Fuels & Combustion

7/30/2019 Fuels & Combustion

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FUELS AND

COMBUSTION


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Atomic MassMolecular Mass


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1 kmol + 1kmol 1 kmol


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1 kmol + 0.5kmol 1 kmol


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1 kmol + 1kmol 1 kmol


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Air 100 m3 100kg


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Example

A sample of coal has 16% ash and 4%moisture. The analysis on dry and ashfree basis is C = 0.88, H=0.06. O=0.04,

N=0.01, S=0.01. Determine the minimumamount of air needed to burn 3 kg ofcoal. Find the composition of productgas.


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Element DAF Actual =DAF*0.8

C : 0.88 : 0.704

H : 0.06 : 0.048

O : 0.04 : 0.032

S : 0.01 : 0.008

N : 0.01 : 0.008

Ash : 0.160

M : 0.040

Total : 1.000


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(A/F)s = 9.73 kg/kg

Air required for 3 kg of fuel = 29.18kg


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mass of products / kg of fuel

CO2 = 0.704 *(11/3) = 2.5813 kg (24.44%)

H2O = 0.048*9 + 0.04 = 0.472 kg (4.47 %)

SO2 = 0.008*2 = 0.016 kg (0.15 %)

N2 = 0.008+9.73*0.77 = 7.500 kg (71.02%)

Total = 10.56 kg


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mass

mass = number of moles * Molecular Mass

m= nM

n = m/M n is proportional to Volume


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Volume of products / kg of fuel

CO2 = 2.5813/44 =0.0586 kmol (16.60%)

H2O = 0.472/18 =0.026 kmol (07.42%)

SO2 = 0.01/64 =0.00025 kmol (0.07%)

N2 = 7.500/28 = 0.267kmol (75.88%)

TOTAL 0.35296 kmol 100.00%


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Volumetric (Volume based)

Analysis

CO2 = 2.5813/44 =0.0586 kmol (16.71%)

H2O = 0.432/18 =0.024 kmol (6.8571%)

SO2 = 0.01/64 =0.00025 kmol (0.0714%)

N2 = 7.500/28 = 0.267kmol (76.457%)

TOTAL 0.35065 kmol 100%


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Equivalence Ratio (or)

Relative Fuel to Air Ratio

mixturesrichfor1

mixturestricStoichiomefor1

mixturesleanfor1)(

)(

s

a

AF

AF


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Chemical Reaction

Equations forStoichiometric combustion

(OR)Theoretical Combustion

(OR)Ideal Combustion


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Example

Calculate the theoretical A/F for burning

gasoline in an engine if it is assumed that

the chemical formula of the gasoline is

C8H18 .


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Combustion with oxygen

C8H18 + O2 = CO2 + H2O (unbalanced)

C8H18 + 12.5O2 = 8CO2 + 9H2O (balanced)


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Combustion with Air

nN2 / nO2 = 79/21 = 3.7619

C8H18 + 12.5 O2 + (12.5*3.7619) N2= 8CO2 + 9H2O + (12.5*3.7619) N2

C8H18 + 12.5 O2 + 47.023 N2= 8CO2 + 9H2O + 47.023 N2


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mass of fuel = 8*12 +18*1 = 114 kg

mass of air = 12.5*32 + 47.0237*28

= 1716.6 kg

(A/F)s = 15.05 kg/kg


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Gravimetric Compositions

CO2 : 19.22%

H2O : 8.85 %

N2 : 71.92%


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Volumetric Compositions

Total Volume = 64.02 kmol

CO2 : 12.49%

H2O : 14.057%

N2 : 73.44%


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A fuel has the following gravimetriccomposition

Hexane (C6H14) 40%

Octane (C8H18) 30%

Cyclohexane (C6H12) 25%

Benzene (C6H6) 05%

Determine the stoichiometric air-fuel ratio.

example


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Molecular mass of

Hexane (C6H14) 86 kg/kmol

Octane (C8H18) 114Cyclohexane (C6H12) 84

Benzene (C6H6) 78


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n = m/M

Molar volume of 100 kg of fuel

Hexane (C6H14) 0.4651 kmol

Octane (C8H18) 0.2631Cyclohexane (C6H12) 0.2976

Benzene (C6H6) 0.0641


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0.465(C6H14)+0.2631(C8H18)+0.2976(C6H12)+0.064(C6H6) + a O2+3.76a N2

= bCO2 + d H2O + e N2

C : 0.465*6 +0.2631*8 + 0.2976*6+0.064*6 = b

b = 7.064

H : 0.465*14 +0.2631*18 + 0.2976*12+0.064*6 = 2d

d = 7.6029O : 2a = 2b + d a = 10.8683

N : 3.76*2*a = 2e e = 40.86

Chemical Reaction for 100 kg of

fuel


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0.465(C6H14)+0.2631(C8H18)+0.2976(C6H12)+0.064(C6H6) + 10.8683 O2 + 40.86 N2

= 7.064CO2 + 7.6029 H2O + 40.86 N2

mass of air = 1491.86 kg

mass of fuel = 100 kg

(A/F)s = 14.9186 kg/kg

Chemical Reaction for 100 kg of

fuel


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Excess Air combustion


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Need for Excess Air


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Example

Octane is burnt with 30% excess air.

Calculate the amount of air needed and

the volumetric analysis of resultant dry

flue gases


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Stoichiometric Combustion with

oxygen

C8H18 + O2 = CO2 + H2O

C8H18 + 12.5 O2 = 8CO2 + 9H2O


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Stoichiometric Combustion with Air

nN2 / nO2 = 79/21 = 3.7619

C8H18 + 12.5 O2 + (12.5*3.7619) N28CO2 + 9H2O + (12.5*3.7619) N2

C8H

18+ 12.5 O

2+ 47.023 N

2

8CO2 + 9H2O + 47.023 N2


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mass of fuel = 8*12 +18*1 = 114 kg

mass of air = 12.5*32 + 47.0237*28

= 1716.6 kg

(A/F)s = 15.05 kg/kg


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Combustion with Excess Air

C8H18 + (1.3) 12.5 O2 + (1.3) (47.0237)N2

aCO2 + bH2O + d N2 + eO2.

C : 8=a

H : 18=2b

N : (1.3)(47.0237)(2)=2dO : (1.3)(12.5)(2)=2e+2a+b


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C8H18 + (1.3) 12.5 O2 + (1.3) (47.0237)N2

8CO2 + 9H2O + 61.13 N2 + 3.75O2.

C : a = 8

H : b = 9

N : d = 61.13O : e = 3.75


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Volumetric Compositions of total

Product gas

CO2 : 9.77 %

H2O : 10.99 %

N2 : 74.66 %

O2 : 4.50%


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Volumetric Compositions of Dry

Product gas

CO2 : 10.97 %

N2 : 83.88 %

O2 : 5.146 %


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Example

Octane is burnt with 30% insufficient air.

Calculate the amount of air needed and

the volumetric analysis of resultant dry

flue gases


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oxygen

C8H18 + O2 = CO2 + H2O

C8H18 + 12.5 O2 = 8CO2 + 9H2O


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nN2 / nO2 = 79/21 = 3.7619

C8H18 + 12.5 O2 + (12.5*3.7619) N28CO2 + 9H2O + (12.5*3.7619) N2

C8

H18

+ 12.5 O2

+ 47.023 N2

8CO2 + 9H2O + 47.023 N2


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mass of fuel = 8*12 +18*1 = 114 kg

mass of air = 12.5*32 + 47.0237*28

= 1716.6 kg

(A/F)s = 15.05 kg/kg


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Combustion with Insufficient Air

C8H18 + (0.7) 12.5 O2 + (0.7) (47.0237)N2

aCO2 + bH2O + d N2 + eCO.

C : 8=a+e

H : 18=2b

N : (0.7)(47.0237)(2)=2dO : (0.7)(12.5)(2)=2a+b+e


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Combustion with insufficient Air

C8H18 + (1.3) 12.5 O2 + (1.3) (47.0237)N2

8CO2 + 9H2O + 32.9166 N2 +7.5O2.

C : a = 0.5

H : b = 9

N : d = 32.9166O : e = 7.5

V l t i C iti f T t l


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Volumetric Compositions of Total

Product gas

CO2 : 1.00%

H2O : 18.03%

N2 : 65.943%

CO : 15.025%

V l t i C iti f D


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Volumetric Compositions of Dry

Product gas

CO2 : 1.22%

N2 : 80.44%

O2 : 18.33%


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The fuel supplied to petrol engine isassumed to have the composition C2H16.

Calculate (a) the stoichiometric air-fuel

ratio by mass, (b) the percentage

volumetric composition of the products of

combustion if 50% excess air is supplied

and combustion is complete. Assume air

contains 21% O2 by volume

Example


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Combustion with oxygen

C2H16 + O2 = CO2 + H2O

C2H16 + 6 O2 = 2CO2 + 8H2O


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Combustion with Stoichiometric Air

nN2 / nO2 = 79/21 = 3.7619

C2H16 + 6 O2 + 6*3.7619 N2

= 2CO2 + 8H2O + 22.57 N2

mass of Air = 823.68 kg

mass of fuel = 40 kg

(A/F)s= 20.592


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(A/F)a = 1.5 (A/F)s

= 30.89 kg/kg


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C2H16 + (1.5) 6 O2 + (1.5) 6*3.7619 N2

= aCO2 + bH2O + d N2 + eO2.

C : a = 2

H : b = 8

N : d = 33.85O : e = 3


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C2H16 + 9 O2 + 33.85 N2

= 2CO2 + 8H2O + 33.85 N2 + 3O2.


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mass of Air = kg

mass of fuel = kg

(A/F)s =


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Example

Find the percentage volumetric compositionof the pre combustion and combustionmixtures when heptane (C7H16) is burnt

with 20% excess air. Find also molecularweight, the specific volume and the gasconstant at S.T.P of the post combustionmixture. Assume that air contains 21%

oxygen and 79% Nitrogen by volume. Thevolume of the kg mole at S.T.P. is 22.41m3. [8].


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In this case :

Std Temperature = 273 K


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Stoichiometric Combustion

C 7H16+aO2+3.76aN2 bCO2+dH2O+eN2


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a = 11

b = 7

d = 8

e = 41.36

C 7H16+ 11O2+ 41.36N2 7CO2+8H2O+41.36N2


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(A/F)s = 15.10

(A/F)a

= 1.2*15.10 = 18.10


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C7H16+aO2+3.76aN2

bCO2+dH2O+eO2+fN2

(A/F)a =(32a+3.76*28*a)/(7*12+16) = 18.10

a = 13.2

Actual Combustion


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a = 13.200

b = 7.000

d = 8.000

e = 2.200

f = 49.632

Volumetric composition of Pre


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Volumetric composition of Pre

Combustion Mixture

Total volume = 1+a +3.76a = 63.83 kmol

C7H16 1.56%

O2 20.67%

N2 77.77%


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Post Combustion Mixture

Total volume= b+d+e+f = 66.83 kmol

CO2 10.49%

H2O 11.99%

O2 3.27%

N2 74.25%


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M= 28.6

R=0.29

Given T = 273

v= 0.7826 m3/kg


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Example

One kg of Octane burns in 12 kg of air. If

there is no free oxygen in the products,

calculate the percentage of CO2 in the

dry flue gases



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oxygen

C8H18 + O2 = CO2 + H2O

C8H18 + 12.5 O2 = 8CO2 + 9H2O


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nN2 / nO2 = 79/21 = 3.7619

C8

H18

+ 12.5 O2

+ (12.5*3.7619) N2

8CO2 + 9H2O + (12.5*3.7619) N2

C8H18 + 12.5 O2 + 47.023 N2

8CO2 + 9H2O + 47.023 N2


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mass of fuel = 8*12 +18*1 = 114 kg

mass of air = 12.5*32 + 47.0237*28

= 1716.6 kg

(A/F)s = 15.05 kg/kg


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C8H18 + a O2 + 3.76aN2

bCO2 + dH2O + e N2 + f CO.

(A/F)a = (32a+3.76a*28)/(8*12 + 18) = 12

a = 9.968


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C8H18 + a O2 + 3.76aN2

bCO2 + dH2O + e N2 + fCO.

C : 8 = b + f

H : 18 = 2d

N : 2*3.76a = 2eO : 2a = 2b + d + f


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Combustion with Insufficient AirC8H18 + a O2 + 3.76aN2

bCO2 + dH2O + e N2 + fCO.

C : 8 = b + f

H : 18 = 2d

N : 2*3.76a = 2eO : 2a = 2b + d + f

b = 2.93

d = 9e = 37.468

f = 5.07


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C8H18 + 9.968 O2 +37.468N2

2.93CO2 + 9 H2O + 37.468 N2 + 5.07 CO.


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C8H18 + 9.968 O2 +37.468N2 2.93CO2 + 9 H2O + 37.468 N2 + 5.07 CO.

Volume of CO2 = 2.93 kmol

volume of dry flue gases = 2.93+37.468+5.07 = 45.45 kmol

Volumetric % of CO2 = 2.93/45.45 = 6.45%


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Example

A Diesel engine is supplied with fuel of

average composition of C12H16. The air

coefficient (actual A/F / Stoichiometric A/F)

during idling is 4.5 and during full loadrunning is 1.35. Determine stoichiometric

fuel air ratio. Also Determine the fuel air

ratio for each of the above cases and thegravimetric analysis of the products of

combustion at full load.


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Combustion with St. Air

C12H16 + 16O2 + 16*3.76 N2 12 CO2 + 8 H2O + 16*3.76 N2

mass of air = ma = 16*32 + 16*3.76*28 = 2196.4 kg

mass of fuel = mf = 12*12 + 16*1 = 160 kg

(A/F)s = 13.72


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Idling

Idling(A/F)a / (A/F)s = 4.5

(A/F)a =4.5*13.728 = 61.776

(F/A)a =0.016

% of excess air = (61.776 13.72)/13.72

= 350%

Combustion with 350% Excess Air


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or

Combustion with 450% St. Air

C12H16 + (4.5)16O2 + (4.5) 16*3.76 N2

a CO2 + b H2O + d N2 + eO2

C ( )16O ( ) 16*3 6


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C12H16 + (4.5)16O2 + (4.5) 16*3.76 N2

a CO2 + b H2O + d N2 + eO2

C : 12 = a

H : 16 = 2b

N : 4.5*16*3.76*2 = 2d

O : 4.5*16*2 = 2a + b + 2e

a = 12

b = 8

d = 270.72e = 56

F ll O L d


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Full O Load

Idling

(A/F)a / (A/F)s = 1.35

(A/F)a = 1.35 *13.728 = 18.53

(F/A)a =0.052

% of excess air = (18.53 13.72)/13.72= 35%



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or

Combustion with 135% St. Air

C12H16 + (1.35)16O2 + (1.35) 16*3.76 N2

a CO2 + b H2O + d N2 + eO2

C H (1 35)16O (1 35) 16*3 76 N


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C12H16 + (1.35)16O2 + (1.35) 16*3.76 N2

a CO2 + b H2O + d N2 + eO2

C : 12 = a

H : 16 = 2b

N : 1.35*16*3.76*2 = 2d

O : 1.35*16*2 = 2a + b + 2e

a = 12

b = 8

d = 81.22e = 5.6

C H (1 35)16O (1 35) 16*3 76 N


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C12H16 + (1.35)16O2 + (1.35) 16*3.76 N2

12 CO2 + 8 H2O + 81.22 N2 + 5.6O2

Proportional

mass of CO2 = 12*44 = 528 kg 16.89%

mass of H2O = 8*18 = 144 kg 4.60

mass of N2 = 81.22*28 = 2275.182 kg 72.76%mass of O2 = 5.6*32 = 179.22 kg 5.73%

TOTAL =3125.36 kg 100%


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Molar coeff Mass mass%

O2 21.59

Co2 12 528 16.89

H2o 8 144 4.6

O2 5.6 179.2 5.73

N2 81.216 2274.048 72.76

total 3125.248

P1 D98 6


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P1D98 6

Fuel for a SI engine contains 84% C and16% H2. The F/A is 1:14. Assuming

complete combustion of H2 and no

Carbon residue Calculate(i) The mass of C burning to CO

(ii) The mass of C burning to CO2

(iii) The mass of individual constituent gasesin the product

E l


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Example

Fuel for a SI engine contains 84% C and16% H2. The F/A is 1:14. Assuming

complete combustion of H2 and no

Carbon residue Calculate(i) The mass of C burning to CO

(ii) The mass of C burning to CO2

(iii) The mass of individual constituent gasesin the product


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167827

2H

c

HCHC

8n

7n

nMm

kg16H2ofmass

84KgCofmass


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304.15F

A

0016.08)84.0(3

8

23

100

F

A

S8

OH8C

3

8

23

100

F

A

s

s

s


89/154

(A/F)s =15.304 kg/kg

Given value < 15.304

Insufficient Air combustion

A i 100 k f f l


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Assuming 100 kg of fuel,

mc = 84 kg 1

mH2 = 16 kg 2

nM = m

n1 = 7 kmoln2 = 8 kmol


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7C + 8 H2 + a O2 + 3.76 a N2

b CO2 + d H2O +e N2 + f CO

(A/F)a = [32 a + 3.76 a *28] /100 = 14

a = 10.19


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7C + 8 H2 + a O2 + 3.76 a N2 b CO2 + d H2O +e N2 + f CO

C : 7 = b + f

H : 16 = 2d

O : 2a = 2b +d +f

N : 2*3.76a = 2e


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7C + 8 H2 + a O2 + 3.76 a N2 b CO2 + d H2O +e N2 + f CO

a = 10.19 (from Actual A/F)b = 5.39

d = 8

e = 38.31

f = 1.6


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7C + 8 H2 + 10.19 O2 + 38.31 N2

5.39 CO2 + 8 H2O +38.31 N2 + 1.6 CO


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Mass of C CO2 : [(Vco2)/(Vco2+Vco)]. mc

: {(5.39)/(5.39+1.6)}*0.84

: 0.647 kg of C/kg of Fuel

Mass of C CO : {(Vco)/(Vco2+Vco)}. mc

: {(1.6)/(5.39+1.6)}*0.84

: 0.192 kg of C/kg of Fuel

The mass of individual constituent


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gases in the product

CO2 : b*44/100 : 2.3716 kg/kg

H2O : d*18/100 : 1.44 kg/kg

N2 : e*28/100 : 10.78

CO : f* 28/100 : 0.448

ComBUSTOR

1` kg of fuel

14 kg of fuel

15 kg of fuel

Alternative method


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Alternative method

Based on 100 kmol of fuel


98/154

Vol of C = 7 kmol

Vol of H = 16 kmol

Vol % of C = 30.43%Vol % of H = 69.57%



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30.43C+69.56H+aO2+3.76aN2==>bCO2+dH2O+eN2

a = 47.8225b= 30.43

d= 34.785

e=179.81

(A/F)s= 15.1

Actual Combustion


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Actual Combustion

30.43C+69.56H+aO2+3.76aN2==>bCO2+dH2O+eN2+fCO

a = 44.26b= 23.41

d= 34.785

f =6.965e = 166.68


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Mf mco = 0.1926

Mf mco2 = 0.6740

Mass of combustion gas


102/154

Mass of combustion gas

Mfg = 14.94 kg/kg

mco2 = 2.376 kgof co2/kg of F

Mh2o = 1.4

Mco = 0.4494

Mn2 = 10.73

LOSS DUE TO INCOMPLETE COMBUSTION


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HHV

kJ/Kg

LHV

C 33900 33900

CO 10170 10170

S 9295 9295

H2 144000 121500

LOSS DUE TO INCOMPLETE COMBUSTION

For 1 kg of Fuel


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For 1 kg of Fuel

Heat loss due to incomplete combustion0.1926 (33900-10170)= 4570.39kJ

Total heat released= 0.647*33900

+0.1926*10170+0.16*144000 = 46932.04kJ

% of heat loss = 9.73%

???????????


105/154

???????????

Q = 51516 kJ/kg

Q = 49577

Loss = 3.76%


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A/F from EG Composition

Example


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The Orsat analysis of the exhaust gasesfrom a diesel engine using a hydrocarbon

fuel is 7.4% CO2, 8.8O2, 1.1 CO. The

pressure of the exhaust gases is 10Pa(bar). Calculate (a) A/F (b) the mass

analysis of fuel the wet product analysis.

Take molecular mass of air as 29. Aircontains 21% O2 and 79% N2.

Example

C bH2 dO2 3 6dN2


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aC+bH2+dO2+3.76dN27.4CO2+8.8O2+1.1CO+82.7N2+eH2O

C a = 7.4 + 1.1

H 2b = 2eO 2d = 2*7.4 +2*8.8+1.1+e

N 2*3.76*d=2*82.7

Molar coeff Mass Mass%

C a= 8 5 Co2 325 6 10 36


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C a 8.5 Co2 325.6 10.36

H2 b=10.489 O2 281.6 8.96

O2 d=21.99 CO 30.8 0.98

N2 82.7 H2O 189 6.014

e = 10.48 N2 2315.6 73.68

Total 3142.6 100

mass of air = (21.99*32)+ (3.7621.99*28)=3019.28kg

mc=102 kg

mH2 =20.96 kg

mf =122.96

(A/F)a =24.55

C 82.95%, H 17.05%

NOTE 1


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NOTE 1


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170569105.1791.652.8291.6

2H

c

HCHCHC

52.8n

91.6n

nMm

kg17.05H2ofmass

82.95KgCofmass


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5.15F

A

S8OH8C

38

23100

FA

s

s

Dew Point Temperature


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Dew Point Temperature

Volumetric % of H2O = 9.5 % = 0.095

the partial pressure of H2O = 0.095*10 bar

= 0.95 bar

dew point temperature

98.2 deg C

Example


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Example

Octane (C8H18) is burnt with dry air. Thevolumetric analysis of the dry products of

combustion is: CO2=10.02%;

CO = 0.88% and N2 = 83.48%;O2=5.62% Determine : (a) AirFuel ratio

(b) % of (theoretical) air used.

For 100 kmol of dry product gas


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For 100 kmol of dry product gas

aC8H18+ bO2 + 3.76b N210.02CO2 + 0.88 CO+5.62 O2 + 83.48N2 + dH2O

C a=1.36O b = 22.2

H d = 12.26

N To check


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(A/F)a = 19.61

(A/F)s = 15.05

% of actual air = 19.61/15.05*100= 130.3%

Example


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Example

Coal having ultimate analysis of 0.65 C,0.05H, 0.05 O, 0.05 H2O (rest ash) is

burnt with 10% excess air. The Orsat

analysis shows no CO 9% CO2, 10.5%O2 and rest Nitrogen. Determine the air

fuel ratio and percent excess air

Enthalpy of Reaction


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Enthalpy of Reaction

Generally, a chemical reaction is performedin a steady flow device at a constant

pressure with no work transfer, as shown

in fig.

FUEL

AIR

Gases

Q1 2


119/154

For this steady flow process, neglecting thechanges in kinetic and potential energies,

we have

QR = HP - HR= H2 - H1 = (H)RWhere

H2 = Enthalpy of products at section 2

H1 = Enthalpy of reactants at section 1

Chemical Reactions


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Chemical Reactions

C + O2 CO2

H2 + 0.5 O2 H2O

CO+0.5O2 CO2

CaCO3 CaO + CO2

If the reactants and products are both at the


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same temperature, the quantity (H)R is

called the enthalpy of react ion.

The enthalpy of react ion or heat ofreaction of any chemical reaction atconstant pressure, is the heat released(per unit quantity of fuel) in a steady-flowprocess in which the reactants and

products are both at the sametemperature.


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When a chemical reaction takes place there isusually l iberat ion o r absorp t ion o f heat


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usually l iberat ion o r absorp t ion o f heat.

Chemical reactions which liberate energy in theform of heat are known as exothermicreact ionsand those which absorb energy arecalled endothermic react ions.

In the above equation, if Q is positive, heat isadded to the system and Hp2 > HR1, the reactionis endothermic reaction.

If Q is negative, heat is released by the system,the reaction is exothermic and Hp2 < HR1,.

Combustion Reaction


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Combustion Reaction

It is a fast, exothermic, oxidation, chemicalreaction.

Enthalpy of combust ion


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py

The enthalpy of Combus t ion or heat ofCombustion of any combustion reactionat constant pressure, is the heat released(or enthalpy change) per quantity unit offuel in a steady-flow combustion processin which the reactants and products areboth at the same temperature.

Units : kJ/kg or kJ/kmol

Combustion Reactions


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C + O2 CO2

H2 + 0.5 O2 H2O

CO+0.5O2 CO2

CaCO3 CaO + CO2


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gOHgO21gH

lOHgO2

1gH

222

222


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During combustion reactions energy isliberated in the form of heat.

Here Q is negative, because heat isreleased by the system, the reaction is

exothermic and Hp2 < HR1,.

Enthalpy of Formation


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py

The enthalpy of formation of a compound issimply the enthalpy of reaction for the

formation of one unit of compound from its

most stable elements at the standardconditions (at 25C and 1 atm).

Formation Reactions


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C + O2 CO2

H2 + 0.5 O2 H2O

CO+0.5O2 CO2

CaCO3 CaO + CO2


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Consider the reaction of burning carbon with

oxygen to yield carbon dioxide as the product


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oxygen to yield carbon dioxide as the product.

The chemical equations is

C(g) + O2 (g) CO2(g)

Although this is a combustion process, this may

also be considered as forming carbon dioxide

from its elements carbon and oxygen.

For this carbon oxygen reaction at constant pressure of 1atm and 25C the heat transferred from the system is


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atm and 25 C, the heat transferred from the system isstandard enthalpy of reaction.

If the enthalpies of the elements are set equal to zero, theenthalpy of reaction (from the table) for this process isfound to be (393741 kJ/kg-mole.

Thus the enthalpy of formation of carbon dioxideis393741kJ/kg-mole.

That is, ( ) CO2 = -393741 kJ/kg-mole of CO2.

22 OCCO0R0pC25 hhhHHHQ

fh

NOTE


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The enthalpy of a reaction can be calculatedif the enthalpies of formation of all the

reactants and products are known.

The enthalpy of formation data can often be

used to compute enthalpy of combustion .

NOTE


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Enthalpy of formation of elements atStandard conditions is taken as zero

ADIABATIC FLAME

TEMPERATURE


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TEMPERATURE

It is the temperature of the products if no heat is transferredfrom the combustion system and all the energy madeavailable is used to increase the temperature of theproducts.

It is thus the maximum possible temperature of theproducts.

There are a number of systems of industrial interest inwhich the energy released during a reaction is stored in

the products.

In these it is important to know the maximum resultingtemperature.


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Features of AFT (Tf)


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f

Example


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Determine the adiabatic flame temperature that

can be obtained by the combustion of n-Pentane(C5H12) with stoiciometric air. Both the fuel and

air enter at 25 C and 1 atm. Assume complete

combustion. Given the following data :

[PTO]


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Given the following data :

Average molar specific heat of

CO2 (g) : 62.75 J/mol.K

H2O (g) : 52.96 J/mol.K

O2(g) : 38.67 J/mol.K

N2(g) : 37.13 J/mol.K

Standard heat of formation of

CO2 (g) : -393510 J/mol

H2O (g) : -241820 J/mol

C5H12 (g) : -146760 J/mol

O2 and N2 : 0.00 J/mol

[PTO]


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C5H12+8O2+(8*3.76)N2

5CO2+6H2O+30.08N2


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Enthalpy of the Reactants

HR= 1 (-146760) + 8 (0) + (8*3.76)(0)


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Enthalpy of the Products:Hp = 5*(-393510) +5*(62.75) (T-25)

+6*(-241820)+6*(52.96)(T-25)+30.06*(0)+30.06*(37.13)(T-25)

= (-3462179.5)+1748.38T


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T =1897 deg.C

Example


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Determine the adiabatic flame temperaturethat can be obtained by the combustion of

n-Pentane (C5H12) with 25 % excess air.

Both the fuel and air enetr at 25 C and 1atm. Assume complete combustion. Given

the following data :

[PTO]


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Types of Heating Values


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Higher Heating Value (HHV) (or)Higher Calorific Value (HCV)

(OR)

Gross Heating Value (GHV) (or)

Gross Calorific Value (GCV)

Lower Heating Value (LHV) (or)Lower Calorific Value (LCV)

(OR)

Net Heating Value (NHV) (or)Net Calorific Value (NCV)

FUEL (1 k )


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FUEL (1 kg)

Products

With H2O

as Liquid

HHV

Products

With H2O

as vapour

LHV

Air (15 kg or more)48 MJ/kg

4 MJ

44 MJ/kg

Higher Heating Value (HHV) (or)

Higher Calorific Value (HCV)


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Higher Calorific Value (HCV)

HHV of a fuel is the amount of heat released duringthe complete combustion of unit quantity of fuelwhen the water formed during the combustion is

present in the form of Liquid in the products ofCombustion

lOH9)g(CO8gO5.12lHC 222188

Lower Heating Value (LHV) (or)

Lower Calorific Value (LCV)


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Lower Calorific Value (LCV)

LHV of a fuel is the amount of heat releasedduring the complete combustion of unit quantity

of fuel when the water formed during the

combustion is present in the form of vapour

in the products of Combustion

gOH9)g(CO8gO5.12lHC 222188


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Fuels & Combustion