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// 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")
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