Dear Andreas,

Do a simple mass balance using and sthchiometric equation.

0.8C + 0.11H + 0.08O2 + a(O2+3.76N2) = 0.8CO2 + 0.055H2O

So the value of "a" is calculated as

0.08*2 +2a = 0.8*2 + 0.055

From this equation the value of "a" = 0.747

So 1 mole of oil requires 0.747*4.76 moles of air.

Thanks for the reply Syed Mughees Ali

Can you elaborate how you do (or think) to get to the equation?

0.8C + 0.11H + 0.08O2 + a(O2+3.76N2) = 0.8CO2 + 0.055H2O

The reactants I understand, and I presume you leave out the nitrogen part for simplicity? To be more specific, how do you get the products:

0.8CO2 + 0.055H2O?

In terms of excess air, I assume you adjust the factor a, i.e., 3.5% excess O2 gives you a factor x, which for 3.5% excess O2 becomes 1.035 and then you get:

0.08*2 +x2a = 0.8*2 + 0.055?

Hi,

I am going to answer this question in the first instance with some questions. Do you have a viscosity and SG of the liquid? Is the ambient temperature ever sufficiently low for it to solidify? Do you have any medium like low pressure steam to help the atomisation process? Can you heat the oil without any breakdown? If no external help for atomisation, can you pressure atomise (hp pumps) and accept the limited turndown? Are the components listed given as mol. percent or weight percent? For heavy liquids, the use of weight percent is not unusual. What is the ambient temperature range?

My reason for asking these questions is that there will be a lower acceptable limit for excess air/oxygen. It is highly unlikely that you will ever get to zero percent excess air without making a lot of smoke. On that issue are there any pollution emission limits for the products of combustion? You can easily form a lot of NOx with this fuel depending on how the nitrogen is bound into the fuel molecule.

Do you have any details of the burner? If you know the supplier they will want to help as they will want to sell to you spare/change parts as dictated by this fuel. The initial advise is usually free. They may even offer to do a burner test run for you in their test facilities if the burner is not too big. Do you know for what fuel the burner was originally designed? For example, if it was originally a gas burner then it is highly likely that you will need a total replacement but there were certain burners that were designed for combination gas and oil fuels but were being used with gas only.

I'll leave my questions there at this moment. Depending on the accuracy of the given constituents of the fuel, determining the excess oxygen stoichiometry is straightforward and an outline is given above by others. Don't forget that there is oxygen in the fuel already to make life interesting.

Regards Stephen.

Hi!

I will try to answer your questions. N/A means that I'm unable to provide an answer. Numbered from top to bottom in your answer.

As you may have read above, I don't comprehend the above given answer fully, and it may well be that thats the right answer to the question. However, if I don't understand it completely, its not much use for me. Perhaps needless to say, but I'm far from a chemist, and need help to understand from you guys.

Regards,

Andreas

Hi all,

we'll assume that the percent components are weight percent since if they were mole percent the molecule will look more like coal, assuming that such a beast exists.

You have N2 = 0.1%. Can you confirm this since 1.0 percent would give a balance. Unfortunately 1.0 % as fuel bound nitrogen is asking for trouble with emissions as this could easily produce lots of NOx.

With this fuel I would guess that you would need a minimum of 30% excess air (roughly 6.5 % oxygen). On this basis 1 000 kg/hr fuel would need some 13 900 kg/h air or roughly 6 680 acfm. Do not use these numbers yet as I have not checked them since I would like to talk to a couple of burner suppliers to get their opinions. Where are you located and what is the application as I would like to know if there is any back radiation or cold spots. If you want to generate steam, please advise me of the style of boiler, the steam conditions (pressure, superheat, etc.), number of burners, etc.

I am not in the UK at this moment so can you be patient?

Best wishes, Stephen.

Stephen Lloyd,

To further comment on the fuel composition: the analysis I've seen says

[C (80%), H(10-12%), N(

Andreas Volden,

As it is not mentioned in the question , I have assumed that the given composition is on volumetric basis (80% C, 11%H, 8 % O2 and 1% N2) and wrote an stoichiometric equation. (Note: The stoichiometric equation can also be written if the composition is given in mass basis).

The assumption of products (CO2 and H20) is based on the fact that, the combustion process efficiency is very high (0.97-0.99 is most practical cases) and very small amount of products like CO, H2,NO, N2O .....etc. are formed compared to CO2 and H2O. By this simplified approach it was calculated that 3.55 moles of air are required to burn 1 mole of fuel. By converting the moles into mass we get that, 102.81 gm of air required to burn 12.41 gm of fuel at stoichiometry. 102.81 gm of air consist of 23.86 gm (102.81 * 0.233 = 23.86) of O2.

Now let us say we require 3.5 % excess O2, that means 12.41 gm of fuel requires 24.69 gm (23.86*1.035= 24.69) of O2, hence 105.96 gm (24.69/0.233) of air. In terms of given fuel flow rate, 1000 Kg/h fuel requires 8538.7 Kg/h of air.

If the above composition is given on mass basis. On mass basis the empirical formula will be derived as C6.67H11O0.5N0.071.

For this case also you can write the stoichiometric equation and perform the same analysis as above.

Syed Mughees Ali,

thanks for the thorough reply.

I'm starting to understand.

However, from the equation you wrote in the very first answer

"From this equation the value of "a" = 0.747

So 1 mole of oil requires 0.747*4.76 moles of air."

Where does this 4.76 factor come from?

Br,

Andreas

Dear Andreas Volden,

In terms of mole fraction air composition is given as (0.21 O2 + 0.79 N2).

By using basic upper-primary class mathematics. Apply unitary method.

So for 0.21 Mole of O2 requires = 1 mole of air

Now 1 mole of O2 requires = 1/0.21 =4.761 mole of air

So 0.746 mole of O2 requires = 4.76*0.746 =

@ Stephen Lloyd and @ Andreas Volden,

This composition is provided on mass basis.

Reason: Let us say we consider it to be volumetric basis, by this the emprical formula for the fuel we get will be CH13.6O0.1N0.01. So, this compound cannot exist is reality. As the maximum number of hydrogen a single carbon atom can hold is 4. But in this case one carbon atom is able to hold upto 14 atoms.

On mass basis on the empirical formula is C6.67H11O0.5N0.07. In this case each carbon atom is holding two hydrogen on average. This is quite realistic as higher hydrocarbons exist in the from of liquid and have similar H2/C ratio. So, it has to be mass basis. Andreas, I suggest you perform the exact same analysis in terms of mass basis which I have performed on volumetric/molar basis.

Dear Mr. Andreas Volden,

The amount of air you need to combust fully at around 3.5 % excess oxygen is about 20000 to 22000 m3/h of air for 1000 kg/h of pitch. You do not need to talk to any person. Just ensure that the blower is able to deliver at least this amount of flow ( perhaps up to 25000 m3/h would be better). Run the combustion system and if the system is enclosed with no additional leakage points of air into the system and there is an exhaust, just measure the oxygen fraction at the exhaust. You may need to go raise the air flow to get about 5 % oxygen to ensure clean combustion because of mixing limitations of practical systems). That is all

Andreas Volden ,

Good morning,

not too many people make these burners. I was advised to try Metso. They operate in many countries. They have a website so enquiries should be straightforward.

For the fan, you will need to know the RDL (register draught loss) of the burner as well as other system losses to get the necessary fan head. You have the approx. flow rate as given above. You have steam for atomisation so you will need approx. 0.5 to 1.0 kg steam/kg fuel with this fuel - just ensure it is not wet. With this knowledge you should be able to get clean combustion.

Regards Stephen.

Hi all!

Just to sort of make my contribution to this discussion.

Assuming the percentage of each component of the fuel is given in mass:

C0.0667H0.11O20.0025 in kmole/kg

In terms of O2 required (no excess O2) from air, I then get 0.1741 (0.066+0.11+0.0025) kmoles O2/kg fuel.

For an excess of 3.5% O2, I get 0.18021 (1.035*0.1741) kmoles O2/kg fuel required.

By assuming fuel flow of 1300kg/h, I then get 234,27 kmoles/h of O2 required.

This translates to 7496,53 kg/h O2 required, and a total air flow of 35697 kg/h.

In terms of flow, considering air density at sea level and 15 deg.C, I get a total air amount required of 29141 Nm3/h.

Then considering that primary air should be 15%, we have the required amount at 4371 Nm3/h.

For 1000kg/h fuel flow, I get a total air amount needed (3.5% O2 excess) at 22416 Nm3/h and for 15%, I get 3362 Nm3/h.

So, this does coincide with Hanasoge Mukunda's answer.

And, I'm somewhat convinced that this is the right track.

Any comments on the above is of course very welcome.

Thanks for all the valuable inputs to you all!

Best regards,

Andreas