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
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
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
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.
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."
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)
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
#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
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)
#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
#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
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.