# Chapter 2 : Load Characteristics¶

## Example 2.1 Page No : 39¶

In [7]:
%matplotlib inline
from numpy import linspace,array
from matplotlib.pyplot import bar,suptitle,xlabel,ylabel

# Variables
t =  linspace(0,24,25)
SL = array([100,100,100,100,100,100,100,100,0,0,0,0,0,0,0,0,0,0,100,100,100,100,100,100,100]);
R = array([200,200,200,200,200,200,200,300,400,500,500,500,500,500,500,500,500,600,700,800,1000,1000,800,600,300]);
C = array([200,200,200,200,200,200,200,200,300,500,1000,1000,1000,1000,1200,1200,1200,1200,800,400,400,400,200,200,200]);

# Calculations
Tl = SL+R+C;

# Results
#To print lay the Load bar curve diagram
bar(t,Tl,color='red')#,0.5,'red')
suptitle('Example 2.1')
xlabel("Time in hrs")

Out[7]:
<matplotlib.text.Text at 0x1098963d0>

## Example 2.2 Page No : 46¶

In [9]:
# Variables
Fls = 0.15;
Ppl = 80*(10**3);         #Power Loss at peak load.

# Calculations
Avgpl = Fls*Ppl;         #Average Power Loss
TAELCu = Avgpl*8760;         #Total annual energy loss

# Results
print 'a) The average annual power loss  =  %g kW'%(Avgpl/1000)
print ' b) The total annual energy loss due to the copper losses of the feeder circuits  =  %g kWh'%(TAELCu/1000)

a) The average annual power loss  =  12 kW
b) The total annual energy loss due to the copper losses of the feeder circuits  =  105120 kWh


## Example 2.3 Page No : 47¶

In [10]:
# Variables
TCDi = [9,9,9,9,9,9];         #Load for each house all in kilowatt
DFi = 0.65;         #Demand factor
Fd = 1.1;         #Diversity factor

# Calculations
Dg = sum(TCDi)*DFi/Fd;

# Results
print 'The diversified demand of the group on the distribution transformer is %g kW'%(Dg)

The diversified demand of the group on the distribution transformer is 31.9091 kW


## Example 2.4 Page No : 48¶

In [11]:
# Variables
SP = 3000*(10**3);         #System peak in kVA per phase
Cl = 0.5/100;         #Percentage of copper loss

# Calculations
I2R =  Cl*SP;         #Copper loss of the feeder per phase
I2R3 = 3*I2R;         #Copper losses of the feeder per 3 phase

# Results
print 'a) The copper loss of the feeder per phase  =  %g kW'%(I2R/1000)
print ' b) The total coper losses of the feeder per three phase  =  %g kW'%(I2R3/1000)

a) The copper loss of the feeder per phase  =  15 kW
b) The total coper losses of the feeder per three phase  =  45 kW


## Example 2.5 Page No : 48¶

In [12]:
# Variables
Pi = 2000.*(10**3);         #Peak for industrial load
Pr = 2000.*(10**3);         #Peak for residential load
Dg = 3000.*(10**3);         #System peak load as specified in the diagram
P = [Pi,Pr];         #System peaks for various loads

# Calculations
Fd =  sum(P)/Dg;         #Diversity factor
LD =  sum(P)-Dg;         #Load diversity factor
Fc = 1/Fd;         # Coincidence factor

# Results
print 'a) The diversity factor of the load is %g'%(Fd)
print ' b) The load diversity of the load is %g kW'%(LD/1000)
print ' c) The coincidence factor of the load is %g'%(Fc)

a) The diversity factor of the load is 1.33333
c) The coincidence factor of the load is 0.75


## Example 2.6 Page No : 50¶

In [14]:
from numpy import array,transpose

# Variables
#Refer diagram of the first example of this chapter
Ps = 100.;         #Peak load for street lighting in kW
Pr = 1000.;         #Peak load for Residential in kW
Pc = 1200.;         #Peak Commercial load in kW
P = array([Ps,Pr,Pc])         #Peaks of various loads

Ls5 = 0.;         #Street lighting load at 5.00 PM in kW
Lr5 = 600.;         #Residential load at 5.00 PM in kW
Lc5 = 1200.;         #Commercial Load at 5.00 PM in kW

# Calculations
Cstreet = Ls5/Ps;
Cresidential = Lr5/Pr;
Ccommercial = Lc5/Pc;
C = array([Cstreet,Cresidential,Ccommercial]);         #Class distribution for various factors

Fd = (sum(P))/(sum(P* transpose(C)));
Dg = (sum(P* transpose(C)));
Fc = 1/Fd;

print 'a The class distribution factors factor of:'
print ' i) Street lighting  =  %g \
\nii) Residential  =  %g \
\niii) Commercial  = %g'%(C[0],C[1],C[2])
print ' b) The diversity factor for the primary feeder  =  %g'%(Fd)
print ' c) The diversified maximum demand of the load group  =  %g kW'%(Dg)
print ' d) The coincidence factor of the load group  =  %g'%(Fc)

a The class distribution factors factor of:
i) Street lighting  =  0
ii) Residential  =  0.6
iii) Commercial  = 1
b) The diversity factor for the primary feeder  =  1.27778
c) The diversified maximum demand of the load group  =  1800 kW
d) The coincidence factor of the load group  =  0.782609


## Example 2.7 Page No : 55¶

In [15]:
print 'Assuming a monthly load curve as shown in the figure attached\
to this code'

# Variables
TAE = 10.**7;         # Total annual energy in kW
APL = 3500.;         #Annual peak load in kW

# Calculations
Pav =  TAE/8760;         #Annual average power demand
Fld =  Pav/APL;         #Annual load factor

# Results
print 'a) The annual power demand is %g kW'%(Pav)
print 'b) The annual load factor is %g'%(Fld)
print 'The unsold energy, as shown in the figure is a measure of capacity and investment math.cost. Ideally it should be kept at a minimum'

Assuming a monthly load curve as shown in the figure attached to this code
a) The annual power demand is 1141.55 kW
b) The annual load factor is 0.326158
The unsold energy, as shown in the figure is a measure of capacity and investment math.cost. Ideally it should be kept at a minimum


## Example 2.8 Page No : 57¶

In [16]:
print 'Assuming a monthly load curve as shown in the figure attached to this code'

# Variables
Nl = 100.;         #100% percent load to be supplied
TAE = 10.**7;         # Total annual energy in kW
APL = 3500.;         #Annual peak load in kW

# Calculations
Pav =  TAE/8760;         #Annual average power demand
Fld =  (Pav+Nl)/(APL+Nl);         #Annual load factor
Cr = 3;        #Capacity math.cost
Er = 0.03;        #Energy math.cost
ACC = Nl*12*Cr;         #Additional capacity math.cost per kWh
AEC = Nl*8760*Er;         #Additional energy math.cost per kWh
TAC = ACC+AEC;         #Total annual math.cost

# Results
print 'a) The new annual load factor on the substation is %g'%(Fld)
print 'b) The total annual additional costs to NL&NP to serve this load is $%g'%(TAC)  Assuming a monthly load curve as shown in the figure attached to this code a) The new annual load factor on the substation is 0.344876 b) The total annual additional costs to NL&NP to serve this load is$29880


## Example 2.9 Page No : 58¶

In [18]:
# Variables
TAE = 5.61*(10**6);         #Total annual energy in kW
APL = 2000.;         #Annual peak load in kW
Lc = 0.03;         #Cost of energy per kWh in dollars
Plp = 100.;         #Power at peak load in kW

# Calculations
Fld = TAE/(APL*8760);
Fls =  (0.3*Fld)+(0.7*(Fld**2));
AvgEL = Fls*Plp;         #Average energy loss
AEL = AvgEL*8760;         #Annual energy loss
Tlc = AEL*Lc;         #Cost of total annual copper loss

# Results
print 'a) The annual loss factor is %g'%(Fls)
print ' b) The annual copper loss energy is %g kWh and the cost of total annual copper loss is $%g'%(AEL,Tlc)  a) The annual loss factor is 0.167834 b) The annual copper loss energy is 147022 kWh and the cost of total annual copper loss is$4410.67


## Example 2.10 Page No : 59¶

In [27]:
import math
from sympy import Symbol,solve

Fd = 1.15;
Pi = [1800,2000,2200];         #Demands of various feeders in kW (Real Power)
PF = [0.95,0.85,0.90];         #Power factor of the respective feeders
Dg = sum(Pi)/Fd;
P = Dg;
theta = arccos(PF);

S = math.sqrt((P**2)+(Q**2));
LD = sum(Pi)-Dg;

#Transformer sizes
Tp = array([2500,3750,5000,7500]);
Ts = array([3125,4687,6250,9375]);

Ol = 1.25;         #Maximum overload condition
Eol = Ts*Ol;         #Overload voltages of the transformer
Ed = abs(Eol-S);         # Difference between the overload values of the transformers and the P value of the system

A = sorted(Ed);         # To sort the differences and choose the best match

T = [Tp[1],Ts[1]];         #Suitable transformer

g = Symbol('g');
X = (1+g)-pow(2,1./10);         #To find out the fans on rating
R = solve(X)[0];
g = R*100;

n = Symbol('n');
Sn = 9375.;         # Rating of the to be installed transformer
# Equation (1+g)**n * S  =  Sn
# a = (1+g)
# b = Sn/S

a = 1+R;
b = Sn/S;
n = math.log(b)/math.log(a);

print 'a) The 30 mins annual maximum deman on the substation transformer are %g kW and %g kVA respectively'%(P,S)
print ' b) The load diversity is %g kW'%(LD)
print ' c) Suitable transformer size for 25 percent short time over loads is %g/%g kVA'%(T[0],T[1])
print ' d) Fans on rating is %g percent and it will loaded for %g more year if a 7599/9375 kVA transformer is installed'%(g,math.ceil(n))

a) The 30 mins annual maximum deman on the substation transformer are 5217.39 kW and 5217.53 kVA respectively
b) The load diversity is 782.609 kW
c) Suitable transformer size for 25 percent short time over loads is 3750/4687 kVA
d) Fans on rating is 7.17735 percent and it will loaded for 9 more year if a 7599/9375 kVA transformer is installed


## Example 2.11 Page No : 62¶

In [34]:
from numpy import array

print 'NOTE'
print 'The figure 1 attached along with this code is the Maximum diversified 30- min demand characteristics\
of various residential loads;  A  =  Clothes dryer; D  =  range; E  =  lighting and miscellaneous appliances; \
J  =  refrigerator Only the loads required for this problem have been mentioned '

Ndt = 50.;         #Number of distribution transformers
Nr = 900.;         #Number of residences

PavMax6 = array([1.6,0.8,0.066,0.61]);         #Average Maximum diversified demands (in kW) per house for dryer, range, refrigerator, for lighting and misc aapliances respectively according to the figure 1 attached with code.

Mddt =  sum(6*PavMax6);         #30 min maximum diversified demand on the distribution transformer

PavMax900 = array([1.2,0.53,0.044,0.52]);         #        #Average Maximum diversified demands (in kW) per house for dryer, range, refreigerato, for lighting and misc aapliances respectively according to the figure 1 attached with code.

Mdf = sum(Nr*PavMax900);         #30 min maximum diversified demand on the feeder

#From the figure 2 attached to this code
Hdd4 = array([0.38,0.24,0.9,0.32]);         #Hourly variation factor at time 4 PM for dryer, range, refrigerator, lighting and misc appliances
Hdd5 = array([0.30,0.80,0.9,0.70]);         #Hourly variation factor at time 5 PM for dryer, range, refrigerator, lighting and misc appliances
Hdd6 = array([0.22,1.0,0.9,0.92]);         #Hourly variation factor at time 6 PM for dryer, range, refrigerator, lighting and misc appliances

Thdd4 = (6*PavMax6)*(Hdd4.T);         #Gives the total hourly diversified demand in kW at time 4 PM
Thdd5 = (6*PavMax6)*Hdd5;         #Gives the total hourly diversified demand in kW at time 5 PM
Thdd6 = (6*PavMax6)*Hdd6;         #Gives the total hourly diversified demand in kW at time 6 PM

print ' a) The 30 min maximum diversified demand on the distribution transformer  =  %g kW'%(Mddt)
print ' b) The 30 min maximum diversified demand on the distribution transformer  =  %g kW'%(Mdf)
print ' c The total hourly diversified demands at:'
print ' i) 4.00 PM  =  %g kW'%(sum(Thdd4))
print ' ii) 5.00 PM  =  %g kW'%(sum(Thdd5))
print ' iii) 6.00 PM  =  %g kW'%(sum(Thdd6))

NOTE
The figure 1 attached along with this code is the Maximum diversified 30- min demand characteristicsof various residential loads;  A  =  Clothes dryer; D  =  range; E  =  lighting and miscellaneous appliances; J  =  refrigerator Only the loads required for this problem have been mentioned
a) The 30 min maximum diversified demand on the distribution transformer  =  18.456 kW
b) The 30 min maximum diversified demand on the distribution transformer  =  2064.6 kW
c The total hourly diversified demands at:
i) 4.00 PM  =  6.3276 kW
ii) 5.00 PM  =  9.6384 kW
iii) 6.00 PM  =  10.6356 kW


## Example 2.12 Page No : 72¶

In [40]:
import math

T = 730;         #Average monthly time in hrs
Pla = 22;         #Peak Load for consumer A in kW
Plb = 39;         #Peak load for consumer B in kW
MEC = array([0.025,0.02,0.015]);         #Monthly Energy charges in cents/kWh according to the units consumed
Uc = array([1000,3000,3000]);         #Units consumption according to the Energy charges
MDC = 2;         #Monthly demand charge in dollars/kW

Pa = 7000.;         #Units served to Consumer A in kWh
Pb = 7000.;         #Units served to Consumer B in kWh

#Power factors
Pfa = 0.9;         # Lag
Pfb = 0.76;         #Lag

Flda = Pa/(Pla*T);
Fldb = Pb/(Plb*T);

#Continous kilovoltamperes for each distribution transformer
Sa = Pla/Pfa;
Sb = Plb/Pfb;

#Ratings of the distribution transformers needed
Ta = round(Sa/5)*5;
Tb = round(Sb/5)*5;

#Billing Charges
#For Consumer A
Mbda = Pla*(0.85/Pfa);         # Monthly billing demand
Mdca = Mbda*MDC;         #Monthly demand charge
#Since the units served are 7000 it can be split according to the rates  as 1000, 3000, 3000 excess units.
Uca = Uc;         #Units consumption by A
Meca = MEC*Uca.T;         #Monthly energy charge
Tmba = Meca+Mdca;         #Total monthly bill

#For Consumer B
Mbdb = Plb*(0.85/Pfb);         # Monthly billing demand
Mdcb = Mbdb*MDC;         #Monthly demand charge
#Since the units served are 7000 it can be split according to the rates  as 1000, 3000, 3000 excess units.
Ucb = Uc;         #Units consumption by B
Mecb = MEC*Ucb.T;         #Monthly energy charge
Tmbb = Mecb+Mdcb;         #Total monthly bill

#To find the capacitor size
Q1 = Pb*math.radians(arctan(math.acos(Pfb)));         #For original power factor
Q2 = Pb*math.radians(arctan(math.acos(0.85)));         #For new power factor

dQ = (Q1-Q2)/T;         #Capacitor size

#For new power factor
#For Consumer B
Mbdbn = Plb*(1);         # Monthly billing demand
Mdcbn = Mbdbn*MDC;         #Monthly demand charge
#Since the units served are 7000 it can be split according to the rates  as 1000, 3000, 3000 excess units.
Ucbn = Uc;         #Units consumption by B
Mecbn = MEC*Ucbn.T;         #Monthly energy charge
Tmbbn = Mecbn+Mdcbn;         #Total monthly bill

Saving = abs(Tmbbn-Tmbb);         #Saving due to capacitor installation
Ci = 30;         # Cost of capacitor in dollar per kVAr
Cc = Ci*dQ;         #The math.cost of the installed capacitor
PP = Cc/Saving;         #Payback Period
PPr = 90/Saving;         #Realistic Payback period

print 'a Monthly load factor for :'
print ' i) Consumer A  =  %g'%(Flda)
print ' ii) Consumer B  = %g'%(Fldb)
print ' b Rating of the each of the distribution transformer:'
print ' i) A  =  %g kVA'%(Ta)
print ' ii) B  =  %g kVA'%(Tb)
print ' c Monthly bil for:'
print ' i) Consumer A  =  $%g'%(sum(Tmba)) print ' ii) Consumer B =$%g'%(sum(Tmbb))
print ' d) The capacitor size required is %g kVAr'%(dQ)
print ' e Payback period:'
print ' i) Calculated : %g months'%(sum(ceil(PP)))
print ' ii) Realistic as capacitor size available is 3 kVAr : %g months'%(sum(ceil(PPr)))

a Monthly load factor for :
i) Consumer A  =  0.435866
ii) Consumer B  = 0.245873
b Rating of the each of the distribution transformer:
i) A  =  25 kVA
ii) B  =  50 kVA
c Monthly bil for:
i) Consumer A  =  $254.667 ii) Consumer B =$391.711
d) The capacitor size required is 0.018276 kVAr
e Payback period:
i) Calculated : 3 months
ii) Realistic as capacitor size available is 3 kVAr : 30 months


## Example 2.13 Page No : 84¶

In [41]:
# Variables
Kh = 7.2;         #Meter consmath.tant
Kr1 = 32;         #Revolutions of the disk in the first reading
Kr2 = 27;         #Revolutions of the disk in the second reading
T1 = 59;         #Time interval for revolutions of disks for the first reading
T2 = 40;         #Time interval for revolutions of disks for the second reading

# Calculations
# Self contained watthour meter; D  =  (3.6*Kr*Kh)/T

def  Id1(a,b):
return ((3.6*Kh*a)/b)         #Function to calculate the instaneous demand

D1 = Id1(Kr1,T1);
D2 = Id1(Kr2,T2);
Dav = (D1+D2)/2;

# Results
print 'The instantenous demands are %g kW and %g kW for reading 1 and 2 and the average demand is %g kW'%(D1,D2,Dav)

The instantenous demands are 14.0583 kW and 17.496 kW for reading 1 and 2 and the average demand is 15.7772 kW


## Example 2.14 Page No : 84¶

In [42]:
# Variables
#For a transformer type watthour meter; D  =  (3.6*Kr*Kh*CTR*PTR)/T
CTR = 200.;
PTR = 1.;
Kh = 1.8;
Kr1 = 32.;         #Revolutions of the disk in the first reading
Kr2 = 27.;         #Revolutions of the disk in the second reading
T1 = 59.;         #Time interval for revolutions of disks for the first reading
T2 = 40.;         #Time interval for revolutions of disks for the second reading

# Calculations
def Id1(a,b):
return ((3.6*Kh*a*CTR*PTR)/b)         #Function to calculate the instaneous demand

D1 = Id1(Kr1,T1);
D2 = Id1(Kr2,T2);
Dav = (D1+D2)/2;

# Results
print 'The instantenous demands are %g kW and %g kW for reading 1 and 2 and the average demand is %g kW'%(D1,D2,Dav)

The instantenous demands are 702.915 kW and 874.8 kW for reading 1 and 2 and the average demand is 788.858 kW


## Example 2.15 Page No : 85¶

In [43]:
# Variables
Kh = 1.2;
CTR = 80.;
PTR = 20.;
#Revolutions of the disk in a watthour meter for 1 and 2 readings respectively
Kr1 = 20.;
Kr2 = 30.;
#Revolutions of the disk in a VArhour meter for 1 and 2 readings respectively
Kr3 = 10.;
Kr4 = 20.
#Time interval for revoltion of disks in watthour meter for 1 and 2 readings respectively
T1 = 50.;
T2 = 60.;
#Time interval for revoltion of disks in VArhour meter for 1 and 2 readings respectively
T3 = 50.;
T4 = 60.;

def Id1(a,b):
return ((3.6*Kh*a*CTR*PTR)/b)         #Function to calculate the instaneous demand

#instanmath.taneous kilowatt demands for readings 1 and 2
D1 = Id1(Kr1,T1);
D2 = Id1(Kr2,T2);

#instanmath.taneous kilovar deamnds for readings 1 and 2
D3 = Id1(Kr3,T3);
D4 = Id1(Kr4,T4);

Davp = (D1+D2)/2;         #Average kilowatt demand
Davq = (D3+D4)/2;         #Average kilovar demand

Dav = math.sqrt((Davp**2)+(Davq**2));         #Average kilovoltampere demand

# Results
print 'a) The instanmath.taneous kilowatt hour demands for readings 1 and 2 are %g kW and %g kW respectively'%(D1,D2)
print ' b) The average kilowatt demand is %g kW'%(Davp)
print ' c) The instanmath.taneous kilovar hour demands for readings 1 and 2 are %g kVAr and %g kVAr respectively'%(D3,D4)
print ' d) The average kilovar demand is %g kVAr'%(Davq)
print ' e) The average kilovoltampere demand is %g kVA'%(Dav)

a) The instanmath.taneous kilowatt hour demands for readings 1 and 2 are 2764.8 kW and 3456 kW respectively
b) The average kilowatt demand is 3110.4 kW
c) The instanmath.taneous kilovar hour demands for readings 1 and 2 are 1382.4 kVAr and 2304 kVAr respectively
d) The average kilovar demand is 1843.2 kVAr
e) The average kilovoltampere demand is 3615.52 kVA