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diff --git a/3472/CH24/EX24.1/Example24_1.sce b/3472/CH24/EX24.1/Example24_1.sce new file mode 100644 index 000000000..6ca53dff1 --- /dev/null +++ b/3472/CH24/EX24.1/Example24_1.sce @@ -0,0 +1,41 @@ +// 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 17: ELECTRIC POWER SUPPLY SYSTEMS
+
+// EXAMPLE : 17.1 :
+// Page number 422-423
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+no_phase = 3.0 // Number of phases in ac transmission system
+V = 380.0*10**3 // Voltage b/w lines(V)
+load = 100.0 // Load(MW)
+PF = 0.9 // Power factor
+l = 150.0 // Line length(km)
+n = 0.92 // Efficiency
+r = 0.045 // Resistance(ohm/km/sq.cm)
+w_cu_1 = 0.01 // Weight of 1 cm^3 copper(kg)
+
+// Calculations
+// Case(i)
+P_loss = (1-n)*load // Power loss in the line(MW)
+I_L = load*10**6/(3**0.5*V*PF) // Line current(A)
+loss_cu = P_loss/no_phase*10**6 // I^2*R loss per conductor(W)
+R = loss_cu/I_L**2 // Resistance per conductor(ohm)
+R_km = R/l // Resistance per conductor per km(ohm)
+area = r/R_km // Conductor area(Sq.cm)
+volume = area*100.0 // Volume of copper per km run(cm^3)
+W_cu_km = volume*w_cu_1 // Weight of copper per km run(kg)
+W_cu = no_phase*l*1000*W_cu_km // Weight of copper for 3 conductors of 150 km(kg)
+// Case(ii)
+W_cu_dc = 1.0/2*PF**2*W_cu // Weight of copper conductor in dc(kg)
+
+// Results
+disp("PART II - EXAMPLE : 17.1 : SOLUTION :-")
+printf("\nWeight of copper required for a three-phase transmission system = %.f kg", W_cu)
+printf("\nWeight of copper required for the d-c transmission system = %.f kg \n", W_cu_dc)
+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision")
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