How does a vacuum ejector using air work?

[sturle]

Hello

I`m working on a vacuum project, and I am trying to understand how the vacuum ejector I am using works.
I understant the basic principle of how it works, but can anybody here try to explain the concept in detail,
taking into consideration that I am using air? Most of the stuff i find when i search for this is about water, and lower velocity, or they just say: and vacuum occures..
Can i use Bernoulli`s equation when i work with air at supersonic speed?

https://www.festo.com/wiki/en/Function_of_a_vacuum_generator

Excuse my english, thanks.

 

[dRic2]

I venture an explanation. This is what I understood so far reading your post and the link you attached, but I didn't know anything about vacuum generators before so I hope someone more competent will come and help us out.

First of all let's consider the air coming in. its velocity is very low so it is not a supersonic flow.
According to the equation of conservation of mass-flow if the section of the pipe gets smaller the velocity of the fluid increases (that's quite intuitive). So the first part of the vacuum generator consist in a pipe whose section gets smaller and smaller so that the velocity of the air increase until it reaches sound's speed. Now let's consider Bernoulli's equation:

$\frac p {\gamma} + \frac {v^2} {2g} + h = const$

If $v$ (velocity) gets bigger, $p$ (pressure) has to get smaller in order to the equation to remain constant. So pressure decreases. Now if you put some holes in the pipe (this is the part where I'm not 100% sure, but I think it is very reasonable) where the pressure is very low the air from outside will be attracted inside the pipe and voilà: you have a vacuum generator.

What's the point of reaching the speed of sound? Well you can't reduce the section of the pipe too much mainly for practical reason so this could be a problem. Here it comes in hand a very useful property of supersonic fluids: they behave the opposite of non-supersonic fluids meaning that when the section of the pipe gets smaller also their velocity gets smaller; on the other hand when the section gets bigger also the velocity increases. But If the velocity increase then the pressure decrease! It sounds good!

To recap:

1) You accelerate the air in the pipe by reducing the section of the pipe (non-supersonic fluid)
2) The pressure drops consequently
3) You reach sound's speed -> change from non-supersonic to supersonic fluid
4) You put holes in the pipe (see point 6)
5) You accelerate the air in the pipe by increasing the section of the pipe (supersonic fluid) and pressure keeps decreasing
6) the air from outside is attracted inside because of the pressure gradient

This how I understood it, hope it will help you.

 

[boneh3ad]

I'd be surprised if this system is incompressible. I'd honestly wager that it is supersonic, as most similar vacuum ejector systems are.

In other words, I'd be surprised if Bernoulli is valid here.

 

[dRic2]

Why should Bernoulli be not valid? Maybe I'm wrong but I though Bernoulli worked also for compressible fluids... you just have to take it into account

BTW when the fluids becomes supersonic I didn't use Bernoulli. I just used it when the air was still sub-sonic

 

[sturle]

I'd be surprised if this system is incompressible. I'd honestly wager that it is supersonic, as most similar vacuum ejector systems are.

In other words, I'd be surprised if Bernoulli is valid here.

Thats what I was questioning, cause i read somewhere that you can`t use bernoullis on compressible fuids when the velocity is higher than 0.3 mach.
In that case, do any of you know how i can calculate the pressure and velocity dealing with this kind of speed?

 

[boneh3ad]

It's fairly easy to do the compressible calculations as well. It's just different, and if you've never seen in before it will look strange. You should look into the concept of isentropic flow and the corresponding relationships between Mach number and every other variable.

 

[russ_watters]

Thats what I was questioning, cause i read somewhere that you can`t use bernoullis on compressible fuids when the velocity is higher than 0.3 mach.

It's the simplest form of Bernoulli's equation works only up to about mach 0.3, then there is a compressible flow equation that works for a while above that. I'm not certain the limit, but I think it is as you approach Mach 1.

In that case, do any of you know how i can calculate the pressure and velocity dealing with this kind of speed?

There was some good discussion in this thread (...and a bit of tooth-pulling...):
https://www.physicsforums.com/threads/blower-fitted-with-de-laval-nozzle.943335/

 

[boneh3ad]

There are "compressible versions" of the Bernoulli equation, though they aren't particularly useful. They are honestly harder to use, in my opinion, than just doing it using isentropic flow relations.

The problem here is also that this flow is likely to see Mach number ranging from near zero to something greater than one if it operates like a typical vacuum ejector system. The flow is choked through the construction and then expands to a supersonic Mach number as it leaves and expands into the outlet port. This expansion not only increases the Mach number, but dramatically lowers the pressure, drawing whatever gas you are "pumping" up into that outlet chamber as well. It's a fairly easy problem to analyze up until the point where it leaves the converging section of the nozzle. Once it starts that supersonic expansion, it's more difficult because you will have a mix of species.

 

[sturle]

Okey, I am not sure if I am able to calculate this stuff because i don't have a lot of information about the properties along the way. I don't have any equipment to messure the temperature at given states, or won't there be a big change in temperature? its not a jet engine were talking about

 

[boneh3ad]

Any flow that is compressible will experience meaningful temperature change. Think about the little aerosol gas cans you use to blow dust off of a keyboard of computer components. If you hold the button down long enough, the can gets cold. This is effectively the same process.