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I would like to estimate the time needed to open a linear single-acting pneumatic actuator. The actuator's piston is loaded with a spring pushing downwards and the operating fluid (air or nitrogen) pushes upwards; other forces like weight and friction are present.

The operating fluid is provided by a large tank (ideally infinite) with stagnation pressure $P_\infty$ and stagnation temperature $T_\infty$.

If the tank is connected directly to the actuator, balance of forces, with fluid force depending on actuator inlet mass flow rate, leads to an ODE which I can solve numerically.

Since tank and actuator are actually connected by a long pipe, $\frac{length}{diameter}\approx 1000$, I need to estimate the time to reach the equilibrium state in the pipe ("pressurise the pipe"). Initial conditions for pipe and actuator are $P|_{t=0}<P_\infty$ and $T|_{t=0}=T_\infty$.

Does anyone know a way to estimate the time required to reach the equilibrium state in the pipe, maybe based on some dimensional argument or some typical time related to the pipe?

If not, I must determine the evolution of pressure, temperature and mass flow rate inside the pipe.

I have been reading literature for compressible flow in pipes, usually coming from natural gas distribution applications, but the problem of transient compressible flow in a pipe, be it isothermal or non-isothermal, leads to a system of PDEs.

The model verification examples I found led me to suppose that the PDE was

requiring some sort of boundary condition also at the exit of the pipe. Since I do not know the pressure / temperature / mass flowrate entering the actuator in advance, I am missing something.

but since the PDE is hyperbolic, it actually requires boundary conditions at the inlet only, besides initial conditions.

Any help or suggestion would be appreciated.

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  • $\begingroup$ What is in the pipe just before the valve opens? Even air will need some pushing out especially if the pipe is long - ie there is a pressure drop relationship. $\endgroup$ – Solar Mike Mar 16 at 20:35
  • $\begingroup$ Initially the pipe is filled with operating gas at the initial pressure, lower than the tank pressure. How can I use the initial pressure drop? $\endgroup$ – Giovanni Mariani Mar 16 at 20:38
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Flow rate of the gas into the cylinder will change according to the pressure drop. So first you have to model flow rate through the pipe to the cylinder, which you can probably model as a plenum. I would assume this is a differential equation version of head loss flow. So as the mass of air enters the cylinder you have the pressure increase in the cylinder, which reduces the flow through the pipe. If you want precision here then you should also model opening of the valve.

Now as pressure builds to greater than the load side of the cylinder you need to model cylinder movement against spring, load, friction, and inertia. So another differential equation, except now it effects the cylinder pressure as well since expansion of the cylinder changes the pressure increase. Depending on pressures and pipe sizes, the inertia of the load might be important, or the pipe might be the limiting factor.

Sounds like a matrix solution to me. Or you could build a couple and time them.

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  • $\begingroup$ Why pressure increase in the cylinder would reduce the flow through the pipe? I assumed that the upstream tank, considered as a plenum, would prevent that. $\endgroup$ – Giovanni Mariani Mar 18 at 15:38
  • $\begingroup$ @GiovanniMariani, being a plenum just means we don't have to model flow in the cylinder and pressure is the smae everywhere in the cylinder. But as pressure increases in the cylinder, flow will reduce going in to it just due to increased resistance to flow. $\endgroup$ – Tiger Guy Mar 19 at 0:42
  • $\begingroup$ OK for stagnation pressure inside the cylinder; I meant that the upstream plenum, providing constant upstream pressure, would compensate a pressure drop at the connection between pipe and cylinder, keeping a constant pressure at the cylinder inlet. $\endgroup$ – Giovanni Mariani Mar 20 at 7:26
  • $\begingroup$ @GiovanniMariani if you mean the pressure on the non-pressurized side of the cylinder, that is usually just atmospheric or vented to the supply tank. $\endgroup$ – Tiger Guy Mar 21 at 9:42

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