Diagram for CH4, C2H6, C2H2, H2 1

[291432]

Does anyone know where I can find diagrams N2/O2/flammable substance for CH4, C2H6, C2H2, H2?
Thank you
LB

 

[Guidoo]

Method I used myself is when you have literature data of flammable limits both in air (being 80% N2 and 20% O2) and in pure oxygen. When plotting these upper and lower flammable limits in a composition triangle (with O2, N2 and the flammable gas in the three corners), and connecting the points by straight lines, you get a pretty good idea of what the flammability envelope looks like.

For your components (at 20°C and 101.325 kPa):
CH4:
LFL(air) 4.4 vol% UFL(air) 16.5 vol%
LFL(O2) 5.1 vol% UFL(O2) 61.0 vol%

C2H6:
LFL(air) 3.0 vol% UFL(air) 12.5 vol%
LFL(O2) 3.0 vol% UFL(O2) 66.0 vol%

C2H2:
LFL(air) 2.5 vol% UFL(air) 80/100 vol%
LFL(O2) ~2.5 vol% UFL(O2) ~100 vol%


H2:
LFL(air) 4.0 vol% UFL(air) 75.0 vol%
LFL(O2) 4.0 vol% UFL(O2) 94.0 vol%

 

[Guidoo]

The flammability diagram (for CH4/N2/O2 system) in following document can serve as a good illustration to what I wrote in my previous post:

http://www.engr.mun.ca/~fkhan/EN-8671/fire&explosion.doc

By the way, note that C2H4 (acetylene) can decompose explosively, even at 100% concentration.

 

[Guidoo]

Some more searching gave me following values for acetylene:

C2H2:
LFL(air) 2.2 vol% UFL(air) 80 vol%
LFL(O2) 2.8 vol% UFL(O2) 93 vol%

Source:

http://www.krona.srv.br/display05_engl.htm#t6

Again, note that acetylene can decompose explosively even when concentration is outside these flammibility limits (e.g. 100%).

 

[25362]

To Guidoo, the first site you mentioned (by fkhan) apparently contains a printing error in the equations for the dependence of LFL and UFL on temperature.

Given:

LFL_t=LFL₂₅-0.75(T-25)/[Δ]H_c
HFL_t=HFL₂₅+0.75(T-25)/[Δ]H_c


Should be:

LFL_t=LFL₂₅(1-0.75(T-25)/[Δ]H_c)
HFL_t=HFL₂₅(1+0.75(T-25)/[Δ]H_c)

Can you confirm ?

 

[Guidoo]

25362,

I checked the standard work on this subject:
M.G. Zabetakis, Flammability Characteristics of combustible gases and vapors, Bureau of Mines Bulletin 627 (1965)
I found (on pages 22 and 23 of the bulletin) that equations as given by fkhan are correct. The equations are called the modified Burgess-Wheeler laws. You can check for yourself by doing a check calculation for methane at 300 °C. When you use your proposed equation you will find a negative value for LFL...

Zabetakis even gives some more simplified equations for paraffin hydrocarbons:

LFL_t=LFL₂₅(1-0.000721(T-25))
HFL_t=HFL₂₅(1+0.000721(T-25))

These simplifications are based on the assumptions that LFL₂₅*?H_c = 1,040 and heat release at upper limit is equal to that at lower limit.

 

[Guidoo]

It seems like there is quite some confusion around these equations. For example, in their book "Chemical Process Safety: Fundamentals with Applications", Daniel A. Crowl and Joseph F. Louvar show the equations as given by 25362 in previous post, and they refer to a lecture by M.G. Zabetakis in 1959. Same is true for Frank P. Lees in his book "Loss prevention in the Process Industries" (equations 16.2.2 and 16.2.3 in 2nd edition of his book).

However, when looking at the curves and equations in the Bureau of Mines Bulletin 627 my conclusion remains that equations as shown in this Bulletin and on site from fkhan are correct, and equations given by 25362, Crowl&Louvar and Lees are not correct.

When we write both possibilities (for LFL) in a similar form, the both look like:

LFL_t/LFL₂₅=1-term·(T-25)

What it all boils down to is whether the term before (T-25) should be (for LFL):

0.75/?H_c

or

0.75/(?H_c·LFL₂₅)

Both Perry's and Lees state that this term has a value of about 0.0008/°C, where Bulletin 627 uses 0.000721.

When we fill in values for methane (from table 2 in Bulletin 627): ?H_c= 191.8 kcal/mole and LFL₂₅=5.0 vol% we get:

0.00391 vs. 0.000782

From this I conclude that the term should be:

0.75/(?H_c·LFL₂₅)

this one corresponds to the equation from Bulletin 627 and used on fkhan site.

Note that for UFL the term would be 0.000261 since UFL₂₅ for methane is 15.0 vol%, where Perry, Lees and Bulletin 627 say that term should be about the same as for LFL (so about 0.0008). So my conclusion is that for UFL the equation should actually be:

UFL_t/UFL₂₅=1+0.75/(?H_c·LFL₂₅)(T-25)

 

[25362]

To Guidoo, thank you. I'm now convinced that your assumption is right. The equations taken from Lees, and Crowl and Louvar, quoted from Zabetakis as empirically derived, are incorrect, and the term, as you call it, should be [Δ]H_c multiplied by the LFL.

I think Lees speaks of a flame temperature of about 1300^oC, making the "well-correlated" formula:

LFL_t/LFL₂₅ = 1 - [(t-25)/(1300-25)]
= 1 - 0.00078 (t-25)

The highlighted term following your "correction" for gaseous flammables, is equal to:

the term compound

0.75/(5*212.8) = 0.000705 methane
0.75/(3*372.8) = 0.000671 ethane
0.75/(2.1*530) = 0.000674 propane
0.75/(1.6*687.3) = 0.000682 butane
0.75/(7*136.3) = 0.000786 formaldehyde
0.75/(2.7*337.1) = 0.000824 ethylene
0.75/(15*91.4) = 0.000547 ammonia
0.75/(12.5*67.6) = 0.000888 carbon monoxide

Heats of combustion taken from the CRC Handbook.

Your analysis merits a star.