# Chapter 14: A.C. Distribution¶

### Example 14.1, Page Number: 359¶

In :
from __future__ import division
import math

#Variable declaration:
l = 300                      #line length(m)
magI1 = 100                     #current at load 1(A)
pf1 = 0.707                  #power factor at load 1
l1 = 200                      #line length till load 1(m)
magI2 = 200                     #current drawn at load 2(A)
pf2 = 0.8                   #power factor at laod 2
l2 = 300                    #line length till load 2(m)
R = 0.2                      #total resistance of line(ohm/km)
X = 0.1                      #total reactance of the line(ohm/km)

#Calculation:
phy1 = math.acos(pf1)
phy2 = math.acos(pf2)
Zac = (R+X*1j)*200/1000       #Impedance of section AC(ohm)
Zcb = (R+X*1j)*100/1000       #Impedance of section CB(ohm)
#Taking voltage at the far end B as the reference vector, we have
I2 = magI2*(pf2-1j*math.sin(phy2))         #Load current at point B(A)
I1 = magI1*(pf1-1j*math.sin(phy1))         #Load current at point C(A)
Icb = I2                                   #A
Iac = I1+I2                              #A
Vcb = Icb*Zcb                           #V
Vac = Iac*Zac                            #V
V = Vac+Vcb                            #Voltage drop in the distributor(V)

#Result:
print "Voltage drop in the distributor is",round(abs(V),2),"V"

Voltage drop in the distributor is 17.85 V


### Example 14.2, Page Number: 359¶

In :
from __future__ import division
import math

#Variable declaration:
l = 2000                      #line length(m)
magI1 = 80                     #current at load 1(A)
pf1 = 0.9                 #power factor at load 1
l1 = 1000                      #line length till load 1(m)
magI2 = 120                     #current drawn at load 2(A)
pf2 = 0.8                   #power factor at laod 2
l2 = 2000                    #line length till load 2(m)
R = 0.05                      #total resistance of line(ohm/km)
X = 0.1                      #total reactance of the line(ohm/km)
magVb = 230                     #voltage maintained at point B(V)

#Calculation:
phy1 = math.acos(pf1)
phy2 = math.acos(pf2)
Zac = (R+X*1j)*1000/1000       #Impedance of section AC(ohm)
Zcb = (R+X*1j)*1000/1000       #Impedance of section CB(ohm)
#Taking voltage at the far end B as the reference vector, we have
I2 = magI2*(pf2-1j*math.sin(phy2))         #Load current at point B(A)
I1 = magI1*(pf1-1j*math.sin(phy1))         #Load current at point C(A)
Icb = I2                                   #A
Iac = I1+I2                              #A
Vcb = Icb*Zcb                           #V
Vac = Iac*Zac                            #V
V = Vac+Vcb+magVb*(1+0j)                            #Voltage drop in the distributor(V)
theta = math.atan(V.imag/V.real)

#Result:
print "(i) Voltage drop in the distributor is",round(abs(V),2),"V"
print "(ii)The phase difference between Va and Vb is ",round(math.degrees(theta),2),"degrees"

(i) Voltage drop in the distributor is 261.67 V
(ii)The phase difference between Va and Vb is  3.83 degrees


### Example 14.3, Page Number: 360¶

In :
from __future__ import division
import math

#Variable declaration:
magI1 = 100                     #current at load 1(A)
pf1 = 0.6                 #power factor at load 1
magI2 = 100                     #current drawn at load 2(A)
pf2 = 0.8                   #power factor at laod 2
R = 0.1                      #total resistance of line(ohm/km)
X = 0.15                      #total reactance of the line(ohm/km)
magVb = 200                     #voltage maintained at point B(V)

#Calculation:
phy1 = math.acos(pf1)
phy2 = math.acos(pf2)
Zam = (R+X*1j)              #Impedance of section AM(ohm)
Zmb = (R+X*1j)             #Impedance of section MB(ohm)
#Taking voltage at the far end B as the reference vector, we have
I2 = magI2*(pf2-1j*math.sin(phy2))         #Load current at point B(A)
Imb = I2
Vb = magVb*(1+0j)                       #V
Vmb = Imb*Zmb                           #V
Vm = Vb+Vmb                              #V
alpha = math.atan(Vm.imag/Vm.real)         #V
#The load current I1 has a lagging p.f. of 0·6 w.r.t. VM. It lags
#behind Vm by an angle phy1.
#Phase angle between I1 and Vb
phy11 = phy1-alpha
I1 = magI1*(math.cos(phy11)-math.sin(phy11)*1j)         #A
Iam = I1+I2                              #A
Vam = Iam*Zam                            #V
Va = Vm+Vam                             #V
theta = math.atan(Va.imag/Va.real)

#Result:
print "(i)Voltage at mid-point is",round(abs(Vm),1),"V"
print "(ii) Sending end voltage Va is",round(abs(Va),2),"V"
print "(iii)The phase difference between Va and Vb is ",round(math.degrees(theta),2),"degrees"

(i)Voltage at mid-point is 217.1 V
(ii) Sending end voltage Va is 252.33 V
(iii)The phase difference between Va and Vb is  3.07 degrees


### Example 14.4, Page Number: 362¶

In :
from __future__ import division
import math

#Variable declaration:
Zab = 1+1j                    #ohm
Zbc = 1+2j                    #ohm
Zac = 1+3j                    #ohm
Ib = 20                       #load current at B(A)
pfb = 0.8                      #power factor at A
Ic = 15                       #load current at C(A)
pfc = 0.6                      #power factor at B

#Calculation:
Iab = Ib*(pfb-1j*math.sin(math.acos(pfb)))        #Current in section AB(A)
Iac = Ic*(pfc-1j*math.sin(math.acos(pfc)))        #Current in section AB(A)
Vab = Iab*Zab                     #Voltage drop in section AB(V)
Vac = Iac*Zac                     #Voltage drop in section AC(V)
#point B is at higher potential than point C. The p.d. between B and C
#is Thevenin’s equivalent circuit e.m.f. Eo i.e.
Eo = Vac-Vab                     #volt
Zo = Zab+Zac                    #Thevenin’s equivalent impedance(ohm)
Ibc = Eo/(Zo+Zbc)                #A
Iab1 = Iab+Ibc                   #A
Iac1 = Iac-Ibc                   #A
Ia = Iab+Iac                     #Current fed at A(A)

#Result:
print "The total current fed at A is",Ia,"A"
print "\nCurrent in AB is",Iab1.real+round(Iab1.imag,2)*1j,"A"
print "\nCurrent in BC is",Ibc.real+round(Ibc.imag,2)*1j,"A"
print "\nCurrent in AC is",Iac1.real+round(Iac1.imag,2)*1j,"A"

The total current fed at A is (25-24j) A

Current in AB is (18.6-13.53j) A

Current in BC is (2.6-1.53j) A

Current in AC is (6.4-10.47j) A


### Example 14.5, Page Number: 363¶

In :
from __future__ import division
import math

#Variable declaration:
l = 1000                      #line length(m)
magI1 = 5                     #current at load 1(A)
pf1 = 0.8                 #power factor at load 1
l1 = 600                      #line length till load 1(m)
pf2 = 0.85                   #power factor at motor load B
Po = 10                      #power output at B(H.P)
n = 0.9                     #efficiency
l2 = 400                    #line length till load 2(m)
R = 1                      #total resistance of line(ohm/km)
X = 0.5                      #total reactance of the line(ohm/km)
magVb = 400                     #voltage maintained at point B(V)

#Calculation:
Zac = (R+X*1j)*l1/l                 #ohm
Zcb = (R+X*1j)*l2/l                 #ohm
magVbp = magVb/3**0.5               #volt per phase
Vbp = magVbp*(1+0j)                 #V
magIb = Po*746/(3**0.5*magVb*pf2*n)           #Line current at B(A)
magI2p = magIb
I2p = magI2p*(pf2-1j*math.sin(math.atan(pf2)))    #A
I1p = magI1*(pf1-1j*math.sin(math.atan(pf1)))       #A
Iac = I1p+I2p                      #Current in section AC(A)
Icb = I2p                          #Current in section CB(A)
Vcb = Icb*Zcb                      #V
Vac = Iac*Zac                      #V
Va = Vbp+Vcb+Vac                    #V

#Result:
print "Line voltage at A is",round(abs(Va)*3**0.5),"V"

Line voltage at A is 434.0 V


### Example 14.6, Page Number: 364¶

In :
from __future__ import division
import math
from sympy import *

#Variable declaration:
magVa = 11000                      #volt
magIb = 50                            #load current at B(A)
pf2 = 0.8                           #power factor(lagging)
magIc = 120                            #load current at C(A)
pf3 = 1.0                           #power factor(lagging)
magId = 70                             #load current at D(A)
pf4 = 0.866                         #power factor(lagging)
Zab = 1+0.6j                        #ohm
Zbc = 1.2+0.9j                      #ohm
Zcd = 0.8+0.5j                        #ohm
Zda = 3+2j                          #ohm

#Calculation:
#Let current in section AB be (x + j*y).
x,y = symbols('x,y')
Iab = x+1j*y
x,y = symbols('x,y')
Ibc = Iab-magIb*(pf2-math.sin(math.atan(pf2)))     #A
Icd = (x-40+1j*(y+30))-(120+0j)                   #A
Ida = ((x-160)+1j*(y+30))-(70*(0.866-0.5j))          #A
Vab = Iab*Zab                         #Drop in section AB(V)
Vbc = Ibc*Zbc                         #Drop in section BC(V)
Vcd = Icd*Zcd                          #Drop in section CD(V)
Vda = Ida*Zda                         #Drop in section DA(V)
#  Vab+Vbc+Vcd+Vda = 0
#  As the real and imaginary parts have to be separately zero,
#  6*x-4*y-1009.8 = 0
#  4*x+6*y-302.2 = 0
x1 = solve(6*x-4*y-1009.8,x)
y1 = round(solve(4*x1+6*y-302.2,y),1)
x11 = round(solve(6*x-4*y1-1009.8,x),1)
#now putting the values of x11 and y1 in above equationa,
Iab1 = x11+1j*y1                                     #A
Ibc1 = (x11-40)+1j*(y1+30)                           #A
Icd1 = (x11-40+1j*(y1+30))-(120+0j)                   #A
Ida1 = ((x11-160)+1j*(y1+30))-(70*(0.866-0.5j))          #A
magVap = round(magVa/3**0.5)                     #Voltage at supply end A(V)
Vb = magVap*(1+0j)-Iab1*Zab               #Voltage at station B(V/phase)
Vc = Vb-Ibc1*Zbc               #Voltage at station C(V/phase)
Vd = Vc-Icd1*Zcd               #Voltage at station D(V/phase)

#Result:
print "Current in section AB is",Iab1,"A"
print "Current in section BC is",Ibc1,"A"
print "Current in section CD is",Icd1,"A"
print "Current in section DA is",Ida1,"A"
print "Voltage at A is",magVap*(1+0j),"V/phase"
print "Voltage at B is",Vb,"V/phase"
print "Voltage at C is",Vc,"V/phase"
print "Voltage at D is",Vd,"V/phase"

Current in section AB is (139.8-42.8j) A
Current in section BC is (99.8-12.8j) A
Current in section CD is (-20.2-12.8j) A
Current in section DA is (-80.82+22.2j) A
Voltage at A is (6351+0j) V/phase
Voltage at B is (6185.52-41.08j) V/phase
Voltage at C is (6054.24-115.54j) V/phase
Voltage at D is (6064-95.2j) V/phase


### Example 14.7, Page Number: 368¶

In :
from __future__ import division
import math

#Variable declaration:
Pr = 10                        #load connected to line R(kW)
Py = 8                        #load connected to line Y(kW)
Pb = 5                        #load connected to line B(kW)
Vl = 400                      #line voltage(V)

#Calculation:
Vp = round(Vl/3**0.5)                    #phase voltage(V)
Ir = Pr*1000/Vp                 #A
Iy = Py*1000/Vp                  #A
Ib = Pb*1000/Vp                  #A

#Resolving the three currents along x-axis and y-axis, we have,
Ih = Iy*math.cos(math.pi/6)-Ib*math.cos(math.pi/6)         #Resultant horizontal component(A)
Iv = Ir-Iy*math.cos(math.pi/3)-Ib*math.cos(math.pi/3)       #Resultant vertical component(A)
In = (Ih**2+Iv**2)**0.5          #current in neutral wire(A)

#Result:
print "(i) Ir =",round(Ir,1),"A"
print "    Iy =",round(Iy,1),"A"
print "    Ib =",round(Ib,2),"A"
print "(ii) Current in neutral wire is",round(In,1),"A"

(i) Ir = 43.3 A
Iy = 34.6 A
Ib = 21.65 A
(ii) Current in neutral wire is 18.9 A


### Example 14.8, Page Number: 369¶

In :
from __future__ import division
import math

#Variable declaration:
Vl = 400                          #line voltage(V)
Vp = 230                          #voltage across lamp(V)
I1 = 70                           #current in load RN(A)
I2 = 84                           #current in load YN(A)
I3 = 33                           #current in load BN(A)
Im = 200                         #current taken by the motor(A)
pf = 0.2                        #power factor(lagging)

#Calculation:
Ih = I2*math.cos(math.pi/6)-I3*math.cos(math.pi/6)         #Resultant H-component(A)
Iv = I1-I3*math.cos(math.pi/3)-84*math.cos(math.pi/3)      #Resultant V-component(A)
In = math.sqrt(Ih**2+Iv**2)                          #Neutral current(A)

Ir = Im*pf                             #Active component of motor current(A)
Ix = Im*math.sin(math.acos(pf))        #Reactive component of motor current(A)
IR = ((Ir+I1)**2+Ix**2)**0.5          #A
IY = ((Ir+I2)**2+Ix**2)**0.5          #A
IB = ((Ir+I3)**2+Ix**2)**0.5          #A
P = Vp*(I1+I2+I3)*1                   #Watt       #( cos phy_L = 1)
Pm = 3**0.5*Vl*Im*pf                  #Power supplied to motor(W)

#Result:
print "Lamp load alone: neutral curent is",round(In,2),"A"
print "The current components are:"
print "Neutral current is",round(In,2),"A"
print "IR =",round(IR,1),"A;\tIY =",round(IY,0),"A;\tIB =",round(IB,2),"A"
print "Power supplied to the lamp is",P,"W"
print "Power supplied to the motor is",round(Pm),"W"

Lamp load alone: neutral curent is 45.64 A

The current components are:
Neutral current is 45.64 A
IR = 224.7 A;	IY = 232.0 A;	IB = 209.11 A
Power supplied to the lamp is 43010 W
Power supplied to the motor is 27713.0 W


### Example 14.9, Page Number: 370¶

In :
from __future__ import division
import math

#Variable declaration:
Prn = 20                          #kW
pf1 = 1                          #power factor of loaf RN
kVAyn = 28.75                   #kVA of load YN
kVAbn = 28.75                   #kVA of load BN
pf2 = 0.866                     #power factor of laod YN & BN each.(lagging)
Vl = 400                        #line voltage(V)
Vp = 230                         #phase voltage(V)

#Calculation:
phy1 = math.acos(pf1)
phy2 = math.acos(pf2)
phy3 = phy2
Ir = Prn*1000/Vp                   #A
Iy = kVAyn*1000/Vp                 #A
Ib = kVAbn*1000/Vp                 #A
Ih = Ir-Iy*math.cos(phy2)-Ib*math.cos(phy2)          #A
Iv = 0+Iy*math.sin(phy2)-Iy*math.sin(phy3)           #A
In = math.sqrt(Ih**2+Iv**2)                               #A
#When load from B to N removed.:
#When the load from B to N is removed, the various line currents are:
#Ir in phase with Vrn; Iy lagging by 30 deg.; Ib = 0.
Ir1 = Ir
Iy1 = Iy;     Ib1 = 0                    #A
Ih1 = Ir1-Iy1*math.cos(math.pi/6)             #A
Iv1 = 0-Iy1*math.sin(math.pi/6)               #A
In1 = math.sqrt(Ih1**2+Iv1**2)               #A

#Result:
print "When no changes were made, the various currents are:"
print "Ir =",round(Ir,2),"A;\tIy =",Iy,"A;\tIb =",Ib,"A;\tIn =",round(In,2),"A"
print "\nWhen load from B to N removed, the various currents are:"
print "Ir =",round(Ir1,2),"A;\tIy =",Iy1,"A;\tIb =",Ib1,"A;\tIn =",round(In1,2),"A"

When no changes were made, the various currents are:
Ir = 86.96 A;	Iy = 125.0 A;	Ib = 125.0 A;	In = 129.54 A

When load from B to N removed, the various currents are:
Ir = 86.96 A;	Iy = 125.0 A;	Ib = 0 A;	In = 66.03 A


### Example 14.10, Page Number: 371¶

In :
from __future__ import division
import cmath
import math

#Variable declaration:
Vl = 400                       #line voltage(V)
Vp = 230                         #phase voltage(V)
Ir = 30                        #load current at R-phase(A)
pf1 = 0.866                    #power factor for R-phase(lagging)
Iy = 30                        ##load current at Y-phase(A)
pf2 = 0.866                    #power factor for R-phase(lagging)
Ib = 30                         ##load current at R-phase(A)
pf3 = 1.0                    #power factor for R-phase(lagging)
R = 0.2                       #resistance of each line conductor(ohm)

#Calculation:
phy1 = math.acos(pf1)
phy2 = math.acos(pf2)
phy3 = math.acos(pf3)

VR = Vp*(1+0j)                #V
VY = Vp*(math.cos(-2*math.pi/3)+math.sin(-2*math.pi/3))        #V
VB = Vp*(math.cos(2*math.pi/3)+math.sin(-2*math.pi/3))          #V

#the line currents can be expressed as :
IR = cmath.rect(30,-math.pi/6)                    #A
IY = cmath.rect(30,-math.pi/2)                     #A
IB = cmath.rect(30,2*math.pi/3)                   #A
IN = IR+IY+IB                                #A

#Since, the area of X-section of neutral is half of any line conductor.
Rn = 2*R                                 #resistance of neutral(ohm)
#ER = VR + Drop in R phase + Drop in neutral
ER = VR+R*IR+IN*2*R                    #V

#Result:
print "The supply end voltage for R phase is",round(ER.real,3)+1j*round(ER.imag,3),"V"

The supply end voltage for R phase is (239.588-10.608j) V


### Example 14.11, Page Number: 371¶

In :
from __future__ import division

#Variable declaration:
Vl = 400                    #line voltage(V)
Vp = 230                      #phase voltage(A)
Pln = 100                   #load connected b/n LN(W)
Pyn = 150                   #load connected b/n YN(W)

#Calculation:
#before disconnecting the neutral wire,
R1 = Vp**2/Pln                        #Resistance of lamp L1(ohm)
R2 = Vp**2/Pyn                        #Resistance of lamp L2(ohm)

#When the neutral wire is disconnected,
EL = 400                              #V
I = EL/(R1+R2)                       #A
V1 = I*R1                            #Voltage across lamp L1(V)
V2 = I*R2                            #Voltage across lamp L2(V)

#Result:
print "The voltage across the lamps are:"
print "Lamp 1, Voltage =",V1,"V ;\tLamp 2, voltage =",V2,"V"

The voltage across the lamps are:
Lamp 1, Voltage = 240.0 V ;	Lamp 2, voltage = 160.0 V