Evaporation from Tank (Theory Question)

[Jieve]

Assume I have an enclosure, say 4ft x 3ft x 3ft which is closed but not completely airtight. There is a small cylindrical tank of ethanol, say 1.5ft D x 1ft H, with a 1.5 inch pipe sticking out the top, open to the atmosphere (enclosure space). The liquid ethanol level in the tank is about 8” high. The surroundings are at 70F and 14.5psia.

Ethanol vapor density is higher than air, so evaporated ethanol should settle at the bottom of the enclosure. As far as I understand, the ethanol should continue to evaporate until equal concentrations of ethanol exist in the liquid and vapor state inside of the enclosure such that equilibrium vapor pressure is achieved (although this likely will never completely happen, as the enclosure is mostly, but not completely airtight).

But what is the driving force pushing the ethanol vapor out of the container? Pressure build-up that overcomes the surrounding air pressure? Diffusion due to concentration difference? I’ve heard argued that ethanol vapor will just sit in the tank and not build up in the enclosure, but we know intuitively that this isn’t correct (leaving an open gas can in a closed garage will fill the garage with gas fumes).

Also, what would be the effect of pressurizing the air in the enclosure (low pressure, say a fan blowing in atmospheric air from outside). Would this reduce the escape of ethanol from the tank in that case?

Thanks.

 

[Compositepro]

The main cause of evaporation from a closed, but not sealed, container is due to "breathing" of the container. Any cyclical temperature or pressure change inside or outside of the container causes air to flow into and out of the container. This carries solvent vapors out. Even the tiniest pinhole in the container will allow this to occur.

 

[LittleInch]

Always good to draw these things as it can get lost in words when trying to describe it, but I think in this case your issue is mainly diffusion form the ethanol rich atmosphere in the tank which is connected to the "enclosure".

The 1.5" pipe is what will limit this diffusion rate and may be that it never gets equal if the "enclosure" is not air tight as the ethanol vapour will then try to diffuse into the atmosphere

Pressure in the enclosure will make little difference in terms of diffusion as the pressure in the tank will be the same as the pressure in the enclosure. What this could do is reduce the concentration of ethanol in the vapour space in the tank due to different partial pressures, hence reduce the diffusion rate.

 

[Snickster]

This is the way I see it:

The vapor pressure of ethanol is only about 1 psia at P/T indicated. Assuming initially 1.5 inch pipe is closed with complete tank vapor space filled with ethanol vapor at 1 psia. The pipe is then opened. Air at higher pressure 14.7 psia will flow into the tank and push the ethanol vapor out. Ethanol will then evaporate into the tank vapor space until the partial pressure of the ethanol in the air is equal to the vapor pressure of the ethanol or 1 psia approx. The air in the room enclosure will still have very little ethanol vapor except for the vapor initially pushed out. As the tank vapor space of air and ethanol mixture becomes saturated with ethanol the only way it gets out of the tank is by a concentration density difference diffusion process, I believe, between the ethanol concentration in the vapor space in the tank and the concentration of ethanol in the air in the enclosure. This will continue until the concentration/psia of ethanol in the air in the enclosure equals that in the tank or 1 psia.

 

[IRstuff]

But what is the driving force pushing the ethanol vapor out of the container?

The root cause is your temperature not being 0 kelvins and not being a closed system. Fick's Law describes diffusion as driven by concentration gradients; in this case, there is always a concentration maximum at the surface of the liquid because the thermally driven evaporation to achieve the partial pressure. However, since there is a concentration gradient of gas in the system, there will ALWAYS be a net flow of gas away from the surface, given that the system is not closed. So, the simple answer is Fick's Law https://en.wikipedia.org/wiki/Fick's_laws_of_diffusion. See the equation J = -D * (d_psi/d_x)

You can only achieve the cited partial pressure in a closed system; if it's open, there will always be a flow of gas outward, which is replaced by additional evaporation from the liquid.

Also, what would be the effect of pressurizing the air in the enclosure (low pressure, say a fan blowing in atmospheric air from outside). Would this reduce the escape of ethanol from the tank in that case?

I think you need to rethink this proposition; a fan, or pump, blowing in air wiithout some air leaving will result in infinite pressure; therefore, in an open system, there will be a comparable outflow, exacerbating the evaporation, since the fan increases the air flow inward, which increases the outward flow which entrains the ethanol gas thereby forcing more ethanol to evaporate.

The bottom line is that if you want to stop the evaporative loss, you have to seal the system; you could slow down the evaporation by cooling the system, since that reduces the partial pressure. Freezing would be the ideal, but there's likely some level of sublimation, particularly if the air is not comparably cooled.

 

[pierreick]

Hi,
Consider this material to support your work (examples &theory)