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// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
// TURAN GONEN
// CRC PRESS
// SECOND EDITION
// CHAPTER : 3 : FUNDAMENTAL CONCEPTS
// EXAMPLE : 3.3 :
clear ; clc ; close ; // Clear the work space and console
// GIVEN DATA
// For case (c)
I_normal = 1000 ; // Normal full load current in Ampere
// CALCULATIONS
// For case (a) equation is (1.5pu)*I_rated = (2 pu)*I_normal
// THEREFORE
// I_rated = (1.333pu)*I_normal ; // Rated current in terms of per unit value of the normal load current
// For case (b)
Mvar = (1.333)^2 ; // Increase in Mvar rating in per units
// For case (c)
I_rated = (1.333)*I_normal ; // Rated current value
// DISPLAY RESULTS
disp("EXAMPLE : 3.3 : SOLUTION :-") ;
printf("\n (a) Rated current , I_rated = (1.333 pu)*I_normal \n") ;
printf("\n (b) Mvar rating increase = %.2f pu \n",Mvar) ;
printf("\n (c) Rated current value , I_rated = %.f A \n",I_rated) ;
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