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I have a U- like pipe. Its inlet has atmospheric pressure $p_o=10^{5} \, Pa$. Vacuum is applied to the other end with a pressure gradient $\nabla p_v=-30 \cdot 10^{3} \, kPa/s$. The total time of the simulation is $t=0.25$ s.

I assume that the highest velocity would be reached at the end of the simulation.

After 0.25 s the vacuum pressure is $p_v=-7.5 \cdot 10^{3} \, Pa$. I am interested in knowing the maximum speed of air ($\rho=0.146$ $kg/m^{3}$) at the outlet (assuming this is where it is highest).

One idea that comes to mind is to use the Bernoulli equation: \begin{equation} p_0 + \frac{1}{2} \rho v_{0}^{2} +\rho g h_0 = p_v + \frac{1}{2} \rho v_{v}^{2} +\rho g h_v \end{equation} The $h$ value in both cases is the same, hence the equation simplifies. Assuming the speed of air at the inlet $v_0=0$ m/s, further simplifying the equation. \begin{equation} p_0=p_v+\frac{1}{2} \rho v_{v}^{2} \end{equation}

\begin{equation} v_v=\sqrt{\frac{2}{\rho}(p_0-p_v)} \approx 1213.5 \, m/s \end{equation}

However that seems unrealistically high. My rough computational simulation gives a value of $v_v=64 \, m/s$. How can I calculate $v_v$? Should I not be taking into account properties of the geometry? I am not really sure I can consider $v_0=0$, because surely the speed will be almost as high at the inlet as at the outlet due to the long term effects of the pressure gradient.

The purpose of this is to give me a rough value of the maximum speed of air in such a scenario, so I could use $v_v$ value in a computational simulation.

A U-shaped pipe.

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Your title is very misleading. If you really knew the velocity at the inlet $v_{in}$, you could just use mass conservation and the areas of the inlet $A_{in}$ and of the outlet $A_{out}$. Then you would have $v_{out} \, A_{out} = v_{in}\, A_{in}$, or $v_{out} = v_{in}\, A_{in} / A_{out}$ (plus corrections for density). But you claim that $v_{in}$ (which you call $v_0$) is zero, so the velocity at the outlet would necessarily be zero -- unless you were creating mass inside your pipe. Instead, you know the velocity of the ambient air near the inlet. – Mike Jun 20 '13 at 16:31
@Mike Except for compressible gasses :) – Bernhard Jun 20 '13 at 17:45
@Bernhard Well, I'll still file that one under "plus corrections for density" (by which I meant density changes from inlet to outlet)... – Mike Jun 20 '13 at 18:11
True, overlooked that in your comment. Putting one end at vacuum makes this more than just a correction, I think. – Bernhard Jun 20 '13 at 18:15
I don't think continuity is of much use here, since areas are identical, and an unsteady boundary condition is present. – A.L. Verminburger Jun 21 '13 at 6:58

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