// 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 5: MECHANICAL DESIGN OF OVERHEAD LINES // EXAMPLE : 5.6 : // Page number 199-200 clear ; clc ; close ; // Clear the work space and console // Given data a = 0.484 // Area of conductor(sq.cm) d = 0.889 // Overall diameter(cm) w_c = 428/1000.0 // Weight(kg/m) u = 1973.0 // Breaking strength(kg) s = 2.0 // Factor of safety L = 200.0 // Span(m) t = 1.0 // Ice thickness(cm) wind = 39.0 // Wind pressure(kg/m^2) // Calculations // Case(i) l = L/2.0 // Half span(m) T = u/s // Allowable maximum tension(kg) D_1 = w_c*l**2/(2*T) // Maximum sag due to weight of conductor(m) // Case(ii) w_i = 913.5*%pi*t*(d+t)*10**-4 // Weight of ice on conductor(kg/m) w = w_c+w_i // Total weight of conductor & ice(kg/m) D_2 = w*l**2/(2*T) // Maximum sag due to additional weight of ice(m) // Case(iii) D = d+2.0*t // Diameter due to ice(cm) w_w = wind*D*10**-2 // Wind pressure on conductor(kg/m) w_3 = ((w_c+w_i)**2+w_w**2)**0.5 // Total force on conductor(kg/m) D_3 = w_3*l**2/(2*T) // Maximum sag due to (i), (ii) & wind(m) theta = atand(w_w/(w_c+w_i)) // θ(°) vertical_sag = D_3*cosd(theta) // Vertical sag(m) // Results disp("PART II - EXAMPLE : 5.6 : SOLUTION :-") printf("\nCase(i) : Maximum sag of line due to weight of conductor, D = %.2f metres", D_1) printf("\nCase(ii) : Maximum sag of line due to additional weight of ice, D = %.2f metres", D_2) printf("\nCase(iii): Maximum sag of line due to (i),(ii) plus wind, D = %.2f metres", D_3) printf("\n Vertical sag = %.2f metres", vertical_sag)