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291432 (Chemical) (OP)
21 Jul 04 3:07
Does anyone know where I can find diagrams N2/O2/flammable substance for CH4, C2H6, C2H2, H2?
Thank you
LB
Helpful Member!  Guidoo (Chemical)
30 Jul 04 9:02
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 (Chemical)
30 Jul 04 10:02
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:
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 (Chemical)
30 Jul 04 10:20
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 (Chemical)
6 Oct 04 2:25
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:

LFLt=LFL25-0.75(T-25)/ΔHc
HFLt=HFL25+0.75(T-25)/ΔHc


Should be:

LFLt=LFL25(1-0.75(T-25)/ΔHc)
HFLt=HFL25(1+0.75(T-25)/ΔHc)

Can you confirm ?
Guidoo (Chemical)
18 Oct 04 4:37
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:

LFLt=LFL25(1-0.000721(T-25))
HFLt=HFL25(1+0.000721(T-25))


These simplifications are based on the assumptions that LFL25*?Hc = 1,040 and heat release at upper limit is equal to that at lower limit.
Guidoo (Chemical)
21 Oct 04 4:23
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:

LFLt/LFL25=1-term·(T-25)

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

0.75/?Hc

or

0.75/(?Hc·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): ?Hc= 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/(?Hc·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:

UFLt/UFL25=1+0.75/(?Hc·LFL25)(T-25)
25362 (Chemical)
22 Feb 05 8:05

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 ΔHc multiplied by the LFL.

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

LFLt/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.

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