Electromechanical Transmission Control fourth week simulation operation

Last Update:2016-03-20 Source: Internet

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Electromechanical Transmission Control fourth week simulation operation

Title Requirements:

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.

Simulation Analysis:

In order to realize the good starting characteristics of high starting torque and low starting current, the paper chooses the stepwise cut-off starting resistance method of winding asynchronous motor to realize the starting of the motor, and selects the rotor connection resistance method as the speed-adjusting method, and takes the reverse braking as the braking way to control the whole process relay moment size is not greater than 85n m, 2, stator resistance: 0.531 ohm, rotor resistance: 0.408 ohm, stator leakage sense: 2.52 MH, rotor leakage sense: 2.52 MH, Mutual inductance: 8.47 MH, the simulation process using auxiliary parameters to achieve the process of time-point positioning.

Simulation Code and results:

Model SACIM "A simple AC Inductionmotor 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");

Typeangularvelocity=real (unit= "rad/s");

Constant Real Pi = 3.1415926;

Real Y1;

Real Y2;

Real Z;

Current I_a "A Phase current Ofstator";

Current I_b "B Phase current Ofstator";

Current I_c "C Phase current Ofstator";

Voltage u_a "A Phase Voltage ofstator";

Voltage u_b "B Phase Voltage ofstator";

Voltage u_c "C Phase Voltage of Stator";

Current I_a "A Phase current Ofrotor";

Current I_b "B Phase current Ofrotor";

Current I_c "C Phase current Ofrotor";

Frequency f_s "Frequency ofstator";

Torque Tm "Torque of the Motor";

Speed n ' Speed of the ' motor ';

Flux psi_a "A Phase flux-linkage ofstator";

Flux psi_b "B Phase flux-linkage ofstator";

Flux psi_c "C Phase flux-linkage ofstator";

Flux psi_a "A Phase flux-linkage ofrotor";

Flux Psi_b "b Phase flux-linkage ofrotor";

Flux Psi_c "C Phase flux-linkage ofrotor";

Angle Phi "Electrical Angle ofrotor";

Angle phi_m "mechnical Angle ofrotor";

angularvelocity W "Angular Velocity ofrotor";

Resistance R1;

Parameter Torque Tl =15 "Loadtorque";

Parameter Torque Tmax = 85;

Parameter Torque Tmin =-85;

Parameter current IMAX = 25;

Parameter Current imin =-25;

Parameter resistance Rs = 0.531 "Statorresistance";

Parameter resistance Rr = 0.408 "Rotorresistance";

Parameter inductance Ls =0.00252 "stator leakage inductance";

Parameter inductance Lr = 0.00252 "Rotorleakage inductance";

Parameter inductance Lm =0.00847 "Mutual inductance";

Parameter Frequency f_n = "ratedfrequency of stator";

Parameter Voltage u_n = "Ratedphase Voltage of Stator";

Parameter Real P =3 "number of polepairs";

Parameter Inertia Jm = 0.1 "Motorinertia";

Parameter Inertia Jl = 1 "Loadinertia";

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+R1) * i_a + $ * der (psi_a);

0 = (RR+R1) * I_b + $ * der (Psi_b);

0 = (RR+R1) * 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);

If time <=

u_a = 0;

U_b = 0;

U_c = 0;

f_s = 0;

ElseIf time <= 2430 Then

f_s = f_n;