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authorpriyanka2015-06-24 15:03:17 +0530
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+// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN
+// TURAN GONEN
+// CRC PRESS
+// SECOND EDITION
+
+// CHAPTER : 6 : DIRECT CURRENT POWER TRANSMISSION
+
+// EXAMPLE : 6.2 :
+clear ; clc ; close ; // Clear the work space and console
+
+// GIVEN DATA
+K = 3 ; // factor
+
+// CALCULATIONS
+// For case (a)
+V_d = poly(0,'V_d') ; // defining a polynomial
+E_p = poly(0,'E_p') ;
+V_d = K*2*E_p ; // From equ 6.18
+
+// For case (b)
+P_dc = poly(0,'P_dc') ;
+P_ac = poly(0,'P_ac') ;
+P_dc = 2*P_ac ;
+
+// For case (c)
+P_ld = poly(0,'P_ld') ; // P_loss(dc)
+P_la = poly(0,'P_la') ; // P_loss(ac)
+P_ld = (2/3)*P_la ;
+
+// DISPLAY RESULTS
+disp("EXAMPLE : 6.2 : SOLUTION :-") ;
+printf("\n (a) Maximum operating V_d in terms of voltage E_p , V_d = \n") ; disp(V_d) ;
+printf("\n (b) Maximum power transmission capability ratio,i.e,ratio of P_dc to P_ac , P_dc/P_ac = \n") ; disp(P_dc/P_ac) ;
+printf("\n (or) P_dc = \n") ; disp(P_dc) ;
+printf("\n (c) Ratio of total I^2*R losses , i.e ,Ratio of P_loss(dc) to P_loss(ac),which accompany maximum power flow = \n") ; disp(P_ld/P_la) ;
+printf("\n (or) P_loss(dc) = \n") ; disp(P_ld) ;