Dry Air Calculations for Adiabatic Application 1

[tc7]

I am sizing up air flasks for a piston accumulator. The purpose of the accumulator is to capture a large volume of liquid (glycol) which is exausted through an upstream valve in our process. The gas side of the piston will be charged with dry air. The initial charge on the air (and the entire system) will be approximatly 400 psig. During the fluid accumulation, the pressure will rise to about 650 psig in a matter of 3 seconds. I have been told the air quality must be based on a dewpoint of -10 degrees F or lower. I am assuming an adiabatic compression on the air side but I have no idea how the air dryness works in to any calculations I need to make, either in sizing the flasks or in estimating temp rise. Based on simple equations I am using, it doesn't seem to matter if the air is dry or saturated! Can anyone help explain how air dryness may be necessary in this application and what calculations should be used? Standard thermo or fluids texts aren't helping me much.
Thanks in advance,
TC7

 

[cbiber]

Sounds to me like the dewpoint won't affect your calculation -- you just don't want any condensation going on in the process.

I'd go back to the person who told you about the -10 degrees F, and ask them.

 

[hacksaw]

the dew point requirement is an alternative way of specifying air purity-clean dry air. you'll need the requirement when the cylinder gas is purchased.

it has little effect on the calc. if you are worried, then use dry nitrogen.

 

[tc7]

cbiber & hacksaw-
I appreciate your thoughts on this, I started to believe the air quality requirement was only precautionary to minimize pitting or corrosion in the equipment and this may be true, HOWEVER, I've now been told that we'll need to do energy balance calculations at some time in the future. With this in mind, my revised question is "how does air quality (moisture content) affect calculations for work required to compress air in an accumulator"?
Thanks in advance for advice or references.
TC7

 

[hacksaw]

grab you stoichiometry text book or Perry's and sort it out.

non-condensible gases are relatively easy to model as long as the pressure is not extreme.

 

[fjd]

FYI, Have you considered the effects of the solubility of air / nitrogen in glycol??? You are working with might be a "super champagne" bottle on your hands.

I know of three situations of gas in liquid:
air in water in a chiller application (resulted in pump problem)
helium in water in a pressure test application (never attained expected pressure)
carbon dioxide in beer in a "cost effective" distribution line to the bar ( unsatisfactory foaming at the draft valve)

Frank

 

[tc7]

Frank-
No I have not concerned myself with air solubilities in the glycol. I will not be using nitrogen for the gas charge.
The fluid we use is ethylene glycol to be more precise and we buy it with anti-foaming additives. Although I have other applications where foaming does occur, but these applications are in accumulators which have no piston or membrane seperation between the liquid and the air charge. In these cases foaming occurs when our tempo of operations increase. In fact, now that you make me think about it, air entrainment is very evident throughout the fluid but has not caused any obvious performance degradation. What advice would you offer if I did need to worry about air entrainment or foaming of the EG?
Thanks,
TC7

 

[fjd]

tc7 - If you do have to worry about air entrainment or foaming, I would suggest a "liquid / vapor" seperator with some sort of secured demister pad. Be certain the demister pad cannot come loose and become entrained with the flow. Note the demister geometry should be cylindrical with a vertical centerline. The liquid / gas input stream should enter about 1/3 the way up the vessel, liquid discharge from the bottom and gas discharge from the top.

fj

 

[kenvlach]

My understanding is that dewpoint values are for the air at 1 atm.
When the pressure is raised to 650 psi (44.2 atm), the water vapor increases accordingly. The –10[sup]o[/sup]F dewpoint corresponds to P[sub]H[sub]2[/sub]O[/sub] = 383 Pa (0.00378 atm) at 1 atm total pressure, but the partial pressure increases to 16.9 kPa (0.167 atm) at 44.2 atm total pressure. The corresponding dewpoint (saturation T) becomes 56.5[sup]o[/sup]C (133.7[sup]o[/sup]F). So, it would appear as though condensation would occur (unless at elevated temperature).

However, this condensation would not occur; the ethylene glycol would absorb nearly all of the moisture out of the air. Perry’s has a chart for vapor pressures over aqueous solutions of diethylene glycol [not for ethylene glycol] which shows that at about the saturation temperature of 56.5[sup]o[/sup]C, the vapor pressure of a 1 vol% water solution is about 4 Torr (0.005 atm) while the v.p. of a 5 vol% water solution is about 30 Torr (0.039 atm). (Essentially all of the vapor above the solution is water, by the way). An ethylene glycol-water solution has a similar strong interaction (which is why it helps prevent freezing and boilovers in automobile radiators).

So, if the condensation of water vapor is not a problem, why worry? One reason for using ethylene glycol as a hydraulic fluid (besides its very low v.p.) is its low compressibility,

&[ignore]kappa[/ignore]; = 3.64 x 10[sup]-4[/sup]/MPa at 20[sup]o[/sup]C.

Some variation in compressibility would be expected if the ethylene glycol began absorbing water. I remember when the brake fluid (pre-silicone type) used in automobiles would absorb water when exposed to air. This was a problem because of corrosion and because it caused a ‘mushy’ brake pedal when the fluid got hot. So, perhaps the logic in using ~dry air is to keep water out of the ethylene glycol.

 

[tc7]

Ken-
In the accumulator applcation I am presently working with there is a physical separation between the glycol and the air charge, so I don't need to consider the absoption by the glycol.
I still havent figured a good answer to my initial posting, i.e. what is the work required to compress dry air (-10 degree dewpoint) as compared to the work required to compress saturated air. I'm am thinking that the best way to approach this is to determine the entahlpy of the the vapor at the partial pressure of the saturted mixture and then use Work = (delta h for the vapor + delta h for the pure air at its partial pressure). What do you think?
tc7

 

[kenvlach]

OK, I was reading the later posts more intensely; a good reminder to always check the original!

Beginning with the 400 psi (27.2 atm) conditions, the initial air (with dewpoint –10oF for 1 atm total pressure) will have a water partial pressure of 10.4 kPa (0.1028 atm). This means you will liquid water will be present for temperatures less than about 46.7[sup]o[/sup]C (116[sup]o[/sup]F). Two questions:
1) What is the temperature with the system at 400 psi?
2) Do you have any means to drain liquid water from the compressed air?

The water will cause P & T transients in your system upon increase of pressure to 600 psi. A quick increase in P will give an ~adiabatic increase in T, but then as T slowly declines (to ambient?), condensation of water will decease P. This necessitates additional air to maintain piston position.
Do some iterations to follow the behavior throughout, or do a calculation for the final T while holding at 600 psi, figure out how much water will condense out and work backwards (doing the dry air & water separately).