How To Wire A Three Phase, Single Phase, Wye Transformer

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2 Three-Phase Transformers When more power is needed - three transformers can be tied together. This is called three-phase. Here s a simple way of comparing single-phase to threephase power. Single-Phase Three-Phase Think of single-phase as one guy driving a stake into the ground; Three-phase as three guys, working together, driving the same stake into the ground. 25

3 Generating Three-Phase Three-phase power is delivered from a generator with three separate windings. These windings are equally spaced around a rotating cylinder (rotor) with each winding occupying one-third of the rotor circumference. Three-phase generator windings are spaced 120 apart for a total of Because the three source voltages timed at three different intervals, they can be transmitted over just three conductors (rather than the six conductors needed to transmit three separate single-phase loads) A three-phase system can produce 173% higher effective voltage than a single-phase system. This is very useful for higher power loads such as large motors. 26

4 We calculate the total voltage (of all three lines of a three-phase transformer) by multiplying the voltage of each single winding (phase) by the square root of 3 (!3). The reason for this is the angles of each phase are always 120 apart By using this trigonomic formula (don t get scared) we can see where we derive the square root of three. sin = =!3 sin Wye Connection One way of tying three transformers together is in parallel (+ to +) and (- to -). This is called a Wye connection. Here s how a Wye connection is wired: Line Phase eutral 27

5 We call each of the single-phase transformer windings Phases.The combination (hooked together) of the three single-phase transformers (-- ) is called the Line. We really only draw three-phase from the Line side. Also, Line Amps on a Wye connected transformer equals Phase Amps. It would seem that current would not remain the same in a parallel circuit. But, three transformers wired in parallel are not like three resistors in parallel. Transformers produce, instead of consume voltages, so the opposite rules would would apply in this case. A more simplified representation of a Wye looks like this: A B C If we tie together three single-phase, olt, transformers in a Wye configuration our output voltage would look like this: 208 v 3Ø Why do we get 208 volts, three-phase on the Line side (--)? Just multiply olts times the square root of 3 (!3). Remember phase angles of three olt sources. olts x (!3) = 208 volts 28

6 We can also get 208 volts, single-phase on the Line side of this transformer by connecting (-); "A" 208 v 1Ø "C" "B" As you see, we are connecting only two of the three single-phase transformers available. We still only get 208 volts (120 x!3) from this connection, and it is just single-phase. But, we have the capability of using two transformers ( A & B ) to help share the load. Also, we can use two other possible connections (-) and (-) to feed other 208 volt, single-phase loads. Guess what! Yes, it s even possible to get olts, single-phase out of the same transformer. "A" "C" "B" 1Ø Remember, we have the benefit of a neutral conductor and, by connecting (-) we can feed olt, single-phase loads. And, seeing that we are only using one single-phase transformer ( B ), that leaves two other possibilities (- ) and (-) for serving other olt circuits. What s also very useful is that we can balance three olt circuits on just one neutral. In other words, A 208/olt (Wye configured) transformer could supply olt, single-phase equipment (like fluorescent lights), 208 volt, single-phase 29

7 equipment (like range or oven), and 208 volt, three-phase equipment (like heavy machinery). This is what makes three-phase so great! Delta Connections Another way of tying three transformers together is in series (+ to -). This is called a Delta connection. Here s how a Delta connection is wired: Line Phase A more simplified representation of a Delta configuration looks like this: A B C 30

8 By connecting three (240 volt) single-phase transformers (Delta) we get 240 volts, three-phase. We don t multiply by!3 because Line volts equals Phase volts on Delta transformers. It would seem that voltage, would not remain the same in a series type circuit. But, three transformers wired in series are not like three resistors in series. Transformers produce, (instead of consume) voltages, so the opposite rules would would apply in this case. 240 v 240 v 240 v 240 v 3Ø Connect (-) and you ll get 240 volts, single-phase. Connecting (-) or (-) will also give us 240 volts, single-phase. otice how three different 240 volt (1Ø) loads utilize three different transformers. 240 v 240 v 240 v 240 v 1Ø We can still get olts, (single-phase) out of a 240 volt Delta system, but it s going to take some additional work. We are going to have to center-tap one of the 240 volt, single-phase transformers. It s customary to center-tap transformer C... "C" 240 v, 1Ø, 1Ø 31

9 As you can see, we can get olts, (single-phase) from a Delta configuration (-) (-). Unfortunately, all olt loads must be carried on transformer C. Types Of Three-Phase Transformers With the possibility of wiring both the primary and secondary of threephase transformers, there are four possible configurations: (1) Wye/Wye is commonly used for interior wiring systems. (2) Wye/Delta is used to step-down utilities high line voltages. (3) Delta/Delta is often used for industrial applications. (4) Delta/Wye is popular for stepping down transmission lines to fourwire services when neutrals are needed. The most commonly used three-phase transformer is a Delta/Wye configuration. It s used primarily for power distribution from utilities to residential and commercial services. 480 v 3Ø Primary 480 v 480 v 480 v Secondary 208 v 1Ø 208 v 3Ø 1Ø The given voltage designations for a Delta/Wye three-phase transformer is: Primary-Line/Secondary-Line/Secondary-Phase Or in this case: Delta/Wye (480/208/120) 32

10 Phase-To-Phase Let s not forget that three-phase is a product of three individual singlephase transformers. When calculating the Phase-To-Phase relationship the same rules apply as any single-phase transformer. In other words, Our Phase-To- Phase voltage below is 480 to 120 which is a ratio of 4:1. Line Phase Phase Line "A" 480 v "A" "B" 480 v R A T I O "A" eutral "C" 480 v "A" We can also use the the same ladder chart we used for a typical single-phase transformer. Primary Phase Secondary Phase x VA R 480 A 120 T I O x Don t forget that this is just one of three single-phase transformers. There may be efficiency or power factor losses between the primary and secondary phases. But, we ll assume that this transformer is 100% efficient with no power factor loss so the Primary va will equal the Secondary Watts. 33

11 Calculate the Primary va, Primary Amps and Secondary Watts assuming that our Secondary Amps = 100. Primary Phase Secondary Phase x 480 VA R A T I O x Did you get 12,000 Primary va? Multiplying 12,000 by 3 (three transformers) will give us 36 kva (36,000 va). We ll call this our Line kva, which is the sum total of all three-phases. In other words, our transformer rating is 36 kva. It s important to understand the difference between Phase and Line. Phase is the single-phase transformer relationship. Line is the result of the combination of all three transformers hooked together. Line-To-Line There are a few of rules we must remember. (1) Line Watts = Phase Watts x 3 (or) Phase Watts = Line Watts 3 (2) Delta Phase Volts = Delta Line Volts (3) Wye Phase Amps = Wye Line Amps Yes, we can use the ladder Line-To-Line on three-phase transformers Check this out... Primary Secondary! 3 VA 36, ! 3 36, x! 3 x! 3 34

12 For example, knowing the values of Secondary Line Amps (100) and Secondary Line Volts (208) we would multiply going up the ladder to find Secondary Line Watts. In this case, because we are dealing with three-phase, we ll multiply our answer by!3 (1.732). Here s what it looks like... Line Amps (100) x Line Volts (208) x!3 (1.732) = Line Watts (36,000) I know! You get 36, watts. The problem is that 208 volts (3Ø) is not exactly 208 volts. olts x = So we round up! Remember, whenever we see 3Ø, the square root of three!3 must be involved. On the Primary side. stepping down the ladder... Primary va (36,000) Primary Volts (480)!3 (1.732) = Primary Amps (43.3) The key to using the ladder in three-phase is multiplying the answer by!3 going up the ladder, and divide the answer by!3 going down the ladder. Here s a chart that may be useful.!3 va e i x!3 delta line line volts=phase volts amps phase to phase va e i line phase!3 watts line phase 3 w e i volts line phase!3 wye line w e i line amps=phase amps!3 x!3 On the bottom of this chart are a few helpful tools: For the Delta side Amps... line amps phase amps x!3 For the Wye side Amps... line volts phase volts x!3 For the Watts... line watts phase watts x 3 Just cover the value you need to know. For example: line amps = phase amps x!3 (or) phase amps = line amps/!3 35

13 Here s one for you to try; Primary Secondary! 3 VA ! x! 3 x! 3 Delta/Delta A Delta/Delta connected transformer looks like this... Primary Secondary! 3 VA! 3 x! 3 x! 3 There are the rules we must remember. (1) Line Watts = Phase Watts x 3 (or) Phase Watts = Line Watts 3 (2) Delta Phase Volts = Delta Line Volts Phase-To-Phase Although our transformer is connected Delta/Delta the Phase-To-Phase situation has not changed. We still have three single-phase transformers. Let s do some calculations for a Delta/Delta 480/240/120 transformer. First we must find the Secondary Watts, then the Primary va and Primary Amps... 36

14 Primary Phase Secondary Phase x 480 VA R A T I O x Then, we can calculate the values for Line... Primary Secondary! 3 VA 480! 3 7, x! 3 x! 3 3Ø Transformer Size Chart Full-Load Current In Amps At The Line Voltages Listed Below kva Rating olts 240 volts 480 volts 600 volts 2400 volts 4160 volts kva = ( full load current x voltage x!3) 1,000 37

15 Wye Transformer Balancing We might have forgotten the real purpose of a three-phase transformer which is to serve loads. Wye connected 208/olt secondaries can feed loads at three different levels: (1) 208 volts, 3Ø (--) (2) 208 volts, 1Ø (-) or (-) or (-) (3) olts, 1Ø (-) or (-) or (-) "A" "B" "C" A 208 volt (3Ø) load would be distributed on all three Phases... 1/3 on Phase A 1/3 on Phase B 1/3 on Phase C A 208 volt (1Ø) load would be distributed on two Phases... 1/2 on one Phase 1/2 on one Phase A olt (1Ø) load would be distributed on one Phase % on any one Phase For example, Let s balance the following loads Ø, 208 volt, Motor at 9,000 watts 3-1Ø, 208 volt, Ovens at 6,000 watts each 3-1Ø, olt, Light Circuits at 1,800 watts each 38

16 The 208 volt (3Ø) Motor can be distributed on all three Phases... Load Phase A Phase B Phase C Motor (3Ø) 3,000 w (1/3) 3,000 w (1/3) 3,000 w (1/3) The three 208 volt (1Ø) Ovens can be distributed on two Phases each... Load Phase A Phase B Phase C Oven (1Ø) 3,000 w (1/2) 3,000 w (1/2) - Oven (1Ø) 3,000 w (1/2) 3,000 w (1/2) Oven (1Ø) 3,000 w (1/2) 3,000 w (1/2) The three olt (1Ø) Light Circuits can be distributed on one Phases each... Load Phase A Phase B Phase C Light Ckt (1Ø) 1,800 w (100%) - - Light Ckt (1Ø) - 1,800 w (100%) - Light Ckt (1Ø) - - 1,800 w (100%) Totals 10,800 w 10,800 w 10,800 w So, our total demand is 32,400 watts (10,800 x 3). We re lucky, our loads balanced evenly, otherwise, we would have to multiply our largest Phase by three. Looks like we need a 45 kva (standard size) Wye/Wye or Delta/Wye connected transformer. Delta Transformer Balancing What if we had a Delta connected (240/) secondary Delta connected 240/olt secondaries can also feed loads at three different levels: (1) 240 volts, 3Ø (--) (2) 240 volts, 1Ø (-) or (-) or (-) (3) olts, 1Ø (-) or (-) A 240 volt (3Ø) load would be distributed on all three Phases... 1/3 on Phase A 1/3 on Phase B 1/3 on Phase C A 240 volt (1Ø) load would be distributed 100% on one Phase only... 39

17 100% on Phase A 100% on Phase B 100% on Phase C A olt (1Ø) load would be distributed on Phase C only % on Phase C only Let s balance the same loads on a Delta Ø, 208 volt, Motor at 9,000 watts 3-1Ø, 208 volt, Ovens at 6,000 watts each 3-1Ø, olt, Light Circuits at 1,800 watts each The 240 volt (3Ø) Motor can be distributed on all three Phases... Load Phase A Phase B Phase C Motor (3Ø) 3,000 w (1/3) 3,000 w (1/3) 3,000 w (1/3) The three 240 volt (1Ø) Ovens can be distributed on one Phase each... Load Phase A Phase B Phase C Oven (1Ø) 6,000 w (100%) - - Oven (1Ø) - 6,000 w (100%) - Oven (1Ø) - - 6,000 w (100%) The three olt (1Ø) Light Circuits can be distributed on Phase C only... Load Phase A Phase B Phase C Light Ckt (1Ø) - - 1,800 w (100%) Light Ckt (1Ø) - - 1,800 w (100%) Light Ckt (1Ø) - - 1,800 w (100%) Totals 9,000 w 9,000 w 14,400 w Poor old Phase C has to carry 1/3 of all three-phase loads, 100% of any 240 volt (1Ø) loads that distributed on it, and 100% of all olt loads. Our total demand is 43,200 watts (14,400 x 3). This time we re not so lucky, our loads did not balanced evenly. But, we can still use a 45 kva (standard size) Wye/Delta or Delta/Delta connected transformer. 40

18 Open Delta One benefit of a Delta three-phase configuration is that it can still work with just two single-phase transformers. An Open-Delta or V-connected transformer can be used in a emergency or temporary situation.the only problem is that there is a power reduction to just 57.7% of full power. Primary Secondary A third single-phase transformer can be added later if needed. Or, if one original transformer should fail, it can be removed temporarily for repair. In this situation, partial service can be maintained temporarily. Three-Phase Transformer Sample Problem Problem: Calculate the three-phase 480/208/olt transformer required to feed the following loads: one 10 hp, 208 volt, 3Ø Motor (11,096 w) three 3 kw, 208 volt, 1Ø Heaters three 1 kw, olt, 1Ø Light Banks Solution: Loads Balanced: Phase A Phase B Phase C 3Ø Motor (1/3 per phase) 3,699 w 3,699 w 3,699 w 1Ø Heaters (1/2 per phase) 1,500 w 1,500 w - - 1,500 w 1,500 w 1,500 w - 1,500 w 1Ø Light Banks (100% per phase) 1,000 w ,000 w w 7,699 w 7,699 w 7,699 w 41

19 !3 va e i x!3 delta line line volts=phase volts amps phase to phase va e i line phase!3 watts line phase 3 w e i volts line phase!3 wye line w e i line amps=phase amps!3 x!3 Voltages: Primary Line = 480 volts Secondary Line = 208 volts Primary Phase = 480 volts Secondary Phase = olts Amps: Primary Line = 27.7 amps Secondary Line = 64 amps Primary Phase = 16 amps Secondary Phase = 64 amps Wattage: Primary Line = 23,096 watts Secondary Line = 23,096 watts Primary Phase = 7,699 watts Secondary Phase = 7,699 watts Use (25 kva) rated, 3Ø, 480/208/olt transformer. 42

20 Transformer Problems (1) The phase current of a fully loaded 36 kva, 208/olt, three-phase transformer would be amps. (a) 300 (b) 175 (c) 100 (d) 75 (2) If ten single phase 208 volt, 5000 watt office machines were on the same transformer, the total current with all the machines operating would be amps. (a) 120 (b) 208 (c) 240 (d) 416 (3) With a three-phase, delta-wye transformer bank having a 480 volt primary and a 208/olt secondary, the line current of the secondary is 120 amps. The primary line current will be amps. (a) 480 (b) 208 (c) 52 (d) 30 (4) The voltage from either leg, excluding the hi-leg, to neutral of a threephase, delta-delta hi-leg transformer system with a secondary of 240/120 volts is volts. (a) 120 (b) 190 (c) 208 (d) 240 (5) If all the load on a three-phase 480/208/olt transformer is single-phase olt, the size of the neutral on a 24 kva transformer would have a load current rating of amps. (a) 24 (b) 67 (c) 75 (d) 100 (6) If a three-phase wattmeter reads 12,970 watts on a three-phase 208/120 volt system and each ammeter reads 36 amps, the power factor would be. (a) 1 (b).96 (c).62 (d) none of these (7) With a three-phase delta-delta, hi-leg transformer system, the primary is 480 volts, and the secondary is 240/olts, the turns ratio for the transformer is. (a) 2 to 1 (b) 4 to 1 (c) 1 to 4 (d) 1 to 2 43

21 (8) If three 115 volt, 1,800 watt lighting circuits and two 2 H.P. single-phase 115 volt pumps are balanced as closely as possible on a three-wire 230 volt system. The conductor amp rating will be approximately amps. (a) 15 (b) 45 (c) 48 (d) 54 (9) The secondary load of a three-phase 480/208/olt includes; 1- Motor (3 hp, 3Ø, 208 volt) 4- Heat Strips (2500 va, 3Ø, 208 volt) 1- Induction Heater (2000 va, 1Ø, 208 volt) 3- Light Banks (1500 va, 1Ø, olt) Calculate- Primary Feeder? Secondary Phase Current? Transformer Size Required? (10) Determine the three-phase transformer requirements based on the following loads: (voltages are 480/208/olts connected Delta-Wye) 1- Motor (15 hp, 3Ø, 208 volt) 5- Water Heaters (2500 va, 1Ø, 208 volt) 2- Dryers (1500 va, 1Ø, 208 volt) 1- Cooktop (1600 va, 1Ø, 208 volt) 4- Light Circuits (3000 va, 1Ø, olt) Calculate- Primary Feeder? Secondary Phase Current? Transformer Size Required (kva)? 44