# chapter 2, Layout of transmission systems¶

## Exa 2.1 : page 32¶

In [245]:
from sympy import symbols
from __future__ import division
a=symbols('a')
#Given Data :
CableCost=20+400*a #in Rs./meter (a=cross section in cm**2)
Cable_cost=(20+400*a)*1000 #in Rs./km
l=1 #in Km
P=1 #in MW
V=11 #in KV
cosfi=0.8 #powerfactor
h=3000 #hours
i=10 #in %
E_cost=15 #in paisa/kwh
rho=1.75*10**-6 #sp. resistance in ohm-cm
C1=CableCost*1000 #in Rs./km
R=rho*l*10**3/(a*10**-2) #in ohm
Ifl=(P*10**6)/(V*10**3*cosfi) #in Ampere
Ploss=2*Ifl**2*R #in Watts
Annual_cost=Ploss*10**-3*h*E_cost/100 #in Rs.
AnnualCost2=400*10**3*a*i/100 #in Rs.
annually_energy_lost = 40000/a  # Rs.
#comparing Annual_cost = AnnualCost2
a=Annual_cost/AnnualCost2*a**2
print "Most economical cross section area, a = ",round(a,4), "cm2 "

Most economical cross section area, a =  0.0508 cm2


## Exa 2.2 : page 33¶

In [246]:
from math import sqrt
#Given Data :
Pt=30*10**6 #in watt
V=220*10**3 #in volt
l=275*10**3 #in meter
R=0.173 #in ohm/km
Eta=90 #in %
density=8.9 #copper density
Loss=100-Eta #in %
cosfi=0.8 #powerfactor
print "3-phase 3 wire :"
IL=Pt/(sqrt(3)*V*cosfi) #in Ampere
LineLosses=(Loss/100)*Pt #in watts
rho=R*10**-4/(1*10**3) #in ohm-meter
a=3*IL**2*rho*l/(LineLosses) #in m**2
Volume=3*a*l #in m**3
Cu_weight=Volume*density #in Tones
print "Weight of copper = %0.f Tones" %Cu_weight
print "Single phase 2 wire :"
IL=Pt/(V*cosfi) #in Ampere
a=2*IL**2*rho*l/(LineLosses) #in m**2
Volume=2*a*l #in m**3
Cu_weight=Volume*density #in Tones
print "Weight of copper = %0.1f Tones" %Cu_weight
#Note : answer is not accurate in the book.

3-phase 3 wire :
Weight of copper = 338 Tones
Single phase 2 wire :
Weight of copper = 451.1 Tones


## Exa 2.3 : page 35¶

In [247]:
from sympy import symbols
from __future__ import division
from math import sqrt
a=symbols('a')
#Given Data :
l=1 #in Km
l=l*10**5 #in cm
I=200 #in Ampere
CableCost=50*a #in Rs./meter (a=cross section in cm**2)
E_cost=5 #in paisa/kwh
i=10 #in %
rho=1.72*10**-6 #resistivity in ohm-cm
R=rho*l/a #in ohm
Eloss=2*I**2*R*24*365/1000 #in kwh
AnnualCost2=(E_cost/100)*2*I**2*rho*l*24*365/1000/a  #in Rs.
C1=CableCost*1000 #in Rs./km
AnnualCharges=C1*i/100 #in Rs.
a=sqrt(AnnualCost2/AnnualCharges*a**2)  # cm2
print "Most economical cross section area a = %0.3f cm2" %a

Most economical cross section area a = 1.098 cm2


## Exa 2.4 : page 36¶

In [248]:
from sympy import symbols
from __future__ import division
from math import sqrt
a=symbols('a')
#Given Data :
l=4*10**5 #in cm
VL=30 #in KV
LineCost=40000*a+7500 #in Rs/km
i=8 #in %
E_cost=4 #in paisa/kwh
rho=1.72*10**-6 #in ohm-cm
R=rho*l/a #in ohm
P1=3*10**6 #in watt
h1=10 #in hours
cosfi1=0.8#unitless
I1=P1/(sqrt(3)*VL*10**3*cosfi1) #in Ampere
P2=1.5*10**6 #in watt
h2=6 #in hours
cosfi2=0.9#unitless
I2=P2/(sqrt(3)*VL*10**3*cosfi2) #in Ampere\
P3=0.5*10**6 #in watt
h3=8 #in hours
cosfi3=0.9 #unitless
I3=P3/(sqrt(3)*VL*10**3*cosfi3) #in Ampere
Etot=3*(I1**2*h1+I2**2*h2+I3**2*h3)*R*365/1000 #in kwh
Ccost_line=40000*a*4 #in Rs.
AnnualCharges=Ccost_line*i/100 #in Rs.
AnnualCost2=(E_cost/100)*Etot  #in Rs.
a=sqrt(AnnualCost2/AnnualCharges*a**2)  # cm2
print "Most economical cross section area a = %0.2f cm2" %a

Most economical cross section area a = 0.37 cm2


## Exa 2.5 : page 37¶

In [249]:
from sympy import symbols
from __future__ import division
from math import sqrt
a, l=symbols('a l')
#Given Data :
P=5*10**6 #in watt
VL=33*10**3 #in volt
cosfi=0.8#unitless
LineCost=31250*a+4000 #in Rs/km
rho=10**-6 #in ohm-cm
i=8 #in %
E_cost=4 #in paisa/kwh
IL=P/(sqrt(3)*VL*cosfi) #in Ampere
Line_length=l*10**5 #in cm
R=rho*l*10**5/a #in ohm
E_lost=3*IL**2*R*365/1000 #in kwh
Ccost_line=31250*a*l #in Rs.
AnnualCharges=Ccost_line*i/100 #in Rs.
a=sqrt(E_lost/AnnualCharges*a**2)
print "Most economical cross section area a = %0.2f cm2" %a

Most economical cross section area a = 0.72 cm2


## Exa 2.6 : page 38¶

In [250]:
#Given Data :
P=50*10**6 #in watt
VL=220*10**3 #in volt
cosfi=0.8#unitless
Eta=90 #in %
l=200*10**3 #in meter
rho=1.75*10**-8 #in ohm-cm
W=P*(100-Eta)/100 #in Wats(Line losses)
#Part (i) : 3 phase 3 wire with Cu condutor
gravity=8.9 #specific gravity
IL=P/(sqrt(3)*VL*cosfi) #in Ampere
a=3*IL**2*rho*l/W #in m**2
Vol3=3*a*l #volume of 3 lines(in m**3)
CuWeight=Vol3*gravity #in Tones
print "Weight of copper = %0.2f Tones" %CuWeight
#Part (ii) : When Al conductor is used.
gravity=2.7 #specific gravity
rho=3*10**-8 #in ohm-meter
a=3*IL**2*rho*(l/W) #in m**2
Vol=3*a*l #volume of 3 lines(in m**3)
AlWeight=Vol*gravity #in Tones
print "Weight of Alluminium = %0.3f Tones" %AlWeight

Weight of copper = 301.69 Tones
Weight of Alluminium = 156.896 Tones


## Exa 2.7 : page 39¶

In [251]:
from sympy import symbols
from __future__ import division
from math import sqrt
P2, V, cosfi, R = symbols('P2 V cosfi R')
#Given Data :
Vcon=V #in volt
pf=cosfi #unitless
Rcon=R #in ohm
P1=15*10**6 #in watt
I1=P1/(V*cosfi) #in Ampere
W1=2*I1**2*Rcon #in Wats(Line losses)
Lloss_percent=W1/P1*100 #in %         eqn(1)

I2=P2/(V*cosfi*sqrt(3)) #in Ampere
W2=3*I2**2*Rcon #in Wats(Line losses)
Lloss_percent=W2/P2*100 #in %         eqn(2)
# Equating eqn(1) & eqn(2)
P2=W1/P1*100/W2/100*P2**2   # W
P2/=10**6  # MW
print "3 phase load = %0.f MW " %P2

3 phase load = 30 MW


## Exa 2.8 : page 40¶

In [252]:
#Given Data :
from sympy import symbols
P, V, rho, l, a = symbols('P V rho l a')
I=P/V #in Ampere
R=rho*l/a
a=2*I**2*2*rho*l/W
K=a*V**2
#(i) When V=220 volts
V1=220 #in volts
vol1=K/V1**2
#(ii) When V=500 volts
V2=500 #in volts\
vol2=K/V2**2
saving=(vol1-vol2)/vol1*100 # in %
print "Saving in copper =",round(saving,2),"%"
# Answer not accurate in the textbook.

Saving in copper = 80.64 %


## Exa 2.9 : page 41¶

In [253]:
#Given Data :
P=30*10**6 #in watts
V=220*10**3 #in Volt
l=250*10**3 #in meter
Eta=85 #in %
rho=3*10**-8 #in ohm-meter
cosfi=0.8 #power factor
W=P*(100-Eta)/100 #in watts
I=P/(sqrt(3)*V*cosfi) #in Ampere
a=3*I**2*rho*l/W #in m**2
Volume=3*a*l #in m**3
print "Volume of the conductor material = %0.1f m3" %Volume

Volume of the conductor material = 36.3 m3


## Exa 2.10 : page 42¶

In [254]:
#Given Data :
P=20*10**6 #in VA
cosfi=0.75 #power factor
P=20*10**6*cosfi #in watts
V=33*10**3 #in Volt
l=20*10**3 #in meter
Eta=85 #in %
rho=3*10**-8 #in ohm-meter
W=P*(100-Eta)/100 #in watts
#For single phase system :
I=P/(V*cosfi) #in Ampere
a1=2*I**2*rho*l/W #in m**2
V1=2*a1*l #in m**3
print "For single phase system :\nVolume of the conductor material = %0.3f m3" %(V1)
#For 3 phase 3 wire system :
I=P/(sqrt(3)*V*cosfi) #in Ampere
a2=3*I**2*rho*l/W #in m**2
V2=3*a2*l #in m**3
print "For three phase 3-wire system :\nVolume of the conductor material = %0.2f m3" %V2

For single phase system :
Volume of the conductor material = 7.836 m3
For three phase 3-wire system :
Volume of the conductor material = 5.88 m3


## Exa 2.11 : page 43¶

In [255]:
from sympy import symbols
from __future__ import division
from math import sqrt
a=symbols('a')#Given Data :
l=1*10**3 #in meter
rho=1.85*10**-6 #in ohm-cm
R=rho*l*100/a #in ohm
IL=300 #in Ampere
Rate=10 #in Rs/kwh
Eloss=2*IL**2*R*365*24/1000 #in kwh
CableCost=100*a #in Rs/meter : a=cross sectional area(in cm**2)
i=10 #in %
AnnualCost=Eloss/Rate #in Rs
Ccost=100*a*l #in Rs
Ccost=Ccost*i/100  # with effet of depreciation
#a=sqrt(29170.8/10000)
a=sqrt(AnnualCost/Ccost*a**2)  # cm2
print "Most economical cross section = %0.3f cm2" %a

Most economical cross section = 1.708 cm2


## Exa 2.12 : page 44¶

In [256]:
from sympy import symbols
pf, Vcon, Rcon = symbols('cosfi V R')
#Given Data :
#Part (i) : single phase system
P1=5*10**6 #in watt
I1=P1/(Vcon*pf) #in Ampere
W1=2*I1**2*Rcon #in Watts(Line losses)
Lloss_percent=W1/P1*100 #in %         eqn(1)
#Part (ii) : 3 phase 3 wire system
P2 = symbols('P2') # MW
I2=P2/(Vcon*pf*sqrt(3)) #in Ampere
W2=3*I2**2*Rcon #in Watts(Line losses)
Lloss_percent=W2/P2*100 #in %         eqn(2)
# Equating two eqn
P2=(W1/P1*100)/(W2*100)*P2**2
P2/=10**6   # MW
print "3 phase load = %0.f MW " %P2

3 phase load = 10 MW


## Exa 2.13 : page 45¶

In [257]:
import math
#Given Data :
from sympy import symbols
P, V, rho, l, a = symbols('P V rho l a')
I=P/V #in Ampere
R=rho*l/a
a=2*I**2*2*rho*l/W
K=a*V**2
#(i) When V=200 volts
V1=200 #in volts
vol1=K/V1**2
#(ii) When V=600 volts
V2=600 #in volts\
vol2=K/V2**2
saving=(vol1-vol2)/vol1*100 # in %
print "Saving in copper =",math.floor(saving),"%"

Saving in copper = 88.0 %