Chapter 7 : Voltage Drop and Power Loss Calculations¶

Example 7.2 Page No : 327¶

In :
import math

# Variables
Vll = 416.;         #Voltage Line to Line
Vph = Vll/(math.sqrt(3));         #Phase Voltage and Base Voltage
Ia = 30.;
Ib = 20.;
Ic = 50.;

pfa = 1.;
pfb = 0.5;
pfc = 0.9;

#Impedances of the Sections
ZA = 0.05+(1j*0.01);
ZAB = 0.1+(1j*0.02);
ZBC = 0.05+(1j*0.05);
#Impedance upto the point of load
ZB = ZA+ZAB;
ZC = ZB+ZBC;

# Calculations
#Function to Calculate Voltage Drop
def VD(a,b,c):
return a*((b.real*c)+(b.imag*math.sin(math.acos(c))))

#Voltage Drops at A,B and C
VDA = VD(Ia,ZA,pfa);
VDB = VD(Ib,ZB,pfb);
VDC = VD(Ic,ZC,pfc);

TVD = VDA+VDB+VDC;         #Total Voltage Drop

TVDpu = TVD/Vph;         # In Per Unit

def Real(y,z):
return  Vph*y*z         #Function to Calculate Real Power
def Reactive(y,z):
return  Vph*y*math.sin(math.acos(z))         #Funtion to Calculate the Reactive power

#Real Powers
Pa = Real(Ia,pfa);
Pb = Real(Ib,pfb);
Pc = Real(Ic,pfc);
P = Pa+Pb+Pc;         #Total Real Power

#Reactive Powers
Qa = Reactive(Ia,pfa);
Qb = Reactive(Ib,pfb);
Qc = Reactive(Ic,pfc);
Q = Qa+Qb+Qc;         #Total Reactive Power

S = math.sqrt((P**2)+(Q**2));         #Total output from the Transformer
PF = P/S;         #Load Power Factor

# Results
print 'a) The Total Voltage drop is %g pu'%(TVDpu)
print 'b) The Real Power per Phase is %g kW'%(P/1000)
print 'c) The Reactive Power per Phase is %g kVAr'%(Q/1000)
print 'd) The Kilovoltampere output and load factor is %g kVA and %g lagging'%(S/1000,PF)
a) The Total Voltage drop is 0.0593859 pu
b) The Real Power per Phase is 20.4151 kW
c) The Reactive Power per Phase is 9.39455 kVAr
d) The Kilovoltampere output and load factor is 22.473 kVA and 0.90843 lagging

Example 7.3 Page No : 329¶

In :
import math

# Variables
pf = 0.9;         #Power Factor
Vb = 120.;         #Base Voltage
#From The Tables
r = 0.334;         #Resismath.tance per thousand feet
x = 0.0299;         #Reacmath.tance per thousand feet
K1 = 0.02613;         #Voltage Drop

#Assumed Cable
I = 100.;         #Secodary line Current
Ls = 100.;         #Length of Secondary line in feet

# Calculations
R = r*Ls/1000;         # Resismath.tance Value for a 100 feet Line
X = x*Ls/1000;         # Reacmath.tance Value for a 100 feet Line

VDpu = VD/Vb;         #Per unit value

# Results
print 'The Cable Selected is of 100 feet, carrying 100A and cable size         #2 AWG'
print 'The Voltage drop for the above cable is %g pu V'%(VDpu)
print 'The Above Value is Close to the Value%g pu V) in the table given.'%(K1)
The Cable Selected is of 100 feet, carrying 100A and cable size         #2 AWG
The Voltage drop for the above cable is 0.0250696 pu V
The Above Value is Close to the Value0.02613 pu V) in the table given.

Example 7.4 Page No : 333¶

In :
import math

# Variables
Sts = (10+(11*4.4));         #Load Selected on the transformer
V = 240.;         #Voltage
Sta = 50.;         #Available Unit
pf = 0.9;         #Load Power Factor
I = (Sts/V)/(Sta/V);

# Calculations
def VD(a,b,c):
return a*b*c/(10**4)         #Function to find Voltage Drop Per unit

VDSL = VD(0.0088,116.7,150);
VDSD = VD(0.01683,41.76,70);

TVD = VDT+VDSL+VDSD;

Is = 80.;
Smotor = Is*V/1000;
pf1 = 0.5;

VDIPSL = VD(0.00636,80,150);
VDIPSD = VD(0.01089,80,70);
TVDIP = VDIPT+VDIPSL+VDIPSD;

VDSL1 = VD(0.00769,116.7,150);
VDSD1 = VD(0.0136,41.6,70);
TVD1 = VDT+VDSL1+VDSD1;

# Results
print 'a The Voltage drops are:'
print 'Transformer is %g pu V'%(VDT)
print 'Secondary Lines is %g pu V'%(VDSL)
print 'Service Drops is %g pu V'%(VDSD)
print 'Total is %g pu V'%(TVD)
print 'The Above Value doesn''t meet the required criterion'
print 'b The Voltage dip for motor starting are:'
print 'Transformer is %g pu V'%(VDIPT)
print 'Secondary Lines is %g pu V'%(VDIPSL)
print 'Service Drops is %g pu V'%(VDIPSD)
print 'Total is %g pu V'%(TVDIP)
print 'The Above Value does meet the required criterion'
print 'C The Voltage drops after changing the conductors are:'
print 'Transformer is %g pu V'%(VDT)
print 'Secondary Lines is %g pu V'%(VDSL1)
print 'Service Drops is %g pu V'%(VDSD1)
print 'Total is %g pu V'%(TVD1)
print 'The Above Value doesn''t meet the required criterion'
print 'Thus 350 kcmilcable size for the secondary lines and        \
#2/0 AWG cable size for service drops to meet the criteria'
a The Voltage drops are:
Transformer is 0.0113756 pu V
Secondary Lines is 0.0154044 pu V
Service Drops is 0.00491975 pu V
Total is 0.0316998 pu V
The Above Value doesnt meet the required criterion
b The Voltage dip for motor starting are:
Transformer is 0.00215195 pu V
Secondary Lines is 0.007632 pu V
Service Drops is 0.0060984 pu V
Total is 0.0158824 pu V
The Above Value does meet the required criterion
C The Voltage drops after changing the conductors are:
Transformer is 0.0113756 pu V
Secondary Lines is 0.0134613 pu V
Service Drops is 0.00396032 pu V
Total is 0.0287973 pu V
The Above Value doesnt meet the required criterion
Thus 350 kcmilcable size for the secondary lines and         #2/0 AWG cable size for service drops to meet the criteria

Example 7.5 Page No : 336¶

In :
import math

# Variables
#Length in feet
Lsd = 100.;         #Service Drop Line
Lsl = 200.;         #Secondary Line

SB = 75.;         #Transformer Capacity in kVA
pf = 0.9;         #Load Power Factor

#From the Tables
ISL = 129.17;         #Secondary Line Current
ISD = 41.67;         #Service Drop Current
KSD = 0.01683;         #Service Drop Cable Consmath.tant
KSL = 0.0136;         #Secondary Cable Consmath.tant

#for Transformer
R = 0.0101;         #Resismath.tance in per unit
X = 0.0143;         #Reamath.tance in per unit

# Calculations
#From the Diagram
ST = (3.+2+8+6)+(5+6+7+4)+(6+7+8+10);         #Total Load on transformer

STpu = ST/SB;         #In Per unit;

#The Above value also corresponds to the Current as Well

I = STpu;         #Current in Per Unit

VDT = I*((R*pf)+(X*math.sin(math.radians(math.acos(pf)))));         #Voltage Drop in the Transformer
VDSL = KSL*(Lsl*ISL/(10**4));         #Secondary Line
VDSD = KSD*(Lsd*ISD/(10**4));         #Service Drop Line

VD = VDT+VDSD+VDSL;         #Total Voltage Drop

# Results
print 'a)The Load in transformer is %g kVA or %g pu'%(ST,STpu)
print 'b) The Total Voltage Drop is %g pu V'%(VD)
a)The Load in transformer is 72 kVA or 0.96 pu
b) The Total Voltage Drop is 0.0509818 pu V

Example 7.7 Page No : 340¶

In :
import math
from numpy import exp

# Variables
An = 4.;         #Service Area
l = 1.;         #Length of 0a
#Voltages in kV
Vll = 13.2;         #Line to line
Vln = 7.62;         #Line to neutral

Sl = 2000.;         #Lumped Load in kVA

VB = 7620.;         #Base Voltage

Vs = 1.025;         #Substation Voltages

Sn = Dp*An;         #Load Connected to the square shaped service area

K = 0.0003;         #Cable Consmath.tant

# Calculations
VD0a = K*Sm*l;         #Voltage Drop between substation and a
lab = 2;          #Dismath.tance from a to b
VDab = (K*Sn*lab/2)+(K*Sl*lab);        #Voltage drop from b to c
lbc = 2;         #Dismath.tance from b to c
VDbc = 3*K*Sl*lbc;         #Voltage drop from b to c         #Change in Consmath.tant

I = Sl/(math.sqrt(3)*(0.947*Vll));
Ib = Sl/(math.sqrt(3)*(Vll));        #BAse Current

MIpu = I/Ib;         #Per Unit Current

Ztpu = complex(0,0.05);
pf = 0.9;         #Load Power Factor

Ipu = MIpu*exp(1j*math.pi*math.acos(pf)/180);

#The Voltage has been tapped up 5 percent

VDcd = puVDcd*100;
def volt(a,b):
return  (a-(b/100.))         #funtion to find out voltages

#per unit Voltages
puVa = volt(Vs,VD0a);
puVb = volt(puVa,VDab);
puVc = volt(puVb,VDbc);
puVd = volt(puVc,VDcd);

#Voltages in V
Va = puVa*VB;
Vb = puVb*VB;
Vc = puVc*VB;
Vd = puVd*VB;

# Results
print 'a The Percentage drops are'
print ' Substation to a is %g percent'%(VD0a);
print ' a to b is %g percent'%(VDab);
print ' b to c is %g percent'%(VDbc);
print ' c to d is %g percent'%(VDcd);
print 'b The Per unit voltages are:'
print ' Point a is %g pu V'%(puVa)
print ' Point b is %g pu V'%(puVb)
print ' Point c is %g pu V'%(puVc)
print ' Point d is %g pu V'%(puVd)
print 'c The Line to Neutral voltages are:'
print ' Point a is %g  V'%(Va)
print ' Point b is %g  V'%(Vb)
print ' Point c is %g  V'%(Vc)
print ' Point d is %g  V'%(Vd)
a The Percentage drops are
Substation to a is 1.8 percent
a to b is 2.4 percent
b to c is 3.6 percent
c to d is -4.95844 percent
b The Per unit voltages are:
Point a is 1.007 pu V
Point b is 0.983 pu V
Point c is 0.947 pu V
Point d is 0.996584 pu V
c The Line to Neutral voltages are:
Point a is 7673.34  V
Point b is 7490.46  V
Point c is 7216.14  V
Point d is 7593.97  V

Example 7.8 Page No : 340¶

In :
import math
#To determine the percent drop from the substation to various points
#Page 340

# Variables
V0op = 1.;         #At off Peak
An = 4.;         #Service Area
l = 1.;         #Length of 0a
#Voltages in kV
Vll = 13.2;         #Line to line
Vln = 7.62;         #Line to neutral

Sl = 2000.;         #Lumped Load in kVA

VB = 7620.;         #Base Voltage

Vs = 1.025;         #Substation Voltages

Sn = Dop*An;         #Load Connected to the square shaped service area

K = 0.0003;         #Cable Consmath.tant

# Calculations
VD0a = K*Sm*l;         #Voltage Drop between substation and a
lab = 2.;          #Dismath.tance from a to b
VDab = (K*Sn*lab/2)+(K*Sl*lab);        #Voltage drop from b to c
lbc = 2.;         #Dismath.tance from b to c
VDbc = 3*K*Sl*lbc;         #Voltage drop from b to c         #Change in Consmath.tant

I = Sl/(math.sqrt(3)*(0.947*Vll));
Ib = Sl/(math.sqrt(3)*(Vll));        #BAse Current

MIpu = I/Ib;         #Per Unit Current

Ztpu = complex(0,0.05);
pf = 0.9;         #Load Power Factor

Ipu = MIpu*exp(1j*math.pi*math.acos(pf)/180);

#The Voltage has been tapped up 5 percent

VDcd = puVDcd*100;
def volt(a,b):
return (a-(b/100))        #funtion to find out voltages

#per unit Voltages
puVa = volt(V0op,VD0a);
puVb = volt(puVa,VDab);
puVc = volt(puVb,VDbc);
puVd = volt(puVc,VDcd);

#Voltages in V
Va = puVa*VB;
Vb = puVb*VB;
Vc = puVc*VB;
Vd = puVd*VB;

# Results
print 'a The Percentage drops are'
print ' Substation to a is %g percent'%(VD0a);
print ' a to b is %g percent'%(VDab);
print ' b to c is %g percent'%(VDbc);
print ' c to d is %g percent'%(VDcd);
print 'b The Per unit voltages are:'
print ' Point a is %g pu V'%(puVa)
print ' Point b is %g pu V'%(puVb)
print ' Point c is %g pu V'%(puVc)
print ' Point d is %g pu V'%(puVd)
print 'c The Line to Neutral voltages are:'
print ' Point a is %g  V'%(Va)
print ' Point b is %g  V'%(Vb)
print ' Point c is %g  V'%(Vc)
print ' Point d is %g  V'%(Vd)
a The Percentage drops are
Substation to a is 0.9996 percent
a to b is 1.5996 percent
b to c is 3.6 percent
c to d is -4.95844 percent
b The Per unit voltages are:
Point a is 0.990004 pu V
Point b is 0.974008 pu V
Point c is 0.938008 pu V
Point d is 0.987592 pu V
c The Line to Neutral voltages are:
Point a is 7543.83  V
Point b is 7421.94  V
Point c is 7147.62  V
Point d is 7525.45  V

Example 7.9 Page No : 344¶

In :
import math

# Variables
Vll = 13.2;         #Voltage in kV (Line voltage)
TCDi = 0.45;         #Load Density in kVA per feet
FD = 1.08;         #Diversity Factor for all loads
FLS = 0.2;         #Annual Loss Factor
DFi = 0.6;         #30 min Annual Demand Factor

Dg = TCDi*DFi/FD;         #Divesified Maximum Demand of the Group

L = 30000;         #Length of the Whole Feeder in Feet

#To Achieve Minimum Voltage Drop, The Substation must be located at the centre of the line
Ln = 15000.;         #NEW Length of the Feeder

# Calculations
SPK = Dg*Ln;         #Peak Load on each main of the substation trnasformer
I = (SPK/(math.sqrt(3)*Vll));         #Current in the Line

K = 0.0009;         #For the Assumed Conductor
VD = K*SPK*Ln/(2*5280);         #Voltage Drop, Is divided by 5280, to convert the length in miles

# Results
print 'a) To Achieve Minimum Voltage Drop, The Substation is Placed at the Centre of the Line,\
and For a Current of %g A following in it,         #4 AWG or         #2 AWG ACSR conductors are used'%(I)
print 'b) For a         #4 AWG Copper Conductor, The Percentage Voltage drop at annual peak load is %g pecent'%(VD)

#Calculation Mistake in the TextBook
a) To Achieve Minimum Voltage Drop, The Substation is Placed at the Centre of the Line, and For a Current of 164.02 A following in it,         #4 AWG or         #2 AWG ACSR conductors are used
b) For a         #4 AWG Copper Conductor, The Percentage Voltage drop at annual peak load is 4.79403 pecent

Example 7.10 Page No : 346¶

In :
# Variables
Vll = 13.2;         #Voltage in kV (Line voltage)
TCDi = 0.45;         #Load Density in kVA per feet
FD = 1.08;         #Diversity Factor for all loads
FLS = 0.2;         #Annual Loss Factor
DFi = 0.6;         #30 min Annual Demand Factor

# Calculations
Dg = TCDi*DFi/FD;         #Divesified Maximum Demand of the Group
L = 30000;         #Length of the Whole Feeder in Feet
I = 164.2;         #Current

r = 0.592;         #Resismath.tance Per Mile
R = r*L/(3.*5280);         #Total Resismath.tance

CL = 3*(I**2)*R;         #Total Power Loss in entire length

TAEL = CL*FLS*8760/(10**3);         #Total Annual Energy Loss

# Results
print 'The Total Annual Eddy Current Loss is %g kWhr'%(TAEL)
The Total Annual Eddy Current Loss is 158887 kWhr

Example 7.11 Page No : 347¶

In :
import math

# Variables
Sbc = 3000.;         #Load Along bc
Sd = 2000.;         #Load At Point d
Sab = Sbc/2;         #Load along ab

#Cable Consmath.tants
K0a = 0.0005;         #For 0 to a
Kab = 0.0010;         #For a to b
Kac = 0.0010;         #For a to c
Kad = 0.0010;         #For a to d

#Length
l0a = 1.0;         #From 0 to a
lab = 2.;        #From a to b
lad = 2.;         #From a to d

# Calculations
#Voltage Drops At Specific Points
VDa = K0a*S0a*l0a;
VDb = (Kab*Sab*lab/2)+VDa;
VDc = VDb;

#To determine the Voltages at Point a
Vll = 12650.;         #Line to Line Voltage
Vln = 7300.;         #Line to Neutral Voltage

Valn = Vln-(VDa*Vln/100);
Vall = Vll-(VDa*Vll/100);

# Results
print 'a The Voltage Drops at:'
print 'Point a is %g percent'%(VDa)
print 'Point b is %g percent'%(VDb)
print 'Point c is %g percent'%(VDc)
print 'Point d is %g percent'%(VDd)
print 'b) The Voltages VaL-N is %g V and VaL-L is %g V'%(Valn,Vall)
a The Voltage Drops at:
Point a is 2.5 percent
Point b is 4 percent
Point c is 4 percent
Point d is 6.5 percent
b) The Voltages VaL-N is 7117.5 V and VaL-L is 12333.8 V