# Chapter 3 - TRANSMISSION LINES¶

## Example E1 - Pg 66¶

In [1]:
#calculate Weight of copper required,Alluminium required
#Given data :
import math
P=30.*10.**6.##W
pf=0.8##lagging power factor
VL=132.*1000.##V
l=120.*1000.##m
Eta=90./100.##Efficiency
rho_Cu=1.78*10.**-8.##ohm-m
D_Cu=8.9*10.**3.##kg/m**3
rho_Al=2.6*10.**-8.##ohm-m
D_Al=2.*10.**3.##kg/m**3
IL=P/(math.sqrt(3.)*VL*pf)##A
#W=3.*IL**2.*rho*l/a=(1-Eta)*P
a_Cu=(3.*IL**2.*rho_Cu*l)/(1.-Eta)/P##m**2
V_Cu=3.*a_Cu*l##m**3
Wt_Cu=V_Cu*D_Cu##kg
print '%s %.2f' %("Weight of copper required(kg)",Wt_Cu)#
a_Al=(3.*IL**2.*rho_Al*l)/(1.-Eta)/P##m**2
V_Al=3.*a_Al*l## m**3
Wt_Al=V_Al*D_Al##kg
print '%s %.2f' %("Weight of Alluminium required(kg)",Wt_Al)#
#Answer in the textbook is not accurate.

Weight of copper required(kg) 184114.15
Weight of Alluminium required(kg) 60433.88


## Example E2 - Pg 70¶

In [2]:
#calculate Most economical cross sectional area
#Given data :
import math
a=100.#
cost=90.*a+20.##Rs./m
i=10.##%(interest and depreciation)
l=2.##km
cost_E=4.##paise/unit
Im=250.##A
a=1.##cm**2
rho_c=0.173##ohm/km/cm**2
l2=1.*1000.##km
R=rho_c*l/a##ohm
W=2.*Im**2.*R##W
Eloss=W/1000.*365.*24./2.##per annum(kWh)
P3BYa=cost_E/100.*Eloss##Rs
Cc=90.*a*l*1000.##Rs(capital cost of feeder cable)
P2a=Cc*i/100.##Rs
#P2a=P3BYa##For most economical cross section
a=math.sqrt(P3BYa*a/(P2a/a))##cm**2
print '%s %.3f' %("Most economical cross sectional area in cm**2 : ",a)#

Most economical cross sectional area in cm**2 :  0.649


## Example E3 - Pg 71¶

In [3]:
#calculate Best current density
#Given data :
import math
t=2600.##hour
Con_Cost=3.##Rs/kg(conductor cost)
R=1.78*10.**-8.##ohm-m
D=6200.##kg/m**3
E_Cost=10./100.##Rs/unit(energy cost)
i=12.##%(interest and depreciation)
a=100.##mm**2    ##cross sectional area
W=a*1000.*D/1000./1000.##kg/km(Weight of conductor of 1km length)
cost=Con_Cost*W##Rs./km(cost of conductor of 1km length)
In_Dep=cost*i/100.##Rs(Annual interest and depreciation per conductor per km)
In_DepBYa=In_Dep/a#
I=100.##A
E_lost_aBY_Isqr=R*1000./10.**-6.*t/1000.##Energy lost/annum/km/conductor
E_lost_cost_aBY_Isqr=E_Cost*E_lost_aBY_Isqr##Rs/annum
#In_Dep=E_lost_cost##For most economical cross section
IBYa=math.sqrt((In_DepBYa))/(E_lost_cost_aBY_Isqr)##cm**2
print '%s %.2f' %("Best current density in A/mm**2 : ",IBYa)#
#Answer in the textbook is not accurate.

Best current density in A/mm**2 :  0.32


## Example E4 - Pg 71¶

In [4]:
#calculate Diameter of conductor,Most economical current density
#Given data :
import math
V=11.##kV
P=1500.##kW
pf=0.8##lagging power factor
t=300.*8.##hours
a=100.##cross section area
Cc=8000.+20000.*a#Rs/km
R=0.173/a##ohm/km
E_lost_cost=2./100.##Rs/unit
i=12.##%(interest and depreciation)
Cc_var=20000.*a#Rs/km(variable cost)
P2a=Cc_var*i/100.##Rs/km
P2=P2a/a#
I=P/math.sqrt(3.)/V/pf##A
W=3.*I**2.*R##W
E_loss=W/1000.*t##kWh
P3BYa=E_lost_cost*E_loss##Rs
#P2a=P3BYa##For most economical cross section
a=math.sqrt((P3BYa))/(P2)##cm**2
d=math.sqrt(4.*a/math.pi)##cm
dela=I/a;##A/cm**2
print '%s %.2f' %("Diameter of conductor in cm : ",d)#
print '%s %.2f' %("Most economical current density in A/cm**2 : ",dela)#

Diameter of conductor in cm :  0.03
Most economical current density in A/cm**2 :  152057.18


## Example E5 - Pg 72¶

In [5]:
#Exa 3.5
#Given data :
import math
a=100.##cross section area
I=100.##Current
Cc=500.+2000.*a#Rs/km
i=12.##%(interest and depreciation)
E_lost_cost=5./100.##Rs/kWh
rho=1.78*10.**-8.##ohm-cm
Cc_var=2000.*a#Rs/km(variable cost)
P2a=Cc_var*i/100.##Rs/km
P2=P2a/a#
R_into_a=rho*1000./(10.**-4.)##ohm
W_into_a=I**2*R_into_a##W
P3BYIsqr=E_lost_cost*E_loss_into_a/I**2##Rs
#P2a=P3BYa##For most economical cross section
IBYa=math.sqrt((P2))/(P3BYIsqr)##cm**2
print '%s %.2f' %("Most economical current density in A/cm**2 : ",IBYa)#

Most economical current density in A/cm**2 :  1655.89


## Example E6 - Pg 73¶

In [6]:
#Exa 3.6
#Given data :
import math
A=100.##cross section area
I=100.##Current
Cc=500.+2000.*A#Rs/km
i=12.##%(depreciation)
E_lost_cost=0.05##Rs/kWh
R=0.17/A##ohm/km
Cc_var=2000.*A#Rs/km(variable cost)
P2A=Cc_var*i/100.##Rs/km
P2=P2A/A#
R_into_A=R*A##ohm
W_into_A_BY_Isqr=R_into_A##W
P3BYIsqr=E_lost_cost*E_loss_into_A_BY_Isqr##Rs
#P2a=P3BYa##For most economical cross section
IBYa=math.sqrt((P2))/(P3BYIsqr)##cm**2
print '%s %.2f' %("Most economical current density in A/cm**2 : ",IBYa)#
#Answer in the textbook is wrong.

Most economical current density in A/cm**2 :  1733.81


## Example E7 - Pg 73¶

In [7]:
#calculate Most economical cross sectional area
#Given data :
import math
P1=1000.##kW
pf1=0.8##
t1=10.##hours
P2=500.##kW
pf2=0.9##
t2=8.##hours
P3=100.##kW
pf3=1.##
t3=6.##hours
a=100.##cross section area
I=100.##Current
L=100.##length in km
CcBYL=(8000.*a+1500.)#Rs/km(variable cost)
i=10.##%(depreciation)
E_lost_cost=80./100.##Rs/kWh
rho=1.72*10.**-6.##ohm-cm
Cc_varBYL=8000.*a*i/100.#Rs/km(variable cost)
I1=P1*1000./math.sqrt(3.)/10000./pf1##A
I2=P2*1000./math.sqrt(3.)/10000./pf2##A
I3=P3*1000./math.sqrt(3.)/10000./pf3##A
R_into_a_BY_L=rho*1000.*100.##ohm
W_into_A_BY_Isqr=R_into_a_BY_L##W
E_loss_into_A_BY_L=3*R_into_a_BY_L*(I1**2*t1+I2**2*t2+I3**2*t3)*365./1000.##kWh
E_loss_cost_into_A_BY_L=E_loss_into_A_BY_L*E_lost_cost##Rs
#Cc_var=E_loss_cost##For most economical cross section
a=math.sqrt((E_loss_cost_into_A_BY_L))/(Cc_varBYL/a)##cm**2
print '%s %.2f' %("Most economical cross sectional area in cm**2 : ",a)#

Most economical cross sectional area in cm**2 :  0.12