Work week Four

Work week Four

This week reviewed the fourth chapter of the three-phase asynchronous motor working principle and mechanical characteristics, based on the consolidation of the three-phase induction motors start, speed and braking.

The extra trivia of this week's extracurricular games did not add to the additional Reading.

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Simulation Jobs

Basic requirements:

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

1 ) to achieve 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 rest and remain stationary 0.5s ,
• take a heavy load down and reach from the standstill 600r/min
• 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 rotor shaft of the motor is 1kg.m^2, and the equivalent load torque on the rotor shaft of the motor is converted to 15n.m .

3 ) using a unified motor model , If the control strategy uses a rotor string resistor, it is permissible to change the rotor of the motor into a winding rotor (the parameter is unchanged).

4 refer to the AC motor start, speed and braking methods given in the textbook, Design control strategies, Modelica implement the control strategy and realize the joint simulation with the motor model.

5 the stator string resistance, the rotor string resistance, the stator voltage regulation, the stator frequency modulation and so on, but must have the engineering to be possible to implement.

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

7 ) Program best students get this week " Star of Control " title.

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Simulation process:

First adjust the load inertia value, at the same time, considering the starting torque can not be too large, according to the inherent mechanical characteristics of the asynchronous motor, when applied to the stator per phase winding voltage reduction, the starting torque will be significantly reduced, when the rotor resistance increases properly, the starting torque will increase, start to take Buck start.

Braking, for the pursuit of efficiency, the use of reverse braking, due to the reverse braking when the current is very large, so as appropriate in the stator circuit to connect additional resistors.

When adjusting the speed, according to

It is known that the ratio of speed change and frequency and voltage change rate is the same, the motor original model speed R/min, by adjusting the frequency f and voltage can achieve speed regulation.

When the set speed is R/min, it is noted that the direct multiplication coefficient 800/1500 can not obtain the final stability to meet the r/min speed curve, due to the rated speed, take the coefficient 800/(1500*0.985), the result curve is good. But for the speed of the reverse r/min, the adjustment effect is inferior to the original value, so the direct take coefficient 600/1500.

Stop section, since the motor is still subject to the n.m torque, it is still necessary to set a small frequency and voltage value.

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The code is as follows:

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");

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Constant Real Pi = 3.1415926;

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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";

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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";

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Angle Phi "Electrical Angle of Rotor";

Angle phi_m "mechnical Angle of Rotor";

angularvelocity W "Angular Velocity of Rotor";

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Torque Tl "Load Torque";

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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";

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Initial equation

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psi_a = 0;

Psi_b = 0;

Psi_c = 0;

psi_a = 0;

Psi_b = 0;

Psi_c = 0;

phi = 0;

w = 0;

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Equation

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u_a = Rs * i_a + $ * der (psi_a);

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

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

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

0 = Rr * i_b + $ * der (Psi_b);

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

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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;

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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;

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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));

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w = * der (phi_m);

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Phi_m = phi/p;

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

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

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If time <=

u_a = 0;

U_b = 0;

U_c = 0;

f_s = 0;

Tl = 0;

Rs = 0.531;

Else

If time <=

f_s = f_n * 800/(1500 * 0.985);

u_a = u_n * 1.414 * sin (2*pi*f_s*time/1000) * 800/(1500 * 0.985) * 0.85;

U_b = u_n * 1.414 * sin (2*PI*F_S*TIME/1000-2*PI/3) * 800/(1500 * 0.985) * 0.85;

U_c = u_n * 1.414 * sin (2*PI*F_S*TIME/1000-4*PI/3) * 800/(1500 * 0.985) * 0.85;

Tl = 15;

Rs = 0.531;

Else

If time <= 1850 then

f_s = f_n * 800/(1500 * 0.985);

u_a = u_n * 1.414 * sin (2*pi*f_s*time/1000) * 800/(1500 * 0.985);

U_b = u_n * 1.414 * sin (2*PI*F_S*TIME/1000-2*PI/3) * 800/(1500 * 0.985);

U_c = u_n * 1.414 * sin (2*PI*F_S*TIME/1000-4*PI/3) * 800/(1500 * 0.985);

Tl = 15;

Rs = 0.531;

Else

If time <= 4300 Then

f_s = f_n;

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

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);

Tl = 15;

Rs = 2.531;

Else

If time <= 4800 then

f_s = f_n * 0.0705;

u_a = u_n * 1.414 * sin (2*pi*f_s*time/1000) * 0.0705;

U_b = u_n * 1.414 * sin (2*PI*F_S*TIME/1000-2*PI/3) * 0.0705;

U_c = u_n * 1.414 * sin (2*PI*F_S*TIME/1000-4*PI/3) * 0.0705;

Tl = 15;

Rs = 0.531;

Else

If time <= 4900 then

f_s = f_n * 600/1500;

u_a = u_n * 1.414 * sin (2*PI*F_S*TIME/1000-4*PI/3) * 600/1500 * 0.85;

U_b = u_n * 1.414 * sin (2*PI*F_S*TIME/1000-2*PI/3) * 600/1500 * 0.85;

U_c = u_n * 1.414 * sin (2*pi*f_s*time/1000) * 600/1500 * 0.85;

Tl = 15;

Rs = 0.531;

Else

If time <= 6450 then

f_s = f_n * 600/1500;

u_a = u_n * 1.414 * sin (2*PI*F_S*TIME/1000-4*PI/3) * 600/1500;

U_b = u_n * 1.414 * sin (2*PI*F_S*TIME/1000-2*PI/3) * 600/1500;

U_c = u_n * 1.414 * sin (2*pi*f_s*time/1000) * 600/1500;

Tl = 15;

Rs = 0.531;

Else

If time <= 9150 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;

Rs = 1.531;

Else

f_s = f_n * 0.0705;

u_a = u_n * 1.414 * sin (2*pi*f_s*time/1000) * 0.0705;

U_b = u_n * 1.414 * sin (2*PI*F_S*TIME/1000-2*PI/3) * 0.0705;

U_c = u_n * 1.414 * sin (2*PI*F_S*TIME/1000-4*PI/3) * 0.0705;

Tl = 15;

Rs = 0.531;

End If;

End If;

End If;

End If;

End If;

End If;

End If;

End If;

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End SACIM;

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Simulate (sacim,starttime=0,stoptime=10000)

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Plot (N)

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Plot (Tm)

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The result diagram is as follows:

Work week Four