Electromechanical drive control fourth week operation

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.

Design Analysis:

The actual analysis shows that in order to achieve the required starting torque and minimize the starting current, the use of the self-coupling transformer buck start, in order to reduce the energy consumption of small stepless speed regulation and a larger speed range using variable frequency speed control, and finally adopt reverse braking mode of the AC motor of the above process analysis and simulation.

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 ';

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 = "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";

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 + $ * 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/p;

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

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

Tl = 15;

If time <=

u_a = 0;

U_b = 0;

U_c = 0;

f_s = 0; Rs = 0.531;

ElseIf time<=120 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<=1810 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<=1920 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<=2767 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<=3250 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<=3600 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<=4879 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<=4980 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<=5660 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;

The whole process needs 5400~5500ms, the maximum torque of the motor is not more than 180n.m current peak is relatively high.

Electromechanical drive control fourth week operation