oPe", l"'~"\ Ii o c 5~~;~ ~U~ 1";~' q. Thosar (j)r. K Chairperson II " rwJ C01..1..EGE AURANGABAD "e~ Cl , 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 - n•.•• ,- 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""", ~-~~----~--- __~----oLI 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