Eductor Questions - How much air can be circulated for each liter of water? 1

[LiquidGravity]

My apologies ahead of time since I'm a newbie. I am wondering if someone can help me with an equation to calculate gas circulation based on motive water flow. The scenario is a Venturi eductor or injector whichever works better for liquid circulating a gas.

What is a ballpark ratio of liquid to gas? For example, for every 1 liter of water (motive liquid) circulates how many kilograms of air?

Is there an equation to calculate this?

If you require specific variables, let me know.

 

[LittleInch]

"Circulating a gas"?

Why not use a fan?

This sounds like a hard way to do it.

You need to describe the system with a sketch.

Yes there are ways to calculate this, but you need to know water flow or velocity and pressure of the gas and motive liquid.

[Edit] It seems to be quite variable depending on size, pressure of the water and pressure at the outlet.

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

 

[pierreick]

Hi,
Consider this link to support your work.

https://www.cebeco.com.au/wp-content/uploads/2018/11/Eductors-to-pump-gases-and-dry-solids.pdf

Pierre

 

[Artisi]

Did you bother doing an internet search, I got a number of interesting hits from the simple entry of "water / gas educator".

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

 

[bimr]

The amount of air recirculated will depend on the depth of water and the pressure of the motive stream, neither of which you have addressed. Refer to the brochure for information:

Brochure

 

[LiquidGravity]

Thank you to everyone who responded. Yes, I did do an Internet search, but I couldn't find exactly what I was looking for. I would like either a max ratio of gas to liquid that is possible or an equation that can predict the ratio.

I found one manufacturer of eductors that claims a 20:1 gas to water ratio, but it is unclear exactly what this means. I take it to mean for every 1 liter of the motive liquid (water), 20 kilograms of air would be circulated. However, I cannot verify this. If anyone knows the max ratio as a "ballpark" that would be good to know. Or, of course, the equation to predict the ratio.

I am trying to figure out the amount of heat energy that could potentially be transferred from air with a specific heat of 1,005 Joules per degree per kilogram, so 1 kilogram of 60 C air, for example, injected into 1 liter of water at 30 C, could impart -- 60 - 30 = 30 Degrees difference between air and water so the air could potentially impart 30 x 1,005 = 30,150 Joules per kilogram.

This link calculates the final temperature, but I have to know the ratio of gas to air as inputs to determine the amount of heat.

https://www.engineeringtoolbox.com/mixing-fluids-temperature-mass-d_1785.html

Use these variables for the water flow:

Water Temp = 30 C
Air Temp = 60 C
Water Flow rate = 44 l/s
Upstream Water Pressure = 202,650 Pascals

 

[LittleInch]

They will mean volume of air to water, so 1 m3 of liquid flow could pull in 20 m3 of gas (not kg). Just look at the figures in the chart linked by pierre, but remember the liquid is quoted in GPM and the vapour in scf/min.

But it is massively affected by what pressure the gas is at when it comes into the educator and the flowrate and pressure of the liquid and the pressure of the liquid as it exits the eductor.

For your issue above remember the density of air is ~1.25kg/m3. So 1 kg of air is ~800 litres of air (volume ) of air at STP. So about 40 times more air than the best eductor can manage.

So basically your air at 60C being "injected" through an eductor at a 20: 1 ratio will not heat up the water by more than 0.2 C at most.

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

 

[LiquidGravity]

LittleInch - Thank you. Ok. Since the density of air is 1.25kg/m3, roughly 800 liters of air is 1 kg. If the eductor ratio is 20:1 then for every cubic meter of water (1,000 liters), the eductor will circulate 20 cubic meters of air (25 liters of air) which is only ~.032 kg of air. So the cubic meter ratio of the eductor as expressed in liters and kilograms would be .032 kg of air : 1,000 liters of water. That's not very much at all. No wonder the temperature difference is negligible.

 

[LittleInch]

Yup. sorry to burst your bubble so to speak.

Always make sure you're using the same units (volume, mass, whatever) when doing ratios or comparing such different things as air and water.

Water is also one of the highest heat capacity fluids you can find (other than ammonia for some reason) so takes a lot of energy to heat it up or cool it down.

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