Download Three-Phase Transformers

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.
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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 360˚.
120°
120°
120°
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).
120°
120°
120°
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.
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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.
120°
30°
By using this trigonomic formula (don’t get scared) we can see where we
derive the square root of three.
sin 120°
----------sin 30°
0.866
--------0.500
=
1.732
=
!3
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:
L1
L2
Neutral
Phase
L3
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Line
We call each of the single-phase transformer windings Phases.The
combination (hooked together) of the three single-phase transformers (L1-L2L3) 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:
L1
A
L2
B
N
C
L3
If we tie together three single-phase, 120 volt, transformers in a Wye
configuration our output voltage would look like this:
L1
L2
120 v
208 v
3Ø
120 v
N
120 v
L3
Why do we get 208 volts, three-phase on the Line side (L1-L2-L3) ? Just
multiply 120 volts times the square root of 3 (!3). Remember phase angles of
three 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 transformer
by connecting (L1-L2);
L1
"A"
120 v
L2
"B"
120 v
"C"
120 v
208 v
1Ø
N
L3
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 (L2-L3) and (L1-L3) to feed other 208 volt, single-phase
loads.
Guess what ! Yes, it’s even possible to get 120 volts, single-phase out of the
same transformer.
L1
"A"
120 v
L2
"B"
120 v
N
"C"
120 v
120 v
1Ø
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 only
using one single-phase transformer (“B”), that leaves two other possibilities (L1N) and (L3-N) for serving other 120 volt circuits. What’s also very useful is that
we can balance three 120 volt circuits on just one neutral.
In other words, A 208/120 volt (Wye configured) transformer could supply
120 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 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:
L1
Line
L2
Phase
L3
A more simplified representation of a Delta configuration looks like this:
L1
A
B
C
L2
L3
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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.
L1
240 v
240 v
240 v
240 v
3Ø
L2
L3
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 different
240 volt (1Ø) loads utilize three different transformers.
L1
240 v
240 v
240 v
1Ø
240 v
L2
L3
We can still get 120 volts, (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”...
L1
"C"
240 v
L2
120 v, 1Ø
N
120 v, 1Ø
L3
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As you can see, we can get 120 volts, (single-phase) from a Delta
configuration (L2-N) (L3-N). Unfortunately, all 120 volt 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.
Primary
Secondary
L1
480 v
480 v
3Ø
L2
208 v
1Ø
480 v
L2
120 v
480 v
L1
208 v
3Ø
120 v
N
L3
120 v
120 v
1Ø
L3
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 singlephase transformers. When calculating the Phase-To-Phase relationship the same
rules apply as any single-phase transformer. In other words, Our Phase-ToPhase voltage below is 480 to 120 which is a ratio of 4:1.
Line
Phase
Phase
Line
L1
"A"
480 v
"A"
120 v
L1
R
A
T
I
O
"B"
480 v
L2
L3
L2
"A"
120 v
Neutral
L3
"C"
480 v
"A"
120 v
We can also use the the same ladder chart we used for a typical single-phase
transformer.
Primary
Phase
÷
480
x
VA
Secondary
Phase
R
A
T
I
O
÷
120
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.
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Calculate the Primary vA, Primary Amps and Secondary Watts assuming
that our Secondary Amps = 100.
Primary
Phase
÷
480
Secondary
Phase
VA
÷
R
A
T
I
O
x
120
100
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
L1
L1
÷!3
÷!3
V A 36,000
480
L2
L2
x!3
N
L3
L3
34
x!3
36,000
208
100
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 36,0256.6 watts. The problem is that 208 volts (3Ø) is not
exactly 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 going
up the ladder, and divide the answer by !3 going down the ladder.
Here’s a chart that may be useful.
phase to phase
delta line
÷!3
x!3
va
e
i
line volts=phase volts
va
e
i
amps
watts
line
phase !3
line
phase 3
wye line
w
e
i
volts
line
phase !3
w
e
i
line 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:
!3
line amps = phase amps x ! 3 (or) phase amps = line amps/!
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÷!3
x!3
Here’s one for you to try;
Primary
Secondary
L1
L1
÷!3
÷!3
VA
L2
480
10 L2
x!3
N
L3
208
x!3
L3
Delta/Delta
A Delta/Delta connected transformer looks like this...
Primary
Secondary
L1
÷!3
L1
÷!3
VA
L2
L2
N
x!3
L3
x!3
L3
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...
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Primary
Phase
÷
480
VA
x
Secondary
Phase
÷
R
A
T
I
O
240
10
x
Then, we can calculate the values for Line...
Primary
Secondary
L1
÷!3
VA
L1
÷!3
480
L2
7,200
240
L2
N
x!3
L3
x!3
L3
3Ø Transformer Size Chart
Full-Load Current In Amps At The Line Voltages Listed Below
kVa Rating
3
6
9
15
30
120 volts
8.3
16.6
25.0
41.6
83.0
240 volts
7.2
14.4
21.6
36
72
480 volts
3.6
7.2
10.8
18
36
600 volts
2.9
5.8
8.7
14.4
28.8
2400 volts
0.7
1.4
2.2
3.6
7.2
4160 volts
0.4
0.8
1.3
2.1
4.2
45
75
112.5
150
225
300
500
750
1000
125
208
312
415
625
830
1380
2082
2776
108
180
270
360
540
720
1200
1804
2406
54
90
135
180
270
360
600
902
1203
43
72
108
144
216
288
480
722
962
10.8
18
27
36
54
72
120
180
241
6.3
10.4
15.6
20.8
31.2
41.6
69.4
104
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 transformer
which is to serve loads. Wye connected 208/120 volt secondaries can feed loads
at 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)
L1
"A"
L2
"B"
120 v
N
120 v " C "
L3
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 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 watts
3- 1Ø, 208 volt, Ovens at 6,000 watts each
3- 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
Motor (3Ø)
Phase “A”
3,000 w (1/3)
Phase “B”
3,000 w (1/3)
Phase “C”
3,000 w (1/3)
The three 208 volt (1Ø) Ovens can be distributed on two Phases each...
Load
Oven (1Ø)
Oven (1Ø)
Oven (1Ø)
Phase “A”
3,000 w (1/2)
Phase “B”
3,000 w (1/2)
3,000 w (1/2)
3,000 w (1/2)
Phase “C”
3,000 w (1/2)
3,000 w (1/2)
The three 120 volt (1Ø) Light Circuits can be distributed on one Phases
each...
Phase “A”
Phase “B”
Phase “C”
Load
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 (240/120 v) secondary Delta connected
240/120 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 watts
3- 1Ø, 208 volt, Ovens at 6,000 watts each
3- 1Ø, 120 volt, Light Circuits at 1,800 watts each
The 240 volt (3Ø) Motor can be distributed on all three Phases...
Load
Motor (3Ø)
Phase “A”
3,000 w (1/3)
Phase “B”
3,000 w (1/3)
Phase “C”
3,000 w (1/3)
The three 240 volt (1Ø) Ovens can be distributed on one Phase each...
Load
Oven (1Ø)
Oven (1Ø)
Oven (1Ø)
Phase “A”
6,000 w (100%)
-
Phase “B”
Phase “C”
6,000 w (100%)
6,000 w (100%)
The three 120 volt (1Ø) Light Circuits can be distributed on Phase “C”
only...
Phase “A”
Phase “B”
Phase “C”
Load
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 120 volt 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.
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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.
Secondary
Primary
L1
L1
L2
L2
L3
L3
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 480/208/120 volt 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, 120 volt, 1Ø Light Banks
Solution:
Loads Balanced:
Phase ‘A’
3Ø Motor (1/3 per phase)
1Ø Heaters (1/2 per phase)
3,699 w
1,500 w
1,500 w
1Ø Light Banks (100% per phase) 1,000 w
7,699 w
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Phase ‘B’
Phase ‘C’
3,699
1,500
1,500
1,000
7,699
3,699
1,500
1,500
1.000
7,699
w
w
w
w
w
w
w
w
w
w
phase to phase
delta line
÷!3
x!3
va
e
i
va
e
i
line volts=phase volts
amps
watts
line
phase !3
line
phase 3
wye line
w
e
i
volts
line
phase !3
w
e
i
÷!3
x!3
line amps=phase amps
Voltages: Primary Line = 480 volts
Primary Phase = 480 volts
Secondary Line = 208 volts
Secondary Phase = 120 volts
Amps:
Primary Line = 27.7 amps
Primary Phase = 16 amps
Secondary Line = 64 amps
Secondary Phase = 64 amps
Wattage:
Primary Line = 23,096 watts
Primary Phase = 7,699 watts
Secondary Line = 23,096 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 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 480/208/120 volt transformer is single-phase
120 volt, 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 240/120 volts, the turns ratio for the
transformer is _____.
(a) 2 to 1
(b) 4 to 1
(c) 1 to 4
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(d) 1 to 2
(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)
CalculatePrimary 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)
CalculatePrimary Feeder ?
Secondary Phase Current ?
Transformer Size Required (kVA) ?
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