# Chapter7 - Distribution systems¶

## Problem 7.1 - page 174¶

In [19]:
from __future__ import division
import cmath
#Given data :
l=1 #in km
I=100 #in Ampere
cosfi=0.8 #Power factor(lag) unitless
VC=200 #in volt
IL=60 #in Ampere
R=0.6 #in ohm
XL=0.08 #in ohm
IC=I*complex(0.8,-0.6) #in Ampere
z=complex(0.06,0.08)/2 #in ohm
VD_BC=z*IC #in volt
VB=VC+VD_BC #in volt
IB=IL*complex(0.9,-0.4357)+IC #in Ampere
VD_AB=z*IB #in volt
VD_AB=round(VD_AB.real,2)+round(VD_AB.imag,2)*1J
print "V.D. from sending end to mid point =" ,VD_AB,"Volt"
print "V.D. from mid point to the far end =",VD_BC,"Volt"

V.D. from sending end to mid point = (7.47+2.78j) Volt
V.D. from mid point to the far end = (4.8+1.4j) Volt


## Problem 7.2 - page 175¶

In [26]:
from __future__ import division
from sympy import symbols
#Given data :
l=500 #in meter
i=1 #in Ampere/meter
IL1=200;IL2=150;IL3=50;IL4=100 #in Ampere
l1=100;l2=200;l3=300;l4=400 #in meter
r=0.1 #in ohm/km
Vd=250 #in volt
I=symbols('I')
Drop_AC=100*(r/10**3)*(I-i*l1/2)
Drop_CD=I
Drop_DE=100*r*(I-550)-I*100/2
Drop_EF=100*r*(I-700-I*100/2)
Drop_FB=100*r*(I-900-I*100/2)
VD_tot=0.05*I-27 #in volts
#both ends are fed with same voltage,
#VD_tot should be equal to zero."
I=27/0.05 #in Ampere
print "Curent = %0.2f Ampere" %I
print "Value at minimum potential at D = %0.2f V" %(Vd-Drop_AD)
#Note : Ans in the book is wrong as 27/0.05 gives 540 instead of 54.

Curent = 540.00 Ampere
Value at minimum potential at D = 243.20 V


## Problem 7.3 - page 176¶

In [32]:
from __future__ import division
#Given data :
l=250 #in meter
VA=230 #in volt
VB=232 #in volt
r=0.5 #in ohm/km
r=0.5/10**3 #in ohm/m
RAC=r*50*2 #in ohm
RCD=RAC;RDE=RAC;REF=RAC;RFB=RAC #in ohm
#VA-VB=VAC+VCD+VDE+VEF+VFB #in volt
Ia=(VA-VB+15)/(5*RAC) #Ampere
IAC=Ia;ICD=IAC-20;IDE=IAC-60;IED=-IDE;IEF=IAC-100;IFE=-IEF;IFB=IAC-120;IBF=-IFB #in Ampere
print "IAC = %0.f A" %IAC
print "ICD = %0.f A" %ICD
print "IDE = %0.f A" %IDE
print "IED = %0.f A" %IED
print "IEF = %0.f A" %IEF
print "IFE = %0.f A" %IFE
print "IFB = %0.f A" %IFB
print "IBF = %0.f A" %IBF
VAC=IAC*RAC #in volt
VCD=ICD*RCD #in volt
VD=VA-VAC-VCD #in volt
print "The minimum potential = %0.1f Volt" %(VD)

IAC = 52 A
ICD = 32 A
IDE = -8 A
IED = 8 A
IEF = -48 A
IFE = 48 A
IFB = -68 A
IBF = 68 A
The minimum potential = 225.8 Volt


## Problem 7.4 - page 177¶

In [43]:
from __future__ import division
#Given data :
VA=235 #in volt
VB=236 #in volt
l=200 #in meter
IL1=20;IL2=40;IL3=25;IL4=30 #in Ampere
l1=50;l2=75;l3=100;l4=50 #in meter
r=0.4 #in ohm/km
r=0.4/10**3 #in ohm/m
RAC=r*l1*2 #in ohm
RCD=r*(l2-l1)*2*RAC;RDE=r*(l2-l1)*2*RAC;REF=r*l1*2*RAC;RFB=r*l1*2*RAC #in ohm
#VA-VB=VAC+VCD+VDE+VEF+VFB #in volt
IA=(VA-VB+9.6)/(0.16) #in Ampere
IAC=IA;ICD=IA-IL1;IDE=IA-IL1-IL2;IEF=IA-IL1-IL2-IL3;IFB=IA-IL1-IL2-IL3-IL4 #in Ampere
print "IAC = %0.2f A" %IAC
print "ICD = %0.2f A" %ICD
print "IED = %0.2f A" %(-IDE)
print "IFE = %0.2f A" %-IEF
print "IFB = %0.2f A" %-IFB
VAC=IAC*RAC #in volt
VCD=ICD*RCD #in volt
VD=VA-VAC-VCD #in volt
print "The minimum potential = %0.3f Volt" %VD
# Answer wrong in the textbook due to accuracy.

IAC = 53.75 A
ICD = 33.75 A
IED = 6.25 A
IFE = 31.25 A
IFB = 61.25 A
The minimum potential = 232.823 Volt


## Problem 7.5 - page 179¶

In [46]:
from __future__ import division
#Given data :
VA=400 #in volt
r=0.03 #in ohm/km
r=0.03/1000 #in ohm/m
RAB=r*500*2 #in ohm
RBC=r*300*2 #in ohm
RAB=r*700*2 #in ohm
RAB=r*500*2 #in ohm
#VA-VB=VAC+VCD+VDE+VEF+VFB #in volt
IA=(17.4)/(0.09) #in Ampere
VAB=(RAB)*IA #in volt
VB=VA-VAB #in volt
print "Voltage at B = %0.2f Volts" %VB
VBC=(RBC)*(IA-150) #in volt
VC=VB-VBC #in volt
print "Voltage at C = %0.2f Volts" %VC
IBC=IA-150 #in A
print "Current in section BC = %0.2f A "%IBC
#Note : Answer of VB is wrong in the book.

Voltage at B = 394.20 Volts
Voltage at C = 393.42 Volts
Current in section BC = 43.33 A


## Problem 7.6 - page 180¶

In [47]:
from __future__ import division
#Given data :
VA=240 #in volt
MAxVDrop=VA*5/100 #in volt
rho=2.87*10**-6 #in ohm-cm
#VAB+VBC+VCA=0 #in volt
IA=(3200)/(26) #in Ampere
IAB=IA #in Ampere
IBC=IA-100 #in Ampere
#Allowed voltage drop: IAB*RAB+IBC*RBC=12
R=12/(1015.26) #in ohm
RAB=R*300*2/100 #in ohm
RBC=R*600*2/100 #in ohm
RCA=R*400*2/100 #in ohm
#formula : R=rho*l/a
a=rho*(100*100)/R #in cm**2
print "Cross section area = %0.2f cm2 " %a

Cross section area = 2.43 cm2


## Problem 7.7 - page 182¶

In [50]:
from __future__ import division
from numpy import sqrt
#Given data :
R=0.2 #in ohm/km
X=0.1 #in ohm/km
ZAM=((R+X*1J)/1000)*200 #in ohm
ZMB=((R+X*1J)/1000)*100 #in ohm
I1=100*(0.707-0.707*1J) #in A
I2=200*(0.8-0.6*1J) #in A
IAM=I1+I2 #in Ampere
VAM=ZAM*IAM #in volts
VMB=ZMB*I2 #in volts
VAB=VAM+VMB #in volts
magVAB=sqrt(VAB.real**2+VAB.imag**2)
print "Total voltage drop = %0.2f Volts" %magVAB

Total voltage drop = 17.85 Volts


## Problem 7.8 - page 183¶

In [63]:
from __future__ import division
from numpy import sqrt, sin, pi, arctan, arccos, cos
import cmath
#Given data :
VB=200 #in volts
R=0.2 #in ohm/km
X=0.3 #in ohm/km
I=100 #in Ampere
ZAB=(R+X*1J) #in ohm
ZMB=ZAB/2 #in ohm
ZAM=ZMB #in ohm
cosfi_1=0.6 #unitless
cosfi_2=0.8 #unitless
IMB=I*(cosfi_2-cosfi_1*1J) #in A
I2=IMB #in Ampere
VMB=IMB*ZMB #in volts
VM=VB+VMB #in volts
print "Voltage at M = %.2f∠%.2f°" %(abs(VM),cmath.phase(VM)*180/pi)
#print('Sending end voltage , V_S = %.2f∠%.2f° kV/phase' %(abs(V_S*10**-3),cmath.phase(V_S)*180/math.pi))
fi=arctan(VM.imag/VM.real)*180/pi #in degree
fi_1=arccos(cosfi_1)*180/pi #in degree
fi_VBandI1=fi_1-fi #in degree
I1=I*(cos(fi_VBandI1*pi/180)-sin(fi_VBandI1*pi/180))*1J #in Ampere
IAM=I1+I2 #inA Ampere
VAM=ZAM*IAM #in volts
VA=VM+VAM #in volts
magVA=sqrt(VA.real**2+VA.imag**2)
print "Voltage at A, standing end voltage = %0.2f Volts" %magVA

Voltage at M = 217.08∠1.58°
Voltage at A, standing end voltage = 236.65 Volts


## Problem 7.9 - page 184¶

In [64]:
#Given data :
l=500 #in meter
VA=200 #in volt
MAxVDrop=6 #in % of declared voltage
rho=0.014 #in ohm/m
#VD in the distributor=53*10**3*r
AllowedVD=VA*(6/100) #in volts
r=AllowedVD*10**6/(53*10**3) #in ohm/meter
#formula : R=rho*l/a
a=rho*(2*l)/r #in m**2
print "Cross section area = %0.2f m2 " %(a)

Cross section area = 0.06 m2


## Problem 7.10 - page 185¶

In [66]:
#Given data :
l=300 #in meter
I=0.75 #in A/m
R=0.00018 #in ohm/m
x=200 #in meter
Vs=250 #in volt
VD=I*R*(l*x-x**2/2) #in volt
V_A=Vs-VD #in volt(Voltage at 200m from end A)
print "Voltage as 200m from supply end A = %0.1f Volts" %V_A

Voltage as 200m from supply end A = 244.6 Volts


## Problem 7.11 - page 185¶

In [69]:
#Given data :
l=600 #in meter
VA=440 #in volt
VB=400 #in volt
R=0.01 #in ohm/100m
RAC=(R/100)*300 #in ohm
RCD=(R/100)*300 #in ohm
RDE=(R/100)*100 #in ohm
REF=(R/100)*200 #in ohm
RFB=(R/100)*300 #in ohm
#VA-VB=VAC+VCD+VDE+VEF+VFB #in volt
IA=(VA-VB+42.5)/(0.12) #in Ampere
IAC=IA;ICD=IA-100;IDE=IA-300;IFE=IA-550;IFB=IA-850 #in Ampere
print "Current fed at A, IA = %0.1f A  "%IAC
print "Current fed at B, IB = %0.1f A  "%-IFB

Current fed at A, IA = 687.5 A
Current fed at B, IB = 162.5 A


## Problem 7.12 - page 187¶

In [2]:
#Given data :
VA=220 #in volt
VB=200 #in volt
R=0.1 #in ohm/km
I=1 #in A/m
l=500 #in meter
R=2*R/1000 #in ohm/m
x=(VA-VB)/(I*R*l)+l/2 #in meter
Vmin=VA-I*R*x**2/2 #in volts
print "Value of minimum potential = %0.2f V  "%Vmin
IA=I*x #in A
print "Current supplied from end A = %0.f A   " %IA
IB=I*(l-x) #in A
print "Current supplied from end B = %0.f A"%IB

Value of minimum potential = 199.75 V
Current supplied from end A = 450 A
Current supplied from end B = 50 A


## Problem 7.13 - page 187¶

In [4]:
#Given data :
VL=240 #in volt
Router=0.2 #in ohm
I1=VL/5 #in Ampere
I2=VL/6 #in Ampere
Ineutral=I1-I2 #in Ampere
#Applying KVL on +ve side
V1=VL+I1*0.2+8*0.4 #in volt
print "Voltage at +ve side = %0.1f V" %V1
#Applying KVL on +ve side
V2=VL-(8*0.4)+I2*0.2 #in volt
print "Voltage at -ve side = %0.1f V" %V2

Voltage at +ve side = 252.8 V
Voltage at -ve side = 244.8 V


## Problem 7.14 - page 188¶

In [7]:
#Given data :
#Applying KVL on +ve side
V1=200-(600*0.015)-(100)*0.03 #in volt
print "Voltage at +ve side = %0.f V" %V1
#Applying KVL on -ve side
V2=200-(-100*0.03)-500*0.0015 #in volt
print "Voltage at -ve side = %0.1f V" %V2
#Note : answer of 2nd part is wrong in the book.

Voltage at +ve side = 188 V
Voltage at -ve side = 202.2 V


## Problem 7.15 - page 189¶

In [10]:
from sympy import symbols
x=symbols('x')
#Given data :
#VD in section AC from RHS:
VD1=(40+x)*0.02+0.17*x
#VD in section AC from LHS:
VD2=(350-x)*0.015+(150-x)*0.03
#Equating two VDs we get
#x*0.02+0.17*x+0.015*x+x*0.03=350*0.015+150*0.03-40*0.02
x=(350*0.015+150*0.03-40*0.02)/0.082 #in A
VB=500-(x+40)*0.02 #in volts
print "Potential at point B = %0.2f V" %VB
VC=VB-(x*0.017) #in volts
print "Potential at point C = %0.2f V" %VC
VD=500-(350-x)*0.015 #in volts
print "Potential at point D = %0.2f V" %VD
#Note : Answer of 3rd part is given wrong in the book.

Potential at point B = 497.02 V
Potential at point C = 495.16 V
Potential at point D = 496.39 V


## Problem 7.16 - page 190¶

In [14]:
from sympy import symbols
x=symbols('x')
#Given data :
#Applying KVL in loop AFEDA:
p1=((0.016*x)+0.09*(x-30)+0.14*(x-17)-0.1*y) #eqn(1)
p2=(0.1*y-0.12*(95-x-y)-.01*(145-x-y)-0.008*(165-x-y)) #eqn(2)
#Equating two equtions we get
#3.9*x-125=97.75-0.75*x
x=(97.75+125)/(3.9+0.75) #in A
y=97.75-0.75*x #in A
print "x = %0.f A" %x
print "y = %0.2f A"%y
print "Point of minimum ppotential is E."

x = 48 A
y = 61.82 A
Point of minimum ppotential is E.


## Problem 7.17 - page 190¶

In [15]:
#Given data :
V=200 #in volt
I=1 #in A/m
R=2*0.05/1000 #in ohm/m
l=1*1000 #in meter
IT=I*l #in Ampere
RT=R*l #in ohm
VD=IT*RT/8 #in volt
Vmin=V-VD #in volt
print "Minimum potential occurs at mid point. It is %0.2f V" %Vmin

Minimum potential occurs at mid point. It is 187.50 V


## Problem 7.19 - page 191¶

In [20]:
#Given data :
VB=400 #in volt
ZAC=0.04+0.08*1J #in ohm
ZCB=0.08+1J*0.12 #in ohm
I1=60*(0.8-1J*0.6)
I2=120*(0.8-1J*0.6)
VCB=I2*ZCB #in Volt
VAC=(I1+I2)*ZAC #in volt
VC=VB+I2*ZCB #in Volt
print "Voltage at C =",VC,"Volt"
VA=VC+(I1+I2)*ZAC #in volt
print "Voltage at A =",VA,"Volt"
#Answer not accurate in the textbook.

Voltage at C = (416.32+5.76j) Volt
Voltage at A = (430.72+12.96j) Volt