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Air Entrainment into a Venturi

Air Entrainment into a Venturi

Air Entrainment into a Venturi

(OP)
(this is a cross-post from Gas Compression Engineering)

This is basically part 2 of a previous project: a vapor stream is directed into the mouth of a venturi, which entrains air into the stream, resulting in a mixture. My problem comes from the fact that I don't understand the pressure behavior inside the venturi:
on one hand, the pressure needs to be below ambient atmospheric to draw in air (since the air will move down the pressure gradient from high P to low P).
on the other hand, the outlet pressure of the venturi is significantly above ambient (~ 5 psi --> storage tank). so the pressure gradient inside the venturi appears to be against the flow direction (higher P at outlet than at inlet), which would indicate to me that really flow should be in reverse.
But I know that it all works out, since the dang thing is sitting outside, doing its job just fine. Obviously, i'm missing something.
I tried to do an energy balance, taking into account the pressures and the velocities of the vapor and the air (initially, Vvapor ~ 260 m/s, Vair ~ 0, Pvapor ~ Pair ~ 101 kPa. end, Vmix ~ 50 m/s, Pmix ~ 135 kPa) but it all fell apart.  
Is a momentum/energy balance the way to go, or am I just totally on the wrong track here? (btw, I'm looking more for the theory, rather than just an answer, hence no real precise numbers - but if it helps, I'll post later.) thanks for your help in advance

cheers,
rad
"Remember, if you leave it to the last minute, it'll only take a minute"

RE: Air Entrainment into a Venturi

I think this is more of a Bernoulli equation thing; the higher velocity the lower the pressure.  This effect creates the pressure differential in venturi type flow meters, and in the venturis in carburetors.  
I don’t remember enough to call up the formula, somebody else may be able to help with that. You also might look up ejectors or eductors, as I think they operate on a similar principal.

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