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+// Calculating the upper and lower limits of current during starting and resistance of each section

+clc;

+disp('Example 18.1, Page No. = 18.3')

+// Given Data

+// d.c. shunt motor

+P = 37;// Power rating (in kW)

+V = 250;// Voltage rating (in Volts)

+e = 0.84;// Full load efficiency

+rm = 0.2;// Armature circuit resistance (in ohm)

+ns = 8;// Number of studs

+// Maximum torque is 150% of full load torque

+// Calculation of the upper and lower limits of current during starting

+Ifl = P*10^(3)/(V*e);// Full load current (in Ampere)

+I1 = 1.5*Ifl;// Maximum current (in Ampere).  Since torque is proportional to current

+n = ns-1;// Number of sections

+alpha = (rm*I1/V)^(1/n);

+I2 = alpha*I1;// Lower limit of current (in Ampere)

+disp(I1,'Upper limit of current (Ampere) =');

+disp(I2,'Lower limit of current (Ampere) =');

+// Calculation of the resistance of each section

+R1 = V/I1;// Total resistance at starting (in ohm)

+r1 = (1-alpha)*R1;

+r2 = alpha*r1;

+r3 = alpha*r2;

+r4 = alpha*r3;

+r5 = alpha*r4;

+r6 = alpha*r5;

+r7 = alpha*r6;

+disp(R1,'Total resistance at starting (ohm) =');

+disp('Resistance of each section')

+disp(r1,'r1 (ohm) =');

+disp(r2,'r2 (ohm) =');

+disp(r3,'r3 (ohm) =');

+disp(r4,'r4 (ohm) =');

+disp(r5,'r5 (ohm) =');

+disp(r6,'r6 (ohm) =');

+disp(r7,'r7 (ohm) =');

+//in book answers are I1 = 264 ampere, I2 = 211 ampere, R1 = 0.947 ohm, r1 = 0.189 ohm,, r2 = 0.151 ohm, r3 = 0.121 ohm, r4 = 0.097 ohm, r5 = 0.077 ohm, r6 = 0.062 ohm, r7 = 0.050 ohm.    The answers vary due to round off error