1
$\begingroup$

I'm trying to design a venturi tube that will have a pressure drop of 91525 Pascals where the inlet pressure is at 1 atm and the throat pressure is at 9800 Pascals. I calculated the minimum throat velocity need to have this drop assuming my inlet velocity is super small by doing the following.

$p_1-p_2=\frac{ρ}{2}(v_2^2-v_1^2)$
$91525=\frac{1.204}{2}(v_2^2-0), v_2>>>v_1$\

What I get for $v_2$ is around 389 m/s.

This is over the speed of sound and I have been told that venturi tubes cant have supersonic flow at the throat. I have also been told that the velocity should stay under mach 0.3 as above mach 0.3 compressibility starts to become a problem.

I guess my question is, Is it possible to have a venturi tube that has a pressure drop of 91525 Pascals, why can't supersonic flow happen in the venturi tube, and if a venturi tube won't work what can I use to drop the pressure.

PS: The fluid in my tube is air

$\endgroup$
4
  • $\begingroup$ Hi, when askiing a question, please provide references for "I have been told" or "I read somewhere" so we can evaluate the reliability of your sources. $\endgroup$ Aug 8, 2022 at 14:57
  • $\begingroup$ engapplets.vt.edu/fluids/CDnozzle/cdinfo.html $\endgroup$ Aug 8, 2022 at 15:41
  • $\begingroup$ I was told this on Reddit so credibility isn't the best. $\endgroup$ Aug 9, 2022 at 2:28
  • $\begingroup$ Compressibility is not a problem, the only problem is that theory for incompressible flow starts to be more than a few percent off when the velocity exceeds Mach 0.3. $\endgroup$
    – Orbit
    Aug 9, 2022 at 10:40

1 Answer 1

1
$\begingroup$

Supersonic flow is possible in a Venturi tube, just not at the throat. In order for the flow at the throat to be supersonic, there would have to be a location somewhere between the upstream (stagnation) end and the throat where the flow was exactly at the (local) speed of sound. But that's not possible, because if you combine the continuity equation, the steady flow energy equation, and the Poisson adiabat, and differentiate with respect to downstream distance, you can rearrange to get an expression for the derivative of cross-sectional area with respect to downstream distance in terms of (among other things) the ratio of the local flow velocity to the local speed of sound. It turns out that, if the local flow velocity is exactly equal to the local speed of sound, the derivative of cross-sectional area with respect to downstream distance has to be zero, i.e. the only place where the local flow velocity can be exactly equal to the local speed of sound is at the throat. Hence, there can be no upstream location where the local flow velocity is exactly equal to the speed of sound, and no supersonic flow at the throat.

There's a good exposition of this argument in Fluid Mechanics by Douglas et al..

$\endgroup$
4
  • $\begingroup$ Can you expand a bit with an example showing where in the system supersonic speeds would be reached? $\endgroup$ Aug 8, 2022 at 14:59
  • $\begingroup$ @CarlWitthoft Since the throat is the only place where the flow can be exactly at the speed of sound, it follows that if the flow is going to be supersonic anywhere, it will be immediately downstream of the throat. The flow then has to continue being supersonic as one travels downstream, until either one meets another extremum of the cross-sectional area, or something happens to make one of the physical assumptions break down. $\endgroup$ Aug 8, 2022 at 16:01
  • 1
    $\begingroup$ What actually happens is that, some distance downstream of the throat, a stationary shock forms, at which the Poisson adiabat breaks down, because a shock has temperature variation over a small enough distance for there to be significant conduction of heat. Downstream of the shock, the flow is back in the subsonic regime. $\endgroup$ Aug 8, 2022 at 16:06
  • $\begingroup$ do you know precisely what section in Fluid Mechanics I could find this in? $\endgroup$ Aug 9, 2022 at 2:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.