DESIGN AND DEVELOPMENT OF A EFFICIENT COIL FOR A RESONANT .

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DESIGN AND DEVELOPMENT OF A EFFICIENT COIL FOR A RESONANTHIGH FREQUENCY INVERTER FOR INDUCTION HEATING.Umar Shami.University of Engineering and Technology, Lahore. Pakistan.umarshami 99@yahoo.comABSTRACT.Induction heating systems employ non-contact heating.Inducing heat electromagnetically rather than using aheating element in contact with a part to conduct heat,as does resistance heating. This paper presents atechnique to calculate the parameters of the coil to beused to heat an iron piece of given parameters, to 850oC.A design of the full bridge power resonant inverter ispresented followed by the results.I. BACKGROUND.When a ferromagnetic material is placed in a time varyingmagnetic field , an E.M.F is induced in the ferromagneticmaterial. This induced E.M.F causes eddy currents to flowalong the surface of the material. The major factors thatcontrol the magnitude of these eddy currents are (i) themagnitude of the time varying magnetic field , (ii)frequency of the time varying magnetic field, and (iii) theresistivity of the material.These eddy currents are beneficial in the case of inductionheating because they flow inside the material and produceI2R losses which causes the material to heat up. So bycontrolling the amount of eddy current we can control thetemperature of the material.This requires a coil suitably dimensioned to heat up aworkpiece to a given temperature.III. CALCULATIONS FOR FINDING THE COILNUMBER OF TURNS AND AMPERES REQUIREDFOR INDUCTION HEATING OF A GIVENCYLINDER.For the material to be heated let us visualize it as shown infigure 1. The coil assembly consists of a multi turn coil andan ironwork piece as the core. [1],[2],[3],[4],[5],[6].The ampere-turn must be equal on the both primary i.e., thecoil it self and the secondary i.e., the work piece (core).Figure 1: A workpiece in coil, the coil is energized from an ACsource.Figure 2 shows the electrical circuit analogy betweeninduction heating and transformer principle.II. PROBLEM DEFINITIONTo design and develop a coil to heat steel metal AISI 1045cylinder from 25 oC to 800 oC 850oC in 30 seconds.Parameters of the workpiece:Diameter ( D)Length(l)RelativePermeability( µr )Specific Heat ( c )Density ( γ )Resistivity ( ρ ) 1 cm. 2 cm. 50. 486J/Kg.oC. 7870 Kg/m3. 1.62 x 10 –7Ω.m.Figure 2:Equivalent model of the workpiece placed in the coil.Where : Vp primary voltage (V); Ip primary current (A); Np number of primary turns; Is secondary current (A); Ns number of secondary turns; Vs secondary voltage (V); RL load resistance(Ω)From the equivalent model shown in figure 2, we canapply the transformer equation as follows.I p N p Is NsBut in our case .

Ns 1The magnetizing force is:soI p N p IsThe heat generated in the workpiece is given as :P I s 2 R wattsWhere R is the resistance path of the eddy currents.Again(IP N p ) πρ D2pδlwattsWhereD is the diameter of workpiece.l is the length of workpiece.δ is the skin depth .The power density or surface power gives the amount ofpower entering unit area of the cylinder.AsArea of cylinder π D lPsur 2π ( H )Equation 1 plays a vital role in calculating the magnetizingforce required for the correct power density. From themagnetizing force we can calculate the number of turnsrequired and amperes for the design of the coil.IV. RELATIONSHIP BETWEEN POWER, MEANTEMPERATURE AND TIME.[5],[6],[7],[8],[9],[10],[11],[12].If a cylinder is heated uniformly at a constant surfacepower density and there is no losses , then all the heat isstored in the cylinder, i.e.,Energy in Energy storedEnergy in Psur .t. ( Surface area )tθmcγr2δ l.π Dlwatts / m 2which finally becomes :Psur (INow,Np ) ρ2pδ l2watts / m2Now the skin depth is given as :δ 2ρµωµr f ρ 10-7 watts / m 2 (1)where :orPsur 2Energy stored θ m cγπ r 2lP watts / m 2π Dl( I p N p ) πρ Dampere.turn / mlconsidering l to be equal to one meter then;So Surface Power will bePsurIpNpH meterswhere;δ is the skin depth.ρ is the resistivety of the workpiece.µ is the magnetic permeability of workpiece.ω is the angular frequency of the varying magnetic field. time. temperature rise. specific heat. density. radius.Psur .t. (π Dl ) θ m cγπ r 2lPsur θ m cγ R2twatts / m 2V. SOLUTION TO OUR PROBLEMBy using figure 2.6 of chapter 2 of reference [2], we selectfrequency to be 10kHz.Heat up time 30 Seconds.Rise in Temperture 25 oC To 800 oC 850oCNow first we find the skin depth.So,2Psur I N µr fwatts / m 2 p p 2πρ7ρ .10 l (2)δ 2ρµωmeters 0.000286 meters 0.286mm.

From equation (2) the surface Power entering the cylinderis :Psur 254988 watts / m 2 (3)To find number of turns and amperes of the primary coil,we make use of equation 1.Psur 2π ( H )2µr f ρ 10-7 watts / m2Psur 0.000565 H 2 watts / m 2 (4)Equating Eq.3 and Eq.4dv/dt rating 50 V/µsecTurn On Time 12 µsecTurn On Gate Current 50 mA.Working Temperature -40 to 120o CRecommended Snubber Resistance 20Ω.VII. SELECTION OF FAST DIODE .The fast diode must support a frequency of 10 KHz .Thediode selected is YG912S6.The characteristics of this diode are as follows : 254988 0.000565H 2H 21234 ampere.turn./meterAnd for a length of 0.02m the H required will be :H required 21234 x 0.02VIII.H NI 425I 14 amperes.N 30 turnsNow we are concerned to design the inverter that cansupply the required amount of current that is needed asconcluded above. Our power circuit will be a VOLTAGEFED INVERTER as in figure 3.For this we have to design the following : Proper SCR selection.Proper Fast Recovery DiodesSnubber Circuit.Resonant Inverter.Resonant Inductor .Resonant Capacitor.VI. SELECTION OF SCR .Required frequency that the SCR must maintain is 10 KHz.Required current that will pass through the SCR is 14 Amp.The SCR selected for this purpose is ACR25U12.It's datasheet is as follows :Maximum controllable current 25 Amp.Maximum breakdown voltage 1200 VoltsMaximum operable frequency 12 KHz.di/dt rating 10 A/µsecdiPROTECTION OF THE SCR :dt[13],[14],[15],[16].For the di/dt protection we have to design an inductor .Theequation that will help us in designing is as follows:Vdi LdtH required 425 ampere.turnSo One Possibility is :High Speed Switching.Maximum Breakdown Voltage 1000VMaximum Current 10 A.in our caseV 150 volts.di 10 A / µ secdtHenceLdi 15 µ HdtWe choose L 33µH as it is easy to build soLdi 33 µ HdtIX. THE SNUBBER DESIGN. [14],[15],[16],[17],[18].The snubber circuit is connected in parallel to the powerSCR . It consists of a series RC circuit. The snubber circuitdesign comprises of the design of the resistance Rsnub andthe capacitance Csnub .X. DESIGN OF RESISTANCERsnubThe value of the snubber resistance is recommended by themanufacture. The manufacture of the SCR used in thisproject i.e, ACR25U12 recommends the snubber resistanceto be 20 ohms A snubber resistance of 22 ohms has beenused as it is easily available.SoRsnub 22 Ω

XI. DESIGN OF CAPACITANCELtotal 101 µ HCsnub[14],[16],[22],[24],[23].By calculations from the reference text , we get ,The resonant frequency is given as ;f resonant Csnub 0.086 µ Fwe choose,Hzf resonant 10 kHzXII. DESIGN OF RESONANT sonant circuit consists ofLresonantLresonant Ltotal 101 µ Hso,Cresonant & Rreflected . The Rreflected is the reflectedresistance of the workpiece when placed in the coil .XIII. DESIGN OF RESONANT INDUCTOR2π LCNowCsnub 0.1 µ FThe1LresonantInstead of designing a separate resonant inductor we willmake use of the inductance of the workpiece in the coil andinductance of the di/dt protection coil .We will consider theresultant of the two inductances as our total resonantinductance of the resonant circuit.We measure the inductance of the workpiece in the coil. Itsvalue is :Cresonant 2.5 µ FXV. TEST RESULTSGiven below the results of various tests conducted atdifferent frequencies on the test piece. Efforts have beenmade to have as correct measurements as possible. The testcircuit has been removed away from any large piece ofmetals to avoid any induction effects. The measurementstaken are as shown in tabular form in table 1 , 2 ,3 and ingraphical form in figure 4.XVI. CONCLUSIONS.The design of a Resonant High Frequency inverter hasbeen presented and a device based on this design has beenLcoil 35 µ Hdeveloped using normal commercial components. The testpiece of ferrous metals had been heated to the point that itdiprotection coil is :The inductance of thewas red hot , using the inverter as of figure 3 and thedtmeasurements made to determine various input and outputLdi 33 µ Hparameters.dtFrom the measurements it can be seen that for steelThe total inductance becomes :cylinders (workpiece) with small diameters the frequencyof the pulsating electromagnetic field in which the cylinderLtotal Lcoil 2. Ldiis placed ,must be high.dtAs the diameter of the cylinder of the workpiece increasesLtotal 101 µ Hthe frequency of the pulsating electromagnetic field shouldbe decreased . This is obvious because at low frequencyXIV. DESIGN OF THE RESONANT CAPACITANCE more eddy currents penetrate the cylinder i.e., skin depthincreases with the decrease in frequency and hence moreCresonantI2R loses are produced inside the workpiece.The resonant inductance is ;

Figure 3: Full bridge high frequency resonant inverter for induction [9].[10].[11].[12].[13].[14].V.L.Rudnev, R.L.Cook, D.L.Loveless, " InductionHeat Treatment", Marcel Dekker Inc. 1997.CA.Tudbury, " Basics of Induction Heating“ Rider , New York, 1960.E.J.Davis " Induction Heating Handbook",McGraw Hill ,1979.M.Orfeuil," Electric Process Heating". BatteuePress. 1987.S.Zinn, "Elements of Induction Heating, Design,Control and Applications" ,ASM Int. 1988.J.M.Jin." The finite Element Method inElectromagnets", Wiley, 1993.G.F.Golovin. M.M.Zamjartin." High FrequencyInduction Heating Treating”. Mashinstroenie.Russia 1990.P.Hammond, “Electromagnetism For Engineers",Peragamon. 1978.Heat treating, Vol 4, “Metals Handbook,” 9thed., American Society for Metals, 1981S.L. Semiatin and D.E. Stutz, “Induction HeatTreating of Steel”, American Society for Metals,1986S.L. Semiatin and S. Zinn, “Induction HeatTreating”, ASM International, 1988V. Rudnev et al., “SteelHeat TreatmentHandbook”, Marcel Dekker, Inc., 1997R.E. Haimbaugh, Induction Heat TreatingCorp., personal research.F.F. Mazda, "Power Electronics Handbook",Butterworth-Heinemann ltd. 1990.2nd 3].[24].Bimal K. Bose, ''Modern Power Electronics AndAC Drives", Pearson Education,Inc.2002.Muhammad H. Rashid and Tower “ElectronicsCircuits, Devices and applications". PrenticeHall International Editions. 1993.2nd Edition.P.C Sen, "Power Electronics", Tata-MacgrawHill Publications. 1990.Don L.Loveless, R.A.Cook and V.I Rudnev ,"Considering Nature And Parameters Of Power Supplies For Efficient Induction Heating'” Ind.heating, June 1995.pp 33-37.S.VaItchev, B.V.Borheg and J.B.Klaassens,"Series Resonant Converter Applied ToContactless Energy Transmission". InstituteDe Telecomunicacoes, Institute SuperiorTecnico. Email: bborges@lx.it.pt.Rex M. Davis, " Power Diode and ThyristorCircuits " IEE monograph series 7, 1976.Biswanath Paul, "Industrial Electronics andcontrol ",Prentice Hall of Inidia,2002.Mitsubishi corp. "Design Method of theInduction Heating Coil for Maximum. PowerDissipation under Voltage Constraint",www.mitsubishi.com.Discrete Application Power Device Division."Induction Heating System Topology Review ",Fairchild Semiconductor. July, 2000.Discrete Application Power DeviceDivision ”Induction Heating System TopologyReview”, Fairchild Semiconductor, July 2000.

Table 1Table 2Resonant Frequency 10 KHzResonant Frequency 5 KHzTest Piece:1Test Piece:1Diameter:0.5cmDiameter:0.5cmResonant Frequency:10 KHzResonant Frequency:5 KHzVin (AC):210 V(AC)Vin (AC):210 V(AC)Iin (AC):14AIin (AC):6.6APower Factor:0.77 laggingPower Factor:0.755 laggingPower Consumed from AC Source:1.5 KWPower Consumed from AC Source:1.4 KWKVAR consumed from AC Source:1.5 KVARKVAR consumed from AC Source:0.933 KVARVout (DC):170 V (DC)Vout (DC):192 V (DC)Iout (DC):7.3 AIout (DC):6.7 AIntial Temperature:25 o CelsiusIntial Temperature:25oCelsiusFinal Temperature:850 CelsiusFinal Temperature:850oCelsiusTime to heat up:05 secondsTime to heat up:11 secondsResonant Frequency 3 KHzTest Piece:1Diameter:0.5cmResonant Frequency:3 KHzVin (AC):210 V(AC)Iin (AC):7.7APower Factor:0.745 laggingPower Consumed from AC Source:1.2 KWKVAR consumed from AC Source:1.125 KVARVout (DC):200 V (DC)Iout (DC):6 AIntial Temperature:25oCelsiusFinal Temperature:850oCelsiusTime to heat up:13 secondsPower (kW)1.51.414131.2Input Current (Amp)11Time (sec)Power (kW)1.05Input Current (Amp)Table 3o76Time(sec)13510Frequency (kHz)Figure 4: Measurements of Power, current etc as a function of frequency.

HIGH FREQUENCY INVERTER FOR INDUCTION HEATING. Umar Shami. University of Engineering and Technology, Lahore. Pakistan. umarshami_99@yahoo.com ABSTRACT. Induction heating systems employ non-contact heating. Inducing heat electromagnetically rather than using a heating element in contact with a part to conduct heat, as does resistance heating.

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