Yes, sort of......
Example Methane
LEL (LFL) = 5.6% in air
= Gas 5.6 : Oxygen 3.54 : Nitrogen 13.3
Stoic = 9.5%
= Gas 9.5 : Oxygen 19 : Nitrogen 71.5
UEL = 14.3% in air
= Gas 14.3 : Oxygen 18 : Nitrogen 67.7
When you dilute the flammable with inert the LEL stays
roughly at the same total dilution (ie. 5.6 in 94.4) but if the some of the dilution is provided by the mixture itself, (say 1 gas: 4 nitrogen) then that nitrogen becomes part of the overall dilution number (94.4)
This gives the following ratios
Gas 5.6 : Diluent N2 22.4 :Oxygen 15.1: Atmos N2 56.9
The stoichiometric ratio of oxygen to gas doesn't change so the stoic mixture is
Gas 6.9 : Diluent N2 27.6 : Oxygen 13.8 : Atmos N2 51.7
The UEL also changes (here it's a bit more iffy and this is just an guide for estimations) in
roughly the same proportion as the original
excess air LEL/stoic/UEL ratio which was 1.77:1

1/1.59) Applying that proportioning (1.77

1/1.59) to the diluted stream becomes 1.249 : (1/1.122) which gives
Gas 7.7 : Diluent N2 30.8 : Oxygen 12.9 : Atmos N2 48.6
I know this is all a bit convoluted. It works a lot better and is easier to see graphically but it means that the oxygen at UEL for the gas only is normally 18% in air and the total oxygen at UEL for the diluted stream in air is 12.9%. For purging, you are concerned with the oxygen concentration, not the gas concentration (which is the LEL or UEL value)
Added to this is the fact that all diluents affect the LEL in proportion to the specific heat of the diluent so the values for CO2 or Water as diluents are different. This is all (in brief) in my paper (I think you have it).
The flame speed, like the LEL, depends largely on the radiative heat transfer from the burning flame to the non-burning gas and will also slow down as the LEL/UEL envelope narrows.
I hope that it's clear. Remember that these are rules of thumb, not a scientific presentation .. which would need to delve into heats of formation and free radical formation .. so it's an engineer's solution not a chemist's.
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David