A The neutral current, I.C. 2 H.C. 2 amp hence, I N I Y N A Ans 1.C φ tan tan 1 H.C. 26. hence, φ relative to I Ans 3.11 Advantages of Three-ase Systems In previous sections, some of the advantages of three-ase systems, compared with single-ase systems, have been outlined. These advantages, together with others not yet described, are listed below. The advantages may be split into two distinct groups: those concerned with the generation and distribution of power, and those concerning a.c. motors. The last four of the advantages listed refer to the second group, and their significance will be appreciated when you have completed Chapter 5, which deals with a.c. machines. 1 The whole of the stator of a three-ase machine is utilised. A single ase alternator utilises only two thirds of the stator slots. 2 A three-ase alternator has a better distribution factor. 3 A three-ase, four-wire system provides considerable savings in cable costs, for the distribution of an equivalent amount of power. 4 A three-ase, four-wire system provides alternative voltages for industrial and domestic users. 5 For a given machine frame size, the power output from a three-ase machine is greater than that from a single-ase machine. 6 A three-ase supply produces a rotating magnetic field; whereas a single-ase supply produces only a pulsating field. 7 Three-ase motors are inherently self-starting; whereas single-ase motors are not. 8 The torque produced by a three-ase motor is smooth; whereas that produced by a single-ase motor is pulsating.

27 Three-Phase A.C. Circuits 111 Summary of Equations Phase and line quantities: In star, I I ; 3 In delta, ; I 3I Power dissipation: P 3 I cos φ watt P 3 P 3 I cos φ 3 I2 and for the same load, P (delta) 3 P (star) watt Power measurement: For two wattmeter method: P P 1 P 2 watt oad ase angle, φ tan 3 P P P P

28 112 Further Fundamental Electrical Electrical and Electronic and Electronic Prinicples Principles Assignment Questions 1 A three-ase load is connected in star to a 400, 50 Hz supply. Each ase of the load consists of a coil, having inductance 0.2 H and resistance 40 Ω. Calculate the line current. 2 If the load specified in Question 1 above is now connected in delta, determine the values for ase and line currents. 3 A star-connected alternator stator generates 300 in each of its stator windings. (a) Sketch the waveform and asor diagrams for the ase voltages, and (b) calculate the p.d. between any pair of lines. 4 A balanced three-ase, delta-connected load consists of the stator windings of an a.c. motor. Each winding has a resistance of 3.5 Ω and inductance H. If this machine is connected to a 415, 50 Hz supply, calculate (a) the stator ase current, (b) the line current drawn from the supply, and (c) the total power dissipated. 5 epeat the calculations for Question 4, when the stator windings are connected in star configuration. 6 Three identical coils, connected in star, take a total power of 1.8 kw at a power factor of 0.35, from a 415, 50 Hz supply. Determine the resistance and inductance of each coil. 7 Three inductors, each of resistance 12 Ω and inductance 0.02 H, are connected in delta to a 400, 50 Hz, three-ase supply. Calculate (a) the line current, (b) the power factor, and (c) the power consumed. 8 The star-connected secondary of a three-ase transformer supplies a delta-connected motor, which takes a power of 80 kw, at a lagging power factor of If the line voltage is 600, calculate (a) the current in the transformer secondary windings, and (b) the current in the motor windings. 9 A star-connected load, each ase of which has an inductive reactance of 40 Ω and resistance of 25 Ω, is fed from the secondary of a three-ase, delta-connected transformer. If the transformer ase voltage is 600, calculate (a) the p.d. across each ase of the load, (b) the load ase current, (c) the current in the transformer secondary windings, and (d) the power and power factor. 10 Three coils are connected in delta to a 415, 50 Hz, three-ase supply, and take a line current of 4.8 A at a lagging power factor of 0.9. Determine (a) the resistance and inductance of each coil, and (b) the power consumed. 11 The power taken by a three-ase motor was measured using the two-wattmeter method. The readings were 850 W and 260 W respectively. Determine (a) the power consumption and, (b) the power factor of the motor. 12 Two wattmeters, connected to measure the power in a three-ase system, supplying a balanced load, indicate 10.6 kw and 2.4 kw respectively. Calculate (a) the total power consumed, and (b) the load ase angle and power factor. State the significance of the negative reading recorded. 13 Using the two-wattmeter method, the power in a three-ase system was measured. The meter readings were 120 W and 60 W respectively. Calculate (a) the power, and (b) the power factor. 14 Each branch of a three-ase, star-connected load, consists of a coil of resistance 4 Ω and reactance 5 Ω. This load is connected to a 400, 50 Hz supply. The power consumed is measured using the two-wattmeter method. Sketch a circuit diagram showing the wattmeter connections, and calculate the reading indicated by each meter. 15 A three-ase, 415, 50 Hz supply is connected to an unbalanced, star-connected load, having a power factor of 0.8 lagging in each ase. The currents are 40 A in the red ase, 55 A in the yellow ase, and 62 A in the blue ase. Determine (a) the value of the neutral current, and (b) the total power dissipated.

29 Fundamental Electrical Three-Phase and Electronic A.C. Principles Circuits 113 Suggested Practical Assignments Assignment 1 Apparatus: To determine the relationship between line and ase quantities in three-ase systems. ow voltage, 50 Hz, three-ase supply 3 1 k Ω rheostats 3 ammeters 3 voltmeters (preferably DMM) Method: 1 Using a digital meter, adjust each rheostat to exactly the same value (1 kω ). 2 Connect the rheostats, in star configuration, to the three-ase supply. 3 Measure the three line voltages and the corresponding ase voltages, and record your results in Table 1. 4 Measure the three line currents, and record these in Table 1. 5 Check that the neutral current is zero. 6 Carefully unbalance the load by altering the resistance value of one or two of the rheostats. ENSUE that you do not exceed the current ratings of the rheostats. 7 Measure and record the values of the three line currents and the neutral current, in Table 2. 8 Switch off the three-ase supply, and disconnect the circuit. 9 Carefully reset the three rheostats to their original settings, as in paragra 1 of the Method. 10 Connect the rheostats, in delta configuration, to the three-ase supply, with an ammeter in each line. 11 Switch on the supply, and measure the line voltages and line currents. ecord your values in Table Switch off the supply, and connect the three ammeters in the three ases of the load, i.e. an ammeter in series with each rheostat. This will involve opening each ase and inserting each ammeter. 13 Switch on and record the values of the three ase currents. ecord values in Table From your tabulated readings, determine the relationship between line and ase quantities for both star and delta connections. 15 From your readings in Table 2, calculate the neutral current, and compare this result with the measured value. 16 Write an assignment report. Include all circuit and asor diagrams, and calculations. State whether the line and ase relationships measured conform to those expected (allowing for experimental error). Assignment 2 Apparatus: To measure the power in three-ase systems, using both single and twowattmeter methods. ow voltage, 50 Hz, three-ase supply 3 1 k Ω rheostats 2 wattmeters 1 DMM

30 114 Further Electrical and Electronic Principles Method: 1 As for Assignment 1, carefully adjust the three rheostats to the same value (1 k Ω ). 2 Connect the circuit shown in Fig. 3.33, and measure the power in the red and yellow ases. W 1 N Y B W 2 Fig Transfer one of the wattmeters to the blue ase, and measure that power. 4 Add the three wattmeter readings to give the total power dissipation. Check to see whether this is three times the individual ase power. 5 Switch off the power supply, and reconnect as in Fig W 1 B Y W 2 Fig ecord the two wattmeter readings, and check whether their sum is equal to the total power recorded from paragra 4 above. 7 Switch off the power supply and reconnect the circuit as in Fig ecord the two wattmeter readings. 9 Switch off the supply, and transfer one of the wattmeters to the third ase. 10 ecord this reading. Add the three readings to give the total power, and check that this is three times the ase power. 11 Switch off the supply and connect the circuit of Fig ecord the two wattmeter readings, and check that their sum equals the total power obtained from paragra 10 above.

31 Three-Phase A.C. Circuits 115 W 1 W 2 Y B Fig W 1 B Y W 2 Fig Assignment 3 Show how the two-wattmeter method of power measurement can be accomplished by means of a single wattmeter, together with a suitable switching arrangement. You may either devise your own system, or discover an existing system through library research.

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