# Chapter 2:Generating Stations¶

### Example 2.1, Page Number: 16¶

In :
from __future__ import division

#Variable declaration:
n=20                        #overall efficiency of plant
h=860                       #kcal
m=0.6           #Mass of fuel burnt(kg) per KW of electrical energy generated

#Calculations:
x=h*100/(m*n)           #Calorific value of fuel(kcal/kg)

#Results:
print "Calorific value of fuel =",round(x,2),"kcal/kg"

Calorific value of fuel = 7166.67 kcal/kg


### Example 2.2, Page Number: 17¶

In :
from __future__ import division

#Variable declaration:
M = 20000                           #maximum demand(kW)
n_b= 85                             #boiler efficiency(%)
m=0.9                             #coal consumption(kg/kWh)
n_t=90                          #turbine efficiency(%)
c=300                           #cost of 1 tonne of coal(Rs)

#Calculations:
n_th = n_b * n_t/100            #in %
cb = LF*M*m*c*24*365/(1000*100)

#Results:
print "Thermal efficiency = ",n_th,"%"
print "Coal bill per annum = Rs",cb

Thermal efficiency =  76.5 %
Coal bill per annum = Rs 18921600.0


### Example 2.3, Page Number: 17¶

In :
from __future__ import division

#Variable declaration:
ct=3000000                     #annul cost of coal(Rs)
cv=5000                         #Calorific value of coal(kcal/kg)
c=300                          #cost of coal per tonne(Rs)
n_th=33                        #thermal efficiency(%)
n_elec=90                      #electrical efficiency(%)

#Calculations:
n_t=n_th*n_elec/100                 #overall efficiency(%)
h=ct*cv*1000/c                  #heat of combustion(kcal)
ho=n_t*h/(100*860)             #heat output(kWh)
L=ho/8760                      #kW

#Results:
print "Avg load on the station=",round(L),"kW"

Avg load on the station= 1971.0 kW


### Example 2.4, Page Number: 17¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
kWh= symbols('kWh')
W = 13500 + 7.5 * kWh               #Water evaporated in kg
C = 5000 + 2.9 * kWh                #coal cumsumption in kg

#Calculations:
#part (i):
#As the station output (i.e., kWh) increases towards infinity,
#the limiting value of W/C approaches
L1= 7.5/2.9                        #in kg

#part (ii):
kWh=0
c=(5000+2.9*kWh)/8                             #coal per hour in kg
#Results:
print "Limiting value of water/kg of coal=",round(L1,1),"kg"
print "Required Coal per hour",c,"kg"

Limiting value of water/kg of coal= 2.6 kg
Required Coal per hour 625.0 kg


### Example 2.5, Page Number: 18¶

In :
from __future__ import division

#Variable declaration:
C=100                            #Capacity of station in MW
cv=6400                          #kcal/kg
n_th=0.3                         #thermal efficiency
n_elec=0.92                      #electrical efficiency

#Calculations:
n_t=n_th*n_elec                   #overall efficiency
U=C*1*10**3                      #units generated/hr in kWh
H=U*860/n_t                    #total heat of combustion(kcal)
w=H/cv                          #Coal consumption in kg

#Results:
print "The coal consumption per hour =",round(w),"kg"

The coal consumption per hour = 48687.0 kg


### Example 2.6, Page Number: 23¶

In :
from __future__ import division

#Variable declaration:
C=5*10**6                   #reservoir capacity in m^3
n_t=0.75                    #overall efficiency
d=1000                      #density of water in kg/m^3

#Calculations:
W=C*d*9.81                   #weight of water in Newton
E=W*H*n_t/(3600*1000)        #electrical energy available(kWh)

#Results:
print "The total energy available=",round(E/10**6,3),"* 10^6 kWh"

The total energy available= 2.044 * 10^6 kWh


### Example 2.7, Page Number: 23¶

In :
from __future__ import division

#Variable declaration:
V=94                    #volume of water in m^3/sec
d=1000                  #density of water in kg/m^3
H=39                    #head of water in m
nt=0.80                 #overall efficiency

#Calculations:
W=V*d                   #weight of water in kg/sec
w=W*H*9.81/1000         #work done per sec in kW
FC=nt*w                 #firm capacity in kW
YGO=FC*8760             #yearly gross capacity in kWh

#Results:
print "Firm capacity=",FC,"kW"
print "Yearly gross output",round(YGO/10**6),"* 10^6 kWh"

Firm capacity= 28770.768 kW
Yearly gross output 252.0 * 10^6 kWh


### Example 2.8, Page Number: 23¶

In :
from __future__ import division

#Variable Declaration:
Q=1                        #discharge, m^3/sec
nh=0.86                    #hydraulc efficiency
nelec=0.92                 #electrical efficiency
d=1000                     #density of water, kg/m^3

#Calculations:
W=Q*d*9.81                 #weight of water in N
Po=W*H*nh*nelec/1000        #power produced, kW
E=Po*1                     #in kWh

#Results:
print "Electrical energy generated per hr=",round(E),"kWh"

Electrical energy generated per hr= 776.0 kWh


### Example 2.9, Page Number: 23¶

In :
from __future__ import division

#Variable declaration:
CA=5*10**9                      #Catchment area in m^2
F=1.25                          #Annual rainfall in m
K=0.80                          #yeild factor
n=0.70                          #overall efficiency
d=1000                          #density of water(kg/m^3)

#Calculations:
V=CA*F*K                              #volume of water utilised per annum(m^3)
W=V*d*9.81                                   #Weight of water available per annum (N)
E=round(W*H*n/(10**11*3600),2)*10**8         #Electrical energy available per annum(kWh)
Pav=E/8760                                   #average power(kW)
Dmax=Pav/LF                                 #Maximum demand

#Results:
print "Average power generated is ",round(Pav),"kW"
print "Rating of generators is",round(Dmax),"kW"

Average power generated is  32648.0 kW
Rating of generators is 81621.0 kW


### Example 2.10, Page Number: 24¶

In :
from __future__ import division

#Variable Declaration:
A=2.4                    #Area of reservoir(km^2)
V=5*10**6                #Capacity of reservoir(m^3)
H=100                    #in m
np=0.95                  #penstock efficiency
nt=0.90                  #turbine efficiency
ng=0.85                  #generation efficiency

#Calculations:
W=V*1000*9.81             #in Newton
n=int(np*nt*ng*1000)/1000               #overall efficiency
E=W*H*n/(1000*3600)       #Elecctrical energy generated(kWh)
x=L*3*3600/(A*10**6*9.81*H*n)

#Results:
print "Total electrical energy generated is ",round(E),"kWh"
print "Fall in reservoir level is",round(x*100,3),"cm"

Total electrical energy generated is  989175.0 kWh
Fall in reservoir level is 9.478 cm


### Example 2.11, Page Number: 25¶

In :
from __future__ import division

#Variable declaration
H=25                       #head of reservoir in m
Pr=400                    #power required by factory(kW)
n=0.80                    #overall efficiency of plant

#Calculations:
#part (i):
#(a):
d1 = 10                     #discharge in m^3/sec
w1 = d1*1000*9.81           #weight of water in N
P1 = w1*H*n/1000             #power developed(kW)

#(b)
d2 = 6                      #in m^3/sec
P2 = P1*d2/d1               #kW

#(c)
d3 = 1.5                    #in m^3/sec
P3 = P1*d3/d1               #kW

Ps = Pr-P3                  #standby power(kW)

#part(ii):
Dav = (d1*4+d2*2+d3*6)/12   #avg discharge(m^3/sec)
P = P1*Dav/d1               #power developed(kW)
Pex = P-Pr                  #Excess power available(kW)

#Results:
print "(i) Standby power is ",round(Ps),"kW"
print "(ii) Excess power available is ",round(Pex,1),"kW"

(i) Standby power is  106.0 kW
(ii) Excess power available is  597.4 kW


### Example 2.12, Page Number: 25¶

In :
from __future__ import division

#Variable declaration:
#for part (i):
C = 10                          #Installed capacity(MW)
H = 20                          #head of reservoir(m)
n = 0.80                        #overall efficiency
#for part (ii):
Q2 = 20                         #discharge

#Calculations:
#for part(i):
U = C*LF*24*7*10**3             #units generated per week(kWh)
Q = U/(H*n*9.81*24*7)           #Discharge in m^3/sec

#for part(ii):
U2 = Q2*9.81*1000*n*H*24/1000     #units generated per day(kWh)
LF2 = U2/(C*10**3*24)

#Results:
print "(i) The river discharge is ",round(Q,2),"m^3/sec"
print "(ii) The load factor is ", round(LF2*100,1),"%"

(i) The river discharge is  25.48 m^3/sec
(ii) The load factor is  31.4 %


### Example 2.13, Page Number: 26¶

In :
#Variable declaration:
H = 15                            #head of reservoir(m)
n = 0.85                          #efficiency

#Calculations:
Qavg = (500+520+850+800+875+900+546)/7    #in m^3/sec

#It is clear from graph that on three dyas
#(viz., Sun, Mon. and Sat.), the discharge is less than
#the average discharge.

V1 = (500+520+546)*24*3600  #Actual volume available in these 3 days(m^3/s)
V2 = 3*Qavg*24*3600         #Vol. of water required in these 3 days(m^3/s)
Pr = V2-V1                  #Pondage required(m^3/sec)
Po = Qavg*9.81*1000*H*n     #Avg output produced(W)
C = Po/L                    #Capacity of plant(W)

#Results:
print "(i) The average daily discharge is ",Qavg,"m^3/sec"
print "(ii) Pondage required is (",round(Pr/10**5),"* 10^5)  m^3"
print "(iii)Installed capacity of plant is ",round(C/10**6),"MW"

(i) The average daily discharge is  713.0 m^3/sec
(ii) Pondage required is ( 495.0 * 10^5)  m^3
(iii)Installed capacity of plant is  223.0 MW


### Example 2.14, Page Number: 29¶

In :
from __future__ import division

#Variable declaration:
w = 0.28                        #fuel consumption in kg/kWh
C = 10000                       #calorific value of fuel(kcal/kWh)
na = 0.95                        #efficiency of alternator

#Calculations:
H = w*C/860                     #heat produced by 0.28 kg/kWh of fuel

no = 1/H                        #Overall efficiency
ne = no/na                      #Efficiency of engine

#Results:
print "(i) The overall efficiency is ",round(no*100,1),"%"
print "(ii)The efficiency of the engine",round(ne*100,1),"%"

(i) The overall efficiency is  30.7 %
(ii)The efficiency of the engine 32.3 %


### Example 2.15, Page Number: 30¶

In :
from __future__ import division

#Variable declaration:
w = 1000                      #fuel consumption in kg/day
E = 4000                      #Units generated in kWh/day
C = 10000                     #calorific value in kcal/kg
na = 0.96                     #Alternator efficiency
nem = 0.95                     #engine mechanical efficiency

#Calculations:
s = w/E                       #specific fuel consumption(kg/kWh)
E2 = w*C                      #energy input per day(kcal/day)
no = E*860/E2                 #overall efficiency
ne = no/na                    #engine efficiency
net = ne/nem                  #engine thermal efficiency

#Results:
print "Specific fuel consumption is ",s,"kg/kWh"
print "Overall efficiency is ",round(no*100,1),"%"
print "Thermal efficiency of the engine is ",round(net*100,2),"%"

Specific fuel consumption is  0.25 kg/kWh
Overall efficiency is  34.4 %
Thermal efficiency of the engine is  37.72 %


### Example 2.16, Page Number: 30¶

In :
from __future__ import division

#Variable declaration:
C1 = 700                   #capacity of plant 1(kW)
C2 = 2*500                 #capacity of plant 2(kW)
pcf = 0.40                 #plant capacity factor
w = 0.28                   #fuel cunsumption in kg/kWh
H = 10200                  #specific heat of fuel in kcal/kg

#Calculatios:
M = (C1+C2)*30*24          #max energy can be produced in 30 days(kWh)
E = pcf*M                  #Actual energy produced in 30 days(kWh)
W = E*w                    #actual fuel consumption in kg

Po = E*860                 #output energy in kWh
Pin = W*H                  #Input energy in kWh
n = Po/Pin                 #Overall efficiency

#Results:
print "The fuel oil required is ",W,"kg"
print "Ovreall efficiency is",round(n*100),"%"

The fuel oil required is  137088.0 kg
Ovreall efficiency is 30.0 %


### Example 2.17, Page Number: 34¶

In :
from __future__ import division

#Variable declaration
M = 300                   #Energy received from reactor(MW)
E1 = 200                  #Energy released fron each atom(MeV)

#Calculations:
E2 = M*10**6*3600            #Energy released per hour(J)
E3 = E1*1.6*10**-19*10**6      #Energy released per fission(J)
N = E2/E3                    #No of atoms fissioned
m = N*235/(6.022*10**23)    #mass of uranium fissioned per hr(g)

#Results:
print "Mass of Uranium fissioned per hour is",round(m,2),"g"

Mass of Uranium fissioned per hour is 13.17 g


### Example 2.18, Page Number: 35¶

In :
from __future__ import division

#Variable declaration:
t = 30                         #days
w = 2                          #weight of uranium(kg)
Eo = 200                     #energy released per fission(MeV)

#Calculations:
N = 2*1000*6.022*10**23/235     #No of atoms fissioned in 2kg of fuel
Po = N*Eo*(1.6*10**-19)*10**6/(24*60*60*30)    #Watt

#Results:
print "Power output is",round(Po*10**-6,1),"MW"

Power output is 63.3 MW