# CHAPTER 2.17: ELECTRIC POWER SUPPLY SYSTEMS¶

## Example 2.17.1, Page number 422-423¶

In :
#Variable declaration
no_phase = 3.0          #Number of phases in ac transmission system
V = 380.0*10**3         #Voltage b/w lines(V)
PF = 0.9                #Power factor
l = 150.0               #Line length(km)
n = 0.92                #Efficiency
r = 0.045               #Resistance(ohm/km/sq.cm)
w_cu_1 = 0.01           #Weight of 1 cm^3 copper(kg)

#Calculation
#Case(i)
P_loss = (1-n)*load                  #Power loss in the line(MW)
loss_cu = P_loss/no_phase*10**6      #I^2*R loss per conductor(W)
R = loss_cu/I_L**2                   #Resistance per conductor(ohm)
R_km = R/l                           #Resistance per conductor per km(ohm)
area = r/R_km                        #Conductor area(Sq.cm)
volume = area*100.0                  #Volume of copper per km run(cm^3)
W_cu_km = volume*w_cu_1              #Weight of copper per km run(kg)
W_cu = no_phase*l*1000*W_cu_km       #Weight of copper for 3 conductors of 150 km(kg)
#Case(ii)
W_cu_dc = 1.0/2*PF**2*W_cu           #Weight of copper conductor in dc(kg)

#Result
print('Weight of copper required for a three-phase transmission system = %.f kg' %W_cu)
print('Weight of copper required for the d-c transmission system = %.f kg' %W_cu_dc)
print('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision')

Weight of copper required for a three-phase transmission system = 32462 kg
Weight of copper required for the d-c transmission system = 13147 kg

NOTE: Changes in the obtained answer from that of textbook is due to more precision


## Example 2.17.2, Page number 423¶

In :
#Variable declaration
P_1 = 1.0    #Assume P1 to be 1

#Calculation
P_2 = (3.0*2)**0.5            #3-phase power transmitted in terms of P_1
inc_per = (P_2-P_1)/P_1*100   #Increase in power transmitted(%)

#Result
print('Percentage increase in power transmitted = %.f percent' %inc_per)

Percentage increase in power transmitted = 145 percent


## Example 2.17.3, Page number 424¶

In :
#Variable declaration
PF = 0.95    #Lagging power factor

#Calculation
P_1 = 1.0                                     #Power in terms of V*I_1
P_2 = 2.0*PF**2                               #Power in terms of V*I_1
P_additional_percentage = (P_2-P_1)/P_1*100   #Percentage additional power transmitted in a 3-phase 3-wire system

#Result
print('Percentage additional power transmitted in a 3-phase 3-wire system = %.f percent' %P_additional_percentage)

Percentage additional power transmitted in a 3-phase 3-wire system = 80 percent


## Example 2.17.4, Page number 424-425¶

In :
#Variable declaration
n = 3.0       #3-phase 4 wire ac system

#Calculation
a2_a1 = 1.0/6            #Ratio of cross-sectional area of 2 wire dc to 3-phase 4-wire system
ratio_cu = 3.5/2*a2_a1   #Copper for 3 phase 4 wire system to copper for 2 wire dc system

#Result
print('Copper for 3-phase 4-wire system/Copper for 2-wire dc system = %.3f : 1' %ratio_cu)

Copper for 3-phase 4-wire system/Copper for 2-wire dc system = 0.292 : 1


## Example 2.17.5, Page number 425¶

In :
#Variable declaration
L = 60.0            #Line length(km)
PF = 0.8            #Lagging power factor
V = 33.0*10**3      #Voltage(V)
n = 0.85            #Transmission efficiency
rho = 1.73*10**-8   #Specific resistance of copper(ohm-mt)
density = 8900.0    #Density(kg/mt^3)

#Calculation
I = P*10**6/(3**0.5*V*PF)               #Line current(A)
line_loss = (1-n)*P*1000/n              #Line loss(kW)
line_loss_phase = line_loss/3.0         #Line loss/phase(kW)
R = line_loss_phase*1000/I**2           #Resistance/phase(ohm)
a = rho*L*1000/R                        #Area of cross section of conductor(m^2)
volume = 3.0*a*L*1000                   #Volume of copper(m^3)
W_cu = volume*density                   #Weight of copper in 3-phase system(kg)
I_1 = P*10**6/V                         #Current in single phase system(A)
R_1 = line_loss*1000/(2*I_1**2)         #Resistance in single phase system(ohm)
a_1 = rho*L*1000/R_1                    #Area of cross section of conductor in single phase system(m^2)
volume_1 = 2.0*a_1*L*1000               #Volume of copper(m^3)
W_cu_1 = volume_1*density               #Weight of copper in 1-phase system(kg)
reduction_cu = (W_cu-W_cu_1)/W_cu*100   #Reduction in copper(%)

#Result
print('Weight of copper required for 3-phase 2-wire system = %.2e kg' %W_cu)
print('Reduction of weight of copper possible = %.1f percent' %reduction_cu)
print('\nNOTE: ERROR: Calculation mistakes in the textbook solution')

Weight of copper required for 3-phase 2-wire system = 6.76e+04 kg
Reduction of weight of copper possible = 14.7 percent

NOTE: ERROR: Calculation mistakes in the textbook solution


## Example 2.17.6, Page number 427-428¶

In :
from sympy import Symbol

#Variable declaration
L = 250.0                 #Cable length(m)
V = 400.0                 #Voltage(V)
PF = 0.8                  #Lagging power factor
time = 4000.0             #Time of operation(hours/annum)
a = Symbol('a')           #Area of each conductor(Sq.cm)
cost_instal = 15.0*a+25   #Cost of cable including installation(Rs/m)
interest_per = 0.1        #Interest & depreciation
cost_waste_per = 0.1      #Cost of energy wasted(Rs/unit)
r = 0.173                 #Resistance per km of 1 cm^2(ohm)

#Calculation
I = P/(3**0.5*V*PF)                                     #Line current(A)
energy_waste = 3.0*I**2*r/a*L*10**-3*time*10**-3        #Energy wasted per annum(kWh)
cost_energy_waste = cost_waste_per*energy_waste         #Annual cost of energy wasted as losses(Rs)
capitaL_cost_cable = cost_instal*L                      #Capital cost of cable(Rs)
annual_cost_cable = capitaL_cost_cable*cost_waste_per   #Annual cost on cable(Rs)
area = (1081.25/375)**0.5                               #Area = a(Sq.cm)

#Result
print('Economical cross-section of a 3-core distributor cable, a = %.1f cm^2' %area)

Economical cross-section of a 3-core distributor cable, a = 1.7 cm^2


## Example 2.17.7, Page number 428¶

In :
from sympy import Symbol

#Variable declaration
V = 110.0*10**3               #Voltage(V)
t_1 = 6.0                     #Time(hours)
t_2 = 6.0                     #Time(hours)
t_3 = 12.0                    #Time(hours)
PF = 0.8                      #Lagging power factor
a = Symbol('a')               #Cross-section of each conductor(Sq.cm)
cost_line = 12000.0+8000*a    #Cost of line including erection(Rs/km)
R = 0.19/a                    #Resistance per km of each conductor(ohm)
cost_energy = 8.0/100         #Energy cost(Rs/unit)
interest_per = 0.1            #Interest & depreciation. Assumption

#Calculation
annual_charge = interest_per*cost_line     #Total annual charge(Rs)
I_1 = l_1/(3**0.5*V*PF)                    #Line current for load 1(A)
I_2 = l_2/(3**0.5*V*PF)                    #Line current for load 2(A)
I_3 = l_3/(3**0.5*V*PF)                    #Line current for load 3(A)
I_2_t = I_1**2*t_1+I_2**2*t_2+I_3**2*t_3   #I^2*t
annual_energy = 3.0*R*365/1000*I_2_t       #Annual energy consumption on account of losses(kWh)
cost_waste = annual_energy*cost_energy     #Cost of energy wasted per annum(Rs)
area = (2888.62809917355/800.0)**0.5       #Economical cross-section = a(Sq.cm)

#Result
print('Most economical cross-section, a = %.2f cm^2' %area)

Most economical cross-section, a = 1.90 cm^2


## Example 2.17.8, Page number 428-429¶

In :
#Variable declaration
cost_km_cu = 2800.0         #Cost per km for each copper conductor of sq.cm(Rs)
LF_loss = 65.0/100          #Load factor of losses
interest_per = 10.0/100     #Rate of interest and depreciation
cost_energy = 5.0/100       #Cost of energy(Rs/kWh)
rho = 1.78*10**-8           #Resistivity(ohm-m)

#Calculation
P_2 = cost_km_cu*interest_per                   #Cost in terms of L(Rs)
time_year = 365.0*24                            #Total hours in a year
P_3 = cost_energy*rho*10**4*time_year*LF_loss   #Cost in terms of I^2 & L(Rs)
delta = (P_2/P_3)**0.5                          #Economical current density for the transmission line(A/sq.cm)

#Result
print('Most economical current density for the transmission line, δ = %.f A/sq.cm' %delta)

Most economical current density for the transmission line, δ = 74 A/sq.cm


## Example 2.17.9, Page number 429¶

In :
#Variable declaration
MD = 1000.0                 #Maximum demand(kW)
energy_cons = 5.0*10**6     #Annual energy consumption(kWh)
PF = 0.85                   #Power factor
capital_cost = 80000.0      #Capital cost of cable(Rs/km)
cost_energy = 5.0/100       #Energy cost(Rs/kWh)
interest_per = 10.0/100     #Rate of interest and depreciation
r_specific = 1.72*10**-6    #Specific resistance of copper(ohm/cubic.cm)
V = 11.0                    #Voltage(kV)

#Calculation
I = MD/(3**0.5*V*PF)                                             #Line current corresponding to maximum demand(A)
hours_year = 365.0*24                                            #Total hours in a year

Most economical cross-section of the conductor, a = 2.04 cm^2