# Chapter2, Economics of Generation¶

## Ex1 : Page 73¶

In [1]:
from __future__ import division
#To Determine the Demand and Supply Parameters for 15 bulbs

W=60  #Wattage of the bulb
N=15  #No. of bulbs
Wih=2*(10**3)  #Wattage of immersion heater
Wh=2*(10**3)  #Wattage of heater

#Usage of Bulbs at different time periods
N1=5
N2=10
N3=6

#Time periods for bulbs
T1=2  #6pm - 8pm
T2=2  #8pm - 10pm
T3=2  #10pm - 12pm
#Time Periods for heaters
T4=4  #1pm - 5pm
T5=3  #8pm - 11pm

#CASE 1
MD1=W*N2  #Maximum Demand
DF=MD1*100/CL  #Demand Factor
EC1=(N1*W*T1)+(N2*W*T2)+(N3*W*T3)  #Energy Consumed

#CASE 2
MD2=(W*N2)+Wh  #From 8pm - 10pm
EC2=(T4*Wih)+(T5*Wh)+EC1  #Energy Consumed

print '''i)a) Connected Load is %g W
b) The Maximum Demand is %g W
c) The Demand Factor is %0.2f percent
d) The Daily Load Factor is %g percent'''%(CL,MD1,DF,DLF1)
print 'ii) The Improved Daily Load Factor is %0.2f %%'%DLF2

i)a) Connected Load is 900 W
b) The Maximum Demand is 600 W
c) The Demand Factor is 66.67 percent
d) The Daily Load Factor is 17.5 percent
ii) The Improved Daily Load Factor is 26.47 %


## Ex2 : Page 74¶

In [2]:
#To determine the Demand and supply parameter of four consumers
#Maximum Demands of various users
MD1=2*(10**3)  #9pm
MD2=2*(10**3)  #12 noon
MD3=8*(10**3)  #5pm
MD4=4*(10**3)  #8pm
MDT=MD1+MD2+MD3+MD4  #Sum of all Maximum Demands

#Demands of various users
D1=1.6*(10**3)  #8pm
D2=1*(10**3)  #8pm
D3=5*(10**3)  #8pm

#The Number after the Alphabets represents the Consumer

#Maximum Demand of the System arises at 8.00 PM
MDS = D1+D2+D3+MD4

TDF=MDT/MDS  #Diversity Factor
#Given Values
AL2=500
AL4=1000
LF1=15/100
LF3=25/100
#Calculated Values
AL1=LF1*MD1
AL3=LF3*MD3
LF2=AL2*100/MD2
LF4=AL4*100/MD4

print 'i) The Diversity Factor is %0.3f '%TDF
print ' Consumer 1 : %g kW and %g %%'%(AL1/1000,LF1*100)
print ' Consumer 2 : %g kW and %g %%'%(AL2/1000,LF2)
print ' Consumer 3 : %g kW and %g %%'%(AL3/1000,LF3*100)
print ' Consumer 4 : %g kW and %g %%'%(AL4/1000,LF4)
print 'iii) The Combined Load Factor and The Combined Average Load is %0.2f percent and %0.2f kW respectively'%(LFS,ALS/1000 )

i) The Diversity Factor is 1.379
Consumer 1 : 0.3 kW and 15 %
Consumer 2 : 0.5 kW and 25 %
Consumer 3 : 2 kW and 25 %
Consumer 4 : 1 kW and 25 %
iii) The Combined Load Factor and The Combined Average Load is 32.76 percent and 3.80 kW respectively


## Ex3 : Page 75¶

In [3]:
#To Determine the Yearly Cost of the substation

Teff=95/100  #Transmission Efficiency
Deff=85/100  #Distribution Efficiency
DFT=1.2  #Diversity Factor For Transmission
DFD=1.3  #Diversity Factor For Distribution
MDGS=100*(10**6)  #Maximum Demand of Generating Station
ACCT=2.5*(10**6)  #Annual Capital Charge for Transmission
ACCD=2*(10**6)  #Annual Capital Charge for Distribution
GCC=100  #Generating Cost per KW demand
GCCU=5/100  # Per Unit Cost
#Fixed Charges from Supply to Substation Annually
GFC=GCC*MDGS/1000  #Generating
TFC=ACCT  #Transmission
TotFCS=GFC+TFC  #Total
#Fixed Charges for supply upto Consumer Annually
DFC=ACCD  #Distribution
TotFCC=TotFCS+DFC  #Total

AMDS= DFT*MDGS/1000  #Aggregate of Maximum Demand at Supply
AMDC= DFD*AMDS  #Aggregate of Maximum Demand for Consumers

FCS=TotFCS/AMDS  #Fixed Charges Per KW at substation
CES=GCCU/Teff  #Cost of energy at the substation

FCC=TotFCC/AMDC  #Fixed Charges per KW at the consumer premises
CEC=CES/Deff  #Cost of Energy at the consumer premises

print 'The Yealy Cost per KW demand and the cost per KWhr at:'
print 'a) The substation is %0.2f rupees per KW and %0.2f paise per KWhr'%(FCS,CES*100)
print 'b) The consumer premises is %g rupees per KW and %0.1f paise per KWhr'%(FCC,CEC*100)

The Yealy Cost per KW demand and the cost per KWhr at:
a) The substation is 104.17 rupees per KW and 5.26 paise per KWhr
b) The consumer premises is 92.9487 rupees per KW and 6.2 paise per KWhr


## Ex4 : Page 78¶

In [4]:
#To determine the Load factor and suitable units for 24 hr operation of the plant

#Demands at Various Time Periods starting from 12PM to 12PM
D1=500*(10**3)
D2=800*(10**3)
D3=2000*(10**3)
D4=1000*(10**3)
D5=2500*(10**3)
D6=2000*(10**3)
D7=1500*(10**3)
D8=1000*(10**3)

MD=D5  #Maximum Demand
#Time Periods of demands ststing from 12PM
T1=5
T2=5
T3=2
T4=2
T5=3
T6=3
T7=2
T8=2

#Total Energy Demand in 24hrs
TED=(T1*D1)+(T2*D2)+(T3*D3)+(D4*T4)+(T5*D5)+(D6*T6)+(D7*T7)+(T8*D8)

LF=TED*100/(24*MD)

print '''Since Maximum Demand is 2500 kW,
2 units 0f 1000W and one unit of 500W is required.
Also for continuity of supply, A reserve of 1000W unit is required.'''

C1000=3*1000*(10**3)  #1000 unit
C500=1*500*(10**3)  #500 Unit

TCP=C1000+C500  #Total capacity of the plant
PCF=TED*100/(24*TCP)  #Plant Capacity Factor

#Operating Schedule, Units operated can be seen in the textbook
G1=500*(10**3)
G2=1000*(10**3)
G3=2000*(10**3)
G4=1000*(10**3)
G5=2500*(10**3)
G6=2000*(10**3)
G7=1500*(10**3)
G8=1000*(10**3)

TEG=(T1*G1)+(T2*G2)+(T3*G3)+(G4*T4)+(T5*G5)+(G6*T6)+(G7*T7)+(T8*G8) #Total Energy Generated
PUF=TED*100/(TEG)  #Plant Use Factor

print 'a) The Reserve Capacity is a 1000kW Unit and Load Factor is %0.2f percent'%LF
print 'b) The Plant Capacity Factor is %0.2f percent'%PCF
print 'c) The Plant Use Factor is %0.2f percent'%PUF

Since Maximum Demand is 2500 kW,
2 units 0f 1000W and one unit of 500W is required.
Also for continuity of supply, A reserve of 1000W unit is required.
a) The Reserve Capacity is a 1000kW Unit and Load Factor is 51.67 percent
b) The Plant Capacity Factor is 36.90 percent
c) The Plant Use Factor is 96.88 percent


## Ex5 : Page 80¶

In [5]:
#To determine the Plant use factore of each unit

MDS=25*(10**6)  #Maximum Demand on the System
U1=15*(10**6)  #Load Supplied By Unit 1
U2=12.5*(10**6)  #Load Supplied By Unit 2
#Running Time Factor of the Unit
T1=1
T2=40/100

#Energy generated by each unit
E1=1*(10**8)
E2=1*(10**7)
Et=E1+E2  #Total Energy

#Maximum Demands on Each Units
MD1=U1
MD2=MDS-U1

#Annual Load Factor for the Units
ALF1=E1*1000*100/(MD1*8760)
ALF2=E2*1000*100/(MD2*8760)

#Since Unit runs all through the year without any reserve, Load Factor ,Plant Use Factor and Plant Capacity Factor are the same
PUF1=ALF1  #Plant Use Factor
PCF1=ALF1  # Plant Capacity Factor

PCF2=E2*1000*100/(U2*8760)  #Plant Capacity Factor for Unit 2
PUF2=E2*1000*100/(U2*0.4*8760) #Plant Use Factor for Unit 2

LFP=Et*100*1000/(MDS*8760)  #Annual Load Factor of the Complete Plant

print 'The Load Factor, Plant Capacity Factor, Plant Use Factor of: '
print 'Unit 1 : %g percent, %g percent, %0.2f %%'%(ALF1,PCF1,PUF1)
print 'Unit 2 : %g percent, %g percent, %0.2f %%'%(ALF2,PCF2,PUF2)
print 'The Annual Load Factor of the Entire Plant is %0.2f %%'%(LFP)

The Load Factor, Plant Capacity Factor, Plant Use Factor of:
Unit 1 : 76.1035 percent, 76.1035 percent, 76.10 %
Unit 2 : 11.4155 percent, 9.13242 percent, 22.83 %
The Annual Load Factor of the Entire Plant is 50.23 %


## Ex6 : Page 82¶

In [6]:
#To determine the Overall cost per kWhr

# C1 =(100,000 rupees + 100 rupees/kW + 6 paise /kWhr) #Base Load Station
# C2 =(80,000 rupees + 60 rupees/kW + 8 paise /kWhr) #Peak Load Station

MaxD=15*(10**6)
MinD=5*(10**6)

BLS=lambda b,c:100000+(b*100)+((6/100)*c)  #Function to Find Annual Cost of Base Load Station.

PLS = lambda b,c:80000+(b*60)+((8/100)*c)  #Function to Find Annual Cost of Peak Load Station.

a1=100
a2=60
b1=6/100
b2=8/100

Tpeak=(a1-a2)/(b2-b1)  #Number of hours the peak plant should operate
#From the straight line annual load duration curve, Maximum Demand at Peak Load Station can be calculated
MDP=Tpeak*(MaxD-MinD)/8760

TotEG=(MaxD+MinD)*8760/(2*1000)  #Total Energy Generated
EGP=MDP*Tpeak/(2*1000)  #Energy Generated by Peak Load
EGB=TotEG-EGP  #Energy Generated by Base Load

MDB=MaxD-MDP  #Maximum Demand at the Base Load

C1=BLS((MDB/1000),EGB)
C2=PLS((MDP/1000),EGP)

TotC=C1+C2  #Total Cost of both the plants
CE=TotC*100/TotEG  #Cost of energy in paise per kWhr

print 'The Operating Scedule of Peak Load Station for Minimum Annual Cost is %g hours'%Tpeak
print 'The Overall Cost per kWhr is %0.1f paise'%CE

The Operating Scedule of Peak Load Station for Minimum Annual Cost is 2000 hours
The Overall Cost per kWhr is 7.9 paise


## Ex7 : Page 90¶

In [7]:
#To determine the amount saved to replace the equipment

Pc=80000  #Plant Cost
UL= 15  #Useful Life of the Plant
SVE=5000  #Salvage Value of the Equipment
r=5/100  #Compound Interest Rate

A1=(Pc-SVE)/UL  #Annual Amount to be saved using straight line method
A2=(Pc-SVE)*r*100/(100*(((1+r)**UL)-1))# Annual Amount to be saaved using Sink Fund Method

print 'i) The Amount to be Saved Annually according to straight line method is %g Rupees'%(A1)
print 'ii) The Amount to be Saved Annually according to sink fund method is %g Rupees'%(A2)

i) The Amount to be Saved Annually according to straight line method is 5000 Rupees
ii) The Amount to be Saved Annually according to sink fund method is 3475.67 Rupees


## Ex8 : Page 90¶

In [8]:
#To Obatin a two part tariff for the consumers

EG=390*(10**6)  #Energy Generated in kWhr
MD=130*(10**6)  #Maximum Demand of the Supply
SCeff=90/100  #The Amount of energy transferred from Substation to Consumer
#Total Cost for Each Division
Fuel=5*(10**6)
Gen=2.4*(10**6)
Trans=5*(10**6)
Dist=3.4*(10**6)
Tot=Fuel+Gen+Trans+Dist  #Total Cost

Runcost = lambda y,z:(y*z/100) #Function to Find out the Running Costs
#Running Costs
Fuelr=Runcost(90,Fuel)
Genr=Runcost(10,Gen)
Transr=Runcost(5,Trans)
Distr=Runcost(7,Dist)
Totr=Fuelr+Genr+Transr+Distr  #Total Cost

FixCost=Tot-Totr # Fixed Cost
FixChar=FixCost*1000/MD  #Fixed Charges per KW
EnChar=Totr*100/(EG*SCeff)  #Energy Charges in Paise for Consumer
OverCost=Tot*100/(EG*SCeff)  #Overall Energy Charges

LF=40/100  #Load Factor Raised to 40%
EG1=LF*MD*8760/1000   #Energy Generated for Different Load Factor
Totr1=Totr*(EG1/EG)  #Cost of Energy Generated
Tot1=FixCost+Totr1  #Total Cost for the New Load Factor
OverCost1=Tot1*100/(EG1*SCeff)  #Overall Energy Charges

Saving=(OverCost-OverCost1)*100/OverCost  #Percentage Saving in the Overall Cost per kWhr

print 'The Fixed Charges is %0.2f rupees per kW'%FixChar
print 'The Energy Charges for the Consumer is %0.3f paise per kWhr '%(EnChar)
print '''IF the Load Factor is raised to 40percent of the Same Maximum Demand,
then the percentage saving in the overall costs is %0.2f percent'''%Saving

The Fixed Charges is 81.32 rupees per kW
The Energy Charges for the Consumer is 1.489 paise per kWhr
IF the Load Factor is raised to 40percent of the Same Maximum Demand,
then the percentage saving in the overall costs is 9.62 percent


## Ex9: Page 91¶

In [9]:
import numpy as np
#To determine the most economic power factor

P=200*(10**3)  #Maximum Demand
pf=0.707  #Power Factor Lagging

a=100  #Tariff per kVA per year

b=200  #Power factor improvement cost Per kVA.
r=20  #Interest Depriciation, maintenance and cost of losses amount to  20% of capital cost per year

# Economic PF = sqrt(1-((b1/a)**2))

b1=r*b/100 # b' term accrding to the equation above

pfeco=np.sqrt(1-((b1/a)**2))  #Economic Power Factor

print 'The Economic Power Factor is %0.4f'%pfeco

The Economic Power Factor is 0.9165