《机电传动控制》学习笔记04
胡恒谦 机卓1301
一、设计要求
结合本周学习的交流电机原理及启动、调速、制动特性，用Modelica设计和仿真一个用三相交流异步电机带动起重机起升机构运行。具体要求如下：
1）实现如下机械运动周期：
控制电机带重物上升，从静止加速到800r/min
保持800r/min匀速运动0.5s，
减速到静止，保持静止状态0.5s，
带重物下降，从静止达到600r/min
保持600r/min匀速运动0.6s，
减速到静止。
(为了便于仿真，匀速和静止持续时间较短)
2) 升降机构和重物折算到到电机转子轴上的等效负载惯量为1Kg.m^2，折算到到电机转子轴上的等效负载转矩是15N.m。
3）使用统一的电机模型,如果控制策略中用到转子串电阻，允许将该电机的转子改为绕线式转子（参数不变）。
4）参照教材中给出的交流电机启动、调速和制动方法，设计控制策略，用Modelica实现控制策略并与电机模型实现联合仿真。
5）可以采用定子串电阻、转子串电阻、定子调压、定子调频等手段，但必须具备工程上的可实施性。
6）评价指标：快速启动、制动，冲击转矩和冲击电流小，能耗小，兼顾实施的经济性。
二、过程分析
本例中采用了自耦变压器降压启动，变频调速，反接制动的方式对交流电机的上述过程进行分析和仿真。
三、参数计算
本例中未给的数据都是通过一次次调试而得到的，没有进行理论计算和推倒，一步步调试的效率比较低，如果可以通过计算得到可以大大提高效率。
四、源代码
model SACIM "A Simple AC Induction Motor Model" type Voltage=Real(unit="V"); type Current=Real(unit="A"); type Resistance=Real(unit="Ohm"); type Inductance=Real(unit="H"); type Speed=Real(unit="r/min"); type Torque=Real(unit="N.m"); type Inertia=Real(unit="kg.m^2"); type Frequency=Real(unit="Hz"); type Flux=Real(unit="Wb"); type Angle=Real(unit="rad"); type AngularVelocity=Real(unit="rad/s"); constant Real Pi = 3.1415926; Current i_A"A Phase Current of Stator"; Current i_B"B Phase Current of Stator"; Current i_C"C Phase Current of Stator"; Voltage u_A"A Phase Voltage of Stator"; Voltage u_B"B Phase Voltage of Stator"; Voltage u_C"C Phase Voltage of Stator"; Current i_a"A Phase Current of Rotor"; Current i_b"B Phase Current of Rotor"; Current i_c"C Phase Current of Rotor"; Frequency f_s"Frequency of Stator"; Torque Tm"Torque of the Motor"; Speed n"Speed of the Motor"; Resistance Rs"Stator Resistance"; Torque Tl"Load Torque"; Flux Psi_A"A Phase Flux-Linkage of Stator"; Flux Psi_B"B Phase Flux-Linkage of Stator"; Flux Psi_C"C Phase Flux-Linkage of Stator"; Flux Psi_a"a Phase Flux-Linkage of Rotor"; Flux Psi_b"b Phase Flux-Linkage of Rotor"; Flux Psi_c"c Phase Flux-Linkage of Rotor"; Angle phi"Electrical Angle of Rotor"; Angle phi_m"Mechnical Angle of Rotor"; AngularVelocity w"Angular Velocity of Rotor"; parameter Resistance Rr=0.408"Rotor Resistance"; parameter Inductance Ls = 0.00252"Stator Leakage Inductance"; parameter Inductance Lr = 0.00252"Rotor Leakage Inductance"; parameter Inductance Lm = 0.00847"Mutual Inductance"; parameter Frequency f_N = 50"Rated Frequency of Stator"; parameter Voltage u_N = 220"Rated Phase Voltage of Stator"; parameter Real p =2"number of pole pairs"; parameter Inertia Jm = 0.1"Motor Inertia"; parameter Inertia Jl = 1"Load Inertia"; parameter Real K=0.8"starting rate"; parameter Real a=0.544"frequency rate"; parameter Real b=0.0683"stable frequency rate"; parameter Real c=0.3893"another frequency rate"; parameter Real P=0.7"stoping rate"; initial equation Psi_A = 0; Psi_B = 0; Psi_C = 0; Psi_a = 0; Psi_b = 0; Psi_c = 0; phi = 0; w = 0; equation u_A = Rs * i_A + 1000 * der(Psi_A); u_B = Rs * i_B + 1000 * der(Psi_B); u_C = Rs * i_C + 1000 * der(Psi_C); 0 = Rr * i_a + 1000 * der(Psi_a); 0 = Rr * i_b + 1000 * der(Psi_b); 0 = Rr * i_c + 1000 * der(Psi_c); Psi_A =(Lm+Ls)*i_A+(-0.5*Lm)*i_B+(-0.5*Lm)*i_C+(Lm*cos(phi))*i_a+(Lm*cos(phi+2*Pi/3))*i_b+(Lm*cos(phi-2*Pi/3))*i_c; Psi_B =(-0.5*Lm)*i_A+(Lm+Ls)*i_B+(-0.5*Lm)*i_C+(Lm*cos(phi-2*Pi/3))*i_a+(Lm*cos(phi))*i_b+(Lm*cos(phi+2*Pi/3))*i_c; Psi_C =(-0.5*Lm)*i_A+(-0.5*Lm)*i_B+(Lm+Ls)*i_C+(Lm*cos(phi+2*Pi/3))*i_a+(Lm*cos(phi-2*Pi/3))*i_b+(Lm*cos(phi))*i_c; Psi_a =(Lm*cos(phi))*i_A+(Lm*cos(phi-2*Pi/3))*i_B + (Lm*cos(phi+2*Pi/3))*i_C + (Lm+Lr)*i_a + (-0.5*Lm)*i_b + (-0.5*Lm)*i_c; Psi_b =(Lm*cos(phi+2*Pi/3))*i_A+(Lm*cos(phi))*i_B + (Lm*cos(phi-2*Pi/3))*i_C + (-0.5*Lm)*i_a + (Lm+Lr)*i_b + (-0.5*Lm)*i_c; Psi_c =(Lm*cos(phi-2*Pi/3))*i_A + (Lm*cos(phi+2*Pi/3))*i_B + (Lm*cos(phi))*i_C + (-0.5*Lm)*i_a + (-0.5*Lm)*i_b + (Lm+Lr)*i_c; Tm =-p*Lm*((i_A*i_a+i_B*i_b+i_C*i_c)*sin(phi)+(i_A*i_b+i_B*i_c+i_C*i_a)*sin(phi+2*Pi/3)+(i_A*i_c+i_B*i_a+i_C*i_b)*sin(phi-2*Pi/3)); w = 1000 * der(phi_m); phi_m = phi/p; n= w*60/(2*Pi); Tm-Tl = (Jm+Jl) * 1000 * der(w); Tl = 15; if time <= 10 then u_A = 0; u_B = 0; u_C = 0; f_s = 0; Rs = 0.531; elseif time<=1580 then f_s = f_N*a; Rs = 0.531; u_A = u_N * 1.414 * sin(2*Pi*f_s*time/1000)*a; u_B = u_N * 1.414 * sin(2*Pi*f_s*time/1000-2*Pi/3)*a; u_C = u_N * 1.414 * sin(2*Pi*f_s*time/1000-4*Pi/3)*a; elseif time<=1630 then f_s = f_N*a; Rs = 5; u_A = u_N * 1.414 * sin(2*Pi*f_s*time/1000-4*Pi/3)*a; u_B = u_N * 1.414 * sin(2*Pi*f_s*time/1000-2*Pi/3)*a; u_C = u_N * 1.414 * sin(2*Pi*f_s*time/1000)*a; elseif time<=2550 then f_s = f_N*a; Rs = 0.531; u_A = u_N * 1.414 * sin(2*Pi*f_s*time/1000-4*Pi/3)*a; u_B = u_N * 1.414 * sin(2*Pi*f_s*time/1000-2*Pi/3)*a; u_C = u_N * 1.414 * sin(2*Pi*f_s*time/1000)*a; elseif time<=3075 then f_s = f_N*b; Rs = 0.531; u_A = u_N * 1.414 * sin(2*Pi*f_s*time/1000)*b; u_B = u_N * 1.414 * sin(2*Pi*f_s*time/1000-2*Pi/3)*b; u_C = u_N * 1.414 * sin(2*Pi*f_s*time/1000-4*Pi/3)*b; elseif time<=3120 then f_s = f_N*K*c; Rs = 0.531; u_A = u_N * 1.414 * sin(2*Pi*f_s*time/1000-4*Pi/3)*K*c; u_B = u_N * 1.414 * sin(2*Pi*f_s*time/1000-2*Pi/3)*K*c; u_C = u_N * 1.414 * sin(2*Pi*f_s*time/1000)*K*c; elseif time<=4410 then f_s = f_N*c; Rs = 0.531; u_A = u_N * 1.414 * sin(2*Pi*f_s*time/1000-4*Pi/3)*c; u_B = u_N * 1.414 * sin(2*Pi*f_s*time/1000-2*Pi/3)*c; u_C = u_N * 1.414 * sin(2*Pi*f_s*time/1000)*c; elseif time<=4430 then f_s = f_N*P*a; Rs = 4; u_A = u_N * 1.414 * sin(2*Pi*f_s*time/1000)*a*P; u_B = u_N * 1.414 * sin(2*Pi*f_s*time/1000-2*Pi/3)*a*P; u_C = u_N * 1.414 * sin(2*Pi*f_s*time/1000-4*Pi/3)*a*P; elseif time<=5395 then f_s = f_N*a; Rs = 0.531; u_A = u_N * 1.414 * sin(2*Pi*f_s*time/1000)*a; u_B = u_N * 1.414 * sin(2*Pi*f_s*time/1000-2*Pi/3)*a; u_C = u_N * 1.414 * sin(2*Pi*f_s*time/1000-4*Pi/3)*a; else f_s = f_N*b; Rs = 0.531; u_A = u_N * 1.414 * sin(2*Pi*f_s*time/1000)*b; u_B = u_N * 1.414 * sin(2*Pi*f_s*time/1000-2*Pi/3)*b; u_C = u_N * 1.414 * sin(2*Pi*f_s*time/1000-4*Pi/3)*b; end if;end SACIM;
五、仿真结果
在Modelica中，根据一步步调试得到的参数，编写程序，并绘制Tm、n、i_A、i_a随时间变化的曲线，分别如下图所示。
图1 Tm和n随时间变化的曲线
图2 i_A随时间变化的曲线
图2 i_a随时间变化的曲线
六、结果分析
从曲线中可以看出，整个过程需要约5400ms，电机转矩最大值不超过190N.m。缺点是定子绕组和转子绕组的电流峰值比较大。