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diff --git a/3472/CH10/EX10.24/Example10_24.sce b/3472/CH10/EX10.24/Example10_24.sce new file mode 100644 index 000000000..9ef0b9568 --- /dev/null +++ b/3472/CH10/EX10.24/Example10_24.sce @@ -0,0 +1,33 @@ +// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES
+
+// EXAMPLE : 3.24 :
+// Page number 156-157
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+A = 0.94*exp(%i*1.5*%pi/180) // Constant
+B = 150.0*exp(%i*67.2*%pi/180) // Constant(ohm)
+D = A // Constant
+Y_t = 0.00025*exp(%i*-75.0*%pi/180) // Shunt admittance(mho)
+Z_t = 100.0*exp(%i*70.0*%pi/180) // Series impedance(ohm)
+
+// Calculations
+C = (A*D-1)/B // Constant(mho)
+A_0 = A*(1+Y_t*Z_t)+B*Y_t // Constant
+B_0 = A*Z_t+B // Constant(ohm)
+C_0 = C*(1+Y_t*Z_t)+D*Y_t // Constant(mho)
+D_0 = C*Z_t+D // Constant
+
+// Results
+disp("PART II - EXAMPLE : 3.24 : SOLUTION :-")
+printf("\nA_0 = %.3f∠%.f° ", abs(A_0),phasemag(A_0))
+printf("\nB_0 = %.f∠%.1f° ohm", abs(B_0),phasemag(B_0))
+printf("\nC_0 = %.6f∠%.1f° mho", abs(C_0),phasemag(C_0))
+printf("\nD_0 = %.3f∠%.1f° \n", abs(D_0),phasemag(D_0))
+printf("\nNOTE: Changes in obtained answer from that of textbook is due to more precision")
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