Gas MW or H:C Ratio vs Ringelmann Number 2

[Ridgways]

Hi All!

I'm trying to find a relationship between the composition of a gas going to flare and the predicted amount of black smoke that would be produced. Not necessarily the quantity of smoke, just what opacity I could expect.

I know there are some relationships between the H:C ratio and smoke potential, but I don't know what those are.

Can anyone point me in the right direction?

Thanks!

 

[mbeychok]

Ridgways:

I would think that the opacity is a function of the unburnt hydrocarbons exiting the flare ... and that is more of a function of the individual flare's combustion efficiency than a function of the composition of the flared gas.

In other words, the same flared gas might smoke in flarestack A and and not smoke in flarestack B ... because flarestack B is more efficient than flarestack A.



Milton Beychok
(Visit me at www.air-dispersion.com)
.

 

[Flareman]

Ridgeways,

Wouldn't it be nice if it were that easy !
The carbon (smoke) produced in a flare flame is a very complex feature of the stability of the hydrocarbon, the size of the flare, the exit velocity, the crosswind, and the adiabatic flame temperature range between UEL and LEL.
The carbon forms largely in the core of the flame before it reaches any air due to the radiant load from the burning envelope and in the air lean zone as concentration pass through the UEL. I wish that I thought an easy relationship to be practical but I'm afraid I don't.

The techniques of smoke suppression mainly involve flame temperature reduction, changing the chemical balance of the mixture by adding steam, and enhanced air mixing.

External thermal radiation from the flame is characterised by the amount of free carbon in the burning envelope. I have (an imperfect) formula for emissive fraction in the paper "Making the Flare Safe" at

http://www.geocities.com/flareman_xs,

navigate to main index|downloads. You could build on something like that formula to get an estimate of the potential carbon outfall from the flame. My baseline would be that incipent smoke formation starts around e = 0.15 and maxes out at 0.4 (disastrous black smoke). Then you have to estimate the size of the downwind plume around the end of the flame and distribute the carbon across that expanding cone to fudge a number which represents the total obscuration across that section. (I didn't say it would be easy). Ringlemann is essentially % obscuration. 1 = 20%, 2 = 40% etc.

Having done all that however, what does it get you? My experience with Ringlemann readers on flares is as varied as there are flares. Some readers are happy to read the plume off the flame at the end, others are so picky that they "call" the smallest eddy and wisp of soot even if it disappears almost instantly. At the risk of sounding cynical, why do you even care? Ringlemann numbers were developed for real chimneys where all the combustion occurs before you see the smoke, akin to the reading after the flame is ended.

regards
David

 

[Ridgways]

Thanks all,

What I was looking for was much simpler, just an indicator.

For example, flaring a gas that is 90% C1, 5% CO2, and 5% other hydrocarbons is not likely to produce black smoke, even under non-ideal combustion circumstances, whereas a gas that is 40% C3 and 60% C4 likely will. This is speaking strictly of open pipe flares.

This is largly experience driven: I've never seen black smoke, or even grey, from a light-gas flare (such as sales quality or leaner), but I have seen it from tank vents, LPG bullets, and "waste" flares. Very rarely have I seen it from a gas plant's inlet stream. This led me to wonder if there was any relationship that has been "proven out".

I understand that wind speed, combustion efficiency, flare gas speed, and anything else that affects residence time of the gas, temperature of the flame, the formation of carbon particles, and their subsequent cooldown, affects the potential for smoke production. Surely what is being burned also has an effect along with how?

 

[25362]

Ridgways,

Thread124-187849 and thread124-202007 may interest you.

Please note that of the fuel characteristics influencing soot formation one is the carbon-to-hydrogen (C-to-H) ratio as you indicated, the other is the molecular structure of the gases to be burned.

Unsaturated hydrocarbons tend more toward soot formation than do saturated ones.

However, branched-chain paraffins smoke more readily than the corresponding normal isomers (same C-to-H ratio). The more highly branched the paraffin, the greater the tendency to smoke.

I hope Flareman agrees.

 

[Flareman]

You're on the ball 25362.
Generalizing relative to composition, paraffins are less smoke producing than olefins, which are less than aromatics with acetylenes at the top of the list.
This goes along with the idea that spare arms on the molecule easily get "knocked off", and the more weak bonds or weaker bonds you have, the worse smoke you will make.
However, C-H isn't the whole story. Olefins have a constant C-H but ethylene is worse than propylene. That points us to the real culprit which is the relationship between the thermal capacity of the molecule relative to the free energy of formation. Warm up a paraffin until you get an energy balance and the molecule cracks open (pyrolisis), Easy to see for paraffins which all start off in negative territory, but the other types are already positive at ambient temperature and only held together by the notion that they have nowhere to go if they decompose (excuse my non-scientific paraphrasing) or the heat released during a single molecular decomposition is insufficient to kick off a violent reaction (or quickly absorbed by the containment vessel or pipe as in the case of acetylenes).
This is all suitable when comparing roughly equal heat releases at the same velocity in the same flare tip in coparable ambient conditions (and would make a good experimental paper if anyone wanted to fund it), but the whole thing gets really complicated when you throw in tip diameter, flame size and discharge momentum. This latter actually works to enhance the air mixing in the base of the diffusion flame and is a key to the operation of most high velocity (or "sonic") flare tips.
Most flare manufacturers have some sort of handle on the above but generally just relate it to the amount of steam you need in a given tip to suppress the smoke at source.
Ridgways' original question directed us to true quantity of carbon released from a flame. On that, I'm out in left field other than my prior "stab".
The work we did through DGMK in the '70s and with the EPA in the early '80s showed that most dirty flames are better than 98% efficient and that the inefficient part is approx 60% carbon. Really bad flames can go down to 95% but that's an exception. I would stick with the 60% as carbon. The reports are available as EPA-600.

Regards
David

 

[Ridgways]

Thanks all!

I think I've found my direction. All of your inputs have been greatly appreciated!

 

[CHEMANG]

Hello all!!

Now that Ridgeways has found his direction.
I just want to know a small thing as quoted by Flareman -"Generalizing relative to composition, paraffins are less smoke producing than olefins, which are less than aromatics with acetylenes at the top of the list."

Based on the above quote, i am just curious to know why a oxy-acetylene flame do not smoke even though acetylenes are on top of the so called smoky list, Is it because of the presence of oxygen. If so,why?

 

[Flareman]

CHEMANG
Acetylene is very unstable and decomposes easily. When the oxygen needed for combustion is premixed (which of course is not the case in a flare flame) the combustion reaction proceeds very quickly and completely. The speed of combustion also concentrates the reaction into a small and very hot flame, which is why oxy-acetylene is able to melt steel.
David