Three-Phase Transformers

When more power is needed - three transformers can be tied together. Thisis called three-phase. Here’s a simple way of comparing single-phase to three-phase power.

Single-Phase

Three-Phase

Think of single-phase as one guy driving a stake into the ground; Three-phaseas three guys, working together, driving the same stake into the ground.

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Generating Three-Phase

Three-phase power is delivered from a generator with three separatewindings. These windings are equally spaced around a rotating cylinder (rotor)with each winding occupying one-third of the rotor circumference. Three-phasegenerator windings are spaced 120˚ apart for a total of 360˚.

120°

120° 120°

Because the three source voltages timed at three different intervals, they canbe transmitted over just three conductors (rather than the six conductorsneeded to transmit three separate single-phase loads).

120°120°120°

A three-phase system can produce 173% higher effective voltage than asingle-phase system. This is very useful for higher power loads such as largemotors.

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We calculate the total voltage (of all three lines of a three-phase trans-former) by multiplying the voltage of each single winding (phase) by the squareroot of 3 (!3). The reason for this is the angles of each phase are always 120°apart.

30°

120°

By using this trigonomic formula (don’t get scared) we can see where wederive the square root of three.

sin 120° 0.866----------- --------- = 1.732 = !3sin 30° 0.500

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:

Neutral

L1

L2

L3

Line

Phase

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We call each of the single-phase transformer windings Phases.Thecombination (hooked together) of the three single-phase transformers (L1-L2-L3) 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 seemthat current would not remain the same in a parallel circuit. But, threetransformers wired in parallel are not like three resistors in parallel.Transformers produce, instead of consume voltages, so the opposite ruleswould would apply in this case.

A more simplified representation of a Wye looks like this:

N

L1L2

L3

A B

C

If we tie together three single-phase, 120 volt, transformers in a Wyeconfiguration our output voltage would look like this:

N

L1

L2120 v

120 v

120 v

208 v3Ø

L3

Why do we get 208 volts, three-phase on the Line side (L1-L2-L3) ? Justmultiply 120 volts times the square root of 3 (!3). Remember phase angles ofthree 120 volt sources.

120 volts x 1.732 (!3) = 208 volts

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We can also get 208 volts, single-phase on the Line side of this transformerby connecting (L1-L2);

N

L1

L2

L3

" A "120 v

208 v1Ø

" B "120 v

"C"120 v

As you see, we are connecting only two of the three single-phasetransformers available. We still only get 208 volts (120 x !3) from thisconnection, and it is just single-phase. But, we have the capability of using twotransformers (“A” & “B”) to help share the load. Also, we can use two otherpossible connections (L2-L3) and (L1-L3) to feed other 208 volt, single-phaseloads.

Guess what ! Yes, it’s even possible to get 120 volts, single-phase out of thesame transformer.

N

L1

L2" A "120 v

120 v1Ø

" B "120 v

"C"120 v

L3

Remember, we have the benefit of a neutral conductor and, by connecting(L2-N) we can feed 120 volt, single-phase loads. And, seeing that we are onlyusing one single-phase transformer (“B”), that leaves two other possibilities (L1-N) and (L3-N) for serving other 120 volt circuits. What’s also very useful is thatwe can balance three 120 volt circuits on just one neutral.

In other words, A 208/120 volt (Wye configured) transformer could supply120 volt, single-phase equipment (like fluorescent lights), 208 volt, single-phase

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equipment (like range or oven), and 208 volt, three-phase equipment (like heavymachinery). This is what makes three-phase so great !

Delta Connections

Another way of tying three transformers together is in series (+ to -). This iscalled a Delta connection. Here’s how a Delta connection is wired:

L1

L2

L3

PhaseLine

A more simplified representation of a Delta configuration looks like this:

L1

L2

L3

A

C

B

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By connecting three (240 volt) single-phase transformers (Delta) we get 240volts, three-phase. We don’t multiply by !3 because Line volts equals Phasevolts on Delta transformers. It would seem that voltage, would not remain thesame in a series type circuit. But, three transformers wired in series are not likethree resistors in series. Transformers produce, (instead of consume) voltages,so the opposite rules would would apply in this case.

L1

L2

240 v 240 v

240 v

L3

240 v3Ø

Connect (L1-L2) and you’ll get 240 volts, single-phase. Connecting (L2-L3)or (L1-L3) will also give us 240 volts, single-phase. Notice how three different240 volt (1Ø) loads utilize three different transformers.

L1

L2

240 v 240 v

240 v

L3

240 v1Ø

We can still get 120 volts, (single-phase) out of a 240 volt Delta system, butit’s going to take some additional work. We are going to have to center-tap oneof the 240 volt, single-phase transformers. It’s customary to center-taptransformer “C”...

L3

L1

L2"C "240 v

120 v, 1ØN

120 v, 1Ø

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As you can see, we can get 120 volts, (single-phase) from a Deltaconfiguration (L2-N) (L3-N). Unfortunately, all 120 volt loads must be carriedon transformer “C”.

Types Of Three-Phase Transformers

With the possibility of wiring both the primary and secondary of three-phase 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 four-

wire services when neutrals are needed.

The most commonly used three-phase transformer is a Delta/Wyeconfiguration. It’s used primarily for power distribution from utilities toresidential and commercial services.

L1

L2

L3

480 v 480 v

480 v

L3

N

L1

L2120 v

120 v

120 v

208 v3Ø

208 v1Ø

120 v1Ø

480 v3Ø

Primary Secondary

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)

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Phase-To-Phase

Let’s not forget that three-phase is a product of three individual single-phase transformers. When calculating the Phase-To-Phase relationship the samerules 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.

L1

L2

L3

Neutral

L1

L2

L3

Phase PhaseLine Line

RATIO

"A"480 v

"B"480 v

"C"480 v

"A"120 v

"A"120 v

"A"120 v

We can also use the the same ladder chart we used for a typical single-phasetransformer.

VA RATIO

PrimaryPhase

SecondaryPhase

x

÷

x

÷480 120

Don’t forget that this is just one of three single-phase transformers. Theremay be efficiency or power factor losses between the primary and secondaryphases. But, we’ll assume that this transformer is 100% efficient with no powerfactor loss so the Primary vA will equal the Secondary Watts.

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Calculate the Primary vA, Primary Amps and Secondary Watts assumingthat our Secondary Amps = 100.

VA RATIO

PrimaryPhase

SecondaryPhase

x

÷

x

÷480 120

100

Did you get 12,000 Primary vA ? Multiplying 12,000 by 3 (threetransformers) will give us 36 kVa (36,000 vA). We’ll call this our Line kVa, whichis the sum total of all three-phases. In other words, our transformer rating is 36kVa.

It’s important to understand the difference between Phase and Line. Phaseis the single-phase transformer relationship. Line is the result of the combinationof 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 Checkthis out...

Primary SecondaryL1

L2

L3

L3

N

L1

L2

x ! 3

÷ ! 3VA

x ! 3

÷ ! 3

48036,000

20836,000

100

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For example, knowing the values of Secondary Line Amps (100) andSecondary Line Volts (208) we would multiply going up the ladder to findSecondary 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 36,0256.6 watts. The problem is that 208 volts (3Ø) is notexactly 208 volts. 120 volts x 1.732 = 207.84. 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 goingup the ladder, and divide the answer by !3 going down the ladder.

Here’s a chart that may be useful.

vaei

wei

phase to phasedelta line wye line

weix!3

÷!3 vaei

x!3

÷!3

linephase 3

wattsline

phase !3line

phase !3

voltsampsline volts=phase voltsline amps=phase amps

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

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Here’s one for you to try;

Primary SecondaryL1

L2

L3

L3

N

L1

L2

x ! 3

÷ ! 3VA

x ! 3

÷ ! 3

48010

208

Delta/Delta

A Delta/Delta connected transformer looks like this...

Primary SecondaryL1

L2

L3 L3N

L1

L2

x ! 3

÷ ! 3VA

x ! 3

÷ ! 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-Phasesituation has not changed. We still have three single-phase transformers. Let’sdo some calculations for a Delta/Delta 480/240/120 transformer. First we mustfind the Secondary Watts, then the Primary vA and Primary Amps...

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VA RATIO

PrimaryPhase

SecondaryPhase

x

÷

x

÷480 240

10

Then, we can calculate the values for Line...

Primary SecondaryL1

L2

L3 L3N

L1

L2

x ! 3

÷ ! 3VA

x ! 3

÷ ! 3

2404807,200

3Ø Transformer Size Chart

Full-Load Current In Amps At The Line Voltages Listed BelowkVa Rating 120 volts 240 volts 480 volts 600 volts 2400 volts 4160 volts

3 8.3 7.2 3.6 2.9 0.7 0.46 16.6 14.4 7.2 5.8 1.4 0.89 25.0 21.6 10.8 8.7 2.2 1.3

15 41.6 36 18 14.4 3.6 2.130 83.0 72 36 28.8 7.2 4.245 125 108 54 43 10.8 6.375 208 180 90 72 18 10.4

112.5 312 270 135 108 27 15.6150 415 360 180 144 36 20.8225 625 540 270 216 54 31.2300 830 720 360 288 72 41.6500 1380 1200 600 480 120 69.4750 2082 1804 902 722 180 104

1000 2776 2406 1203 962 241 139

kVa = ( full load current x voltage x !3) ÷ 1,000

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Wye Transformer Balancing

We might have forgotten the real purpose of a three-phase transformerwhich is to serve loads. Wye connected 208/120 volt secondaries can feed loadsat three different levels:

(1) 208 volts, 3Ø (L1-L2-L3)(2) 208 volts, 1Ø (L1-L2) or (L2-L3) or (L1-L3)(3) 120 volts, 1Ø (L1-N) or (L2-N) or (L3-N)

N

L1L2

L3

" A " " B "

"C"

120 v

120 v

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 Phase1/2 on one Phase

A 120 volt (1Ø) load would be distributed on one Phase...

100% on any one Phase

For example, Let’s balance the following loads...

1- 3Ø, 208 volt, Motor at 9,000 watts3- 1Ø, 208 volt, Ovens at 6,000 watts each3- 1Ø, 120 volt, Light Circuits at 1,800 watts each

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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 120 volt (1Ø) Light Circuits can be distributed on one Phaseseach...

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 loadsbalanced evenly, otherwise, we would have to multiply our largest Phase bythree. Looks like we need a 45 kVa (standard size) Wye/Wye or Delta/Wyeconnected transformer.

Delta Transformer Balancing

What if we had a Delta connected (240/120 v) secondary Delta connected240/120 volt secondaries can also feed loads at three different levels:

(1) 240 volts, 3Ø (L1-L2-L3)(2) 240 volts, 1Ø (L1-L2) or (L2-L3) or (L1-L3)(3) 120 volts, 1Ø (L2-N) or (L3-N)

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...

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100% on Phase “A”100% on Phase “B”100% on Phase “C”

A 120 volt (1Ø) load would be distributed on Phase “C” only...

100% on Phase “C” only

Let’s balance the same loads on a Delta...

1- 3Ø, 208 volt, Motor at 9,000 watts3- 1Ø, 208 volt, Ovens at 6,000 watts each3- 1Ø, 120 volt, 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 120 volt (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 wPoor 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 120 volt loads. Ourtotal demand is 43,200 watts (14,400 x 3). This time we’re not so lucky, ourloads did not balanced evenly. But, we can still use a 45 kVa (standard size)Wye/Delta or Delta/Delta connected transformer.

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Open Delta

One benefit of a Delta three-phase configuration is that it can still work withjust two single-phase transformers. An Open-Delta or V-connected transformercan be used in a emergency or temporary situation.The only problem is thatthere is a power reduction to just 57.7% of full power.

Primary SecondaryL1

L2

L3 L3

L1

L2

A third single-phase transformer can be added later if needed. Or, if oneoriginal transformer should fail, it can be removed temporarily for repair. In thissituation, partial service can be maintained temporarily.

Three-Phase Transformer Sample ProblemProblem:

Calculate the three-phase 480/208/120 volt transformer required to feed thefollowing loads:

one 10 hp, 208 volt, 3Ø Motor (11,096 w)three 3 kW, 208 volt, 1Ø Heatersthree 1 kW, 120 volt, 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 w1Ø Heaters (1/2 per phase) 1,500 w 1,500 w -

- 1,500 w 1,500 w1,500 w - 1,500 w

1Ø Light Banks (100% per phase) 1,000 w - -- 1,000 w -

- - 1.000 w7,699 w 7,699 w 7,699 w

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vaei

wei

phase to phasedelta line wye line

weix!3

÷!3 vaei

x!3

÷!3

linephase 3

wattsline

phase !3line

phase !3

voltsampsline volts=phase voltsline amps=phase amps

Voltages: Primary Line = 480 volts Secondary Line = 208 voltsPrimary Phase = 480 volts Secondary Phase = 120 volts

Amps: Primary Line = 27.7 amps Secondary Line = 64 ampsPrimary Phase = 16 amps Secondary Phase = 64 amps

Wattage: Primary Line = 23,096 watts Secondary Line = 23,096 wattsPrimary Phase = 7,699 watts Secondary Phase = 7,699 watts

Use (25 kVa) rated, 3Ø, 480/208/120 volt transformer.

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Transformer Problems

(1) The phase current of a fully loaded 36 kVA, 208/120 volt, 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 208/120 volt 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 three-phase, 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 480/208/120 volt transformer is single-phase120 volt, the size of the neutral on a 24 kVA transformer would have a loadcurrent 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 240/120 volts, the turns ratio for the transformer is _____.

(a) 2 to 1 (b) 4 to 1 (c) 1 to 4 (d) 1 to 243

(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 480/208/120 volt 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, 120 volt)

Calculate-Primary Feeder ?Secondary Phase Current ?Transformer Size Required ?

(10) Determine the three-phase transformer requirements based on the following loads: (voltages are 480/208/120 volts 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, 120 volt)

Calculate-Primary Feeder ?Secondary Phase Current ?Transformer Size Required (kVA) ?

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