經(jīng)過第四周關(guān)于交流電機(jī)的學(xué)習(xí)，我們通過對(duì)起調(diào)速特性的了解，進(jìn)行仿真：

• 控制電機(jī)帶重物上升，從靜止加速到800r/min
• 保持800r/min勻速運(yùn)動(dòng)0.5s，
• 減速到靜止，保持靜止?fàn)顟B(tài)0.5s，
• 帶重物下降，從靜止達(dá)到600r/min
• 保持600r/min勻速運(yùn)動(dòng)0.6s，
• 減速到靜止。
  (為了便于仿真，勻速和靜止持續(xù)時(shí)間較短)

分析：這里，由于自耦降壓?jiǎn)?dòng)具有K值方便直接調(diào)節(jié)的特點(diǎn)，所以我們采用此種啟動(dòng)方法。由于變頻調(diào)速具有調(diào)速范圍廣、平滑性好、能耗小無(wú)級(jí)調(diào)速，故采用變頻調(diào)速。制動(dòng)采用相對(duì)方便的反接制動(dòng)。

參數(shù)：我們需要設(shè)置K，a,b,c 4個(gè)參數(shù)，由于不方便計(jì)算，都是通過觀察圖像進(jìn)行調(diào)試，從而得出近似值。我是將K先固定，通過調(diào)節(jié)a,使轉(zhuǎn)速穩(wěn)定時(shí)接近800，調(diào)節(jié)b,使其接近制動(dòng)，調(diào)節(jié)c，使轉(zhuǎn)速穩(wěn)定時(shí)接近600.

從而我這里得出的參數(shù)K=0.8，a=0.54,b=0.07,c=0.405.

代碼：

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=0.54"frequency rate";
parameter Real b=0.07"stable frequency rate";?
parameter Real c=0.405"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<=1706 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<=1756 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<=2679 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<=3196 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<=3310 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;

得出的仿真圖像：

?

轉(zhuǎn)載于:https://www.cnblogs.com/cyykhyql/p/5305839.html

總結(jié)

以上是生活随笔為你收集整理的第四周仿真作业的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。