• Abstract and Figures
• Public Full-text

DESIGN METHOD OF AN OPTIMAL INDUCTION HEATER CAPACITANCE FOR MAXIMUM POWER DISSIPATION AND MINIMUM POWER LOSS CAUSED BY ESR Jung-gi Lee, Sun-kyoung Lim, Kwang-hee Nam ¤ and Dong-ik Choi ¤¤ ¤ Department of Electrical Engineering, POSTECH University, Hyoja San-31, Pohang, 790-784 Republic of Korea. Tel:(82)54-279-2218, Fax:(82)54-279-5699, E-mail:[email protected] ¤¤ POSCO Gwangyang Works, 700, Gumho-dong, Gwangyang-si, Jeonnam, 541-711, Korea. Abstract: In the design of a parallel resonant induction heating system, choosing a proper capacitance for the resonant circuit is quite important. The capacitance a®ects the resonant frequency, output power, Q-factor, heating e±ciency and power factor. In this paper, the important role of equivalent series resistance (ESR) in the choice of capacitance is recognized. Without the e®ort of reducing temperature rise of the capacitor, the life time of capacitor tends to decrease rapidly. This paper, therefore, presents a method of ¯nding an optimal value of the capacitor under voltage constraint for maximizing the output power of an induction heater, while minimizing the power loss of the capacitor at the same time. Based on the equivalent circuit model of an induction heating system, the output power, and the capacitor losses are calculated. The voltage constraint comes from the voltage ratings of the capacitor bank and the switching devices of the inverter. The e®ectiveness of the proposed method is veri¯ed by simulations and experiments. Keywords: Induction heating, Equivalent series resistance, Design 1. INTRODUCTION four thyristors and a parallel resonant circuit com- prised capacitor bank and heating coil. Thyris- tors are naturally commutated by the ac current Induction heating is widely used in metal industry flowing through the resonant circuit. Therefore, for melting or heating of thin slab in a contin- this type of inverter called as a load commutated uous casting plant because of good heating e±- inverter (Y. kwon, 1999; F.P. Dawson, 1991; R. ciency, high production rate, and clean working Bonert 1994). environments. A typical parallel resonant inverter circuit for induction heater is shown in Fig. 1. The The induction heater was kept up the metal slab of o phase controlled recti¯er provides a constant DC 1100 C for the next milling process. In the parallel current source. The H-bridge inverter consists of resonant inverter, if the switching frequency is 3-phase H-bridge Heating Rectifier Idc Inverter is iL Coil DC link Inductor vC Capacitor Heated Bank Slab (a1, a2, a3, a4, a5, a6 ) Power VoltageLimiter Reference Firing angle a, P* Vmax vC S DC link current Idc control + -V - max Output Power P Output Power vC Calulator 1 i P = V I cosa s out 2 C s Fig. 1. Block diagram of induction heating system. closer to the resonant frequency for high power, elled as an inductance plus a series resistance, and higher voltage is generated at capacitor bank the capacitor bank is modelled as a pure capac- (F.P. Dawson, 1991). However, due to the limit in itance with an equivalent series resistance(ESR) the voltage tolerance of the capacitor bank, the (Dec.2002 issue of Practical Wireless). Utilizing inverter output voltage vC needs to be limited this model, the output power of the induction below the rated voltage Vmax. One method of heater and the power loss in the capacitor are limiting vC is to reduce the DC-link current Idc derived as functions of capacitance. An optimal by increasing the ¯ring angle of the recti¯er. value of the capacitance under the voltage con- However, it results in the decrease of Idc, thereby straint is found with the help of Lagrange multi- the decrease of the output power. plier (D.G. Luenberger, 1989). At the mini-mill in a POSCO steel plant, eight This procedure looks very useful in the design induction heaters of load commutated type were of the induction heating system, speci¯cally, in installed to heat thin slabs for next milling pro- determining the size of capacitor bank. All the cess in the continuous casting plant. The rated parameters used in this work come from the output power of each induction heater is 1.5 MW. induction heater operating in POSCO. However, there were two problems: One was the insu±ciency in the output power, and the other was the frequent damages of the capacitor bank. 2. EQUIVALENT CIRCUIT AND POWER Insu±ciency in output power was caused by a EQUATIONS poor power factor of the inverter. On the other hand, the damage to the capacitor bank was due In general, the heating coil and the load are to a little voltage margin between v and V , C max modelled as a transformer with a single turn sec- and it resulted in a large power dissipation in the ondary winding as shown in Fig. 3(a). Almost capacitor causing a high temperature rise. all magnetic flux generated by the induction coil The capacitance of the capacitor bank a®ects the (primary winding) penetrates into the slab (sec- overall operating factors of induction heater such ondary winding). Hence, in the secondary circuit as resonant frequency, Q-factor, e±ciency, and no leakage inductance appears and the coupling power factor (P. Jain,1988; E. J. Davis,1979; E. coe±cient is equal to one. The secondary circuit J. Davies, 1990). Hence, in this work, we pro- can be moved to the primary part as shown in pose a method of choosing an optimal capacitance Fig. 3(b) (E. J. Davis, 1979). Denoting the mutual value Copt, which maximizes the output power, inductance by LM , it follows that and at the same time, minimizes the capacitor loss. Firstly, an equivalent model of the induction heater is developed based on previous works (Y. kwon, 1999; T. Imai, 1997; E. J. Davis, 1979; E. J. L1 = Ll + NLM ; (1) L Davies, 1990). The heating coil and slab is mod- L = M ; (2) 2 N 1:2 10¡6 ­ F to 1:5 10¡6 ­ F (Dec.2002 issue of Practical£ Wireless).¢ £ ¢ Fig. 2. Model of slab for one turn coil. where L1 and Ll denote the inductance of the heating coil and the leakage inductance, re- spectively. Using Wheeler's formula (Harold A. Wheeler, 1942), the inductance of the heating coil is calculated as follows: r2 N 2 L (¹H) = £ ; (3) 1 9r + 10h where r is the radius, N is the turn number and h is the height of the coil. The slab resistance RL for one turn coil is given by L 2(w + 2b) R = ½ = ½ ; (4) L A l± where L and A are length and area of eddy cur- rent, l is the e®ective length of the slab occupied by one turn coil and b and w are de¯ned in Fig. 2 (S. Zinn, 1988), ± and ½ are skin depth almost distributed over the surface of slab and electrical resistivity of the material. Simpli¯ed equivalent Fig. 3. (a) Based circuit of the induction heater. model for a transformer can be represented in (b) Equivalent circuit of the induction heater Fig. 3(c) by an equivalent inductance Leq and based on a transformer concept. (c) Simpli- resistance Req (Y. kwon, 1999; E. J. Davis, 1979). ¯ed equivalent circuit of the induction heater. These equivalent parameters depend on several variables including the shape of the heating coil, A useful variable to calculate the power is a total the spacing between the coil and slab, the elec- impedance seen from the impedance of equivalent trical conductivity and magnetic permeability of circuit in Fig. 3(c) (H. Kojima, 1990). The total the slab, and the angular frequency of the varying impedance is given by current !s (E. J. Davis, 1979). (ZC + ZESR) (ZL + ZR) Zt = ¢ 2 Z + Z + Z + Z Leq = L1 A L2; (5) L R C ESR ¡ 2 2 (!skReq + !sLeq) + j(!s Leqk Req) Req = R1 + A RL; (6) = ¡ ;(7) !s(CReq + k) + j(!sLeqC 1) ¡ where R1 denotes the resistance of the heating Z Z Z coil, RL denotes the resistance of the heated slab, where L = j!sLeq, C = 1 j!sC, R = Req, 2 2 2 and ZESR = RESR = k=C. and A = !sLM !s L2 + RL . It is noted that ± the inductance of heating coil L1 is not a®ected by The recti¯er and H-bridge inverter of the induc- ±p the existence of the slab in the heating coil, since tion heater are represented by a square waved at about 1100oC temperature the permeability of current source whose magnitude is equal to the the iron slab is equal to that of air, i.e., ¹ = 4¼ DC-link current I . Therefore, the current source £ dc 10¡7 (H/m) (E. J. Davis, 1979). expanded in a fourier series is described as follows: To represent the power dissipation in the capaci- 1 4I tor bank, it is modelled by a pure capacitance C i (t) = dc sin n! t n = 1; 3; 5; : : : :(8) s n¼ s and an equivalent series resistance (ESR) R . n¡1 ESR X It is noted that RESR is inversely proportional to the capacitance, hence, it is modelled as RESR = The fundamental component of the square waved k C, where k is a coe±cient of ESR ranged from current source is given by ± 4 where VC is the peak of vC , Vmax is rated voltage is(t) = Idc sin !st = Is sin !st; (9) ¼ of the capacitor bank, and Zt is total impedance of capacitor bank and heating parts. One can see where Is = 4Idc=¼ is a peak of is.

Load more

  7

Copy Link