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diff --git a/3472/CH24/EX24.5/Example24_5.sce b/3472/CH24/EX24.5/Example24_5.sce new file mode 100644 index 000000000..6975ac3c5 --- /dev/null +++ b/3472/CH24/EX24.5/Example24_5.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 : 17.5 :
+// Page number 425
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+L = 60.0 // Line length(km)
+P = 5.0 // Load(MW)
+PF = 0.8 // Lagging power factor
+V = 33.0*10**3 // Voltage(V)
+n = 0.85 // Transmission efficiency
+rho = 1.73*10**-8 // Specific resistance of copper(ohm-mt)
+density = 8900.0 // Density(kg/mt^3)
+
+// Calculations
+I = P*10**6/(3**0.5*V*PF) // Line current(A)
+line_loss = (1-n)*P*1000/n // Line loss(kW)
+line_loss_phase = line_loss/3.0 // Line loss/phase(kW)
+R = line_loss_phase*1000/I**2 // Resistance/phase(ohm)
+a = rho*L*1000/R // Area of cross section of conductor(m^2)
+volume = 3.0*a*L*1000 // Volume of copper(m^3)
+W_cu = volume*density // Weight of copper in 3-phase system(kg)
+I_1 = P*10**6/V // Current in single phase system(A)
+R_1 = line_loss*1000/(2*I_1**2) // Resistance in single phase system(ohm)
+a_1 = rho*L*1000/R_1 // Area of cross section of conductor in single phase system(m^2)
+volume_1 = 2.0*a_1*L*1000 // Volume of copper(m^3)
+W_cu_1 = volume_1*density // Weight of copper in 1-phase system(kg)
+reduction_cu = (W_cu-W_cu_1)/W_cu*100 // Reduction in copper(%)
+
+// Results
+disp("PART II - EXAMPLE : 17.5 : SOLUTION :-")
+printf("\nWeight of copper required for 3-phase 2-wire system = %.2e kg", W_cu)
+printf("\nReduction of weight of copper possible = %.1f percent \n", reduction_cu)
+printf("\nNOTE: ERROR: Calculation mistakes in the textbook solution")
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