Air removal before thermal process

[sentenza]

Hello everyone, I need some advice on how to quickly remove air from a 400-liter tank before performing a thermal process. I was thinking of creating a separate tank where I can create a vacuum using a pump, and then use an automatic valve to connect the two tanks. My question is: how do I calculate the volume of the vacuum tank? What level of vacuum should I achieve inside it for efficient air removal from the system?
Thanks for help

 

[LittleInch]

First define "remove".

You need to look at pressure, whether bara, psia, mm H, Torr, whatever.

"Vacuum" is very vague - again define.

So you can look in a number of way, but try this.

400l of air at 1 bara,
Vacuum tank is at 0.1 bara, say 4000l

So you have 400 l air in your tank and 400l of air in your "vacuum" tank.
Connecting the two at the same temp, you have 800l in a total volume of 4,400l This will then be at a pressure of 800 / 4400 = 0.18 bara.

So to use a large tank / small tank thing, your large vacuum tank needs to be at a lower pressure than you need. Or at least get you closer to where you want to be.

Vacuum tends to reduce faster at the start then takes longer and longer per unit pressure reduction.

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

 

[sentenza]

thanks for help, I supposed to have a vacuum tank of 800 l with 65cmHg pressure. After valve opening, the vacuum in the full system of 800l+400l should be 43cmHg. I used Boyle-Mariotte for this calc.

 

[EdStainless]

Tip, when doing calculations for vacuum applications always use absolute pressures.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed

 

[Snickster]

I get that the resulting pressure after opening valve is 68.73cmHg.

I will work it out in Imperial units:

First find total mass in system before valve is opened:

For 800 L (28.25 ft3) tank at 65cmHg (12.57 psia), 80 F (assumed) using ideal gas equation PV=mRT:

12.57(144)(28.25) = m(1545/29)(540)

m = 1.78 lbs

For 400 L (14.13 ft3) tank at atmospheric 14.7 psia (assumed):

14.7(144)(14.13) = m(1545/29) (540)

m = 1.04 lbs

Total mass = 2.82 pounds.

Since mass is constant use ideal gas equation again for total mass and total volume and solve for resulting pressure after valve is opened:

P(144)(42.38) = 2.82(1545/29)(540)

P = 13.29 psia = 68.73cmHg

 

[IRstuff]

Simpler math says that 400L of air expanded into 1200L volume still results in the equivalent mass of 400L/3 of air, i.e., you "removed" only 2/3 of the air, regardless of what pressure you end up with.

Seems to me that pumping out 800L volume to remove 2/3*400L doesn't sound particularly efficient, compared to just pumping out 400L volume to start with. It'll take less than half the time, compared to pumping out the 800L tank, and you get ~99% of the air removed instead of 2/3.

The only efficiency gain would be if you were pumping the 800L tank while doing some processing in the 400L tank, but you'd still only be removing 2/3 of the air in that approach.

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[shvet]

4 options are available: pressure swing; vacuum swing; flow through; displacement. See CEN/TR 15281 for details. Each of those has procs and cons and any of those can be preferred depending on utilities available on site.

 

[sentenza]

thanks for suggestion. What I'm interested in is speed and removing at least 70% of the air. Do you think that connecting the vacuum pump directly to the 400-liter tank would give me an advantage in terms of speed? Or would it be better to have an additional tank where I create the vacuum in advance? I would like to complete this phase in 5-6 seconds

 

[shvet]

Any option above can provide 5 secs depending on utilities available on site and money spent.
You need all those things called 'feasibility study' and 'value engineering'. You are seeking for a shortcut like 'that guy from the internet has said...', shortcuts lead to overcosts or off-specs.

 

[LittleInch]

It would give you an advantage if you can find a vacuum pump large enough to do this without being ridiculously sized. That's very doubtful.

If you have time to create the vacuum down to about 0.3 bara, then it depends on how big your second tank is and how low a vacuum you can develop.

How long do you have between emptying will give you the size of your vacuum pump, but as said before, the pressure drop slows over time for the same speed of the vacuum pump.

Also if this is normal air then sudden reduction in pressure you could easily end up with water vapour mist in the vessel. Is that what you want?

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

 

[EdStainless]

Build you second tank to take full vacuum.
And being larger would help.
Pump it as low as you can.
A typical two stage rotary vane pump will reach <1 mTorr.
If your cycle time requires faster pumping of your second tank, then it is usually better to add a blower ahead of your vacuum pump rather than go to a larger vane pump.
And yes you will be making a lor of condensate.
If you want this fast the piping between the vessels needs to be large.
And large vacuum valves are expensive.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed

 

[IRstuff]

Unlikely that any pump can realistically pump out your tank in 6 seconds. Best bet is something more like 1600L tank pumped to vacuum state and cross connect. That should hypothetically get to 80% air removal, although you're limited by your plumbing at that point, so you still might not achieve 6 seconds.

TTFN (ta ta for now)
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[sentenza]

after pumping air, i'll put 15bar of steam in the first vessel. I have enough time, about 50 sec, to empty the buffer tank for vacuum. This is a machine to peel fruit and vegetables. I want to be sure that I don't have air that create a insulating barrier for product.

 

[LittleInch]

" I have enough time, about 50 sec, to empty the buffer tank for vacuum. "

Are you serious?

If you've only got 50 seconds to empty your bigger tank then you're going to need a humungous vacuum pump.

15 bar of steam will reduce the air content to less than 6 to 7%. Is that not enough?

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

 

[MintJulep]

If you have steam...

But really, you think that air will "insulate" your vegetables from 15 bar steam?

 

[EdStainless]

Mint, you beat me to it.
Steam eductors (also called ejectors) will do a great job of pulling a lot of air quickly.
You can get them designed to not draw a deep vacuum but simply to remove something like 1/2 of the air.
If they are going to cycle quickly you could have a small buffer tank to have them draw against when not connected to your main tank.

https://croll.com/sb/how-steam-ejec...create,the system at a high velocity as well.

https://graham-mfg.com/solutions/products/ejectors/

https://www.jetvactechnologies.com/product/vacuum-systems/

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed

 

[IRstuff]

60 seconds is practically possible, but probably absurdly expensive. I randomly picked components in Leybold's calculator and got under 60 seconds using 3 of everything.

https://calc.leybold.com/en/proj/66f5da5097482dbad7a0ca2e/edit

[segue] I can't find references, but I remember that McDonnell Douglas Huntington Beach has a HUGE vacuum chamber, 30-ft diameter, 39-ft height, I think, and the diffusion pumps are something like 6-ft tall and there were 3(?) of them. Still took gross amounts of time to pump down.

Oh, I found one picture; the things hanging off the left side are the diffusion pumps, I think

[/segue]


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[Compositepro]

The OP has not said anything about what level of vacuum is required. The numbers mentioned above are over 0,5 bara,[which most engineers would not call a vacuum, more a suction). Generally you cannot store a vacuum. Any extra volume just requires more pumping time to get to the final pressure, if you are talking about less than 10 cmhg.

 

[MintJulep]

you cannot store a vacuum

That's up there with "you can't push on a rope."

There's plenty of vacuum "stored" all around us - just need a really tall pipe to get access to it.

The OP's plan is to "dilute" the air in the tank by adding some vacuum. That works, to a degree.

OP appears to lack appreciation for exponential decay and other fundamentals.

 

[IRstuff]

The OP has not said anything about what level of vacuum is required.

That's partly true; the OP stated that they wanted to remove 2/3 of the air, so something like 1 psi would be a good compromise, which the system I codged together in the calculator would only need one copy of each pump and blower to get to 1 psi in 54 seconds.

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