Combustion Thermodynamics

COMBUSTION THERMODYNAMICS Definitions :-

(a)

Theoretical air (stoichiometric air)

:-

The amount of air required to react with the reactants to  produce complete combustion.

(b)  Enthalpy of Formation

:-

The amount of energy absorbed or released as a compound is formed from its stable elements during a steady flow  process, both the stable elements and the compound being at standard reference state.

(c)  Heating value

:-

It is the amount of energy released when a fuel is burnt completely in a steady flow process and the products are returned to state of the reactant i.e., the standard reference state

(d)

Combustion Efficiency

:-

The ratio of ideal air-fuel ratio to actual air-fuel ratio.

(e)  Adiabatic Temperature :It is the temperature of the frame for complete combustion of  fuel with theoretical air when the combustion chamber is completely insulated. (f)

 Dew Point Temperature :-

The temperature at which the water vapor starts condensing (or dew starts forming). (g)  Internal Energy of Combustion :It is the difference of internal energies of products and reactants at the standard reference state.

Orsat Apparatus :-

Valve Flue Gas

P

X

Y

Eudiometer 

Z

 p  NaOH Solution

Aspirator  Bottle Water 

CuCl Solution 2

Pyrogallic Acid

atm

L

Flexible Hose

Q.1) Methane Methane is whirled whirled with atmosphe atmospheric ric air. The analysi analysiss of  the products on dry basis is as follows. CO2 – 10 % ; O2 – 2.37 % ; CO – 0.53 % ; N2 – 87.1 % Find the actual combustion equation. Also, calculate the air- fuel ratio ratio for for th this is combu combustion stion.. Find Find th thee perce percentage ntage of theoretical air. Find the air-fuel ratio on molar basis as well as the mass basis.

Solution :Actual combustion equation is :aCH4 + b(O2 + 3.76 N2) = 10CO2 + 2.37O2 + 0.53CO + 87.1N2 + dH2O

C : a = 10 + 0.53 = 10.53 H : 4a = 2c ; c = 21.06 O : 2b = 20 + 4.74 + 0.53 + c ; b = 23.16 Actual combustion equation is :

10.53CH4 + 23.16(O2 + 3.76 N2) = 10CO2 + 2.37O2 + 0.53CO + 87.1N2 + 21.06H2O For 1 mole of CH4, the equation is :

CH4 + 2.2(O2 + 3.76 N2) = 0.95CO2 + 0.22O2 + 0.05CO + 8.27N2 + 2H2O Theoretical combustion equation is :

CH4 + 2(O2 + 3.76 N2) = CO2 + 2H2O + 7.52 N 2 (A / F)molal = [2.2 (1 + 3.76)] / 1

= 10.47

(A / F)mass = [(2.2*32) + (2.2*3.76*28)] / [(1*2.2) + (2*3.76*28)] = 1.1 = 110 % theoretical air  =

10 % excess air 

Q.2) The mass mass analysis analysis of gaseous gaseous mixtur mixture e is given given below :

Gas :- CO2

CO

O2

N2

16.1

0.9

7.7

75.3

Determine the percentage of individual gases on volume basis.

Solution :Constituent Gases

Mass of  Gases

Molecular  Weight, M 44

Moles / 100 Kg of  mixture 0.366

Volumetric Percentage (%) 11

CO2

16.1

CO

0.9

28

0.032

0.964

O2

7.7

32

0.241

7.21

 N2

75.3

28

2.68

80.57

Total :-

100.0

3.326

100.0

xa = Va /V = 0.365 / 3.326 = 0.11 na = ma / Ma = 16.1 / 44 = 0.365

Q.3) The volumet volumetric ric analysis analysis of dry dry flue flue gas is given given as : Gas :- CO2 10

CO

O2

N2

1.5

8

80.5

Determine : (a) (a)

perc pe rcen enta tage ge of each each gas gas by mass mass

(b)

mass of O2 per kg of dry flue gas

Solution :Constituent Percentage Mole Molecular  Mass / Mass Gases By Fraction, Weight, mole of  Percentvolume(%) x M mixture age CO2

10

0.1

44

4.4

14.7

CO

1.5

0.015

28

0.42

1.4

O2

8

0.08

32

2.56

8.56

 N2

80.5

0.805

28

22.54

75.33

Total :-

100.0

1.0

29.92

100.0

Q.4) Ethane Ethane is burned burned with with 200 % excess excess air air during during a complete combustion process. The total pressure of the products is 100 kPa. Determine the air-fuel ratio and the dew point temperature of the products.

Solution :The theoretical combustion equation is : C2H6 + 3.5 (O2 + 3.76 N2 ) -------- 2 CO2 + 3 H2O + 13.16 N2 Actual combustion equation is : C2H6 + 3*3.5 (O2+3.76 N2) -------- 2CO2 + 3H2O + 39.48N2+ 7O2

A/F = ma / mf  = [ 3 * 3.5 * ( 1 * 32 + 3.76 * 28 ) ] / [ 1 * (2 * 12 + 6)] = 48.05

 pwv,prod / p prod = nwv,prod / n prod Assuming that the products of combustion are ideal gases,  pwv,prod = 3 / 51.48 * 100 = 5.83 kPa From the saturated steam table, for 5.83 kPa, the saturated temperature = 36.18 ◦C So, dew point temperature = 36.18 ◦C.

Q.5) The products of combustion of hydrocarbon fuel of 

unknown composition have the following following percentage of  products on dry basis : CO2 – 8 % ; O2 – 0.8 % ; CO – 8.8 % ; N2 – 82.4 % Calc Calcul ulat atee th thee

(a) (a)

Air Air fu fuel el rati ratio o on mass mass basi basiss

(b)) (b

Comp Compos osit itio ion n of of the the fu fuel el on mass mass basi basiss

(c) (c)

Perc Percen enta tage ge of th theo eore reti tica call air air on on mas masss basis

Solution :CaH b + d ( O2 + 3.76 N2 ) -------0.08 CO2 + 0.008 CO + 0.088 O 2+ 0.824 N2 + e H2O C:

a = 0.08 + 0.008 = 0.088

H:

b = 2e

O:

2d = 2 * 0.0 0.08 8 + 0.00 0.008 8 + 2 * 0.08 0.088 8 +e

N:

2d * 3. 3.76 = 2 * 0.824

So,

d = 0.219 e = 0.093  b = 0.185

(a )

( A / F )mass = ma/mf  =

0.219 * ( 32 + 3.76 * 28) / ( 12 * 0.089 + 0.185 ) =

(b) (b)

24

% of C = 12 * 0.08 0.089 9 / ( 12 * 0.08 0.089 9 + 0.18 0.185 5) = 85 % % of H2 = 0.185 / ( 12 * 0.089 + 0.185 ) = 15 %

(c) (c)

Theo Theore reti tica call comb combus usti tion on equa equati tion on is :

C0.089H0.185 + .1355 ( O2 + 3.76 N2 ) -------- 0.089 CO2 + 0.088 O2 + (0.1355 * 3.76) N2 + 0.093 H2O

So , % Theoretical air = (ma)actual / (ma)theoretical = 0.219 * ( 32 + 3.76 * 28 ) / 0.1355 ( 32 + 3.76 * 28) = 162 % = 62 % excess air 

Q.6) Determin Determinee the enthalpy enthalpy of combustion combustion of liquid liquid octane (C8C18) at 25 ◦C and 1 atm. using enthalpy of formation data. Assume that water in the products is in the liquid form. What is the HHV of the same fuel ?

Solution :-

The combustion equation is : C8H18 + 12.5 ( O2 + 3.76 N2 ) --------- 8 CO2 + 9 H2O + 47 N2 ħc = H products - Hreactants = nCO2 * ħf,CO2 + nH2O * ħf,H2O – nC8H18 * ħf,C8H18 = 8 * ( -393796 ) + 9 ( -286043 ) – 1 ( -250131 )

ħc = - 5474624

kJ / kg-mol of fuel

So, HHV = | ħc | = 5474624

kJ / kg-mol of fuel

Molecular wt. of fuel = 12 * 8 +18 = 114 kg / kg-mol

So, HHV = 5474624 / 114

=

48003.6 kJ / kg of fuel