VOC concentration and LEL problem

[stepp]

Hello everybody,

I have following problem:
I have a combustion chamber for 6000m3/h of gas with concentration 10g VOC/m3. Question is, what amount of gas with VOC concentration 36g VOC/m3 (96% LEL) I can let flow to combustion chamber?

My opinion is, that I can make a mixture:
4400m3/h of fresh air and 1600 m3/h of gas (90%LEL).

But somebody told me that I can have only 800 m3/h of gas (90%LEL).

So, I would like to know who has true?
Thak you very much...

 

[jwysmooth]

No one seems to have input here so I'll give it a shot.

First of all there are many design parameters that affect the sizing of a combustion chamber; nozzle sizes, temperatures (insulation characteristics), heat content of the gases to be burned and probably a dozen others. Without knowing the original design conditions for the combustion chamber in question, it is impossible to tell what alternate conditions might be acceptable.

Obviously, if the VOC gas is below the LEL it will not burn without the assistance of a catalyst or the addition of a fuel gas. You might try supplying a few more details and see if you get better responses.
jwy

 

[stepp]

Thank you for answer, but maybe I was wrong in problem specification.

1)We have a tank with gas (VOC concentration 36 g/m3).
2)Thare is combustion chamber designed for gas with maximal concentration 10g VOC/m3 and for 6000m3/h.

Problem is, how to calculate a diluting of gas by fresh air (in duct before chamber).

I don´t know, if I can use only rule of three sum. Or if there are any others restrictive parameters for gas/air mixture...

 

[quark]

If the fresh air is devoid of the VOC under consideration then you can use weighted average method. Your answer is more correct. The issue is how you maintain the balance.

 

[jwysmooth]

Yes, your answer in your original post was correct. I actually calculate 1667 m3/hr of the higher concentration VOC gas. I look at the problem as an VOC balance between the 2 conditions...
36(x) = 10(6000) ==> x = 1667 m3/hr
If the compontents are the two VOC gas streams are the same.

I suspect your combustion chamber is an enclosed flare device similar to what John Zink Co. calls a terminal combustor, VCU or ZTOF. If so, the combustion chamber or stack is sized for a specific heat release so that the stack can enclose or hide the flame. Again, if this is the device you are referring to, there is no need to add additional air to make up the volume of the VOC gas back to 6000 m3/hr. Just reduce the flow to the combustion chamber to 1667 m3/hr. The "flame volume" will be approximately the same whether its 6000 of 10 gas or 1667 of 36 gas.
I hope that helps.
jwy

 

[Flareman]

Stepp
VOC simply means volatile Organic Compound. That could be almost anything. We can suppose that it IS a hydrocarbon so measuring by weight we can guess at a usual heating value of approx 20,000 Btu/lb. (say 11,100 kcal/kg)
As a general approximation, 100 Btu needs 1 ft^3 of air to burn, allow 1.3 ft^3 to be on the safe side. That is approx 7.3 m^3 of air per lb or 16 m^3 of air per kg of flammable gas. You don't say what the VOC is carried in. It could be air? or perhaps it's inert (like nitrogen). Either way you need air in roughly the amounts mentioned above. By indicating a fractional LEL you suggest that the incoming gas is supported by a supplementary fuel. That has to figure into the calculations. Then you have to think about the furnace temperature and the draft to be able to ensure that you can get in all the air you need and get out all the flue gas you make based on whatever temperature it balances out at. It's a delicate balance and depends on the height and diameter of the furnace and the chimney. You give no information on any of that. A combustion chamber isn't like a garbage dump with no limitations. You have to be prepared to do the math. How do you input the mixture? Is there a dedicated burner, is there an existing configuration. How big is the combustion chamber? You may be able to squeeze heat into there at between 25,000 Btu/h per cu.ft and 50,000 Btu/h per cu.ft of furnace volume but it depends on how you are shaping the flame and bringing in the combustion air. Does anyone care about the quality of the combustion or what trace compounds you are putting out (CO, NOx, unburned VOC(s))? That depends on the residence time and the control temperature.
Are you hoping someone else will solve your problem without you having to slog the calculations????
Did you think of asking the original designer for help?
Think long and hard
Good luck
David