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。
2、过程分析：
由于自耦变压器启动时的K是可以调节的因此启动转矩可调，而且可以得到较小的启动电流，故采用自耦变压器降压启动。由于变频调速具有调速范围广、平滑性好、能耗小无级调速，故采用变频调速。而制动则采用反接制动，因为反馈制动用于电机的运行速度高于其同步转速与此情景不符合，能耗制动在制动的后阶段，随着转速的降低，能耗制动转矩也很快减小，因此制动效果比反接制动差。
3、参数确定：
由于理论推导能力的不足，仿真过程中需要的时间以及K,a,b,c只能通过修改数据一步步通过图像调试出来的。而本人调试能力有限，因此只能给出代码的模板，而里面的参数还没有确定下来。
4、仿真代码：
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";
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";
Torque Tl"Load Torque";
Resistance Rs"Stator Resistance";
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 = 0.1"Load Inertia";
parameter Real K=0.8"starting rate";
parameter Real a=0.5438"frequency rate";
parameter Real b=0.06891"stable frequency rate";
parameter Real c=0.4079"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);
= Rr * i_a + 1000 * der(Psi_a);
= Rr * i_b + 1000 * der(Psi_b);
= 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 = 10;
if time <= 100 then
u_A = 0;
u_B = 0;
u_C = 0;
f_s = 0;Rs = 0.531;
elseif time<=106 then
f_s = f_N*a;
Rs = 0.531;
u_A = u_N * 1.414 * sin(2*Pi*f_s*time/1000)*K*a;
u_B = u_N * 1.414 * sin(2*Pi*f_s*time/1000-2*Pi/3)*K*a;
u_C = u_N * 1.414 * sin(2*Pi*f_s*time/1000-4*Pi/3)*K*a;
elseif time<=1705 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<=1755 then
f_s = f_N*a;Rs = 3;
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<=2680 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<=3195 then
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;
f_s = f_N*b;Rs = 0.531;
elseif time<=3315 then
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;
f_s = f_N*K*c;Rs = 0.531;
elseif time<=4560 then
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;
f_s = f_N*c;Rs = 0.531;
elseif time<=4595.5 then
f_s = f_N*P*a;Rs = 2;
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<=5560 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
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;
f_s = f_N*b;Rs = 0.531;
end if;
end SACIM;
5、仿真图像
6、仿真结果：
整个过程用时8秒，转矩的变化较剧烈，最大转矩不超过185，最后阶段制动过程不理想，需要进一步调试