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Sucking air using water outfall

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Hilal1984

Mechanical
Nov 7, 2015
13
Hello Guys,

I am not sure whether this forum is the right one to ask my question or not.

My question is simple. I want to suck air by making side opening in a pipe carrying water from outfall of more than 6 meters high and discharging to nearby canal. The air will be discharged later from the pipe by making another opening in the bottom. I want to calculate how much air can be sucked. Please have a look on the attached sketch.

I need equations for doing that.

Thank you in advance,

 
 http://files.engineering.com/getfile.aspx?folder=31c0fbc0-c4dc-4baa-b366-819eaed83547&file=air_sucker.png
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Based on your sketch it will not work!

Sometimes its possible to do all the right things and still get bad results
 
You need to study the venturi effect.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
As you have drawn it, it will not work.

You need to study the Bernoulli equation with the venture application.

What are you trying to accomplish with this?

Aerate the water? A stair step drop will provide aeration if that is what you are after.
 
I am trying to use trompe technique for compressing air
 
Having had a look at this, I can't see anywhere a set of calculations like you are looking for. However if you start with a few first principles:

You need a velocity of > 1m/sec to actually prevent the air bubbles from rising.

All the somewhat vague diagrams I can see show the air entering at the top of the tube either via a venture or a reverse pitot tube.

At atmospheric conditions I would not expect the volume of air right at the start to be more than 10% of the volume as otherwise the change in overall density would be too much to still allow flow when you only have a few metres (if that) of head forcing it down the pipe.

I think you'll have to do a few expoeriments to see how it works.

Depends what you're trying to do - Is it just suck air? or is it to take the air and do something with it? If the latter, you need to have quite a high return point so that the air in the bottom is actually under pressure.

I managed to glean this with about 20 minutes searching. What have you found out / know?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
let the water drive a generator with compressor, isn't that easier?
 
Basically what this compressor does is that it uses the venturi effect AND a separator to compress air (a little - in the wikipedia example enough to ventilate an blacksmiths fire).

So a more modern example would be to have a venturi and then a separator vessel afterwards. Put the ventiri in the low part and let the water/air stream eject into the separator. Your downstream air conditions will determine your pressure and in turn your water flow rate (increase downstream pressure) and then your total air volume flow.

The venturi can be bought in a shop or you can maybe find a drawing somewhere how to size the water and air inlets. Normally these are used to create a vacuum where you draw air from - but in your case you will just draw air from ambient. So this is not the "normal" sizing regime and you may have to read carefully when selecting the venturi dimensions.
 
@europipe, i have seen some interest in very low tech (no moving parts) pumping (and now compressing) i guess its for 3. world applications where e.g. the inability to get spare parts has a tendency to stop all projects sooner or later - especially if the equipment or parts of it e.g. cobber wire) can be sold or has another use somewhere else.
 
The maximum amount of air is limited by the rising air. It has been found empirically that, for a void fraction larger than 0.3, the flow pattern changes from bubbly flow to
irregular flow patterns with large "chunks" of air. This gives rise to unstable flow conditions and may resulting in a backflow of the air (flooding).




books


 
From the drawing in IRstuff's Wikipedia link, it is apparent that a trompe does not use the venturi principle to entrain air. It uses the siphon principle. It is basically a siphon where a negative pressure is created by a restriction at the suction end of the pipe, rather than a rise in elevation. Once you have this negative pressure, any orifice connected to the atmosphere will suck in air.

You are not going to find a an "equation" to design this. But as bimr noted a 0.3 volume fraction of air in water (at the suction) is probably an upper limit to how much air can be entrained and still have stable flow. The design and arrangement of the nozzles to inject air into the water flow is the critical design issue. You can reinvent the wheel or learn from the experience of others (copy previous designs).
 
Compo - year 200 years ago, you can still use the principle but upgrade it with modern knowledge and materials - if you have access to water at one elevation, can "dispose" it at another and need an air stream at a (slightly) elevated pressure.

I dont think the "air fraction" is of any impotance - you dont _need_ to entrain the air in a vertical pipe as shown in the old drawing. You can use a more modern approach. Attached schematics shows my suggestion. The venturi will determine how much air that can be moved (along with the design parameters, elevation difference, needed air pressure, water volume flow and temperature etc)


 
Since, the venturi is also over 200 years old, the use of a venturi is not exactly an updating of technology. Neither technology appears to be feasible.

The trompe device has been extensively studied. The amount of air possible with the trompe deveice is limited by the void fraction.

 
Design of MortenA is exactly I wanted to do earlier. However, I had difficulty to find good literature about ejector design calculations
 
Ejectors are a much older technology than venturi's.

Venturi's are different than ejectors. In a venturi, most of the energy is recovered. Ejectors lose most of the energy.

The reason ejectors are rarely used is energy efficiency is lower than 30%. The available outlet pressure will be less than 30% of the inlet.

See the attached paper for design.

Why don't you just explain what you are trying to accomplish with this device?
 
 http://files.engineering.com/getfile.aspx?folder=5fbf53c1-ed58-4ed5-9014-8ffbf51e03ba&file=Eductor-Pumps.pdf
I am trying to produce slightly compressed air (0.5 bar) for our plant which require big volume of air. I am trying in the same time to reduce operation cost and get rid of normal compressing machines which require spare parts and labor for maintenance.
 
To get a big volume of air you'll need a HUGE volume of water. My estimate is 8 to 10 times the standard volume of air in water flow.

Let us know how it goes.

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