I have a question that relates to an arrangement of natural gas pipelines but for simplicity I will be describing a less complex problem.

Imagine that I have two large pressure vessels connected by a closed 6” diameter valve. The smaller pressure vessel is 119.23 cu-ft in volume and contains natural gas at 50 degrees Fahrenheit and a pressure of 450.1 psig. The other and much larger pressure vessel also contains natural gas also at 50 degrees Fahrenheit but has a pressure of 587.5 psig. The larger pressure vessel has an unknown volume. At time zero, the valve separating the pressures was cracked open slightly and the pressures begin to equalize. 12 minutes later, the pressure in the smaller vessel is 524.7 psig and the pressure in the larger vessel is 566.9 psig. Based on the data it appears that when the pressure difference across the valve was 137.4 (587.5-450.1) the rise in pressure in the smaller vessel was approximately 5.00 psig/min. After the 12 minutes, with only a pressure difference of 42.2 psig (566.9-524.7) it appeared that the rise in pressure in the smaller vessel was approximately 3.75 psig/min. We know that at if valve were left cracked open the pressure vessels would eventually equalize and with 0 pressure difference across the valve, there would be no flow of gas.

It certainly appears and would seem logical that the greater the difference in pressure across the valve would cause the pressure in the smaller vessel to rise more quickly. As the pressure difference across the valve decreased, so would the rate at which the pressure in the smaller vessel would rise. Similarly, I would assume that there would also be an asymptote that would be approached as the pressure difference were to grow very large. As greater and greater pressure differences across the valve were experienced, the amount of pressure increase per minute in the smaller vessel would be negligibly increasing.

My question is: is there enough information to generate a formula to estimate the amount of gas transferred based on pressure difference? Also, can the amount of gas transferred (cu-ft) in this 12 minute window be calculated?

  • $\begingroup$ The answer to your questions is "yes there is (almost)" and "yes it can" . The only thing missing is: is the rate of flow slow enough that the temperature in the vessels remained constant? Or do we assume adiabatic flow - no heat in and out of the vessels - in which case the higher pressure vessels cools down and the smaller one heats up. $\endgroup$
    – Floris
    Nov 7, 2014 at 22:30
  • $\begingroup$ You probably need to further simplify your question into a conceptual one about gas / fluid flow in and out of a container to get much help. $\endgroup$ Nov 7, 2014 at 22:54
  • $\begingroup$ One problem you run into when you ask about cu-ft of gas transferred - at what pressure. The volume of the smaller container at the end of 12 minutes is the same as at the start (I assume - you call it a "pressure vessel") - and if you measure the "volume" that flowed through the pipe, the answer is different depending on whether you measure upstream or downstream of the valve (different pressure, so different rate). So can you clarify what you really want to quantify? $\endgroup$
    – Floris
    Nov 7, 2014 at 23:25
  • $\begingroup$ The title of this question is pretty broad/vague given what you're actually asking. Please take a look at this post offering guidelines for how to write good titles. $\endgroup$
    – DanielSank
    Jan 11, 2015 at 17:51

1 Answer 1



Two vessels $1$ and $2$ with volumes $V_1$ < $V_2$ and pressure $P_1$ < $P_2$ contain methane at temperature $T$. A small valve is opened between them so gas starts to flow from $2$ to $1$. After time $t$ the pressures are $P_1'$ and $P_2'$. The rate of change of pressure at $t=0$ is estimated at $R_1$, and at $t=t_1$ it is estimated at $R_2$.

Writing the pressure in each vessel as a function of time, and assuming that the mass flow rate $M$ through the valve is linearly dependent on the pressure difference, we get

$$M = \alpha (P_2 - P_1)\tag1$$

We also know from the ideal gas law that the pressure scales with the amount of material:

$$PV = nRT\tag{2}$$ $$M = \frac{dn}{dt}\tag{3}$$

Now with $M$ expressed in moles / second (see $(3)$), we can combine $(1)$, $(2)$ and $(3)$ and write an expression for both $P_1$ and $P_2$:

$$\frac{dP_1}{dt}V_1 = \alpha(P_1-P_2)RT$$ $$\frac{dP_2}{dt}V_2 = \alpha(P_2-P_1)RT$$ $$\frac{dP_1}{dt}V_1 = -\frac{dP_2}{dt}V_2$$

It follows that the rate of change in pressure for the two vessels is inversely proportional to its volume. Integrating both sides with respect to time,

$$\Delta P_1 V_1 = -\Delta P_2 V_2$$

So we can now compute $V_2$ from the data given: $$V_2 = -\frac{\Delta P_1}{\Delta P_2}V_1 = \frac{524.7-450.1}{566.9-524.7}\cdot 119.23 \text{cu ft} = 210.8 \text{cu ft}$$

if we can assume that the temperature is constant (low flow rate, isothermal process).

With the volume and the pressure known, we can calculate the mass of material that was added - it is simply given by rearranging $(2)$:

$$\Delta n = \frac{\Delta P V}{R T} = 23.5 \cdot \frac{74.6\cdot 119.23}{283} = 739 \text{ mol}$$

You can use the above equations to convert to cubic feet at whatever pressure you would like. Note that $R=8.31 \text{J/mol K}$ - I converted it to $1/23.5 \text{cu ft} \cdot \text{psi} / \text{mol K}$. You can't work in Fahrenheit with the absolute gas law, so I converted 50F to 283 K.

Now I did not, in the above, use your estimated change in pressure (which I called $R_1$ and $R_2$ in the simplified version). It is hard to see how the rate could change so little when the pressure difference changed so much. And while it might change the equation for the time evolution of the mass flow a little bit (apparently it changes less than I would predict from the pressure drop) it should not change the expression for the amount of gas in the smaller container, or the calculation of the volume of the larger container.

Helpful link: http://www.sensortechnics.com/cms/upload/appnotes/AN_Massflow_E_11153.pdf


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