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diff --git a/1092/CH9/EX9.8/Example9_8.sce b/1092/CH9/EX9.8/Example9_8.sce new file mode 100755 index 000000000..d84dc04d8 --- /dev/null +++ b/1092/CH9/EX9.8/Example9_8.sce @@ -0,0 +1,90 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 9: POLYPHASE INDUCTION (ASYNCHRONOUS) DYNAMOS
+// Example 9-8
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data (Exs.9-5 through 9-7)
+P = 8 ; // Number of poles in the SCIM
+f = 60 ; // Frequency in Hz
+R_r = 0.3 ; // Rotor resistance per phase in ohm
+X_lr = 1.08 ; // Locked rotor reactance in ohm
+S_r = 650 ; // Speed in rpm at which motor stalls
+E_lr = 112 ; // Induced voltage per phase
+
+disp("Example 9-8 : ");
+printf(" \n The new and the original conditions may be summarized in the following table\n");
+printf(" \n _________________________________________________________");
+printf(" \n Condition \t R_r \t\t X_lr \t\t T_starting ");
+printf(" \n \t ohm \t\t ohm \t ");
+printf(" \n _________________________________________________________");
+printf(" \n Original : \t %.1f \t\t %.2f \t\t T_o = 2*T_n ",R_r,X_lr);
+printf(" \n New :\t(%.1f+R_x) \t %.2f \t\t T_n = 2*T_n ",R_r,X_lr);
+printf(" \n _________________________________________________________\n");
+
+// Calculating
+// case a
+// Neglecting constant Kn_t ,since we are equating torque T_o and T_n
+T_o = ( R_r / ((R_r)^2 + (X_lr)^2) ); // Original torque
+
+// T_o = K_n_t*( 0.3 / ((0.3)^2 + (1.08)^2) );
+// T_n = K_n_t*( 0.3 + R_x) / ( (0.3 + R_x)^2 + (1.08)^2 );
+// T_n = T_o
+// Simplyifing yields
+// 0.3 + R_x = 0.24[(0.3+R_x)^2 + (1.08)^2]
+// Expanding and combining the terms yields
+// 0.24*(R_x)^2 - 0.856*R_x = 0
+// This is a quadratic equation having two roots,which may be factored as
+// R_x*(0.24*R_x - 0.856) = 0,yielding
+// R_x = 0 and R_x = 0.856/0,24 = 3.57
+R_x = poly(0,'R_x'); // Defining a polynomial with variable 'R_x' with root at 0
+a = 0.24 ; // coefficient of x^2
+b = -0.856 ; // coefficient of x
+c = 0 ; // constant
+
+// Roots of p
+R_x1 = ( -b + sqrt (b^2 -4*a*c ) ) /(2* a);
+R_x2=( -b - sqrt (b^2 -4*a*c ) ) /(2* a);
+// Consider R_x>0 value,
+R_x = R_x1;
+
+R_T = R_r + R_x ; // Total rotor resistance in ohm
+
+// case b
+Z_T = R_T + %i*X_lr ; // Total impedance in ohm
+Z_T_m = abs(Z_T);//Z_T_m = magnitude of Z_T in ohm
+Z_T_a = atan(imag(Z_T) /real(Z_T))*180/%pi;//Z_T_a=phase angle of Z_T in degrees
+
+cos_theta = R_T / Z_T_m ; // Rotor PF that will produce the same starting torque
+
+// case c
+Z_r = Z_T_m ; // Impedance in ohm
+I_r = E_lr / Z_r ; // Starting current in A
+
+// Display the results
+disp("Solution : ");
+
+printf(" \n a: T_o = %.2f * K_n_t ",T_o );
+printf(" \n T_n = %.2f * K_n_t \n",T_o );
+printf(" \n Simplyifing yields");
+printf(" \n 0.3 + R_x = 0.24[(0.3+R_x)^2 + (1.08)^2]");
+printf(" \n Expanding and combining the terms yields");
+printf(" \n 0.24*(R_x)^2 - 0.856*R_x = 0");
+printf(" \n This is a quadratic equation having two roots,which may be factored as");
+printf(" \n R_x*(0.24*R_x - 0.856) = 0,yielding");
+printf(" \n R_x = 0 ohm and R_x = 0.856/0.24 = 3.57 ohm\n\n This proves that ");
+printf(" \n Original torque is produced with an external resistance of either ");
+printf(" \n zero or 12 times the origianl rotor resistance.Therefore,\n");
+printf(" \n R_T = R_r + R_x = %.2f ohm \n",R_T);
+
+printf(" \n b: Z_T in ohm = ");disp(Z_T);
+printf(" \n Z_T = %.2f <%.1f ohm ",Z_T_m,Z_T_a);
+printf(" \n cosÓ¨ = R_T / Z_T = %.3f or \n cosÓ¨ = cosd(%.1f) = %.3f\n",cos_theta,Z_T_a,cosd(Z_T_a));
+
+printf(" \n c: I_r = E_lr / Z_r = %.f A \n\n This proves that,",I_r);
+printf(" \n Rotor current at starting is now only 28 percent of the original");
+printf(" \n starting current in part(a) of Ex.9-7");
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