// 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 : 5.11 : // Page number 202-203 clear ; clc ; close ; // Clear the work space and console // Given data a = 120.0 // Area(mm^2) ds = 2.11 // Diameter of each strand(mm) W = 1118.0/1000 // Weight of conductor(kg/m) L = 200.0 // Span(m) stress = 42.2 // Ultimate tensile stress(kg/mm^2) wind = 60.0 // Wind pressure(kg/m^2) t = 10.0 // Ice thickness(mm) // Calculations n = 3.0 // Number of layers d = (2*n+1)*ds // Overall diameter of conductor(mm) u = stress*a // Ultimate strength(kg) T = u/4.0 // Working stregth(kg) // Case(a) S_a = W*L**2/(8*T) // Sag in still air(m) // Case(b) area = d*100*10.0*10**-6 // Projected area to wind pressure(m^2) w_w = wind*area // Wind load/m(kg) w_r = (W**2+w_w**2)**0.5 // Resultant weight/m(kg) S_b = w_r*L**2/(8*T) // Total sag with wind pressure(m) w_i = 0.915*%pi/4*((d+2*t)**2-(d**2))/1000.0 // Weight of ice on conductor(kg/m) area_i = (d+2*t)*1000.0*10**-6 // Projected area to wind pressure(m^2) w_n = wind*area_i // Wind load/m(kg) w_r_c = ((W+w_i)**2+w_n**2)**0.5 // Resultant weight/m(kg) S_c = w_r_c*L**2/(8*T) // Total sag with wind pressure and ice coating(m) S_v = S_c*(W+w_i)/w_r_c // Vertical component of sag(m) // Results disp("PART II - EXAMPLE : 5.11 : SOLUTION :-") printf("\nCase(a) : Sag in still air, S = %.2f m", S_a) printf("\nCase(b) : Sag with wind pressure, S = %.2f m", S_b) printf("\n Sag with wind pressure and ice coating, S = %.2f m", S_c) printf("\n Vertical sag, S_v = %.2f m \n", S_v) printf("\nNOTE: ERROR: calculation mistake in the textbook")