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diff --git a/2777/CH7/EX7.6/Ex7_6.sce b/2777/CH7/EX7.6/Ex7_6.sce new file mode 100755 index 000000000..4bf4f07df --- /dev/null +++ b/2777/CH7/EX7.6/Ex7_6.sce @@ -0,0 +1,54 @@ +
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 7 : SPECIAL MOTORS AND INTRODUCTION TO GENERALIZED MACHINE THEORY
+
+// EXAMPLE : 7.6
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+R = 1.4; // Total Resistance of the AC series motor in Ohms
+V = 115; // supply voltage in Volts
+f = 50; // Frequency in Hertz
+N = 5000; // Rotating speed in RPM
+X = 12; // Total reactance in Ohms
+P = 250; // Electrical power output in Watts
+loss = 18; // Rotational losses in Watts
+
+
+// CALCULATIONS
+
+Pd = P + loss; // Mechanical power developed in Watts
+// We know that Er = Pd/I and from phasor diagram in figure 7.11 page no. 501 V^2 = (Er+I*R)^2+(I*X)^2, 115^2 = (268/I-1.4*I)^2+(12*I)^2, 13225*I^2 = 71824+2.036*I^4-750.4*I^2+144*I^2, solving this we get 2.036*I^4-13831.4*I^2+71824 = 0, I^4-6793.42*I^2+3577 = 0 this gives I = 2.28A or 82.38A (The above calculation part is wrong )
+i = poly ([3577 0 -6793.42 0 1],'x','coeff'); // Expression for the Current in Quadratic form
+a = roots (i); // 4-Value of the current in Amphere
+I = a(4,1); // Curent in Amphere neglecting higher value and negative value
+
+pf_a = sqrt(1-((I*X)/V)^2); // Power factor lagging
+Er_a = sqrt(V^2-(I*X)^2)-(I*R); // Rotational Voltage in Volts
+T_a = (Er_a*I)/(2*%pi*N/60); // Developed torque in Newton-meter
+Ih = I/2; // Current halved in Amphere
+pf_b = sqrt(1-((Ih*X)/V)^2); // Power factor lagging when load current halved
+Er_b = sqrt(V^2-(Ih*X)^2)-(Ih*R); // Rotational Voltage in Volts when load current halved
+N2 = (N*Er_b*I)/(Er_a*Ih); // New speed in RPM when load current halved
+T_b = (Er_b*Ih)/(2*%pi*N2/60); // Developed torque in Newton-meter when load current halved
+eta = 100*(Er_b*Ih)/(V*Ih*pf_b); // Efficiency when load current halved
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 7.6: SOLUTION :-");
+printf("\n At rated condition, \n\n (a.1) Current, I = %.2f A \n",I)
+printf("\n (a.2) Power factor = %.3f lagging \n",pf_a)
+printf("\n (a.3) Developed torque = %.2f N-m \n",T_a)
+printf("\n When load current halved (reduced to half), \n\n (b.1) Speed, N2 = %.f RPM \n",N2)
+printf("\n (b.2) Power factor = %.4f lagging \n",pf_b)
+printf("\n (b.3) Developed torque = %.2f N-m \n",T_b)
+printf("\n (b.4) Efficiency = %.1f percenatge \n",eta)
+printf("\n From Calculation of the Current(I), rest all the Calculated values in the TEXT BOOK is WRONG because of the Current equation and its value both are WRONGLY calculated and the same used for the further Calculation part, so all the values are in the TEXT BOOK IS WRONG \n")
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