Liquid Ring Vacuum Pump - will it remove ethanol vapor

Hi,
Use your favorite search engine and check for solubility of Ethanol in water, ETOH is very soluble.
For the gas, you will need a gas scrubber or an adsorber with activated carbon.
Perry's chemical engineer's handbook will help.
Good luck.
Pierre

So, assuming the EtOH is in equilibrium with water at the pump exit at a pressure of 1atm and 70degF water temp, EtOH will split out into the water and the vapor in accordance with the vapor pressure of the EtOH / water solution leaving the pump. Obviously, the more water you dispose of, the more dilute it will be (ie the less EtOH in the water) and correspondingly, the less the EtOH in the vapor ejected from the pump. See if you can find a vapor pressure curve at 1atm, 20degC for the EtOH / water binary - I dont find this on the net, and it is not in the physical - chemical props chapter in Perry Chem Eng Handbook either.

But would it be at 1atm?

The pressure inside the pump will be the real one to look for?

The liquid ring will be subject to the"vacuum" - What sort of absolute pressure are we talking here? - so it's the equibrium solution at xx mbara. Flow rate of water and vapour and also contact time matter.

The will be some ethanol in the seal water but a lot of the ethanol vapour will be expelled I think, more so as the pressure drops.

Why do you want to know?
You have to assume there will be some vapour coming out of the pump so either send it through a scrubber or a water contact tower to pickup the rest.

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

I agree with the analysis by georgeverghese. The ethanol-water VLE curve is available at
The controlling factor in my opinion will be the flow of incondensible gases through the pump. If there is no flow of gas there will be very little flow of ethanol vapor. Once you know the flow of gas into the pump you can calculate how much ethanol is entrained into the pump, and a mass balance using the VLE data together with the seal water flow rate and the incondensible gas flow rate will give you the split between the ethanol caught by the water and the ethanol escaping to the atmosphere.

Katmar Software - AioFlo Pipe Hydraulics

"An undefined problem has an infinite number of solutions"

I've just realised that this water - EtOH mix at pump exit does not have a total vapor pressure of 1atm. It is a subcooled liquid pumped up to 1atm.

You could assume that EtOH in water is a small fraction, assume ideal behavior and get an approx answer. Pure EtOH vapor pressure at 20degC = 5.3kpa abs. Assume all EtOH drawn in as vapor is dissolved into water rejected, and work out the partial pressure of EtOH in the exit, which will then convert to vapor fraction.

PP of EtOH in vapor = 5.3 x mole fraction of EtOH in liquid.

Mole fraction EtOH in vapor exit = PP of EtOH in vapor / 101.3

This ideal behavior assumption at low EtOH concentration in water is most likely okay since the DDBST graph @katmar has posted shows a straight line at low "x" values.

Hi,
To add to your query a resource from Sihi about liquid ring vacuum pump. The set up at the end of the document should help to visualize the process, in particular the separation pot with the release of gas and liquid.
Just curious why do you pull vacuum on this storage tank? why not blanketing it?
note:
In water, miscible /1X10+6 mg/L/ at 25 °C
Riddick, J.A., W.B. Bunger, Sakano T.K. Techniques of Chemistry 4th ed., Volume II. Organic Solvents. New York, NY: John Wiley and Sons., 1985., p. 192

Good luck,
Pierre

Hi,
Several thermo models are available in the spreadsheet attached for binary system (water-EtOH).
Manually you change the system pressure. On each model you resolve the equations line per line as explained.
Good luck
Pierre

@birkATO
All sources above are based on ideal conditions. In reality some amount of EtOH will go to atm because of entrained gas, concentration of EtOH in water and non-ideality of absorption in pump vanes. Vapors and water contact inside of vanes lasts during milliseconds - you do not believe it will reach equilibrium, do you? Amount released depends on vanes design and local conditions and varies significantly. Only practice can guide you.

If in your particular case EtOH vapors can be totally absorbed by water then you do not need vacuum pump at all. In this case an absorber any kind of can work as a vacuum source and vacuum pump would be an overkill. Am I right? Think.

The other question is concentration of EtOH in vapors. You can reduce capacity of vacuum pump absorbing water-soluble components in an absorber upstream of a pump with a little increase of vacuum in pump.

Thank you all for the responses. I need to digest this and decide what I need to do. I am an air quality engineer and I am trying to identify a justifiable means of estimating emissions of EtOH from this process. I believe there will be very little entrained, non-condensible gasses in the system. There probably will be some when the process starts, but once any air gets removed from the system, the headspace in the container being pulled on will be saturated with EtOH vapor with very little "flow" through the system/vacuum pump. I am considering potentially just assuming that once volume of headspace gets evacuated per batch. As long as there are no significant leaks, this should be a reasonable estimate. I will also check out the thermo models and see if I can justify another method of estimating.

To answer someone's question; this is not my process, it is a client's. I can't answer why they are pulling vacuum constantly, but it is connected to a larger extraction process and what I described is just the end process. The whole process must be under vacuum or else the product they are extracting would be degraded.

-Ryan Birkenholz, Licensed PE in MN and IA

It depends on your process conditions and the level of vacuum.

I saw a reply above suggesting that if the flow of non-condensables is small, the EtOH lost would be minimal. I do not think that is the case.

Liquid ring vacuum pumps are often used for fairly intensive vacuums - i.e. rotary vac drum applications to get to 26-28 in.Hg. If you pull vacuum on the vessel and the vapor pressure of the liquid mixture inside is greater than the headspace pressure, you will constantly pull of vapors.

This sounds like a solvent stripping activity, where, after a reactions is complete, vacuum is pulled to strip the solvent from the heavier products. Vacuum stripping is very effective, and I would expect a tank with pure or near pure EtOH that has a Nash pump pulling on it will pull of a large amount of EtOH very quickly.

Why do you think that flow will be low if the vacuum line gets to 100% EtOH?[highlight #FCE94F][/highlight]