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// FUNDAMENTALS OF ELECTICAL MACHINES
// M.A.SALAM
// NAROSA PUBLISHING HOUSE
// SECOND EDITION
// Chapter 11 : SINGLE-PHASE MOTORS
// Example : 11.5
clc;clear; // clears the console and command history
// Given data
f = 50 // supply frequency in Hz
V_nl = 100 // no-load voltage in v
I_nl = 2.5 // no-load current in A
P_nl = 60 // no-load power in W
V_br = 60 // Block rotor test voltage in v
I_br = 3 // Block rotor test current in A
P_br = 130 // Block rotor test power in W
R_1 = 2 // main windning resistance in ohm
// caclulations
Z_br = V_br/I_br // impedance due to blocked rotor test
R_br = P_br/I_br^2 // resistance due to blocked rotor test in ohm
X_br = sqrt(Z_br^2-R_br^2) // reactance under blocked condition in ohm
X_1 = X_br/2 // reactance in ohm X_1=X_2
R_2 = R_br-R_1 // resistance in ohm
Z_nl = V_nl/I_nl // impedance due to no-load in ohm
R_nl = P_nl/I_nl^2 // resistance due to no-load in ohm
X_nl = sqrt(Z_nl^2-R_nl^2) // reactance due to no-load in ohm
X_m =2*(X_nl-X_1-0.5*X_1) // magnetizing reactance in ohm
P_rot = P_nl-I_nl^2*(R_1+((R_2)/4)) // rotational loss in W
// display the result
disp("Example 11.5 solution");
printf(" \n Magnetizing reactance \n X_m = %.1f ohm \n", X_m );
printf(" \n Rotational loss \n P_rot = %.0f W \n", P_rot );
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