# Chapter7-Energy utilisation and conversion systems¶

## Ex1-pg201¶

In [1]:
import math
## Example 7.1
print('Example 7.1');
print('Page No. 201');

## given
C = 220.*10**3;##Original annual cost of fuel in Pound
O_E = 73.;## Original Efficiency
Fl_i = 20.;## Initial Flue loss
Fl_f = 18.7;## Final Flue loss
N_E = O_E + (Fl_i - Fl_f);## New Efficiency
F_save = C*((N_E-O_E)/N_E);
print'%s %.2f %s '%('Fuel saving is',F_save,' Pound')
##Deviation in answer is due to some wrong calculation the book, instead of new efficiency in the denominator in line 13, the book has taken original efficiency

Example 7.1
Page No. 201
Fuel saving is 3849.26  Pound


## Ex2-pg201¶

In [14]:
import math
## Example 7.2
print('Example 7.2');
print('Page No. 201');
import numpy
##From Example 2.1

## given
C= 35000.;## cost of boiler  in Pound
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 = numpy.array([0, E, Fs, 0, E+Fs, r_i*(E+Fs), 0 ])

bal_1 = (E+Fs)+(r_i*(E+Fs))-0## 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 = C_bal+ r_i*C_bal- r_t*(Fs+(r_i*C_bal))  ##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= (bal_2+Fs)+r_i*(bal_2+Fs)-r_t*(Fs+(r_i*(bal_2+Fs))) ## 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

print('The data in example 2.1 indicated that:-  Saving could be made by replacing exising oil-fired burners by new burners requiring considerably less atomising steam. The financial saving are 15.25*10**3 Pound per year for an insulation and capital cost of 35*10**3 Pound.')

Example 7.2
Page No. 201
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
The data in example 2.1 indicated that:-  Saving could be made by replacing exising oil-fired burners by new burners requiring considerably less atomising steam. The financial saving are 15.25*10**3 Pound per year for an insulation and capital cost of 35*10**3 Pound.


## Ex3-pg203¶

In [15]:
import math
## Example 7.3
print('Example 7.3');
print('Page No. 203');

## given
C = 250.*10**3;##Original annual cost of fuel in Pound
O_E = 71.5;## Original Efficiency
Fl_i = 20.;## Initial Flue loss
Fl_f = 17.5;## Final Flue loss
N_E = O_E + (Fl_i - Fl_f);## New Efficiency
F_save = C*((N_E-O_E)/N_E);## in Pound
print'%s %.2f %s '%('Fuel saving is ',F_save,' Pound per year')
##Deviation in answer is due to some calculation approximation the book

Example 7.3
Page No. 203
Fuel saving is  8445.95  Pound per year


## Ex4-pg204¶

In [16]:
import math
## Example 7.4
print('Example 7.4\n\n');
print('Page No. 204\n\n');

## This question doesnot contain any calculation part.
##Refer figure 7.3, 7.4, 7.5
T_max = 200.;## Flue gas exit temperature in degree celcius
print(' The company investigate four alternative methods of heat abstraction using the flue gas.\n\n  System-1   The efficiency of the furnace without any air preheater is 79.2 per cent.\n  System-2   The efficiency of the furnace, with the air preheater only in the system operating as shown in figure 7.3, is increased to 86.6 per cent.\n  System-3   By the incorporation of the heat exchanger,the furnace efficiency is increased to 93.3 per cent using the arrangement shown in figure 7.4.\n  System-4   Using no preheating,finally achievied an overall thermal  efficiency of 93.7 per cent.\n  \t     The new air preheater scheme is shown in figure 7.5.\n\n The pay-back period in all instances is less than 3.5 years.')

Example 7.4

Page No. 204

The company investigate four alternative methods of heat abstraction using the flue gas.

System-1   The efficiency of the furnace without any air preheater is 79.2 per cent.
System-2   The efficiency of the furnace, with the air preheater only in the system operating as shown in figure 7.3, is increased to 86.6 per cent.
System-3   By the incorporation of the heat exchanger,the furnace efficiency is increased to 93.3 per cent using the arrangement shown in figure 7.4.
System-4   Using no preheating,finally achievied an overall thermal  efficiency of 93.7 per cent.
The new air preheater scheme is shown in figure 7.5.

The pay-back period in all instances is less than 3.5 years.


## Ex5-pg205¶

In [17]:
import math
# Example 7.5
print('Example 7.5\n\n');
print('Page No. 205\n\n');

##The temperature difference is not given the question.
##Refer Table 7.1
T1 = 1000.;## Furnace operating temperature in degree celcius
##T2 is  back calculated by the first condition given in table 7.1 and applying Fourier,s law of condition
T2 = 997.9545;## in degree Celcius()
dT = T1 - T2;## in degree celcius
t = 120.;## Continuous cycle time in h
K1 = 44.;## Thermal conductivity (W/m-K)
K2 = 11.;## Thermal conductivity (W/m-K)
K3 = 4.;## Thermal conductivity (W/m-K)
x1 = 250.*10**-3;## mm converted into m
x2 = 50.*10**-3;## mm converted into m
dT = T1 - T2;## in K

##By Fourier,s law of heat conduction- Q = (dT *K)/x in W/sq.m

##For 250 mm firebrick
Q1 = (dT *K1)/x1;## im W/sq.m
print'%s %.2f %s'%('Energy losses by 250 mm firebrick is ',Q1,' W/sq.m \n')

##For 250 mm hot-face insulation
Q2 = (dT *K2)/x1;## im W/sq.m
print'%s %.2f %s'%('Energy losses by 250 mm hot-face insulation is ',Q2,' W/sq.m \n') ##Deviation in answer is due to assumption of T2 as its not mentioned in the question

##For 250 mm hot-face insulation backed by 50 mm insulation
##As the resistances are in series - R = (x1/K1)+ (x2/K2) and Q = dt/R in W/sq.m
R = (x1/K2)+ (x2/K3);## in ohm
Q3 = dT/R;## in W/sq.m
print'%s %.2f %s'%('Energy losses by 250 mm hot-face insulation backed by 50 mm insulation is ',Q3,' W/sq.m \n\n') ##Deviation in answer is due to assumption of T2 as its not mentioned in the question

Example 7.5

Page No. 205

Energy losses by 250 mm firebrick is  360.01  W/sq.m

Energy losses by 250 mm hot-face insulation is  90.00  W/sq.m

Energy losses by 250 mm hot-face insulation backed by 50 mm insulation is  58.07  W/sq.m



## Ex6-pg209¶

In [18]:
import math
## Example 7.6
print('Example 7.6\n\n');
print('Page No. 209\n\n');

## given
P = 150*10*83;## Power of compressor in W
F_load = .78;## full load percentage of the time
Re = .7;##  Heat Recovery
T = 2200.;##Compressor operating time in h/year
C = 20.*10**-6;## Energy cost in Pound/Wh

H_Re = P*F_load*Re;## in W
print'%s %.2f %s'%('Heat recovered is ',H_Re,' W \n')
E_save = H_Re*T*C;## in Pound/year
print'%s %.2f %s'%('Economic Saving is ',E_save,' Pound per year')
##Deviation in answer is due to some calculation approximation the book

Example 7.6

Page No. 209

Heat recovered is  67977.00  W

Economic Saving is  2990.99  Pound per year


## Ex7-pg212¶

In [19]:
import math
## Example 7.7
print('Example 7.7\n\n');
print('Page No. 212\n\n');

## given
C_S = 1./10**3;## Cost of steam production in p/Wh
P = 75.*10**3;## Power required in W
T = 4.*10**3;## Production time in h/year
C_T = 7.*10**3;## Cost of turbine in Pound
R_T = 4.*10**3;## Annual running cost of turbine in W
C_M = 1.5*10**3;## Cost of electric motor in Pound
R_M = 14.*10**3;## Running cost of electric motor in Pound
C_M_A = 3.5/10**3;## Auunal running cost of electic motor in p/Wh
Save_R = R_M - R_T;## in Pound per year
print'%s %.e %s'%('The saving in running costs would be ',Save_R,' Pound per year')

Example 7.7

Page No. 212

The saving in running costs would be  1e+04  Pound per year


## Ex8-pg214¶

In [20]:
import math
## Example 7.8
print('Example 7.8\n\n');
print('Page No. 214\n\n');

## given
m_s = 5.3;## Factory requirement of process steam in kg/s
Pr_s_1 = 2.5;## Pressure of process steam at bar absolute
E_load_1 = 1.10*10**3;## Electrical load requirement in W
E_load_2 = 1.5*10**3;## Electrical load requirement in W
m_e = 6.0;## Mass flow rate of generated electricity in kg/s
Pr_e = 14.;## Pressure of generated electricity at bar absolute
T_heat = 2.790*10**6;## Total heat content in J/kg

##The 14 bar absolute steam would undergo an adiabatic heat drop and the steam will be expanded
h_drop = 306*10**3;## Adiabatic heat drop in J/kg
Pr_2 = 2.5;## Expanded pressure at bar absolute
Ex_stm = 0.11;## Exhaust steam percent
Ef_T = 0.65;## Tubine efficiency
R_h_drop = h_drop * Ef_T;## Real heat drop in J/kg
P_T = m_e * R_h_drop;## Power generated by turbine in W
Ef_G = 0.94;## Generator efficiency
P_G = 1.13*10**6;## Output of generator in W

##(a) Combined heat and power system
Eq_Eva = 8;## Equivalent evaporation of steam per kg coal in kg
C_req = m_e/Eq_Eva;## in kg/s
print'%s %.2f %s'%('Coal Required is ',C_req,' kg/s\n')
print('If the plant operates on a 140-h week for 50 weeks per annum the coal consumption is 18.9*0^6 kg per year.\nAt an average price of, for example, 35 Pound per tonne, the aanual cost is 660*10^3 Pound.\n\n')

##(b) Coal required for process steam
## for low pressure steam
Eq_Eva_2 = 8.25;## Equivalent evaporation of steam per kg coal in kg
Coal_req = m_s/Eq_Eva_2;## in kg/s
print'%s %.2f %s'%('Coal Required is ',Coal_req,' kg/s\n\n')
print('Assuming similar operating conditions for the plant the total coal consumption is 16.2*10^6 kg per year,\nand the annual cost is 556*10^6 Pound.\n')

##(c)Electrical Power
print('The cost of 1.15*10^6 W of electricity for the same period of time is,assuming a cost of 23 Pound per 10^6 Wh,177*10^3Pound.\nThe coal equivalent to generate 1.15*10^6 W of power for the grid would be about 5.0*10^6 kg per year.\n\nThe C.H.P. unit saves a coal equivalent of 2.3*10^6kg per year,\nover the system generating process steam and utilizing grid electricty.\nThe economic savings are 83*1063 Pound per year illustrating the benefits of a C.H.P. syatem in this case.')

Example 7.8

Page No. 214

Coal Required is  0.75  kg/s

If the plant operates on a 140-h week for 50 weeks per annum the coal consumption is 18.9*0^6 kg per year.
At an average price of, for example, 35 Pound per tonne, the aanual cost is 660*10^3 Pound.

Coal Required is  0.64  kg/s

Assuming similar operating conditions for the plant the total coal consumption is 16.2*10^6 kg per year,
and the annual cost is 556*10^6 Pound.

The cost of 1.15*10^6 W of electricity for the same period of time is,assuming a cost of 23 Pound per 10^6 Wh,177*10^3Pound.
The coal equivalent to generate 1.15*10^6 W of power for the grid would be about 5.0*10^6 kg per year.

The C.H.P. unit saves a coal equivalent of 2.3*10^6kg per year,
over the system generating process steam and utilizing grid electricty.
The economic savings are 83*1063 Pound per year illustrating the benefits of a C.H.P. syatem in this case.