Deign Calculation and Performance Teting of Heating Coil in Induction Surface Hardening Machine Soe Sandar Aung, Han Phyo Wai, and Nyein Nyein Soe Abtract The induction hardening machine are utilized in the indutrie which modify machine part and tool needed to achieve high ware reitance. Thi paper decribe the model of induction heating proce deign of inverter circuit and the reult of induction urface hardening of heating coil. In the deign of heating coil, the hape and the turn number of the coil are very important deign factor becaue they decide the overall operating performance of induction heater including reonant frequency, Q factor, efficiency and power factor. The performance will be teted by experiment in ome cae high frequency induction hardening machine. Keyword Induction Heating, Reonant Circuit, Inverter Circuit, Coil Deign, Induction Hardening Machine. T I. INTRODUCTION HE principle of induction heating i hown in Fig., there an electric conductor uch a iron or teel placed in the inductor i heated rapidly by induced eddy current caued by electromagnetic induction, and hyteretic heat lo, which i generated by vibration and friction of each molecule in magnetic material under AC magnetic flux. In induction heating, a the frequency of the heating current tend to concentrate cloe to the metal urface (work ).Thi i referred to a the kin effect. The kin effect i the phenomenon, which electric current flow only in the limited area near urface of conductive material, and proximity effect i the phenomenon, which the primary current in the inductor and the econdary current in the conductive material pull each other becaue the direction of current i oppoite each other, and flow in the limited area near urface where ditance i nearet each other. The depth depend upon the frequency and a the frequency i higher, the depth become maller. [] The penetration depth δ i calculated a follow; δ = (m) () πfµσ δ = penetration depth, m µ = pecific permeability f = frequency, Hz Thi formula how that a the frequency i higher, δ will be maller and the heating will be concentrated a the urface in cae the material are ame. However in actual heating, the heated tend to become bigger becaue of heat conduction in the heated material. Alternating current Work coil Fig. Baic Induction Type Heating Sytem Single phae power upply II. SYSTEM CONFIGURATION Non controlled rectifier circuit Inverter circuit Triggering cirrcuit Work coil Induced current Alternating magnetic field Work Fig. how the general block diagram of the induction heating ytem. The AC power ource i ingle phae and it applie line frequency and line voltage. The non controlled rectifier convert the AC voltage to the DC value and applie the deired DC current to the inerter circuit. The inverter change the DC ignal to the AC ignal with deired frequency to apply the work coil. When the work ha been heated for a time, the quenching ytem i applied to the work.[] Fig. General Block Diagram Soe Sandar Aung i with the Electrical Power Engineering Department, Mandalay Technological Univerity, Mandalay, Myanmar (correponding author to provide phone: 095-067-3; e-mail: oeandarag@gmail.com). Han Phyo Wai i with the electrical Power Engineering Department, Mandalay Technological Univerity, Mandalay, Myanmar (e-mail: hanphyowai@007.com). Nyein Nyein Soe i with the Electrical Power Engineering Department, Mandalay Technological Univerity, Mandalay, Myanmar (e-mail: nyeinnoe@gmail.com). A. Equivalent Circuit III. SYSTEM ANALYSIS The work coil and work have the pecial property of reitance and reactance value due to their reitivity and inerted flux. Uing Wheeler formula, the inductance of the work coil can be calculated a follow. PWASET VOLUME 3 AUGUST 008 ISSN 070-3740 446 008 WASET.ORG
r out N L c = () 0.054( 9r out + 0l wc ) Where L c = inductance of work coil, µh r out = outer radiu of work coil, m l wc = length of work coil, m Fig. 3 Impedance Circuit of Work Coil and Work Piece The work coil and work can be repreented by an equipment erie inductance and reitance model a hown in Fig. 4. L eq = L c + M (3) R = R + R (4) eq c w Where M = magnetizing inductance, H L eq =equivalent inductance of work coil and work R eq = equivalent reitance of work coil and work C. Serie Parallel Reonant Inverter Thi configuration ha the deirable characteritic of erie and parallel reonant inverter. The load hort circuit and the no load regulation are poible. High part-load efficiency i poible with the proper choice of reonating component. A reonant inverter can be operated either below or above reonance frequency. Thi inverter contain impedance matching ytem. The tank circuit incorporating the work coil (L w ) and it capacitor (C w ) can be though of a a parallel reonant circuit Thi ha a reitance (R) due to the lo work coupled into the work coil due to the magnetic coupling between the two conductor. In practice, the reitance of the work coil, the reitance of the tank capacitor and the reitance of the work all introduce a lo into the tank circuit and damp reonance. Therefore, it i ueful to combine all of thee cae into a ingle lo reitance. In the cae of parallel reonant circuit thi lo reitance appear directly acro the tank circuit. Thi reitance repreent the only component that can conume power and therefore it can be though of reitance a the load that it i being tried to drive power into a efficiently a poible. C W L W R Fig. 6 Circuit Diagram of Tank Circuit A CS LS L w Fig. 4 Equivalent Circuit of Work Coil and Work Piece B. Reonant Circuit A hown in Fig. 4, the equipment inductance and reitance of work coil and work are in erie connection. To reonate the circuit a capacitor i connected in parallel reonant circuit and it i hown in Fig. 5. B R w Cp Cw R L Fig. 7 Diagram of Matching Network I T C I c L eq R eq Fig. 5 Reonant Circuit for the Load If the capacitor i charged to a upply voltage, the energy T tored in CV. And thi energy tranfer to the inductance L eq and return again to the capacitor o the frequency of the ocillation depend on the value of inductance and capacitance. In the circuit, the diipated energy in reitance R eq, and after each cycle of ocillation the tore of energy in the capacitor i reduced. IV. REQUIRED SPECIFICATIONS FOR INDUCTION SURFACE HARDENING MACHINE The pecification for operating are the ambient temperature i aumed 300.5 K, the deired hardened temperature i 6.48 K, the duration of hardened time i 0 ec, the output power i 5 kw and the ue of apply frequency i 35 khz. Table I i for the pecification of conductor ued a work coil. TABLE I SPECIFICATIONS OF CONDUCTOR Unit Specification value - material copper Ωm reitivity.7 0-8 (at 93.5 K) Hm permeability kg/m denity 786.3 PWASET VOLUME 3 AUGUST 008 ISSN 070-3740 447 008 WASET.ORG
TABLE II SPECIFICATIONS OF WORK PIECE Unit Specification value - Material 040 carbon teel Ωm Reitivity.7 0-8 (at 93.5 K) 5.6 0-8 (at 53.5 K) Hm Permeability J/kg.K Specific heat 434 (at 300 K ) 69 (at 000 K ) K Melting temperature 794.6 K Hardened temperature 6.48 _ 7.03 kg/m Denity 786.3 V. CALCULATIONS OF INDUCTION SURFACE HARDENING MACHINE A. Calculation of Work Coil The number of turn of work coil i mainly baed on the length of work and the pitch of coil winding. Thu, l w N= (5) d c +C p N = number of turn of work coil L w = length of work to be hardened, m And the inner diameter of work coil i D in =d w +C p (6) The outer diameter of work coil i D out =D in +d c (7) d w = diameter of work coil, m d c = diameter of conductor, m The total length of conductor for work coil i l c =l lead +N (π m ) +(.5 d c ) (8) l c = length of conductor, m l lead = length of work coil lead, m r m = inner radiu of work coil, m The minimum thickne of conductor mut be at leat two time of depth of current penetration in conductor itelf. Therefore, the minimum thickne of conductor i t c =δ c t c = minimum thickne of conductor, m δ c = depth of current penetration in conductor, m The depth of current penetration in conductor i δ c = (9) πfµ c µ o σ c µ c = permeability of conductor, H/m µ o = permeability of free pace, H/m σ c = electric conductivity of conductor, mho/m f = applied frequency, Hz B. Calculation of Impedance Matching Sytem L C Q = (0) R L ω F = () ω o From Equation (0) and (), L S = 0.03385 mh C S = 0.753953 µf C P = 0.753953 µf The capacitor in the matching net work (C P ) and tank capacitor (C w ) are both in parallel. In practice, both of thee function are uually accomplihed by a ingle capacitor. C pw = C p + C w =.796509 µf A B + V AB _ C L S C pw Fig. 8 Circuit Diagram of Matching Sytem Zcpw =R - jx cpw = - j = - j.5378 ωc pw I c = V c Z cpw I Ic 9 = -j.5378 = j47amp I t - I = I c = 6.0496 + j.655880 = 7.760837(θ =54.70 ) V AB =I Z - V c Z =jx l - jx c = L jω - jω C = j7.978 j6.03583 = j.66564 V AB = -47.69606 +j0.37976 =49.087088(θ =7.7 ) Required voltage for matching ytem i V AB = 49.087088Volt Required current for matching ytem i I = 7.760837Amp (θ =54.70 ) The elected erie capacitor C S i 0.8 µf, 600 Volt. The elected erie inductor L S i 0.03 mh, 600 Volt, Amp. V C I t L w R w R L PWASET VOLUME 3 AUGUST 008 ISSN 070-3740 448 008 WASET.ORG
The elected parallel capacitor C pw i.796507 µf, 600 Volt. C. Calculation of Voltage and Current Rating for Inverter Device voltage and current rating mut to be atified upply bu voltage and the load impedance o that power can be delivered to the load. The required voltage for the load i V AB =49.087088 Volt. The upply dc voltage i 49.087088 volt. Peak of upply voltage = 49.087088 = 0.84098 Volt The inverter i driven high frequency witching. Thi i upplied by inductance load. D. Calculation of Single Phae Rectifier Circuit Inverter input voltage E d = 49.087088 Volt Inverter input current I d = 7.760837 Amp So, required dc voltage E d = 49.087088 49 Volt Required dc current I d = 7.760837 8 Amp Average load voltage V 0(avg) = 0.636 V m V m i peak load voltage. V= V RMS V RMS i upplied voltage RMS value. E d = V 0 (avg) V m = V 0(avg) / 0.636 = 49.087088 / 0.636 = 34.76730 Volt Supply voltage for ytem = 34.76730 / = 65.658664 66 Volt Required upply voltage i 66 Volt to 0 Volt RMS value of load current = average load current = 7.760837 Amp Average current in each diode I D(avg) = I 0(avg) / = 3.88049 4 Amp I o(avg) Peak load current, I m = =43.6495Amp 0.636 Supply current for ytem, 43.6495 Irm = = 30.864585 3Amp Required power =VI =5.597 kw A.C 0V, + TABLE III RESULT FOR WORK PIECE - Material 040 carbon teel - hape cylindrical - Nature of urface uniform m Depth of hardne 0.0009587 m Diameter 0.067008 m Length 0.033504 m Cro ectional area 0.00099 m Surface area 0.007053 µm 3 Volume 6.66507 TABLE IV RESULTS FOR WORK COIL - hape round - number of turn 4 m inner diameter 0.07084 m outer diameter 0.08884 m Length 0.038 m coil pitch 0.00375 m coupling ditance 0.00588 TABLE V RESULTS FOR CONDUCTOR - material copper - hape round m thickne 0.00070 m diameter 0.00635 m length.878 TABLE VI RESULT FOR ELECTRICAL PROPERTIES OF THE SYSTEM Ω Reitance of work coil 0.0034 Ω Reitance of work 0.0 µh Inductance of work coil.434858 µh Magnetizing inductance 0.553 µf Reonated capacitance.04355 - Power factor 0.7379 - Quality factor 3.5809 Ω Total impedance.658596 A Supply current 7 V Supply voltage 9 Fig. 9 Circuit Diagram of Rectifier Circuit VI. DESIGN RESULTS D.C - The reult for work, conductor, work coil and electrical propertie of the ytem are calculated. The reult are hown in table repectively. VII. PERFORMANCE TESTING A. Teting of Control Circuit Wave hape, frequency and voltage value at the input and output of control circuit are meaured with ocillocope. Reulting wave are quare wave and the wave hape are hown in Fig. 0. PWASET VOLUME 3 AUGUST 008 ISSN 070-3740 449 008 WASET.ORG
PROCEEDINGS OF WORLD ACADEMY OF SCIENCE, ENGINEERING AND TECHNOLOGY VOLUME 3 AUGUST 008 ISSN 070-3740 Fig. 3 Wave Shape of Inverter Output with Tank Circuit ACKNOWLEDGMENT Fig. 0 IGBT gate driver circuit (for tart heating) Firtly, the author would like to expre her deepet great thank to her parent. The author deeply want to expre her pecial appreciation to Dr. Ni Ni Win, Department of Electrical Power Engineering Department, Mandalay Technological Univerity, for her invaluable upport and advice. REFERENCES [] [] [3] Fig. IGBT gate driver circuit (after heating) B. Performance Teting of Inverter Firt, the inverter output i meaured without tank circuit a hown in Fig. and reulting wave hape i quare wave with pite. Then, the inverter i concerned with tank capacitor and meaured. The reulting wave hape i pure ine wave. The wave hape are hown in Fig. 3. Curit,F.W.944. High Frequency Induction Heating. ted. New York: McGraw-Hill Book Company, Inc. Zinn S., and Semiatin, S.L. 988. Coil Deign and Fabrication: Baic Deign and Modification. July 005. Available: http://www.ameritherm.com Bhattachrya, S.K and Chute, R.D. 97. Indutrial Electronic and Control. New Delihi; Tata MC Graw-Hill Publihing Company Ltd. Soe Sandar Aung tudied in Electrical Power Engineering Major and held B.E degree in 004 from Mandalay Technological Univerity, Mandalay, Myanmar. Then I wa awarded M.E degree of Electrical Power Engineering in 006 from Yangon Technological Univerity, Yangon, Myanmar. I am now tudying and making induction heating reearch in my Univerity.. Fig. Wave Shape of Inverter Output without Tank Circuit PWASET VOLUME 3 AUGUST 008 ISSN 070-3740 450 008 WASET.ORG