We've been doing the following experiment in our Physics Lab course and are coming up with some strange results which we're at a loss to explain.

The purpose of the experiment is to determine the conductance of various narrow pipes.

The outline of the experiment is as follows:

  • Seal off a chamber from the atmosphere by closing tap 2 (See diagram below).
  • Depressurise the chamber with a rotary vane pump.
  • Seal off the chamber from the rotary vane pump by closing tap 1 and detach from the rotary pump.
  • Open tap 2 and record how the pressure in the chamber varies as a function of time.

Experimental Setup

According to the theory we've been given, the flow through the narrow pipe will be viscous and thus the throughput will be given by:

$$Q = -V\frac{dP}{dt} = A(P+P_a)(P-P_a)$$

where $V$ is the volume of the chamber, $P$ is the pressure of the chamber, $P_a$ is the atmospheric pressure and $A$ is a constant which depends on the dimensions of the narrow pipe.

Solving this equation, one would expect that:

$$ P = P_a\mbox{Tanh}\left(\frac{AP_a}{V}t + C\right) $$

The following plot shows one of our actual results:


You won't be surprised to learn that fitting the above equation to this graph produces a terrible fit. What is really baffling us however is the way the pressure actually noticeably decreases a little after reaching the peak and then slowly starts to increase again. We've repeated the experiment with different equipment and the phenomena isn't going away so it doesn't appear to be a case of equipment failure. We've also noticed that the pressure decrease is more pronounced when the narrow pipe has a higher conductivity (i.e. wider or shorter pipes).

So what is causing this inexplicable pressure drop? Once the chamber has filled with air, why would its pressure drop to below atmospheric and then start to rise towards it again?

One possibility we've thought of is that our experiment involved turbulent flow as opposed to laminar flow (we've done the calculations and this is certain). The theory we've managed to find describes the difference but is ambiguous as to whether the equations for pipe conductivity are applicable to only laminar flow or also turbulent flow. Is this relevant or are we barking up the wrong tree? If it is relevant, how would this explain the pressure drop? You should bear in mind that we haven't taken any classes on fluid dynamics yet so we have very little knowledge to work with.

Thanks in advance.

  • $\begingroup$ Excuse me, but I don't think your formula $Q=-V\frac{dP}{dt}$ is good. Its RHS unit is watts. Shouldn't it be m³/s? $\endgroup$ Commented Nov 13, 2014 at 17:01
  • $\begingroup$ Your expression for a viscous pipe flow -- what speed regime do you think that is valid to use? And what speed do you think the air entering the pipe will be at given your pressure ratios? $\endgroup$
    – tpg2114
    Commented Nov 13, 2014 at 17:43
  • $\begingroup$ Also, "pressure drop" relative to what? What is your final expected pressure? Does your pressure peak overshoot the expected pressure or not? $\endgroup$
    – tpg2114
    Commented Nov 13, 2014 at 17:45
  • $\begingroup$ Quite frankly, wasn't it supposed to fill up much more slowly than that? Seems too fast. Intuitively, I don't think such a fast flow would be laminar (air isn't very viscous right?). $\endgroup$ Commented Nov 13, 2014 at 19:28
  • $\begingroup$ @AndréNeves Since this is a homework question, let's be careful about how much we guide the answer. Not that we've crossed that line yet, but I have to keep reminding myself not to answer the question (and your previous comment as the answer to one is sort of the answer to the other). $\endgroup$
    – tpg2114
    Commented Nov 13, 2014 at 20:35

2 Answers 2


I'm not a fluid mechanics expert, but my mechanical systems knowledge suggests it might be simply a natural oscillatory behavior, which is always present but in this case is more noticeable due to the aggressive initial response (i.e fast influx of air) your chamber experiences.

So what is causing this inexplicable pressure drop? Once the chamber has filled with air, why would its pressure drop to below atmospheric and then start to rise towards it again?

It works like this, in a nutshell: initially, your chamber is in a vacuum. Air loves vacuum, so lots of it enter your chamber very quickly. Now your chamber has more pressure than the outside environment, so the dynamic is reversed: air leaves, until the inside pressure is lower than the outside. That happens a couple times until the system stabilizes. The graph below shows this behavior in a general manner.

typical transient

I must say, however, that your graph isn't of a very typical oscillatory behavior. But it might be that.

More info I might use:

  • What is your atmospheric pressure? Use your altitude to find it, if you don't have it. We need it to find if the chamber pressure has indeed exceeded atmospheric pressure. That is called overshoot (see picture above) and will confirm our hypothesis.
  • Post a new graph zooming in on the period where the chamber is filling. The 200-225 seconds window should be good.
  • What were your expectations about the experiment? How long should it take to fill the chamber? Was filling up in less than ten seconds expected?
  • 1
    $\begingroup$ And I would have put the pressure gauge on the chamber itself, not on the pipe (with laminar/turbulent flow) leading to the chamber. If the diagram is accurate, you are looking at the pressure in the pipe while the chamber is filling, which will depend on where the gauge is along the pipe. $\endgroup$
    – Jon Custer
    Commented Nov 13, 2014 at 18:14
  • $\begingroup$ I concur, but, given that sometimes it is difficult to adapt the gauge on the chamber itself, putting it close to the end of the pipe should be good enough for didatic purposes, right? $\endgroup$ Commented Nov 13, 2014 at 18:19
  • 1
    $\begingroup$ I'd say if it is further from the chamber than a few pipe internal diameters the limited conductance will make sure it is more affected by the inlet air pressure rather than chamber pressure. I.e., most of the pressure drop is in that last bit of the small tube. $\endgroup$
    – Jon Custer
    Commented Nov 13, 2014 at 18:26
  • 1
    $\begingroup$ @ThinkingSkeptically You may also want to plot your pressure on a log scale for the Y axis. Since you are looking at something that spans several decades, that may help you see the variation with time better. Also note how rapid the rise is compared to your sampling -- the "peak" you have is atmospheric, but do you think it's possible the sensor missed the real peak? What the sampling frequency of the sensor? How does sampling frequency relate to the values you are measuring? $\endgroup$
    – tpg2114
    Commented Nov 13, 2014 at 21:55
  • 1
    $\begingroup$ It's been long enough that it would be okay to give more detailed answers I think. I tried to ping you in chat to explain things back on 13 Nov but you had never been in chat so that didn't work. At any rate, my explanation is here. $\endgroup$
    – tpg2114
    Commented Nov 24, 2014 at 21:44

One environmental variable that is being ignored, but which is significant at the 5L volume you are dealing with is Temperature!

As you evacuate that much air, you’re going to have a measurable cooling effect on the air and the whole container.

As the air rushes in, it changes this temperature, but not fast enough that you can simply hand waive it’s existence. The effect of your container equilibrating to room temp could take minutes.

I see this all the time with compressor tanks where it’s the opposite effect wrt Temp. (Heating). What I observe is that one can fully discharge a tank leaving valve 100% open (let’s say half inch valve), and come back 30 to 60 seconds after there is no audible or tangible flow, close off the valve, and in a few minutes there is enough pressure built up in the system to discharge another blast of air. Sometimes 2 or 3 times hence.

The answers above about damped oscillation after exponential rise are correct, and I offer consideration of internal temp to refine this idealistic solution.


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