Saturday, September 5, 2009
Display problem ? Click HERE
Recommended :
Subscribes to FREE Hydrocarbon Processing
Minimum oxygen concentration (MOC) is minimum quantity of oxygen required present in hydrocarbon mixture so that a fire can be initiated and propagated. Below this limit, a fire will not form. More discussion in "Minimum Oxygen Concentration (MOC) for Flare Purge". A mixture is combustible / flammable if and only if the hydrocarbon composition is within mixture LFL/LEL and UFL/UEL as discussed in "Estimate Mixture Flammability & Explosivity At Reference P & T One shall remember, as the operating pressure (P) and temperature (T) change (from reference P & T, the mixture LFL/LEL and UFL/UEL at P & T will change accordingly. This post will discuss the relationship between MOC with LFL/LEL.LFL Relate to MOC
LFL is minimum hydrocarbon (HC) concentration in air which a mixture will burn when an ignition source is present. LFL can be written as follow
MOC is minimum oxygen present in hydrcarbon mixture which a mixture will burn. MOC can be written as follow
Combining [1] & [2],
Combustion of Hydrocarbon
Above equation may be used to estimate MOC if you know the LFL of hydrocarbon.
Example
1) A Ethane (C2) having LFL of 3.0 vol% (Refer to "Estimate Mixture Flammability & Explosivity At Reference P & T". Estimate MOC of Ethane.
Combustion of C2H6,
C2H6 + d.O2 ==> 2CO2 + 3H2O
a = 2
b = 6
c = 0
d = 2 + 6 /4 - 0/2 = 3.5
MOC = 3.5 x 3.0 = 10.5 Vol%, close to 11.2 vol% in literature.
2) A n-butane (nC4) having LFL of 1.86 vol% (Refer to "Estimate Mixture Flammability & Explosivity At Reference P & T". Estimate MOC of n-butane.
Combustion of C4H10,
C4H10 + d.O2 ==> 4CO2 + 5H2O
a = 4
b = 10
c = 0
d = 4 + 10 /4 - 0/2 = 6.5
MOC = 6.5 x 1.86 = 12.1 Vol%. Close to 12.3 vol% in literature.
Related Topic
posted by Webworm, 2:56 AM
-->
0 Comments:
Post a Comment
Let us know your opinion !!! You can use some HTML tags, such as <b>, <i>, <a>
Subscribe to Post Comments [Atom]
Links to this post:
Create a Link
Home:
<< Home