# Chapter 5 : Design Considerations of Primary Systems¶

## Example 5.1 Page No : 254¶

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))

a) Vr is 0.79784/_0 pu, del is -38.3625 degrees, Ir  =  Is  =  1.56673/_-0.643501 pu
b) Vr is 0.841577/_-10.4293 pu, which is almost equal to the previous case.


## Example 5.2 Page No : 259¶

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'

a The percent voltage drops at :
i 3Phase
Lateral End is 4.23818 percent
Main End is 6.475 percent
ii 1Phase
Lateral End is 16.9527 percent
Main End is 6.475 percent
b) Conductors with Ampacities of 480A and 270A are used to find the Percent voltage drop of the Main and Lateral as 1.9425 percent and 1.69527 percent respectively
The Above Drops meet the required criterion of 4 percent voltage drop


## Example 5.3 Page No : 263¶

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'

a The percent voltage drops at :
Lateral End is 6.08569 percent
Main End is 4.35998 percent
b) Conductors with Ampacities of 278A and 174A are used to find the Percent voltage drop of the Main and Lateral as 1.75389 percent and 1.51958 percent respectively
The Above Drops meet the required criterion of 4 percent voltage drop


## Example 5.4 Page No : 265¶

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)

a The percent voltage drops at :
Lateral End is 0.494455 percent
Main End is 1.9425 percent
b The Above Drops meet the required criterion of 4 percent voltage drop
c) The Difference in Voltage drop is 0.0435921 ohm/mile, which is a smaller VD valuue that it really is.


## Example 5.5 Page No : 268¶

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'

a)From Table A5, 300-kcmilACSR conductors, with 500A Ampacity is used for mainand AWG         #2 XLPE Al URD cable with 168A Ampacity
b) The Total Voltage Drop in Percent for Normal Operation is 1.1836 percent
c) The Total Voltage Drop in Percent for Worst Condition is 4.08313 percent
d The Voltage drop is met for Normal operation and NOT for emergency operation