# Chapter2-Energy Sources¶

## Ex1-pg44¶

In :
import math
## Example 2.1
print('Example 2.1\n\n');
print('Page No. 44\n\n');

## given
C= 35000.;## cost of boiler
C_grant=.25;## Capital grant available from goverment
E= -(C-(C_grant*C));## Net expenditure
Fs= 15250.;## Fuel Saving
r_i = 0.15;## interest
r_t = 0.55;## tax

a = ([0, E ,Fs ,0 ,E+Fs, r_i*(E+Fs) ,0 ])
bal_1 = a+a-a## Total Balance after 1st year

c_all = 0.55;## capital allowance in 2nd year
C_bal= (bal_1+0+Fs+(-(c_all*E)));## Cash Balance after 2nd year
b = ([bal_1 ,0 ,Fs ,-(c_all*E) ,C_bal, r_i*C_bal, r_t*(Fs+(r_i*C_bal))]);
bal_2 = b+b-b##Total  Balance after 2nd year

c = ([bal_2, 0 ,Fs ,0 ,bal_2+Fs ,r_i*(bal_2+Fs) ,r_t*(Fs+(r_i*(bal_2+Fs)))])
bal_3= c+c-c## Total  Balance after 3rd year

if(bal_2>0):
print('Pay back period is of two year')
else:
print('Pay back period is of three year')

print'%s %.2f %s'%('Total saving at the end of second year is ',bal_2,' Pound\n');
print'%s %.2f %s'%('Total saving at the end of third year is ',bal_3,' Pound\n');
## Deviation in answer due to direct substitution

Example 2.1

Page No. 44

Pay back period is of two year
Total saving at the end of second year is  9800.03  Pound

Total saving at the end of third year is  18353.41  Pound



## Ex2-pg45¶

In :
import math
## Example 2.2
print('Example 2.2');
print('Page No. 45');

## given
C= 35000.;## cost of boiler
C_grant=0.;## Capital grant available from goverment
E= -(C-(C_grant*C));## Net expenditure
Fs= 15250.;## Fuel Saving
r_i = 0.15;## interest
r_t = 0.55;## tax

a = ([0, E ,Fs ,0, E+Fs, r_i*(E+Fs) ,0 ])
bal_1 = a+a-a## Total  Balance after 1st year

c_all = 0.55;## capital allowance in 2nd year
C_bal= (bal_1+0+Fs+(-(c_all*E)));## Cash Balance after 2nd year
b = ([bal_1 ,0 ,Fs ,-(c_all*E), C_bal ,r_i*C_bal, r_t*(Fs+(r_i*C_bal))]);
bal_2 = b+b-b##Total  Balance after 2nd year

c = ([bal_2, 0 ,Fs ,0 ,bal_2+Fs, r_i*(bal_2+Fs) ,r_t*(Fs+(r_i*(bal_2+Fs)))])
bal_3= c+c-c## Total  Balance after 3rd year

if(bal_2>0):
print('pay back period is of two year')
else:
print('pay back period is of three year')

print'%s %.2f %s'%('Total saving at the end of second year is ',bal_2,' Pound');
print'%s %.2f %s'%('Total saving at the end of third year is ',bal_3,' Pound');
## Deviation in answer due to direct substitution

Example 2.2
Page No. 45
pay back period is of two year
Total saving at the end of second year is  4195.66  Pound
Total saving at the end of third year is  12370.74  Pound


## Ex3-pg46¶

In :
import math
## Example 2.3
print('Example 2.3');
print('Page No. 46');

## given
F= 350*10**3.;## fuel oils in gallons
Ci= 5000.;## cost of insulation of tanks

As= 7500.;##Annual Saving in Pound

if(As> Ci):
print("The investment has a pay-back period of less than 1 year");
else:
print("The investment has not a pay-back period of less than 1 year");

## Note- Since here pack back period is less than 1 year and the company is in profit so they can go with this fuel oil,
## although it can be noted that there are more problems handling heavy fuels oils
##and that the pay-back increases considerably the smaller the installation.
##So the company can changeover from oil to coal as a fuel.

Example 2.3
Page No. 46
The investment has a pay-back period of less than 1 year


## Ex4-pg47¶

In :
import math
## Example 2.4
print('Example 2.4');
print('Page No. 47');

## given
F1= 500.*10**3;## fuel oil in gallons
F2= 500.*10**3;## coal in gallons in Pound
C1= 165.*10**3;## cost of oil per year in Pound
C2= 92.*10**3;## cost of an equivalent of coal in Pound
Ce= 100.*10**3;## capital cost of extra handling eqiupment

Cm= (Ce*0.2);## Maintenance , interest costs per year
As= C1-C2;## Annual Saving in Pound
print'%s %.2f %s'%('Annual Saving is ',As,' Pound')

if((2*As)> Ce):
print("Replacing an obsolete boiler plant is considerable");
else:
print("Replacing an obsolete boiler plant is not considerble");

Example 2.4
Page No. 47
Annual Saving is  73000.00  Pound
Replacing an obsolete boiler plant is considerable


## Ex5-pg49¶

In :
import math
## Example 2.5
print('Example 2.5\n\n');
print('Page No. 49\n\n');

## given
F= 10.*10**3;## fuel oils in gallons
Cs= 2200.;## cost of maintaining tanks per year in Pound
Ci= 1850.;## cost of insulation of pipe in Pound

As= (Cs*.85);##company saving is 85 per cent to the cost
print'%s %.2f %s'%('Annual Saving on heating is ',As,'  Pound\n')

if(As> Ci):
print("The investment has a pay-back period of less than 1 year");
else:
print("The investment has not a pay-back period of less than 1 year");

Example 2.5

Page No. 49

Annual Saving on heating is  1870.00   Pound

The investment has a pay-back period of less than 1 year


## Ex6-pg52¶

In :
import math
## Example 2.6
print('Example 2.6\n\n');
print('Page No. 52\n\n');

## given
P1= 50.;## Dry saturated steam pressure in bar
P2= 0.5;## condenser  pressure in bar

##By using the steam tables saturation temperature is obtained at given pressures
T1= 537.##The saturation temperatue in K at 50 bar
T2= 306.##The saturation temperatue in K at 0.5 bar

## For Carnot Cycle
n=(1.-(T2/T1))*100.;
print'%s %.2f %s'%('Efficiency percentage of Carnot Cycle is ',n,' \n')

## For Rankine Cycle
## By usins steam tables, the total heat and the sensibles heat  and other remaining parameter has been calculated
h1= 2794.*10**3;##the total heat in dry steam at 50 bar in J/kg
d= 0.655;## dryness fraction
h2= 1725.*10**3;## the entropy at state 2 in J/kg
h3= 138.*10**3;## the sensible heat at 0.5 bar in J/kg
Vf= 1.03*10**-3;## volume of fluid im m^3,calculated from steam table
W= (Vf*(P1-P2))*10**5;## pump work in J/kg
E=(((h1-h2)-(W))/((h1-h3)-(W)))*100.;
print'%s %.2f %s'%('Efficiency percentage of Rankine Cycle is ',E,' \n')

Example 2.6

Page No. 52

Efficiency percentage of Carnot Cycle is  43.02

Efficiency percentage of Rankine Cycle is  40.13