H2 LFL in oxygen deficient atmospheres 2

[HazGas]

I have a vacuum system which contains a flammable gas. I am trying to determine the explosion risks if this system were to be vented to atmosphere. I have calculated that once the pressure equilibrates (assuming no diffusion), the atmosphere is roughly:

10% O2
37% N2
48% He
5% H2

I guess I need the LFL of H2 in 10% O2. Any ideas where to get this information?

Thanks.

 

[Guidoo]

You not only have less oxygen, you also have a lot of Helium in the mixture. Note that flammability limits changes when the inert composition changes (e.g. it is different for CO2 and for N2). The correct answer would be that you will have to test your specific mixture.

According to Bureau of Mines Bulletin 627, LFL of hydrogen at 25°C and atmospheric pressure both in oxygen and air is 4.0 vol%. Therefore, LFL does not (or hardly) vary with oxygen concentration.

 

[25362]

Try thread798-33615. You may find that it contains some useful information.

 

[Flareman]

HazGas
It seems as though you have two different conditions - a gas before it meets the air and a gas after it meets the air.
You need to look at this as the H2/He mixture (ratio = 1:9.6) and find its LEL and UEL.

BOM 503 (p 22) gives the following data.
H2 100 : He 0 LEL 4.2 UEL 71.5 % vol
H2 58.1 : He 41.9 LEL 71. UEL 76.2
H2 27.9 : He 72.1 LEL 16.6 UEL 79.2
H2 19.3 : He 80.7 LEL 24.9 UEL 79.2
H2 10.7 : He 89.3 LEL 51.3 UEL 80.3
H2 8.7 : He 91.3 LEL 69.8 UEL 69.8
Draw the envelope and see how your conditions can fit into it. The final equilibrium mixture which you estimated in your question is not the only condition you may have.

Your original mixture looks to me like
H2 10.2 : He 89.8 (?)

These data were from experimental tests for upward flame propagation in a 2" tube.
As with any calculation of flammable limits - don't believe the numbers and apply a safety factor to whatever you interpret.

 

[Guidoo]

Flareman,

There appears to be a typo in your table:
H2 58.1:He 41.9 LEL 7.1 UEL 76.2

Note that:
58.1/100 * 7.1 = 4.1, which is about same as LEL for pure H2
just as:
27.9/100 * 16.6 = 4.6
19.3/100 * 24.9 = 4.8
10.7/100 * 51.3 = 5.5
8.7/100 * 69.8 = 6.1

So an increasing amount of He results in a (slight) increase of LEL for hydrogen. As long as H2 conc is more than 60 vol% there is hardly any effect.

 

[Flareman]

Thanks Guidoo, my fingers run away with me fom time to time.
You are correct about the ratio. The reason there is a slight increase in the LEL as the He increases is because the specific heat of the diluent Helium is less than that of Nitrogen, which is the diluent in air. As LEL is basically a measure of the minimum flame temperature which will transfer enough energy to propagate the dissociation reaction of the flammable components, the LEL with any diluent is based on the specific heat of that diluent.
If you draw the entire envelope however you also see the effect on the UEL, which is not constant, and my point to HazGas was that there are intermediate cases to examine,not just the final equilibrium.