# Chapter 3, Constructional Mechanical Features of line¶

## Exa 3.1 : page 70¶

In [5]:
from __future__ import division
from numpy import sqrt
#Given Data :
m=1/10 #unitless
EL=66 #in KV
E=EL/sqrt(3) #in KV
#Formula : E=E1+(11/10)*E1+(131/100)*E1+(1651/1000)*E1=(5061/1000)*E1
E1=E*(1000/5061) #in KV
print "E1 = %0.2f KV" %E1
E2=E1*(11/10) #in KV
print "E2 = %0.3f KV" %E2
E3=E1*(131/100) #in KV
print "E3 = %0.2f KV" %E3
E4=E1*(1651/1000) #in KV
print "E4 = %0.2f KV" %E4
Eta=(E/(4*E4))*100 #in %
print "String Efficiency = %0.1f %%" %Eta

E1 = 7.53 KV
E2 = 8.282 KV
E3 = 9.86 KV
E4 = 12.43 KV
String Efficiency = 76.6 %


## Exa 3.2 : page 71¶

In [7]:
from __future__ import division
from numpy import sqrt
#Given Data :
W=0.85 #in Kg/meter
L=250 #in meter
Ww=1.4 #in Kg
SafetyFactor=5 #unitless
UTS=10128 #Ultimate tensile strength in Kg
T=UTS/SafetyFactor #in Kg
Wi=0 #there is no ice
Wr=sqrt((W+Wi)**2+Ww**2) #in Kg
S=Wr*L**2/(8*T) #in meter
Sv=(W/Wr)*S #in meter
print "Horizontal sag = %0.3f m" %S
print "Vertical sag = %0.3f m" %Sv

Horizontal sag = 6.317 m
Vertical sag = 3.278 m


## Exa 3.3 : page 72¶

In [8]:
from __future__ import division
from numpy import sqrt
#Given Data :
L=150 #in meter
A=2 #in cm**2(cross sectional area)
US=5000 #in Kg/cm**2(ultimate strength)
g=8.9 #specific gravity
Ww=1.5 #in Kg/m(wind pressure)
SafetyFactor=5 #unitless
B_strength=2*US #in Kg
T=B_strength/SafetyFactor #in Kg
Volume=A*100 #in cm**2
Wc=1.78 #in Kg/m
Wr=sqrt(Wc**2+Ww**2) #in Kg
Sag=Wr*L**2/(8*T) #in meter
print "Sag = %0.2f m" % Sag

Sag = 3.27 m


## Exa 3.4 : page 73¶

In [11]:
from __future__ import division
from numpy import sqrt, pi
#Given Data :
L=160 #in meter
d=0.95 #in cm
A=pi*d**2/4 #in cm**2(cross sectional area)
US=4250 #in Kg/cm**2(ultimate strength)
g=8.9 #specific gravity
Ww=1.5 #in Kg/m(wind pressure)
SafetyFactor=5 #unitless
B_strength=2*US #in Kg
T=B_strength/SafetyFactor #in Kg
Volume=A*100 #in cm**2
Wc=1.78 #in Kg/m
Wr=sqrt(Wc**2+Ww**2) #in Kg
Sag=Wr*L**2/(8*T) #in meter
print "Sag = %.f m" % Sag
#Note : Answer in the book is not accurate.

Sag = 4 m


## Exa 3.5 : page 73¶

In [17]:
from __future__ import division
#Given Data :
m=75-45 #in meter
L=300 #in meter
T=2500 #in Kg
w=0.9 #in kg/meter
x=L/2-T*m/(w*L) #in meters
print "x = %0.2f m" %x
x=L/2-x #in meter
print "Centre point P from O is %0.2f m" %(x)
y=w*x**2/(2*T) #in meter
print "Height of point P, y= %0.2f m" %y
x=L/2-T*m/(w*L) #in meters
z=w*(L-x)**2/(2*T) #in meters
print "Height of B above O is, z = %0.2f m" %z
print "The mid point of the line is ",(z-y)," meter below point B, i.e., ",(75-(z-y))," meter above water level."

x = -127.78 m
Centre point P from O is 277.78 m
Height of point P, y= 13.89 m
Height of B above O is, z = 32.94 m
The mid point of the line is  19.05  meter below point B, i.e.,  55.95  meter above water level.


## Exa 3.6 : page 74¶

In [18]:
from __future__ import division
#Given Data :
L=60 #in meter
S=25*10**-2 #in meter
A=61.36 #in mm**2(cross sectional area)
W=0.5445 #in Kg/m
UTS=42.20 #in Kg/mm**2
T=W*L**2/(8*S) #in Kg
B_strength=UTS*A #in Kg
SafetyFactor=B_strength/T #unitless
print "Factor of safety =", round(SafetyFactor,2)

Factor of safety = 2.64


## Exa 3.7 : page 75¶

In [20]:
from __future__ import division
#Given Data :
L=220 #in meter
W=0.604 #in Kg/m
T_strength=5758 #in Kg
SafetyFactor=2 #unitless
T=T_strength/SafetyFactor #in Kg
S=W*L**2/(8*T) #in meter
print "Sag = %0.2f m " %S

Sag = 1.27 m


## Exa 3.8 : page 75¶

In [21]:
#Given Data :
W=850/1000 #in Kg/m
US=7950 #in kg
L=275 #in meter
h=8 #in meter(ground clearance)
SafetyFactor=2 #unitless
T=US/SafetyFactor #in Kg
S=W*L**2/(8*T) #in meter
H=h+S #in meter
print "Height above the ground = %0.2f m " %H

Height above the ground = 10.02 m


## Exa 3.9 : page 76¶

In [27]:
from math import floor
#Given Data :
m=1/9 #unitless
EL=33 #in K
EbyE1=1+(1+m)+(1+3*m+m**2) #assumed
E=EL/sqrt(3) #in KV
E1=E/EbyE1 #in KV
print "E1 = %0.2f kV" %E1
E2=(1+m)*E1 #in KV
print "E2 = %0.2f kV" %E2
E3=(1+3*m+m**2)*E1 #in KV
print "E3 = %0.2f kV" %E3
E=E1+E2+E3 #in KV
Eff=E/(3*E3)
Eff*=100  # %
print "String Efficiency = %.f %%" %floor(Eff)

E1 = 5.51 kV
E2 = 6.12 kV
E3 = 7.42 kV
String Efficiency = 85 %


## Exa 3.10 : page 77¶

In [29]:
#Given Data :
#Applying KCL we get I1+i1=I2+ix and I2+i2=I3+iy
#On solving we get : 1*2*E1=1*1*E2+0*1*E3 and 0*2*E1=-1*2*E2+1*3*E3
E1byE=1/(1+(154/155)+(166/155)) #assumed
E2byE=(154/155)*E1byE #assumed
E3byE=(166/155)*E1byE #assumed
Eff=1/((3*(166/155)*E1byE))
Eff*=100  # %
print "String Efficiency = %.f %%" %Eff

String Efficiency = 95 %


## Exa 3.11 - page : 78¶

In [30]:
#Given Data :
L=200 #in meter
W=684/1000 #in Kg/m
T=1450 #in Kg
S=W*L**2/(8*T) #in meter
print "Sag = %0.2f m" %S

Sag = 2.36 m


## Exa 3.12 : page 78¶

In [34]:
from math import sqrt
#Given Data :
L=220 #in meter
T=586 #in Kg
Wc=0.62 #in Kg
Ww=39.2*0.94/100 #in Kg
Wr=sqrt(Wc**2+Ww**2) #in Kg
cos_theta=Wc/Wr #unitless
Sv=Wr*L**2*cos_theta/(8*T) #in meter
print "Vertical Sag = %0.2f m" %Sv
# Answer is not accurate in the textbook.

Vertical Sag = 6.40 m