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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Mike
    Jun 20, 2013 at 16:31
  • $\begingroup$ @Mike Except for compressible gasses :) $\endgroup$
    – Bernhard
    Jun 20, 2013 at 17:45
  • $\begingroup$ @Bernhard Well, I'll still file that one under "plus corrections for density" (by which I meant density changes from inlet to outlet)... $\endgroup$
    – Mike
    Jun 20, 2013 at 18:11
  • $\begingroup$ True, overlooked that in your comment. Putting one end at vacuum makes this more than just a correction, I think. $\endgroup$
    – Bernhard
    Jun 20, 2013 at 18:15
  • $\begingroup$ I don't think continuity is of much use here, since areas are identical, and an unsteady boundary condition is present. $\endgroup$ Jun 21, 2013 at 6:58

2 Answers 2


You have a U tube open to atmosphere at one end and to a vacuum pump at the other. You turn the vacuum pump. The vacuum pump is specified in terms of a negative pressure gradient per second.You then specify the total simulation time.Then you give the vacuum pressure reading at the end of simulation. Then you gove the density of air and ask what is the air speed at the vacuum pump outlet of the tube.

If you want to solve as a unsteady state problem then you can't .. there are not enough information.. you need the size, material and detailed shape of the pipe and you have to solve the Nvier-Stokes equation for the answer.

If you want to solve the steady state problem, then you need the pressure head of the vacuum pump(from specs) which is not given. You've given pressure per second which is not usual.. you need just pressure. Then you need the flow rate of the vacuum pump which would be given in the specs too, in addition to the vacuum pressure head. If you know the mass flow rate=ρVA, then use the pipe diameter/area A, and density ρ to find the velocity V. That's all, and you don't need to know the pressure- accept in determining the flow rate at the vacuum pump- if the characteristic curve of the pump is not flat- which is common.


In this problem, it will be easiest to assume that air is flowing out at a given density at the inlet and outlet.Thus, we set $\rho _{in}=1.225 kg/m^3$ and $\rho _{out}=0.146 kg/m^3$. We use the Navier-Stokes equations for a compressible flow. The corresponding code written for Mathematica v 11-12 is discussed in two topics on this site



Figure 1 shows the velocity magnitude at the time of the maximum flow velocity and at the moment when the rarefaction wave reaches the bottom of the channel. The flow velocity is normalized to the speed of sound at the input (about 340 m/s). The maximum speed is reached at the moment $t=0.8d/c$ and is about $1.3c=442 m/s$. Here $d$ is the diameter of the channel, $c$ is the speed of sound at the input. Figure 1


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