Octane + air combustion equation

In summary, to calculate the flame temperature of normal octane burning in air at an equivalence ratio of 0.5, you would use the equation C8H18 + 2*12.5 O2 + 2*47N2 --> 8CO2 + 9H2O + 2*47N2 + 12.5 O2 and equate the enthalpy of the reactants to the enthalpy of the products at the adiabatic flame temperature. The adiabatic flame temperature will be lower than if using the theoretical stoichiometric amount of air due to the excess moles of air that have to be heated.
  • #1
engineer23
75
0

Homework Statement


Calculate the flame temperature of normal octane (liquid) burning in air at an equivalence ration of 0.5. All reactants are at 298 K and the system operates at the pressure of 1 atm.


Homework Equations



Hp = Hr


The Attempt at a Solution


I know that the combustion of liquid octane in the theoretical amount of air is
C8H18 + 12.502 + 47N2 --> 8C02 + 9H20 + 47N2
Does the equivalency ratio affect this equation? I know equivalency ratio is the actual fuel/oxidizer ratio divided by the stoichiometric fuel/oxidizer ratio, but what does this mean for the above equation? If we assume the stoichiometric fuel/oxidizer ratio is 1, this means that the actual fuel/oxidizer ratio is 1/2...

If I can get the equation right, I know how to solve for flame temp. using Hp = Hr, JANAF tables, and iteration.

Thanks!
 
Physics news on Phys.org
  • #2
well, by your definition of equivalence ratio, that would mean that you're inputting twice as much air as is theoretically needed, so your equation would go:

C8H18 + 2*12.5 O2 + 2*47N2 --> 8CO2 + 9H2O + 2*47N2 + 12.5 O2

and all the extra moles of air that form part of the combustion gases have to be heated too, so the adiabatic flame temperature will be less than when you're using the theoretical stoichiometric amount of air. To find the adiabatic flame temp, just equate the enthalpy of the reactants at the input temp. to the enthalpy of the produts at the adiabatic flame temp and solve for the adiabatic flame temp (of course finding the flame temp is a recursive process because you don't know the enthalpy of the products without first assuming the temp they're at).
 
  • #3


I would start by clarifying the question and making sure I have all the necessary information. In this case, I would ask for the units of the equivalence ratio (is it mass or mole basis?) and also for the heat of formation values for all the species involved in the reaction.

Once I have all the information, I would proceed to use the combustion equation provided to calculate the heat of reaction (Hr) using the heat of formation values. Then, I would use the given equation Hp = Hr to solve for the flame temperature (Tp).

In order to incorporate the equivalence ratio, I would need to use the specific heat of the mixture (Cp) instead of the specific heat of individual species. This is because the equivalence ratio affects the overall composition of the reactants and therefore the specific heat of the mixture.

After obtaining a value for Tp, I would use JANAF tables or other relevant data to iteratively solve for the actual temperature of the reactants and products (298 K) and the pressure (1 atm) to get a more accurate result.

In summary, the equivalence ratio affects the overall composition and specific heat of the mixture, and it must be considered in the calculations for the flame temperature.
 

FAQ: Octane + air combustion equation

What is the chemical equation for the combustion of octane and air?

The chemical equation for the combustion of octane and air is C8H18 + 12.5O2 -> 8CO2 + 9H2O.

What are the reactants and products in the combustion of octane and air?

The reactants are octane (C8H18) and oxygen (O2), while the products are carbon dioxide (CO2) and water (H2O).

What is the energy released during the combustion of octane and air?

The energy released during the combustion of octane and air is approximately 47.8 kJ/mol.

What is the stoichiometric ratio for the combustion of octane and air?

The stoichiometric ratio for the combustion of octane and air is 1:12.5, meaning that for every 1 mole of octane, 12.5 moles of oxygen are needed for complete combustion.

What are some real-life applications of the octane and air combustion equation?

The combustion of octane and air is the basis for the operation of internal combustion engines in vehicles, which allows for the conversion of chemical energy into mechanical energy. It is also used in industrial processes such as power generation and heating systems.

Back
Top