Study notes (fourth week)

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.

Analysis

⑴ start: Using the self-coupling transformer buck start, take k=0.8

⑵ Speed regulation: Adopt frequency conversion speed regulation. Three frequencies for each of the two

N=800,F=0.54*FN;

N=0,F=0.069*FN;

N=-600,f=0.39*fn.

These three frequencies should be calculated using the formula, but the individual ability is limited, after many tests, reference to the parameters of Chong students to obtain the three frequency. The use of variable frequency speed regulation is because of wide speed range, good speed and smoothness, high economic benefits.

⑶ braking: According to the book, "Hoist down the heavy, in order to make the descent speed is not too fast, it is commonly used in this state of work", here "this state" refers to the reverse pull brake. The resistors in series here are set to r=2 ohm.

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";

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 = "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=0.54 "frequency rate";

Parameter Real b=0.069 "stable frequency rate";

Parameter Real c=0.39 "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<=180 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<=1870 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<=1930 Then

f_s = F_n*a; Rs = 2;

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<=2841 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<=3450 Then

f_s = f_n*b; Rs = 0.531;

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;

ElseIf time<=3571 Then

f_s = F_n*k*c; Rs = 0.531;

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;

ElseIf time<=4980 Then

f_s = F_n*c; Rs = 0.531;

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;

ElseIf time<=5050 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<=6000 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

f_s = f_n*b; Rs = 0.531;

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;

End If;

End SACIM;

Conclusion

0-1000 the speed increases uneven, the stability is general, the whole impact torque does not exceed 200n.m.

Study notes (fourth week)