In [31]:

```
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"
```

In [32]:

```
from __future__ import division
#Variable declaration:
M = 20000 #maximum demand(kW)
n_b= 85 #boiler efficiency(%)
m=0.9 #coal consumption(kg/kWh)
LF=40 #load factor(%)
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
```

In [33]:

```
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"
```

In [34]:

```
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):
#at no load
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"
```

In [35]:

```
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"
```

In [36]:

```
from __future__ import division
#Variable declaration:
C=5*10**6 #reservoir capacity in m^3
H=200 #water head in m
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"
```

In [37]:

```
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"
```

In [38]:

```
from __future__ import division
#Variable Declaration:
H=100 #Water head in m
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"
```

In [39]:

```
from __future__ import division
#Variable declaration:
CA=5*10**9 #Catchment area in m^2
H=30 #head in m
F=1.25 #Annual rainfall in m
K=0.80 #yeild factor
n=0.70 #overall efficiency
LF=0.40 #Load factor
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"
```

In [20]:

```
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
L=15000 #load supplied in kW
#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"
```

In [41]:

```
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"
```

In [42]:

```
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
LF = 0.40 #load factor
#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),"%"
```

In [43]:

```
#Variable declaration:
H = 15 #head of reservoir(m)
n = 0.85 #efficiency
L = 0.40 #load factor
#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"
```

In [44]:

```
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),"%"
```

In [45]:

```
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),"%"
```

In [46]:

```
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),"%"
```

In [47]:

```
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"
```

In [48]:

```
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"
```