In [2]:

```
import math
from numpy import exp,sqrt
# Variables
Z = 0.1+(0.1*1j); #Feeder Impedance per unit
R = Z.real; #Resismath.tance
X = Z.imag; #Reacmath.tance
Vs = 1.; #Sending End Voltage
Pr = 1.; #Consmath.tant Power Load
pfr = 0.8; #Power Factor at recieving end
tr = math.acos(pfr); #Power FActor angle
# Calculations
def angle(y):
return math.degrees(math.atan(y.imag/y.real))
K = (Vs**2)-(2*Pr*(R+(X*(math.atan(tr)))));
Vr = math.sqrt((K/2)*(1+math.sqrt(1-((2*abs(Z)*Pr/(K*pfr))**2)))); #Recieving End Voltage
C = Pr*(X-(R*math.degrees(math.atan(tr))))/((Vr**2)+(Pr*(R+(X*math.degrees(math.atan(tr))))));
del1 = math.degrees(math.atan(C));
Ir = (Pr/(abs(Vr)*pfr))*exp(-1*math.pi*1j*tr/180) #Recieving End Current
Is = Ir; #Sending End Current
Tir = angle(Ir);
Vr1 = Vs-(Z*Ir);
# Results
print 'a) Vr is %g/_%g pu, del is %g degrees, Ir = Is = %g/_%g pu'%(abs(Vr.real),angle(Vr),del1,abs(Ir),Tir)
print 'b) Vr is %g/_%g pu, which is almost equal to the previous case.'%(Vr1.real,angle(Vr1))
```

In [6]:

```
import math
# Variables
Sl = 518.; #Total Load on Lateral
Sm = 1036.; #Total Load on Main
Vll = 4.16; #Line to Line voltage
# Calculations
#Currents in the respective current
Ilateral = Sl/(math.sqrt(3)*Vll);
Imain = Sm/(math.sqrt(3)*Vll);
C = 5280.; #Length Consmath.tant
Ll = 5760./C; #Lateral Length
Lm = 3300./C; #Main Length
#Consmath.tant for the cables
Kl = 0.015;
Km = 0.01;
#Voltage Drop Percents for 3 phase
VDlateral3 = Ll*Kl*Sl/2;
VDmain3 = Lm*Km*Sm;
TVD3 = VDmain3+VDlateral3;
#Voltage Drop Percents for 1 phase according to Morrisoncfor laterals
VDlateral1 = VDlateral3*4;
VDmain1 = VDmain3;
TVD1 = VDlateral1+VDmain1;
#CASE B
#To meet the maximum primary voltage drop criterion of 4.00 percent
#Conductors with ampacities of 480A and 270A for Main and laterals
#Consmath.tants from the table
Klb = 0.006;
Kmb = 0.003;
#Voltage Drop Percents
VDlateralb = Ll*Klb*Sl/2;
VDmainb = Lm*Kmb*Sm;
TVDb = VDmainb+VDlateralb;
# Results
print 'a The percent voltage drops at :'
print 'i 3Phase'
print 'Lateral End is %g percent'%(VDlateral3)
print 'Main End is %g percent'%(VDmain3)
print 'ii 1Phase'
print 'Lateral End is %g percent'%(VDlateral1)
print 'Main End is %g percent'%(VDmain1)
print 'b) Conductors with Ampacities of 480A and 270A are used to find the Percent voltage drop of the \
Main and Lateral as %g percent and %g percent respectively'%(VDmainb,VDlateralb)
print 'The Above Drops meet the required criterion of 4 percent voltage drop'
```

In [7]:

```
import math
# Variables
#Terms taken from Example two
Il = 72.;
Im = 144.;
C = 5280.; #Length Consmath.tant
Ll = 5760./C; #Lateral Length
Lm = 3300./C; #Main Length
#From Tables
#Lateral
rl = 4.13; #Resismath.tance per mile
xLl = 0.258; #Reacmath.tance per mile
#Main
rm = 1.29; #Resismath.tance per mile
xLm = 0.211; #Reacmath.tance per mile
pf = 0.9; #Power Factor
Vb = 2400.; #Base Voltage
# Calculations
#Voltage Drops
VDlateral = Il*((rl*pf)+(xLl*math.sin(math.radians(math.acos(pf)))))*Ll/2;
VDmain = Im*((rm*pf)+(xLm*math.sin(math.radians(math.acos(pf)))))*Lm;
#Percent Voltage Drop
perVDlateral = VDlateral*100/Vb;
perVDmain = VDmain*100/Vb;
TVD = perVDlateral+perVDmain; #Total Percent Voltage drop
#Case B
#Conductors With Ampacities of 268A and 174A for Main and Laterals
#From Tables
#Lateral
rlb = 1.03; #Resismath.tance per mile
xLlb = 0.207; #Reacmath.tance per mile
#Main
rmb = 0.518; #Resismath.tance per mile
xLmb = 0.191; #Reacmath.tance per mile
Vb = 2400; #Base Voltage
#Voltage Drops
VDlateralb = Il*((rlb*pf)+(xLlb*math.sin(math.radians(math.acos(pf)))))*Ll/2;
VDmainb = Im*((rmb*pf)+(xLmb*math.sin(math.radians(math.acos(pf)))))*Lm;
#Percent Voltage Drop
perVDlateralb = VDlateralb*100/Vb;
perVDmainb = VDmainb*100/Vb;
TVDb = perVDlateralb+perVDmainb; #Total Percent Voltage drop
# Results
print 'a The percent voltage drops at :'
print 'Lateral End is %g percent'%(perVDlateral)
print 'Main End is %g percent'%(perVDmain)
print 'b) Conductors with Ampacities of 278A and 174A are used to find the Percent voltage drop of \
the Main and Lateral as %g percent and %g percent respectively'%(perVDmainb,perVDlateralb)
print 'The Above Drops meet the required criterion of 4 percent voltage drop'
```

In [9]:

```
import math
# Variables
Sl = 518.; #Total Load on Lateral
Sm = 5180.; #Total Load on Main
Vll = 12.47; #Line to Line voltage
#Currents in the respective current
Ilateral = Sl/(math.sqrt(3)*Vll);
Imain = Sm/(math.sqrt(3)*Vll);
C = 5280.; #Length Consmath.tant
Ll = 5760./C; #Lateral Length
Lm = 3300./C; #Main Length
#Consmath.tant for the cables
Km = 0.0008;
Kl = 0.00175;
# Calculations
#Voltage Drop Percents for 3 phase
VDlateral = Ll*Kl*Sl/2;
#Due to peculiarity of this new problem, one half of the main has to considered as express feeder and the other connected to a uniformly distributed load of 5180kVA
VDmain = Lm*Km*Sm*3/4;
TVD = VDmain+VDlateral;
#Since the inductive reacmath.tance of the line is
Cd = 12.; #Consmath.tant to find the dismath.tance in terms of feet
#Diameters of the Conductors
Dmi = 37.;
Dmn = 53.;
#Drops per mile
xdi = 0.1213*math.log(Dmi/Cd);
xdn = 0.1213*math.log(Dmn/Cd);
Dxd = xdn-xdi; #Difference in Drops
# Results
print 'a The percent voltage drops at :'
print 'Lateral End is %g percent'%(VDlateral)
print 'Main End is %g percent'%(VDmain)
print 'b The Above Drops meet the required criterion of 4 percent voltage drop'
print 'c) The Difference in Voltage drop is %g ohm/mile, which is a smaller VD valuue that it really is.'%(Dxd)
```

In [10]:

```
import math
# Variables
Vb = 7200.; #Base Voltage in V
pf = 0.9; #Power Factor
Sm = 10360.; #Load on Main Feeder in kVA
Vll = 12.47; #Line to Line voltage in kV
Imain = Sm/(math.sqrt(3)*Vll); #Current in Main Feeder
#Note Suffix l means lateral and m means main
Vph = 7.2; #Phase Voltage in kV
Sl = 2*518.; #Load on Lateral Feeder in kVA
Ilateral = Sl/Vph; #Current in Laterals
#Length of the Feeder
#Length Consmath.tant
Cm = 5280.; #Main
Cl = 1000.; #Lateral
Ll = 5760./Cl; #Lateral Length
Lm = 3300./Cm; #Main Length
#Consmath.tants for the particular cables from the tables
rl = 0.331;
xLl = 0.0300;
rm = 0.342;
xam = 0.458;
xdm = 0.1802;
xLm = xam+xdm;
# Calculations
#Voltage Drops for Normal Condition
VDmainn = (Imain/2)*((rm*pf)+(xLm*math.sin(math.radians(math.acos(pf)))))*Lm/2;
VDlateraln = (Ilateral/2)*((rl*pf)+(xLl*math.sin(math.radians(math.acos(pf)))))*Ll/2;
perVDmainn = VDmainn*100/Vb;
perVDlateraln = VDlateraln*100/Vb;
TVDn = perVDmainn+perVDlateraln;
#Voltage Drops for Worst Conditions
VDmainw = (Imain)*((rm*pf)+(xLm*math.sin(math.radians(math.acos(pf)))))*Lm/2;
VDlateralw = (Ilateral)*((rl*pf)+(xLl*math.sin(math.radians(math.acos(pf)))))*Ll;
perVDmainw = VDmainw*100/Vb;
perVDlateralw = VDlateralw*100/Vb;
TVDw = perVDmainw+perVDlateralw;
# Results
print 'a)From Table A5, 300-kcmilACSR conductors, with 500A Ampacity is used for mainand AWG #2 XLPE Al\
URD cable with 168A Ampacity'
print 'b) The Total Voltage Drop in Percent for Normal Operation is %g percent'%(TVDn)
print 'c) The Total Voltage Drop in Percent for Worst Condition is %g percent'%(TVDw)
print 'd The Voltage drop is met for Normal operation and NOT for emergency operation'
```