# Chapter 4: Economics of Power Generation¶

### Example 4.1, Page Number: 74¶

In :
#Variable Declaration:
P = 90000                  #initial cost of transformer(Rs)
n = 20                     #20 years
S = 10000                  #Salvage value(Rs)

#Calculation:
D = (P-S)/n                #Annual depreciation charge(Rs)

print "The annual depreciation charge is Rs",D

The annual depreciation charge is Rs 4000.0


### Example 4.2, Page Number: 74¶

In :
from __future__ import division

#Variable declaration
P = 200000                        #Initial cost of transformer(Rs)
S = 10000                         #Salvage value of transformer(Rs)
n = 20                            #Useful life(years)
r = 0.08                              #annual interest rate(%)

#Calculation:
q = (P-S)*r/(((1+r)**n)-1)        #Annual payment(Rs)

#Results:
print "The annual amount to be saved is Rs",round(q)

The annual amount to be saved is Rs 4152.0


### Example 4.3, Page Number: 75¶

In :
#Variable declaration:
P = 1560000                 #Initial cost of equipment(Rs)
S = 60000                    #salvage value of equipment(Rs)
n = 25                      #useful life(years)
t = 20                      #years
#Calculation:

#(i) Straight line method:
D1 = (P-S)/n                 #Annual depreciation(Rs)
V1 = P-D1*t                   #Value of equipment after 't' years

#(ii)Diminishing value method:
D2 = 1-(S/P)**(1/n)         #Annual unit depreciation(Rs)
V2 = P*(1-round(D2,3))**t            #Value of equipment after 't' years

#(iii)Sinking fund method:
r = 0.05                    #rate of interest
q = (P-S)*(r/((1+r)**n-1))   #Annual deposit in the sinking fund(Rs)
q1 = 31433*(((1+r)**20-1)/r)        #Sinking fund at the end of 20 years(Rs)
V = P-q1                    #Value of plant after 20 years(Rs)

#Results:
print "The depreciated values using:"
print "(i) Straight line method: Rs",V1
print "(ii)Diminishing value method Rs:",round(V2)
print "(iii)Sinking fund method: Rs",round(V)

The depreciated values using:
(i) Straight line method: Rs 360000.0
(ii)Diminishing value method Rs: 115615.0
(iii)Sinking fund method: Rs 520638.0


### Example 4.4, Page Number: 76¶

In :
#Variable declaration:
M = 50000                  #Max demand(kW)
CC = 95*10**6              #Capital cost(Rs)
LF = 0.4                   #annual load factor
C1 = 9*10**6               #Annual cost of fuel and oil(Rs)
C2 = 7.5*10**6             #taxes wages salaries etc(Rs)
i = 12                    #interest & depreciation(%)

#Calculation:
E = M*LF*8760             #units generated(kWh/year)
AFC = i*CC/100            #Annual fixed charges(Rs)
T = C1+C2                #Total annual running charges(Rs)
TAC = AFC+T               #Total annual charges(Rs)
c = TAC/E                 #cost per unit(Rs)

#Results:
print "Cost per unit generated = ",round(c,2)*100,"paise"

Cost per unit generated =  16.0 paise


### Example 4.5, Page Number: 76¶

In :
#Variable declaration:
C = 50000                 #Installed capacity(kW)
E = 220*10**6             #units generated(kWh/year)
AFC1 = 160               #annual fixed charges per kW of C(Rs/kW)
RC = 0.04                 #running charges(Rs)

#Calculation:
AFC = AFC1*C               #annual fixed charges(Rs)
ARC = RC*E                  #annual running charges(Rs)
c = (AFC+ARC)/E           #cost per unit(Rs)

#Results:
print "Cost per unit generated is ",round(c,4)*100,"paise"

Cost per unit generated is  7.64 paise


### Example 4.6, Page Number: 76¶

In :
from __future__ import division

#Variable declaration:
C = 160000                 #cost of plant(Rs)
r = 12                    #annual fixed charges(%)
r1 = 5                    #interest(%)
r2 = 5                     #depreciation(%)
r3 = 2                     #taxes(%)
M = 100                    #max demand(kW)

#Calculation:
AFC = C*r/100             #annual fixed charges(Rs)

E1 = M*1*8760              #kWh/year
c1 = AFC/E1                   #Rs

E2 = M*0.5*8760            #kWh/year
c2 = AFC/E2                   #Rs

#Results:
print "Fixed charges when:"
print "(i) load factor is 100% :",round(c1*100,2),"paise"
print "(ii)load factor is 50% :",round(c2*100,2),"paise"

Fixed charges when:
(i) load factor is 100% : 2.19 paise
(ii)load factor is 50% : 4.38 paise


### Example 4.7, Page Number: 77¶

In :
#Variable declaration:
PC = 50                    #plant capacity(MW)
LF = 0.4                  #annual load factor
C1 = 1.2*10**7            #capital cost(Rs)
C2 = 4*10**5              #annual cost of wages, taxation etc.(Rs)
C3 = 1.0                  #cost of fuel,lubrication, maintenance etc.(paise/kWh)
r1 = 5                    #interest rate(%)
r2 = 6                    #depreciation(%)

#Calculation:
T1 = (C1*(r1+r2)/100)+C2    #Total annual fixed charges(Rs)
E = PC*10**3*LF*8760        #units generated(kWh/year)
C4 = E*C3/100               #Cost of fuel, lubrication etc.(Rs)
T = T1+C4                 #total annual charges(Rs)
c = T/E                   #generating cost(Rs/kWh)

#Results:
print "Cost per kWh is ",round(c*100),"paise"

Cost per kWh is  2.0 paise


### Example 4.8, Page Number: 77¶

In :
#Variable declaration:
PC = 300                  #plant capacity(MW)
CF = 0.5                  #capacity factor
LF = 0.6                  #annual load factor
C1 = 9*10**7              #Annual cost of fuel, oil etc(Rs)
C2 = 10**9                #capital cost(Rs)
r1 = 10                   #annual interest and depreciation(%)

#Calculation:
M = L/LF                 #max demand(MW)
RC = PC-M                #Reserve capacity(MW)
TC = C1+C2*r1/100        #total cost(Rs)
E = L*10**3*8760         #units generated(kWh/year)
c = TC/E                 #cost per kWh

#Results:
print "The minimum reserve capacity of the station is",RC,"MW"
print "Cost per kWh generated is ",round(c*100),"paise"

The minimum reserve capacity of the station is 50.0 MW
Cost per kWh generated is  14.0 paise


### Example 4.9, Page Number: 78¶

In :
from __future__ import division
#Variable declaration:
PC = 50               #plant capacity(MW)
CC = 1000              #capital cost(Rs/kW)
r1 = 10               #annual depreciation charges(%)
r2 = 20                #part of salaries, maitenance to fixed charges(%)
M = 40                #max demand(MW)
C1 = 700000      #Annual cost of salaries, maintenance charges etc.(Rs)
R1 = 1                #royalty(Re/(kW*year))
R2 = 0.01       #royalty paid for using the river water for generation(Re/kWh)

#Calculation:
#for annual fixed cost
E = M*10**3*LF*8760                #kWh/year
C = CC*PC*10**3             #capital cost(Rs)
T1 = r1*C/100+r2*C1/100     #total annual fixed charges(Rs)

C2 = T1/(M*10**3)+R1                #Cost per kW(Rs)

#for running cost:
C3 = ((1-r2/100)*C1)/E+R2           #Cost per kW(Rs)

#Results:
print "Total generation cost in two part form is given by:"
print  "Rs (",C2,"* kW +",round(C3,4),"* kWh)"

Total generation cost in two part form is given by:
Rs ( 129.5 * kW + 0.0127 * kWh)


### Example 4.10, Page Number: 78¶

In :
#Variable declaration:
M = 60                 #plant capacity(MW)
C1 = 5*10**6            #capital cost of building and equipment(Rs)
C2 = 900000        #annual cost of fuel,oil,taxation and wages(Rs)
r1 = 10                 #interest and depreciation(%)
C3 = 5000000             #annual cost of organisation and
#interest on cost of site etc.

#Calculation:
E = M*10**3*LF*8760     #kWh/year
a = C3                  #Rs
T2 = r1*C3/100          #Annual semi-fixed cost(Rs)
b = T2/(M*10**3)        #cost per kW
c = C2/E                #cost per kWh

#Results:
print "The required values are:"
print "a = Rs",a,", b = ",round(b,2),", c = Re",round(c,4)

The required values are:
a = Rs 5000000 , b =  8.33 , c = Re 0.0034


### Example 4.11, Page Number: 79¶

In :
#Variable declaration:
PC = 100                 #installed capacity(kW)
CC = 3000                #plant cost(Rs/kW of PC)
r1 = 5                   #interest(%)
r2 = 2                   #depreciation(2%)
r3 = 2                   #operation & maintenance(%)
r4 = 1.5                 #insurance, rent(%)
r = 12.5              #losses in transmission and distribution(%)
DF = 1.25                #diversity factor
M = 0.8*PC               #max demand(kW)

#Calculation:
L = M*LF                  #avg demand(kW)
C1 = CC*PC                #Capital cost(Rs)
T1 = (r1+r2)*C1/100       #annual fixed charges(Rs)
T11 = T1/(DF*M)            #annual fixed charges(Rs/kW)
RC = C1*(r3+r4)/100        #annual running charges(Rs)
E = L*8760                #kWh/year
E1 = E*(1-r/100)           #units reaching the consumer(kWh)
RC1 = RC/E1               #annual running charges(Rs/kWh)
T = T1+RC                 #total charges(Rs)
C2 = T/E1                 #cost per kWh(Rs)

#Results:
print "Total generation cost in two part form is given by:"
print  "Rs (",T11,"* kW +",round(RC1,3),"* kWh)"
print "Overall cost of generation per kWh is ",round(C2*100,1),"paise"

Total generation cost in two part form is given by:
Rs ( 210.0 * kW + 0.043 * kWh)
Overall cost of generation per kWh is  12.8 paise


### Example 4.12, Page Number: 80¶

In :
#variable declaration:
M = 1000                    #max demand(kW)
#for (i)a private oil engine generating plant:

CC1 = 12*10**5               #capital cost(Rs)
c1 = 0.005                   #Cost of repair and maintenance(Rs/kWh)
c2 = 1600                    #Cost of fuel(Rs/1000kg)
r1 = 10                      #Interest and depreciation (%)
w = 0.3                      #fuel consumption(kg/kWh)
c3 = 50000                   #wages(Rs)

#for (i)Public supply company:
#Rs 150 per kW of maximum demand plus 15 paise per kWh

#Calculation:
E = M*LF*8760               #kWh/year

#for(i) Private oil engine generating plant:
W = w*E                     #annual fuel consumption(kg)
C1= W*c2/1000               #cost of fuel(Rs)
T1 = C1+c1*E+r1*CC1/100+c3      #total annual cost(Rs)

#for(ii)Public supply company:
T2 = 150*M+.15*E           #total annual cost(Rs)

#Results:
print "(i)For Private oil engine generating plant,"
print  "   total annual charges is Rs",T1
print "(ii)Public supply company, total annual charges is Rs",T2

(i)For Private oil engine generating plant,
total annual charges is Rs 2294300.0
(ii)Public supply company, total annual charges is Rs 807000.0


### Example 4.13, Page Number: 80¶

In :
from __future__ import division

#Variable declaration:
M  = 100                     #max demand(MW)

#steam:
cc1 = 1250                 # Capital cost/kW installed(Rs)
r11 = 12                   # Interest and depreciation(%)
c11 = 5                    # Operating cost/kWh, paise
tc11 = 0                   # Transmission cost/kWh

#hydro:
cc2 = 2500                 # Capital cost/kW installed(Rs)
r21 = 10                   # Interest and depreciation(%)
c21 = 1.5                  # Operating cost/kWh,paise
tc21 = 0.2                 # Transmission cost/kWh,paise

#Calculation:
E = M*LF*8760*10**3                  #kWh/year

#(i)steam station in conjunction with a hydro station:
Eh = 100*10**6              #units supplied by hydro station(kWh)
Es = E-Eh                   #units supplied by steam station(kWh)
Pmh = 40                    #max o/p of hydro stn.(MW)
Pms = 60                    #max o/p of steam stn(MW)

#(i)(a)for steam station:
Cs1 = cc1*Pms*10**3              #capital cost(Rs)
Ts1 = r11*Cs1/100+c11*Es/100+0         #total cost(Rs)

# (b)for hydro station:
Ch1 = cc2*Pmh*10**3          #capital cost(Rs)
Th1 = r21*Ch1/100+c21*Eh/100+tc21*Eh/100   #total cost(Rs)

Ta = Ts1+Th1                 #total annual cost(Rs)
OC1 = Ta/E                   #kWh

#(ii)Steam station:
Pm2 = 100*10**3            #max o/p of steam plant(kW)
Cs2 = cc1*Pm2                   #capital cost(Rs)
Ts2 = (Cs2*r11/100)+c11*E/100   #Rs
OC2 = Ts2/E                     #overall cost(Rs)

#(iii)Hydro station:
Ch3 = cc2*Pm2                   #capital cost(Rs)
Th3 = Ch3*r21/100+c21*E/100     #Rs
OC3 =  Th3/E+tc21/100                    ##overall cost(Rs)

#Results:
print "(i)steam station in conjunction with a hydro station:"
print "    for steam stn., overall cost is ",round(OC1*100,2),"paise"
print "(ii) Steam station, overall cost is ",round(OC2*100,2),"paise"
print "(ii) Hydro station, overall cost is ",round(OC3*100,2),"paise"

(i)steam station in conjunction with a hydro station:
for steam stn., overall cost is  10.97 paise
(ii) Steam station, overall cost is  10.71 paise
(ii) Hydro station, overall cost is  11.21 paise


### Example 4.14, Page Number: 81¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
M = 150*10**3                 #maximum demand(kW)

#Steam plant:
cc1 = 1600                     #Rs/kW
oc1 = 0.06                     #operating cost(Rs/kWh)
r1 = 7                         #interest(%)

#Hydro plant:
cc2 = 3000                     #Rs/kW
oc2 = 0.03                     #operating cost(Rs/kWh)
r2 = 7                         #interest(%)

#Calculation:
x = symbols('x')               #total no. of units generated
#for steam plant:
cs = cc1*M                      #capital cost(Rs)
Ts = r1*cs/(x*100)+oc1               #total cost(Rs)
Ts1 = Ts                       #Rs/kWh

#for hydro plant:
ch = cc2*M                     #capital cost(Rs)
Th = r2*ch/(100*x)+oc2               #total cost(Rs)
Th1 = Th                       #Rs/kWh
L = solve(Ts1-Th1,x)
LF = (L)/(M*8760)


Load factor is  37.3 %


### Example 4.15, Page Number: 82¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
Eo = 40*10**6                               #units needed to generate

#Hydro                     Steam
cc1 = 2100;           cc2 = 1200             #Capital cost(Rs/kW)
rc1 = 3.2;            rc2 = 5                #Running cost(paise/kWh)
r1 = 7.5;              r2 = 9                #Interest & depreciation(%)
RC1 = 33;             RC2 = 25             #Reserve capacity(%)

#Calculation:
#Let x kW be the maximum demand.
#Let y be the annual load factor at which cost/unit
#of steam and hydro stations is the same.

x,y = symbols('x y')
E = x*y*8760                   #units generated per annum(kWh)
C2 = x+RC2*x/100               #installed capacity of steam station(kW)
C1 = x+RC1*x/100               #installed capacity of hydro station(kW)

#steam station:
CC2 = cc2*C2                      #capital cost(Rs)
OC2 = (r2*CC2/100+rc2*E/100)/E       #Overall cost/kWh

#hydro station:
CC1 = cc1*C1                      #capital cost(Rs)
OC1 = (r1*CC1/100+rc1*E/100)/E       #Overall cost/kWh

E2 = 8760*x*LF                    #kWh
M = solve(E2-Eo,x)             #max. demand(kW)
C = (135*M + 438*M*LF)            #cost of generation(Rs)

#Results
print "Required cost of generation is Rs (",round(C/10**6,1),"* 10**6 )"

Load factor is  47.23 %
Required cost of generation is Rs ( 3.3 * 10**6 )


### Example 4.16, Page Number: 83¶

In :
from __future__ import division
from sympy import *

#Variable declaration:

#Let x = Installed capacity of station B in kW
#y = Hours of operation of station B

x,y = symbols('x y')
M = 50000                   #max demand(kW)

#Calculation:
PCa = M-x                   #installed capacity of stationA(kW)
Eb = (1/2)*x*(8760*x/M)      #Units generated/annum by station B
Ea = (1/2)*8760*M-Eb         #Units generated/annum by station A
Cb = 50000+50*x+0.03*Eb      #Rs
Ca = 75000+80*(50000-x)+0.02*Ea    #Rs
C = Ca+Cb                       #total operating cost(Rs)
#After differentiating C w.r.t x, we get C1 as
C1 = -30+0.00174*x
x1 = solve(C1,x)

PCb = x1                     #kW
PCa = M-PCb                  #kW
t = 8760*PCb/M               #hours of operation of station B

#Results:
print "Installed capacity of station A is ",round(PCa),"MW"
print "Installed capacity of station B is ",round(PCb),"MW"
print "No. of hours of operation of plant B is",round(t/10)*10,"hrs"

Installed capacity of station A is  32759.0 MW
Installed capacity of station B is  17241.0 MW
No. of hours of operation of plant B is 3020.0 hrs