From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001 From: prashantsinalkar Date: Tue, 10 Oct 2017 12:27:19 +0530 Subject: initial commit / add all books --- 3472/CH20/EX20.4/Example20_4.sce | 34 ++++++++++++++++++++++++++++++++++ 1 file changed, 34 insertions(+) create mode 100644 3472/CH20/EX20.4/Example20_4.sce (limited to '3472/CH20/EX20.4/Example20_4.sce') diff --git a/3472/CH20/EX20.4/Example20_4.sce b/3472/CH20/EX20.4/Example20_4.sce new file mode 100644 index 000000000..654ae14ea --- /dev/null +++ b/3472/CH20/EX20.4/Example20_4.sce @@ -0,0 +1,34 @@ +// 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 13: WAVE PROPAGATION ON TRANSMISSION LINES + +// EXAMPLE : 13.4 : +// Page number 366 +clear ; clc ; close ; // Clear the work space and console + +// Given data +R_1 = 60.0 // Surge impedance of underground cable(ohm) +R_2 = 400.0 // Surge impedance of overhead line(ohm) +e = 100.0 // Maximum value of surge(kV) + +// Calculations +i = e*1000/R_1 // Current(A) +k = (R_2-R_1)/(R_2+R_1) +e_ref = k*e // Reflected voltage(kV) +e_trans = e+e_ref // Transmitted voltage(kV) +e_trans_alt = (1+k)*e // Transmitted voltage(kV). Alternative method +i_ref = -k*i // Reflected current(A) +i_trans = e_trans*1000/R_2 // Transmitted current(A) +i_trans_alt = (1-k)*i // Transmitted current(A). Alternative method + +// Results +disp("PART II - EXAMPLE : 13.4 : SOLUTION :-") +printf("\nReflected voltage at the junction = %.f kV", e_ref) +printf("\nTransmitted voltage at the junction = %.f kV", e_trans) +printf("\nReflected current at the junction = %.f A", i_ref) +printf("\nTransmitted current at the junction = %.f A\n", i_trans) +printf("\nNOTE: ERROR: Calculation mistake in textbook in finding Reflected current") -- cgit