Work week Four

Work week Four

Design content:

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:

L control motor with heavy lifting, from stationary acceleration to 800r/min

L Keep 800r/min Uniform motion 0.5s,

L slow down to standstill and remain stationary for 0.5s,

L with heavy weight drop, reach 600r/min from standstill

L Keep 600r/min Uniform motion 0.6s,

L 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 .

Solution Ideas:

Directly run the given program, you can get the starting curve, according to the control design requirements of simple control time and the stator of the magnetic field of the steering can be preliminarily obtained simulation curve, the main code is as follows:
If time <=

u_a = 0;

U_b = 0;

U_c = 0;

f_s = 0;

Tl = 0;

ElseIf Time <= 280 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;

ElseIf Time <=

f_s = f_n;

u_a = 0;

U_b = 0;

U_c = 0;

Tl = 0;

ElseIf time <= 974 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/1000-4*PI/3);

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

Tl = 15;

ElseIf Time <=1500 Then

f_s = f_n;

u_a = 0;

U_b = 0;

U_c = 0;

Tl = 0;

ElseIf Time <=1625 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/1000-4*PI/3);

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

Tl = 0;

ElseIf Time <=2125 Then

f_s = f_n;

u_a = 0;

U_b = 0;

U_c = 0;

Tl = 0;

ElseIf Time <= 2300 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

f_s = f_n;

u_a = 0;

U_b = 0;

U_c = 0;

Tl = 0;

End If;

Simulation Image:

Appeal graphics can meet the first requirement, but the analysis can be clearly seen in the motor speed changes in the current amplitude is large, so in order to reduce the starting current, here is considered in the stator Rs based on series resistance to achieve buck start.

Work week Four