# Chapter 8 : Application of Capacitors to Distribution Systems¶

## Example 8.1 Page No : 390¶

In :
import math

# Variables
SL = 700.;         #Load in kVA
pf1 = 65./100;         #Power Factor
PL = SL*pf1;         #Real Power
#From the Table of Power Factor Correction
CR = 0.74;         #Co-relation factor
CS = PL*CR;         #Capacitor Size

CSr = 360.;         #Next Higher standard Capacitor Size

# Calculations
CRn = CSr/PL;         #New Co-Relation Factor

#From the table by linear interpolation
pfr = 93.;         #In Percentage
pfn = pfr+(172./320);

# Results
print 'a) The Correction Factor is %g'%(CR)
print 'b) The Capacitor Size Required is %g kVAr'%(CS)
print 'c) Resulting power factor if the next higher standard capacitor size is used is %g percent'%(pfn)

a) The Correction Factor is 0.74
b) The Capacitor Size Required is 336.7 kVAr
c) Resulting power factor if the next higher standard capacitor size is used is 93.5375 percent


## Example 8.2 Page No : 393¶

In :
import math

# Variables
Vll = 4.16;         #Line to Line Voltage in kV
Pr = (500*0.7457);         #Rating of motor in kW
pf1 = 0.75;         #Initial Power Factor
pfn = 0.9;         #Improved Power Factor
eff = 0.88;         #Efficiency
P = Pr/eff;         #Input Power of Induction Motor

# Calculations
Q1 = P*math.degrees(math.atan(math.acos(pf1)));         #Reactive Power
Q2 = P*math.degrees(math.atan(math.acos(pfn)));         #REactive power of motor after power factor improvement
f = 60.;         #Frequency of supply
w = 2.*math.pi*f;         #Angular Frequency
Qc = Q1-Q2;         #Reactive Power of Capacitor
Il = Qc/(math.sqrt(3)*Vll);

#Capacitor Connectd in Delta
Ic1 = Il/(math.sqrt(3));
Xc1 = Vll*1000/Ic1;         #Reacmath.tance of each capacitor
C1 = (10**6)/(w*Xc1);         #Capacimath.tance in Micro Farad

#Capacitor Connected in Wye
Ic2 = Il;
Xc2 = Vll*1000/(math.sqrt(3)*Ic2);         #Reacmath.tance of each capacitor
C2 = (10**6)/(w*Xc2);         #Capacimath.tance in Micro Farad

# Results
print 'a) Rating of Capacitor Bank is %g kVAr'%(Qc)
print 'b) The Value of Capacimath.tance if there are connected in delta is %g micro F'%(C1)
print 'c) The Value of Capacimath.tance if there are connected in wye is %g micro F'%(C2)

a) Rating of Capacitor Bank is 4906.48 kVAr
b) The Value of Capacimath.tance if there are connected in delta is 250.687 micro F
c) The Value of Capacimath.tance if there are connected in wye is 752.06 micro F


## Example 8.3 Page No : 396¶

In :
import math

# Variables
V = 2.4;         #Voltage in kV
P = 360;         #Real Load in kW
S1 = V*I;         #Total Load in kVA
pf1 = P/S1;         #Power Factor

# Calculations
Qc = 300;         #Capacitor Size

Q2 = Q1-Qc;         #The New Reactive Load
pf2 = P/math.sqrt((P**2)+((Q1-Qc)**2));         #Improved Power Factor

# Results
print 'a) The Uncorrected power factor and reactive load is %g and %g kVAr'%(pf1,Q1)
print 'b) The New Corrected factor after the introduction of capacitor unit is %g'%(pf2)

a) The Uncorrected power factor and reactive load is 0.75 and 6.0546 kVAr
b) The New Corrected factor after the introduction of capacitor unit is 0.77459


## Example 8.4 Page No : 398¶

In :
import math

# Variables
S1 = 7800.;         #Peak Load in kVA
T = 3*2000.;         #Total Rating of the Transformer
pf1 = 0.89;         #Load Power Factor
TC = 120./100;         #Thermal Capability
Qc = 1000.;         #Size of capacitor

# Calculations

Q2 = Q1-Qc;         #The New Reactive Load
pf2 = P/math.sqrt((P**2)+((Q1-Qc)**2));         #Improved Power Factor

S2 = P/pf2;         #Corrected Apprarent power

ST = T*TC;         #Transformer Capabilty

pf3 = P/ST;         #New Corrected Power factor required

Q2new = P*math.degrees(math.atan(math.acos(pf3)));         #Required Reactive Power

# Results
print 'a) Since the Apparent Power%g kVAr is more than Transformer Capability %g kVAr),\

a) Since the Apparent Power7004.76 kVAr is more than Transformer Capability 7200 kVAr), Hence Additional Capacitors are required
b) The Rating of the Addtional capacitor is -105274 kVAr


## Example 8.5 Page No : 411¶

In :
import math
from numpy import array

# Variables
# 1 is Total Loss Reduction due to Capacitors
# 2 is Additional Loss Reduction due to Capacitor
# 3 is Total Demand Reduction due to capacitor
# 4 is Total required capacitor additions

C90 = array([495165,85771,22506007,9810141]);         #Characteristics at 90% Power Factor
C98 = array([491738,75343,21172616,4213297]);         #Characteristics at 98% Power Factor

#Responsibility Factors
RF90 = 1;
RF98 = 0.9;

SLF = 0.17;         #System Loss Factor
FCR = 0.2;         #Fixed Charge Rate
DC = 250;         #Demand Cost
ACC = 4.75;         #Average Capacitor Cost
EC = 0.045;         #Energy Cost
Cd = C90-C98;         #Difference in Characteristics

# Calculations
TAS = Cd+Cd;         #Total Additional Kilowatt Savings

ASDR1 = Cd*RF90*DC*FCR;
ASDR2 = Cd*RF98*DC*FCR;
TASDR = ASDR1+ASDR2;         #Total Annual Savings due to demand
x = 27;         # Cost for Per kVA of the capacity
TASTC = Cd*FCR*x;         #Annual Savings due to Transmission Capacity
TASEL = TAS*SLF*EC*8760;         #Savings due to energy loss reduction
TACAC = Cd*FCR*ACC;         #Annual Cost of Additional Capacitors
Savings = TASEL+TASDR+TASTC;         #Total Savings

# Results
print 'a) The Resulting additional savings in kilowatt losses due to power factor improvement at the\
substation buses is %g kW'%(Cd)
print 'b) The Resulting assitional savings in kilowatt  losses due to the power factor improvement for feeders is %g kW'%(Cd)
print 'c) The Additional Kilowatt Savings is %g kW'%(TAS)
print 'd) The Additional savings in the system kilovoltampere capacity is %g kVA'%(Cd)
print 'e) The Additional Capacitors required are %g kVAr'%(Cd)
print 'f) The Annual Savings in demand reduction due to capacitors applied to distribution substation buses\
is approximately is %g dollars/year'%(TASDR)
print 'g) The Annual Savings due to the additional released transmission capacity is %g dollars/year'%(TASTC)
print 'h) The Total Anuual Savings due to the energy loss reduction is %g dollars/year'%(TASEL)
print 'i) The Total Annual Cost of the additional capacitors is %g dollars/year'%(TACAC)
print 'j) The Total Annual Savings is %g dollars/year'%(Savings)
print 'k) No Since the total net annual savings is not zero'

a) The Resulting additional savings in kilowatt losses due to power factor improvement at the substation buses is 3427 kW
b) The Resulting assitional savings in kilowatt  losses due to the power factor improvement for feeders is 10428 kW
c) The Additional Kilowatt Savings is 13855 kW
d) The Additional savings in the system kilovoltampere capacity is 1.33339e+06 kVA
e) The Additional Capacitors required are 5.59684e+06 kVAr
f) The Annual Savings in demand reduction due to capacitors applied to distribution substation buses is approximately is 640610 dollars/year
g) The Annual Savings due to the additional released transmission capacity is 7.20031e+06 dollars/year
h) The Total Anuual Savings due to the energy loss reduction is 928479 dollars/year
i) The Total Annual Cost of the additional capacitors is 5.317e+06 dollars/year
j) The Total Annual Savings is 8.7694e+06 dollars/year
k) No Since the total net annual savings is not zero