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In my example, water is in a tube and is being accelerated by a compressed air. I already know that $P = \frac{F}{A}$ so if you know the pressure and you know the surface area its acting on, you can calculate the force acting on the water. Then I can use $F=m\dot{}a$ to find the acceleration of the water if the mass is known. This is easy, but it only works with a constant pressure. How would you come to the same result if the tube was closed at the end with the compressed air and open on the other? The compressed air would accelerate the water, but also expand at the same time, decreasing the pressure as it goes. How do I take this into account and how would I model the acceleration of the water?

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You use the ideal gas law to get the gas pressure as a function of time, based on the current location of the trailing edge of the water. The gas is acting like a spring accelerating the water. As the compressed spring extends, its force decreases. P=nRT/V, where $V=V_0+Ax$, with x representing the displacement of the trailing edge of the water from its initial location. This is very similar to an air gun.

SUPPLEMENT: I'm going to assume that the gas expansion is adiabatic (insulated pipe) and that the expansion is nearly reversible. So the pressure-volume behavior of the gas is described by: $$PV^{\gamma}=P_0V_0^{\gamma}\tag{1}$$where the 0 subscripts represent the initial values and $\gamma$ is the ratio of the heat capacity at constant pressure to the heat capacity at constant volume.

The gas volume is related to the initial volume by $$V=V_0+Ax\tag{2}$$, where A is the cross sectional area of the tube and x is the displacement of the liquid.

I assume that there is a massless piston in front and behind the fluid so that the gas cannot blow past by splitting through the liquid.

The force balance on the liquid is given by Newton's 2nd law:$$PA-P_DA=M\frac{d^2x}{dt^2}\tag{3}$$where M is the mass of the liquid and $P_D$ is the downstream pressure in the open end of the tube.

If we combine Eqns. 1-3, we obtain: $$\frac{M}{A^2}\frac{d^2V}{dt^2}=P_0\left(\frac{V_0}{V}\right)^{\gamma}-P_D\tag{4} $$ Now, all you need to do is solve this equation for V as a function of time.

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  • $\begingroup$ But the displacement is based on velocity time and area. Velocity is based on the acceleration and time. And the acceleration is what I'm trying to solve for, which leads me back to where I started $\endgroup$ – Ryan Feb 29 '16 at 20:43
  • $\begingroup$ So you are really saying that you don't know how to do the math, not that you don't understand the physics, correct? $\endgroup$ – Chet Miller Feb 29 '16 at 21:02
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    $\begingroup$ Right @ChesterMiller $\endgroup$ – Ryan Feb 29 '16 at 21:37
  • $\begingroup$ OK. I'll make an addition to my answer to help. Give me a few minutes. $\endgroup$ – Chet Miller Feb 29 '16 at 21:40
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You must be looking for Bernoulli's equations. You need $ \frac{ d \int (\rho dV)}{dt} + \int( \rho V n dA)$ where $\rho$ is density, V is vectorial speed of flux, n is perpendicular surface, A is area, V is volume and t is time. Good luck

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  • $\begingroup$ You should use different symbols for Volume and "vectorial speed of flux". Also, try using mathjax (latex formatted math type) for your equations. It will make it easier to read. $\endgroup$ – tmwilson26 Mar 1 '16 at 13:56
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Have you considered using energy of the system? The column of water will have a kinetic energy, and the air that you have will also have a certain amount of energy. This can be related to the potential energy stored initially in the gas.

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