From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001 From: priyanka Date: Wed, 24 Jun 2015 15:03:17 +0530 Subject: initial commit / add all books --- 1092/CH8/EX8.14/Example8_14.sce | 69 +++++++++++++++++++++++++++++++++++++++++ 1 file changed, 69 insertions(+) create mode 100755 1092/CH8/EX8.14/Example8_14.sce (limited to '1092/CH8/EX8.14/Example8_14.sce') diff --git a/1092/CH8/EX8.14/Example8_14.sce b/1092/CH8/EX8.14/Example8_14.sce new file mode 100755 index 000000000..abb791b02 --- /dev/null +++ b/1092/CH8/EX8.14/Example8_14.sce @@ -0,0 +1,69 @@ +// Electric Machinery and Transformers +// Irving L kosow +// Prentice Hall of India +// 2nd editiom + +// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS +// Example 8-14 + +clear; clc; close; // Clear the work space and console. + +// Given data +P_o = 2000 ; // load in kW drawn by a factory +cos_theta_o = 0.6 ; // PF lagging +sin_theta_o = sqrt( 1- (cos_theta_o)^2 ); +cos_theta_f = 0.85 ; // final PF lagging required +sin_theta_f = sqrt( 1- (cos_theta_f)^2 ); +P_a = 275 ; // Losses in the synchronous capacitor in kW + +// Calculations +// case a +S_o_conjugate = P_o / cos_theta_o ; // Original kVA drawn from the utility + +// case b +Q_o = S_o_conjugate * sin_theta_o ; // Original lagging kvar + +// case c +P_f = P_o + P_a ; // Final system active power consumed from the utility in kW + +// case d +S_f_conjugate = P_f / cos_theta_f ; // Final kVA drawn from the utility +S_f_conjugate_a = acosd(cos_theta_f); // Phase angle of S_f_conjugate in degrees + +// case e +jQ_f = S_f_conjugate * sin_theta_f ; // Final lagging kvar +jQ_a = %i*(jQ_f) - %i*(Q_o); // Correction kvar produced by the synchronous capacitor +Q_a = abs(jQ_a); // Magnitude of jQ_a in kvar + +// case f +P = P_a ; +S_a_conjugate = P -%i*(abs(jQ_a)); // kVA rating of the synchronous capacitor +S_a_conjugate_m = abs(S_a_conjugate);//S_a_conjugate_m = magnitude of S_a_conjugate in kVA +S_a_conjugate_a = atan(imag(S_a_conjugate) /real(S_a_conjugate))*180/%pi; +//S_a_conjugate_a=phase angle of S_a_conjugate in degrees +PF_f = cosd(S_a_conjugate_a); // PF + +// Display the results +disp("Example 8-14 Solution : "); +printf(" \n a: S*o = %.1f kVA \n",S_o_conjugate ); + +printf(" \n b: Q*o in kvar = " );disp(%i*Q_o); + +printf(" \n c: P*f = %.f kW \n",P_f ); + +printf(" \n d: S*f = %.1f <%.1f kVA\n ",S_f_conjugate,S_f_conjugate_a ); + +printf(" \n e: jQ_f in kvar = ");disp(%i*jQ_f); +printf(" \n -jQ_a in kvar = ");disp(jQ_a); + +printf(" \n f: S*a = %.f <%.2f kVA ", S_a_conjugate_m , S_a_conjugate_a ); +printf(" \n (cos(%.2f) = %.3f leading)\n",S_a_conjugate_a,PF_f); + +printf(" \n g: Power tabulation grid : \n "); +printf(" \n \t\t P \t ±jQ \t S* "); +printf(" \n \t\t(kW) \t(kvar) \t(kVA) \t cosӨ "); +printf(" \n ___________________________________________"); +printf(" \n Original : \t %d \t +j%.f %.1f %.1f lag",P_o ,Q_o ,S_o_conjugate,cos_theta_o); +printf(" \n Added : \t %d \t -%.fj %.f \t %.3f lead",P_a ,Q_a,S_a_conjugate_m,cosd(S_a_conjugate_a) ); +printf(" \n Final : \t %d \t +j%.f %.1f %.2f lag",P_f ,jQ_f ,S_f_conjugate,cos_theta_f); + -- cgit