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+
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 7 : SPECIAL MOTORS AND INTRODUCTION TO GENERALIZED MACHINE THEORY
+
+// EXAMPLE : 7.7
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+V = 220;                                            // supply voltage in Volts
+f = 50;                                             // Frequency in Hertz
+p = 4;                                              // Number of poles
+Xm = 50;                                            // Mutual reactance in Ohms
+Rs = 0.4;                                           // Resistance of stator windings in Ohms
+Xs = 2.5;                                           // Leakage reactance of stator windings in Ohms
+Ra = 2.2;                                           // Resistance of Armature windings in Ohms
+Xa = 3.1;                                           // Leakage reactance of armature windings in Ohms
+loss = 30;                                          // Rotational losses in Watts
+N = 2000;                                           // Motor running speed in RPM
+
+
+// CALCULATIONS
+
+Ns = (120*f)/p;                                                        // Synchronous speed in RPM
+s = N/Ns;                                                              // Speed ratio
+I1 = V/(2*Rs + 2*%i*Xs + 2*%i*Xm + (%i*Xm^2)*((s-%i)/(Ra+%i*Xa+%i*Xm)));// line current in Amphere
+pf = cosd(atand(imag(I1),real(I1)));                                   // Power factor lagging
+I2 = (s-%i)*(%i*Xm*I1)/(Ra+%i*Xa+%i*Xm);                                  // line current in Amphere
+P1 = V*abs(I1)*cosd(atand(imag(I1),real(I1)));                         // Input power in Watts
+Pm = P1 - 2*(abs(I1)^2)*Rs - (abs(I2)^2)*Ra;                           // Mechanical power developed in Watts
+Po = Pm - loss;                                                        // output power in Watts
+eta = 100*(Po/P1);                                                     // Efficiency 
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 7.7: SOLUTION :-");
+printf("\n (a) Line currents, I1 = %.2f < %.2f A and I2 = %.2f < %.2f A \n",abs(I1),atand(imag(I1),real(I1)),abs(I2),atand(imag(I2),real(I2)))
+printf("\n (b) Power factor = %.2f lagging \n",pf)
+printf("\n (c) Efficiency = %.2f percentage \n",eta)
+printf("\n\n    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n      WRONGLY PRINTED ANSWERS ARE :- (a) I1 = 3.37 < -42.78 A instead of %.2f < j(%.2f) A \n ",abs(I1),atand(imag(I1),real(I1)));
+printf("\n                                     (b) I2 = 5.26 < -77.34 A instead of %.2f < j(%.2f) A \n ",abs(I2),atand(imag(I2),real(I2)));
+printf("\n                                     (b) eta = 81.53 percent instead of %.2f percent  \n ",eta)
+printf("\n From Calculation of the I1, rest all the Calculated values in the TEXT BOOK is WRONG because of the I1 value is WRONGLY calculated and the same used for the further Calculation part \n")