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Contents Introduction 3 Interconnected systems 4 Fault conditions 14 Blown supply fuse (star load) 14 Blown supply fuse (delta load) 21 Reversed phase winding (star load) 27 Reversed phase winding in delta 29 Open delta connection 31 Summary 34 Answers 40 EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 1 2 EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 Introduction The power source used to supply loads, whether single-phase, two-phase or three-phase, is typically a star connected system. This system allows for any of the three types of loads to be connected. For single-phase and two-phase loads, the relationship of line to phase values is as described for a star system. For three-phase star connected loads, the relationship of line to phase values is the same for the load as for the supply. However, for three-phase delta connected loads, there is a complication, as the load phase values will not equal the supply phase values. This is addressed in the first part of this section. Later in the section we will look at various fault conditions that may arise. At the end of this section you should be able to: draw the typical combinations of three phase interconnected systems using star-connection and delta-connection show the relationship between line and phase voltages and line and phase currents in the typical interconnected systems using starconnections and delta-connections give example of loads in typical power systems state the limitations and uses of open delta connections state the effect of a reversed phase winding of a delta connected transformer EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 3 Interconnected systems This term applies to systems which have one type of arrangement, star or delta, for power delivery, and the other type at the load. Figure 1 shows star supply–delta load. Figure 2 shows delta supply–star load. Figure 1: Star supply-delta load Figure 2: Delta supply-star load 4 EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 As you can see from Figures 1 and 2, there is no connection point for a neutral. The first impression is that the loads connected in this type of system must be balanced, and usually they are, but there are some instances where the load is unbalanced. Typically, the star system is used for the supply, with the three-phase loads being in either delta or star. If the supply is in star, any type of load may be connected. If the phase load is unbalanced, the supply must be in star. The types of loads that can be connected to a star supply were discussed in Section 3. They are: single-phase loads where the supply is one active and a neutral (eg appliances such as toasters) single-phase loads where the supply is two actives (eg welders) two-phase loads, where the supply to the load is two actives and a neutral (eg cooking ranges) balanced three-phase loads, where the supply is three actives (eg three-phase motors) unbalanced three-phase loads, where the supply is three actives and a neutral (eg three-phase and one-phase load combinations). Some of these loads are shown in Figure 3. Figure 3: Connection of loads to three-phase four-wire supply To determine the phase and line values of current and voltage for either the load or the supply, we use the methods already discussed in this module. EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 5 Example 1 A three-phase star connected transformer develops 240 V in each phase winding and supplies a three-phase delta connected motor. The impedance of each stator windings is 24 . Determine the: (a) line potential of the supply (b) phase potential at the load (c) phase currents in the load (d) line currents from the supply (e) phase currents in the supply. Solution Step 1 Sketch the circuit (Figure 4). Figure 4: Circuit for example 1 Step 2 Determine the line potential of the supply. Remember that in star, VLine 3 Vphase As the supply is always balanced, VAB VBC VCA 415 V 6 EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 Step 3 Determine the phase potential at the load. Remember that in delta, VLine V phase and that as the voltages supplied to the load are balanced, the phase voltages must also be balanced. VAB VA 415 V VBC VB 415 V VCA VC 415 V Step 4 Determine the phase currents using Ohm’s law. I phase V phase Z phase 415 24 17.3 A As each phase has the same impedance, I A I B IC 17.3 A Step 5 Determine the line currents from the supply. Remember that in a balanced delta load I line 3 I phase , and that as the phase currents in the load are equal, the line currents supplied to the load are also equal. I A 3 I AB 3 17.3 30 A I B 3 I BC 3 17.3 30 A I C 3 I CA 3 17.3 30 A Step 6 Determine the phase currents in the supply. Remember that in star I line I phase , and that as the line currents supplied to the load are equal, the phase currents in the supply are also equal. I A phase I A line 30 A I B phase I B line 30 A I C phase I C line 30 A EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 7 If the delta connected load is unbalanced the problem is still solved in the same way, the only difference being that the phase currents will not be equal and the line currents will not be equal. Example 2 In this example we will consider an extremely unbalanced load where each phase has a different impedance and a widely different phase angle. You will not be examined with this type of problem, but it will help you to understand what the effect of an unbalanced load is. Work through the example step by step. The end result provides a good reason why threephase loads should be as balanced as possible. A three-phase star connected transformer develops 240 V, 50 Hz in each phase winding and supplies a three-phase delta connected load. The load is made up of a 20 resistor, a 25 impedance with a power factor of 0.8 lagging, and a 160 µF capacitor. Determine the: (a) line potential of the supply (b) phase potential at the load (c) phase currents in the load (d) line currents from the supply (e) phase currents in the supply. Solution Step 1 Sketch the circuit (Figure 5). Step 2 Determine the line potential of the supply. Recall that in star, VLine 3 Vphase VAB 3 VA 3 240 415 V As the supply is always balanced, VAB VBC VCA 415 V 8 EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 Figure 5: Circuit for example 2 Step 3 Determine the phase potential at the load. Remember that in delta, VLine V phase , and that as the voltages supplied to the load are balanced, the phase voltages must also be balanced. VAB VA 415 V VBC VB 415 V VCA VC 415 V Step 4 Determine the phase impedances. Z A RA2 X A2 202 02 20 Z B 25 ZC RC 2 X C 2 Now: XC 1 2 fC 1 2 50 160 10 –6 19.9 Z 02 19.92 C 19.9 EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 9 Step 5 Determine the phase angle of the current using the equation I V – Z : VBC ZBC 00 0 IBC VBC – ZBC –120–36.8 –156.8 ICA VCA –ZCA 120–(–90) 210 Step 6 Determine the phase currents in the load using Ohm’s law. I phase V phase Z phase As each phase has the same voltage, I AB VAB Z Ab 415 20 20.75 A V BC Z BC I BC 415 25 16.6 A V CA Z CA I CA 415 19.9 20.8 A Step 7 Determine the line currents from the supply. Remember that in an unbalanced delta load, we have I A I AB – I CA I B I BC – I AB I C I CA – I BC 10 EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 and that the calculations must be done using a phasor diagram. Select a suitable scale—we will use 5 A10 mm. The scaled values are 20.75 10 5 41.5 mm at 0 16.6 10 5 33.2 mm at –156.8 20.8 10 5 41.6 mm at 210 I AB I BC I CA Step 8 Draw the phasor and add the phase currents to determine the line current. This is shown in Figure 6. Figure 6: Phasor diagram for example 2 Step 9 Measure the line currents. I A 80 mm I B 73 mm I C 9.5 mm Convert the scaled values of line current to real current values in amperes: EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 11 80 5 10 40 A 73 5 IB 10 36.5 A 9.5 5 IC 10 4.7 A IA Step 10 Determine the phase currents in the supply. Remember that in star, I Line I phase , and that as the line currents supplied to the load are equal, the phase currents in the supply are also equal. I A phase I A line 40 A I B phase I B line 36.5 A I C phase I C line 4.7 A The results show two obvious problems with connecting this load to the supply. 12 There are two heavily loaded phases in the supply, while the third phase has very little load. This gives rise to problems at the point of supply. The line current in line A and line B (the two most heavily loaded lines) exceeds the nominal amount of 3 times the phase current. This can cause extra heating of the supply cables and will increase voltage drop in those two lines, so that the line terminal voltage at the load could drop at one or two points. This may cause the load to malfunction. EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 Student exercise 1 A three-phase star connected transformer develops 415 V between the line terminals and supplies a three-phase delta connected motor. The impedance of the stator windings is 15 . Determine the: (a) phase potential of the supply _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ (b) phase potential at the load _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ (c) phase currents in the load _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ (d) line currents from the supply _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ (e) phase currents in the supply. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ Check your answers with those given at the end of the section. EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 13 Fault conditions There are a number of fault conditions that can arise within a three-phase system. They can be loosely classified as follows: blown supply fuse reversed phase winding in star reversed phase winding in delta open delta condition. Blown supply fuse (star load) When looking at this problem, we must begin by determining what type of load—star or delta—is connected to the supply system. They react differently when the supply loses one of the three phases. How the load reacts also depends on whether the supply is three-phase three wire or threephase four wire. We will consider how to deal with each case, beginning with star loads. If a three-phase three wire supply loses one line to a star connected load, the load will appear to the supply as a single phase load, with one phase of the load having no power supplied to it. This is illustrated in Figure 7, where line A is open circuit to the load. Figure 7: Supply fault to star load (three-phase three-wire) 14 EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 If we change the diagram in Figure 7 so that only the phases supplied with voltage are shown, we can see that the result of an open circuit in the line, or a blown supply fuse, is that two phases of the load are in series with each other. This is shown in Figure 8. Figure 8: Simplified diagram of fault condition In the circuit in Figure 8 the B phase and the C phase in the load are connected in series with each other, and there is line voltage applied to this combination. The line voltage will be divided proportionally across the two load impedances, and both load phases will have the same current flowing in them, regardless of whether they are balanced or unbalanced. The result is that phase voltages across the loads may be different, depending on whether the load is balanced or unbalanced. In this text, we will use balanced loads. Example 3 Three 24 heating elements are connected in star configuration to a threephase three wire 415 V, 50 Hz supply. If the fuse in line B blows, determine the: (a) voltage across each phase of the load (b) phase currents in the load (c) line currents supplied to the load. Solution Step 1 Draw the circuit as in Figure 9, circling or highlighting the open circuit fuse. EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 15 Figure 9: Circuit diagram for example 3 Step 2 Redraw the circuit as in Figure 10 to show the phases still supplied with voltage. Figure 10: Redrawn circuit for example 3 Step 3 Determine the total impedance of the load. For a series circuit use the equation Zt Z1 Z 2 . Zt Z A ZC As the load is resistive, Z equals R. Z A RA 24 Z C RC 24 Zt 24 24 48 16 EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 Step 4 Determine the phase voltages using the voltage divider equation: VA Vt Z A Zt 415 24 48 207.5 V VC Vt Z C Zt 415 24 48 207.5 V VB = 0 V, as there is no current path. Step 5 Determine the phase currents using Ohm’s law. I phase IA V phase Z phase VA ZA 207.5 24 8.6 A As the two phases are in series, the current in the A phase is the same as the current in the C phase. I A IC 8.6 A Step 6 Determine the line currents. In a series circuit there is only one current: I line I phase 8.6 A If the supply to the load is a three-phase four-wire star system and one line becomes open circuit, the load will not be classed as a single phase load. There will be a potential from each line terminal of the load still supplied with power to the neutral point in the load. This is shown in Figure 11. EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 17 Figure 11: Fault in supply to star load (three-phase four-wire) The two phases of the load still connected to the supply will each react individually as if there was no problem. If we look at the circuit of Figure 11, we can see that the phase voltages across the two phases supplied with power are unchanged by the loss of one phase. Since the phase voltages in those phases are unchanged, the current flowing in them is also unchanged. However, the current in the neutral conductor will now be the phasor sum of two currents, not three. This means that the neutral current will acquires a substantial value. Example 4 Three 24 heating elements are connected in star configuration to a threephase four wire 415 V, 50 Hz supply. If the fuse in line C blows, determine the: (a) voltage across each phase of the load (b) phase currents in the load (c) line currents supplied to the load (d) neutral current. Solution Step 1 18 Draw the circuit as in Figure 12 and circle or highlight the open circuit fuse. EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 Figure 12: Circuit for example 4 Step 2 Determine the phase voltages using the equation Vline 3 V phase VAB 3 240 VA VB VA 240 V Step 3 Determine the phase currents using Ohm’s law. I phase IA V phase Z phase VA ZA 240 24 10 A IB VC ZC 240 24 10 A IC 0 A EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 19 Step 4 Determine the line currents. As the load is star connected, the line currents must equal the phase currents. I line I phase 10 A Step 5 Determine the neutral current. Remember that this must be done using a phasor diagram. I N I A I B IC Because the impedance of each phase is purely resistive, there is no phase separation between the phase currents and their corresponding phase voltages. Therefore: A 0 B –120 C 120 The currents may therefore be written as: I A 10 A at 0 I B 10 A at –120 I C 0 A at 120 Convert the current values to lengths, using a scale of 2 A10 mm. 10 10 2 50 mm at 0 IA 10 10 2 50 mm at –120 IB 0 10 2 0 mm at 120 IC Draw the phasor diagram as in Figure 13, and add the phase quantities to determine the neutral current value. 20 EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 Figure 13: Phasor circuit drawn for example 4 Measure the length of the neutral phasor: I N 50 mm and convert it to a real value. 50 2 10 10 A IN As you can see, the phase values and the line values of current and voltage remain the same, but the neutral current increases from 0 A (because it was a balanced load initially) to 10 A. Blown supply fuse (delta load) As there is no connection point for a neutral conductor in a delta load, the only type of supply available to it is three-phase three-wire. If the supply loses one line to a delta connected load, the load, like the star load just discussed, will appear to the supply as a single phase load. This is shown in Figure 14, where line A is open circuit to the load. EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 21 Figure 14: Fault in supply to delta load If we draw the circuit in Figure 14 so that only the lines supplying voltage to the load are shown, as in Figure 15(a), we can see that the result of an open circuit in the line or a blown supply fuse is that two phases of the load are placed in series with each other, and that this series combination is in parallel with the third phase. This is shown more clearly in Figure 15(b). Figure 15: Simplified diagram of delta load fault condition As you can see, if the fuse in line A blows, the A phase and the C phase in the load become connected in series with each other and this combination is 22 EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 connected in parallel with the B phase. The circuit has line voltage applied to the combination. The result of this is that the load phase voltages for the A phase and the C phase are equal, and that the load phase current in the A phase is the same as that in the C phase. Remember a series circuit is also called a voltage divider circuit, because the line voltage will be divided proportionally across the load impedances. As a delta load (a three-phase motor for example) is typically balanced, VA VC 0.5 VB . For the B phase, the phase voltage is equal to the line voltage, and the phase current is the normal load current. The line current will decrease from 3I phase to 1.5 I phase in two lines and drop to zero in the line with the blown fuse. Example 5 Three 24 heating elements are connected in delta to a three-phase three wire 415 V, 50 Hz supply. If the fuse in line B blows, determine the: (a) voltage across each phase of the load (b) phase currents in the load (c) line currents supplied to the load. Solution Step 1 Draw the circuit as in Figure 16, and circle or highlight the open circuit fuse. EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 23 Figure 16: Circuit for example 5 Step 2 Redraw the circuit as in Figure 17 to show the situation more clearly. Figure 17: Redrawn circuit for example 5 Step 3 Determine the impedances of the load. For the series circuit use the equation Zt Z1 Z 2 As the load is a heating load, Z is equal to R, since the inductive reactance (XL) is 0 . Z A RA Z C RC Z AC RA RC 24 24 48 Z B 24 Step 4 Determine the phase voltages for the series phases of the load using the voltage divider equation: 24 EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 ZA Vt Z A Zt 415 24 48 207.5 V VC Vt Z C Zt 415 24 48 207.5 V VB 415 V Determine the phase currents using Ohm’s law. Step 5 I phase IA V phase Z phase VA ZA 207.5 24 8.6 A As the two phases are in series, the current in the C phase of the load is the same as the current in the A phase of the load. I C I A 8.6 A IB VB ZB 415 24 17.3 A Step 6 Determine the line currents. As the load now appears to be a single phase load, and in a parallel circuit the total current is the sum of the branch currents, the line currents must equal the sum of the branch (phase) currents: I line I A I B 8.6 17.3 25.9 A EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 25 Student exercise 2 1 Three 41.5 heating elements are connected in delta configuration to a three-phase three-wire 415 V, 50 Hz supply. If the fuse in line C blows, determine the: (a) voltage across each phase of the load _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ (b) phase currents in the load _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ (c) line currents supplied to the load. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 2 Three 41.5 heating elements are connected in star configuration to a three-phase four wire 415 V, 50 Hz supply. If the fuse in line B blows, determine the: (a) voltage across each phase of the load _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ (b) phase currents in the load _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ (c) line currents supplied to the load 26 EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ (d) neutral current. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ Check your answers with those given at the end of the section. Reversed phase winding (star load) We know that the voltages developed in a three phase star winding are displaced by 120°. However, this will only occur if the correct method of connection is used. That method is to connect either all the starts or all the finishes of the windings together to form the star point. This is shown in Figure 18. Figure 18: Correct star connections This connection allows each of the phase voltages to be displaced by 120°, as shown in Figure 19. EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 27 Figure 19: Phase voltages for correct connection If, however, one of the phases is reversed in an alternator or supply transformer, the resultant phasor causes the line voltages to change in both magnitude and phase displacement. This is illustrated in Figure 20, which shows the C phase as the reversed phase, and in Figure 21, which shows the phasor addition to determine the line voltages. Figure 20: Phase voltages with C phase winding reversed 28 EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 Figure 21: Line voltages with reversed winding As you can see from Figure 21, only one line potential is correct (VBC), with the other two line potentials (VAB and VCA) lower than normal. This can prevent the load from functioning as intended. A polarity check must be performed on the windings before connection. This is covered in the resource module Transformers. Reversed phase winding in delta The voltages developed in a three phase delta winding are also displaced by 120°, provided again that the correct method of connection is used. The method is to connect the start of one winding to the finish of the previous winding to form a closed loop called a delta connection. This is shown in Figure 22. Figure 22: Correct delta connection This connection allows each of the phase voltages to be displaced by 120° as shown in Figure 23. The phasor sum of these voltages is 0 V. EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 29 Figure 23: Phase voltages for correct connection If, however, one of the phases is reversed in a supply transformer, the phase voltages, when added on the phasor, will total twice the normal phase voltage value. This is illustrated in Figure 24, which shows the reversed phase, and Figure 25, which shows the phasor addition to determine the total voltage. Figure 24: One winding reversed 30 EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 Figure 25: Phase voltages with reversed winding The large resultant voltage only exists within the delta connection, and as the windings are of low impedance, this will create a very large circulating current in the transformer. The result will be that the windings will burn out. To find out whether the connection has been made correctly, a simple but effective method is to break the delta loop and place a voltmeter across the break. If the voltmeter reads zero, the connection is correct. If the voltmeter reads twice Vphase, then one phase in the delta connection is reversed. This is illustrated in Figure 26. Figure 26: Checking for correct connections Open delta connection In the open delta connection, one phase of a delta connected load is open circuited, either accidentally or deliberately. This is shown in Figure 27. It should not be mistaken for a blown line fuse. EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 31 Figure 27: Open delta load A three phase supply to a load can be obtained using two single phase transformers. The circuit for the open delta using two single phase transformers is shown in Figure 28. In this circuit, the two primaries are connected in series. Three terminals are thus created, one at the unconnected end of each primary and one at the junction of the primaries. The load is connected in a similar fashion to series connected secondaries. Figure 28: Open delta supply This type of connection has the disadvantage of not producing a symmetrical supply, and so it does not have the same power capabilities as a star or delta system. The available power from the open delta system is 57.7% of what would be available from a supply connected in delta with three windings in the supply transformer. For this reason, the system is rarely used. More often an open delta connection is produced as the result of an accident; for example, where one winding of a motor burns out. If the load becomes open delta connected, the supply (three phase three wire) is unaffected. The 32 EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 line potential is also unaffected, but the value of the line currents now equals that of the phase currents, and the normal relationship of I line 3 I phase , does not apply. As one phase of the load is not functioning, the load output is reduced to 57.7% of the rated delta connected output. The result of this is that the machine is placed into an overload condition with normal full load. The machine therefore heats up, with the possibility that further damage will occur. EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 33 Summary Interconnected systems are systems with either a star supply and a delta load or a delta supply and a star load. The most common supply system is star connected. The types of loads that can be connected to a star connected system are: – – – – – single-phase loads where the supply is one active and a neutral single-phase loads where the supply is two actives two-phase loads where the supply is two actives and a neutral balanced three-phase loads where the supply is three actives unbalanced three-phase loads where the supply is three actives and a neutral In star, VLine 3 Vphase and I Line I phase . In delta, VLine V phase and I Line 3 I phase . To determine the current drawn from the supply, the load current value(s) must first be determined, using Ohm’s law. There are a number of fault conditions: – – – – 34 blown supply fuse reversed phase winding in star reversed phase winding in delta open delta condition For three-wire star connected supplies, if the load is star connected and a line fuse blows, the two remaining phases of the load revert to a single phase load supplied with line potential. The voltage across each load component will be one half of line voltage. For four-wire star connected supplies, if the load is star connected and a line fuse blows, the load will become a two phase load. The voltage across each phase of the load will be normal for two phases and zero for the third. If the load is delta connected and a line fuse blows, the load will be referred to as a single phase load, with two original phases of the load in series and this series arrangement in parallel with the third original phase of the load. If the load is balanced, the voltage across EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 one original phase of the load will be line voltage, while the other two original phases will have one half line voltage. A reversed phase winding in star causes the line potential for two lines to decrease, while the third line value is unaltered. This will cause the load to operate at a decreased level or rate. A reversed phase winding in delta causes the total voltage to alter from zero volts to twice line potential. This will result in a very large circulating current in the supply transformer, creating heat and eventually resulting in its destruction. A voltmeter can be used to determine if a supply transformer has a reversed winding. Open delta connections may be created deliberately but are typically the result of a fault in the load. The result of an open delta connection is that the load output is reduced to 57.7% of the rated output. In open delta, the line current equals the phase current. EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 35 Check your progress In questions 1 to 6, place the letter matching your answer in the brackets provided. 1 The resultant voltage around the closed loop of a delta connected transformer with one phase winding reversed is: (a) zero (b) half the phase voltage (c) equal to the phase voltage (d) twice the phase voltage. 2 ( ) The angle between line voltages obtained from a star connected transformer with one phase reversed will be: (a) 0 degrees (b) 30 degrees (c) 120 degrees (d) unequal. 3 ( ) The magnitude of line voltages of a star connected alternator with one winding reversed will be: (a) equal (b) unequal (c) 1.73 times phase voltage (d) the phasor sum of the phase voltages. 4 ( ) If a delta connected load has one supply line open circuited then there will be: (a) zero current in all phases (b) zero current in one phase (c) equal current in all phases (d) equal current in two phases. 5 ( ) The available power from an open delta connected transformer compared to a delta connected transformer is: (a) 33.3% (b) 50% (c) 57.7% (d) 66.6%. 36 ( ) EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 6 The line current of an open delta system is: (a) equal to the phase current (b) 1.73 times the phase current (c) twice the phase current (d) half the phase current. 7 ( ) A three-phase star connected transformer develops 240V in each phase winding and supplies a three-phase delta connected motor. The motor stator winding impedances are each 20 Ω. Determine the line current taken by the motor when it is started. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 8 If the motor in question 7 is reconnected in star determine the new line current taken by the motor at start. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 9 A delta connected heater is supplied from a three-phase, three-wire 415V 50Hz system. Each element has a resistance of 24 Ω. If the fuse in line C blows, determine the current in each of the heating elements. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 37 10 Three 24 ohm heating elements are connected in star to a 415V three phase supply. Assuming the fuse in line C blows then determine the load phase voltages and line currents for a: (a) 4-wire supply system (b) 3-wire supply system _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 11 A heater on low heat has three elements connected in star to a 415 volt 3-phase supply and takes a current of 10 A per phase. Calculate the line current when the elements are connected in delta for high heat. _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 38 EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 12 Draw the phasor diagram (tip to tail) of the phase voltages of a delta connected supply with one winding reversed. 13 Draw the phasor diagram of the line voltages of a star connected system with one winding reversed. Answers to Check your progress are at the end of the section. EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 39 Answers Student exercise 1 (a) VA = 240 V, VB = 240 V, VC = 240 V (b) VAB = 415 V, VBC = 415 V, VCA = 415 V (c) IAB = 27.7 A, IBC = 27.7 A, ICA = 48 A (d) IA = 48 A, IB = 48 A, IC = 48 A (e) IA = 48 A, IB = 48 A, IC = 48 A Student exercise 2 1 (a) VAB = 415 V, VBC = 207.5 V, VCA = 207.5 V (b) IAB = 10 A, IBC = 5 A, ICA = 5 A (c) IA = 15 A, IB = 15 A, IC = 0 A 2 (a) VA = 240 V, VB = 0 V, VC = 240 V (b) IA =5.8 A, IB = 0 A, IC = 5.8 A (c) IA =5.8 A, IB = 0 A, IC = 5.8 A (d) IN = 5.8 A Check your progress 1 (d) 2 (d) 3 (b) 4 (d) 5 (c) 40 EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 6 (a) 7 35.9 A 8 12 A 9 IA and IC both = 8.64 A, IB = 17.29 A 10 (a) VA = 240V, VB = 240V, VC = 0 V IA = 10 A, IB = 10 A, IC = 0 A (b) VA = 207.5 V, VB = 207.5 V, VC = 0 V IA = 8.64 A, IB = 8.64 A, IC = 0 A 11 IL = 30 A 12 13 EGG202A: 13 Analyse interconnected systems NSW DET 2017 2006/060/05/2017 LRR 4977 41