Chapter Three-Phase A.C Circuits Learning Outcomes This chapter introduces the concepts and principles of the three-phase electrical supply, and the corresponding circuits On completion you should be able to: Describe the reasons for, and the generation of the three-phase supply Distinguish between star (3 and 4-wire) and delta connections State the relative advantages of three-phase systems compared with single-phase-systems Solve three-phase circuits in terms of phase and line quantities, and the power developed in three-phase balanced loads Measure power dissipation in both balanced and unbalanced three-phase loads, using the 1, and 3-wattmeter methods, and hence determine load power factor 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-phase supply, let us firstly consider the generation of a singlephase 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 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 CH003-H8747.indd 85 85 4/10/2008 3:24:46 PM 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 field refers to the magnetic field This field 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 I2R 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-phase alternator is illustrated in Fig 3.1 The conductors of the stator winding are placed in slots G E C J L N A stator B rotor M S K D F H Fig 3.1 CH003-H8747.indd 86 4/10/2008 3:24:50 PM Three-Phase A.C Circuits 87 around the inner periphery 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 phase with each other by this same angle In Fig 3.1 there are a total of twelve conductors, so this phase difference must be 30° The total stator winding emf will therefore not be the arithmetic sum of the conductor emfs, but will be the phasor sum, as shown in Fig 3.2 The ratio of the phasor sum to the arithmetic sum is called the distribution factor For the case shown (a fully distributed winding) the distribution factor is 0.644 F E 30Њ J G D H K B C L M A AB, CD etc are conductor (coil) emfs AM is the phasor 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 phase Hence the phasor 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 0.966 In a practical single-phase alternator, the stator winding is distributed over two thirds of the slots Let 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 singlephase alternators in the same space as the original Each of these would CH003-H8747.indd 87 4/10/2008 3:24:50 PM 88 Further Electrical and Electronic Principles also have a good distribution factor of 0.966 The three winding emfs will of course be mutually out of phase 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-phase alternator The term three-phase alternator is in some ways slightly misleading What we have, in effect, are three identical single-phase 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 phase windings, or phases They are identified by the colours red, yellow and blue Thus we have the red, yellow and blue phases The circuit representation for the stator winding of such a machine is shown in Fig 3.3 In this figure, the three phase windings are shown connected, each one to its own separate load This arrangement is known as a three-phase, six-wire system However, three-phase alternators are rarely connected in this way LOAD LOAD LOAD Fig 3.3 Since the three generated voltages are sinewaves of the same frequency, mutually out-of-phase by 120°, then they may be represented both on a waveform diagram using the same angular or time axis, and as phasors The corresponding waveform and phasor diagrams are shown in Figs 3.4 and 3.5 respectively In either case, voltage VY VR 180 VB 360 θ (deg) Fig 3.4 CH003-H8747.indd 88 4/10/2008 3:24:51 PM Three-Phase A.C Circuits 89 VB VR VY Fig 3.5 we need to select a reference phasor By convention, the reference is always taken to be the red phase voltage The yellow phase 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 phase sequence is red, yellow, blue If, for any reason, the rotor was driven in the opposite direction, then the phase sequence would be reversed, i.e red, blue, yellow We shall assume that the normal sequence of R, 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 phasor diagram is redrawn as in Fig 3.6 In this diagram, the three phasors 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 VR VB VY Fig 3.6 3.2 Three-Phase, Four-Wire System It is not necessary to have six wires from the three phase 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 CH003-H8747.indd 89 4/10/2008 3:24:51 PM 90 Further Electrical and Electronic Principles of the three phases 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 F F F S S Fig 3.7 If the three loads were identical in every way (same impedance and phase angle), then the currents flowing in the three lines would be identical If the waveform and/or phasor 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 Let us now consider one of the advantages of this system compared with both a single-phase system, and the three-phase 6-wire system Suppose that three identical loads are to be supplied with 200 A each The two lines from a single-phase alternator would have to carry the total 600 A required If a 3-phase, 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-phase system, but of course, being six lines would entail using the same total amount of conductor material If a 4-wire, 3-phase 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-phase, 4-wire system yields an enormous saving in cable costs This is one of the reasons why the power generating companies use three-phase, star-connected generators to supply the grid system CH003-H8747.indd 90 4/10/2008 3:24:51 PM Three-Phase A.C Circuits 3.3 91 Relationship between Line and Phase Quantities in a Star-connected System Consider Fig 3.8, which represents the stator of a 3-phase alternator connected to a 3-phase balanced load The voltage generated by each of the three phases is developed between the appropriate line and the neutral These are called the phase voltages, and may be referred to in general terms as Vph, or specifically as VRN, VYN and VBN 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 VL, or specifically as VRY, VYB arid VBR respectively IL R Iph Vph IN VL N VL Vph Iph B Iph Vph VL IL Y Fig 3.8 A line voltage is the phasor difference between the appropriate pair of phase voltages Thus, VRY is the phasor difference between VRN and VYN In terms of a phasor diagram, the simplest way to subtract one phasor from another is to reverse one of them, and then find the resulting phasor sum This is, mathematically, the same process as saying that a Ϫ b ϭ a ϩ (Ϫb) The corresponding phasor diagram is shown in Fig 3.9 ϪVYN C VBN B 30Њ 30Њ A VRN VYN Fig 3.9 CH003-H8747.indd 91 4/10/2008 3:24:51 PM 92 Further Electrical and Electronic Principles Note: If VYN is reversed, it is denoted either as ϪVYN or as VNY We shall use the first of these The phasor difference between VRN and VYN is simply the phasor sum of VRN ϩ (ϪVYN) Geometrically this is obtained by completing the parallelogram as shown in Fig 3.9 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 3.10 Since triangle OAB is a 30°, 60°, 90° triangle, – then the ratios of its sides AB:OA:OB will be 1:2:Ί3 respectively C B 30° 60° O A Fig 3.10 Hence, OB ϭ OA OA but OC ϭ ϫ OB ϭ OA so OB ϭ and since OC ϭ VRY , and OA ϭ VRN then VRY ϭ 3VRN Using the same technique, it can be shown that: VYB ϭ 3VYN and VBR ϭ 3VBN Thus, in star configuration, VL ϭ 3Vph (3.1) The complete phasor diagram for the line and phase voltages for a star connection is shown in Fig 3.11 Also, considering the circuit diagram of Fig 3.8, the line and phase currents must be the same Hence, in star configuration, I L ϭ I ph (3.2) We now have another advantage of a 3-phase system compared with single-phase 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 CH003-H8747.indd 92 4/10/2008 3:24:52 PM Three-Phase A.C Circuits VBN 93 VRY VBR 30° 30° VRN 30° VYN VYB Fig 3.11 of about 25 kV By means of transformers, this voltage is stepped up to 400 kV for long distance transmission over the National Grid For more localised distribution, transformers are used to step down the line voltage to 132 kV, 33 kV, 11 kV, and 415 V The last three of these voltages are supplied to various industrial users The phase voltage derived from the 415 V lines is 240 V, and is used to supply both commercial premises and households Worked Example 3.1 Q A 415 V, 50 Hz, 3-phase supply is connected to a star-connected balanced load Each phase of the load consists of a resistance of 25 and inductance 0.1 H, connected in series Calculate (a) phase voltage, (b) the line current drawn from the supply, and (c) the power dissipated A Whenever a three-phase 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 phase of the load The figures for the other two phases and lines will be identical to these VL ϭ 415 V; f ϭ 50 Hz; Rph ϭ 25 ; Lph ϭ 0.1 H IL ϭ Iph 25 Ω Vph 0.1 H VL Fig 3.12 CH003-H8747.indd 93 4/10/2008 3:24:52 PM 94 Further Electrical and Electronic Principles (a) VL 415 ϭ 3 so Vph ϭ 240 V Ans Vph ϭ (b) Since it is possible to determine the impedance of a phase of the load, and we now know the phase voltage, then the phase current may be calculated: X L ϭ fL ohm ϭ ϫ 50 ϫ 0.1 hence X L ϭ 31.42 2 Z ph ϭ Rph ϩ X L ohm ϭ 252 ϩ 31.422 Z ph ϭ 40.15 I ph ϭ Vph Z ph amp ϭ 240 40.15 so I ph ϭ 5.98 A In a star-connected circuit, IL ϭ IPh therefore IL ϭ 5.98 A Ans The power in one phase, Pph ϭ I phRph watt ϭ 5.982 ϫ 25 Pph ϭ 893.29 W and since there are three phases, then the total power is: P ϭ ϫ Pph watt ϭ ϫ 893.29 hence P ϭ 2.68 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 3.13 The delta connection is not reserved for machine windings only, since a 3-phase load may also be connected in this way IL F Iph S Iph VL F IL Iph S F S IL Fig 3.13 CH003-H8747.indd 94 4/10/2008 3:24:53 PM 102 Further Electrical and Electronic Principles P ϭ V1I1 cos φ watt (c) ϭ ϫ 11 ϫ 103 ϫ 0.212 ϫ 0.75 therefore P ϭ 3.02 kW Ans Worked Example 3.7 Q The star-connected stator of a three-phase, 50 Hz alternator supplies a balanced delta-connected load Each phase of the load consists of a coil of resistance 15 and inductance 36 mH, and the phase voltage generated by the alternator is 231 V Calculate (a) the phase and line currents, (b) the load power factor, and (c) the power delivered to the load A f ϭ 50 Hz; R ϭ 15 ; L ϭ 36 mH; Vph ϭ 231 V I1 I2 V1 15 Ω 231 V V2 R L 36 mH Fig 3.23 (a) For the alternator: V1 ϭ Vph ϭ 231 V V2 ϭ VL ϭ Vph V2 ϭ 400 V I ph ϭ I L ϭ I1 For the load: V2 ϭ VL ϭ Vph ϭ 400 V X L ϭ fL ohm ϭ 100 ϫ ϫ 36 ϫ 10Ϫ3 X L ϭ 11.31 Z ph ϭ R ϩ X L ohm ϭ 152 ϩ 11.3 12 Z ph ϭ 18.79 I ph ϭ I ϭ Vph Z ph amp ϭ 400 18.79 I ϭ 21.29 A Ans I L ϭ I1 ϭ I ϭ ϫ 21.29 I1 ϭ 36.88 A Ans (b) p.f ϭ cos φ ϭ R Z 15 18.79 p.f ϭ 0.8 lagging Ans ϭ CH003-H8747.indd 102 4/10/2008 3:24:56 PM Three-Phase A.C Circuits (c) 103 P ϭ 3VL I L cos φ watt ϭ ϫ 400 ϫ 36.88 ϫ 0.8 P ϭ 20.4 kW Ans Alternatively, P ϭ ϫ Pph ϭ ϫ I R watt ϭ ϫ 21.292 ϫ 15 P ϭ 20.4 kW Ans 3.8 Measurement of Three-phase 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 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 phase angle (or power factor) of the load Thus the wattmeter reading indicates the true power, PϭVI cos φ watt If a three-phase load is balanced, then it is necessary only to measure the power taken by one phase 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 3.24 R W1 N B Y Fig 3.24 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 phase in turn The total power P ϭ P1 ϩ P2 ϩ P3, where P1, P2 and P3 represent the three separate readings CH003-H8747.indd 103 4/10/2008 3:24:56 PM 104 Further Electrical and Electronic Principles W1 R R R B Y Fig 3.25 For a delta-connected load, the procedure is not quite so simple The reason is that the phase current is not the same as the line current Thus, if possible, one of the phases must be disconnected to allow the connection of the wattmeter current coil This is shown in Fig 3.26 Again, if the load was unbalanced, this process would have to be repeated for each phase R W1 B Y Fig 3.26 3.9 The Two-Wattmeter Method The measurement of three-phase 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 3.27 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 CH003-H8747.indd 104 4/10/2008 3:24:57 PM Three-Phase A.C Circuits IR W1 R 105 VRB IB B VYB IY Y W2 Fig 3.27 Considering Fig 3.27, the following statements apply: Instantaneous power for W1 , p1 ϭ vRB iR watt and for W2 , p2 ϭ vYB iY watt total instantaneous power ϭ p1 ϩ p2 ϭ vRB iR ϩ vYB iY ………[1] Now, any line voltage is the phasor difference between the appropriate pair of phase voltages, hence: vRB ϭ vRN Ϫ vBN and vYB ϭ vYN Ϫ vBN and substituting these into eqn [1] yields: p1 ϩ p2 ϭ iR (vRN Ϫ vBN ) ϩ iY (vYN Ϫ vBN ) ϭ vRN iR ϩ vYN iY Ϫ vBN (iR ϩ iY ) but, iR ϩ iY ϭ ϪiB i.e the phasor sum of three equal currents is zero therefore p1 ϩ p2 ϭ vRN iR ϩ vYN iY ϩ vBN iB The instantaneous sum of the powers measured by the two wattmeters is equal to the sum of instantaneous power in the three phases Hence, total power, P ϭ P1 ϩ P2 ϭ PR ϩ PY ϩ PB watt (3.7) Consider now the phasor diagram for a resistive-inductive balanced load, with the two wattmeters connected as in Fig 3.27 This phasor diagram appears as Fig 3.28, below CH003-H8747.indd 105 4/10/2008 3:24:57 PM 106 Further Electrical and Electronic Principles IB VBN VBR VR φ VRN φ φ IY (30° ϩ φ) (30° Ϫ φ) VYN IR VRB VYB Fig 3.28 The power indicated by W1, P1 ϭ VL I L cos (30Њ Ϫ φ) (3.8) P2 ϭ VL I L cos (30 Њ ϩ φ) (3.9) and that for W2, From these results, and using Fig 3.29, the following points should be noted: wattmeter ϩ readings Ϫ90 Ϫ30 30 90 150 (30° Ϯ φ) Ϫ Fig 3.29 If the load p.f Ͼ0.5 (i.e φ Ͻ 60°); both meters will give a positive reading If the load p.f ϭ0.5 (i.e φ ϭ 60°); W1 indicates the total power, and W2 indicates zero If the load p.f Ͻ0.5 (i.e φ Ͼ 60°); W2 attempts to indicate a negative reading In this case, the connections to the voltage coil CH003-H8747.indd 106 4/10/2008 3:24:57 PM Three-Phase A.C Circuits 107 of W2 need to be reversed, and the resulting reading recorded as a negative value Under these circumstances, the total load power will be P ϭ P1ϪP2 The load power factor may be determined from the two wattmeter readings from the equation: ⎛P ϪP ⎞ 1⎟ ⎟ φ ϭ tanϪ1 ⎜ ⎜ ⎜P ϩP ⎟ ⎟ ⎜ ⎝ 1⎠ (3.10) hence, power factor, cos φ can be determined Worked Example 3.8 Q The power in a 3-phase balanced load was measured, using the two-wattmeter method The recorded readings were 3.2 kW and kW respectively Determine the load power and power factor A P1 ϭ 3.2 kW; P2 ϭ kW P ϭ P1 ϩ P2 watt ϭ (3.2 ϩ 5) kW therefore, P ϭ 8.2 kW Ans ⎛P ϪP ⎞ 1⎟ ⎟ φ ϭ tanϪ1 ⎜ ⎜ ⎜P ϩP ⎟ ⎟ ⎜ ⎝ 1⎠ ⎛ Ϫ ⎞ ⎟ ϭ tanϪ1 ⎜ ⎟ ⎜ ⎟ ⎜ ⎝ ϩ ⎠ hence, φ ϭ 20.82Њ and p.f ϭ cos φ ϭ 0.935 Ans Worked Example 3.9 Q A 3-phase balanced load takes a line current of 24 A at a lagging power factor of 0.42, when connected to a 415 V, 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 IL ϭ 24 A; cos φ ϭ 0.42; VL ϭ 415 V φ ϭ cosϪ1 0.42 ϭ 65.17Њ P1 ϭ VL I L cos (30 Њ Ϫ φ ) watt ϭ 415 ϫ 24 ϫ cos (Ϫ35.17Њ ) therefore, P1 ϭ 8.142 kW Ans P2 ϭ VL I L cos (30Њ ϩ φ ) watt ϭ 415 ϫ 24 ϫ cos (95.17Њ ) therefore, P2 ϭϪ896.7 W Ans P ϭ P1 ϩ P2 watt ϭ 8142 ϩ (Ϫ 896.7) hence, P ϭ 7.244 kW Ans To obtain the negative reading on W2, the connections to its voltage coil must have been reversed CH003-H8747.indd 107 4/10/2008 3:24:58 PM 108 Further Electrical and Electronic Principles Worked Example 3.10 Q A delta-connected load has a phase impedance of 100 at a phase angle of 55°, and is connected to a 415 V three-phase 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 Zph ϭ 100 ; φ ϭ 55°; VL ϭ 415 V ϭ Vph W1 IL Zph 100 Ω VL 415 V VL 415 V IL W2 Fig 3.30 I ph ϭ Vph Z ph amp ϭ 415 ϭ A 100 I L ϭ I ph ϭ ϫ 15 ϭ 19 A P1 ϭ VL I L c os (30Њ Ϫ φ ) watt ϭ 415 ϫ 7.19 ϫ cos Ϫ25Њ P1 ϭ 2.704 kW Ans P2 ϭ VL I L cos (30Њ ϩ φ ) watt ϭ 415 ϫ 7.19 ϫ cos 85Њ P2 ϭ 260 W Ans P ϭ P1 ϩ P2 watt ϭ 2704 ϩ 260 P ϭ 2.964 kW Ans 3.10 Neutral Current in an Unbalanced Three-phase Load We have seen that the neutral current for a balanced load is zero This is because the phasor sum of three equal currents, mutually displaced by 120°, is zero If the load is unbalanced, then the three line (and phase) 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 phasor sum of the line currents The technique is basically the same as that used previously, by resolving the phasors 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 CH003-H8747.indd 108 4/10/2008 3:24:58 PM Three-Phase A.C Circuits 109 Worked Example 3.11 Q An unbalanced, star-connected load is supplied from a 3-phase, 415 V source The three phase loads are purely resistive These loads are 25 , 30 and 40 , and are connected in the red, yellow and blue phases respectively Determine the value of the neutral current, and its phase angle relative to the red phase current A VL ϭ 415 V; RR ϭ 25 ; RY ϭ 30 ; RB ϭ 40 The circuit diagram is shown in Fig 3.31 IR R 25 Ω Vph VL 40 Ω IN IY 30 Ω Y B IB Fig 3.31 VL 415 ϭ 240 V volt ϭ 3 Vph Vph Vph amp IR ϭ amp; IY ϭ amp; I B ϭ RR RY RB Vph ϭ 240 25 I R ϭ 9.6 A ϭ 240 30 IY ϭ A ϭ ϭ 240 40 IB ϭ A The corresponding phasor diagrams are shown in Fig 3.32 IB 60 ° 60 ° IR H.C φ V.C IY IN Fig 3.32 CH003-H8747.indd 109 4/10/2008 3:24:58 PM 110 Further Electrical and Electronic Principles Horizontal components, H.C ϭ I R Ϫ IY cos 60Њ Ϫ I B cos 60Њ ϭ Ϫ ( ϫ 5) ϭ ( ϫ 5) so, H.C ϭ 2.6 A Vertical components, V.C ϭ I B sin 60Њ Ϫ IY sin 60Њ ϭ (6 ϫ 0.866 ) Ϫ (8 ϫ 0.866 ) so V.C ϭϪ1.732 A The neutral current, I N ϭ V.C.2 ϩ H.C.2 amp ϭ Ϫ1.7322 ϩ 2.62 hence, I N ϭ 3.124 A Ans V.C Ϫ1.732 ϭ tanϪ1 H.C hence, φ ϭϪ33.67Њ relative to I R Ans φ ϭ tanϪ1 3.11 Advantages of Three-phase Systems In previous sections, some of the advantages of three-phase systems, compared with single-phase 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 The whole of the stator of a three-phase machine is utilised A single phase alternator utilises only two thirds of the stator slots A three-phase alternator has a better distribution factor A three-phase, four-wire system provides considerable savings in cable costs, for the distribution of an equivalent amount of power A three-phase, four-wire system provides alternative voltages for industrial and domestic users For a given machine frame size, the power output from a three-phase machine is greater than that from a single-phase machine A three-phase supply produces a rotating magnetic field; whereas a single-phase supply produces only a pulsating field Three-phase motors are inherently self-starting; whereas single-phase motors are not The torque produced by a three-phase motor is smooth; whereas that produced by a single-phase motor is pulsating CH003-H8747.indd 110 4/10/2008 3:24:59 PM Three-Phase A.C Circuits 111 Summary of Equations Phase and line quantities: In star, IL ϭ Iph; VL ϭ 3Vph In delta, VL ϭ Vph; IL ϭ 3I ph Power dissipation: P ϭ 3VL I L cos φ watt P ϭ ϫ Pph ϭ ϫ Vph I ph cos φ ϭ ϫ I ph R ph watt and for the same load, P(delta) ϭ ϫ P (star) Power measurement: For two wattmeter method: P ϭ P1 ϩ P2 watt ⎛P ϪP ⎞ 1⎟ ⎟ Load phase angle, φ ϭ tanϪ1 ⎜ ⎜ ⎜P ϩP ⎟ ⎟ ⎜ ⎝ 1⎠ CH003-H8747.indd 111 4/10/2008 3:24:59 PM 112 Further Electrical and Electronic Prinicples Fundamental Electrical and Electronic Principles Assignment Questions If the load specified in Question above is now connected in delta, determine the values for phase and line currents A star-connected alternator stator generates 300 V in each of its stator windings (a) Sketch the waveform and phasor diagrams for the phase voltages, and (b) calculate the p.d between any pair of lines A balanced three-phase, delta-connected load consists of the stator windings of an a.c motor Each winding has a resistance of 3.5 and inductance 0.015 H If this machine is connected to a 415 V, 50 Hz supply, calculate (a) the stator phase current, (b) the line current drawn from the supply, and (c) the total power dissipated Three identical coils, connected in star, take a total power of 1.8 kW at a power factor of 0.35, from a 415 V, 50 Hz supply Determine the resistance and inductance of each coil Three inductors, each of resistance 12 and inductance 0.02 H, are connected in delta to a 400 V, 50 Hz, three-phase supply Calculate (a) the line current, (b) the power factor, and (c) the power consumed 10 Three coils are connected in delta to a 415 V, 50 Hz, three-phase 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-phase 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-phase system, supplying a balanced load, indicate 10.6 kW and Ϫ2.4 kW respectively Calculate (a) the total power consumed, and (b) the load phase angle and power factor State the significance of the negative reading recorded 13 Using the two-wattmeter method, the power in a three-phase 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-phase, star-connected load, consists of a coil of resistance and reactance This load is connected to a 400 V, 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-phase, 415 V, 50 Hz supply is connected to an unbalanced, star-connected load, having a power factor of 0.8 lagging in each phase The currents are 40 A in the red phase, 55 A in the yellow phase, and 62 A in the blue phase Determine (a) the value of the neutral current, and (b) the total power dissipated Repeat the calculations for Question 4, when the stator windings are connected in star configuration CH003-H8747.indd 112 If the transformer phase voltage is 600 V, calculate (a) the p.d across each phase of the load, (b) the load phase current, (c) the current in the transformer secondary windings, and (d) the power and power factor A three-phase load is connected in star to a 400 V, 50 Hz supply Each phase of the load consists of a coil, having inductance 0.2 H and resistance 40 Calculate the line current The star-connected secondary of a three-phase transformer supplies a delta-connected motor, which takes a power of 80 kW, at a lagging power factor of 0.85 If the line voltage is 600 V, calculate (a) the current in the transformer secondary windings, and (b) the current in the motor windings A star-connected load, each phase of which has an inductive reactance of 40 and resistance of 25 , is fed from the secondary of a three-phase, delta-connected transformer 4/10/2008 3:24:59 PM Fundamental ElectricalThree-Phase A.C Circuits and Electronic Principles 113 113 Suggested Practical Assignments Assignment To determine the relationship between line and phase quantities in three-phase systems Apparatus: Low voltage, 50 Hz, three-phase supply ϫ k rheostats ϫ ammeters ϫ voltmeters (preferably DMM) Method: Using a digital meter, adjust each rheostat to exactly the same value (1 k ) Connect the rheostats, in star configuration, to the three-phase supply Measure the three line voltages and the corresponding phase voltages, and record your results in Table Measure the three line currents, and record these in Table Check that the neutral current is zero Carefully unbalance the load by altering the resistance value of one or two of the rheostats ENSURE that you not exceed the current ratings of the rheostats Measure and record the values of the three line currents and the neutral current, in Table Switch off the three-phase supply, and disconnect the circuit Carefully reset the three rheostats to their original settings, as in paragraph of the Method 10 Connect the rheostats, in delta configuration, to the three-phase supply, with an ammeter in each line 11 Switch on the supply, and measure the line voltages and line currents Record your values in Table 12 Switch off the supply, and connect the three ammeters in the three phases of the load, i.e an ammeter in series with each rheostat This will involve opening each phase and inserting each ammeter 13 Switch on and record the values of the three phase currents Record values in Table 14 From your tabulated readings, determine the relationship between line and phase 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 phasor diagrams, and calculations State whether the line and phase relationships measured conform to those expected (allowing for experimental error) Assignment To measure the power in three-phase systems, using both single and twowattmeter methods Apparatus: Low voltage, 50 Hz, three-phase supply ϫ k rheostats ϫ wattmeters ϫ DMM CH003-H8747.indd 113 4/10/2008 3:24:59 PM 114 Further Electrical and Electronic Principles Method: As for Assignment 1, carefully adjust the three rheostats to the same value (1 k ) Connect the circuit shown in Fig 3.33, and measure the power in the red and yellow phases W1 R N Y W2 B Fig 3.33 Transfer one of the wattmeters to the blue phase, and measure that power Add the three wattmeter readings to give the total power dissipation Check to see whether this is three times the individual phase power Switch off the power supply, and reconnect as in Fig 3.34 W1 R B Y W2 Fig 3.34 Record the two wattmeter readings, and check whether their sum is equal to the total power recorded from paragraph above Switch off the power supply and reconnect the circuit as in Fig 3.35 Record the two wattmeter readings Switch off the supply, and transfer one of the wattmeters to the third phase 10 Record this reading Add the three readings to give the total power, and check that this is three times the phase power 11 Switch off the supply and connect the circuit of Fig 3.36 12 Record the two wattmeter readings, and check that their sum equals the total power obtained from paragraph 10 above CH003-H8747.indd 114 4/10/2008 3:25:00 PM Three-Phase A.C Circuits R 115 W1 W2 Y B Fig 3.35 W1 R B Y W2 Fig 3.36 Assignment 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 CH003-H8747.indd 115 4/10/2008 3:25:00 PM CH003-H8747.indd 116 4/10/2008 3:25:00 PM ... W and 60 W respectively Calculate (a) the power, and (b) the power factor 14 Each branch of a three- phase, star-connected load, consists of a coil of resistance and reactance This load is connected... connected to a star-connected balanced load Each phase of the load consists of a resistance of 25 and inductance 0.1 H, connected in series Calculate (a) phase voltage, (b) the line current drawn from... Electrical and Electronic Principles Worked Example 3.5 Q A balanced delta-connected load takes a phase current of 15 A at a power factor of 0.7 lagging when connected to a 115 V, 50 Hz, three- phase