The four-week simulation and learning experience

Work week Four

Android 1301 Storage Kingrex

First, the design task:

Combined with this week's study of the principle of AC motor and start, speed, braking characteristics, the use of Modelica design and simulation of a three-phase AC induction Motor drive hoist lifting mechanism operation. The specific requirements are as follows:

1) realize the following mechanical motion cycle:

Control motor with heavy lifting, acceleration from standstill to 800r/min

Keep 800r/min uniform motion 0.5s,

Slow down to standstill, stay stationary for 0.5s,

Take heavy weight down and reach 600r/min from standstill

Keep 600r/min uniform motion 0.6s,

Slow down to standstill.
(for simulation, constant and stationary durations are short)

2) The equivalent load inertia of the lifting mechanism and the heavy weight to the motor rotor shaft is 1kg.m^2, and the equivalent load torque on the rotor shaft of the motor is 15n.m.

3) using a uniform motor model, if the control strategy uses a rotor string resistor, allows the rotor of the motor to be changed to a winding rotor (parameter unchanged).

4) refer to the AC motor start, speed regulation and braking method given in the textbook, design control strategy, implement control strategy with Modelica and realize joint simulation with motor model.

5) The stator string resistor, the rotor string resistor, the stator voltage regulation, the stator frequency modulation and so on can be used, but it must have the engineering implementation.

6) Evaluation indicators: Fast start, braking, impact torque and impact current small, low energy consumption, taking into account the implementation of the economy.

7) The best student of the program is entitled "Star of Control" this week.

Second, theoretical deduction of three, simulation results

It can be seen from the simulation results that the scheme satisfies the problem and the time is very short.

Iv. Learning Experience

The calculation of three-phase AC motor is still more troublesome, but the formula on the book is not very troublesome, but to really understand it from its nature is a bit difficult, at least until now I only have a qualitative understanding, but believe that serious deduction, and DC motor analogy, and finally is understandable. At the same time still very admire Tesla, I think he also mastered the minimum principle of action, my understanding is the smallest effect is symmetrical, there are many extreme points are also in the symmetrical position, but for a moment can not find concrete examples.

Five, source code

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";

Parameter resistance Rs = 0.531 "Stator resistance";

Parameter resistance rr=8 "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 = "Rated Frequency of stator";

Parameter Voltage u_n = "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";

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 + $ * der (psi_a);

U_b = Rs * i_b + $ * der (Psi_b);

U_c = Rs * I_c + $ * der (Psi_c);

0 = Rr * i_a + $ * der (psi_a);

0 = rr* I_b + $ * der (Psi_b);

0 = Rr * i_c + $ * 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 = * der (phi_m);

Phi_m = PHI/3;

n= w*60/(2*PI);

TM-TL = (JM+JL) * + * der (W);

If time <= ten then

u_a = 0;

U_b = 0;

U_c = 0;

f_s = 0;

Tl = 0;

else if time<=1050 then

f_s = f_n;

u_a = u_n *1.732* 1.414 * sin (2*pi*f_s*time/1000);

U_b = u_n *1.732* 1.414 * sin (2*PI*F_S*TIME/1000-2*PI/3);

U_c = u_n *1.732* 1.414 * sin (2*PI*F_S*TIME/1000-4*PI/3);

Tl = 15;

else if time <=1550 then

f_s = f_n;

u_a = u_n * 1.414 * sin (2*pi*f_s*time/1000);

U_b = u_n * 1.414 * sin (2*PI*F_S*TIME/1000-2*PI/3);

U_c = u_n * 1.414 * sin (2*PI*F_S*TIME/1000-4*PI/3);

Tl = 15;

else if time <=1875 then

f_s = f_n;

u_a = u_n *1.732* 1.414 * sin (2*pi*f_s*time/1000);

U_b = u_n *1.732* 1.414 * sin (2*PI*F_S*TIME/1000-4*PI/3);

U_c = u_n *1.732* 1.414 * sin (2*PI*F_S*TIME/1000-2*PI/3);

Tl = 15;

else if time <=2375 then

u_a = 0;

U_b = 0;

U_c = 0;

f_s = 0;

Tl = 0;

else if time <=2825 then

f_s = f_n;

u_a = u_n *1.732* 1.414 * sin (2*pi*f_s*time/1000);

U_b = u_n *1.732* 1.414 * sin (2*PI*F_S*TIME/1000-4*PI/3);

U_c = u_n *1.732* 1.414 * sin (2*PI*F_S*TIME/1000-2*PI/3);

Tl = 15;

else if time <=3450 then

f_s = F_N*6/11;

u_a = u_n * 1.414 * sin (2*pi*f_s*time/1000);

U_b = u_n * 1.414 * sin (2*PI*F_S*TIME/1000-4*PI/3);

U_c = u_n * 1.414 * sin (2*PI*F_S*TIME/1000-2*PI/3);

Tl = 15;

else if time <=3700 then

f_s = f_n;

u_a = u_n * 1.414 * sin (2*pi*f_s*time/1000);

U_b = u_n * 1.414 * sin (2*PI*F_S*TIME/1000-2*PI/3);

U_c = u_n * 1.414 * sin (2*PI*F_S*TIME/1000-4*PI/3);

Tl = 15;

else if time <=3895 then

f_s = f_n;

u_a = u_n *1.732* 1.414 * sin (2*pi*f_s*time/1000);

U_b = u_n * 1.732*1.414 * sin (2*PI*F_S*TIME/1000-2*PI/3);

U_c = u_n *1.732* 1.414 * sin (2*PI*F_S*TIME/1000-4*PI/3);

Tl = 15;

Else

u_a = 0;

U_b = 0;

U_c = 0;

f_s = 0;

Tl = 0;

End If;

End If;

End If;

End If;

End If;

End If;

End If;

End If;

End If;

End SACIM;

Simulate (sacim,starttime=0,stoptime=6000)

Plot (N)

Plot (i_a)

Plot (i_a)

The four-week simulation and learning experience