Ch-3, Power Plant Economics

example 3.1 - pg 43

In [3]:
from __future__ import division
totpow=110*10**3 #(kW)
uc1=18000; fcr1=0.1; cf1=0.55; fuelcons1=0.7; fuelcost1=1500/1000; om1=0.2; utilizationf1=1 
uc2=30000 ;fcr2=0.1; cf2=0.60 ;fuelcons2=0.65 ;fuelcost2=1500/1000 ;om2=0.15 ;utilizationf2=1 
#given uck=unit capital cost k fcrk= fixed charge rate of kth unit cfk=capacity factor at k th unit  omk=annual cost of operating labour  totpow=total power rating of units
afc1=fcr1*uc1*totpow; afc2=fcr2*uc2*totpow 
e1=8760*cf1*totpow; e2=8760*cf2*totpow 
annualfuel1=e1*fuelcons1 ;annualfuel2=e2*fuelcons2 
fc1=annualfuel1*fuelcost1 ;fc2=annualfuel2*fuelcost2 
om11=om1*fc1; om22=om2*fc2 
aoc1=fc1+om1 ;aoc2=fc2+om22 
apc1=aoc1+afc1; apc2=aoc2+afc2 
gc1=apc1/fc1 ;gc2=apc2/fc2
print " solution for (a)"
print " afc1=Rs.%0.2e\n e1=%0.4e kWh\n annualfual1=%0.5e kg \n fc1=Rs.%0.5e \n om1=Rs.%0.6e \n aoc1=Rs.%0.6e \n apc1=Rs.%0.5e \n gc1=%0.4f kWh\n"%(afc1,e1,annualfuel1,fc1,om11,aoc1,apc1,gc1)
print " solution for (b)"
print " afc2=Rs.%0.2e \n e2=%0.4e kWh\n annualfual2=%0.5e kg \n fc2=Rs.%d \n om22=Rs.%d \n aoc2=Rs.%.f \n apc2=Rs.%.f \n gc2=%.fkWh\n"%(afc2,e2,annualfuel2,fc2,om22,aoc2,apc2,gc1)
ogc=(apc1+apc2)/(e1+e2)

print " solution of (c)\n ogc=Rs.%0.3f/kWh"%(ogc)
# Ans in the textbook are not accurate.

 solution for (a)
 afc1=Rs.1.98e+08
 e1=5.2998e+08 kWh
 annualfual1=3.70986e+08 kg 
 fc1=Rs.5.56479e+08 
 om1=Rs.1.112958e+08 
 aoc1=Rs.5.564790e+08 
 apc1=Rs.7.54479e+08 
 gc1=1.3558 kWh

 solution for (b)
 afc2=Rs.3.30e+08 
 e2=5.7816e+08 kWh
 annualfual2=3.75804e+08 kg 
 fc2=Rs.563706000 
 om22=Rs.84555900 
 aoc2=Rs.648261900 
 apc2=Rs.978261900 
 gc2=1kWh

 solution of (c)
 ogc=Rs.1.564/kWh

example 3.2 - pg 45

In [4]:
c=2*10**8 #cost
s=0.15 #salvage value
ul=25  #/useful value
i=0.08 #life of plant
print "solution for (a)"
print "annual straight line depreciation reserve =Rs. %.1e per year\n"%(c*(1-s)/ul)
print "solution for (b)"
it=(i+1)**25-1
iit=i/it
asdr=c*(1-s)*iit*100
print "annual sinking fund depreciation reserve is =Rs. %.3e per year"%(asdr)

solution for (a)
annual straight line depreciation reserve =Rs. 6.8e+06 per year

solution for (b)
annual sinking fund depreciation reserve is =Rs. 2.325e+08 per year

example 3.3 - pg 45

In [5]:
cost=2*10**8
sal=0.15
use=25
t=(1-(sal**(1/use)))
print "rate of depretion by fixed percentage method = %0.1f %%"%(t*100)
rd=cost*(1-t)**10
print "remaining depreciation at the end of 10th year = Rs.%0.3e"%(rd)
print "accumulated depreciation at the end of 10 year is Rs.%0.3e"%(cost-rd)

rate of depretion by fixed percentage method = 7.3 %
remaining depreciation at the end of 10th year = Rs.9.364e+07
accumulated depreciation at the end of 10 year is Rs.1.064e+08

example 3.4 - pg 46

In [6]:
from __future__ import division
import numpy as np
%matplotlib inline
from matplotlib.pyplot import plot,show

p=100  #ratring of steam station
fc=3000 #fixed cost of plant per year
rg=0.9 #90 paise per kv generation
uf=1  #utilization factor 1
lf=np.arange(20,101,20) #let load factor be 5 discreate units
lm=uf*lf  #lwt load MW is as same as lf as utilisation factor is 1
n=len(lm)
fc=[fc*1e5]*n
op=[rg*100]*n
negp = range(0,n)
fcgp = range(0,n)
tgc = range(0,n)
for i in range(0,n):
    negp[i]=lm[i]*8760 * 1e3 # kW-hrs/year
    fcgp[i]=fc[i]*10000/negp[i]* 1e2 # Paisa/unit
    tgc[i]=fcgp[i]+op[i]

print "negp  : ",tgc
    
    
plot(lf,tgc)
show()
print "load factor"
print lf
print "load MW\n"
#fcgp=fcgp/100 ;op=op/100; tgc=tgc/100
print "%dMW\t%dMW\t%dMW\t%dMW\t%dMW"%(lm[0],lm[1],lm[2],lm[3],lm[4])
print "fixed cost"
print "Rs%d\tRS%d\tRs%d\tRs%d\tRs%d"%(fc[0],fc[1],fc[2],fc[3],fc[4])
print "number of KW hrs of energy generated in paise per unit of energy"
print "%dkWh\t%dkWh\t%dkWh\t%dkWh\t%dkWh"%(negp[0],negp[1],negp[2],negp[3],negp[4])
print "fixed cost in paise per unit of energy"
print "Rs%.3f\tRS%.3f\tRs%.3f\tRs%.3f\tRs%.3f"%(fcgp[0],fcgp[1],fcgp[2],fcgp[3],fcgp[4])
print "operating cost in paise per unit of energy"
print "Rs%.3f\tRS%.3f\tRs%.3f\tRs%.3f\tRs%.3f"%(op[0],op[1],op[2],op[3],op[4])
print "total generation cost in paise per unit of energy"
print "Rs%.3f\tRS%.3f\tRs%.3f\tRs%.3f\tRs%.3f"%(tgc[0],tgc[1],tgc[2],tgc[3],tgc[4])

negp  :  [1712418.7671232875, 856254.38356164377, 570866.25570776255, 428172.19178082189, 342555.75342465751]

load factor
[ 20  40  60  80 100]
load MW

20MW	40MW	60MW	80MW	100MW
fixed cost
Rs300000000	RS300000000	Rs300000000	Rs300000000	Rs300000000
number of KW hrs of energy generated in paise per unit of energy
175200000kWh	350400000kWh	525600000kWh	700800000kWh	876000000kWh
fixed cost in paise per unit of energy
Rs1712328.767	RS856164.384	Rs570776.256	Rs428082.192	Rs342465.753
operating cost in paise per unit of energy
Rs90.000	RS90.000	Rs90.000	Rs90.000	Rs90.000
total generation cost in paise per unit of energy
Rs1712418.767	RS856254.384	Rs570866.256	Rs428172.192	Rs342555.753

example 3.5 page 47

In [7]:
ic=120 #installed capacity
ccppkw=40000 #/capital cost of plant
iand=0.15 #interest and depreciation
fco=0.64 #fuel consumption
fc=1.5#fuel cost
oc=50*10**6 #operating cost
pl=100#peak load
lf=0.6 #load factor
al=lf*pl#avarrage load
print "average load %dMW"%(al)
eg=al*8760*10**3#energy generated
print "energy generated =%ekWhr"%(eg)
ti=ic*ccppkw #total investiment
print "total investement Rs.%e"%(ti)
ind=ti*iand*10**3#interest and depreciation
print "investement amd depression is Rs.%e"%(ind)
fcons=eg*fco #fual consumption
print "fuel consumtion is %ekgper year"%(fcons)
fcost=fcons*fc#fuel cost
aco=ti+fcost+ind+oc#annual cost
print "fuel cost Rs.%eper year \nannual plant cost Rs%eper  year \ngeneration cost Rs%fper year"%(fcost,aco,aco/eg)

average load 60MW
energy generated =5.256000e+08kWhr
total investement Rs.4.800000e+06
investement amd depression is Rs.7.200000e+08
fuel consumtion is 3.363840e+08kgper year
fuel cost Rs.5.045760e+08per year 
annual plant cost Rs1.279376e+09per  year 
generation cost Rs2.434125per year

example 3.6 page 47

In [8]:
md=50*10**3 #maximum demand in kW
ecy=0
pst=600*md+2.5*ecy#public supply tariff equation
lfr=0.5  #load factor
rc=20*10**3 #reserve capacity
cik=30000  #capital investiment
inad=0.15 #/interest and depreciation
fuc=0.6 ;fuco=1.4; oct=0.8#fuel consumption#fuel cost#other cost
avl=md*lfr #average load
ecy=avl*8760  #energy cosumption per year
print "solution of (a)"
print " average load = %dkW \n energy consumton =%dkWh\n annual expenditure is Rs%dperyear\n"%(avl,ecy,pst)
print "(b) private steam plant"
ict=md+rc  #installed capacity
caint=cik*ict  #capital investiment
iande=inad*caint  #interest and depreciation
fuelcon=ecy*fuc  #fuel consumption
fucost=fuelcon*fuco  #fuel cost
opwe=oct*ecy #other expenditure
totex=iande+fucost+opwe#total expenditure
print " installed capacity is Rs%d \n capital investiment is Rs%d \n interest and depreciation is Rs.%d \n fuel consumption is Rs.%f \n fuel cost is Rs.%f per year \n wage,repair and other expenses are Rs%f per year \n total expenditure is Rs%e per year"%(ict,caint,iande,fuelcon,fucost,opwe,totex)

solution of (a)
 average load = 25000kW 
 energy consumton =219000000kWh
 annual expenditure is Rs30000000peryear

(b) private steam plant
 installed capacity is Rs70000 
 capital investiment is Rs2100000000 
 interest and depreciation is Rs.315000000 
 fuel consumption is Rs.131400000.000000 
 fuel cost is Rs.183960000.000000 per year 
 wage,repair and other expenses are Rs175200000.000000 per year 
 total expenditure is Rs6.741600e+08 per year

example 3.7 page 48

In [9]:
md=500 #given maximum demand
lf=0.5 #load factor
hp=7200 ;he=0.36#operating cost of hydro plant
tp=3600; te=1.56 #operating cost of thermal plant
teg=md*1000*lf*8760 #total energy generated
print "total energy generated per year %2.2eW"%(teg)
t=(hp-tp)/(te-he) #time of operating useing (de/dp)
ph=md*(1-t/8760) #from triangle adf
pt=md-ph
et=pt*t*1000/2
eh=teg-et
co=hp*ph*1000+he*eh+tp*pt*1000+te*et
ogc=co/teg
print "capacity of hydro plant is %dMW \n capacity of thermal plant %dMW\n energy generatede by hydro plant %dkWh\n energy generated by thermal plant %dkWh\n over all generation cost is %.3f/kWh"%(ph,pt,eh,et,ogc)


 

total energy generated per year 2.19e+09W
capacity of hydro plant is 328MW 
 capacity of thermal plant 171MW
 energy generatede by hydro plant 1933150684kWh
 energy generated by thermal plant 256849315kWh
 over all generation cost is 1.863/kWh

Data 3.16 page 52

In [10]:
pu=500*10**3 ;  pc=2*pu #plant unit,plant capacity
land=11.865*10**9
cicost=30.135*10**9
ccost=land+cicost    #capital cost =land cost+civil cost
plife=25     #plant life
ir=0.16     #interest rate
ond=1.5*10**-2 # o and mof capital cost
gr=0.5*10**-2  #grneral reserve of capital cost
calv=4158  #calorific value kj per kg
coalcost=990  #caol cost per ton
heat=2500#heat rate kcal/kWh
retur=0.08 #return
salvage=0
plf=0.69  ;auxcons=0.075 #auxiliary consumption
print "cost calculation "
print "using sinking fund depreciation"
ande=(ir/((ir+1)**(plife)-1))*100
afixcost=ccost*(ir+ond+retur+gr+(ande/100))
afcppc=afixcost/pc
print "annual depretion reserve is %fpersent \n annual fixed cost Rs%f \n annual fixed cost per Rs%dkWh"%(ande,afixcost,afcppc)
fo=(heat*coalcost)/(calv*1000)
engepc=24*365*plf
enavil=engepc*(1-auxcons)
gencost=(afcppc/enavil)+fo
print "fuel cost Rs.%f/kWh \nenergy generated per kW of plant capacity Rs.%fkWh \nenergy available bus bar %fkWh \n generation cost Rs%f perkWh"%(fo,engepc,enavil,gencost)

cost calculation 
using sinking fund depreciation
annual depretion reserve is 0.401262persent 
 annual fixed cost Rs11088529841.947702 
 annual fixed cost per Rs11088kWh
fuel cost Rs.0.595238/kWh 
energy generated per kW of plant capacity Rs.6044.400000kWh 
energy available bus bar 5591.070000kWh 
 generation cost Rs2.578495 perkWh

data 3.17 page 53

In [11]:
pco=120*10**3 #3 units of 40MW
caco=68*10**8 #6 year of consumption
inr=0.16 #intrest rate
de=2.5*10**-2 #depreciation
oanm=1.5*10**-2#OandM
ger=0.5*10**-2#general reserve
pllf=0.6 #plant load facot
aucon=0.5*10**-2 #auxiliary consumption
tac=caco*(inr+de+oanm+aucon) #/total cost
engpy=pco*pllf*24*365 #energy generatedper year
eabb=engpy*(1-ger) #energy available at bus bar
geco=tac/eabb #generation cost
print " total annual costs is Rs%e per year \n energy generated per year =%ekWh/year \n energy available at bus bar %ekWh/year \n generation cost is Rs.%fper kWh"%(tac,engpy,eabb,geco)

 total annual costs is Rs1.394000e+09 per year 
 energy generated per year =6.307200e+08kWh/year 
 energy available at bus bar 6.275664e+08kWh/year 
 generation cost is Rs.2.221279per kWh