Tham khảo tài liệu 'electric vehicles modelling and simulations part 13', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | Minimization of the Copper Losses in Electrical Vehicle Using Doubly Fed Induction Motor Vector Controlled 349 PM rr T 1 Ce 3m Mr 1 ữLs Lr L J 4 The copper losses are giving as Pcl 5 The motion equation is r A _ T d Ce-d 1 e dt 6 In DFIM operations the stator and rotor mmf s magneto motive forces rotations are directly imposed by the two external voltage source frequencies. Hence the rotor speed becomes depending toward the linear combination of theses frequencies and it will be constant if they are too constants for any load torque given of course in the machine stability domain. In DFIM modes the synchronization between both mmf s is mainly required in order to guarantee machine stability. This is the similar situation of the synchronous machine stability problem where without the recourse to the strict control of the DFIM mmf s relative position the machine instability risk or brake down mode become imminent. 3. Nonlinear vector control strategy Double flux orientation It consists in orienting at the same time stator flux and rotor flux. Thus it results the constraints given below by 7 . Rotor flux is oriented on the d-axis and the stator flux is oriented on the q-axis. Conventionally the d-axis remains reserved to magnetizing axis and q-axis to torque axis so we can write Drid et al. 2005a 2005b I sq l s - I I r 7 sd I rq 0 Using 7 the developed torque given by 4 can be rewritten as follows Ce kc s r. 8 PM where kc - c ơLsLr Ộ 5 Appears as the input command of the active power or simply of the developed torque while ộr appears as the input command of the reactive power or simply the main magnetizing machine system acting. 350 Electric Vehicles - Modelling and Simulations Vector control by Lyapunov feedback linearization Separating the real and the imaginary part of 3 we can write d sd _r . a fi usd d sq . .a f2 usq d rd a f3 u d d rq . jdT f4 urq Where fl f2 f3 and f4 are done as follows f1 Y1 sd -Y2 rd - s sq - f2 Y1 sq -Y2 rq s sd - f3 -Y3
