# Rich Flammability Limit

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 December 15, 2009, 10:29 Rich Flammability Limit #1 New Member   Johann V Join Date: Oct 2009 Posts: 19 Rep Power: 8 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)?

 December 16, 2009, 04:49 #2 New Member   Johann V Join Date: Oct 2009 Posts: 19 Rep Power: 8 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.

August 25, 2010, 23:12
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Dong Liang
Join Date: Jan 2010
Location: Shanghai, China
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Quote:
 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.
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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