COMBUSTION
COMBUSTION
THERMODYNAMICS
THERMODYNAMICS
Definitions Definitions :- :-(a)(a) Theoretical air (stoichiometric air)Theoretical air (stoichiometric air) :-
:-The amount of air required to react with the reactants to The amount of air required to react with the reactants to produce complete combustion.
produce complete combustion.
(b)
(b) Enthalpy of Formation Enthalpy of Formation :-
:-The amount of energy absorbed or released as a compound is The amount of energy absorbed or released as a compound is formed from its stable elements during
formed from its stable elements during a steady flowa steady flow
process, both the stable elements and the compound being at process, both the stable elements and the compound being at
standard reference state. standard reference state.
(c)
(c) Heating value Heating value :-
:-It is the amount of energy released when a fuel is burnt It is the amount of energy released when a fuel is burnt completely
completely in in a a steady steady flow flow process process and and the the products products areare returned
returned to to state state of of the the reactant reactant i.e., i.e., the the standard standard referencereference state
(d)
(d) Combustion EfficiencyCombustion Efficiency :- :-The ratio of ideal air-fuel
The ratio of ideal air-fuel ratio to actual air-fuel ratio.ratio to actual air-fuel ratio.
(e)
(e) Adiabatic Temperature Adiabatic Temperature :-
:-It is the temperature of the frame for complete combustion of It is the temperature of the frame for complete combustion of fuel with theoretical air when the combustion chamber
fuel with theoretical air when the combustion chamber isis completely insulated.
completely insulated. (f)
(f) Dew Point Temperature Dew Point Temperature :-
:-The temperature at which the water vapor
The temperature at which the water vapor starts condensingstarts condensing (or dew starts forming).
(or dew starts forming). (g)
(g) Internal Energy of Combustion Internal Energy of Combustion :-
:-It is the difference of internal energies of products and It is the difference of internal energies of products and reactants at the standard
Orsat Apparatus Orsat Apparatus :- :-P P X X YY ZZ Valve Valve Flue Flue Gas Gas NaOH NaOH Solution Solution Eudiometer Eudiometer p patmatm Aspirator Aspirator Bottle Bottle Water Water Flexible Flexible Hose Hose L L CuCl CuCl22 Solution Solution Pyrogallic Pyrogallic Acid Acid Q.1)
Q.1) MethMethane is whirlane is whirled with atmoed with atmosphespheric air. The anric air. The analysialysis of s of the
the products products on on dry dry basis basis is is as as follows.follows. CO
CO22 – 10 % ; O– 10 % ; O22 – 2.37 % ; CO – 0.53 % ; N– 2.37 % ; CO – 0.53 % ; N22 – 87.1 %– 87.1 %
Find the actual combustion equation. Also, calculate the Find the actual combustion equation. Also, calculate the
air-air- fuel fuel ratiratio foo for thr this cis combuombustionstion. Fin. Find thd the pe perceercentagentage of theoretical air. Find the air-fuel ratio
of theoretical air. Find the air-fuel ratio on molar basison molar basis as well as the mass basis.
as well as the mass basis.
Solution Solution
Actual combustion equation is Actual combustion equation is :-aCH
aCH44 + b(O+ b(O22 + 3.76 N+ 3.76 N22) =) =
10CO
C : a = 10 + 0.53 = 10.53 C : a = 10 + 0.53 = 10.53 H : H : 4a = 2c 4a = 2c ; ; c = c = 21.0621.06 O : O : 2b = 2b = 20 + 20 + 4.74 + 4.74 + 0.53 + 0.53 + c c ; ; b = b = 23.1623.16 Actual combustion equation is :
Actual combustion equation is :
10.53CH
10.53CH44 + 23.16(O+ 23.16(O22 + 3.76 N+ 3.76 N22) =) =
10CO
10CO22 + 2.37O+ 2.37O22 + 0.53CO + 87.1N+ 0.53CO + 87.1N22 + 21.06H+ 21.06H22OO
For 1 mole of CH
For 1 mole of CH44, the equation is :, the equation is :
CH
CH44 + 2.2(O+ 2.2(O22 + 3.76 N+ 3.76 N22) =) =
0.95CO
0.95CO22 + 0.22O+ 0.22O22 + 0.05CO + 8.27N+ 0.05CO + 8.27N22 + 2H+ 2H22OO
Theoretical combustion equation is : Theoretical combustion equation is :
CH CH44 + 2(O+ 2(O22 + 3.76 N+ 3.76 N22) = CO) = CO22 + 2H+ 2H22O + 7.52 NO + 7.52 N22 (A / F) (A / F)molalmolal = = [2.2 [2.2 (1 (1 + + 3.76)] 3.76)] / / 1 1 = = 10.4710.47 (A / F) (A / F)massmass = [(2.2*32) + (2.2*3.76*28)] / [(1*2.2) + = [(2.2*32) + (2.2*3.76*28)] / [(1*2.2) + (2*3.76*28)](2*3.76*28)] =
= 1.1 1.1 = = 110 110 % % theoretical theoretical air air =
= 10 10 % % excess excess air air
Q.2)
Gas
Gas :- :- COCO22 COCO OO22 NN22
1
166..11 00..99 77..77 7755..33
Determine the percentage of individual gases on volume Determine the percentage of individual gases on volume basis.
Solution Solution :-Constituent Constituent Gases Gases Mass of Mass of Gases Gases Molecular Molecular Weight, M Weight, M Moles / 100 Moles / 100 Kg of Kg of mixture mixture Volumetric Volumetric Percentage Percentage (%) (%) CO CO22 1166..11 4444 00..336666 1111 C COO 00..99 2288 0..00 03322 00..996644 O O22 77..77 3322 00..224411 77..2211 N N22 7755..33 2288 22..6688 8800..5577 T Toottaal l ::-- 110000..00 33..332266 110000..00 x xaa = V= Vaa /V /V = = 0.365 0.365 / / 3.326 3.326 = = 0.110.11 n naa = m= maa / M/ Maa = 16.1 / 44 = 0.365= 16.1 / 44 = 0.365 Q.3)
Q.3) The volThe volumetumetric anaric analysis of dlysis of dry flury flue gas is give gas is given as :en as : Gas Gas :- :- COCO22 COCO OO22 NN22 1 100 1..51 5 88 8800..55 Determine : Determine : (a
(a)) pepercrcenentatage ge of of eaeach ch gagas s by by mamassss ((bb)) mmaasss s oof f OO22 per kg of dry flue gasper kg of dry flue gas
Solution Solution :-Constituent Constituent Gases Gases Percentage Percentage By By volume(%) volume(%) Mole Mole Fraction, Fraction, x x Molecular Molecular Weight, Weight, M M Mass / Mass / mole of mole of mixture mixture Mass Mass Percent-age age CO CO22 1100 0..10 1 4444 44..44 1144..77 C COO 11..55 00..001155 2288 00..4422 11..44 O O22 88 00..0088 3232 22..5566 88..5566 N N22 8800..55 00..880055 2288 2222..5544 7755..3333 T Toottaal l ::-- 110000..00 11..00 2299..9922 110000..00 Q.4)
Q.4) EthanEthane is bure is burned wined with 200 % exth 200 % excess aicess air durir during ang a complete
complete combustion combustion process. process. The total The total pressure pressure of of thethe products is 100
products is 100 kPa. Determine the air-fuel ratio and thekPa. Determine the air-fuel ratio and the dew point
dew point temperature of the products.temperature of the products.
Solution Solution
:-The theoretical combustion equation is : The theoretical combustion equation is : C
C22HH66 + 3.5 (O+ 3.5 (O22 + 3.76 N+ 3.76 N22 ) --- 2 CO) --- 2 CO22+ 3 H+ 3 H22O + 13.16 NO + 13.16 N22
Actual combustion equation is : Actual combustion equation is : C
A/F A/F = = mmaa/ m/ mf f = [ 3 * 3.5 * ( 1 * 32 = [ 3 * 3.5 * ( 1 * 32 + 3.76 * 28 + 3.76 * 28 ) ] / [ 1 * (2 * 12 + ) ] / [ 1 * (2 * 12 + 6)]6)] = 48.05 = 48.05
p
p
wv,prodwv,prod// p
p
prod prod= n
= n
wv,prodwv,prod/ n
/ n
prod prodAssuming that the products of
Assuming that the products of combustion are ideal gases,combustion are ideal gases, p
pwv,prodwv,prod = = 3 / 3 / 51.48 * 51.48 * 100 = 100 = 5.83 kPa5.83 kPa
From the saturated steam table, From the saturated steam table,
for 5.83 kPa, the saturated temperature = 36.18 ◦C for 5.83 kPa, the saturated temperature = 36.18 ◦C So, dew point temperature = 36.18 ◦C.
Q.5)
Q.5) The products of combustion of hydrocarbon fuel of The products of combustion of hydrocarbon fuel of
unknown
unknown composition have composition have the followthe following percentage ing percentage of of products on dry basis :
products on dry basis : CO
CO22 – 8 % ; O– 8 % ; O22 – 0.8 % ; CO – 8.8 % ; N– 0.8 % ; CO – 8.8 % ; N22 – 82.4 %– 82.4 %
Ca
Calclcululatate e ththe e (a(a)) AiAir r fufuel el raratitio o on on mamass ss babasisiss (b
(b)) CoCompmpososititioion on of tf the he fufuel el on on mamass ss babasisiss (c
(c)) PePercrcenentatage ge of of ththeoeoreretiticacal ail air or on mn masasss basis basis Solution Solution :-C CaaHH b b + d ( O+ d ( O22 + 3.76 N+ 3.76 N22 ) ---) ---0.08 CO 0.08 CO22 + 0.008 CO + 0.088 O+ 0.008 CO + 0.088 O22+ 0.824 N+ 0.824 N22 + e H+ e H22OO C C : : a a = = 0.08 0.08 + + 0.008 0.008 = = 0.0880.088 H H : : b b = = 2e2e O : O : 2d 2d = 2 = 2 * 0* 0.0.08 + 8 + 0.0.00008 + 8 + 2 * 2 * 0.0.08088 + 8 + ee N N :: 22d d * 3* 3..776 = 6 = 2 * 2 * 00..882244 S Soo,, d d = = 00..221199 e = 0.093 e = 0.093 b = 0.185 b = 0.185
((aa)) ( ( A A / / F F ))massmass = = mmaa/m/mf f = = 0.219 0.219 * * ( ( 32 32 + + 3.76 3.76 * * 28) 28) / / ( ( 12 12 * * 0.089 0.089 + + 0.185 0.185 )) = = 2424 (b (b)) % % of of C C = = 12 12 * * 0.0.08089 / 9 / ( ( 12 12 * * 0.0.08089 + 9 + 0.0.18185 5 )) = = 85 85 %% % of H % of H22 = 0.185 / = 0.185 / ( 12 * ( 12 * 0.089 + 0.089 + 0.185 )0.185 ) = = 15 15 %%
(c
(c)) ThTheoeoreretiticacal l cocombmbusustition on eqequauatition on is is ::
C
C0.0890.089HH0.1850.185 + .1355 ( O+ .1355 ( O22 + 3.76 N+ 3.76 N22 ) ) --- --- 0.089 0.089 COCO22 + 0.088 O+ 0.088 O22
+ (0.1355 * 3.76) N
+ (0.1355 * 3.76) N22 + 0.093 H+ 0.093 H22OO
So
So , , % % Theoretical Theoretical air air = = (m(maa))actualactual/ (m/ (maa))theoreticaltheoretical
= 0.219 * ( 32 + 3.76 * 28 ) / 0.1355 ( 32 + 3.76 * 28) = 0.219 * ( 32 + 3.76 * 28 ) / 0.1355 ( 32 + 3.76 * 28) = = 162 162 %% = 62 % excess air = 62 % excess air
Q.6)
Q.6) DeteDeterminrmine the enthale the enthalpy of combuspy of combustion of liqution of liquid octaneid octane (C
(C88CC1818) at 25 ◦C and 1 atm. using enthalpy of formation) at 25 ◦C and 1 atm. using enthalpy of formation
data. data.
Assume that water in the
Assume that water in the products is in the liquid products is in the liquid form.form. What is the HHV of the same fuel ?
What is the HHV of the same fuel ?
Solution Solution
:-The combustion equation is : The combustion equation is : C
C88HH1818 + 12.5 ( O+ 12.5 ( O22 + 3.76 N+ 3.76 N22 ) --- 8 CO) --- 8 CO22 + 9 H+ 9 H22O + 47 NO + 47 N22
ħ
ħcc = H= H products products- H- Hreactantsreactants
=
= nnCO2CO2 * ħ* ħf,CO2f,CO2 + n+ nH2OH2O * ħ* ħf,H2Of,H2O – n– nC8H18C8H18 * ħ* ħf,C8H18f,C8H18
= 8 * ( -393796 ) + 9 ( -286043 ) – 1 ( -250131 ) = 8 * ( -393796 ) + 9 ( -286043 ) – 1 ( -250131 )
ħ
ħcc = = - - 5474624 5474624 kJ kJ / / kg-mol kg-mol of of fuelfuel
So,
So, HHV HHV = = | | ħħcc| | = = 5474624 5474624 kJ kJ / / kg-mol kg-mol of of fuelfuel
Molecular
Molecular wt. of wt. of fuel = fuel = 12 * 12 * 8 +18 8 +18 = 114 = 114 kg / kg / kg-molkg-mol
So,
