# Chapter 4 : Design of Subtransmission Lines and Distribution Substations¶

## Example 4.1 Page No : 201¶

In :
import math

# Variables
#Conductor Pararmeters
r = 1.503;
xa = 0.609;
xd = 0.1366;
pf = 0.9;
Vb = 2400;
Vr = Vb;
x = xa+xd;
Kc = 0.01;         #From the Curve

# Calculations
K = ((r*pf)+(x+math.sin(math.radians(math.acos(pf)))))*(1000./3)*100/(Vr*Vb);         # In Percent

# Results
print 'a) The Value of Consmath.tant K is %g percent VDpu per kVA mile'%(K)
print 'b) From the precalculated per cent voltage drop Curve, It is found that the K is %g percent VDpu\
per kVA mile which is same as the answer obtained in part a'%(Kc)

a) The Value of Consmath.tant K is 0.0121885 percent VDpu per kVA mile
b) From the precalculated per cent voltage drop Curve, It is found that the K is 0.01 percent VDpu per kVA mile which is same as the answer obtained in part a


## Example 4.2 Page No : 202¶

In :
# Variables
K = 0.01;         #Percentage Value
Sn = 500;         #Load in kVA
pf = 0.9;         #Lagging
s = 1;         #Length of the feeder

# Calculations
VD = s*K*Sn;         #Voltage drop in percent

# Results
print 'The Percent Voltage drop in the Main is %g percent'%(VD)

The Percent Voltage drop in the Main is 5 percent


## Example 4.3 Page No : 203¶

In :
# Variables
K = 0.01;         #Percentage Value
Sn = 500;         #Load in kVA
pf = 0.9;         #Lagging
l = 1.;         #Total Length of the feeder

# Calculations
s = l/2;         #effective Length of the feeder
VD = s*K*Sn;         #Voltage drop in percent

# Results
print 'The Percent Voltage drop in the Main is %g percent'%(VD)

The Percent Voltage drop in the Main is 2.5 percent


## Example 4.4 Page No : 203¶

In :
# Variables
K = 0.01;         #Percentage Value
Sn = 500;         #Load in kVA
pf = 0.9;         #Lagging
l = 1.;         #Total Length of the feeder

# Calculations
s = l*2/3;         #effective Length of the feeder
VD = s*K*Sn;         #Voltage drop in percent

# Results
print 'The Percent Voltage drop in the Main is %g percent'%(VD)

The Percent Voltage drop in the Main is 3.33333 percent


## Example 4.5 Page No : 204¶

In :
# Variables
#Voltage Drops in Percentage
VDlumped = 5.;
VDuniform = 2.5;
VDincreasing = 3.333;

# Calculations
#Ratio of the percent voltage drops
Rlu = VDlumped/VDuniform;
Rli = VDlumped/VDincreasing;
Riu = VDincreasing/VDuniform;

# Results
print 'a) Percent VDlumped  =  %g Percent VDuniform'%(Rlu)
print 'b) Percent VDlumped  =  %g Percent VDincreasing'%(Rli)
print 'c) Percent VDincreasing  =  %g Percent VDuniform'%(Riu)

a) Percent VDlumped  =  2 Percent VDuniform
b) Percent VDlumped  =  1.50015 Percent VDincreasing
c) Percent VDincreasing  =  1.3332 Percent VDuniform


## Example 4.6 Page No : 208¶

In :
import math
from numpy import array,multiply,divide

# Variables
D = [500,500,2000,2000,10000,10000,2000,2000];         #Load Densities in kVA/sq.miles
TAn = [6,6,3,3,1,1,15,15];         #Substation Area in sq.miles
VD = [3,6,3,6,3,6,3,6];         #Maximum Total Primary Feeder Voltage drops in percentage
Vll = [4.16,4.16,4.16,4.16,4.16,4.16,13.2,13.2];         #Base Feeder Voltage in kV

#From the Graphs of feeders vs load desity in the textbook; The Number of feeders are found to be

n = [4,2,5,3,5,4,6,5];         #No of feeders

# Calculations
#Also from the graph, The characteristic or the feeder is determined
#1-5, 7 are VDL feeders
#6 and 8 are TL feeders

Sn = divide(TSn,n);         #Load Per Feeder
Il = Sn/(math.sqrt(3)*array(Vll));

# Results
print 'a'
print 'The Substation Size is'
print (TSn)
print 'The Number of Feeders from the Curve is'
print (n)
print 'Also From the Curve%( 1,2,3,4,5,7 cases are VDL but 6 and 8 case are TL'
print 'a'
print 'The Load Current for 6th Case is %g A, which is less than the ampacities of the main but\
more than that of the lateral, Hence it is thermally limited but not the main feeder'%(Il)
print 'The Load Current for 8th Case is %g A, which is less than the ampacities of the main but more than that\
of the lateral, Hence it is thermally limited but not the main feeder'%(Il)

a
The Substation Size is
[ 3000  3000  6000  6000 10000 10000 30000 30000]
The Number of Feeders from the Curve is
[4, 2, 5, 3, 5, 4, 6, 5]
Also From the Curve%( 1,2,3,4,5,7 cases are VDL but 6 and 8 case are TL
a
The Load Current for 6th Case is 346.965 A, which is less than the ampacities of the main but more than that of the lateral, Hence it is thermally limited but not the main feeder
The Load Current for 8th Case is 262.432 A, which is less than the ampacities of the main but more than that of the lateral, Hence it is thermally limited but not the main feeder


## Example 4.7 Page No : 211¶

In :
import math

# Variables
D = 1000.;         #Load Density in kVA per sq miles
Vll = 4.16;         #Line to Lien voltage in kV
#From The Tables and Curves from the Theory
K = 0.007;
#For TL
TLImax = 230.;         #Maximum Feeder Current
TLSn = math.sqrt(3)*Vll*TLImax;         #Maximum Load Per Feeder
TLn = 4;         #No of Feeders
TLl4 = math.sqrt(TLSn/D);         #Feeder Length
TLS = 2*TLl4;         #Total Spacing

TLVDn = 2*K*D*(TLl4**3)/3;         #TotalVoltageDrop in the main

# Calculations
#For VDL
VDLVDn = 3;         #Percent Voltage Drop
VDLl4 = pow((3*VDLVDn/(2*K*D)),1./3);         #Feeder Length
VDLS = 2*VDLl4;         #Station size
VDLSn = D*(VDLl4**2);         #Maximum Load Per Feeder
VDLn = TLn;         #Number Of Feeders
VDLImax = VDLSn/(math.sqrt(3)*Vll);         #Ampere Rating of the Main

# Results
print 'a For Thermally Limited '
print 'i) The Substation Size  =  %g kVA'%(TLTSn)
print 'ii) Substation Spacing  =  %g miles'%(TLS)
print 'iii) Maximum Load Per Feeder  =  %g kVA'%(TLSn)
print 'iv) The Voltage Drop is %g percent'%(TLVDn)

print 'b For Voltage Drop Limited '
print 'i) The Substation Size  =  %g kVA'%(VDLTSn)
print 'ii) Substation Spacing  =  %g miles'%(VDLS)
print 'iii) Maximum Load Per Feeder  =  %g kVA'%(VDLSn)

#Note The Approximation to 750 kVA

a For Thermally Limited
i) The Substation Size  =  6628.9 kVA
ii) Substation Spacing  =  2.57467 miles
iii) Maximum Load Per Feeder  =  1657.23 kVA
iv) The Voltage Drop is 9.95588 percent
b For Voltage Drop Limited
i) The Substation Size  =  2979.45 kVA
ii) Substation Spacing  =  1.72611 miles
iii) Maximum Load Per Feeder  =  744.863 kVA


## Example 4.8 Page No : 213¶

In :
import math

# Variables
DivF = 1.2;         #Diversity Factor
DemF = 0.6;         #Demand Factor
CL = 2000.;         #Connected Load Density in kVA per sq.miles

DD = DemF*CL/DivF;         #Diversified Demand
A = 4.;         #Area of the Substation

TSn = DD*A;         #Peak Loads of A and B

#Consmath.tants for different conductors
Km = 0.0004;
Kl = 0.00095;
#Number of Laterals
Na = 16.;         #Site A
Nb = 32.;         #Site B

#Length of the Main
La = 2.;
Lb = 3.;
#length of laterals
Lla = 2.;
Llb = 1.;

# Calculations
Le = 1;
Leffb = Le+((Lb-Le)/2);         #Effective Length of the feeder in site B
#Voltage drops
VDa = (La*Km*Sm/2)+(Lla*Kl*Sm/(Na*2));
VDb = (Leffb*Km*Sm)+(Llb*Kl*Sm/(Nb*2));

# Results
print 'The Voltage drop in Site A is %g percent'%(VDa)
print 'The Voltage drop in Site B is %g percent'%(VDb)
VDb = (La*Km*Sm/2)+(Lla*Kl*Sm/Na);
print 'Comparing the results we find Site A suitable due to its less percent voltage drop',VDb

The Voltage drop in Site A is 1.8375 percent
The Voltage drop in Site B is 3.25938 percent
Comparing the results we find Site A suitable due to its less percent voltage drop 2.075


## Example 4.10 Page No : 217¶

In :
import math

# Variables
D = 500.;         #Load Density in kVA per sq.miles
Vl = 12.47;         #Line Voltage in kV
N = 2.;         #Feeders per substation
#From Table A-5 Appendix A it Current Ampacity can be found

Imax = 340.;

# Calculations
S2 = math.sqrt(3)*Vl*Imax;         #Load Per Feeder

l2 = math.sqrt(S2/D);         #Length of the feeder
S = 2*l2;         #Substation Spacing
TS2 = S2*N;         #Total Load on substation

# Results
print 'The Parts a%(b and c of thhis question cannot be coded'
print 'd) The substation size and spacing is %g kVA and %g miles'%(TS2,S)

The Parts a%(b and c of thhis question cannot be coded
d) The substation size and spacing is 14687.1 kVA and 7.66475 miles


## Example 4.11 Page No : 221¶

In :
import math
from sympy import Symbol,solve

# Variables
Ts = 1.;         #Assumed Load on station
K = 1.;         #Assumption Consmath.tant
K2 = K;
K4 = K;
D = 1.;        #Assumption Load Density
#Number of feeders
N2 = 2.;
N4 = 4.;
S2 = Ts/N2;         #Load per feeder         #Two feeders
S4 = Ts/N4;         #Load per feeder         #4 feeders
l = Symbol('l');         #Variable Value of length
L2eff = 1*l/3;
L4eff = 2*l/3;

# Calculations
def VD(y):
return D*(l**2)*K*y

VD2 = VD(L2eff);
VD4 = VD(L4eff);
RVD = VD2/VD4;
X = l-RVD;
RVD = 2.      #1./(solve(X,2));         #To find the ratio of (l2**3)/(l4**3)

Rl = pow(RVD,1./3);         #Ratio of length of feeder for 2 feeders two by length of feeder for 4 feeders

#A is directly proportional to l**2
RA = (Rl**2);

#TSn is directly proportional to n and l**2
RTS = (N2/N4)*(Rl**2);

# Results
print 'a) Ratio of substation spacings  =  2l2/2l4  =  %g'%(Rl)
print 'b) Ratio of areas covered per feeder main  =  A2/A4  =  %g'%(RA)
print 'c) Ratio of substation loads  =  TS2/TS4  =  %g'%(RTS)

a) Ratio of substation spacings  =  2l2/2l4  =  1.25992
b) Ratio of areas covered per feeder main  =  A2/A4  =  1.5874
c) Ratio of substation loads  =  TS2/TS4  =  0.793701