% Induction Motor Model

t = 0:0.001:1; % 1sec simulation with time step of 0.001 sec

dt = 0.001;

% Motor Parameters

R1 = 0.205; % Ohm

R2 = 1.408; % Referred value, Ohm

X1 = 1.507; % Ohm

X2 = 1.507; % Ohm

Rc = 315.6; % Shunt branch, Ohm

Xm = 42.5; % Shunt branch, Ohm

Vin = 537.4*cos(2*pi*50*t); % Input voltage,50 Hz, Vphase 220 V

I = zeros(1,1001);

s = 1; % slip=1 at the beginning

w = zeros(1,1001); % rotor speed

Pe = 0; % Electromechanical power output

Te = zeros(1,1001); % Electromechanical torque

Jm = 0.1285; % kg/m2 , Inertia of machine

% Load parameters

Tl = 0; % Load Torque

Jl = 0; % Load Inertia

Xeq = X1 + X2;

Leq = Xeq / 100 / pi;

for n=2:1001

Vth = abs(1i*Xm/(R1 + 1i*(X1 + Xm))*Vin(1,n));

I(1,n) = (Vin(1,n) + I(1,n-1)*Leq/dt) / (R1 + R2/s + Leq/dt); % Motor Current

Te(1,n) = 3*Vth^2*R2/((R1 + R2/s)^2 + (X1 + X2)^2)/s/100/pi; % Electromechanical Torque

% Torque balance equation

w(1,n) = w(1,n-1) + (Te(1,n) - Tl)/(Jm + Jl)*dt; % new speed value

s = (100*pi-w(1,n))/100/pi ; % new slip value

end

After we run this code, w, Te and I vectors against time is created at the workspace. We can plot them to see their variation in time during start-up.

*Simulation Results (NO-LOAD Start-Up)*

*Simulation Results (LOAD with only inertia no torque Start-Up , inertia Jload = 3xJmotor)*

*Summary*
Note that the in rush current of the motor start-up is easily observed in current graphs.

Note that the results are for 2 pole machine

When a load of only inertia that is 3 times larger than motor inertia, the start-up time is increased.

Please investigate the torque graphs, the mean of the torque values shows a typical torque-speed curve, which is expected.

Since there is no load torque and friction&windage torque is neglected, the motor current reduced to zero after rated speed is reached.

*What if the motor is suddenly loaded with a constant torque of 40 Nm during 1.5 < t < 1.7 sec ?*

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