# Chapter 11 : Chemical Thermodynamics¶

### Example 11.1 Page no : 503¶

In :
# Variables
C = 0.88; 			#Fraction of carbon in coal
H = 0.042; 			#Fraction of Hydrogen in coal
w_f = 0.848; 			#gm
w_fw = 0.027; 			#gm
w = 1950.; 			#gm
w_e = 380.; 			#gm
dt = 3.06; 			#0C; Observed temperature rise
tc = 0.017; 			#0C
dt1 = dt+tc; 			#Corrected temperature rise
Cal = 6700.; 			#J/gm; Calorific value of fuse wire

# Calculations
Q_received = (w+w_e)*4.18*dt1; 			#Heat received by water
Q_rejected = w_fw*Cal; 			#Heat given out by fuse wire
Q_produced = Q_received - Q_rejected;
HCV = Q_produced/w_f;

# Results
print ("Higher calorific value = %.3f")% (HCV), ("kJ/kg")

LCV = HCV - 2465*9*H;
print ("Lower Calorific value = %.3f")% (LCV), ("kJ/kg")

Higher calorific value = 35126.455 kJ/kg
Lower Calorific value = 34194.685 kJ/kg


### Example 11.2 Page no : 505¶

In :
# Variables
p1 = 75.882; 			#cm of Hg
T1 = 286.; 			#K
V1 = 0.08; 			#m**3
p2 = 76.; 			#cm of Hg
T2 = 288.; 			#K

# Calculations
V2 = p1*V1*T2/p2/T1;
m = 28.; 			#kg
c = 4.18;
t2 = 23.5; 			#0C
t1 = 10.; 			#0C
Q_received = m*c*(t2-t1);
HCV = Q_received/V2;

# Results
print ("Higher calorific value  = %.3f")% (HCV), ("kJ/m**3")

amt = 0.06/0.08; 			#Amount of vapour formed per m**3 of gas burnt
LCV = HCV-2465*amt;
print ("Lower calorific value  = %.3f")% (LCV), ("kJ/kg")

Higher calorific value  = 19643.843 kJ/m**3
Lower calorific value  = 17795.093 kJ/kg


### Example 11.3 Page no : 508¶

In :
# Variables
C = 0.85; 			#Weight of Carbon present
H2 = 0.06; 			#Weight of Hydrogen present
O2 = 0.06; 			#Weight of Oxygen present

# Calculations
w_required = C*8./3 + H2*8; 			#Weight of O2 required
w_needed = w_required-O2; 			#Weight of O2 to be supplied
w_air = w_needed*100./23;

# Results
print ("Weight of air needed = %.3f")% (w_air), ("kg")

Weight of air needed = 11.681 kg


### Example 11.4 Page no : 508¶

In :
# Variables
C = 0.848; 			#kg
H2 = 0.152; 			#kg
O2_used = C*8./3 + H2*8;

# Calculations and Results
print ("(i) Minimum weight of air needed for combustion")
w_min = O2_used*100/23.;
print ("Minimum weight of air needed for combustion = %.3f")%(w_min), ("kg")

w_excess = w_min*0.15; 			#Excess air supplied
w_O2 = w_excess*23/100; 			#Weight of O2 in excess air
w_total = w_min + w_excess; 			#Total air supplied for combustion
w_N2 = w_total*77/100; 			#Weight of N2 in flue gases

print ("(ii) the volumetric composition of the products of combustion")

#For CO2
x1 = 3.109;
y1 = 44.;
z1 = x1/y1;

#For O2
x2 = w_O2;
y2 = 32.;
z2 = x2/y2;

#For N2
x3 = w_N2;
y3 = 28.;
z3 = x3/y3;
z = z1+z2+z3;

#For CO2
V1 = z1/z*100;
print ("volume of CO2  = %.3f")% (V1), ("%")

#For O2
V2 = z2/z*100;
print ("volume of O2  = %.3f")% (V2), ("%")

#For CO2
V3 = z3/z*100;
print ("volume of N2  = %.3f")% (V3), ("%")

(i) Minimum weight of air needed for combustion
Minimum weight of air needed for combustion = 15.119 kg
(ii) the volumetric composition of the products of combustion
volume of CO2  = 12.504 %
volume of O2  = 2.884 %
volume of N2  = 84.612 %


### Example 11.5 Page no : 509¶

In :
import math

# Variables
C=0.78;
H2=0.06;
O2=0.03;

w_O2=C*8./3 + H2*8;
w_min=(w_O2-O2)*100./23; 			#Minimum wt. of air needed for combustion

#For CO2
x1=0.104;
y1=44.;
z1=x1*y1;

#For CO
x2=0.002;
y2=28.;
z2=x2*y2;

#For N2
x3=0.816;
y3=28.;
z3=x3*y3;

#For O2
x4=0.078;
y4=32.;
z4=x4*y4;

z=z1+z2+z3+z4;

# Calculations
W_CO2=z1/z; 			#Weight per kg of flue gas
W_CO=z2/z; 			#Weight per kg of flue gas
W_N2=z3/z; 			#Weight per kg of flue gas
W_O2=z4/z; 			#Weight per kg of flue gas

amt=3./11*W_CO2 + 3./7*W_CO;

W=C/amt; 			#Weight of dry flue gas per kg of fuel

# Results
print ("(i)Weight of dry flue gas per kg of fuel = %.3f")% (W), ("kg")

m_O2=W_O2-4/7*W_CO; 			#Weight of excess oxygen per kg of flue gas
m_excess=W*m_O2; 			#Weight of excess O2 per kg of fuel

w_excess=m_excess*100/23; 			#Weight of excess air per kg of fuel
print ("(ii)Weight of excess air per kg of fuel= %.3f")% (w_excess), ("kg")

(i)Weight of dry flue gas per kg of fuel = 18.382 kg
(ii)Weight of excess air per kg of fuel= 6.655 kg


### Example 11.6 Page no : 510¶

In :
# Variables
v_CO = 0.05;
v_CO2 = 0.10;
v_H2 = 0.50;
v_CH4 = 0.25;
v_N2 = 0.10;
V_fuel = 1.;

# Calculations
V_O2 = v_CO/2+v_H2/2+2*v_CH4; 			#Volume of O2 needed
V_air = V_O2*100./21; 			#Volume of air required
V_N2 = V_air*79./100; 			#Volume of nitrogen in the air
V = v_CO + v_CO2 + v_CH4 + v_N2 + V_N2; 			#Dry combustion products
O2 = 6.;
V_excess = O2*V/(21-O2);
V_total = V_air+V_excess;

ratio = V_total/V_fuel;

# Results
print ("Air fuel ratio = %.f")%(ratio)

Air fuel ratio = 5


### Example 11.7 Page no : 511¶

In :
import math

# Variables
C = 0.85;
H2 = 0.15;

#For CO2
x1 = 0.115;
y1 = 44;
z1 = x1*y1;

#For CO
x2 = 0.012;
y2 = 28;
z2 = x2*y2;

#For O2
x3 = 0.009;
y3 = 32;
z3 = x3*y3;

#For N2
x4 = 0.86;
y4 = 28;
z4 = x4*y4;

# Calculations
z = z1+z2+z3+z4;

W_CO2 = z1/z; 			#Weight per kg of flue gas
W_CO = z2/z; 			#Weight per kg of flue gas
W_O2 = z3/z; 			#Weight per kg of flue gas
W_N2 = 4/z; 			#Weight per kg of flue gas

W_C = 3./11*W_CO2 + 3./7*W_CO; 			#Weight of carbon per kg of flue gas

W = C/W_C; 			#Weight of dry flue gas per kg of fuel

Vapour = 1.35; 			#kg; Vapour of combustion

W_total = W+Vapour; 			#Total weight of gas

W_air = W_total-1; 			#Air supplied

ratio = W_air/1;

# Results
print ("Ratio of air to petrol  = %.3f")% (ratio)

S_air = (C*8./3 + H2*8)*100./23; 			#Stoichiometric air

W_excess = W_air-S_air; 			#Excess air

Excess = W_excess/S_air*100; 			#Percentage excess air

print ("Percentage excess air %.3f")% (Excess),
print ("%")

Ratio of air to petrol  = 16.951
Percentage excess air 12.461 %


### Example 11.8 Page no : 512¶

In :
# Variables
C = 0.86;
H2 = 0.08;
S = 0.03;
O2 = 0.02;

# Calculations and Results
W_O2 = C*8./3 + H2*8 + S*1;
A = W_O2-O2; 			#Weight of oxygen to be supplied per kg of fuel
W_min = A*100./23;
r_correct = 1./W_min/1; 			#“correct” fuel-air ratio
r_actual = 1./12;

print ("(i) Mixture strength")
s = r_actual/r_correct*100; 			#Mixture strength

richness = s-100;
print ("richness =  %.3f")% (richness), ("%")
print ("This show that mixture is 6.5% rich.")

D = 1/r_correct-1/r_actual;

CO = 0.313; 			#kg
CO2 = 2.662; 			#kg
N2 = 9.24; 			#kg
SO2 = 0.06; 			#kg

print ("(ii) The percentage composition of dry flue gases")

#For CO
x1 = 0.313; 			#kg
y1 = 28.;
z1 = x1/y1;

#For CO2
x2 = 2.662; 			#kg
y2 = 44;
z2 = x2/y2;

#For N2
x3 = 9.24; 			#kg
y3 = 28;
z3 = x3/y3;

#For SO2
x4 = 0.06; 			#kg
y4 = 64;
z4 = x4/y4;

z = z1+z2+z3+z4;

#Let percentage volume be denoted by V
V_CO = z1/z*100;
print ("Percentage volume of CO = %.3f")% (V_CO), ("%")

V_CO2 = z2/z*100;
print ("Percentage volume of CO2 = %.3f")% (V_CO2), ("%")

V_N2 = z3/z*100;
print ("Percentage volume of N2 = %.3f")% (V_N2), ("%")

V_SO2 = z4/z*100;
print ("Percentage volume of SO2 = %.3f")% (V_SO2), ("%")

(i) Mixture strength
richness =  6.643 %
This show that mixture is 6.5% rich.
(ii) The percentage composition of dry flue gases
Percentage volume of CO = 2.776 %
Percentage volume of CO2 = 15.027 %
Percentage volume of N2 = 81.964 %
Percentage volume of SO2 = 0.233 %


### Example 11.9 Page no :513¶

In :
import math

# Variables
A = 992./284*100./23; 			#Air required for complete combustion
B = 13.; 			#kg/kg of fuel; Air actually supplied
D = A-B; 			#Deficiency of air

# Calculations
W_CO2 = 0.466*11./3;
W_CO = 0.379*7./3;
W_H2O = 22./142*9;
W_N2 = 13.*0.77;

#For CO2
x1 = W_CO2
y1 = 44.;
z1 = x1/y1;

#For CO
x2 = W_CO;
y2 = 28.;
z2 = x2/y2;

#For H2O
x3 = W_H2O;
y3 = 18.;
z3 = x3/y3;

#For N2
x4 = W_N2;
y4 = 28.;
z4 = x4/y4;

z = z1+z2+z3+z4;

CO2 = z1/z*100;

# Results
print ("Percentage of CO2 = %.3f")% (CO2), ("%")

CO = z2/z*100;
print ("Percentage of CO = %.3f")% (CO), ("%")

H2O = z3/z*100;
print ("Percentage of H2O = %.3f")% (H2O), ("%")

N2 = z4/z*100;
print ("Percentage of N2 = %.3f")% (N2), ("%")

Percentage of CO2 = 7.684 %
Percentage of CO = 6.249 %
Percentage of H2O = 15.328 %
Percentage of N2 = 70.739 %


### Example 11.11 Page no : 515¶

In :
# Variables
C = 80.;
#Analysis of gas entering the economiser
CO2_1 = 8.3;
CO_1 = 0.;
O2_1 = 11.4;
N2_1 = 80.3;

#Analysis of gas leaving the economiser
CO2_2 = 7.9;
CO_2 = 0.;
O2_2 = 11.5;
N2_2 = 80.6;

# Calculations and Results
A1 = N2_1*C/33./(CO2_1 + CO_1); 			#Air supplied on the basis of conditions at entry to the economiser
A2 = N2_2*C/33./(CO2_2 + CO_2); 			#Air applied on the basis of conditions at exit

leakage = A2-A1; 			#Air leakage
print ("Air leakege  = %.3f")% (leakage), ("kg of air per kg of fuel")

W_fuel = 0.85; 			#kg; Weight of fuel pasmath.sing up the chimney

c = 1.05;
T2 = 410.;
T1 = 0.;

W = A1+W_fuel; 			#Total weight of products
Q1 = W*c*(T2-T1); 			#Heat in flue gases per kg of coal
Q2 = leakage*1.005*(20-0); 			#Heat in leakage air
t = (Q1+Q2)/(1.005*leakage + W*1.05);
dT = T2-t;

print ("Fall in temperature as a result of the air leakage into the economiser %.3f")%(dT),("C")

Air leakege  = 1.280 kg of air per kg of fuel
Fall in temperature as a result of the air leakage into the economiser 18.711 C


### Example 11.12 Page no : 516¶

In :
# Variables
w_O2 = 3.*32./46*100./23; 			#For complete combustion of 1 kg of C2H6O, oxygen required

# Calculations and Results
ratio = w_O2;
print ("A:F ratio = %.3f")%(ratio)

w1 = 88.; 			#kg
w2 = 54.; 			#kg

w = w1+w2; 			#kg
W = 46.; 			#kg

w_CO2 = w1/W*100;
print ("CO2 produced by fuel %.3f")% (w_CO2), ("%")

w_H2O = w2/W*100;
print ("H2O produced by fuel %.3f")% (w_H2O), ("%")

A:F ratio = 9.074
CO2 produced by fuel 191.304 %
H2O produced by fuel 117.391 %


### Example 11.13 Page no : 517¶

In :
# C2H2+xO2---->aCO2+bH2O

# Calculations
Amount =  3.076 + 10.12;

# Results
print ("Hence amount of theoretical air required for combustion of 1 kg acetylene  = "), (Amount), ("kg")

Hence amount of theoretical air required for combustion of 1 kg acetylene  =  13.196 kg


### Example 11.14 Page no : 517¶

In :
# C2H2+2.5O2+2.5*(79/21)N2 --> 2CO2+H2O+2.5*(79/21)N2

# Variables
m_CO2 = 3.38; 			#kg
m_H2O = 0.69; 			#kg
m_O2 = 3.07; 			#kg
m_N2 = 20.25; 			#kg

# Calculations
m_total = m_CO2+m_H2O+m_O2+m_N2;

CO2 = m_CO2/m_total*100;
H2O = m_H2O/m_total*100;
O2 = m_O2/m_total*100;
N2 = m_N2/m_total*100;

# Results
print ("Hence the gravimetric analysis of the complete combustion is : ")
print ("CO2 = %.3f")%(CO2), ("%")

print ("H2O = %.3f")%(H2O), ("%")

print ("O2 = %.3f")% (O2), ("%")

print ("N2 = %.3f")%(N2), ("%")

Hence the gravimetric analysis of the complete combustion is :
CO2 = 12.340 %
H2O = 2.519 %
O2 = 11.208 %
N2 = 73.932 %


### Example 11.15 Page no : 518¶

In :
# Calculations
AF_mole = (12.5+12.5*(79./21))/1;
AF_mass = AF_mole*28.97/(8.*12+1*18);

# Results
print ("Air fuel ratio  = %.3f")%(AF_mass), ("kg air/kg fuel")

Air fuel ratio  = 15.126 kg air/kg fuel


### Example 11.16 Page no : 518¶

In :
# C8H18+12.5*O2+12.5*(79/21)N2 --> 8CO2+9H2O+12.5*(79/21)*N2
# C8H18 + (2) (12.5) O2 + (2) (12.5)*(79/21)N2-->8CO2 + 9H2O + (1) (12.5) O2 + (2) (12.5)*(79/21)*N2

# Calculations and Results
m_fuel = 1*(8*12+1*18);
m_air = 2*12.5*(1+79./21)*28.97;

AF = m_air/m_fuel;
print ("(i) Air-fuel ratio  = %.3f")%(AF)

print ("(ii) Dew point of the products")
n = 8+9+12.5+2*12.5*(79./21);

x = 9./n;
p = 100.*x;

#Hence
t_dp = 39.7; 			#0C

print ("t_dp = %.3f")%(t_dp),("°C")

(i) Air-fuel ratio  = 30.253
(ii) Dew point of the products
t_dp = 39.700 °C


### Example 11.17 Page no : 519¶

In :
# C2H6 + 3.5O2 → 2CO2 + 3H2O

# Calculations
n = 1.3+0.7+0.9*3.5*(79./21);
CO2 = 1.3/n*100;
CO = 0.7/n*100;
N2 = 11.85/n*100;

# Results
print ("Volumetric analysis of dry products of combustion is as follows ")

print ("CO2  = %.3f")% (CO2), ("%")

print ("CO  = %.3f")% (CO), ("%")

print ("N2  = %.3f")% (N2), ("%")

Volumetric analysis of dry products of combustion is as follows
CO2  = 9.386 %
CO  = 5.054 %
N2  = 85.560 %


### Example 11.18 Page no : 520¶

In :
print ("(i) Combustion equation")

# Variables
z = 87.1;
y = z*(79/21.);
x = 10+0.53;
a = 2*x;

#10.53 CH4 + 23.16 O2 + 87.1 N2 → 10.0 CO2 + 0.53 CO + 2.37 O2 + 21.06 H2O + 87.1 N2

print ("CH4 + 2.2 O2 + 8.27 N2 → 0.95 CO2 + 0.05 CO + 2H2O + 0.225 O2 + 8.27 N2")

print ("(ii) Air-fuel ratio ")

AF_mole = 2.2+8.27;
print ("air-fuel ratio on a mole basis  = "), (AF_mole), ("moles air/mole fuel")

AF_mass = AF_mole*28.97/(12+1*4);
print ("air-fuel ratio on a mass basis  = %.3f")%(AF_mass), ("air/kg fuel")

# CH4 + 2O2 + 2*(79/21)N2 → CO2 + 2H2O + (2)*(79/21)N2
AF_theor = (2+2*(79./21))*28.97/(12+1*4);
print ("theoretical air-fuel ratio  = %.3f")% (AF_theor), ("kg air/kg fuel")

theo = AF_mass/AF_theor*100;
print ("(iii) Percent theoretical air  = %.3f")%(theo), ("%")

(i) Combustion equation
CH4 + 2.2 O2 + 8.27 N2 → 0.95 CO2 + 0.05 CO + 2H2O + 0.225 O2 + 8.27 N2
(ii) Air-fuel ratio
air-fuel ratio on a mole basis  =  10.47 moles air/mole fuel
air-fuel ratio on a mass basis  = 18.957 air/kg fuel
theoretical air-fuel ratio  = 17.244 kg air/kg fuel
(iii) Percent theoretical air  = 109.935 %


### Example 11.19 Page no : 521¶

In :
print ("(i) The stoichiometric A/F ratio")

# Variables
a = 0.82/12; 			# Carbon balance
b = 0.10/2; 			#Hydrogen balance
x = (2*a+b)/2; 			# Oxygen balance

# Calculations and Results
Stoichiometric_AF_ratio = 2.976/0.233;

print ("Stoichiometric AF ratio  = %.3f")%(Stoichiometric_AF_ratio)

n = a+b+3.76*x;
CO2 = 0.068/n*100;
H2 = 0.05/n*100;
N2 = 3.76*0.093/n*100;

print ("the analysis of the products is")
print ("CO2  = %.3f")%(CO2),("%")

print ("H2  = %.3f")%(H2),("%")

print ("N2  = %.3f")%(N2),("%")

(i) The stoichiometric A/F ratio
Stoichiometric AF ratio  = 12.773
the analysis of the products is
CO2  = 14.491 %
H2  = 10.655 %
N2  = 74.516 %


### Example 11.20 Page no : 522¶

In :
import math
# C + O2 → CO2
# 2H2 + O2 → 2H2O
# S + O2 → SO2

# Variables
O2_req = 2.636; 			#kg

# Calculations and Results
AF = O2_req/0.233;
print ("The stoichiometric A/F ratio  = %.3f")%(AF)

AF_act = AF+0.3*AF;
print ("(i) Actual A/F ratio  = %.3f")%(AF_act)

print ("(ii) Wet and dry analyses of products of combustion by volume")

n_wet = 0.5208;
n_dry = 0.5008;

print ("Analysis of wet products is as follows :")

CO2 = 0.0734/n_wet*100;
print ("CO2  = %.3f")%(CO2), ("%")

H2O = 0.0200/n_wet*100;
print ("H2O  = %.3f")%(H2O), ("%")

SO2 = 0.0002/n_wet*100;
print ("SO2  = %.3f")% (SO2), ("%")

O2 = 0.0244/n_wet*100;
print ("O2  = %.3f")% (O2), ("%")

N2 = 0.4028/n_wet*100;
print ("N2  = %.3f")%(N2), ("%")

print ("Analysis of dry products is as follows :")

CO2 = 0.0734/n_dry*100;
print ("CO2  = %.3f")% (CO2), ("%")

SO2 = 0.0002/n_dry*100;
print ("SO2  = %.3f")% (SO2), ("%")

O2 = 0.0244/n_dry*100;
print ("O2  = %.3f")%(O2),("%")

N2 = 0.4028/n_dry*100;
print ("N2  =  %.3f")%(N2),("%")

The stoichiometric A/F ratio  = 11.313
(i) Actual A/F ratio  = 14.707
(ii) Wet and dry analyses of products of combustion by volume
Analysis of wet products is as follows :
CO2  = 14.094 %
H2O  = 3.840 %
SO2  = 0.038 %
O2  = 4.685 %
N2  = 77.343 %
Analysis of dry products is as follows :
CO2  = 14.657 %
SO2  = 0.040 %
O2  = 4.872 %
N2  =  80.431 %


### Example 11.21 Page no : 523¶

In :
# Variables
n_O2 = 0.853; 			#total moles of O2

# Calculations and Results
AF = n_O2/0.21;
print ("(i) Stoichiometric A/F ratio  = %.3f")%(AF)

print ("(ii) Wet and dry analyses of the products of combustion if the actual mixture is 30% weak :")
AF_act = AF+0.3*AF;
n_N2 = 0.79*AF_act;
O2_excess = 0.21*AF_act-n_O2;

n_wet = 5.899;
n_dry = 4.915;

print ("Analysis by volume of wet products is as follows :")
CO2 = 0.490/n_wet*100;
print ("CO2  = %.3f")%(CO2),("%")

H2O = 0.984/n_wet*100;
print ("H2O  = %.3f")%(H2O),("%")

O2 = O2_excess/n_wet*100;
print ("O2  = %.3f")%(O2), ("%")

N2 = n_N2/n_wet*100;
print ("N2  = %.3f")%(N2),("%")

print ("Analysis by volume of dry products is as follows :")

CO2 = 0.490/n_dry*100;
print ("CO2  = %.3f")%(CO2), ("%")

O2 = O2_excess/n_dry*100;
print ("O2  = %.3f")%(O2), ("%")

N2 = n_N2/n_dry*100;
print ("N2  = %.3f")%(N2), ("%")

(i) Stoichiometric A/F ratio  = 4.062
(ii) Wet and dry analyses of the products of combustion if the actual mixture is 30% weak :
Analysis by volume of wet products is as follows :
CO2  = 8.306 %
H2O  = 16.681 %
O2  = 4.338 %
N2  = 70.717 %
Analysis by volume of dry products is as follows :
CO2  = 9.969 %
O2  = 5.207 %
N2  = 84.874 %


### Example 11.22 Page no : 525¶

In :
# Calculations and Results
O2_req = 3*32./46;
AF = O2_req/0.233;
print ("Stoichiometric A/F ratio  = %.3f")% (AF)

mix = 0.8; 			#mixture strength

AF_actual = AF/mix;
print ("Actual A/F ratio  = %.3f")%(AF_actual),

# C2H6O + 1.25*(3 O2 + 3*79/21 N2) → 2CO2 + 3H2O + 0.25*3O2 + 1.25*3*79/21 N2

n = 2+3+0.75+14.1;

print ("Hence wet analysis is")

CO2 = 2/n*100.;
print ("CO2  = %.3f")%(CO2), ("%")

H2O = 3/n*100.;
print ("H2O  = %.3f")%(H2O),("%")

O2 = 0.75/n*100;
print ("O2  = %.3f")%(O2), ("%")

N2 = 14.1/n*100;
print ("N2  = %.3f")%(N2),("%")

nd = 2+0.75+14.1; 			#total dry moles

print ("Hence dry analysis is : ")

CO2 = 2/nd*100.;
print ("CO2  = %.3f")%(CO2),("%")

O2 = 0.75/nd*100;
print ("O2  = %.3f")%(O2),("%")

N2 = 14.1/nd*100;
print ("N2  = %.3f")% (N2), ("%")

mix = 1.3;
AF_act = AF/mix;
print ("Actual A/F ratio  = %.3f")% (AF_act)

print "Hence wet analysis is :"
CO2 = .614/13.678 * 100
print "CO2 = %.2f %%"%CO2
CO = 1.386/13.678 * 100
print "CO = %.2f %%" %CO
H2O = 3/13.678 * 100
print "H2O = %.2f %%" %H2O
N2 = 8.678/13.678 * 100
print "N2 = %.2f %%" %N2

print "Hence dry analysis is :"
CO2 = .614/10.678 * 100
print "CO2 = %.2f %%"%CO2
CO = 1.386/10.678 * 100
print "CO = %.2f %%" %CO
N2 = 8.678/10.678 * 100
print "N2 = %.2f %%" %N2

Stoichiometric A/F ratio  = 8.957
Actual A/F ratio  = 11.196 Hence wet analysis is
CO2  = 10.076 %
H2O  = 15.113 %
O2  = 3.778 %
N2  = 71.033 %
Hence dry analysis is :
CO2  = 11.869 %
O2  = 4.451 %
N2  = 83.680 %
Actual A/F ratio  = 6.890
Hence wet analysis is :
CO2 = 4.49 %
CO = 10.13 %
H2O = 21.93 %
N2 = 63.44 %
Hence dry analysis is :
CO2 = 5.75 %
CO = 12.98 %
N2 = 81.27 %


### Example 11.23 Page no : 527¶

In :
# C2H6O + 3O2 + 3*79/21 N2 → 2CO2 + 3H2O + 3*79/21 N2

# Variables
R0 = 8.314*10**3; 			#kJ/kg K
m = 46.; 			#kg

print ("(i) Volume of reacmath.tants per kg of fuel ")

n = 1+3+3*79./21;
T = 323.; 			#K
p = 1.013*10**5; 			#Pa

# Calculations
V = n*R0*T/p;

Vr = V/m;
print ("Vr = %.3f")% (Vr), ("m**3")

print ("(ii) Volume of products per kg of fuel")

n = 2+3+3*79./21;
T = 403.; 			#K
p = 1*10.**5; 			#Pa

V = n*R0*T/p;

Vp = V/m;
print ("Vp = %.3f")% (Vp), ("m**3")

(i) Volume of reacmath.tants per kg of fuel
Vr = 8.809 m**3
(ii) Volume of products per kg of fuel
Vp = 11.862 m**3


### Example 11.24 Page no : 527¶

In :
# 0.506H2 + 0.1CO + 0.26CH4 + 0.04C4H8 + 0.004O2 + 0.03CO2 + 0.06N2 + 0.21 × 7O2 + 0.79 × 7N2 → a CO2 + b H2O + c O2 + d N2

# Variables
a = 0.1*0.26+4*0.04+0.03;
b = (2*0.506+4*0.26+8*0.04)/2;
c = (0.1+2*0.004+2*0.03+0.21*7*2-2*a-b)/2;
d = (2*0.06+2*0.79*7)/2;
n = 0.55+0.411+5.59;

# Calculations and Results
print ("analysis by volume is")
CO2 = 0.55/n*100;
print ("CO2 = %.3f")%(CO2), ("%")

O2 = 0.411/n*100;
print ("O2 = %.3f")%(O2), ("%")

N2 = 5.59/n*100;
print ("N2  = %.3f")%(N2), ("%")

analysis by volume is
CO2 = 8.396 %
O2 = 6.274 %
N2  = 85.330 %


### Example 11.25 Page no : 528¶

In :
# C_aH_bO_cN_dS_e

# Variables
a = 60./12;
b = 20.;
c = 5./16;
d = 10./14;
e = 5./32;
p = 5;
q = 20./2;

# Calculations
r = 0.1562;
x = (2*p+q+2*r-0.3125)/2;
s = (0.7143+2*x*79./21)/2;
air = (9.92*32+x*79./21*28)/100;

# Results
print ("Stoichiometric air required  = %.3f")%(air), ("kg/kg of fuel")

Stoichiometric air required  = 13.708 kg/kg of fuel


### Example 11.26 Page no : 529¶

In :
# C_aH_bO_cN_d

# Variables
a = 84./12;
b = 10.;
c = 3.5/16;
d = 1.5/14;

# C7 H10 O0.218 N0.107 + x O2 + x*(79/21)N2 → p CO2 + q H2O + r N2

p = 7.;
q = 10./2;
x = (2*p+q-c)/2.;
r = (d+2*x*(79./21))/2;

# Calculations and Results
AF = (x*32+x*79/21*28)/100;
print ("(i)Stoichiometric A/F ratio  = %.3f")% (AF)

# C7H10O0.218N0.107 + (1.2)(9.39) O2 + (1.2)(9.39)*(79/21)N2 → 7CO2 + 5H2O + (0.2)(9.39) O2 + (1.2)(35.4) N2

n = 7+0.2*9.39+1.2*35.4;

print ("(ii)Percentage composition of dry flue gases by volume is as follows :")
CO2 = 7/n*100.;
print ("CO2  = %.3f")% (CO2), ("%")

O2 = 1.878/n*100;
print ("O2  = %.3f")% (O2), ("%")

N2 = 42.48/n*100;
print ("N2  = %.3f")% (N2), ("%")

(i)Stoichiometric A/F ratio  = 12.896
(ii)Percentage composition of dry flue gases by volume is as follows :
CO2  = 13.630 %
O2  = 3.657 %
N2  = 82.714 %


### Example 11.27 Page no : 530¶

In :
# a C + b H + c O2 + (79/21)*c N2  =  8CO2 + 0.5CO + 6.3O2 + x H2O + 85.2N2

# Variables
a = 8+0.5;
c = 85.2/(79./21);
x = 2*(c-8-0.5/2-6.3);
b = 2*x;

# Calculations and Results

AF = (c*32+(79./21)*c*28)/(a*12+b*1);
print ("(i) Air-fuel ratio  = %.3f")% (AF), ("kg of air/kg of fuel")

mf_C = 12*a/(12.*a+b);
mf_H2 = b*1/(12.*a+b);
air = mf_C*8./3*100./23.3 + mf_H2*8*100/23.3; 			#air required for complete combustion
percent = AF/air*100.;
print ("(ii)Per cent theoretical air required for combustion  = %.3f")% (percent), ("%")

(i) Air-fuel ratio  = 23.144 kg of air/kg of fuel
(ii)Per cent theoretical air required for combustion  = 136.444 %


### Example 11.28 Page no : 531¶

In :
print ("(i) By a carbon balance")

# Calculations and Results
# a C8H18 + 78.1N2 + 78.1*(21/79)O2 → 8.9CO2 + 8.2CO + 4.3H2 + 0.5CH4 + 78.1N2 + x H2O
a = (8.9+8.2+0.5)/8;
AF1 = (78.1*28+78.1*21./79*32)/a/(8*12+1*18);
print ("Air fuel ratio  = %.3f")% (AF1)

print ("(ii) By a hydrogen-oxygen balance ")

# a C8H18 + b O2 + b*(79/21)N2 → 8.9CO2 + 8.2CO + 4.3H2 + 0.5CH4 + b*(79/21)N2 + x*H2O

a = (8.9+8.2+0.5)/8;
x = (18*a-4.3*2-4*0.5)/2;
b = (8.9*2+8.2+x)/2;

AF2 = (b*32+b*(79./21.)*28.)/a/(8.*12+1*18);
print ("Air fuel ratio  = %.3f")% (AF2)

(i) By a carbon balance
Air fuel ratio  = 11.368
(ii) By a hydrogen-oxygen balance
Air fuel ratio  = 11.089


### Example 11.29 Page no : 532¶

In :
# X(0.88/12 C + 0.12/2 H2) + Y O2 + 79/21*Y N2 → 0.12CO2 + a O2 + (0.88 – a) N2 + b H2O

# Variables
X = 0.12/(0.88/12);
b = 0.06*X;
a = 0.0513;
Y = 0.2203;

# Calculations
Air_supplied = 0.2203*32/0.233;
AF = Air_supplied/X;

# Results
print ("A/F ratio  = %.3f")% (AF)

A/F ratio  = 18.490


### Example 11.30 Page no : 532¶

In :
# X*(x/12 C + y/2 H2) + Y O2 + 79/21*Y/N2 → 0.15CO2 + 0.03CO + 0.03CH4 + 0.01H2 + 0.02O2 + a H2O + 0.76N2

# Variables
Y = 0.76/(79./21);
a = 2*(Y-0.15-0.03/2-0.02);
Xx = 12*(0.15+0.03+0.03);
Xy = 2*(2*0.03+0.01+a);

# Calculations
ratio = Xx/Xy;

# Results
print ("Ratio of C to H2 in fuel  = %.3f")% (ratio)

Ratio of C to H2 in fuel  = 12.109


### Example 11.31 Page no : 533¶

In :
# Variables
h_fg0 = 2441.8; 			#kJ/kg
m = 3.*18;
dH0_liq = -3301000.; 			#kJ/mole

# Calculations
dH0_vap = dH0_liq+m*h_fg0;

# Results
print ("dH0_vapour  = %.3f")% (dH0_vap), ("kJ/mole")

dH0_vapour  = -3169142.800 kJ/mole


### Example 11.32 Page no : 533¶

In :
# Variables
# C6H6 + 7.5O2  → 6CO2 + 3H2O (vapour)
dH0 = -3169100.; 			#kJ
n_R = 1+7.5;
n_P = 6+3;
R0 = 8.314;
T0 = 298.; 			#K

# Calculations
dU0 = (dH0-(n_P-n_R)*R0*T0)/(6*12+1*6);

# Results
print ("dU0  = %.3f")% (dU0), ("kJ/kg")

dU0  = -40645.369 kJ/kg


### Example 11.33 Page no : 534¶

In :
# CO+1/2 O2 → CO2
import math

# Variables
H_R0 = 1*9705+1/2*9696; 			#kJ
H_RT = 1*94080+1/2*99790; 			#kJ
H_P0 = 1*10760; 			#kJ
H_PT = 1*149100; 			#kJ

# Calculations
dH_T = -(285200+(143975-14553)-(149100-10760));

# Results
print ("dH_T  = "), (dH_T), ("kJ/mole")

dH_T  =  -276282 kJ/mole


### Example 11.34 Page no : 535¶

In :
print ("(i) Higher heating value at consmath.tant pressure")

# Variables
m = 4.*18;
h_fg = 2443.; 			#kJ/kg
LHVp = 2044009.; 			#kJ/kg
R0 = 8.3143; 			#kJ/kg K
T = 298.; 			#K

# Calculations and Results
HHVp = LHVp+m*h_fg;
print ("HHVp  = "), (HHVp), ("kJ/kg")

print ("(ii) Higher heating value at consmath.tant volume")
dn = 3-(1+5.);

HHVv = HHVp+dn*R0*T;
print ("HHVv  = "), (HHVv), ("kJ/kg")

(i) Higher heating value at consmath.tant pressure
HHVp  =  2219905.0 kJ/kg
(ii) Higher heating value at consmath.tant volume
HHVv  =  2212472.0158 kJ/kg


### Example 11.35 Page no : 535¶

In :
# Variables
HHV = 5494977.; 			#kJ/kg
m = 9.*18;
u_fg = 2305.; 			#kJ/kg

# Calculations
LHVv = HHV-m*u_fg;

# Results
print ("LHVv  = "), (LHVv), ("kJ/kg")

LHVv  =  5121567.0 kJ/kg


### Example 11.36 Page no : 535¶

In :
print ("(i) Air and benzene vapour ")

# C6H6(g) + 7.5O2(g) + 7.5*(79/21)N2(g)  =  6CO2(g) + 3H2O(g) + 7.5*(79/21)*N2(g)

# Variables
LHVp = 3169500.; 			#kJ/mole
m = 54.; 			#kg/kg mole of fuel
h_fg = 2442.; 			#kJ/kg

# Calculations and Results
LHVv = round(LHVp/((12*6+6*1)+(7.5*32)+7.5*(79./21)*28))

print ("LHVv per kg of mixture  = %d")% (LHVv), ("kJ/kg")

HHVp = (LHVp+m*h_fg)/(78+240+790);

print ("HHVp per kg of mixture  = %.0f")% (HHVp), ("kJ/kg")

print ("(ii) Air and octane vapour ")
LHVp = 5116200.; 			#kJ/mole of C8H18

# C8H18(g) + 12.5O2(g) → 8CO2(g) + 9H2O(g) + 12.5*(79/21)N2(g)

LHVp1 = LHVp/((12*8+18*1)+12.5*32+12.5*79./21*28);
print ("LHVp per kg of mixture  = %d")% (LHVp1), ("kJ/kg")

m = 9*18.;
HHVp = LHVp+m*h_fg;
HHVp1 = HHVp/(114.+400+1317);
print ("HHVp per kg of mixture  = %d")% (HHVp1), ("kJ/kg")

(i) Air and benzene vapour
LHVv per kg of mixture  = 2861 kJ/kg
HHVp per kg of mixture  = 2980 kJ/kg
(ii) Air and octane vapour
LHVp per kg of mixture  = 2794 kJ/kg
HHVp per kg of mixture  = 3010 kJ/kg


### Example 11.37 Page no : 537¶

In :
# Variables
m_CO2 = 44./12*0.88; 			#kg
m_H2O = 18/2*0.12; 			#kg
u_fg = 2304.; 			#kJ/kg
h_fg = 2442.; 			#kJ/kg
HHVv = 45670.; 			#kJ/kg
R0 = 8.3143; 			#kJ/kg K
T = 298.; 			#K

# Calculations and Results

LHVv = HHVv-m_H2O*u_fg;
print ("(i) (LHV)v  = "), (LHVv), ("kJ/kg")

print ("(ii) (HHV)p, (LHV)p")

#1 mole fuel+x/32 O2-->3.23/44 CO2 + 1.08/18 H2O

x = 32*(m_CO2/44+m_H2O/18./2.);

# 1 kg fuel + 3.31 kg O2  =  3.23CO2 + 1.08H2O
dn = (m_CO2/44-x/32);
HHVp = HHVv-dn*R0*T;

print ("HHVp  = %.3f")% (HHVp), ("kJ/kg")

LHVp = HHVp-m_H2O*h_fg;
print ("LHVp  = %.3f")% (LHVp), ("kJ/kg")

(i) (LHV)v  =  43181.68 kJ/kg
(ii) (HHV)p, (LHV)p
HHVp  = 45744.330 kJ/kg
LHVp  = 43106.970 kJ/kg