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Single Phase to Three Phase Converter Using Scott Connection - A Practical Case Study
Single phase to three phase converter using Scott
connection: A Practical case study
IMrs.D.P.Chopade,
2R.M.Ho\mukhe,
3P.S.Chaudhari, 4Mr.Parandkar
1.2Sharati Vidyapeeth University College of Engineering, Pune, India
3Scientist, DRDO, Pune, India, "Pune, India
Email ID:rajeshmholmukhe@hotrnail.com
Abstract- When it comes to Three phase equipment to be used
with one of the three phase failed, what one requires is a system
which will convert the available one and lor two phase supply in
to equivalent & nearly balanced three phase supply. This study is
one of the solutions to design such system. The heart of this
system is nothing but the 'SCOTT CONNECTION'. Here there
are two individual single phase transformers, designed &
connected in such a fashion that the output will be balanced there
phase supply of equal voltage & phase difference. Also it
incorporates the control unit i.e. required to switch over to &
from this system as soon as there is a phase failure & phase
restore respectively. This project is no substitute for the
application where in the three phase motor keeps delivering
power in spite of the phase loss. More over the motor is safe from
the hazards of single phasing, which may cause the damage to
motor winding. This feature of these systems makes it suitable for
rural areas for pumping application, which is of prime
important.
I.
I
- l
I
!
I TRaDUCTION
eed & requirement:
Many installations for economic reasons are only connected to
one of the three phase of utility power available at the nearest
distribution transformer. A single phase supply line is the
cheaper option when the installations are built. Similarly, large
buildings with 3 phases supplies might still have locations
where it is uneconomic or inconvenient to run 3 phase wiring.
If three phase equipment is to be used a simple comparison
can be made: compare the cost for providing three phase
supply (e.g. distribution transformer, power lines, monthly line
charges) with the price of a single to three phase converter.
Three phase supply system is having more advantages rather
than single phase as following,
I. More HP load can be attained.
2. so that more output can be received.
3. Efficiency improves.
4. Loss reduces.
5. Energy saves.
1.2 Single phase to three phase conversion:
Single phase to three phases are indispensable in situation
where three phase induction motors are to be operated from
the available single phase supply. Several types of such
converters are available utilizing various principles and
various types of components. In many situations, only a single
phase supply is available and typical solution for rating, three
phase motor in such a condition is with the help of a single to
three phase converter interphase. This is the situation in
remote and rural areas, where electric supply systems use
single phase line for economy while consumers prefers three
1.1
phase motors or pumps and other forming equipment. In
electric traction application, only a single phase supply is
available inside the train.
Static phase converters using controlled power
semiconductor device have now become common due to
lower cost, size and better performance of the overall system,
consisting of motor and converter. Using semiconductor
converter for conversion from single phase to three phase
voltage can help in saving energy in two ways. First energy
saving can be affected in conversion itself because
semiconductor converter are very efficient. Secondary, three
phase motors can be used for several other applications where
single-phase motor are presently in use.
Scott connection means three phases to two phase
conversion. Reverse Scott connection means two phase to
three phase conversion.
Here we use reverse Scott connection. In this study single
phase supply is converted into two phase supply by using
capacitor. And then two phase supply is converted into three
phase supply by using reverse Scott connection. Then three
capacitor are connected in delta to get balanced voltage
between three phases.
II. DIFFERENT WA YS OF SINGLE PHASE TO THREE PHASE
CONVERSION
2.1 Conversion by Electronics solid state devices.
Motor speeds are constant as with a three phase supply.
Starting motors draw high currents. A booster TM E will
produce these currents in order to maintain output voltages.
This will draw high input currents for a short time. On the
single phase supply side it pays to install heavy cables. This
will minimize voltage drop. Installation instructions and
service schematics are provided with each converter. A simple
block diagram and schematic circuit diagram are shown
below
_lor
G
bloct dlao"'''
Traft1forrntr
-
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,-
SlnqI. photOI.
r
SolifSttt.
Cont,..l1"
011I
~
>->->--
c.poeltor
Bloct
~
Figure I.
367
Proceedings of International Conference on Energy Optimization and Control (ICEOC-20 J 0)
December 28 - 30. 2010. Aurangabad, Maharashtra. India
Single PhISe to Three Pbase Convert
!I<"'""fI"lrrundl""",
~-~~----~---
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III. WORKl G PRI CIPLE & TECH OLOGY
3.1 SCOTT CONNECTION:
This is a connection by which three phases to three phases
transformation is accomplished with the help of2 transformers
as shown in fi ure.
L
R
. £loct•.•III, ooolroll.,
Figure 2.
Booster ™ phase converter hard starts and momentary
ov.erloads when an electric motor starts under load, or when a
heavy load is applied to a motor, a Booster ™ will output up
to 600% of its continuous maximum current for up to 10
seconds.
Booster '1 M E converters cope well with motors starting under
load, with motor accelerating large and with pumps and
compres ors starting against pressure.
A Booster ™ E will operate chillers, cranes, conveyors, hoists
and vehicle lifts/ramps starting under load. Due to the higher
input currents during Boost mode, single phase supply cables
to a Booster ™ should be oversized. The resistance per unit
length of the cable should be multiplied bv the length and by
the maximum current to provide the voltage drop. As a rough
guide, this should not exceed IOV at peak curren.
2.2 Single Phase Induction Motor coupled to 3
Alternator.
By connecting single phase induction motor to three
alternator • also with suitable capacitor bank for
balancing & phase shifting purpose it can convert single
to three at ou ut.
.
mun
ig, 1
c
Figure 4.
Since it was first proposed by Charles F Scott, it is frequently
referred to as a Scott Connection. This connection can also be
used for 3 phase to 2 phase conversion and vice versa. One of
the transformers has centre taps both on primary and
secondary windings and is known as the main transformer. It
forms the horizontal member of the connection.
The other transformer has a 86.66% tap and is known as
teaser transformer. One end of both the primary and secondary
of the teaser transformer res .as shown in fi
aecondary
Teutr
Transfonner
Phase
phase
phase
phase
B
IOOv
Main Trwfonntr
Figure 5
f~~~~~·~·:"···-·""""'~
•....
---
-_..__
.
'I--
I ~_-!= r---;_...:~I:m...~··'-"
r.
L
Figure 3
2.3 Reverse Scott Connection
Two same rating transformer having tapping on secondary to
50% & 86.6%. So that by proper connection with capacitor for
phase shifting & phase balancing get the three phase out put.
The other end 'A' of teaser primary and the two ends 'B' and
'C' of the main transformer primary are connected to three
phase supply.
NotationK: Transformation ratio.
M: Main Transformer
T: Teaser Transformer
I: Turns primary transformer
2: Turns of secondary transformer
3.2 REVERSE SCOTT CO ECTlON:
2 PHASE TO 3 PHASE CONVERSION
This conversion is required to supply two phase furnaces, to
link two-phase circuit with 3 phase systems and also to supply
Organised by - Department of Electrical Engineering. Govt. College of Engineering, Aurangabad - 431 005. Maharashtra, India
368
Single Phase to Three Phase Converter Using Scott Connection - A Practical Case Study
a three phase apparatus from two phase supply sources. For
this Ul ose, Scott connection as shown in fi is em 10 ed.
Figure 6
IV. TECHNICAL DESIGN
4.1 PROBLEM: Desi!:n a single phase to three phase
converter by using Scott connection to run a motor of SHP.
SOLUTION:
Motor Rating: 5H.P. =3730W.
Efficiency of motor = 0.9
Motor Current = Motor rating 1..J3 x v x cos0 x efficiency
Motor Current = 3730 .;. ..J3 x 415 x 0.8 x 0.9
=7.21 A "'7.5A.
r-r-
--
..
-
..
--
0--
_]teaserf?
..--.-.-
..
-.---------.
-..
86.6%
0,-,-·_··-
..- ..- ...
Voltage ratio = 230 1359.4 V
VA = 2590.28 VA
From table 7.1 A.K.Sawhney for transformer up to I MV A the
valve of output coefficient is to be takes as 1.1
Voltage 1turn Et = K x ..JQ
= 1.I..J2.59
= 1.77 V
Flux 0m = Et / 4.44f
= 1.77 /4.44 x 50
= 0.00797 wb
Taking maximum flux density Bm = 1.35 wb/m'
Net iron area = Ai = 0.00797 11.35
=5.91 x 10-3 m2
Width of Middle core = 7.6 em
Taking stacking factor = 0.9
Stack = 5.91 x 10-31 (7.6 x 0.9)
= 0.0864m.
Gross iron area Agi = 5.91 x 10.3/0.9
=6.56 x 10 -3 m2
Width of central limb = ..JAgi
= ..J6.56 x 10-3 m2
=0.081 m.
Taking the ratio b 12a as
2.5(2a/ gross iron area = 2a =..J6.56 x 10 -3/2.5
2a =0.051 m
a = 25.61 mm
Depth of frame b = 2.5 x 2a
= 2.5 x 51.22
= 128.05 mm
therefore,
Selecting standard nearest stamping size E-! table i.e.8B No.
Size.
sO%
-11
main
Pig.9
..
I
I
I
I
-
Figure 7
VA of Main transformer & Teaser transformer
=Motor rating x 746 (HP) 12 x cos0
efficiency
= 5 x 746/
2 x 0.8 x 0.9
= 2590.28 VA
current in Teaser transformer
= VA 10.866V
= 2590.28/259.4
(
I
= 7.21 A'" 7.5 A = 11.26 A
Current in Main transformer
=VA/V
= 2590.28 1230
= 11.26 A
Capacitor to be connected = VA of transformer
= 2590.28 V
4.2 Design of Teaser Transformer:
[
:1_-
I
.....--:r--
11
~
x
:t>
I
I
1
~£-
: ,-
L
-
1---~----
-e:-
-A-- ----
---...,I
-tFt-
J
I
I
r- ...-- -- - Fig 2.
..
_ ..
Figure 8
A = 235 mm, B = 196 .90mm, C = 76.20 mm, 0=120.7 mm,
E=38.10 mm, F = 4!.30mm, G = 38.10 mm.
From fig.2a is nothing but C = 76.20 mm.
Depth of frame = 6560 176.20
= 86.08 mm.
2a =76.20
a = 38.1
From this, Window dimensions are
Width of windows Ww = A - 2E - C 12
369
Proceedings of International Conference on Energy Optimization and Control (ICEOC-20l0)
December 28 - 30,2010, Aurangabad, Maharashtra, lndia
= 235 - 2 x 38.1 - 76.20 / 2
= 41.3 mm.
Height of window Hw = B - 2G
= 196.9 - 2 x 38.1
=120.7 mm
Winding:
Number of turns in HV winding Ts = V / Et
= 359.4 11.77
= 204 turns.
Now, Tp ITs = Vp I Vs.
Therefore, Tp = Vp x Ts I Vs
Tp = 203 x 230 I 359.4
=130 turns.
No. of turns on LV wdg = 130.
Using bobbin of standard size 5 mm x 5 mm as insulation
between core & wdg.
LVWrNDJNG
LV winding current Ip = Va I V
= 2590.28 1230
= 11.26 A.
Taking current density 6 = 1.35 AI mm "2
Area of LV wdg conductor = 11.26 I 1.35
= 8.34 mm"2
Now, 1td2I 4 = 8.34 rnm',
=3.25mm.
Diameter of LV wdg conductor = 3.25mm
Selecting standard round wire synthetic enamel 3.25 mm with
fine covering, dia.of
Base conductor = 3.25 mm. (10 SWG)
Now, height of window = Hw = 120.7 mm & leaving 5mm
from top & bottom & side.
(120.7 - 15) = 105.7mm
Turns /layer = 105.7/3.25
= 32.52 :::33turns.
Therefore, no oflayers = 130/33
= 4layers.
Outside dimension of LV = 3.95" x 3.95"
Mean dimensions of LV = 3.95" x 3.95"
Length of LV wdg = 2 x (3.95 + 3.95) = 15.8"
=395x 130
= 71350mm.
= 71.350m
= 0.071350 Km.
for 10 SWG as copper is 73.771 Kg /Km = 73.771 x 0.051350
= 3.78 Kg.
HVWJNDJNG:
Current in winding = VA I Vs
= 2590.27 I 359.4
= 7.5 A.
Taking current density 0 = 1.35 A /mm'
Area of HV conductor a = I 10
= 7.5 I 1.35
= 5.5 mm2
now:d2 14 = 5.5 mm2
Therefore d = 2.64 mm.
Diameter of HV conductor = 2.64mm.
Diameter of bare conductor = 2.64 mm
Fine conductor =2.64(12SWG)
Hw = 120.7mm.
Leaving 5 mm from top, bottom & side.
Hw = 120.7 - 15 = 105.7mm
Turns /layer = I 05.7/2.64
=40tums.
No. of layers = 204 140
= 5 layers.
Considering wdg as rectangular shape with inside dimensions
as 3" x 3" and outside dimensions as 3.57" x 3.57".
Therefore the main dimensions are 3.57" x 3.57".
Therefore total length = 2(3.57 + 3.57) = 14.28"=357 x 202 =
92114mm
=72.114m
0.092114Km.
for 12SWG as copper is KgIKm =48.72
=48.72 x 0.092114
= 3.5 Kg.
4.3.Design of MAIN Transformer:
VA of Main transformer = 230 I 415 V.
Voltage I turn Et = K x
= 1.1"2.59
= 1.77V.
Flux 0m = Et 14.44f
= 1.77 I 4.44 x 50
= 0.00797 wb
Taking maximum flux density Bm = 14.35 wb/m2
Net iron area = Ai = 0.00797 I 1.35
= 5.91 x 10.3 m2
Gross iron area Agi = 5.91 x 10.3 10.9
= 6.56 x 10.3 m2
taking the ration b I 2a as
2.5(2a)2 = gross iron area = 2a = 6.56 x 10.3 I 2.5
2a=O.05Im.
selecting standard nearest stamping size E-J table i.e.8B
No.Size.
A=235 mm, B=196.90mm, C = 76.20mm, 0=120.7
E=38.I Omm, F=4I.30mm, G=38.IOmm.
From fig.2a is nothing but C=76.20mm.
Depth of frame = 6560 176.20mm
= 86.08mm
2a=76.20
a==38.1
from this, Window dimensions are
Width of window Ww = A -2E-C/2
= 235-2 x 38.1 - 76.20 /2
=41.3mm.
Height of window Hw = B - 2G
= 196.9-2 x 38.1
= 120.7mm.
"Q
Winding Design:
No.of turns in HV winding Ts = 415 I 1.77
Organisedby - Departmentof ElectricalEngineering.Govt.Collegeof Engineering.Aurangabad- 43/005. Maharashtra,India
370
•
Single Phase to Three Phase Converter Using Scoll Connection - A Practical Case Study
= 234 turns.
No. of turns in LV winding Tp =234 x (230/415)
=130 turns.
Using bobbin of standard size 3" x 3" as insulation between
core of winding.
LV winding current Jp =Va N
= 2590.28 1230
=11.26 A.
Taking current density 6= 1.35 Almm2
Area of LV wdg conductor = 11.26 11.35
=8.34mm2
2
2
Now,nd 14=8.34mm ,
D=3.25rnrn.
Diameter of LV wdg conductor = 3.25rnrn.
Selecting standard round wire synthetic enamel 3.25 mm with
fine covering,dia.of
Base conductor=3.25rnrn.(1 0 SWG)
Now, height of window =Hw =120.7rnrn & leaving 5mm from
top & bottom & side.
(120.7 - 15) =105.7mm
Turnsl layer = 105.7 1 3.25
=32.52 = 33 turns.
Therefore, No.of layers = 130/33
=4 layers.
Taking current density5= 1.35 Almm2
Area of LV wdg conductor= 11.26/1.35
=8.34mm2
2
Now, nd2/4 = 8.34mm ,
D=3.25mm.
Diameter of LV wdg conductor =3.25mm
Selecting standard round wire synthetic enamel 3.25 mm with
fine
Covering, dia of base conductor =3.25mm. (I OSWG)
Now, Height of window =120.7mm & leaving 5mm from top
& bottom & side.
(120.7 -15) =105.7rnrn
Turns Ilayer=105.7 13.25
=32.52 =33turns.
Therefore, No oflayers = 130.33
=4 layers.
HVWinding
Current in HV winding =VA 1Vs
=2590.27 1415
= 6.24 =7.5A.
Taking current density 5 =1.35 AI mm'.
Area of HV conductor a = I 15
=7.5/1.35
=5.5mm2
Now, nd2 14 =5.5 mm"
Therefore, d =2.64mm.
Diameter of HV conductor =2.64mm
Diameter of bare conductor =2.64mm
Fine conductor =2.64( 12SWG) Leaving 5mm from top,
bottom & side.
Hw = 120.7 -15=105.7mm.
Turns /layer= I05.7/2.64
= 5.85=6 layers
4.4Design of Capacitor
We have,
Rsc = 0.5 ohm
Xsc = 0.14 ohm
Now for 0.8 lagging power factor,
Load impedance referred to primary,
Z=V II
=415/7.5
= 55.33
Z = ~ (R2 + X2)
Assume R I = 60 ohm
XI = 500hm.
R Total = Rl + Rsc
= 60 + 0.5
= 60.5 ohm
X Total = XI + Xsc
= 50 + 1.23
= 51.23 ohm
in order to have 90 phase difference the Xsc of capacitor
should be
Xsc = X total + R total (tan 53.7)
= 51.23 + 60.5(tan 53.7)
= 133.59 ohm.
i.e. capacitor of value 23.82 microfarad
The KVAR of capacitor at full load
KVAR = (7.5)2 X 133.59
=7.5 KVAR
4.5 Testing
To calculate the regulation & efficiency we need to conduct
O.C, S.C test.
The results of the same is as detailed below:
O.C.TEST.
Transformer
Main (TI)
Teaser (T2)
Primary
Voltage (V)
230
230
Primary
Current (A)
0.8
0.8
Primary
Power(W)
20
20
Secondary
Voltage (V)
410
410
FORTI
Actual transformation ratio:
V2 1 V I = 230/410
= 0.56
Now,
Woc = Voc x loc x cos 00c
20 = 410 x 0.8 x cos 00c
Cos 00c = 86.50
1m= losin0
= 0.8 x sin 86.50
=0.79 Amp.
Iw= locos0
= 0.8cos86.50
= 0.0488 Amp
Woc at 4 I5 volt = 20(0.56)2
Testing
To calculate the regulation & efficiency we need to conduct
O.C, S.C test.
The result of the same is as detailed below:
371
Proceedings of International Conference on Energy Optimization and Control (ICEOC-20 10)
December 28 - 30,2010, Aurangabad, Maharashtra, India
Transfonner
Main (TI)
Teaser (T2)
-
teaser
O.C.TE3T.
Primary
Voltage (V)
230
230
Primary
Current tAl
0.8
0.8
Primary
Powcr(W)
20
20
R Y
Secondary
Voltage (V)
410
410
VI. FUTURE PROSPECTS
a) Polyphase supply: Because of high efficiency, transformers
serve a excellent polyphase transformation devices in
providing higher polyphase systems.
r-~
i
!
-
FORTI
Actual transformation ratio:
V2 1VI = 230/410
= 0.56
Now,
Woc = Voc x loc x cos 00c
20 = 410 x 0.8 x cos 00c
Cos 00c = 86.50
Im = losin0
= 0.8 x sin 86.50
= 0.79 Amp.
Iw= locos0
= 0.8cos86.50
=0.0488 Amp
Woc at 415 volt = 20(0.56)2
V. APPLICA TIO S AND SCOPE
Application:
I. To give supply to an existing two phase system from a new
three phase source
2. To supply two phase furnace from three phase source
3. To supply three phase motors from a two phase source
4. To interlink a two phase system with a three phase system.
Example applications
and scope for a three phase
converter:
a) Three phase converting in farming
Pumps,compressors,irrigationsystems,airconditioning,refrigera
tion unit, convener, crane, hoist, grain mill.
b) Three phase converting in large buildings
Computer installations, print rooms etc
al
~m
]11§
- -1
1
__
J
..
]llc
••••
l
C":1
f-
'~3~
J~~ ttrJ~J
I-
Figure 9
VII. OUTPUT
WAVEFORMS OF OUTPUT VOLTAGES
WA VEFORMS OP OUTPUT' VOLTAOBS
ACKNOWLEDGME T
Bharati Vidyapeeth University College of Engineering, Pune411043
REFERENCES
I]
2]
3]
4]
5]
6]
ELECTRICAL TECHNOLOGY, B.L.THEREJA
ELECTRICAL MECHINE DESIGN: A.KSA WNEY.
THE J & p .ransformer book
www.goog .•e.com
Sales @ isomatic.co.uk
Electrical Design: Estimation & costing by K.B.Raina S.K.Bhattacharya.
A
0---------,
8
Figure 14
b)Balabce 3 phase contactor to the output supply,the 3 phase
supply from main can maintain the output constant.
Because of this unbalancing,low voltage or single phasing or
two phasing can be avoided.
Organised by - Department of Electrical Engineering. Govt. College of Engineering.Aurangabad - 43/ 005. Maharashtra, India
372