--------The code to plot is as follows-----------------------------------------
% M-file create a plot of the torque-speed curve and power speed
curve of the
% induction motor
% First, initialize the values needed in this program.
r1 = 0.15; % Stator resistance
x1 = 0.852; % Stator reactance
r2 = 0.154; % Rotor resistance
x2 = 1,066; % Rotor reactance
xm = 20; % Magnetization branch reactance
f=60; %frequency of the supply given to machine
P=4; % no of poles of machine
v_phase = 460 / sqrt(3); % Phase voltage
n_sync = 120*f/P; % Synchronous speed (r/min)
w_sync = 2*pi*n_sync/60; % Synchronous speed (rad/s)
% Calculate the Thevenin voltage and impedance from Equations
v_th = v_phase * ( xm / sqrt(r1^2 + (x1 + xm)^2) );
z_th = ((j*xm) * (r1 + j*x1)) / (r1 + j*(x1 + xm));
r_th = real(z_th);
x_th = imag(z_th);
% Now calculate the torque-speed characteristic for many
% slips between 0 and 1. Note that the first slip value
% is set to 0.001 instead of exactly 0 to avoid divide-
% by-zero problems.
s = (0:1:50) / 50; % Slip
s(1) = 0.001;
nm = (1 - s) * n_sync; % Mechanical speed of rotor in rpm
w_nm = 2*pi*n_sync/60; % Mechanical speed of rotor in rad/s
% Calculate torque
for ii = 1:51
t_ind1(ii) = (3 * v_th^2 * r2 / s(ii)) / ...
(w_sync * ((r_th + r2/s(ii))^2 + (x_th + x2)^2) );
end
% Plot the torque-speed curve
plot(nm,t_ind1,'Color','k','LineWidth',2.0);
xlabel('\itn_{m}','Fontweight','Bold');
ylabel('\tau_{ind}','Fontweight','Bold');
title ('Induction Motor Torque-Speed
Characteristic','Fontweight','Bold');
%Calculate power at different speeds
p_conv= t_ind1.*w_nm + 400+400 +150;
% Plot the power-speed curve
figure() % to bring the plot in a seperate figure without
overwriting the previous figure
plot(nm,p_conv,'Color','k','LineWidth',2.0);
xlabel('\itn_{m}','Fontweight','Bold');
ylabel('p_conv','Fontweight','Bold');
title ('Induction Motor Power-Speed
Characteristic','Fontweight','Bold');
-----------------------------------------------------------------------------------------------------------------
the outputs obtained are
Problem A 460-V, four-pole, 60-Hz, Y components R,-0.15 Ω Xi=0.852 Ω X2 = 1.066 Ω Pmech 400 w -co...
fundamentals of machines Problem A 460-V, four-pole, 60-Hz, Y-connected induction motor has the following equivalent circuit components Ri-0.15Ω R,-0.154 Ω X-1.066 Ω Xi,-20 Ω Xi-0.852 Ω Pmech 400 w Pmisc 150 Pcore 400 w Write MATLAB script to plot on two separate figures: a) the torque versus speed of the motor. b) the converted power versus speed of the motor. Problem A 460-V, four-pole, 60-Hz, Y-connected induction motor has the following equivalent circuit components Ri-0.15Ω R,-0.154 Ω X-1.066 Ω Xi,-20...
Problem A 460-V, four-pole, 60-Hz, Y-connected induction motor has the following equivalent components R,-0.15Ω R,-0.154 Ω X2-1.066 Ω M.-20 Ω Xi-0.852 Ω /mech-400 W Pmise 150 Pcore 400 W Write MATLAB script to plot on two separate figures: a) the torque versus speed of the motor. b) the converted power versus speed of the motor. Problem A 460-V, four-pole, 60-Hz, Y-connected induction motor has the following equivalent components R,-0.15Ω R,-0.154 Ω X2-1.066 Ω M.-20 Ω Xi-0.852 Ω /mech-400 W Pmise...
(b) A 460-V 60-Hz four-pole Y-connected induction motor is ratedat \(25 \mathrm{hp}\). The equivalent circuit parameters are$$ \begin{array}{ll} R_{1}=0.15 \Omega & R_{2}=0.154 \Omega \quad X_{M}=20 \Omega \\ X_{1}=0.852 \Omega & X_{2}=1.066 \Omega \\ P_{\mathrm{F} \& \mathrm{~W}}=400 \mathrm{~W} & P_{\text {misc }}=150 \mathrm{~W} \end{array} $$\(P_{\text {core }}=400 \mathrm{~W}\) (lumped with rotational losses)[14 points]For a slip of \(0.02\), find(1) The line current \(I_{L}\). [3 points](2) The stator power factor. [1 point](3) The rotor power factor. [1 point](4) The rotor frequency. [1 point](5)...
A 480 V three-phase synchronized motor (line to line), 60 Hz, 6 poles, connected in Y. The equivalent circuit parameters are R1 = 0.15 W R2 = 0.154 W XM = 20 W X1 = 0.852 W X2 = 1.066 W PFW = 400 W PMISC = 150 W PCORE = 400 W For a slip of 0.02, find (1) Line current. [3 points] (2) The power factor of the stator. [1 point] (3) The power factor of the rotor....
. A 460-V 460-V, 25-hp, 60-Hz, six-pole, Y-connected induction motor has the following impedanc ohms per phase referred to the stator circuit. Ri= 0.212 R2 = 0.282 X1 = 1.055 12 X2 = 1.055 22 XM = 33.9 W The total rotational losses are 1100 W and are assumed to be constant. The core loss is lumped with the rotational losses. For a rotor slip of 4% at the rated voltage and rated frequency, a) The input power b) Peony...
1. A 15 hp, 4 pole, 208 V, 60 Hz, Y-connected, three-phase squirrel cage induction motor has the following parameters in Ω/phase referred to the stator: R,-0.220, Xi = 0.430, X,-15.0, R2 0.130 and X2-0.430. Mechanical and core losses amount to 300 W and 200 W respectively for rated voltage. (a) The motor is fed with a 60 Hz, 208 V voltage. It runs with a slip of 4.5 %. Find the speed, shaft torque and power, efficiency and power...
A 4 phase, 460 V, 60 Hz, 26.8 hp, 4 pole, Y-connected induction motor draws 25 A at a power factor of 0.9 lagging. Core loss is 900 W, stator copper loss is 1100W, friction and winding losses are 300 W, and stray loss is neglected. The machine shaft rotates at 1738 rpm. Calculate the air gap power and output load torque - [Induction Motor - 15 points) A three-phase, 460 V, 60 Hz, 26.8 hp, 4-pole, Y-connected induction motor...
6.12 A three-phase, Y-connected, 460-V (line-line), 37-kW, 60-Hz, four-pole induction motor has the following equivalent-circuit parameters in ohms-per-phase referred to the statr: R10.070 R2 0.152 X 0.743 X2 0.764 Xm 40.1 The total friction and windage losses may be assumed constant at 390 W, and the core loss may be assumed to be equal to 325 W. With the motor connected directly to a 460-V source, compute the speed, output shaft torque and power, input power, and power factor and...
A three phase, 60 Hz, 4-pole, Y-connected, 460 V induction motor has the following values: R1=0.54 Ohm R2=0.5 Ohm Xm=51.10 Ohm X1 =2.09 X2=3.21 Ohm It operates with a slip of 0.02 rated input voltage with mechanical losses of 150 W, core loss of 150 W, and stray losses of 50 W. Find the Line current magnitude and the load torque.
6. [Induction Motor - 15 points] A three-phase, 460 V, 60 Hz, 26.8 hp, 4-pole, Y-connected induction motor draws 25 A at a power factor of 0.9 lagging. The core loss is 900 W, stator copper loss is 1100 W, friction and windage loss is 300 W, and stray loss is neglected. The machine shaft rotates at a speed of 1738 rpm. Calculate the air gap power and output load torque.