Air dissolution rate in seawater

[peterzwart]

Hi All,

I face an engineering problem on which I cannot find an answer. Hence I was wondering whether someone on this forum could help me out.

Please find below our situation. Lets image we have a cylindrical vessel with an open bottom. This vessel is submerged in seawater. This situation is represented by the left image. In this case air is trapped within the vessel. The volume of the air is close to identical to the inner volume of the vessel.

Than we submerge the vessel all the way to 3000 meters of seawater. The hydro static pressure increases to about 300 bar and acts on the gas. As a consequence the gas is compressed and will shrink to a volume of about 1/300 of its original volume. This entrapped air resides at the top of the internal vessel volume.

Now I have two questions:
1. At what rate will the air dissolve in the seawater? (e.g. how long will it take before half the air is dissolved in the seawater?)
2. Will the entire air be dissolved into the seawater or will a part of it remain?

It doesn't have to be seawater, also guidance on normal water would already be great.

Our parameters:
- Internal volume of the vessel: 1.3 m3
- Temperature at 3000 meter, 4 degrees Celcius.
- Air temperature when submerged: 20 degrees celcius.

Any guidance would be greatly appreciated.

Thank you.
With kind regards,
Peter

 

[Compositepro]

My guess is that all the air would dissolve before you got to depth.

https://chem.libretexts.org/Bookshe...cts_of_Temperature_and_Pressure_on_Solubility

 

[LittleInch]

I think you need to get your head around Ficks law of diffusion.

Air into water looks like about 2 x 10-5 mols / m2 sec

So the area of your cylinder is important plus how is the water inside the cylinder going to circulate? Otherwise the gas concentration at the top of the cylinder is going to be higher than in the free water. That makes sense - a small surface area only has a limited are for atoms to move from gas into the liquid wheres a larger area it increases diffusion, hence the need for towers and the like to increase the area of the liquid drops.

So the link by composite pro tells you how much air can be dissolved in a unit of seawater and pressure increases that number, but the rate at which is does it is Ficks law (I think).

NOw what the time period is I can't quite get my head around but I think for 1 mol of gas to diffuse over 1 square meter it takes 14 hours.

Which is 28 grams air.

You have 1600 grams of air. Do the maths.

I get 33 days for the whole lot assuming that the water is somehow circulated to give a constant diffusion gradient / gas concentration. But remember that was for one square meter. Don't know what the area of your cylinder looks like.

I'm happy to be totally proved wrong, but it seems to have the right "feel" to me. Sure the air will diffuse into the water, but really not very fast.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Compositepro]

LittleInch your post implies that the pressure has no affect on dissolution rate, which is not be correct. I agree that the rate depends on diffusion, surface area, and turbulence/mixing. Water containing dissolved air is slightly denser than pure water so there will be convection in the water and not only diffusion. At 300 bar water will dissolve 300 times as much air as at sea level. This is one of those cases where the value of a theoretical calculation is of limited value except to get an order of magnitude feel for the answer. Which is why I responded as I did.

 

[LittleInch]

composite - Yes pressure counts, but I think this is simply as a result of the higher solulability of the seawater. How this affects the diffusion coefficient I couldn't find. Or maybe it doesn't impact it directly.

But the actual number relies on both elements. The key is how fast does the now air saturated seawater get replenished at the interface by seawater with less air saturation? That looks too complex to me for a simple calculation.

I suspect most calculations are based on fixed volumes of water and air and the diffusion stops at some point when equilibrium occurs. That doesn't happen here.

But it all seems rather academeic?

Does the same thing happen in submarine ballast tanks?? Hence why I don't think the air diffusion rate is super massive or super fast, more the opposite.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Compositepro]

Submarine ballast tanks used for buoyancy compensation are inside the pressure hull and kept at low pressure. Pumps are used to pump water out. The amount of air and energy to use compressed air for buoyancy control at depth is not very practical or safe. A slight increase in depth will compress the air and reduce the buoyancy, causing more depth.

The main ballast tank, outside of the pressure hull, is used only at the surface to float above the surface. It can be used for emergency buoyancy with compressed air, but would normally be completely flooded when submerged so that depth changes do not change buoyancy.

https://maritime.org/doc/fleetsub/chap4.htm

 

[IRstuff]

Henry's Law explains that gas solubility increases as a function of pressure; However, the Henry's Law coefficients are different for each constituent of air

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529 Entire Forum list

http://www.eng-tips.com/forumlist.cfm

 

[georgeverghese]

Air solubility in seawater is some what different compared to that for non saline water, but you can take a stab at this using the Henry's Law values for air (or N2 and O2 separately)in pure water at 300bar - see Perry Chem Engg Handbook in the chapter on Physical Data. Would imagine any CO2 in the air would have completely dissolved in the water at 300bar. See if you can google H values in seawater also.

 

[LittleInch]

I think there's plenty of solubility in the seawater, it's the diffusion rate that's the main thing here.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[georgeverghese]

Why do you need to know how long it takes to dissolve half the air mass? The solution to this may be fraught with all sorts of errors due to the unknowns re the mass transfer coeff for N2 and O2 at the gas-liquid interface. In any case, how long does it take for this inverted vessel to get down to 3000m?
Have a read of the subsection on Interphase Mass Transfer in the chapter on Heat and Mass Transfer in Perry Chem Engg Handbook 7th edn - the main unknowns here are the individual mass transfer coeffs for the gas and liquid phase (kg and kl). It may be possible to come up with some theoretical answers with some assumptions.

 

[peterzwart]

Dear All,

Thank you very much for your responses and sorry for my late response.

@Littleinch: can you indicate where you get the 2 x 10-5 mols / m2 sec value from?

The purpose is to keep the air in the vessel as long as possible. So we need to minimize the contact area.

We will likely do some experimentation to really get to know the time whereafter the air is dissolved.

Again thanks,

Kind regards,
Peter

 

[LittleInch]

From ficks law of diffusion searches on google.

A few weeks ago now so can't find the search data.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[TiCl4]

What is the temperature down at that depth? I imagine it will play a large role in the diffusion process.

 

[142846]

After a long time because 1. the cylinder is bottom open 2. the cylinder is very small against sea 3. turbulency in depth of 3000m, therefor the total air goes to sea. And about "how long will it take before half the air is dissolved in the seawater" this is very difficult and a lot of parameters should be available. thanks