Diagram for CH4, C2H6, C2H2, H2
Diagram for CH4, C2H6, C2H2, H2
(OP)
Does anyone know where I can find diagrams N2/O2/flammable substance for CH4, C2H6, C2H2, H2?
Thank you
LB
Thank you
LB
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS Member LoginCome Join Us!Are you an
Engineering professional? Join EngTips Forums!
*EngTips's functionality depends on members receiving email. By joining you are opting in to receive email.
Posting GuidelinesPromoting, selling, recruiting, coursework and thesis posting is forbidden.
Link To This Forum! 
Home >
Forums >
Industrial / Manufacturing Engineers >
Activities >
Occupational safety engineering Forum
Diagram for CH4, C2H6, C2H2, H2

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.
Here's Why Members Love EngTips Forums:
Register now while it's still free!
Already a member? Close this window and log in.
RE: Diagram for CH4, C2H6, C2H2, H2
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%
RE: Diagram for CH4, C2H6, C2H2, H2
www.engr.mun.ca/~fkhan/EN8671/fire&explosion.doc
By the way, note that C2H4 (acetylene) can decompose explosively, even at 100% concentration.
RE: Diagram for CH4, C2H6, C2H2, H2
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%).
RE: Diagram for CH4, C2H6, C2H2, H2
Given:
LFL_{t}=LFL_{25}0.75(T25)/ΔH_{c}
HFL_{t}=HFL_{25}+0.75(T25)/ΔH_{c}
Should be:
LFL_{t}=LFL_{25}(10.75(T25)/ΔH_{c})
HFL_{t}=HFL_{25}(1+0.75(T25)/ΔH_{c})
Can you confirm ?
RE: Diagram for CH4, C2H6, C2H2, H2
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 BurgessWheeler 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}=LFL_{25}(10.000721(T25))
HFL_{t}=HFL_{25}(1+0.000721(T25))
These simplifications are based on the assumptions that LFL_{25}*?H_{c} = 1,040 and heat release at upper limit is equal to that at lower limit.
RE: Diagram for CH4, C2H6, C2H2, H2
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}/LFL_{25}=1term·(T25)
What it all boils down to is whether the term before (T25) should be (for LFL):
0.75/?H_{c}
or
0.75/(?H_{c}·LFL_{25})
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 LFL_{25}=5.0 vol% we get:
0.00391 vs. 0.000782
From this I conclude that the term should be:
0.75/(?H_{c}·LFL_{25})
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 UFL_{25} 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}/UFL_{25}=1+0.75/(?H_{c}·LFL_{25})(T25)
RE: Diagram for CH4, C2H6, C2H2, H2
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 ΔH_{c} multiplied by the LFL.
I think Lees speaks of a flame temperature of about 1300^{o}C, making the "wellcorrelated" formula:
LFL_{t}/LFL_{25} = 1  [(t25)/(130025)]
= 1  0.00078 (t25)
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.