# Chapter 8: Mechanical Design of Overhead Lines¶

### Example 8.1, Page Number: 171¶

In :
import math

#Variable declaration:
k = 0.11                        #ratio of shunt-capatance to self capacitance
V = 33/math.sqrt(3)                  #Voltage across string(kV)
n = 3                           #no. of insulators

#Calculation:

#  At Junction A
#        I2 = I1 + i1
# or V2*w*C = V1*w*C + V1*K*ω*C
# or     V2 = V1*(1 + K) = V1*(1 + 0·11)
# or     V2 = 1·11*V1                     ...(i)

#  At Junction B
#          I3 = I2 + i2
#  or  V3*w*C = V2*w*C + (V1 + V2) K*w*C
#  or      V3 = V2 + (V1 + V2)*K
#             = 1·11*V1 + (V1 + 1·11*V1)*0·11
#          V3 = 1·342 V1
#  Voltage across whole unit,
#    V = V1+V2+V3
# or V = V1+1.11*V1+1.342*V1

V1 = V/3.452                    #Voltage across top unit(V)
V2 = 1.11*V1                    #Voltage across top unit(V)
V3 = 1.342*V1                   #Voltage across top unit(V)

e = V/(n*V3)*100               #string efficiency

#Result:
print "(i)\tVoltage across top unit,V1 is",round(V1,2),"kV"
print "   \tVoltage across middle unit, V2 is",round(V2,2),"kV"
print "   \tVoltage across bottom unit, V3 is",round(V3,1),"kV"
print "(ii)\tThe string efficiency is ",round(e,1),"%"

(i)	Voltage across top unit,V1 is 5.52 kV
Voltage across middle unit, V2 is 6.13 kV
Voltage across bottom unit, V3 is 7.4 kV
(ii)	The string efficiency is  85.7 %


### Example 8.2, Page Number: 172¶

In :
from __future__ import division
import math

#Variable declaration:
V1 = 8                        #voltage across top unit(kV)
V2 = 11                       #voltage across middle unit(kV)
n = 3                         #no. of insulators
#Calculation:

#(i)
# Let K be the ratio of capacitance between pin and the earth
# to self capacitance.
# Let C farad be the self capacitance of each unit,
# then, capacitance between pin and earth = K*C.
# Applying Kirchoff’s current law to Junction A,
#        I2 = I1 + i1
# or V2*w*C = V1*w*C + V1*K*w*C
# or     V2 = V1 (1 + K)

k = (V2-V1)/V1

#(ii)
# Applying Kirchoff’s current law to Junction B,
#          I3 = I2 + i2
#  or  V3*w*C = V2*w*C + (V1 + V2)*K*w*C
#  or      V3 = V2 + (V1 + V2)*K = 11 + (8 + 11) * 0·375
V3 = 18.12                        #kV
V = V1 + V2+ V3                   #kV
Vl = math.sqrt(3)*V                    #line voltage(kV)

#(iii)
e = V/(n*V3)*100                     #stirng efficiency(%)

#Result:
print "(i)  Ratio of capacitance b/w pin & earth is",k
print "(ii) The line voltage is",round(Vl,2),"kV"
print "(iii)String efficiency is",round(e,2),"%"

(i)  Ratio of capacitance b/w pin & earth is 0.375
(ii) The line voltage is 64.29 kV
(iii)String efficiency is 68.29 %


### Example 8.3, Page Number: 172¶

In :
from __future__ import division

#Variable declaration:
n = 3                            #no. of insulators
V3 = 17.5                        #voltage across the line unit(V)

#  At Junction A
#   I2 = I1 + i1
#   V2 ω C = V1 ω C + V1 K ω C
# or V2 = V1 (1 + K) = V1 (1 + 0.125)
#  ∴ V2 = 1·125 V1

#  At Junction B
#    I3 = I2 + i2
# or  V3 ω C = V2 ω C + (V1 + V2) K ω C
# or      V3 = V2 + (V1 + V2) K
#            = 1·125 V1 + (V1 + 1·125 V1) × 0.125
#       ∴ V3 = 1·39 V1

V1 = V3/1.39                     #Voltage across top unit(kV)
V2 = 1.125*V1                    #Voltage across middle unit(kV)

V = V1+V2+V3                     #voltage across line unit(kV)
e = V/(n*V3)*100                     #string efficiency(%)

#Result:
print "Line to neutral voltage is",round(V,2),"kV"
print "String efficiency is",round(e,2),"%"

Line to neutral voltage is 44.25 kV
String efficiency is 84.29 %


### Example 8.4, Page Number: 173¶

In :
from __future__ import division
from pylab import *
from sympy import *

#Variable declaration:
V3 = 13.1                      #voltage across the lowest insulator(kV)
V2 = 11                        #voltage across the middle insulator(kV)
K = symbols('K')
#  Applying Kirchhoff’s current law to Junctions A and B, we can easily
#  derive the following equations:

#     V2 = V1 (1 + K)
#  or V1 =  V2/(1 + K)                   ...(i)
# and V3 = V2 + (V1 + V2)*K           ...(ii)
#  Putting the value of V1 = V2/(1 + K) in eq. (ii), we get,

#V3 = V2 + (V2/(1+k)+ V2)*K
K1 = solve(V3-(V2 + (V2/(1+K)+ V2)*K),K)

V1 = V2/(1+K1)
V = V1+V2+V3                  #kV

#Result:
print "The voltage b/w the bus bars is",round(float(math.sqrt(3)*V)),"kV"

The voltage b/w the bus bars is 59.0 kV


### Example 8.5, Page Number: 173¶

In :
from __future__ import division

#Variable declaration:
n = 3                         #no.of insulators
k = 1/8                 #ratio of self-capacitance to shunt capacitance
V3 = 15                    #safe working voltage of each insulator(kV)

#  Applying Kirchhoff’s current law to junction A, we get,
#     V2 = V1*(1 + K)
#  or V1 = V2/(1 + K) = V2/(1 + 0·125) = 0·89*V2

#  Applying Kirchhoff’s current law to Junction B, we get,
#  V3 = V2 + (V1 + V2)*K = V2 + (0.89*V2 + V2) * 0.125
#  V3 = 1.236*V2
V2 = V3/1.236                     #kV
V1 = 0.89*V2                       #kV
V = V1+V2+V3                      #voltage across the string(kV)
e = V/(n*V3) *100                     #string efficiency(%)

#Result:
print "Voltage across the string is",round(V,3),"kV"
print "The string efficiency is",round(e,2),"%"

Voltage across the string is 37.937 kV
The string efficiency is 84.3 %


### Example 8.6, Page Number: 174¶

In :
from __future__ import division

#Variable declaration:
n = 4                        #no. of unit

#Calculation:
# Suppose Xc = 1 Ω. Given, the ratio of self-capacitance to shunt
# capacitance is 10, so, Xc1 = 10 Ω .
# Suppose that potential V across the string is such that 1 A current
# flows in the top insulator.
#Thus,
V1 = 1*1                          #Voltage across top unit(V)
V2 = 1*1.1                        #Voltage across 2nd unit(V)
V3 = 1*1.31                       #Voltage across 3rd unit(V)
V4 = 1*1.65                       #Voltage across 4th unit(V)
V = V1+V2+V3+V4                   #Voltage obtained across the string(V)

#The voltage across each unit expressed as a % of V becomes:
f1 = V1/V*100
f2 = V2/V*100
f3 = V3/V*100
f4 = V4/V*100

e = V/(n*V4)*100                    #string efficiency(%)

#Result:
print "(i) The voltage distributation from top unit is "
print "\t",round(f1,2),"%\t",round(f2,2),"%\t",round(f3,1),"%\t",round(f4,1),"%"
print "\n(ii)String efficiency is",round(e,1),"%"

(i) The voltage distributation from top unit is
19.76 %	21.74 %	25.9 %	32.6 %

(ii)String efficiency is 76.7 %


### Example 8.7, Page Number: 175¶

In :
from __future__ import division
import math

#Variable declaration:
n = 5                       #no.os insulators
Vl = 100                    #line voltage(kV)

#Suppose Xc = 1 Ω.
#Ratio of self capacitance to shunt capacitance is 10, so, Xc1 = 10 Ω.
#Let V be the voltage s.t 1A current flows in top insulator.

#Calculation:
#calculation for fraction of voltage drop:
v1 = 1*1                          #Voltage across top unit(V)
v2 = 1*1.1                        #Voltage across 2nd unit(V)
v3 = 1*1.31                       #Voltage across 3rd unit(V)
v4 = 1*1.65                       #Voltage across 4th unit(V)
v5 = 1*2.16                       #voltage across 5th unit(V)
v = v1+v2+v3+v4+v5                #total voltage(V)
V = 100/math.sqrt(3)                   #string voltage(V)
V1 = round(v1/v*V,3)                       #kV
V2 = v2/v*V                       #kV
V3 = v3/v*V                       #kV
V4 = v4/v*V                       #kV
V5 = v5/v*V                       #kV
e = V/(n*V5)*100                      #string efficiency(%)

#Result:
print "(i)The distribution of voltage on the insulator discs are"
print "V1 = ",round(V1,3),"kV ","\tV2 =",round(V2,2),"kV \n",
print "V3 =",round(V3,1),"kV","\t\tV4 =",round(V4,2),"kV",
print "\nV5 =",round(V5,1),"kV"
print "\n(ii)The string efficiency",round(e,1),"%"

(i)The distribution of voltage on the insulator discs are
V1 =  7.997 kV  	V2 = 8.8 kV
V3 = 10.5 kV 		V4 = 13.19 kV
V5 = 17.3 kV

(ii)The string efficiency 66.9 %


### Example 8.8, Page Number: 176¶

In :
from __future__ import division
import math
from sympy import *

#Variable declaration:
n = 4                       #no.os insulators
V2 = 13.2                   #Voltage across 2th unit
V3 = 18                     #Voltage across 3th unit
#Suppose K is the ratio of shunt-capacitance to self-capacitance
k = symbols('k')

# Referring to Fig.(ii), we have,
# V2/V1 = ( 1 + K)/1
# V2 = V1 (1 + K)
# V3/V1 = (1 + 3K + K2)/1
# V3 = V1 (1 + 3K + K2)
# putting values of V2 & V3

k1 = solve(13.2*k**2+21.6*k-4.8,k)
V1 = V2/(1+k1)
V4 = V1*(1 + k1**33 + 5*k1**2 + 6*k1)
V = V1+V2+V3+V4                   #Voltage between line and earth(kV)
Vl = math.sqrt(3)*V                    #Voltage between conductors(kV)

#Result:
print "Voltage between conductors is",round(float(Vl)),"kV"

Voltage between conductors is 119.0 kV


### Example 8.9, Page Number: 177¶

In :
from __future__ import division

#Variable declaration:
k = 5               #ratio of self-capacitance(C) to pin-earth capacitance(C1)
V = 1+1.2+1.64+2.408                  #voltage obtained across the string(V)

#Calculation:
#(i) The voltage across each unit expressed as a % of V
v1 = 1/V*100
v2 = 1.2/V*100
v3 = 1.6/V*100
v4 = 2.408/V*100

#(ii)String efficiency
e = V/(4*2.408)*100

#Result:
print "(i) The voltage across each unit expressed as a % of V are:"
print "\tTop Unit: ",round(v1),"%"
print "\tSecond from top: ",round(v2,1),"%"
print "\tThird from top: ",round(v3),"%"
print "\tFourth from top: ",round(v4,1),"%"
print "(ii)String efficiency is",round(e,2),"%"

(i) The voltage across each unit expressed as a % of V are:
Top Unit:  16.0 %
Second from top:  19.2 %
Third from top:  26.0 %
Fourth from top:  38.5 %
(ii)String efficiency is 64.87 %


### Example 8.10, Page Number: 177¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
V1,V2,V3,V= symbols('V1 V2 V3 V')          #volts (shown in fig(ii))

#Calculation:

#We know that in an actual string of insulators, 3 capacitances
#exist i.e, self-capacitance of each insulator, shunt capacitance
#and capacitance of each unit to line.
#But, capacitance of each unit to line is very small and
#can be neglected.

#At Junction A
#     I2 + i1 = I1 + i1
#or   V2*w*C + (V2 + V3)*0·1*w*C = V1*ω*C + 0·15*C*V1*w
#or   0·1*V3 = 1·15*V1 − 1·1*V2
#     V3 = 11·5*V1 − 11*V2                     ...(i)

#At Junction B
#     I3 + i′2 = I2 + i2
#or   V3*w*C + V3*0·1*C*w = V2*w*C + (V1 + V2)*w*0·15 C
#or   1·1*V3 = 1·15*V2 + 0·15*V1                ...(ii)
#Putting the value of V3 from exp (i). into exp. (ii), we get,
#     1·1*(11·5*V1 − 11*V2) = 1·15*V2 + 0·15*V1
#or   13·25*V2 = 12·5*V1
#    V2 = 12.5/13.25*V1
#    V3 = 11.5*V1-11*V2
#    V = V1+V2+V3
#    V  = 40.55/13.25*V1
V1 = 13.25/40.55*V
V2 = 12.5/13.25*V1
V3 = 14.8/13.25*V1

#The voltage across each unit expressed as a % of V becomes:
p1 = V1/V*100
p2 = V2/V*100
p3 = V3/V*100

#Results:
print "The voltage across each unit expressed as a % of V becomes:"
print "for top unit, v1 =",round(p1,1),"%",
print "\nfor middle unit, v2 =",round(p2,1),"%"
print "for 3rd unit, v3 =",round(p3,1),"%"

print "String efficiency is",round(V/(3*V3)*100,1),"%"

The voltage across each unit expressed as a % of V becomes:
for top unit, v1 = 32.7 %
for middle unit, v2 = 30.8 %
for 3rd unit, v3 = 36.5 %
String efficiency is 91.3 %


### Example 8.11, Page Number: 179¶

In :
from __future__ import division
from sympy import *

#Variable Declaration:
V1,V2,V3,V= symbols('V1 V2 V3 V')          #volts (shown in fig(i))

#Calculation:

#At Junction A
#    I2 + i′1 = I1 + i1
#or  V2*w*C + (V2+V3)*w*0·1*C  = V1*w*C+V1*0·2*C*ω
#   V3 = 12 V1 − 11 V2                                  # ...(i)

#At Junction B
#      I3 + i′2 = I2 + i2
#or    V3 ω C + V3 × 0·3 C × ω =V2 ω C + (V1 + V2) ω × 0·2 C
#or    1·3 V3 = 1·2 V2 + 0·2 V1 ...(ii)
#Substituting the value of V3 from exp. (i) into exp. (ii), we get,
#      1·3*(12*V1 − 11*V2) = 1·2*V2 + 0·2*V1
#or    15·5*V2 = 15·4*V1
V2 = 15.4*V1/15.5                     #...(iii)
V2 =  0.993*V1

#Substituting the value of V2 from exp. (iii) into exp. (i), we get,
V3 = 12*V1 - 11*0.993*V1
V3 = 1.077*V1

#Voltage between conductor and earth (i.e. phase voltage)
V = V1+V2+V3
n = V/(3*V3)*100

#Result:
print "String efficiency is",round(n),"%"

String efficiency is 95.0 %


### Example 8.13, Page Number: 184¶

In :
from __future__ import division
from pylab import *
import math

#Variable declaration:
d = 100                       #conductor spacing(cm)
go = 30/math.sqrt(2)               #Dielectric strength of air,(rms)(kV/cm)
dl = 0.952                    #air density factor
mo = 0.9                      #irregularity factor

#Calculation:
Vc = mo*go*dl*r*math.log(d/r)     #kV/phase
Vl = Vc*3**0.5                   #kV/line

#Result:
print "The disruptive critical voltage for the line is",round(float(Vl),2),"kV"

The disruptive critical voltage for the line is 144.97 kV


### Example 8.14, Page Number: 184¶

In :
from __future__ import division
from pylab import *

#Variable declaration:
go = round(30/2**0.5,1)                  #Dielectric strength of air,rms(kV/cm)
Vc = 210/3**0.5                 #Disruptive voltage/phase(kV)
mo = 1                          #irregularity factor(for smooth conductor)
dl = 1              #air density factor(at std pressure and temperature)

#Calculation:
d = r*10**(Vc/(2.3*mo*go*dl*r))

#Result:
print "The spacing between the conductors is",round(d),"cm"

The spacing between the conductors is 341.0 cm


### Example 8.15, Page Number: 185¶

In :
from __future__ import division
from pylab import *
import math

#Variable declaration:
t = 40                            #temperature(deg C)
b = 76                           #atmospheric pressure(cm)
mo = 0.85                        #irregularity factor
f = 50                           #Hz
d = 200                         #conductor spacing(cm)
go = 30/2**0.5                  #Dielectric strength of air,rms(kV/cm)
Vl = 220                        #line voltage(kV)

#Calculation:
dl = round(3.92*b/(273+t),3)

Vc = round(mo*go*dl*r*round(math.log(d/r),2),1)  #Critical disruptive voltage per phase(kV)
V = round(Vl/3**0.5)                    #kV/phase
P = 242.2/dl*(f+25)*(r/d)**0.5*(V-Vc)**2*10**-5      #kW/km/phase
Pl = 3*P                       #corona loss(kW)

#Result:
print "Total corona loss per km for three phases is",round(Pl,5),"kW"

Total corona loss per km for three phases is 0.05998 kW


### Example 8.16, Page Number: 185¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
P1 = 53                    #corona loss(kW)
V1 = 106                   #operating voltage(kV)
P2 = 98                   #corona loss(kW)
V2 = 110.9                #operating voltage(kV)
V3 = 113                  #opeating voltage(kV)

#Calculation:
#As f, δ, r and d are the same for the two cases,
#therfore, P ∝ (V − Vc)**2
k,Vc = symbols('k Vc')           #k = proportionality constant
#Vc = critical disuptive voltage(kV)
#for 1st case,
#    V1/3**0.5 = k*(64 - Vc)**2
#    V2/3**0.5 = k*(61·2 - Vc)**2
# Dividing above equations & solving,
Vc1 = round(solve(((64-Vc)/(61.2-Vc))-math.sqrt(98/53),Vc),2)
Vc1 = 54
#Let W kilowatt be the power loss at 113 kV.
#W = k*(113/3**0.5-Vc1)**2
W = P1*((65.2-Vc1)/(61.2-Vc1))**2

#Result:
print "The disruptive critical voltage at 113 kV is",Vc1,"kV"
print "The corona loss at 113 kV is",round(W),"kW"

The disruptive critical voltage at 113 kV is 54 kV
The corona loss at 113 kV is 128.0 kW


### Example 8.17, Page Number: 190¶

In :
from __future__ import division

#Variable declaration:
w = 680                        #weigth of conductor(kg/km)
l = 260                        #length ofspan(m)
u = 3100                       #ultimate strength(kg)
sf = 2                          #safety factor
h = 10                         #ground clearance(m)

#Calculation:
T = u/sf                        #working tension(kg)
s = w*l**2/(8*T*1000)            #span(m)
H = h+s

#Result:
print "Conductor should be supported at a height of ",round(H,1),"m"

Conductor should be supported at a height of  13.7 m


### Example 8.18, Page Number: 190¶

In :
from __future__ import division
import math

#Variable declaration:
l = 150                        #length ofspan(m)
a = 2                          #conductor cross-section(cm**2)
go = 9.9                      #specific gravity of conductor(gm/cm**3)
ww = 1.5                      #Wind force/m length of conductor(kg)
T = 2000                       #working tension(kg)

#Calculation:
w = go*a*1/10              #Wt. of conductor/m length(kg)
wt = round((w**2+ww**2)**0.5,2)      #Total wt of 1m length of conductor(kg)
s = wt*l**2/(8*T)              #sag(m)
#This is the value of slant sag in a direction making
#an angle theta with the vertical

theta = math.atan(ww/w)
Vs = s*math.cos(theta)

#Result:
print "The sag is",round(s,2),"m"
print "The vertical sag is",round(Vs,2),"m"

The sag is 3.49 m
The vertical sag is 2.78 m


### Example 8.19, Page Number: 191¶

In :
from __future__ import division
import math

#Variable declaration:
l = 200                     #span length(m)
w = 1170/1000                #weigth of conductor(kg)
u = 4218                    #Ultimate Strength(kg/cm**2)
sf = 5                      #safety factor
a = 1.29                    #cross-section of conductor(cm**2)
P = 122                        #wind pressure(kg/m**2)

#Calculation:
T = u/sf*a                   #working tension(kg)
d = round((4*a/math.pi)**0.5,2)            #diameter of conductor(cm)
ww = P*d*10**-2             #Wind force/m length,
wt = (ww**2+w**2)**0.5          #total wt of conductor per metre length
S = wt*l**2/(8*T)
theta = math.atan(ww/w)
VS = S*math.cos(theta)

#Result:
print "The sag is",round(S,2),"m"
print "The vertical sag is",round(VS,2),"m"

The sag is 8.97 m
The vertical sag is 5.38 m


### Example 8.20, Page Number: 191¶

In :
from __future__ import division

#Variable declaration:
l = 275                    #span length(m)
w = 0.865                 #weigth of conductor/m(kg)
d = 1.96                    #conductor diameter(cm)
t = 1.27                    #ice coating thickness(cm)
u = 8060                    #Ultimate Strength(kg)
sf = 2                      #safety factor
a = 1.29                    #cross-section of conductor(cm**2)
wo = 0.91                   #Weight of 1 c.c. of ice(gm)
P = 3.9                      #wind pressure(gm/cm**2)

#Calculation:
T = u/sf                    #working tension(kg)
v = 3.14*t*(d+t)*100            #Volume of ice per metre(cm**3)
wi = v*wo/1000                   #kg
ww = P*(d+2*t)*100/1000          #kg
wt = ((w+wi)**2+ww**2)**0.5       #total weight(kg)
S = wt*l**2/(8*T)                  #sag(m)

#Result:
print "Sag is",round(S,1),"m"

Sag is 6.3 m


### Example 8.21, Page Number: 192¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
Sv = 2.35                    #vertical sag(m)
P = 1.5                      #wind pressure(kg/m)
B = 2540                       #breaking stress(kg/cm**2)
w = 1.125                   #wt of conductor(kg/m)
l = 214                     #length of conductor(m)
a = 3.225                    #conductor cross-section

#Calculation:
f = symbols('f')             #safety factor
wt = (w**2+P**2)**0.5              #Total wt. of one m length of conductor
T = B*a/f                   #working stress(kg)
cos_theta = w/wt
S = Sv/cos_theta            #slant sag
T1 =  wt*l**2/(8*S)
f1 = solve(T-T1,f)           #safety factor

#Result:
print "Safety factor is",round(f1)

Safety factor is 3.0


### Example 8.22, Page Number: 192¶

In :
from __future__ import division

#Variable declaration:
l = 150                         #span length(m)
u = 5000                       #ultimate strength(kg/cm**2)
go = 8.9                       #specific gravity of material(gm/cc)
P = 1.5                        #wind pressure(kg/m)
sf = 5                         #safety factor
a = 2                           #cross-section of conductor(cm**2)
h = 7                          #ground clearance(m)

#Calculation:
w = a*go*100/1000                  #weight of conductor per m(kg)
T = u*a/sf                    #working stress per m(kg)
wt = (w**2+P**2)**0.5         #Total wt of 1 m length of conductor(kg)
S = wt*l**2/(8*T)               #slant sag(m)
Sv = S*w/wt                   #conductor vertical sag(m)
H = h+Sv                     #m

#Result:
print "Conductor should be supported at a height of",round(H,1),"m"

Conductor should be supported at a height of 9.5 m


### Example 8.23, Page Number: 193¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
l = 500                                #distance b/w the two towers(m)
T = 1600                               #tension in the conductor(kg)
w = 1.5                                 #weight of the conductor(kg/m)
h1 = 30                                #height of 1st tower(m)
h2 = 90                                #height of 2nd tower(m)

#Calculation:
h = h2-h1                            #difference in levels(m)
x1,x2 = symbols('x1 x2')
x2 = l-x1
S1 = w*x1**2/(2*T)                   #m
S2 = w*x2**2/(2*T)                    #m
x11 = solve(S2-S1-h,x1)
S1 = w*x11**2/(2*T)                   #m
#Let the mid-point P be at a distance x from the lowest point O
x = l/2-x11
Smid = w*x**2/(2*T)                  #m

#Result:
print "The min. clearance of the conductor and water is ",round(h1-S1),"m"
print "The clearance mid-way b/w the supports is",round(Smid+h1-S1,2),"m"

The min. clearance of the conductor and water is  23.0 m
The clearance mid-way b/w the supports is 30.7 m


### Example 8.24, Page Number: 194¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
l = 600                                #distance b/w the two towers(m)
f = 5                                  #safety factor
U = 8000                               #kg/cm**2
a = 2.2                                #area of cross-section(cm**2)
w = 1.925                              #weight of the conductor(kg/m)
h = 15                                  #difference in towers levels(m)
wi = 1                                 #ice weight(kg)

#Calculation:
T = U*a/f                              #tension in the conductor(kg)
wt = wi+w                              #total weight of conductor(kg)
x1,x2 = symbols('x1 x2')
x2 = l-x1
S1 = wt*x1**2/(2*T)                   #m
S2 = wt*x2**2/(2*T)                    #m
x11 = solve(S2-S1-h,x1)
x22 = l-x11
S22 = wt*x22**2/(2*T)                   #m

#Result:
print "The sag from the taller of the two supports is",round(S22,2),"m"

The sag from the taller of the two supports is 45.27 m


### Example 8.25, Page Number: 195¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
l = 400                                #distance b/w the two towers(m)
T = 2000                               #tension in the conductor(kg)
w = 1                                 #weight of the conductor(kg/m)
h1 = 40                                #height of 1st tower(m)
h2 = 90                                #height of 2nd tower(m)

#Calculation:
h = h2-h1                            #difference in levels(m)
x1,x2 = symbols('x1 x2')
x2 = l-x1
S1 = w*x1**2/(2*T)                   #m
S2 = w*x2**2/(2*T)                    #m
x11 = solve(S2-S1-h,x1)
S1 = w*x11**2/(2*T)                   #m
x_mid = 50+400/2           #Horizontal distance of mid-point P from lowest point O
Sp = w*x_mid**2/(2*T)                      #SAg at P(m)
S2 = w*(l-x11)**2/(2*T)                   #m
hc = h2-S2+Sp                       #Clearance of mid-point P above water level(m)

#Result:
print "Clearance of mid-point P above water level is",round(hc),"m"

Clearance of mid-point P above water level is 55.0 m


### Example 8.26, Page Number: 195¶

In :
from __future__ import division
from pylab import *
from sympy import *
import math

#Variable declaration:
T = 1500                             #tension in the conductor(kg)
w = 1                                #weight of the conductor(kg/m)
DE = 300                              #distance b/w tower's top(m)
H = 22                               #height of each tower(m)

#Calculation:
# Suppose, the conductors are supported between towers AD and BE over
# a hillside having gradient of 1 : 20.
# Let the lowest point on the conductor is O and
# sin(theta) = 1/20.
# since, the lowest conductor is fixed 2m below the top of each tower.
theta = math.asin(1/20)
BE = H-2                          #Effective height of each tower(m)
EC = DE*1/20
x1,x2 = symbols('x1 x2')
DC = (DE**2-EC**2)**0.5                  #Horizontal distance b/w two towers(m)
x2 = DC-x1
S1 = w*x1**2/(2*T)                    #m
S2 = w*x2**2/(2*T)                    #m
x11 = solve(S2-S1-EC,x1)
S2 = w*(DC-x11)**2/(2*T)               #m
BC = BE+EC
OG = BC-S2-x11*math.tan(theta)

#Result:
print "Clearance of the lowest point O from the ground is",round(OG,2),"m"

Clearance of the lowest point O from the ground is 14.4 m


### Example 8.27, Page Number: 196¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
l = 300                     #distance b/w tower(m)
S = 10                        #span(m)
h = 8                      #clearance(m)

#Calculation:
#On level ground:
r = 8*S/l**2              #ratio w/T

#On sloping ground:
#The conductors are supported between towers AD and BE
#over a sloping ground having a gradient 1 in 15 as shown above.

DE = 300
sin_theta = 1/15

h = DE*sin_theta             #Vertical distance b/w the two towers(m)
BE = S+h                      #height of tower(m)
x1,x2,x = symbols('x1 x2 x')
x2 = l-x1
S1 = r*x1**2/2                   #m
S2 = r*x2**2/2                   #m
x11 = solve(S2-S1-h,x1)
S11 = r*x11**2/(2)                   #m
x22 = l-x11
S22 = r*x22**2/(2)                   #m
tan_theta = 1/15
GF = x11*tan_theta
BC = h-S22-GF
#Since O is the origin, the equation of slope of ground is given by :
y = x/15-10.5
#C = Equation of conductor curve − y
C = r*x**2/2-y
C1 = diff(C,x)
xm = solve(C1,x)
#putting values of xm in above equations,
ym = xm/15-10.5
Cm = C = r*xm**2/2-ym

#Result:
print "Required minimum clearance is",round(Cm),"m"

Required minimum clearance is 8.0 m