MEL1(3) Fuels&Combustion

8/3/2019 MEL1(3) Fuels&Combustion

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MEL 411:


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Products of Combustion

In a combustion (burning) reaction the fuel is

burnt in oxygen. The oxidizer is oxygen. All combustion reactions are exothermic energy

(mainly heat) is released.

Explosions are forms of combustion.In an explosive combustion reaction, the fuel isexploded

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Energy Value

Energy Value is the heat of combustion of a fuel given pergram of fuel.

The higher the energy value, the more energy is released,the better the fuel. Heat of combustion of hydrogen is 285kJ/mole

1 mole of hydrogen gas (H2) has a mass equal to itsmolecular mass (molecular weight)

= 2 x 1.008 = 2.016 The heat produced per gram of hydrogen gas = 285

2.016 = 141.4kJ/g The energy value for hydrogen gas is 141.4kJ/g

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Ignition Temperature

Ignition Temperature is the minimum

temperature to which the fuel-oxidizer mixture(or a portion of it) must be heated in order forthe combustion reaction to occur. High ignitiontemperature means the fuel is difficult to ignite,low ignition temperature means the fuel ignites

easily making the fuel potentially hazardous. The greater the activation energy of a reaction,

the higher the ignition temperature will be.

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Ignition Temperature

A match and its striking surface contain a fuel and itsoxidizer with a low activation energy and therefore low

ignition temperature, so low that the friction of striking thematch generates enough heat to raise the temperaturesufficiently for ignition to occur.

Petrol and oxygen in a car engine have a higheractivation energy and therefore a higher ignition

temperature. A spark is needed to raise the temperatureof the mixture sufficiently near the spark for the mixture toignite. The heat of reaction generated heats up more ofthe mixture so the reaction becomes self-sustaining.

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Complete combustion of carbon-based fuels produce carbon dioxide and watervapour. Carbon dioxide gas is the main contributor to the greenhouse effect.

Incomplete combustion of carbon-based fuels produces toxic carbon monoxideand solid carbon(soot).

Sulfur and nitrogen are present in fossil fuels. The sulfur burns to produce

oxides which contribute to acid rain, while the nitrogen burns to produce oxidesthat contribute not only to acid rain but also to photochemical smog

Solid fuels such as coal contain incombusitible minerals leading to ash. The ashcan damage machinery and can cause lung disease.

Unburnt fuel can also be released. Unburnt hydrocarbons from cars contribute

to photochemical smog and some are carcinogenic. Some fuels contain additives (such as the lead in leaded petrol) which can be

harmful

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Chemistry Tutorials: Combustion of Hydrocarbons(www.youtube.com)

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Gaseous fuels (Hydrocarbon, CnHm)

- LPG and methane

Liquid Fuels (Hydrocarbon, CnHm)

- Gasoline and Diesel

Solid fuels (Carbon, C)

- coal and wood

(Source: PIPE, Alcorcon 2005)

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Molecular Weight of common combustible elements

C = 12 H = 1 O = 16 N = 14 S= 32

Composition of air by volume

Oxygen in air = 21%

Nitrogen in air = 79%

Composition of air by weight

Oxygen = 23.20%

Nitrogen = 76.80%

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Chemical Formula: (Source: PIPE, Capote & Mandawe 2001)

Air + Fuels = Products of combustion

O2 + a[3.76N2]+ CnHm= b CO2 + cH2O + 3.76 a N2

Ratio by volume of N2 to O2 in air when both gases are thesame temperature.

Moles N2 79%

----------------------- = ------------- = 3.76

Moles O2 21%

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Chemical Formula of Gaseous fuels

CH4 = methane

LPG = Propane (C3H8) + butane (C4H10)

Chemical Formula of Liquid fuels

C4H18 = Octane

C16H32 = Cetene

C16H34 = CetaneC1H16 = Heptane

C6H16 = Isopropyl

C2H6 = Ethane

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Calculating Air-fuel ratio (A/F)

Calculate the theoretical air-fuel ratio of methane (naturalgas) as fuel in kga/kgf?

CH4 + a[02 + 3.76N2] = (b) CO2 + (c) H20 + 3.76 (a) N2

H: 1 (4) b: 1 kg-mol

C: 1 (1) c: 2 kg-mol

O: 2 (a) = b (2) + c (1)O: 2 (a) = 1(2) + 2 (1)

a = 2 kg-mol

N: 3.76 (2) = 7.52 kg-mol

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Then the balanced equation is:

CH4 + 2 O2 + 7.52 N2 = 1 (CO2) + 2 (H20) + 7.52 N2

The theoretical A/F by weight is:(A/F)t = 2 (32) + 7.52 (28) / 1(12) + 4 (1)

(A/F)t = 274.56 / 16

(A/F)t = 17.16 kga/kgf

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Value of A/F for common liquid fuels:

o methane = 17.2

o

gasoline = 14.7o propane = 15.5

o ethanol = 9

o methanol = 6.4

o hydrogen = 34o diesel = 14.6

(Source: Combustion, URL(pdf); Internal Combustion Engine by the use of Pressure Sesonrs, URL(pdf))

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http://www.tech.plym.ac.uk/sme/ther305-web/Combust1.PDFhttp://www.lub.lu.se/luft/diss/tec432.pdfhttp://www.lub.lu.se/luft/diss/tec432.pdfhttp://www.lub.lu.se/luft/diss/tec432.pdfhttp://www.lub.lu.se/luft/diss/tec432.pdfhttp://www.tech.plym.ac.uk/sme/ther305-web/Combust1.PDFhttp://www.tech.plym.ac.uk/sme/ther305-web/Combust1.PDFhttp://www.tech.plym.ac.uk/sme/ther305-web/Combust1.PDF


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Sample Problem:

C8H18 fuel is burned with ideal proportion of air. Calculatethe ideal air-fuel ratio by weight.

Solution:

C8H18+ a[O2+ 3.76N2] = bCO2+ cH20 + 3.76 a N2

Balance:

C: 1(8) = b(1) : b = 8

H: 1(18) = c(2) : c = 9O: a(2) = b(2) + c(1) : a = 8(2) + 9(1)/2 = 12.5 kg-mol

N: 3.76(a) = 47 kg-mol

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Then the theoretical A/F is:

(A/F)t= 12.5(32) + 47(28) / 12(8) + 1(18)

(A/F)t= 15.05 kga/kgf

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Problem:

A steam generator burns fuel oil represented by C14H30.The fuel gas leave the preheater at 0.31 Mpa. Determinethe ideal A/F in kga/kgf?

Solution:

C14H30 + a[O2+ 3.76 N2] = CO2+ H20 + 3.76(a)N2

C: 1(14) = b(1) : b = 14

H: 1(30) = c(2) : c = 15O: 2(a) = b(2) + c(1) : a = 21.5

N: 3.76(a) = 80.84

(A/F)t= 21.5(32) + 80.84(28) / 12(14) + 1(30) =14.9 kga/kgf

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Theoretical air-fuel ratio (Molal basis):

(A/F)t = x + 3.76 (x) mol air / mol fuel

Theoretical air-fuel ratio (Mass basis): (A/F)t = 32(x) + 28(3.76)x / 12(n) + m kg air/kg fuel

Actual air-fuel ratio: (considering the excess air,e)

CnHm + (1+e)a[O2 + 3.76N2] = CO2 + H20 + 3.76(a)N2

(A/F)t (1+e)

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Problem:

A typical industrial cetene (C16H32) is burned with 20%excess air. Calculate the actual air-fuel ratio by weight.

Solution:

CnHm+ a[O2+ 3.76N2] = b(CO2) + c(H20) + 3.76(a)N2

C16H32+a[O2+3.76N2] =b(CO2) +c(H20) +3.76(a)N2

C: 1(16) = b(1) : b = 16H: 1(32) = c(2) : c = 16

O: (a)2 = 2(16) + 16(1) : a = 24

N: 3.76(24) = 90.24

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Considering excess air = 20%:

C16H32+ (1.20)a[O2+3.76N2] =b(CO2) +c(H20) + d(O2) +3.76(a)N2

1.20(24)2 = 16(2) + 16(1) + d(2)

d = 4.8

Now lets solve for the (A/F)actual:

(A/F)a= (1.20)[(24)(32) + 90.24(28)] /12(16) + 1(32)

(A/F)a= 3953.66 / 224

(A/F)a= 17.65 kga/kgf

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Problem:

A steam generator burns fuel oil with 20% excess airrepresented by C14H30. The fuel gas leave the preheater at0.31 Mpa. Determine the actual A/F in kga/kgf?

Ans. 17.89

Problem:

A furnace of a HRT boiler burns fuel oil with 15% excess air.The fuel oil is represented by C14H30. Calculate the actualmolal air-fuel ratio.

Ans. 117.6 mol air/mol fuel

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Assignment:(to be submitted on Monday, October 10, 2011)

1. Calculate the ideal (A/F) for the following fuels:

- hydrogen

- ethanol- cetane

- heptane

2. Calculate also for its actual (A/F) for 20% excess air

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MEL1(3) Fuels&Combustion