# CHAPTER 7: SUBSTATION AND DISTRIBUTION SYSTEM¶

## Example 7.1, Page number 238-239¶

In :
V = 400.0       #Voltage supplied(V)
f = 50.0        #Frequency(Hz)
P_1 = 75.0      #Power of induction motor at middle of distributor(kVA)
pf_1 = 0.8      #Power factor of induction motor at middle of distributor
P_2 = 50.0      #Power of induction motor at far end(kVA)
pf_2 = 0.85     #Power factor of induction motor at far end
demand_f = 1.0  #Demand factor
diver_f = 1.2   #Diversity factor
L = 150.0       #Length of line(m)

import math
import cmath
theta_1 = math.acos(pf_1)                                       #Power factor angle for 75 kVA(radians)
theta_2 = math.acos(pf_2)                                       #Power factor angle for 50 kVA(radians)
load = P_1*cmath.exp(1j*theta_1)+P_2*cmath.exp(1j*theta_2)      #Total connected load(kVA)
pf_r = math.cos(cmath.phase(load))                              #Resultant power factor
I_max = abs(load)*1000/(3**0.5*V*diver_f)                       #Maximum distributor current per phase(A)
L_1 = L/2
V_per = 0.06*V/3**0.5                                           #Permissible voltage drop(V)

R_f = 0.734*10**-3                                              #Resistance(ohm/m)
X_f = 0.336*10**-3                                              #Reactance(ohm/m)
I_2f = P_2*10**3/(3**0.5*V)
I_1f = P_1*10**3/(3**0.5*V)
V_f = I_1f*L_1*(R_f*pf_1+X_f*math.sin(theta_1))+I_2f*L*(R_f*pf_2+X_f*math.sin(theta_2))
d_f = 9.0                                                       #Overall conductor diameter(mm)
area_f = math.pi*d_f**2/4                                       #Area of ferret conductor(mm^2)

R_R = 0.587*10**-3                                              #Resistance(ohm/m)
X_R = 0.333*10**-3                                              #Reactance(ohm/m)
I_2R = P_2*10**3/(3**0.5*V)
I_1R = P_1*10**3/(3**0.5*V)
V_R = I_1R*L_1*(R_R*pf_1+X_R*math.sin(theta_1))+I_2R*L*(R_R*pf_2+X_R*math.sin(theta_2))
d_R = 10.0                                                      #Overall conductor diameter(mm)
area_R = math.pi*d_R**2/4                                       #Area of rabbit conductor(mm^2)

if(V_f > V_per):
print('Overall cross-sectional area of the 7/3.35 mm Rabbit ACSR conductors having overall conductor diameter of 10.0 mm = %.2f mm^2' %area_R)
else:
print('Overall cross-sectional area of the 7/3.00 mm Ferret ACSR conductors having overall conductor diameter of 9.0 mm = %.2f mm^2' %area_f)

Overall cross-sectional area of the 7/3.35 mm Rabbit ACSR conductors having overall conductor diameter of 10.0 mm = 78.54 mm^2


## Example 7.2, Page number 240¶

In :
V = 400.0       #Voltage supplied(V)
i = 0.5         #Current per meter(A)
demand_f = 1.0  #Demand factor
diver_f = 1.0   #Diversity factor
L = 275.0       #Length of line(m)
pf = 0.9        #Power factor lagging

import math
I = i*L                                                 #Current in distributor/phase(A)
theta = math.acos(pf)                                   #Power factor angle
V_per = 0.06*V/3**0.5                                   #Permissible voltage drop(V)

r_w = 0.985                                             #Resistance(ohm/km)
x_w = 0.341                                             #Reactance(ohm/km)
V_w = 0.5*i*(r_w*pf+x_w*math.sin(theta))*L**2*10**-3    #Voltage drop for Weasel(V)
d_w = 7.77                                              #Diameter of weasel conductor(mm)
area_w = math.pi*d_w**2/4                               #Area of weasel conductor(mm^2)

r_f = 0.734                                             #Resistance(ohm/km)
x_f = 0.336                                             #Reactance(ohm/km)
V_f = 0.5*i*(r_f*pf+x_f*math.sin(theta))*L**2*10**-3    #Voltage drop for Ferret(V)
d_f = 9.00                                              #Diameter of Ferret conductor(mm)
area_f = math.pi*d_f**2/4                               #Area of Ferret conductor(mm^2)

r_r = 0.587                                             #Resistance(ohm/km)
x_r = 0.333                                             #Reactance(ohm/km)
V_r = 0.5*i*(r_r*pf+x_r*math.sin(theta))*L**2*10**-3    #Voltage drop for Rabbit(V)
d_r = 10.0                                              #Diameter of Rabbit conductor(mm)
area_r = math.pi*d_r**2/4                               #Area of Rabbit conductor(mm^2)

if(V_w < V_per):
print('Overall cross-sectional area of the 7/2.59 mm Weasel ACSR conductors having overall conductor diameter of 7.77 mm = %.2f mm^2' %area_w)
elif(V_f < V_per):
print('Overall cross-sectional area of the 7/3.00 mm Ferret ACSR conductors having overall conductor diameter of 9.0 mm = %.2f mm^2' %area_f)
else:
print('Overall cross-sectional area of the 7/3.35 mm Rabbit ACSR conductors having overall conductor diameter of 10.0 mm = %.2f mm^2' %area_r)

Overall cross-sectional area of the 7/3.35 mm Rabbit ACSR conductors having overall conductor diameter of 10.0 mm = 78.54 mm^2


## Example 7.3, Page number 240-241¶

In :
V = 400.0       #Voltage supplied(V)
f = 50.0        #Frequency(Hz)
L = 300.0       #Length of line(m)
I_1 = 50.0      #Current at 100 m from feeding point(A)
pf_1 = 0.8      #Power factor at 100 m from feeding point
L_1 = 100.0     #Length of line upto feeding point(m)
I_2 = 25.0      #Current at 100 m from feeding point(A)
pf_2 = 0.78     #Power factor at 100 m from feeding point
L_2 = 200.0     #Length of line from feeding point to far end(m)
i = 0.2         #Distributed load current(A/metre)
v_drop = 15.0   #Permissible voltage drop

import math
import cmath
theta_1 = math.acos(pf_1)               #Power factor angle for 50 A(radians)
theta_2 = math.acos(pf_2)               #Power factor angle for 25 A(radians)

r_f = 0.734*10**-3                      #Resistance(ohm/m)
x_f = 0.336*10**-3                      #Reactance(ohm/m)
V_con_f = I_1*L_1*(r_f*pf_1+x_f*math.sin(theta_1))+I_2*L*(r_f*pf_2+x_f*math.sin(theta_2)) #Voltage drop at B due to concentrated loading(V)
V_dis_f = 0.5*i*r_f*(L_1+L_2)**2        #Voltage drop at B due to distributed loading(V)
V_f = V_con_f+V_dis_f                   #Total voltage drop(V)

r_r = 0.587*10**-3                      #Resistance(ohm/m)
x_r = 0.333*10**-3                      #Reactance(ohm/m)
V_con_r = I_1*L_1*(r_r*pf_1+x_r*math.sin(theta_1))+I_2*L*(r_r*pf_2+x_r*math.sin(theta_2)) #Voltage drop at B due to concentrated loading(V)
V_dis_r = 0.5*i*r_r*(L_1+L_2)**2        #Voltage drop at B due to distributed loading(V)
V_r = V_con_r+V_dis_r                   #Total voltage drop(V)

if(V_f < v_drop):
print('Ferret ACSR conductors of size 7/3.00 mm having an overall conductor diameter of 9.0 mm is to be used')
print('Total voltage drop = %.2f V, which is within limit' %V_f)
else:
print('Rabbit ACSR conductors of size 7/3.35 mm having an overall conductor diameter of 10.0 mm is to be used')
print('Total voltage drop = %.2f V, which is within limit' %V_r)
print('\nNOTE : ERROR : In distributed load : current is 0.2 A/meter and not 0.25 A/meter as given in problem statement')

Rabbit ACSR conductors of size 7/3.35 mm having an overall conductor diameter of 10.0 mm is to be used
Total voltage drop = 13.63 V, which is within limit

NOTE : ERROR : In distributed load : current is 0.2 A/meter and not 0.25 A/meter as given in problem statement


## Example 7.4, Page number 244-245¶

In :
P = 5.0         #Power of substation(MVA)
V_hv = 33.0     #High voltage(kV)
V_lv = 11.0     #Low voltage(kV)
f = 50.0        #Frequency(Hz)
P_1 = 0.5       #Minimum load(MW)
pf_1 = 0.85     #Lagging power factor of minimum load
P_2 = 2.8       #Maximum load(MW)
pf_2 = 0.78     #Lagging power factor of maximum load
pf_i = 0.9      #Lagging power factor of incoming current

import math
theta_1 = math.acos(pf_1)
theta_2 = math.acos(pf_2)
theta_i = math.acos(pf_i)

load_react = P_1*math.tan(theta_1)*1000                      #Load reactive power(kVAR)
line_react = P_1*math.tan(theta_i)*1000                      #Reactive power supplied by line(kVAR)
rating_fix = load_react - line_react                         #kVAR rating of fixed capacitor bank(kVAR)

bank_react = P_2*(math.tan(theta_2)-math.tan(theta_i))*1000  #Reactive power to be supplied by capacitor banks(kVAR)
rating_swi = bank_react - rating_fix                         #Reactive power rating of switched unit(kVAR)

C_fix = rating_fix*10**-3/(3**0.5*V_lv**2*2*math.pi*f)       #Capacitance for fixed bank
C_swi = rating_swi*10**-3/(3**0.5*V_lv**2*2*math.pi*f)       #Capacitance for switched bank

print('kVAR rating of fixed capacitors = %.1f kVAR' %rating_fix)
print('kVAR rating of switched capacitors = %.1f kVAR' %rating_swi)
print('Capacitance of fixed bank , C = %.2e F/phase' %C_fix)
print('Capacitance of switched bank , C = %.2e F/phase' %C_swi)

kVAR rating of fixed capacitors = 67.7 kVAR
kVAR rating of switched capacitors = 822.6 kVAR
Capacitance of fixed bank , C = 1.03e-06 F/phase
Capacitance of switched bank , C = 1.25e-05 F/phase


## Example 7.5, Page number 245¶

In :
V = 400.0       #Voltage of induction motor(V)
f = 50.0        #Frequency(Hz)
I = 40.0        #Line current(A)
pf_1 = 0.78     #Lagging power factor of motor
pf_2 = 0.95     #Raised lagging power factor

import math
import cmath
theta_1 = math.acos(pf_1)                                           #Motor power factor angle(radians)
P_act_m = 3**0.5*V*I*pf_1*10**-3                                    #Active power demand of motor(kW)
P_rea_m = P_act_m*math.tan(theta_1)                                 #Reactive power demand of motor(kVAR)
theta_2 = math.acos(pf_2)                                           #Improved power factor angle(radians)
P_act_l = 3**0.5*V*I*pf_1*10**-3                                    #Active power supplied by line(kW)
P_rea_l = P_act_m*math.tan(theta_2)                                 #Reactive power supplied by line to motor(kVAR)
rating = P_rea_m - P_rea_l                                          #kVAR rating of capacitor bank(kVAR per phase)
I_C = rating*1000/(3**0.5*V)                                        #Current drawn by capacitor bank(A)
I_L = I*cmath.exp(1j*-theta_1)+I_C*cmath.exp(1j*90*math.pi/180)     #Line current(A)
I_phase = I_C/3**0.5                                                #Phase current of delta connected capacitor bank(A)
C = I_phase/(V*2*math.pi*f)                                         #Per phase capacitance of bank(micro-F/phase)

print('kVAR rating of the bank = %.2f kVAR per phase' %rating)
print('Line current = %.2f∠%.2f° A' %(abs(I_L),cmath.phase(I_L)*180/math.pi))
print('Per phase capacitance of the bank , C = %.2e F/phase' %C)

kVAR rating of the bank = 10.24 kVAR per phase
Line current = 32.84∠-18.19° A
Per phase capacitance of the bank , C = 6.79e-05 F/phase


## Example 7.6, Page number 245-246¶

In :
P_1 = 250.0     #Load at unity power factor(kW)
pf_1 = 1        #Power factor
P_2 = 1500.0    #Load at 0.9 power factor(kW)
pf_2 = 0.9      #Lagging power factor
P_3 = 1000.0    #Load at 0.8 power factor(kW)
pf_3 = 0.8      #Lagging power factor
P_4 = 700.0     #Load at 0.78 power factor(kW)
pf_4 = 0.76     #Lagging power factor

import math
import cmath
theta_1 = math.acos(pf_1)
theta_2 = math.acos(pf_2)
theta_3 = math.acos(pf_3)
theta_4 = math.acos(pf_4)
kW_T = P_1+P_2+P_3+P_4                   #Total kW carried by feeder(kW)
kVAR_T = P_1*math.tan(theta_1)+P_2*math.tan(theta_2)+P_3*math.tan(theta_3)+P_4*math.tan(theta_4)
pf_feed = math.cos(math.atan(kVAR_T/kW_T))
feeder_KVA = (kW_T**2+kVAR_T**2)**0.5    #Feeder kVA
feeder_kW = feeder_KVA                   #Load at unity pf(kW)

print('Feeder power factor = %.3f lagging' %pf_feed)
print('Load at unity power factor = %.f kW' %feeder_kW)
print('\nNOTE : ERROR : The load data should be 700 kW at 0.76 pf lagging instead of 700 kW at 0.78 lagging')

Feeder power factor = 0.857 lagging
Load at unity power factor = 4026 kW

NOTE : ERROR : The load data should be 700 kW at 0.76 pf lagging instead of 700 kW at 0.78 lagging


## Example 7.8, Page number 247-249¶

In :
V = 400.0       #Voltage(V)
f = 50.0        #Frequency(Hz)
HP_1 = 75.0     #Power(H.P)
HP_2 = 25.0     #Power(H.P)
HP_3 = 10.0     #Power(H.P)
pf_1 = 0.75     #Power factor at 3/4 load
pf_2 = 0.78     #Power factor at 4/5 load
pf_3 = 0.8      #Power factor at full load
pf_4 = 0.9      #Lagging power factor improved
pf_5 = 0.74     #Power factor of 2nd motor at 2/3 of full load
pf_6 = 0.8      #Power factor of 3rd motor at full load

import math
import cmath

theta_1 = math.acos(pf_1)
theta_2 = math.acos(pf_2)
theta_3 = math.acos(pf_3)
S_1P = (0.75*HP_1*746*10**-3/pf_1)*cmath.exp(1j*theta_1)    #kVA demanded by first motor(kVA)
S_2P = (0.8*HP_2*746*10**-3/pf_2)*cmath.exp(1j*theta_2)     #kVA demanded by second motor(kVA)
S_3P = (HP_3*746*10**-3/pf_3)*cmath.exp(1j*theta_3)         #kVA demanded by third motor(kVA)
S_TP = S_1P + S_2P + S_3P                                   #Total kVA demanded by all loads(kVA)
pf_l_wc = math.cos(cmath.phase(S_TP))                       #Line power factor without capacitive correction
kW_T = S_TP.real                                            #Total kW demanded by load(kW)
kVAR_T = S_TP.imag                                          #Total lagging kVAR demanded by loads(kVAR)
theta_4 = math.acos(pf_4)
P_react = kW_T*math.tan(theta_4)                            #Reactive power supplied by line for 0.9 pf(kVAR)
power = kVAR_T - P_react                                    #Reactive power supplied by capacitor bank(kVAR)

theta_5 = math.acos(pf_5)
theta_6 = math.acos(pf_6)
S_2L = (2*HP_2*746*10**-3/(3*pf_5))*cmath.exp(1j*theta_5)   #kVA demanded by second motor(kVA)
S_3L = (HP_3*746*10**-3/pf_3)*cmath.exp(1j*theta_3)         #kVA demanded by third motor(kVA)
S_TL = S_2L + S_3L                                          #Total kVA demanded during lean period(kVA)
S_line = S_TL.real - complex(0,(power-S_TL.imag))           #kVA supplied by line(kVA)
pf_line = math.cos(cmath.phase(S_line))                     #Line power factor

print('Line power factor with capacitor bank connected during lean period = %.2f leading' %pf_line)

Line power factor with capacitor bank connected during lean period = 0.95 leading