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Combustion

Particle diagram equationCombustionC2 + 2O22CO2+

Hydrocarbon Oxidationmethane (CH4), the primary constituent of liquefied or compressed natural gas propane (C3H8), the primary constituent of liquid petroleum gas isooctane (C8H18), typical of the molecules found in gasoline n-hexadecane (C16H34), typical of diesel fuel

If sufficient oxygen is available, a hydrocarbon fuel can be completely oxidized, the carbon is converted to carbon dioxide (CO2) and the hydrogen is converted to water (H2O).

The overall chemical equation for the complete combustion of one mole of propane (C3H8) with oxygen is:Elements cannot be created or destroyed, soC balance:3 = b b= 3H balance:8 = 2c c= 4O balance: 2a = 2b + c a= 5

Thus the above reaction is:# of molesspeciesHydrocarbon Oxidation

Hydrocarbon OxidationAir CompositionOxygen : 21%Nitrogen : 79%Oxygen - Nitrogen Ratio in Air 1 : 3.76

Combustion StoichiometryAir contains molecular nitrogen N2, when the products are low temperature the nitrogen is not significantly affected by the reaction, it is considered inert.

The complete reaction of a general hydrocarbon CxHy with air is:The above equation defines the stoichiometric proportions of fuel and air.C balance:x = b b = xH balance:y = 2c c = y/2O balance:2a = 2b + c a = b + c/2 a = x + y/4N balance:2(3.76)a = 2d d = 3.76a/2 d = 3.76(x + y/4)

Combustion StoichiometryThe stoichiometric quantity of oxidizer is just that amount needed to completely burn a quality of fuel.If more than a stoichiometric quantity of oxidizer is supplied, the mixture is said to be fuel leanWhile supplying less than the stoichiometric oxidizer result in fuel rich

Combustion StoichiometrySubstituting the respective molecular weights and dividing top and bottom by x one gets the following expression that only depends on the ratio of thenumber of hydrogen atoms to hydrogen atoms (y/x) in the fuel.Example: For Octane (C8H18), y/x = 2.25 (A/F)s = 15.1 Benzene (C8H16), y/x = 2.0 (A/F)s = 14.7The stoichiometric mass based air/fuel ratio for CxHy fuel is:MO2 : 32MN2 : 28MC : 12MH : 1

Fuel Lean Mixture Fuel-air mixtures with more than stoichiometric air (excess air) can burn

With excess air you have fuel lean combustion

At low combustion temperatures, the extra air appears in the products in unchanged form: for a fuel lean mixture have excess air, so g > 1

Fuel-air mixtures with less than stoichiometric air (excess fuel) can burn.

With less than stoichiometric air you have fuel rich combustion, there is insufficient oxygen to oxidize all the C and H in the fuel to CO2 and H2O.

Get incomplete combustion where carbon monoxide (CO) and molecular hydrogen (H2) also appear in the products.where for fuel rich mixture have insufficient air g < 1Fuel Rich Mixture

The equivalence ratio, f, is commonly used to indicate if a mixture is stoichiometric, fuel lean, or fuel rich.

Off-Stoichiometric Mixturesstoichiometric f = 1 fuel lean f < 1 fuel rich f > 1Stoichiometric mixture:Off-stoichiometric mixture:

Example: Consider a reaction of octane with 10% excess air, what is f?

Off-Stoichiometric Conditions 10% excess air is:16 + 9 + 2a = 1.1(12.5)(2) a = 1.25, b = 1.1(12.5)(3.76) = 51.7Other terminology used to describe how much air is used in combustion:

110% stoichiometric air = 110% theoretical air = 10% excess air mixture is fuel leanStoichiometric :

ExampleA small stationary gas turbine engine operates at full load (3950 kW) at an equivalence ratio of 0.286 with an air flow rate of 15.9 kg/s. The equivalent composition of the fuel is C1.16H4.32. Determine the fuel mass flow rate and operating air fuel ratio for the engineSolution

ExampleA natural gas (methane / CH4) fired industrial boiler operates with an oxygen concentration of 3 mole percent in the flue gases. Determine the operating air-fuel ratio and the equivalence ratio.Solution3% of O2 in flue gases Fuel lean mixtureIf all fuel C is found in CO2 and all fuel H is found in H2OCH4 + a(O2 + 3.76N2) 1CO2 + 2H2O + bO2 + 3.76a N2O balance ; 2a = 2 + 2 + 2b b = a - 2Mole fraction of O2a = 2.368

MWair = 21% x 32 + 79% x 28 = 28.84MWfuel = 12 + 4 x 1 = 16Air Fuel Ratio StoichiometryAir Fuel RatioEquivalence ratio

ExampleO2 Concentration in flue gases is 8 mole percent on oxidation of 200 kg/h diesel fuelDetermine the operating air-fuel ratio, the equivalence ratio and the air flow rate, if:

Solution 1O2 Concentration in flue gases is 8 mole percent on oxidation of 200 kg/h diesel fuel8% of O2 in flue gases Fuel lean mixtureIf all fuel C is found in CO2 and all fuel H is found in H2OC16H28 + a(O2 + 3.76N2) 16CO2 + 14H2O + bO2 + 3.76a N2O balance ; 2a = 32 + 14 + 2b b = a - 23Mole fraction of O2a = 30.05

MWair = 21% x 32 + 79% x 28 = 28.84MWfuel = (12 x 16) + (28 x 1) = 220Air Fuel Ratio StoichiometryAir Fuel RatioEquivalence ratio

Air flow ratemair = mfuel x (A/F)mixturemair = 200 x 23.74 = 4748 Kg/h

The maximum amount of energy is released from a fuel when reacted with astoichiometric amount of air and all the hydrogen and carbon contained in thefuel is converted to CO2 and H2O

This maximum energy is referred to as the heat of combustion or the heating value and it is typically given per mass of fuelHeat of Combustion

FuelEnergy density(MJ/L)Air-fuel ratioSpecific energy(MJ/kg air)Heat of vaporizationGasoline and biogasoline3214.72.90.36 MJ/kgButanol fuel29.211.23.20.43 MJ/kgEthanol fuel19.6 9.03.00.92 MJ/kgMethanol16 6.53.11.2 MJ/kg

Heat of Formation

Heat of Formation for 1 Bar and 298.15 K

Heat Transfer in a Chemically Reacting Flow

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Combustion Flame Temperature

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Heat Combustion and Heating Value

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