[aekerr]

Hi All,

I am currently working on a dissolved gas analysis project.  I need to be able to spike mineral oil to a specified (gas in oil) concentration using a chosen gas. 

I have decided to bubble gas of known concentration through degassed mineral oil using custom-made apparatus.  Essentially what I am looking to know is how long each gas will take to reach equilibrium in the mineral oil. The different gases being used are methane, ethane, acetylene, ethylene, carbon monoxide, carbon dioxide and hydrogen.  I assume that not all the gas will dissolve in the oil, therefore I should be taking into account diffusion co-efficients or something similar, so any information on obtaining/calculating these is needed also.

I am sure there is related theory/equations somewhere, but unfortunately cannot find any.  Help of any kind would be greatly appreciated.

Regards,
Adam

N.B the oil is only being spiked with one gas at a time and they are not an explosive mix.

[Stepan]

your approach is good, but: allows you to work with only normal gas pressure and require a following analysis of dissolved gases, which is a problem by itself.

I believe, you can get the same results in one measurements within the broad range of pressures, if you study Pressure-Volume equilibrium in closed system like syringe.  What I mean - If you fill a syringe with oil and top it up with your gas, you can record V1 (gas volume) and P1 (gas pressure). Eventually the gas volume will drop due to gas solubility.

Then you compress the gas by applying pressure on syringe plunger, and fix the gas volume. You have P2 and V2 Now you can record how fast the pressure drops to the equilibrium point. This gives you kinetics of gas solubility in the given pressure range. Deviation from normal compressibility P-V plot allows you to calculate how much gas was dissolved.

Given that you have your equipment airtight and calibrated you can have all you results as fast as one gas per day.

[aekerr]

Thank you very much for the reply Stepan. 

There are a few points which I don't fully understand.  Firstly when you mention broad range of pressures, what do you mean?  Do you mean the pressure which the gases are at in their cylinders?

Secondly, when you speak of P2 and V2 are you referring to the pvt relationship?

When I calcluate how much gas is dissolved what results will I be able to derive?  Does knowing the original amount of gas injected and the amount actually dissolve give me a diffusion co-eff? 


I neglected to mention this earlier but this procedure is going to be used on a relatively large scale (approx 8 liters of oil in a 10 liter container) hence why I thought the bubbling technique might have been more practical.  What do you think?

Thanks again,

Adam

[Stepan]

Let talk about equilibrium first. With sufficient accuracy, the gas solubility is proportional to its pressure. Let say you have a pressurized cylinder, where you have 5L of oil and 3L of gas phase above the oil.  You purge air from the container and pressurize it with Methane at 10 atm. Your pressure starts to drop right away because the gas is dissolved in oil. at the end it will stabilize at let say 4 atm. This means that at P=4 atm 5L of oil can absorb V*(P1-P2)=3L*(10atm-4atm)=18 L of Methane (given at normal pressure) or 18/5=3.6L of methane/L of oil. By running the experiment at different temperature you can estimate even thermodynamic parameters of your process.

In regards to kinetics. How fast the pressure drops during you experiment is a measure of mass transfer which is a function of the surface area, pressure, temperature, depth of the oil level, oil mixing. and so on. By measuring dP/dt you ca study the kinetics as well. 

[aekerr]

Ok Stepan, this is probably going to seem like a stupid question but how can I be sure of what the pressures P1 and P2 are?  There will be a regulator on the gas cylinder, but not on the 10 litre glass bottle which is holding the oil and bubbling apparatus.

I have this apparatus almost set up so initially I would like to try the bubbling method I have mentioned. (even if it is not ideal)

Am I correct to assume that if the 10 liter bottle is left for a long enough period of time, equilibrium will be reached between the gas bubbles and oil and then the oil and the headspace within the closed 10 liter bottle?

If so, what length of time would it be suitable to achieve equilibrium?

Thanks again,
Adam

[Stepan]

Ok Stepan, this is probably going to seem like a stupid question but how can I be sure of what the pressures P1 and P2 are?  There will be a regulator on the gas cylinder, but not on the 10 litre glass bottle which is holding the oil and bubbling apparatus.

I have this apparatus almost set up so initially I would like to try the bubbling method I have mentioned. (even if it is not ideal)

Am I correct to assume that if the 10 liter bottle is left for a long enough period of time, equilibrium will be reached between the gas bubbles and oil and then the oil and the headspace within the closed 10 liter bottle?

If so, what length of time would it be suitable to achieve equilibrium?

Thanks again,
Adam

Of course, your method is as good as any other. Just out of curiosity, how will you analyse oil for dissolved gases? And how will you know that the equilibrium is reached? And how are you going to monitor pressure drop in headspace when the gas starts to dissolve (pressure drops, and reduced pressure will affect solubility)? How will you preserve oil sample from loosing dissolved gas between experiment and analysis (it will behave as an open can of pop)?

I know that professional test for dissolved gases in oil costs several hundred dollars per 1 sample. This is a good indication of how much equipment and labour are involved. 

[aekerr]

Hi Stepan and any others reading,

I will analyse it by photoacoustic spectroscopy.  How will I know it has reached equilibrium is probably the most important question currently and one which I am struggling to find an answer, any ideas where I should start?

The headspace between the oil and gas will always be kept closed and airtight pipes will be extracting the oil and returning the oil, therefore it shouldn't behave as a can of pop does, however thanks for pointing that out.  I don't think there will be a pressure drop except maybe at the very start as I am continually bubbling the gas through the oil.

Thanks again for your help,
Aekerr