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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /1092/CH8/EX8.17 | |
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initial commit / add all books
Diffstat (limited to '1092/CH8/EX8.17')
| -rwxr-xr-x | 1092/CH8/EX8.17/Example8_17.sce | 68 |
1 files changed, 68 insertions, 0 deletions
diff --git a/1092/CH8/EX8.17/Example8_17.sce b/1092/CH8/EX8.17/Example8_17.sce new file mode 100755 index 000000000..04e50959c --- /dev/null +++ b/1092/CH8/EX8.17/Example8_17.sce @@ -0,0 +1,68 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-17
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kW = 40000 ; // Load on a factory in kW
+PF = 0.8 ; // power factor lagging of the load
+cos_theta = PF;
+sin_theta = sqrt( 1 - (cos_theta)^2 );
+hp = 7500 ; // power rating of the induction motor in hp
+PF_IM = 0.75 ; // power factor lagging of the induction motor
+cos_theta_IM = PF_IM;
+sin_theta_IM = sqrt( 1 - (cos_theta_IM)^2 );
+eta = 91*(1/100) ; // Efficiency of IM
+PF_SM = 1 ; // power factor of the synchronous motor
+
+// Calculations
+kVA_original = kW / PF ; // Original kVA
+kvar_original = kVA_original * sin_theta ; // Original kvar
+
+kW_IM = ( hp * 746 ) / ( 1000 * eta ) ; // Induction motor kW
+kVA_IM = kW_IM / PF_IM ; // Induction motor kVA
+kvar_IM = kVA_IM * sin_theta_IM ; // Induction motor kvar
+
+kvar_final = kvar_original - kvar_IM ; // final kvar
+kVA_final = kW + %i*(abs(kvar_final)); // final kVA
+kVA_final_m = abs(kVA_final);//kVA_final_m = magnitude of kVA_final in kVA
+kVA_final_a = atan(imag(kVA_final) /real(kVA_final))*180/%pi;
+//kVA_final_a=phase angle of kVA_final in degrees
+
+PF_final = cosd(kVA_final_a); // Final power factor
+
+// Display the result
+disp("Example 8-17 Solution : ");
+printf(" \n The synchronous motor operates at the same efficiency as the IM");
+printf(" \n that has been replaced, and therefore the total power of the system");
+printf(" \n is unchanged. The solution involves construction of table that shows ")
+printf(" \n the original condition of the system, the change, and the final condition.\n");
+printf(" \n Original kVA = %d kVA \n ", kVA_original );
+printf(" \n Original kvar = \n" );disp(%i*kvar_original);
+
+printf(" \n Induction motor kW = %d kW \n ", kW_IM );
+printf(" \n Induction motor kVA = %.f kVA \n ", kVA_IM );
+printf(" \n Induction motor kvar = ");disp(%i*kvar_IM)
+
+printf(" \n Final kvar = ");disp(%i*kvar_final);
+printf(" \n Final kVA = " );disp(kVA_final);
+printf(" \n Final kVA = %f <%.2f kVA \n ",kVA_final_m,kVA_final_a);
+
+printf(" \n Final PF = %.3f lagging \n ", PF_final );
+
+printf(" \n __________________________________________________________________________");
+printf(" \n Power tabulation grid : \n ");
+printf(" \n \t\t P \t\t ±jQ \t\t S* ");
+printf(" \n \t\t(kW) \t\t(kvar) \t\t(kVA) \t\t cosӨ ");
+printf(" \n __________________________________________________________________________");
+printf(" \n Original : \t%d \t\tj%.f \t\t%.1d \t\t %.1f lag",kW ,kvar_original ,kVA_original,PF);
+printf(" \n Removed : \t-%.f \t\t-(+j%.f) \t%.f \t\t %.2f lag",kW_IM,kvar_IM,kVA_IM,PF_IM);
+printf(" \n Added : \t+%.f \t\t 0 \t%.1f \t\t 1.0 ",kW_IM,kW_IM);
+printf(" \n Final : \t%d \t\tj%.f \t\t%.1f \t %.3f lag",kW ,kvar_final ,kVA_final_m,PF_final);
+printf(" \n __________________________________________________________________________");
+
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