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第四周仿真作业

發布時間:2023/11/29 编程问答 29 豆豆
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經過第四周關于交流電機的學習,我們通過對起調速特性的了解,進行仿真:

  • 控制電機帶重物上升,從靜止加速到800r/min
  • 保持800r/min勻速運動0.5s,
  • 減速到靜止,保持靜止狀態0.5s,
  • 帶重物下降,從靜止達到600r/min
  • 保持600r/min勻速運動0.6s,
  • 減速到靜止。
    (為了便于仿真,勻速和靜止持續時間較短)

分析:這里,由于自耦降壓啟動具有K值方便直接調節的特點,所以我們采用此種啟動方法。由于變頻調速具有調速范圍廣、平滑性好、能耗小無級調速,故采用變頻調速。制動采用相對方便的反接制動。

參數:我們需要設置K,a,b,c 4個參數,由于不方便計算,都是通過觀察圖像進行調試,從而得出近似值。我是將K先固定,通過調節a,使轉速穩定時接近800,調節b,使其接近制動,調節c,使轉速穩定時接近600.

從而我這里得出的參數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:

?

轉載于:https://www.cnblogs.com/cyykhyql/p/5305839.html

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