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I've seen similar problems all over physics.SE and other sites. Most of the time it is about non-compressible fluids, or not about the differential equations. The physical system of interest is comprised as follows:

A rigid container with a gas at temperature $T_1$, pressure $p_1$ and with volume $V_0$. This rigid container has one hole to the surrounding environment. The temperature $T_r$ and pressure $p_r$ of the surrounding environment is constant.

Given that $p_1$ is greater than $p_r$ at $t=0$, what is the change of $p_1$ with $t$?

I assume the following:

  • We have an ideal gas
  • It is an isothermal process (Is this even a good assumption?)

Here is a sketch:

Sketch

My problem is, I can't even construct a differential equation for this system. I thought I could derive one by simply using the following two equations: The ideal gas law, $$ p_1V_0 = n R T_1, $$ and the volume flow rate through the orifice, which I assume is $$ \frac{dV}{dt} = K\Delta p = K ( p_r - p_1 ),$$ with $K$ as a constant which represents the geometry of the orifice.

Edit: I think I made an error here. The last equation is only applicable for incompressible fluids.

I tried to get a differential equation for $n$, the number of molecules in mole which are in the container, by substituting $V$ with $\frac{m}{\rho}$ and $m$ with $Mn$: $$ \frac{d}{dt}V = \frac{d}{dt} \frac{M n}{\rho}$$ But $\rho$ is also depend on $t$ so, there is nothing won here.

Am I on the right track? I do think I have to incorporate the Bernoulli's equation for compressible fluids, but than again, I do not know how.

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    $\begingroup$ I don't know the answer to your question, but, since I find it attractive, I just try to help. So, I don't think that it is a good idea to work an equation for volume. The volume for the gas is always the volume of the container, isn't it? I understand that you try to find the change in the pressure of the gas remaining in the container. Am I wrong? One more thing: no, it doesn't seem that we have a Bernoulli-type problem here. The gas in the container, essentially, is not a flowing gas. $\endgroup$ – Sofia Jan 8 '15 at 16:51
  • $\begingroup$ Yes, since it is rigid, there is no volume change of the container. However - you will have a flow of molecules out of there - which is proportional to the mass flow rate, out which you can calculate a volume flow rate. And yes, the ultimate target is to calculate $p(t)$ for any $t$. $\endgroup$ – John H. K. Jan 8 '15 at 17:52
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    $\begingroup$ I have a few questions/comments. The flow through the hole seems to me a minor leakage. Shall we consider therefore that for not long intervals of time $T$ and $P$ in the container remain stable. Or, alternatively, shall we consider that the leakage is so slow that there is time for $T$ and $P$ to come to a stable value all along the leakage? $\endgroup$ – Sofia Jan 8 '15 at 21:26
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    $\begingroup$ See my answer below, but regarding a more direct way to solve as a 1 dimensional flow problem, you do need Bernoulli's equation since it deals with the energy loss across the orifice (pressure drop) $\endgroup$ – docscience Jan 8 '15 at 21:27
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    $\begingroup$ Regardless of the size of the hole, the relationship between pressure drop and flow through the hole is generally quadratic, coming from the $V^2$ term in Bernoulli's equation. $\endgroup$ – docscience Jan 8 '15 at 21:29
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One very simple way to solve this problem is by using an electrical circuit analogy where your rigid container with a compressed fluid would be represented by a charged capacitor (voltage equivalent to pressure) and the hole in the container as a resistor that's connected to the capacitor with the other end at a voltage equivalent to the pressure on the outside of the container.

Since you're willing to assume ideal gas in an isothermal process as gas expands through the orifice, the relationship between container pressure and the volume of fluid molecules in the container is linear: $P_1 = V/C$. Here $V$ is not $V_o$ ,the container volume, but rather the volume of gas molecules at $p_r$ that were used to charge the pressure in the container to pressure $P_1$. The pneumatic equivalence of capacitance is 'compliance' and that can be determined using the container volume: $C = V_o/p_r$.

With the two elements you have two equations,

The pressure in the container

$$p_1(t)={1\over C} \int Q(t)dt$$

and the pressure drop across the orifice

$$p_1(t) - p_r = Q(t)^2R$$

where $Q$ is the volumetric flow through the orifice, which is equivalent to current in the electrical circuit.

and the volume that leaves the container at pressure $p_r$ is

$$V(t) = \int Q(t)dt$$

Solving the second of these equations for $Q$ and substituting into the first, then solving for $p_1$ gives you the differential equation you need.

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    $\begingroup$ BTW a good reference for solving your problem is "The Analysis and Design of Pneumatic Systems" by Blaine Andersen $\endgroup$ – docscience Jan 8 '15 at 21:31
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    $\begingroup$ I probably should have also noted that the nonlinear first order differential equation you get does not have a closed form solution. But you can solve it numerical or run it in simulation if you have a stiff solver. $\endgroup$ – docscience Jan 8 '15 at 22:27
  • $\begingroup$ Hey! Thank you for the explanation, a question: Shouldn't $C = V_0/p_r$, as you have written a sentence before? $\endgroup$ – John H. K. Jan 9 '15 at 8:01
  • $\begingroup$ @JohnH.K. you are very welcome - and thanks for catching my error! I've fixed it. $\endgroup$ – docscience Jan 9 '15 at 15:32
  • $\begingroup$ @docscience : I saw your answer, it's appealing, but I have doubts. Doesn't the leakage perturb the thermodynamic equilibrium in the container? I think it does. In the region close to the hole, the gas pressure changes locally, and the change propagates deeper and deeper into the container. The temperature looses meaning. As long as the leakage continues, i.e. there is not yet a thermodynamic equilibrium between the gas in the container and the environment, I don't see that we can speak of pressure and temperature in the container. These are concepts valid at equilibrium. (see continuation) $\endgroup$ – Sofia Jan 9 '15 at 21:54

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