import math
## Example 6.1
print('Example 6.1');
print('Page No. 142');
## given
L = 2.5;## Length of tubes in metre
Do = 10*10**-3;## Internal diameter of tubes in metre
m = 3.46;## mass flow rate in kg/s
Th = 120.;## Temperature of condening steam in degree celcius
Tl_i = 20.;## Inlet temperature of liquid in degree celcius
Tl_o = 80.;## Outlet temperature of liquid in degree celcius
Cp = 2.35*10**3;## Specific heat capacity of liquid in J/kg-K
U = 950.;## Overall heat transfer coefficent in W/m**2-K
T1 = Th- Tl_i;## in degree celcius
T2 = Th- Tl_o;## in degree celcius
Tm = ((T2-T1)/math.log(T2/T1));## logarithmic mean temperature of pipe in degree celcius
a = math.pi*Do*L;##Surface area per tube in m**2
A = ((m*Cp*(Tl_o - Tl_i))/(U*Tm));## in m**2
N = A/a;
print'%s %.2f %s'%('The number of tubes required is',N,'')
import math
## Example 6.2
print('Example 6.2');
print('Page No. 142');
## given
v = 1.50;## velocity in m/s
N_t = 100.;## Number of tubes
Do = 10*10**-3;## Internal diameter of tubes in metre
m = 3.46;## mass flow rate in kg/s
p = 1180.;## density in kg/m**3
A = (N_t*math.pi*Do**2)/4.;## otal cross-sectional area in m**2
V = m/p;##Volumetric flow rate in m**3/s
Fv = V/A;## Fluid velocity in m/s
N_p = v/Fv;
print'%s %.2f %s'%('the number of passes is',N_p,'')
import math
## Example 6.3
print('Example 6.3');
print('Page No. 144');
## given
Th_i = 130.;##Inlet temperature of hot liquid in degree celcius
Th_o = 90.;## Outlet temperature of hot liquid in degree celcius
Tc_i = 20.;## Inlet temperature of cold liquid in degree celcius
Tc_o = 50.;## Outlet temperature of cold liquid in degree celcius
##For Couter-current flow
T1 = Th_i - Tc_o;
T2 = Th_o - Tc_i;
Tm_1 = ((T2-T1)/math.log(T2/T1));
print'%s %.2f %s'%('The logarithmic mean temperature difference for counter-current flow ',Tm_1,' degree celcius')
##For Co-current flow
T3 = Th_i - Tc_i;
T4 = Th_o - Tc_o;
Tm_2 = ((T3-T4)/math.log(T3/T4));
print'%s %.2f %s'%('The logarithmic mean temperature difference for co-current flow is ',Tm_2,' degree celcius ')
import math
## Example 6.4
print('Example 6.4');
print('Page No. 147');
## given
F = 1.;## Fuel feed required in kg
##By ultimate analysis of feed
C = 0.86;## Carbon percentage - [%]
H2 = 0.05;## Hydrogen percentage - [%]
S = 0.001;## Sulphur percentage - [%]
O2 = 0.08;## Oxygen percentage - [%]
w_C = 12.; ## mol. weight of C
w_H2 = 2.; ##mol. weight of H2
w_O2 = 32.; ## mol. weight of O2
w_S = 32.; ##mol. weight of S
##Basis- Per kg of fuel
mol_C = C / w_C;## kmol of C
mol_H2 = H2 /w_H2;##kmol of H2
mol_O2 = O2 /w_O2;##kmol of O2
mol_S = S /w_S;##kmol of S
##Calculation of excess air
C_req = mol_C*1.;##O2 required by entering C given by reaction C+O2->CO2 in kmol
H_req = mol_H2*0.5;##O2 required by entering H2 given by reaction H2+(1/2)O2->H20 in kmol
S_req = mol_S*1.;##O2 required by entering S given by reaction S+O2->SO2 in kmol
O2_req = (C_req + H_req + S_req) - mol_O2;## in kmol
print'%s %.2f %s'%('Total number of kmol of O2 required per kg of fuel is',O2_req, 'kmol ')
m_O2 = O2_req*w_O2;## Mass of O2 required per kg of fuel
print'%s %.2f %s'%('Mass of O2 required per kg of fuel is ',m_O2,' kg ')
##Calculation of air
m_air = m_O2/0.232;## in kg
print'%s %.2f %s'%('Mass of air required per kg of fuel is ',m_air,' kg ')
##Considering air as an ideal gas,calculating volume of air by ideal gas equation-P*V = n*R*T
R = 8310;##Universal gas constant in J/kmol-K
T = (273+20);## in K
P = 1.013*10**5;## in N/m**2
n = 1;## 1 kmol of air
V_kmol = (n*R*T)/P;## In m**3/kmol
M_air = 29;## Mol. weight of air
V_kg = V_kmol/M_air;## in m**3/kg
V_air = m_air*V_kg;## in m**3
print'%s %.2f %s'%('Volume of air required is ',V_air,' m^3 ')
##Deviation in answer is due to some approximation in calculation in the book
import math
## Example 6.5
print('Example 6.5');
print('Page No. 148');
## given
F = 1.;## Weight of coal in kg
##By analysis of coal in weight basis
C = 0.74;## Carbon percentage - [%]
H2 = 0.05;## Hydrogen percentage - [%]
S = 0.01;## Sulphur percentage - [%]
N2 = 0.001;## Nitrogen percentage - [%]
O2 = 0.05;## Oxygen percentage - [%]
H20 = 0.09;## Moisture percentage - [%]
Ash = 0.05;## Ash percentage - [%]
w_C = 12.; ## mol. weight of C
w_H2 = 2.; ##mol. weight of H2
w_O2 = 32.; ## mol. weight of O2
w_S = 32.; ##mol. weight of S
##Basis- Per kg of fuel
mol_C = C / w_C;## kmol of C
mol_H2 = H2 /w_H2;##kmol of H2
mol_O2 = O2 /w_O2;##kmol of O2
mol_S = S /w_S;##kmol of S
##Calculation of excess air
C_req = mol_C*1;##O2 required by entering C given by reaction C+O2->CO2 in kmol
H_req = mol_H2*0.5;##O2 required by entering H2 given by reaction H2+(1/2)O2->H20 in kmol
S_req = mol_S*1;##O2 required by entering S given by reaction S+O2->SO2 in kmol
O2_req = (C_req + H_req + S_req) - mol_O2;## Total number of kmol of O2 required per kg of fuel in kmol
m_O2 = O2_req*w_O2;## Mass of O2 required per kg of fuel
print'%s %.2f %s'%('Mass of O2 required per kg of fuel is ',m_O2,' kg ')
##Calculation of air
m_air = m_O2/0.232;## in kg
print'%s %.2f %s'%('Mass of air required per kg of fuel is',m_air,' kg ')
##Considering air as an ideal gas,calculating volume of air by ideal gas equation-P*V = n*R*T
R = 8310.;##Universal gas constant in J/kmol-K
T = (273.+0);## in K
P = 1.013*10**5;## in N/m**2
n = 1;## 1 kmol of air
V_kmol = (n*R*T)/P;## In m**3/kmol
M_air = 29.;## Mol. weight of air
V_kg = V_kmol/M_air;## in m**3/kg
V_air = m_air*V_kg;## in m**3
print'%s %.2f %s'%('Volume of air required is ',V_air,' m^3')
import math
## Example 6.9
print('Example 6.9');
print('Page No. 157');
## given
P = 10.;## Boiler pressure in bar
Ts = 180.;## Steam temperature in degree celcius
Tf = 80.;## Feed water temperature in degree celcius
X = 0.95;## Steam dryness fraction
m_s = 4100.;## steam rate in kg/h
m_f = 238.;## Gas rate in kg/h
G_CV = 53.5*10**6.;## In J/kg
N_CV = 48*10**6.;##in J/kg
##from steam table,AT 10 bar and at temperature T = Ts
h2 = (763.+(X*2013.))*10**3;##Specific enthalpy of steam in J/kg
##At temperature T = Tf
h1 = 335.*10**3;##Specific enthalpy of feed steam in J/kg
E_G = ((m_s*(h2-h1)*100)/(m_f*G_CV));##
print'%s %.2f %s'%('The gross efficiency percentage is',E_G,'')
E_N = ((m_s*(h2-h1)*100)/(m_f*N_CV));##
print'%s %.2f %s'%('The net efficiency percentage is ',E_N,'')
import math
## Example 6.7
print('Example 6.7');
print('Page No. 150');
## given
F = 1;## Weight of fuel in kg
e = 0.5;## excess air percentage
C = 0.74;## Mass of Carbon in kg
H2 = 0.05;## Mass of Hydrogen in kg
S = 0.01;## Mass of Sulphur in kg
N2 = 0.001;##Mass of Nitrogen in kg
O2 = 0.05;## Mass of Oxygen in kg
H2O = 0.09;## Mass of Moisture in kg
Ash = 0.05;## Mass of Ash in kg
w_C = 12.; ## mol. weight of C
w_H2 = 2.; ##mol. weight of H2
w_O2 = 32.; ## mol. weight of O2
w_S = 32.; ##mol. weight of S
w_N2 = 28.;## mol. weight of N2
w_H20 = 18.;## mol. weight of H2O
##Basis- Per kg of fuel
mol_C = C / w_C;## kmol of C
mol_H2 = H2 /w_H2;##kmol of H2
mol_O2 = O2 /w_O2;##kmol of O2
mol_S = S /w_S;##kmol of S
mol_N2 = N2 /w_N2;##kmol of N2
mol_H2O = H2O /w_H20;##kmol of H20
##By kmol of product
CO2 = mol_C*1.;## CO2 formed by the reaction C + O2 -> CO2
H2O_air = mol_H2*1.;## H2O formed by the reaction H2 + (1/2)O2 -> H2O
SO2 = mol_S*1.;## SO2 formed by the reaction S + O2 -> SO2
Pdt = CO2 + H2O_air + SO2 + mol_N2 + mol_H2O;## Total kmol of combustion products in kmol
##Calculation of excess air
C_req = mol_C*1;##O2 required by entering C given by reaction C+O2->CO2 in kmol
H_req = mol_H2*0.5;##O2 required by entering H2 given by reaction H2+(1/2)O2->H20 in kmol
S_req = mol_S*1;##O2 required by entering S given by reaction S+O2->SO2 in kmol
O2_req = (C_req + H_req + S_req) - mol_O2;## Total number of kmol of O2 required per kg of fuel in kmol
Ex_O2 = O2_req*e;## Amount of excess oxygen in kmol
N2_air = (O2_req*(1+e)*79.)/21.;## in kmol (considering air consists of 79% N2 and 21% O2 by moles)
N2_flue = mol_N2 + N2_air;## Total N2 in flue gas in kmol
H2O_flue = mol_H2O+ H2O_air;## Total H2O in flue gas in kmol
T_wet = CO2 + H2O_flue + SO2 + Ex_O2 + N2_flue;##Total components of flue gas on a wet basis in kmol
T_dry = CO2 + SO2 + Ex_O2 + N2_flue;##Total components of flue gas on a dry basis in kmol
H2O_dry = 0;
C_wet = ((CO2 / T_wet)*100.);## in percentage
H_wet = ((H2O_flue/T_wet)*100.);## in percentage
S_wet = ((SO2/T_wet)*100.);## in percentage
N_wet = ((N2_flue/T_wet)*100.);## in percentage
O_wet = ((Ex_O2/T_wet)*100);## in percentage
C_dry = ((CO2 / T_dry)*100.);## in percentage
H_dry = ((H2O_dry/T_dry)*100.);## in percentage
S_dry = ((SO2/T_dry)*100.);## inpercentage
N_dry = ((N2_flue/T_dry)*100.);## in percentage
O_dry =((Ex_O2/T_dry)*100.);## in percentage
T1 = C_wet + H_wet + S_wet + N_wet +O_wet;## in percentage
T2 = C_dry + S_dry + N_dry + O_dry;## in percentage
print('components\t\t\t\t\t\t kmol \t\t\t\t\t\t % Composition by volume' '\n\n\n')
print('\t wet \t\t dry \t\t\t\t wet \t\t dry ')
print'%s %.2f %s %.2f %s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t'and '',CO2,'' '\t\t' '',CO2,'' '\t\t\t\t' '',C_wet,'' and '\t\t\t\t' '\t\t\t\t',C_dry,' \t\t\t\t ')#and '',CO2,'' and '',C_wet,'' and '',C_dry,'')
print'%s %.2f %s %.2f %s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t' and '',H2O_flue,'' '\t\t' '',H2O_dry,'' '\t\t\t\t' '',H_wet,'' '\t\t\t\t' '',H_dry,'')
print'%s %.2f %s %.2f %s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t' and '',SO2,'' '\t\t' '',SO2,'' '\t\t\t\t' '',S_wet,'' '\t\t\t\t' '',S_dry,'')
print'%s %.2f %s %.2f %s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t'and '',N2_flue,' ''\t\t' '',N2_flue,'' '\t\t\t\t' '',N_wet,'' '\t\t\t\t' '',N_dry,'')
print'%s %.2f %s %.2f %s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t' and '',Ex_O2,'' '\t\t' '',Ex_O2,'' '\t\t\t\t' '',O_wet,'' '\t\t\t\t' '',O_dry,'')
print'%s %.2f %s %.2f %s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t'and '',T_wet,'' '\t\t' '',T_dry,'' '\t\t\t\t' '',T1,'' '\t\t\t\t' '',T2,'')
#printf('\t\t kmol \t\t percent composition by volume\n Component \t Wet \t Dry \t\t Wet \t Dry \n CO2 \t %.4f %.4f \t\t %.1f \t %.1f \n H2O \t %.4f %.0f \t\t\t %.1f \t\t %.1f \n SO2 \t %.4f %.4f \t\t %.1f \t\t %.1f \n N2 \t\t %.4f %.4f \t\t %.1f \t %.1f \n O2 \t\t %.4f %.4f \t\t %.1f \t\t %.1f \n TOTAL \t %.4f %.4f \t\t %.0f \t\t %.0f'
# ,CO2,CO2,C_wet,C_dry,H2O_flue, H2O_dry,H_wet,H_dry,
#SO2,SO2,S_wet,S_dry,N2_flue, N2_flue,N_wet,N_dry,Ex_O2,Ex_O2,O_wet,O_dry,T_wet,T_dry,T1,T2)
#//Deviation in answes is due to some calculation approxiamation in the book.
import math
## Example 6.8
print('Example 6.8');
print('Page No. 156');
## given
H = 0.05;## Hydrogen percentage - [%]
O = 0.08;## Oxygen percentage - [%]
C = 0.86;## Carbon percentage - [%]
S = 0.001;## Sulphur percentage - [%]
G_CV = ((33.9*C)+143*(H-(O/8.))+(9.1*S))*10**6;
print'%s %.2f %s'%('The gross calorific value is ',G_CV,' J/kg ')
N_CV = ((33.9*C)+121*(H-(O/8.))+(9.1*S))*10**6.;
print'%s %.2f %s'%('The net calorific value is ',N_CV,' J/kg')
import math
## Example 6.9
print('Example 6.9');
print('Page No. 157');
## given
P = 10.;## Boiler pressure in bar
Ts = 180.;## Steam temperature in degree celcius
Tf = 80.;## Feed water temperature in degree celcius
X = 0.95;## Steam dryness fraction
m_s = 4100.;## steam rate in kg/h
m_f = 238.;## Gas rate in kg/h
G_CV = 53.5*10**6.;## In J/kg
N_CV = 48*10**6.;##in J/kg
##from steam table,AT 10 bar and at temperature T = Ts
h2 = (763.+(X*2013.))*10**3;##Specific enthalpy of steam in J/kg
##At temperature T = Tf
h1 = 335.*10**3;##Specific enthalpy of feed steam in J/kg
E_G = ((m_s*(h2-h1)*100)/(m_f*G_CV));##
print'%s %.2f %s'%('The gross efficiency percentage is',E_G,'')
E_N = ((m_s*(h2-h1)*100)/(m_f*N_CV));##
print'%s %.2f %s'%('The net efficiency percentage is ',E_N,'')
import math
## Example 6.10
print('Example 6.10');
print('Page No. 158');
## given
##for Boiler-1
P_1 = 15.;## Boiler pressure in bar
Ts_1 = 300.;## Steam temperature in degree celcius
Tf_1 = 80.;## Feed water temperature in degree celcius
X_1 = 0.;## Steam dryness fraction
m_s1 = 9000.;## steam rate in kg/h
m_f1 = 700.;## Gas rate in kg/h
G_CV1 = 43.0*10**6;## In J/kg
##from steam table,at P = 15 bar and at given temperatures
h2_1 = 3039.*10**3;##Specific enthalpy of steam in J/kg
h1_1 = 335.*10**3;##Specific enthalpy of feed steam in J/kg
E_G1 = ((m_s1*(h2_1-h1_1)*100.)/(m_f1*G_CV1));##
print'%s %.2f %s'%('The gross efficiency percentage is ',E_G1,'')
Ee_1 = ((m_s1/m_f1)*(h2_1-h1_1))/(2257*10**3);
print'%s %.2f %s'%('the equivalent evaporation for boiler-1 is',Ee_1,' kg ')
##for Boiler-2
P_2 = 10.;## Boiler pressure in bar
Ts_2 = 180.;## Steam temperature in degree celcius
Tf_2 = 60.;## Feed water temperature in degree celcius
X_2 = 0.96;## Steam dryness fraction
m_s2 = 7000.;## steam rate in kg/h
m_f2 = 510.;## Gas rate in kg/h
G_CV2 = 43.0*10**6.;## In J/kg
##from steam table,AT 10 bar and at temperature T = Ts_2
h2 = (763.+(X_2*2013.))*10**3.;##Specific enthalpy of steam in J/kg
##At temperature T = Tf_2
h1 = 251.*10**3.;##Specific enthalpy of feed steam in J/kg
E_G2 = ((m_s2*(h2-h1)*100)/(m_f2*G_CV2));##
print'%s %.2f %s'%('The gross efficiency percentage is',E_G2,'')
Ee_2 = ((m_s2/m_f2)*(h2-h1))/(2257*10**3);
print'%s %.2f %s'%('the equivalent evaporation for boiler-2 is ',Ee_2,' kg')
import math
## Example 6.11
print('Example 6.11\n\n');
print('Page No. 167\n\n');
## given
m = 10.*10**3;## Production of boiler in kg/h
X = 0.95;##Dryness fraction
P = 10.;##Pressure ib bar
T_fw = 95.;## Feed water temperature in degree celcius
T_mf = 230.;## Mean flue gae temperature in degree celcius
T_mb = 25.;## Mean boiler house temperature in degree celcius
Coal_c = 900.;## Coal consumption in kg/h
A = 0.08;## Ash content in coal
C_c = 0.15;##carbon content in coal
CV_coal = 33.50*10**6;## Calorific value of coal in J
M = 28.;## Mass of flue gas per kg coal in kg
Cp = 1.05*10**3;## Mean Specific heat capacity of the flue gas in J/kg-K
CV_c = 34.*10**6;## Calorific value of carbon in J/kg
M_s = m/Coal_c;## Mass of steam produced per kg coal in kg
H_w = (M_s*(763.+(X*2013.) - 398.)*10**3)/10**6;## Heat absorbed by water per kg coal in 10^6 J(from steam table at given pressure and dryness fraction)
H_f = (M*Cp*(T_mf - T_mb))/10**6;## Heat in flue gas in 10^6 J
H_uc = (A*C_c*CV_c)/10**6;##Heat in unburnt carbon in 10^6 J
h_sup = (CV_coal)/10**6;## Heat supplied by coal in 10^6 J
un_acc = (h_sup - (H_w + H_f + H_uc));## unaccounted heat losses in 10^6 J
a = (h_sup/h_sup)*100.;
b = (H_w/h_sup)*100.;
c = (H_f/h_sup)*100.;
d = (H_uc/h_sup)*100.;
e = (un_acc/h_sup)*100.;
T = b + c + d + e;
print(' THERMAL BALANCE SHEET :\n\t\t\t\t\t\t 10^6 J')
print'%s %.2f %s %.2f %s '%('percentage',a,'\t' 'Heat supplied by coal ',h_sup,'')
print'%s %.2f %s %.2f %s '%('percentage',b,'\t' 'Heat absorbed by water ',H_w,'' )
print'%s %.2f %s %.2f %s '%('percentage',c,'\t' 'Heat in flue gas',H_f,'')
print'%s %.2f %s %.2f %s '%('percentage',d,'\t' 'Heat in unburnt carbon',H_uc,'')
print'%s %.2f %s %.2f %s '%('percentage',e,'\t' 'unaccounted heat losses',un_acc,'')
print'%s %.2f %s %.2f %s '%('TOTALpercentage ',T,'\t' 'TOTAL heat supplied',h_sup,'')
import math
## Example 6.12
print('Example 6.12');
print('Page No. 168');
## given
C_Rate = 2920.;## Coal consumption rate in kg/h
S_Rate = 22.5*10**3;## Steam consumption rate in kg/h
Ps = 20.;## Steam pressure in bar
Ts = 350.;## Steam Temperature in degree celcius
Tf_in = 70.;## Feed water temperature inlet economiser in degree celcius
Tf_out = 110.;## Feed water temperature outlet economiser in degree celcius
Tm_b = 25.;## Mean Boiler house temperature in degree celcius
Tm_f = 260.;## Mean exit flue gas temperature in degree celcius
CO2_f = 15.8;## CO2 content of dry exit flue gas by volume
CO_f = 0.;## CO content of dry exit flue gas by volume
C_ash = 0.025;## Carbon in ash in [%]
G = 0.005;## Grit produced in [%]
##Analysis of coal(as fired)
M = 0.105;## Moisture [%]
VM = 0.308;##Volatile matter [%]
FC = 0.497;## FIxed carbon [%]
Ash =0.09;## ASh [%]
C = 0.66;## Carbon percentage - [%]
H2 = 0.042;## Hydrogen percentage - [%]
S = 0.015;## Sulphur percentage - [%]
N2 = 0.012;## Nitrogen percentage - [%]
O2 = 0.076;## Oxygen percentage - [%]
H20 = 0.105;## Moisture percentage - [%]
G_CV = 26.90;## Gross Calorific Value in 10**6 J/kg
CV_C = 33.8*10**6;## Calorif Value of carbon in J/kg
CV_G = 33.8*10**6;## Calorif Value of Grit in J/kg
Ps_l = 20.;## Pressure of steam leaving the boiler in bar
##(a) Calculation of excess air usage
##(a.1) Theoretical oxygen requirement
F = 1.;## Fuel feed required in kg
w_C = 12.; ## mol. weight of C
w_H2 = 2.; ##mol. weight of H2
w_S = 32.; ##mol. weight of S
w_N2 = 28.; ## mol. weight of N2
w_O2 = 32.; ## mol. weight of O2
##Basis- Per kg of fuel
mol_C = C / w_C;## kmol of C
mol_H2 = H2 /w_H2;##kmol of H2
mol_S = S /w_S;##kmol of S
mol_N2 = N2 /w_N2;##kmol of N2
mol_O2 = O2 /w_O2;##kmol of O2
##Calculation of excess air
C_req = mol_C*1;##O2 required by entering C given by reaction C+O2->CO2 in kmol
H_req = mol_H2*0.5;##O2 required by entering H2 given by reaction H2+(1/2)O2->H20 in kmol
S_req = mol_S*1;##O2 required by entering S given by reaction S+O2->SO2 in kmol
O2_req = (C_req + H_req + S_req) - mol_O2;## in kmol
N2_air = (O2_req*76.8)/23.2;## in kmol (considering air consists of 76.8% N2 and 23.2% O2 )
print('(a.1) ')
print'%s %.2f %s'%('Total number of kmol of O2 required per kg of fuel is ',O2_req,' kmol ')
print'%s %.2f %s'%('N2 associated with O2 is ',N2_air,' kmol ')
##(a.2) Theoretical CO2 content of dry flue gas
T = C_req + S_req + mol_N2 + N2_air;## Total flue gas in kmol
CO2 = (C_req/T)*100.;## in [%]
print('(a.2) ')
print'%s %.2f %s'%('Theoretical CO2 content of dry flue gas in percentage is ',CO2,' ')
##(a.3)Excess air based on CO2 content
Ex_air = ((CO2 - CO2_f)/CO2_f)*100.;## in [%]
print('(a.3) ')
print'%s %.2f %s'%('Excess air based on CO2 content in percentage is ',math.floor(Ex_air),'')
##(b) Fuel gas components
##(b.1) Composition per kg fuel
w_CO2 = 44.;## mol. weight of CO2
w_SO2 = 64.;## mol. weight of SO2
## FOR DRY GAS
CO2_d = C_req * w_CO2;## In kg/kg
SO2_d = S_req * w_SO2;## In kg/kg
N2_d = mol_N2 * w_N2;## N2 from fuel In kg/kg
N2_air_d = N2_air * w_N2;## N2 from air In kg/kg
T_N2 = N2_d + N2_air_d;## In kg/kg
T_dry = CO2_d + SO2_d + T_N2;## In kg/kg
print('(b.1) ')
print('Composition of dry gas ')
print'%s %.2f %s'%('CO2 ',CO2_d,'')
print'%s %.2f %s'%('SO2 ',SO2_d,'')
print'%s %.2f %s'%('N2 from fuel ',N2_d,'')
print'%s %.2f %s'%('N2 from air ',N2_air_d,'')
print'%s %.2f %s'%('Total dry air ',T_dry,'kg/kg')
##FOR WET GAS
w_H2O = 18.;## mol. weight of H2O
H2O_f = M;## H2O from fuel
H2O_H2 = mol_H2 * w_H2O;## H2O from H2
T_H2O = H2O_f + H2O_H2;## in kg/kg
print('Composition of wet gas ')
print'%s %.2f %s'%('H2O from fuel ',H2O_f,'')
print'%s %.2f %s'%('H2O from H2 ',H2O_H2,'')
print'%s %.2f %s'%('Total H2O in wet gas ',T_H2O,'kg/kg')
##FOR DRY EXCESS AIR
O2_dry_ex = O2_req * w_O2 *0.3;##in kg/kg
N2_dry_ex = N2_air * w_N2 *0.3;##in kg/kg
T_dry_ex = O2_dry_ex + N2_dry_ex;## in kg/kg
print('Composition of dry excess air ')
print'%s %.2f %s'%('O2 ',O2_dry_ex,'')
print'%s %.2f %s'%('N2 ',N2_dry_ex,'')
print'%s %.2f %s'%('Total dry excess air ',T_dry_ex,'kg/kg')
##(b.2) Enthalpy
## From steam table or from the appendix C.2; at the given pressure and temperatures, the following specific heat capacity for different gases are obtained
Cp_CO2_T1 = 1.04*10**3;## Specific heat Capacity of CO2 at temperature Tm_f in J/kg-K
Cp_CO2_T2 = 0.85*10**3;## Specific heat Capacity of CO2 at temperature Tm_b in J/kg-K
Cp_SO2_T1 = 0.73*10**3;## Specific heat Capacity of SO2 at temperature Tm_f in J/kg-K
Cp_SO2_T2 = 0.62*10**3;## Specific heat Capacity of SO2 at temperature Tm_b in J/kg-K
Cp_N2_T1 = 1.07*10**3;## Specific heat Capacity of N2 at temperature Tm_f in J/kg-K
Cp_N2_T2 = 1.06*10**3;## Specific heat Capacity of N2 at temperature Tm_b in J/kg-K
Cp_O2_T1 = 0.99*10**3;## Specific heat Capacity of O2 at temperature Tm_f in J/kg-K
Cp_O2_T2 = 0.91*10**3;## Specific heat Capacity of O2 at temperature Tm_b in J/kg-K
Cp_dry_T1 = ((CO2_d * Cp_CO2_T1) + (SO2_d * Cp_SO2_T1) + (T_N2 * Cp_N2_T1))/T_dry;## in J/kg-K
Cp_dry_T2 = ((CO2_d * Cp_CO2_T2) + (SO2_d * Cp_SO2_T2) + (T_N2 * Cp_N2_T2))/T_dry;## in J/kg-K
Cp_air_T1 = ((O2_dry_ex * Cp_O2_T1) + (N2_dry_ex * Cp_N2_T1))/T_dry_ex;## in J/kg-K
Cp_air_T2 = ((O2_dry_ex * Cp_O2_T2) + (N2_dry_ex * Cp_N2_T2))/T_dry_ex;## in J/kg-K
print('(b.2) ')
print'%s %.2f %s'%('Specific heat Capacity of dry gas at 260 deg C is ',Cp_dry_T1,' J/kg-K ')
print'%s %.2f %s'%('Specific heat Capacity of dry gas at 25 deg C is ',Cp_dry_T2,' J/kg-K')
print'%s %.2f %s'%('Specific heat Capacity of dry excess air at 260 deg C is ',Cp_air_T1,' J/kg-K ')
print'%s %.2f %s'%('Specific heat Capacity of dry excess air at 25 deg C is ',Cp_air_T2,' J/kg-K ')
## From Steam table or Appendix B.3, Enthalpy of superheated steam is obtained at 260 deg C and 1 bar
E_s = 2995.*10**3;##in J/kg-K
##(c) Heat transferred to water
E_w = S_Rate / C_Rate;## Evaporation of water per kg of fuel in kg
E = (E_w*(461. - 293.)*10**3)/10**6;## in 10**6 J
B = (E_w*(2797. - 461.)*10**3)/10**6;## in 10**6 J
S = (E_w*(3139. - 2797.)*10**3)/10**6;## in 10**6 J
print('(c) ')
print'%s %.2f %s'%('Heat to water in Economiser is ',E,' *10^6 J ')
print'%s %.2f %s'%('Heat to water in Boiler is ',B,' *10^6 J ')
print'%s %.2f %s'%('Heat to water in Superheater is ',S,' *10^6 J ')
##(d) Heat loss in flue gas
hl = 1056*10**3;## Enthalpy of steam at 25 deg C (from steam table) in J/kg-K
loss_dry = T_dry*((Tm_f*Cp_dry_T1) - (Tm_b*Cp_dry_T2))/10**6;## in 10**6 J
loss_wet = T_H2O*(E_s - hl)/10**6;## in 10**6 J
loss_ex_air = T_dry_ex*((Tm_f*Cp_air_T1) - (Tm_b*Cp_air_T2))/10**6;## in 10**6 J
print('(d) ')
print'%s %.2f %s'%('Heat loss in dry flue gas is ',loss_dry,' *10^6 J ')
print'%s %.2f %s'%('Heat loss in wet flue gas is ',loss_wet,' *10^6 J ')
print'%s %.2f %s'%('Heat loss in dry excess air is ',loss_ex_air,' *10^6 J ')
##(e) Heat loss in combustile matter in ash
loss_ash = (Ash * C_ash * CV_C)/10**6;## in 10**6 J
print'%s %.2f %s'%('(e) Heat loss in combustile matter in ash is ',loss_ash,' *10^6 J ')
##(f) Heat loss in grit
loss_grit = (G * CV_G)/10**6;## in 10**6 J
print'%s %.2f %s'%('(f) Heat loss in grit is ',loss_grit,' *10^6 J ')
##(g) Radiation and unaccounted heat loss
h_sup = G_CV;## Heat supplied by the coal in 10**6 J
loss_rad = (h_sup - (E + B + S + loss_dry + loss_wet + loss_ex_air + loss_ash + loss_grit));## Radiation and unaccounted loss in 10**6 J
a = (h_sup/h_sup)*100.;
b = (E/h_sup)*100.;
c = (B/h_sup)*100.;
d = (S/h_sup)*100.;
e = (loss_dry/h_sup)*100.;
f = (loss_wet/h_sup)*100.;
g = (loss_ex_air/h_sup)*100.;
h = (loss_ash/h_sup)*100.;
i = (loss_grit/h_sup)*100.;
j = (loss_rad/h_sup)*100.;
T = b + c + d + e + f + g + h + i + j;
print('(g) THERMAL BALANCE SHEET :\t percentage \t\t\t 10**6 J ')
print'%s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t percentage',a,'\t' 'Heat supplied by coal ',h_sup,'')
print'%s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t percentage',b,'\t' 'Heat absorbed to loss in economiser ',E,'' )
print'%s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t percentage',c,'\t' 'boiler',B,'')
print'%s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t percentage',d,'\t' 'superheater',S,'')
print'%s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t percentage',e,'\t' 'heat loss in :dry flue gas',loss_dry,'')
print'%s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t percentage',f,'\t' 'wet flue gas ',loss_wet,'')
print'%s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t percentage',g,'\t' 'dry eecess air ',loss_ex_air,'' )
print'%s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t percentage',h,'\t' 'heat loss in ash',loss_ash,'')
print'%s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t percentage',i,'\t' 'heat loss in grit',loss_grit,'')
print'%s %.2f %s %.2f %s '%('\t\t\t\t\t\t\t percentage',j,'\t' 'radiation and unaccounted heat losses',loss_rad,'')
print'%s %.2f %s %.2f %s '%('\t\t\t\t\tTOTAL percentage ',T,'\t' 'TOTAL heat supplied',h_sup,'')
import math
## Example 6.13
print('Example 6.13\n\n');
print('Page No. 188\n\n');
## given
P = 1.5;## Pressure in bar
T = 111.;## Temperature in degree celcius
m = 2.;## mass flow rate of process liquid in kg/s
Cp = 4.01*10**3;## Mean Specific heat capacity in J/kg_K
Tl_i = 20.;## Inlet temperature of liquid in degree celcius
Tl_o = 90.;## Outlet temperature of liquid in degree celcius
Ps = 15.;## Pressure of steam in bar
X = 0.97;## Dryness fraction of steam
Pa = 1.5;##Pressure after adiabatic expansion in bar
Ta = 80.;## Temperature of injecting condensate in degree celcius
##(a)
Q = m*Cp*(Tl_o - Tl_i);## in W
L = 2227.*10**3;## Latent heat of 1.5 bar steam in J/kg
m_a = Q/L;
print'%s %.2f %s'%('(a) Mass flow rate of 1.5 bar steam is ',m_a,' kg/s \n')
##(b)
##from steam table, Specific enthalpy of 0.97 dry 15 bar absolute steam
h = ((843.+(X*1946.))*10**3);## in J/kg
##the balance for the desuperheater,when y is the mass flow rate(kg/s) of condensate at 80 deg C is,on the basis of 1kg/s of superheated steam: => (1*2731*10^3)+(335*10^3*y)=(1+y)*2693*10^3
y = (((2731.-2693.)*10**3)/((2693.-335.)*10**3))## in kg/s
m_b = m_a/(1.+y);## in kg/s
print'%s %.2f %s'%('(b) Mass flow rate of 15 bar steam is ',m_b,' kg/s \n')
##(c)
m_c = y*m_b;##in kg/s
print'%s %.2f %s'%('(c) Mass flow rate of condensateis ',m_c,' kg/s\n')
##(d)
v = 30.;## steam velocity in m/s
##from steam table
V = 1.16;## Specific volum of 1.5 bar saturated steam in m^3/kg
V_d = V*m_a;## in m^3/s
d = ((V_d*4.)/(v*math.pi))**0.5;## im m
print'%s %.2f %s'%('(d) The vapour main diameter is ',d,' m \n')
##(e)
l = 2.5;## Length of tubes in m
d_i = 10*10**-3;## Internal Diameter of tube in m
U = 1500.;## Overall heat transfer coefficent in W/m^2-K
a = math.pi*d_i*l;## in m^2
T1 = T - Tl_i;## in degree celcius
T2 = T - Tl_o;## in degree celcius
Tm = ((T2-T1)/math.log(T2/T1));## logarithmic mean temperature of pipe in degree celcius
A = Q/(U*Tm);## in m^2
N = A/a;
print'%s %.2f %s'%('(e) The number of tubes required is ',N,'\n')