# Chapter 9 : Distribution System Voltage Regulation¶

## Example 9.1 Page No : 468¶

In :
import math
from numpy import exp

# Variables
#Base Value
S3phib = 15;         #in MVA
Vllst = 69;         #in kV
Vllp = 13.2;         #in kV
Vrrb = 120.;

Ztpu = 1j*0.08;         #Transformer impedance per unit length
VSTpuop = 1.05*exp(1j*0);         #Per Unit Maximum Subtransmission Voltage Off Peak
VSTpup = 1.00*exp(1j*0);         #Per Unit Maximum Subtransmission Voltage Peak
pftop = 0.95;         #Off Peak kilovoltageamperage power factor
Sop = 0.25;         #Off Peak kilovoltageamperage
pftp = 0.85;         #Off Peak kilovoltageamperage power factor
Sp = 1.0;         #Off Peak kilovoltageamperage
Regpu = 5./(8*100);         #Regulated percent volts for 32 steps
K = 3.88*(10**-6);         #Drop Consmath.tant
S = 4000.;         # Peak Load in kVA
l = 10.;         #Length of the feeder
#Case A
Rset = 0.;
Xset = 0.;
Vpmax = 1.0417;
BW = 0.0083;
VRRpu = (Vpmax-BW);         #Setting of VRR
VRR = (Vpmax-BW)*Vrrb;

# Calculations
#Case B
IPpuop = (Sop/VSTpuop)*exp(1j*math.acos(pftop)*math.pi/180);         #No Load Primary Current at substation Off Peak
VPpuop  = VSTpuop-(IPpuop*Ztpu);         #Highest Allowable Primary Voltage Off Peak
IPpup = (Sp/VSTpup)*exp(-1*1j*math.acos(pftp)*math.pi/180);         #No Load Primary Current at substation Peak
VPpup  = VSTpup-(IPpup*Ztpu);         #Highest Allowable Primary Voltage  Peak

Step1 = (abs(VPpuop)-VRRpu)/Regpu;         #Off Peak Number Steps
#To find the positive step value
Step2 = -1*(abs(VPpup)-VRRpu)/Regpu;         # Peak Number Steps

#Case C
MaxVpp = 1.075;         #Max
MinVpp = 1.000;         #Min

TVDpu = K*S*l/2;         #Total Voltage Drop
MinVPpu = VRRpu-TVDpu;

#At Annual Peak Load Primary Voltages
APMaxVPpu = MaxVpp-BW;         #Max
APMinVPpu = MinVpp+BW;         #Min

NLMaxVPpu = Vpmax-BW;         #Max
NLMinVPpu = APMinVPpu;         #Min

# Results
print 'a)The Setting of the VRR for the highest allowable primary voltage is %g V'%(VRR)
print 'b The Maximum Number of Steps of buck and boost for:'
print 'Off Peak : %g steps'%(math.ceil(Step1))
print 'Peak : %g steps'%(math.ceil(Step2))
print 'c) At Annual Load%( Significant Values on Voltage Curve'
print 'The Total Voltage Drop is %g pu V'%(TVDpu)
print 'The Minimum Feeder Voltage at the end of the feeder is %g'%(MinVPpu)
print 'The Minimum and Maximum Primary Voltages at Peak Load is %g pu V and %g pu V'%(APMaxVPpu,APMinVPpu)
print 'The Minimum and Maximum Primary Voltages at Peak Load is %g pu V and %g pu V'%(NLMaxVPpu,NLMinVPpu)

a)The Setting of the VRR for the highest allowable primary voltage is 124.008 V
b The Maximum Number of Steps of buck and boost for:
Off Peak : 3 steps
Peak : 5 steps
c) At Annual Load%( Significant Values on Voltage Curve
The Total Voltage Drop is 0.0776 pu V
The Minimum Feeder Voltage at the end of the feeder is 0.9558
The Minimum and Maximum Primary Voltages at Peak Load is 1.0667 pu V and 1.0083 pu V
The Minimum and Maximum Primary Voltages at Peak Load is 1.0334 pu V and 1.0083 pu V


## Example 9.2 Page No : 472¶

In :
import math
from sympy import Symbol,solve

# Variables
#Terms from previous example
TVDpu = 0.0776;         #Total Voltage Drop
VRRpu = 1.035;         #Setting Voltage of Regulator
l = 10.;        #Length of the Feeder

#Primary voltages for various cases
VPpua = 1.01;
VPpub = 1.00;

s1 = Symbol('s1');         #Variable Value of Regulator length
#Function to find the equation for the regulator dismath.tance
def dist(y):
return (s1*(2-(s1/l))/l)-((VRRpu-y)/TVDpu)

# Calculations
#Different Cases
Xa = dist(VPpua);
Xb = dist(VPpub);

s1a = solve(Xa);
if((abs(l-s1a)+(l-s1a)) == 0):
s1a = s1a;
else:
s1a = s1a;

s1b = solve(Xb);
if((abs(l-s1b)+(l-s1b)) == 0):
s1b = s1b;
else:
s1b = s1b;

# Results
print 'a) The Regulator must be placed at %g miles from the start of the feeder'%(s1a)
print 'b) The Regulator must be placed at %g miles from the start of the feeder'%(s1b)
print 'c The Advantage of a over b is that it can compensate for future growth'

a) The Regulator must be placed at 1.76693 miles from the start of the feeder
b) The Regulator must be placed at 2.59076 miles from the start of the feeder
c The Advantage of a over b is that it can compensate for future growth


## Example 9.3 Page No : 473¶

In :
import math

# Variables
l = 10.;         #Length of the feeder
S3phi = 4000.;         #Annual Peak Load in kVA
VPpu = 1.01;         #Primary Feeder Voltage
s1 = 1.75;         # Dismath.tance of the Regulator
Rmax = 10./100;         #Regulation Percent

# Calculations
S = S3phi*(1-(s1/l));         #Uniformly Distributed three phase load
Sph = S/3;         #Single Phase Load

Sreg = Rmax*Sph;         #Regulated Size

# Results
print 'The Calculated Circuit Kilovoltampere Size is %g kVA, And The corresponding Minimum kilovoltampere size \
of the regulator size can be found as 114.3 kVA'%(Sreg)

The Calculated Circuit Kilovoltampere Size is 110 kVA, And The corresponding Minimum kilovoltampere size of the regulator size can be found as 114.3 kVA


## Example 9.4 Page No : 474¶

In :
import math
#To specify the best settings for regulation
#Page 474

# Variables
s1 = 1.75;         #As Found in Example 2
VRRpu = 1.035;         #As R and X are zero, the Settings turn out to produce this

#Parameters of Distribution
K = 3.88*(10**-6);
S = 3300.;
l = 10.;         #length of the line

# Calculations
VDpu = K*S*(l-s1)/2;         #Per unit voltage drop

VP = VRRpu-VDpu;         #Primary Feeder Voltage

#We Obtain VDs  =  K*(S3-((S3*s)/l))s+K(S*s/l)s/2;
#We take various values of s and carry out the computation and hence form the table 9-4 given given in the result file

#We Obtain from the voltage drop value for any give point s between the substation and the regulator  station as
#VDs = I(r.math.cos(theta)+ del math.sin(theta))s*(1-(s/(2*l)))

#We finally Get VDs  =  3.88 * (10**-6) * (3300-(3300s/8.25))s+3.88*(10**-6)*(3300s/8.25)*s/2

# Results
#Again for different values of s we get the table 9-5
print 'a)The Best Settings for LDCs R and X, and for the VRR'
print 'The best settings for LDC of the regulator are when settings for both R and X are zero and VRRpu  =  %g pu V'%(VRRpu)
print 'b)The Voltage Drop occuring in the feeder portion between the regulating point and the end of the feeder is %g pu V'%(VDpu)
print 'The Result Files give the Profiles and relevant information about the solution'

a)The Best Settings for LDCs R and X, and for the VRR
The best settings for LDC of the regulator are when settings for both R and X are zero and VRRpu  =  1.035 pu V
b)The Voltage Drop occuring in the feeder portion between the regulating point and the end of the feeder is 0.0528165 pu V
The Result Files give the Profiles and relevant information about the solution


## Example 9.5 Page No : 478¶

In :
import math
#To determine the setting of the regulator so that the voltage criteria is met
#Page 478

# Variables
l = 10.;         #Length of the feeder
s1 = 1.75;
ra = 0.386;
xa = 0.4809;
xd = 0.1802;
xL = xa+xd;
Vb = 120;
pf = 0.85;         #Power Factor
S = 1100.;         #Load in kVA
Vln = 7.62;         #line to neutral voltage in kV
Reff = ra*(l-s1)/2;
Xeff = xL*(l-s1)/2;

#Primary Ratings
CTp = 150;         #Current Tranformer
PTn = 63.5;         #POtential Transformer

# Calculations
#R Value of the dial
Rset = (CTp/PTn)*Reff;
Rsetpu = Rset/Vb;

#X value of the dial
Xset = (CTp/PTn)*Xeff;
Xsetpu = Xset/Vb;

VRP = 1.0138;         #Assumption Made on the Regulating Point
#Output Voltage of the Regulator

# Results
print 'a) The Regulating Voltage is %g pu V'%(Vreg)
print 'As per Table 9-6; the primary voltage criteria are met by using the R and X settings'
print 'b The Voltage Profiles are given in the result file attached'

a) The Regulating Voltage is 1.03994 pu V
As per Table 9-6; the primary voltage criteria are met by using the R and X settings
b The Voltage Profiles are given in the result file attached


## Example 9.6 Page No : 480¶

In :
import math

# Variables
#From Problems 4 and 5 the co-effcients are obtained
VRRpu = 1.035;
Vreg4 = 1.0337;
Vreg5 = 1.0666;
VRP4 = 1.0337;
VRP5 = 1.0138;
Vmin = 1.010;         #For s =  1.75

# Calculations
#Steps
Buck4 = (VRRpu-VRP4)/(0.00625);
Buck5 = (VRRpu-VRP5)/(0.00625);
Boost4 = (Vreg4-Vmin)/(0.00625);
Boost5 = (Vreg5-Vmin)/(0.00625);

# Results
print 'a) The Number of steps of buck and number is steps of boost in example 9-4 is %g and %g respectively'%(Buck4,Boost4)
print 'b) The Number of steps of buck and number is steps of boost in example 9-5 is %g and %g respectively'%(Buck5,Boost5)

a) The Number of steps of buck and number is steps of boost in example 9-4 is 0.208 and 3.792 respectively
b) The Number of steps of buck and number is steps of boost in example 9-5 is 3.392 and 9.056 respectively


## Example 9.8 Page No : 482¶

In :
import math

# Variables
l = 3.;        #Length of the line
Vlc = 2450.;         #Regulated Voltage
Vcp = 12800.;         #Primary of customer transformer
#Base Values
Vlbp = 2400.;         #Primary Bus Voltage of Customer's Bus(Low Voltage)
Vlbs = 4160.;        #Secondary Bus Voltage of Customer's Bus
Sb = 5000.;         #Power in kVA
r = 0.3;         #Line Resismath.tance per mile
x = 0.8;         #Line Reacmath.tance per mile
Vhbp = 7390.;        #Primary Bus Voltage of High Voltage Bus
Vhbs = 12800.;        #Secondary Bus Voltage of High Voltage Bus
PTn = 63.5;         #Potential Transformer Turns Ratio
CTp = 250.;        #Current Transformer Turns Ratio
VRP = Vlc/Vlbp;        #Voltage at RP
Vll = Vhbs/1000;         #Line Voltage
VBsec = Vcp/(math.sqrt(3)*PTn);         #Secondary Reading of the Customer Transformer

# Calculations
VRRset = VRP*VBsec;         #Setting of the voltage-setting dial of VRR

Zb = (Vll**2)*1000/Sb;         #Applicable Impedance Base
Ztpu = 0.05*1j;        #Transformer Impedance per unit
Zt = Ztpu*Zb;        #Transformer Impedance

#Effective Resismath.tances and Reacmath.tances
Reff = (r*l)+Zt.real;
Xeff = (x*l)+Zt.imag;

Rset = CTp*Reff/PTn;         #X Dial Setting of LDCs
Xset = CTp*Xeff/PTn;         #X Dial Setting of LDCs

# Results
print 'a) The Necessary Setting of the voltage-setting dial of the VRR of each single phase regulator in use is %g V'%(VRRset)
print 'b) R and X dial settings of LDS is %g ohm and %g ohm respectively'%(Rset,Xset)

a) The Necessary Setting of the voltage-setting dial of the VRR of each single phase regulator in use is 118.804 V
b) R and X dial settings of LDS is 3.54331 ohm and 15.8992 ohm respectively


## Example 9.9 Page No : 484¶

In :
import math
#To Determine the Design Parameters of a Distributed System
#Page 484

# Variables
VPpu = 1.035;         #Primary Feeder Voltage per unit
TVDpu = 0.0776;         #Total Voltage Drop of Feeder
Vll = 13.2;         #Line Voltage in kV
Vlpuqsw = 1;        #New Voltage at the End of the Feeder due to Qsw at annual peak load
XL = 0.661;         #Inductive Reacmath.tance per mile
Pl = 3400;         #Real Power
Ql = 2100;         #Reactive Power
l = 10.;         #Length of the Feeder in Miles
CR = 0.27;         #Total Capacitor Compensation Ratio For the Above Load Factor
QNSW = CR*Ql;         #Required Size of the Nonswitched capacitor bank
s = 2*l/3;         #Loacation of Nonswitched capacitor bank for Optimum Result
VRpu = QNSW*(2*XL*l/3)/(1000*(Vll**2));         #Per Unit Voltage Rise
VDspu = TVDpu*s*(2-(s/l))/l;         #Voltage drop for the uniformaly distributed load

VSpu = VPpu-VDspu;        #Feeder Voltage at 2l/3 dismath.tance

nVSpu = VSpu+VRpu;         #New Voltage Rise when there is a fixed capacitor bank

Vlpu = VPpu-TVDpu;         #When No Capcacitor bank is there, the voltage at the end of the feeder

nVlpu = Vlpu+VRpu;         #When Capcacitor bank is there, the voltage at the end of the feeder
VRpuqsw = Vlpuqsw-nVlpu;         #Required Voltage Rise

Q3phisw = 1000*(Vll**2)*VRpuqsw/(XL*l);         #Required Size of the Capacitor Bank

# Calculations
#Let us assume the 15 single phase standard 50 kVAr Capacitor Units  =  750 kVAr

SQ3phisw = 750;         #Selected Capacitor Bank

RVRlpu = VRpuqsw*SQ3phisw/Q3phisw;         #Resulmath.tant Voltage Rises at dismath.tance l
RVRspu = RVRlpu*s/l;         #Resulmath.tant Voltage Rises at dismath.tance s

#At Peak Load when both the Non-Switched and Switched Capacitor Banks are on

PVspu = nVSpu+RVRspu;         #Voltage at s
PVlpu = nVlpu+RVRlpu;         #Voltage at l

# Results
print 'a) The NSW Capacitor Rating is %g kVAr, Which means 2 100kVAr Capacitor Banks per phase'%(QNSW)
print 'b) Voltage Rise per unit is %g pu V'%(VRpu)
print 'i When the No Capacitor Bank is Installed '
print 'Voltage at %g miles is %g pu V'%(s,VSpu)
print 'Voltage at %g miles is %g pu V'%(l,Vlpu)
print 'ii When the Fixed Capacitor Bank is Installed '
print 'Voltage at %g miles is %g pu V'%(s,nVSpu)
print 'Voltage at %g miles is %g pu V'%(l,nVlpu)
print 'c) At Annual Peak Load, The Size of Capacitor Bank Required is %g'%(Q3phisw)
print 'The Voltage Rise at the Annual Load for the Required Capacitor Bank is %g pu V'%(VRpuqsw)

#Note That The Calculations are highly accurate, Rounding of Terms hasn't be emplyed

a) The NSW Capacitor Rating is 567 kVAr, Which means 2 100kVAr Capacitor Banks per phase
b) Voltage Rise per unit is 0.0143399 pu V
i When the No Capacitor Bank is Installed
Voltage at 6.66667 miles is 0.966022 pu V
Voltage at 10 miles is 0.9574 pu V
ii When the Fixed Capacitor Bank is Installed
Voltage at 6.66667 miles is 0.980362 pu V
Voltage at 10 miles is 0.97174 pu V
c) At Annual Peak Load, The Size of Capacitor Bank Required is 744.939
The Voltage Rise at the Annual Load for the Required Capacitor Bank is 0.0282601 pu V


## Example 9.10 Page No : 488¶

In :
import math
from numpy import exp
#To Determine the proper 3 phase capacitor bank
#Page 488

# Variables
V = 12.8;         #Voltage in kV
xl = 0.8;         #Reacmath.tance per unit length
l = 3;         #Dismath.tance of the line
Xl = xl*l;         #Effective Reacmath.tance of the the Line
pf = 0.8;         #Initial Power Factor
pfn = 0.88;         #New Improved Power Factor
Qcu = 150;         #Capacity of each unit available
XT = 1.6384;         #Reacmath.tance of the transformer

# Calculations
S3phi = 5000*exp(1j*math.pi*math.acos(pf)/180);         #Presently existing Load

QLnew = (S3phi.real)*math.degrees(math.atan(math.acos(pfn)));         #The Required VAr

S3phinew  = math.sqrt(((S3phi.real)**2)+(QLnew**2));         #New Apparent Power

Qc = (S3phi.imag)-QLnew;         #Minimum Size of capacitor bank;

N = math.ceil(Qc/Qcu);         #Number of Units Required
Qcn = N*Qcu;         #Required VAr

XL = Xl+XT;         #Total Reacmath.tance

VRpu = Qcn*XL/(1000*(V**2));         #Voltage Rise Per unit

# Results
print 'The The Voltage Rise found out is %g pu V, which is greater than the voltage rise criterion.Hence %g Capacitor units\
of %g kVAr must be installed'%(VRpu,N,Qcu)

The The Voltage Rise found out is -3.2425 pu V, which is greater than the voltage rise criterion.Hence -877 Capacitor units of 150 kVAr must be installed


## Example 9.11 Page No : 493¶

In :
# Variables
Skva = 6.3*(10**3);         #Starting kVA per HP of the Motor
HPmotor = 10.;         #Power Rating
Vll = 7.2*(10**3);         #Line Voltage
I3phi = 1438.;         #Fault Current

# Calculations
Sstart = Skva*HPmotor;         #Starting kVA
VDIP = 120*Sstart/(I3phi*Vll);         #Voltage Dip

# Results
print 'a) The Voltage dip due to the motor start is %g V'%(VDIP)
print 'b) From the Permissible voltage flicker limit curve%( The Voltage dip of 0.73 Vwith a frequency of\
15 times per hour is in the satisfactory flicker zone and therefore is not objectionable to the immediate customers'

a) The Voltage dip due to the motor start is 0.730181 V
b) From the Permissible voltage flicker limit curve%( The Voltage dip of 0.73 Vwith a frequency of 15 times per hour is in the satisfactory flicker zone and therefore is not objectionable to the immediate customers


## Example 9.12 Page No : 495¶

In :
# Variables
Skva = 5.6*(10**3);         #Starting kVA per HP of the Motor
HPmotor = 100.;         #Power Rating
Vll = 12.47*(10**3);         #Line Voltage
I3phi = 1765.;         #Fault Current

# Calculations
Sstart = Skva*HPmotor;         #Starting kVA
VDIP = 69.36*Sstart/(I3phi*Vll);         #Voltage Dip

# Results
print 'a) The Voltage dip due to the motor start is %g V'%(VDIP)
print 'b) From the Permissible voltage flicker limit curve, The Voltage dip of 1.72 Vwith a frequency of three\
times per hour is in the satisfactory flicker zone and therefore is not objectionable to the immediate customers'

a) The Voltage dip due to the motor start is 1.76476 V
b) From the Permissible voltage flicker limit curve, The Voltage dip of 1.72 Vwith a frequency of three times per hour is in the satisfactory flicker zone and therefore is not objectionable to the immediate customers