Theoretical Combustion Pressure Calculations 1

If you send me your email I can send a program to do this calculations.

Nagy

How would I calculate it?

For your simplified system, try this approach:

Based on the desired air-fuel ratio,
calculate the mass of air in the chamber before compression
calculate the mass of fuel...

M.total = P1*V1/(R * T1) where R.air ~ 287 m^2/(s^2 * K)
M.air = M.total * AFR / (AFR+1)
M.fuel = M.total / (AFR+1)

Assume a combustion efficiency.

Using the lower heating value of the fuel, calculate the energy released by combustion of all the fuel, then multiply by your efficiency.

for gasoline, Q.HV ~ 43,000 kJ/kg
Q.in = M.fuel * Q.HV * n.comb

Using the gas properties of air (mixture properties should be fairly close to air), calculate the temperature of the mixture after combustion

Q.in = (M.fuel + M.air) * c.v * (T2 - T1)
c.v ~ 821 m^2 / (s^2 * K)
T2 = [Q.in + (M.fuel + M.air) * c.v * T1] /
(M.fuel + M.air) * c.v

Calculate the pressure after combustion using the ideal gas law:

P2 = P1 * (T2/T1)

Thanks
My e-mail is mrgl.llc@verizon.net

ivymike
I assume that an ideal theoretical AFR would be based on the typical % O2 in Air and the reaction equation for the fuel. ie set the molar ratio between oxygen in air and fuel such that (2n)O2 + CH(2n+2) -> CO2 + (n+1)H2O. Or is this being too nieve?

For R.air ~ 287 m^2/(s^2 * K is the K a constant or degrees Kelvin?

To find theoretical Max, I suppose that the combustion efficiency could be assumed at 1 with the further assumption that all reactants are equal, the chamber is static, the reaction goes to completion, there is no heat loss from the chamber and the world is perfect? But just out of cureosity, what would be a typical combustion efficiency for a gasoline engine? Like one that runs within EPA emmition standards.

for gasoline, stoichiometric AFR (mass basis) is about 14.7:1 to 14.9:1. I'm sure you could get there via your approach, but shortcuts are nice. There are tables out there (Taylor probably has some in the back of his book) that will give you stoich AFR for a long list of hydrocarbons.

yep, K is Kelvin

I think it's common to have combustion efficiencies in excess of 98%. I'm not so much of a combustion/emissions guy, though, so perhaps someone else can better answer that.