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《机电传动控制》学习笔记04-1
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。
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原文地址：http://www.cnblogs.com/xiaobaicai05/p/5300125.html
