[JohannV]

Hi everybody,

I hope someone can help me with my problem. Thanking you in anticipation.

I need to calculate a combustion model (non premixed combustion PDF) and therefore I need the Rich Flammability Limit (RFL).

There are, one fuel stream (CH4 85%, N2 11%, C3H8 4%), one oxidiser (O2 100%), and an ambient air stream.

I read in the FLUENT documentation that the RFL for the ambient stream can be set to 1.

But what do I do with the RFL for the fuel stream?

Is this the same as the Lower Explosive Limit (LEL)?

[JohannV]

Good morning friends of CFD.

I yust calculated the rich limit as folows.


f=(Z_i-Z_i_oxi)/(Z_i_fuel-Z_i_oxi)

Z_i : mass fraction of (CH4 + C3H8) in respect to the Lower Explosive Limits.

Z_i_fuel : mass fraction of (CH4 + C3H8) in the fuel
Z_i_oxi : mass fraction of (CH4 + C3H8) in the oxidizer

All values were calculated for a volume of 1 m^3.

And the result is f=0.03
Is this result reasonable?

Any hint and sugestion is velcome.

[davedong]

Originally Posted by JohannV

Good morning friends of CFD.

I yust calculated the rich limit as folows.


f=(Z_i-Z_i_oxi)/(Z_i_fuel-Z_i_oxi)

Z_i : mass fraction of (CH4 + C3H8) in respect to the Lower Explosive Limits.

Z_i_fuel : mass fraction of (CH4 + C3H8) in the fuel
Z_i_oxi : mass fraction of (CH4 + C3H8) in the oxidizer

All values were calculated for a volume of 1 m^3.

And the result is f=0.03
Is this result reasonable?

Any hint and sugestion is velcome.

your calculation is likely incorrect.
the f is the mixture fraction. Z_i is the mass fraction of i element, like C, H, O, etc. the result is only one with any element in the system.
the RFL is a constant that the fluent will compute the chemistry when the f is above it. in most cases, it will be larger than chemical equivalent.