What Is the Stoichiometric Air-Fuel Ratio for a Dual Fuel IC Engine?

[ashishbhagwat]

Homework Statement

a mixture of 0.64 x 100 percent iso-octane and {1-0.64} x 100 percent methanol, based on volume, is mixed with air and combusted. Calculate the stoichiometric air fuel ratio. Note: please not this is mixing on a molar basis, not mass basis. It is necessary to perform a complete chemical analysis to solve this problem. Begin by writing the chemical equation for fuel and air= products of combustion, for a dual fuel problem. Then, calc the mass of the fuel and the mass of the air

Homework Equations

The Attempt at a Solution

 

[herpamad]

2 urls for you

http://www.tech.plym.ac.uk/sme/ther305-web/Combust_2.PDF

and

http://www.brighthub.com/engineering/mechanical/articles/15235.aspx

You can find the chemicals for any fuel pretty easy, and use periodic table to get values.

 

[Mene]

I would approach this problem by first stating that the air fuel ratio is an important factor in the operation of internal combustion engines (IC engines). This ratio refers to the amount of air and fuel that is mixed together for proper combustion. The stoichiometric air fuel ratio is the ideal ratio at which all the fuel is completely burned with the available oxygen in the air.

To solve this problem, I would begin by writing the chemical equation for the combustion of iso-octane and methanol with air as follows:

C8H18 + 2.5O2 + CH3OH + 3O2 → 9CO2 + 10H2O

Next, I would calculate the mass of the fuel and air based on the given volume ratios of iso-octane and methanol. Since the problem specifies a molar basis, I would use the molar masses of iso-octane (114.23 g/mol) and methanol (32.04 g/mol) to convert the given volume ratios to molar ratios. This would result in a molar ratio of 0.64 mol iso-octane to 0.36 mol methanol.

Using the molar ratios, I would then calculate the mass of the fuel and air. The mass of iso-octane would be 0.64 mol x 114.23 g/mol = 73.18 g and the mass of methanol would be 0.36 mol x 32.04 g/mol = 11.53 g.

To calculate the stoichiometric air fuel ratio, I would use the balanced chemical equation to determine the number of moles of oxygen needed for complete combustion of the fuel. From the equation, we can see that 2.5 moles of oxygen are needed for every 1 mole of iso-octane and 3 moles of oxygen are needed for every 1 mole of methanol. Therefore, the total moles of oxygen needed for 0.64 mol iso-octane and 0.36 mol methanol would be (2.5 x 0.64) + (3 x 0.36) = 2.64 moles.

Finally, the stoichiometric air fuel ratio can be calculated by dividing the moles of air (which is 21% oxygen) by the moles of fuel. This would result in a ratio of 2.64 moles of air