# Chapter3, Distribution¶

## Ex1 : Page 103¶

In [1]:
from __future__ import division
from numpy import sqrt
#To Determine the most economical current density

OFC=15  #Cost of single phase overhead feeder per km per unit area
AIDC= 10*OFC/100  #Annual Interest and depriciation charges
R=1/58  #Resistance of 1m length and 1 sq.mm
CLPUL= 2*R*1000  # Copper Loss per unit length per unit area per unit square current
ACL= CLPUL*365*24*0.5  # Annual Copper Loss
GC= 5/100  # Genereating Cost per unit.
ACEL= ACL*GC/1000  # Annual cost of energy loss
CD= sqrt(AIDC/ACEL)  #Current Density
print 'The Most Economical Current Density for this Case is %0.2f A/sq.mm'%CD

The Most Economical Current Density for this Case is 0.45 A/sq.mm


## Ex2 : Page 104¶

In [2]:
#To determine the most economical cross sectional area

OFC=180  #Cost of single phase overhead feeder per km per unit area + 1200
AIDC= 10*OFC/100  #Annual Interest and depriciation charges + 120
R=1/58  #Resistance of 1m length and 1 sq.mm
I=200  # Maximum Current Flowing
CLPUL= 2*(I**2)*R*1000  # Copper Loss per unit length per unit area
ACL= CLPUL*365*24*(8/12)  # Annual Copper Loss
GC= 5/100  # Genereating Cost per unit.
ACEL= ACL*GC/1000  # Annual cost of energy loss
A= sqrt(ACEL/AIDC)  #Cross Sectional Area of the cable
print 'The Most Economical cross sectional area of the cable for this Case is %0.2f sq.mm'%A

The Most Economical cross sectional area of the cable for this Case is 149.58 sq.mm


## Ex3 : Page 105¶

In [3]:
# To determine the Most Economical Cross Sectional Area to supply a 3 Phase Load

LLF = lambda b:(0.25*b)+(0.75*(b**2)) # Function to determine the Loss Load Factor

OFC=0.20  #Cost of single phase overhead feeder per m per unit area + 10
AIDC= 10*OFC/100  #Annual Interest and depriciation charges + 1
TE= 2.5*(10**6)  # Total energy to be supplied per annum
CEW=10/100  # Cost of energy wasted per unit
LFS= TE/(1000*365*24)  # Load factor of supply
R=1/58  # Resistance of the cable per unit length
PF=1  # Unity power factor
MD= 1*(10**6)  # Maximum Demand
V=11*(10**3)  # Voltage of the feeder
FLCL= 3*(I**2)*R  # Full Load Copper Loss per Metre
ACL= Llf*FLCL  # Actual Copper Loss
CCL=ACL*(365*24*CEW/1000)  # Cost of Copper Loss
A=sqrt(CCL/AIDC)
print 'The Most Economical Cross sectional area for this Case is %g A/sq.mm'%A
# Calculation Mistake in the Book. Hence according to the concepts in the book the answer is as calculated. Please Note.

The Most Economical Cross sectional area for this Case is 28.7493 A/sq.mm


## Ex4 : Page 106¶

In [4]:
#To determine the most economical cross section for a 3 Phase line 8 km long

OFC=5000  #Cost of feeder per km per unit area + 6250
AIDC= 8*OFC/100  #Annual Interest and depriciation charges + 625
AIDCPD= AIDC/365  # Annual Interest and depriciation charges per day
R=1/58  #Resistance of 1m length and 1 sq.mm
CEU= 5/100  # Cost of energy per unit
V=33*(10**3)  # Voltage of the feeder
L10=3000*(10**3)  # Load for 10 hrs at unity power factor.
L8=1000*(10**3)  # Load for 8 hrs at unity power factor.
L6=2000*(10**3)  # Load for 6 hrs at unity power factor.

LC = lambda b:b/(sqrt(3)*V) # Funtion to determine the Load Current
I10=LC(L10) # Load Current for 10 hrs at unity power factor.
I8=LC(L8) # Load Current for 8 hrs at unity power factor.
I6=LC(L6) # Load Current for 6 hrs at unity power factor.
ELPD=3*(((I10**2)*10)+((I8**2)*8)+((I6**2)*6))*1000/(100*58)  # Energy Loss Over the DAy per km (Area in sq.cm)
CEL=ELPD*CEU/1000  # Cost of energy loass per km
A=sqrt(CEL/AIDCPD)  # Cross Sectional Area
print 'The Most Economical Cross Sectional Area for a 3 Phase line 8 km Long is %0.4f sq..cm'%A

The Most Economical Cross Sectional Area for a 3 Phase line 8 km Long is 0.9388 sq..cm


## Ex5 : Page 109¶

In [5]:
# To Calculate Voltage between middle wire and outer at each load point

Vs=220  #Supply Voltage
#Resistances of the respective Sections
Rab=0.2
Rbc=0.2
Rdh=0.2
Rhe=0.2
Ref=0.3
Rfl=0.1
Rlg=0.3
Rij=0.25
Rjk=0.2
Ra=0.2
Rd=0.4
Ri=0.3

#Currents following through the respective Section, Found using KCL
Ibe=10
Icl=12
Ihi=15
Ifj=5
Igk=15
Ia=Iad+Ibe+Icl  # Current through the positive wire
Ii=Ihi+Ifj+Igk  # Current through the negative wire
Id=Ii-Ia  #Current through the Middle wire
Ihe=Ihi-Idh
Ief=Ibe-Ihe
Ifl=Ief-Ifj
Ilg=Igk
Iij=Ii-Ihi
Ijk=Ii-Ihi-Ifj

#Voltage drops across each section
Va= Ra*Ia
Vab=Iab*Rab
Vbc=Ibc*Rbc
Vi=Ii*Ri
Vd=Id*Rd
Vdh=Idh*Rdh
Vhe=Ihe*Rhe
Vef=Ief*Ref
Vfl=Ifl*Rfl
Vlg=Ilg*Rlg
Vij=Iij*Rij
Vjk=Ijk*Rjk

#Voltage across the middle wire and the outer load points Using KVL
Vcl=Vbe-Vbc+Vfl+Vef
Vhi=Vs-Vd-Vdh-Vi
Vfj=Vhi-Vef+Vhe-Vij
Vgk=Vfj-Vfl-Vlg-Vjk

print 'The Voltages between middle wire and outer wire at each load point are:'
print ' 2. Vbe = %g V'%(Vbe)
print ' 3. Vcl = %g V'%(Vcl)
print ' 4. Vhi = %g V'%(Vhi)
print ' 5. Vfj = %g V'%(Vfj)
print ' 6. Vgk = %g V'%(Vgk)

The Voltages between middle wire and outer wire at each load point are:
2. Vbe = 215.6 V
3. Vcl = 215.9 V
4. Vhi = 203.7 V
5. Vfj = 196.7 V
6. Vgk = 188.9 V


## Ex6 : Page 110¶

In [6]:
#To obtain the voltages at the far end of the positive and negative wires

Vs=220  # Supply Voltage
#Resistances of the given sections
r1=0.015
r2=0.035
r3=0.02
r4=0.01
r5=0.025
r6=0.015
r7=0.03
r8=0.01
r9=0.02
r10=0.03
r11=0.04

#Currents follwing in between the two wires
I1=10
I2=20
I3=25
I4=5
I5=15
I6=15
I7=15
I8=15
I9=18
I10=30
I11=15

#Effective resistances for the above currents
R1=r1
R2=r1+r2
R3=r1+r2+r3
R4=r1+r2+r3+r4
R5=r1+r2+r3+r4+r5
R6=r1+r2+r3+r4+r5+r6
R7=r7
R8=r7+r8
R9=r7+r8+r9
R10=r7+r8+r9+r10
R11=r7+r8+r9+r10+r11

#Voltage drop in the respective wires
#Outer positive wire
Vop=(I1*R1)+(I2*R2)+(I3*R3)+(I4*R4)+(I5*R5)+(I6*R6)
#Outer Negative wire
Vnp=(I7*R7)+(I8*R8)+(I9*R9)+(I10*R10)+(I11*R11)
# Net drop in neutral wire
Vn=Vop-Vnp

#Effective drop in the respective wires
#Outer positive wire
Veop=Vop+Vn
#Outer negative wire
Venp=Vnp-Vn

#Voltage at far end of the wires
#Positive Wire
Vpf=Vs-Veop
#Negaitve Wire
Vnf=Vs-Venp

print "The voltage drop at the far end of the wires are:"
print "Positive Wire: %g V"%(Vpf)
print "Negative Wire: %g V"%(Vnf)

The voltage drop at the far end of the wires are:
Positive Wire: 213.43 V
Negative Wire: 213.115 V


## Ex7 : Page 111¶

In [7]:
#To determine the cross section of the conductor for a minimum consumer voltage

from sympy import symbols, solve
#Unknown Variable obtained in the equation
Ib, r=symbols("Ib r")

#Voltages at the respective ends
Va=230
Vb=230

#Minimum Consumers's Voltage
Vc=220

#Lenghths of the segments
r1=200
r2=200
r3=100
r4=300
r5=300
r6=100

#Effective length of segments
R1=r1
R2=r1+r2
R3=r1+r2+r3
R4=r1+r2+r3+r4
R5=r1+r2+r3+r4+r5
R6=r1+r2+r3+r4+r5+r6

I1=25
I2=20
I3=25
I4=25
I5=10

#By the law of momemts of currents
Vnet=2*((I1*R1)+(I2*R2)+(I3*R3)+(I4*R4)+(I5*R5)-(Ib*R6))  #Since the equation is equated to zero, r vanishes
X=Vnet-(Va-Vb)  #Polynomial Equation to Find Ib
Ib=float(solve(X, Ib)[0])  # Numerical Value of Ib

Ia=(I1+I2+I3+I4+I5)-Ib

#From the given figure it is clear that point M is the point of Minimum Potential
Vd=Va-Vc
X=(2*r*((Ia*r1)+((Ia-I1)*r2)+((Ia-I1-I2)*r3)))-Vd  #Polynomial Equation to Find r
r=float(solve(X, r)[0])  # Resistance per unit length (Numerical Value)

Rstd=(1/58)  # Resistance for 1m and 1 sq.mm

A=Rstd/r  # Cross Section of Conductor required.

print 'The Cross Section of the Conductor to provide minimal consumer''s voltage is %0.f sq.mm '%A

The Cross Section of the Conductor to provide minimal consumers voltage is 67 sq.mm


## Ex8 : Page 111¶

In [8]:
#To determine the cross section of the conductor for a minimum consumer voltage

from sympy import symbols, solve
#Unknown Variable obtained in the equation
x,r = symbols("x r")

#Voltages at the respective ends
Va=235
Vb=230

#Minimum Consumers's Voltage
Vc=220

#Lenghths of the segments
r1=200
r2=200
r3=100
r4=300
r5=300
r6=100

#Effective length of segments
R1=r1
R2=r1+r2
R3=r1+r2+r3
R4=r1+r2+r3+r4
R5=r1+r2+r3+r4+r5
R6=r1+r2+r3+r4+r5+r6

I1=25
I2=20
I3=25
I4=25
I5=10

#The Minimum Point assumed is N, Hence the current following to Point N is given by 'x'

#The Effective Drops
Van=Va-Vc
Vbn=Vb-Vc
V=Van/Vbn
A=2*(((x+I1)*r1)+(r2*x))  #Wrt to Van
B=2*(((I2-x)*r3)+((I2+I3-x)*r4)+((I2+I3+I4-x)*r5)+((I2+I3+I4+I5-x)*r6))  #Wrt to Vbn
C=A/B  # The 'r' term gets eliminated
X=C-V  # Polynomial Eqaution to find x
x=solve(X,x)[0]
Ia=I1+x  #Current Supplied at end A
#It is clear the above assumed Minimal Point is wrong and it has to Point M
#Therefore finding drop at Point M

X=(2*r*(((x+I1)*r1)+(r2*x)+((x-20)*r3)))-Van  #Polynomial Equation to find r
r=solve(X,r)[0]  # Numerical Value of r
Rstd=(1/58)  # Resistance for 1m and 1 sq.mm

A=Rstd/r  # Cross Section of Conductor required.

print 'The Cross Section of the Conductor to provide minimal consumer''s voltage is %0.2f sq.mm'%A

# Please note the calculation mistake in the book. The value of r found out is wrong in the text book.

The Cross Section of the Conductor to provide minimal consumers voltage is 51.26 sq.mm


## Ex9 : Page 114¶

In [9]:
#To Determine the location and magnitude of minimum voltage

Vs=220  #Supply Voltage at End A and B
#Different Conductor Lenghts
#From End A
L1=100
L2=50
L3=50

r=0.05  # Resistance of Conductor per Km
#Different Currents drawn by various loads
I1=50
I2=75
I3=A*L4

#Taking moments of all currents at A
Ib=((I1*L1)+((L1+L2)*I2)+((L1+L2+L3+(L4/2))*I3))/(L1+L2+L3+L4)
Ia=I1+I2+I3-Ib

#Minimum Potential Point in this case is the point where All the current from B is drawn
X=Ib/A  # Distance from B
Y=(L1+L2+L3+L4)-X  #Distance from A

#Minimum Potential Drop
Vmind=(2*r/1000)*((Ia*L1)+((Ia-I1)*L2)+((Ia-I1-I2)*L3)+(((Y-L1-L2-L3)*A)*((Y-L1-L2-L3)/2)))

Vmin=Vs-Vmind  #Minimum Potential
print 'The Location of The Minimum Voltage is %g m from side A and its magnitude is %0.1f V'%(Y,Vmin)

The Location of The Minimum Voltage is 279.167 m from side A and its magnitude is 217.4 V


## Ex10 : Page 115¶

In [10]:
from __future__ import division
#To determine the currents supplied to the ring main from A and B

from numpy import array, linalg
#Currents in the ring scheme going clockwise
I1=50
I2=15
I3=25
I4=10
It = I1+I2+I3+I4  # Total Current
#Resistances of the respective segments going clockwise
R1=0.1
R2=0.12
R3=0.04
R4=0.08
R5=0.06
R6=0.02

Ra=0.04  #Resistance at A
Rb=0.06  #Resistance at B

# Va=Vb  Net Voltage is Zero
#Dividing the total current from A as x and y
#Taking voltage drops in clockwise direction and anticlockwise directions
# We get two simultaneous equations
# 3.2x + y = 120
# x + 3y = 114
R=array([[3.2,1],[1,3] ])
V=array([[120],[114] ])
I=linalg.inv(R)*V

#To calculate and seperate the respective currents from the above matrix eqution
x=I[0]
y=I[1]
Ia=x+y
Ib=It-Ia
print 'a) When Va = Vb  Ia = %g A and Ib =%g'%(Ia[0],Ib[0])

# Va+5=Vb  Net Voltage is 5V
#Dividing the total current from A as x and y
#Taking voltage drops in clockwise direction and anticlockwise directions
# We get two simultaneous equations
# 3.2x + y = 120 -(5*NetVoltage) = 70
# x + 3y = 114 -(5*NetVoltage) = 64
R=array([[3.2,1],[1,3]] )
V=array([[70],[64] ])
I=linalg.inv(R)*V

#To calculate and seperate the respective currents from the above matrix eqution
x=I[0]
y=I[1]
Ia=x+y
Ib=It-Ia
print 'b) When Va + 5 = Vb  Ia = %g A and Ib =%g'%(Ia[0],Ib[0])


a) When Va = Vb  Ia = 28.6047 A and Ib =71.3953
b) When Va + 5 = Vb  Ia = 16.9767 A and Ib =83.0233


## Ex11 : Page 116¶

In [11]:
#To Compare the volume of copper required
from sympy import symbols, solve
#Unknown Resistances
r, r1, x=symbols('r r1 x')

#Lengths of the segements of the ring scheme
L1=100
L2=200
L3=200
L4=150
L5=150
I1=40
I2=20
I3=100
I4=40
It=I1+I2+I3+I4 #Total Current

#Without Interconnector
#Let x be the current flowing through the entire ring

Eq=(L1*x)+(L2*(x-I1))+(L3*(x-I1-I2))+(L4*(x-I1-I2-I3))+(L4*(x-I1-I2-I3-I4))  #Polynomial Equation to find x

x=float(solve(Eq, x)[0])
x1=It-x  #Current flowing in the other direction

Vac1=((x1*L5)+((x1-I4)*L4))*r  # Voltage across AC without the connector
MVac1=((x1*L5)+((x1-I4)*L4))  # Magnitude of Vac1

#With Interconnector
#Considering x amount of current to flow clockwise through segment AE
#Considering y amount of current to flow anticlockwise through segment AB
#Considering 200-(x+y) amount if current to flow through the segment AC
# Mesh Analysis of ABCDE gives 5x - 3y = 140
# Mesh Analysis of ABC gives 5x + 11y = 1120
R=array([[5,-3],[5,11]])
V=array([[140],[1120] ])
I=linalg.inv(R)*V
x=I[0]
y=I[1]
Vac2=(It-(x+y))*250*r1  # Voltage across AC with connector
MVac2=(It-(x+y))*250  # Magnitude of Vac2
print 'The Voltage drop across AC in both case is the same.'
print Vac2[0],'Is Equal to',Vac1

#To Compute the Numerical Values of the Ratio of resistances
RatioA = MVac1/MVac2

print RatioA[0],'is',r1,'divided by',r

#Effective Length of both the cases
Leff=L1+L2+L3+L4+L5
LeffC=Leff+250
#Volume is Length * Area
RatioV=Leff*RatioA/LeffC
print 'The Volume of copper without the connector is %g times the volume required with connector'%RatioV[0]

The Voltage drop across AC in both case is the same.
64500.0*r1 Is Equal to 26250.0*r
0.406976744186 is r1 divided by r
The Volume of copper without the connector is 0.310078 times the volume required with connector


## Ex12 : Page 117¶

In [12]:
#To Determine the Voltage at the far end
from math import sin, acos
r=0.3  # Loop Resistance per km
x=0.15  # Loop Reactance per km

reactive = lambda y,z,v:(sin(acos(y)))*z*v*x/1000 #To find the reactive drop of the current

active = lambda y,z,v:y*z*v*r/1000 #To find the reactive drop of the current

#Power Factors of the loads from left to right
pf1=0.707
pf2=1
pf3=0.8
#Currents delivering the respective loads from left to right
I1=50
I2=60
I3=40
#Length of the conductors in metres
l1=200
l2=300
l3=300
#Effective length of the conductors in metres
L1=l1
L2=l1+l2
L3=l1+l2+l3
#Active component drops of the respective currents
Va1=active(pf1,I1,L1)
Va2=active(pf2,I2,L2)
Va3=active(pf3,I3,L3)
Vat=Va1+Va2+Va3  #Total Active Component Drop
#Reactive component drops of the respective currents
Vr1=reactive(pf1,I1,L1)
Vr2=reactive(pf2,I2,L2)
Vr3=reactive(pf3,I3,L3)
Vrt=Vr1+Vr2+Vr3  #Total Reactive Component Drop

Vt=Vrt+Vat  # Total voltage drop

print 'The voltage drop at the far end is %g V '%(Vt)

The voltage drop at the far end is 22.7418 V


## Ex13 : Page 118¶

In [13]:
#To determine line currents and neutral currents
from math import sqrt, acos, exp, atan, degrees, pi
from cmath import exp
V=400  # Supply Voltage
Vph=400/sqrt(3)  #Phase Voltage
pf=0.8  #Power Factor lagging
theta=-1 * acos(pf)  #Angle in radians of the balanced LAGGING current

I=L/(sqrt(3)*V*pf)*exp(1J*theta)  #Balanced Current Magnitude
Ir=Lr/Vph  #Magnitude of R phase voltage
Iy=Ly/Vph  #Magnitude of Y phase voltage
Ib=Lb/Vph  #Magnitude of B phase voltage

#Vr is taken as reference
#Angles of the phase voltages wrt to the reference
Thetay=-120
Thetab=120
Thetar=0
#Net Currents in the respective phases (RYB)
Irf=Ir+I
Iyf=Iy+I
Ibf=Ib+I
#Angle of the above currents in degrees
r=degrees(atan((Irf.imag)/(Irf.real)) )
y=degrees(atan((Iyf.imag)/(Iyf.real)) )
b=degrees(atan((Ibf.imag)/(Ibf.real)) )
#Angles of the above currents with respect to the reference
rf=Thetar+r
yf=Thetay+y
bf=Thetab+b
#Effective Currents wrt to the reference voltage
Irn=abs(Irf)*exp(1J*pi*rf/180)
Iyn=abs(Iyf)*exp(1J*pi*yf/180)
Ibn=abs(Ibf)*exp(1J*pi*bf/180)
In=Irn+Iyn+Ibn  # Neutral Current
#Note Take Vr as reference
print 'The Net Current in phase R is %g/_%g A'%(abs(Irf),rf)
print 'The Net Current in phase Y is %g/_%g A'%(abs(Iyf),yf)
print 'The Net Current in phase B is %g/_%g A '%(abs(Ibf),bf)
print 'The Net Neutral Current is %g/_%g A\n'%(abs(In),degrees(atan((In.imag)/(In.real))))

The Net Current in phase R is 1047.32/_-29.7449 A
The Net Current in phase Y is 1439.4/_-141.161 A
The Net Current in phase B is 1643.17/_101.565 A
The Net Neutral Current is 572.822/_-19.1066 A



## Ex14 : Page 123¶

In [14]:
#To determine the best location of the substation for a given set of loads

#Various loads and how they are positioned on the corners of a square of length 25km
L1=5000  #(0,25)
L2=8000  #(25,25)
L3=3000  #(25,0)
L4=6000  #(0,0)

L=25 #Length of the square


The Susbstation must be located at (12.5 km,14.77 km)