# Chapter 07 : Mechanical Design of Transmission Lines¶

## Example 7.1, Page No 161¶

In [5]:
import math
#initialisation of variables
sf=5.0	   #Factor of safety
d=0.95	   # conductor dia(cm)

#Calculations
Ws=4250.0/sf   # working stress(kg/cm_2)
A=math.pi*(d**2)/4.0   # area (cm_2)
Wp=40.0*d*(10**-2)   #wind pressure (kg/cm)
W=math.sqrt((.65**2)+(0.38**2))   # Total effective weight(kg/m)
T=850.0*A   # working tension (kg)
c=T/W
l=160.0
d=l**2/(8*800)

#Results
print("sag, d=%.0f metres " %d)

sag, d=4 metres


## Example 7.2, Page No 161¶

In [6]:
import math
#initialisation of variables
D=1.95 + 2.6			# overall diameter(cm)
A=4.55*(10**-2)			# area(m_2)
d=19.5					#diameter of conductor(mm)

#Calculations
Wp=A*39    #wind pressure(kg/m_2)
t=13       #ice coating(mm)
US=8000.0  # ultimate strength(kg)
Aice=math.pi*(10**-6)*((r+t)**2 - r**2)#area section of ice (m_2)
Wice=Aice*910
W=(math.sqrt((.85+Wice)**2 + Wp**2))# total weight of ice (kg/m)
T=US/2.0    # working teansion (kg)
c=T/W
l=275       #length of span(m)
Smax=l*l/(8*c)

#Results
print("Maximum sag=%.1f metres\n" %Smax)

Maximum sag=6.4 metres



## Example 7.3, Page No 162¶

In [7]:
import math
import numpy
#initialisation of variables
A=13.2				# cross section of conductor (mm_2)
Ar=4.1*(10**-3)		# projected area
w=0.115

#Calculations
q1=W/0.115
b=w/A
f1=21.0		#working stress
T1=f1*A
c=T1/W
l=45.7
S=l*l/(8*c)
dT=32.2-4.5# difference in temperature
E=1.26*(10000)
a=16.6*(10**-6)
d=8.765*(10**-3)
K=f1-((l*d*q1)**2)*E/(24*f1*f1)
p=numpy.polynomial.polynomial.polyval3d(-84.23,0,-14.44,1)
r=numpy.roots(p)
f2= 14.823332# accepted value of f2
T=f2*A
c=T/w
d1=l*l/(8*c)

#Results
print("sag at 32.2 Celsius , d=%.4f metres" %d1)

sag at 32.2 Celsius , d=0.1534 metres


## Example 7.4 Page No 165¶

In [8]:
import math
#initialisation of variables
T=2000.0        # working tension (kg)
w=1.0
c=T/w
h=90-30
l=300.0        #span(m)

#Calculations
a=(l/2)-(c*h/l)
b=550.0
d1=a*a/(2*c)
d2=(400**2)/(2*c)        # sag at 400 metres(m)
Hm=d2-d1        #height of mid point with respect to A
Cl=30+Hm

#Results
print("The clearance between the conductor and water level midway between the towers= %.3f metres " %Cl)

The clearance between the conductor and water level midway between the towers= 54.375 metres