Study notes (third week)

Design Tasks

Combined with the mechanical characteristics of DC Motors studied this week, a DC motor string resistance Start-up process is designed and simulated with Modelica, with the following specific requirements:

1) The motor works in the rated voltage and rated magnetic flux, using a series of three or four resistor start, the entire starting process armature current can not exceed the rated current of 3 times times.

2) Select the appropriate resistance value, select the optimal resistance cutting strategy, so that under the premise of satisfying the condition 1, the motor as fast as possible to achieve the correct rating point.

3) All students use the following unified DC motor model, the parameters of the motor are:

Rated Voltage: 240V

Rated Current: 16.2A

Rated Torque: 29.2n.m

Rated speed: 1220 r/min

Moment of inertia: 1 kg.m^2

Armature Resistance: 0.6 Ohm

Torque constant (rated magnetic flux): 1.8

EMF constant (rated magnetic flux): 0.189

Main equation

Parameter calculation

Resistance R1 short-circuit process: short-circuit before the current is I1, torque is Tm1, the speed of N1, short-circuit after the current of I1 ';

Resistance R2 short-circuit process: short-circuit before the current is I2, torque is Tm2, the speed of N2, short-circuit after the current of I2 ';

The short circuit of the Resistance R3: The current is i3 before the short circuit, the torque is Tm3, the speed is N3, the current after the short circuit is i3 '.

R1 before short Circuit:

Tm1-tl=j (DN1/DT) (2PI/60)

Tm1=kt*i1

U=KE*N1+I1 (R+R1+R2+R3)

R1 after short circuit:

Tm1=kt*i1 '

U=ke*n1+i1 ' (R+R2+R3)

Since the voltage U is constant, it is known that i1 (R+R1+R2+R3) = I1 ' (R+R2+R3), i.e.

I1 '/i1= (R+R1+R2+R3)/(R+R2+R3); 1

Then: R2,R3 short-circuit before and after the difference can be:

I2 '/i2= (R+R2+R3)/(R+R3); 2

I3 '/i3= (R+R3)/r;3

The i1,i2,i3 is equal to the rated current and the T2 (I2) of each switching resistor is basically the same as the test instructions

Make I1 '/i1=i2 '/i2=i3 '/i3=a;4

By 1234:

R1= (A-1) r+2r* (A-1) ^2+r* (A-1) ^3;

R2= (A-1) r+r* (A-1) ^2;

R3= (A-1) R;

The r+r1+r2+r3=r*a^3.

Just start the motor when the speed n=0, it is known

u= I (R+R1+R2+R3)

i=u/(R+R1+R2+R3) =16.2a

R+r1+r2+r3=u/16.2a=r*a^3

The a=2.229 can be solved.

r1=3.664;r2=1.644;r3=0.737

T2= (1.1~1.2) Tn per switch resistor, the current value before switching should be (1.1~1.2) x16.2a=17.82~19.44a,

Therefore, by adjusting the switching time to meet the current minimum value of not less than 17.82A requirements

Simulation programming

After adjusting the switching time several times, the following simulation program is finally obtained

Model Motor1 "An DC motor model"

Type Voltage=real (unit= "V");

Type Current=real (unit= "A");

Type Resistance=real (unit= "Ohm");

Type Speed=real (unit= "r/min");

Type Torque=real (unit= "n.m");

Type Inertia=real (unit= "kg.m^2");

Torque Tm "Torque of the Motor";

Speed n ' Speed of the ' motor ';

Current I "armature";

Voltage u "Voltage Source";

Resistance r_ad "External resistance";

Resistance R1 "Start-up resistance";

Resistance R2 "Start-up resistance";

Resistance R3 "Start-up resistance";

Parameter Real J = 1 "total inertia";

Parameter Real R = 0.6 "armature resistance";

Parameter Real Kt = 1.8 "Torque Constant";

Parameter Real Ke = 0.189 "EMF Constant";

Parameter Real Tl = 29.2 "Load Torque";

Parameter Real i1=48.6 "Maximum current";

Parameter Real i2=17.8 "Minimum current";

Equation

TM-TL = J * der (n) * 6.28/60;

tm= Kt * i;

u= i * (R+R_AD+R1+R2+R3) + Ke * N;

If time <= 0.1 Then

u = 0;

R_ad = 0;

Else

u = 240;

R_ad = 0;

End If;

If time <= 4 then

r1=3.664;

Else

r1=0;

End If;

If time <= 6.9 then

r2=1.644;

Else

r2=0;

End If;

If time <= 8 then

r3=0.737;

Else

r3=0;

End If;

End Motor1;

Simulate (MOTOR1,STARTTIME=0,STOPTIME=15)

Study notes (third week)