ee153_lab4_exp2_induction_motor_squirrel_cage_init.m f=60; % electrical (synchro
ID: 2084224 • Letter: E
Question
ee153_lab4_exp2_induction_motor_squirrel_cage_init.m
f=60; % electrical (synchronous) angular frequency in Hz
Rs=1.77;
Rr=1.34;
Xls=5.25;
Xlr=4.57;
Xm=139;
Jeq=0.025;
p=4;
% Steady State Operating Condition
VLLrms = 460; % phase-a voltage is at its positive peak at t=0
s = 0.0172; % slip = wm/wsyn where wm - rotor speed in electrical rad/s
Wsyn=2*pi*f; % synchronous speed in electrical rad/s
Wm=(1-s)*Wsyn; % rotor speed in electrical rad/s
% Phasor Calculations
Va = VLLrms * sqrt(2)/ sqrt(3); % Va phasor
Zrotor = j*Xlr + Rr/s; % Equivalent Rotor Branch Impedance
Zm = j*Xm; % Magnetizing Impedance
Zeq = (Rs + j*Xls) + (Zm * Zrotor) / (Zm + Zrotor); % Equivalent Impedance
Ia = Va / Zeq; % Ia phasor
Ema = Va - (Rs + j*Xls) * Ia; % Voltage across the magnetizing branch
Iraprime = Ema / Zrotor; % Rotor branch current phasor
% Space Vectors at time t=0 with stator a-axis as the reference
Vs_0 = (3/2) * Va; % Vs(0) space vector
Is_0 = (3/2) * Ia; % Is(0) space vector
Theta_Is_0 = angle(Is_0); % angle of Is(0) space vector
Ir_0 = (-1) * (3/2) * Iraprime; % Ir(0) space vector; notice factor of (-1); see Fig. 3-11
Theta_Ir_0 = angle(Ir_0); % angle of Ir(0) space vector
% We will assume that at t=0, d-axis is aligned to the stator a-axis. Therefore, Theta_da_0=0
Theta_da_0 = 0;
Isd_0 = sqrt(2/3) * abs(Is_0) * cos(Theta_Is_0 - Theta_da_0); % Eq. 3-64
Isq_0 = sqrt(2/3) * abs(Is_0) * sin(Theta_Is_0 - Theta_da_0); % Eq. 3-65
Ird_0 = sqrt(2/3) * abs(Ir_0) * cos(Theta_Ir_0 - Theta_da_0);
Irq_0 = sqrt(2/3) * abs(Ir_0) * sin(Theta_Ir_0 - Theta_da_0);
% Calculation of machine inductances
Ls = (Xls + Xm) / (2*pi*f);
Lm = Xm / (2*pi*f);
Lr = (Xlr + Xm) / (2*pi*f);
% Inductance matrix M in Eq. 3-61
M = [Ls 0 Lm 0 ;...
0 Ls 0 Lm;...
Lm 0 Lr 0 ;...
0 Lm 0 Lr];
% Flux Linkages
fl_dq_0 = M * [Isd_0; Isq_0; Ird_0; Irq_0]; % dq-winding fluxes in vector form, Eq. 3-61
fl_sd_0 = fl_dq_0(1)
fl_sq_0 = fl_dq_0(2)
fl_rd_0 = fl_dq_0(3)
fl_rq_0 = fl_dq_0(4)
% Electromagnetic Torque, which equals Load Torque in Initial Steady State
Tem_0 = (p/2) * Lm * (Isq_0 * Ird_0 - Isd_0 * Irq_0) % Eq. 3-47
TL_0 = Tem_0
% Wmech = rotor speed in actual rad/s
Wmech_0=(2/p)*Wm %Eq. 3-34
1. Use the Simulink's Step block to simulate a suddenly changed line-to-line voltage. Note that the voltage must have an initial value of vLLrms set up in the initialization code. Set the final value to 3/4*vLLrms at time 0.1 s. Keep the Load Torque constant at value TL o 2. Provide the outputs for Tem and armech (convert to rpm). 3. Repeat 1. for the change of line-to-line voltage to 1/2 vLLrms 4. Discuss the observations.Explanation / Answer
For the given VLLrms = 460 output --> fl_sd_0 = 0.017372 fl_sq_0 = -1.1951 fl_rd_0 = -0.12370 fl_rq_0 = -1.1363 Tem_0 = 12.644 rad/sec TL_0 = 12.644 Wmech_0 = 185.25 rad/sec
Tem_0(rpm) = Tem_0(rad/sec)*(60/2pi) = 12.644*(60/2pi) = 120.741 rpm Wmech_0(rpm) = Wmech_0(rad/sec)*(60/2pi) = 185.25*(60/2pi) = 1769 rpm
For VLLrms = (3/4)*460 = 345 output--> fl_sd_0 = 0.013029 fl_sq_0 = -0.89632 fl_rd_0 = -0.092777 fl_rq_0 = -0.85223 Tem_0 = 7.1125 TL_0 = 7.1125 Wmech_0 = 185.25
Tem_0(rpm) = Tem_0(rad/sec)*(60/2pi) = 7.1125*(60/2pi) = 67.92 rpm Wmech_0(rpm) = Wmech_0(rad/sec)*(60/2pi) = 185.25*(60/2pi) = 1769 rpm
For VLLrms = (1/2)*460 = 230 output--> fl_sd_0 = 0.0086862 fl_sq_0 = -0.59755 fl_rd_0 = -0.061851 fl_rq_0 = -0.56816 Tem_0 = 3.1611 TL_0 = 3.1611 Wmech_0 = 185.25
Tem_0(rpm) = Tem_0(rad/sec)*(60/2pi) = 3.1611*(60/2pi) = 30.186 rpm Wmech_0(rpm) = Wmech_0(rad/sec)*(60/2pi) = 185.25*(60/2pi) = 1769 rpm
In the given condition slip is fixed , so Wmech only depends on the (poles, frequency,slip) ,it is constant irrespective of voltage changes ,(eventhough it is not the case in real time)
Tem depends on voltage input , it is proportional to (v)2 ,Tem decreases with VLLrms.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.