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-   -   Rich Flammability Limit (https://www.cfd-online.com/Forums/fluent/71096-rich-flammability-limit.html)

JohannV December 15, 2009 09:29

Rich Flammability Limit
 
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 December 16, 2009 03:49

Good morning friends of CFD.:D

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 August 25, 2010 23:12

Quote:

Originally Posted by JohannV (Post 240160)
Good morning friends of CFD.:D

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.


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