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How can I prove that the rate of which water leaves a vertical cylindrical container (through a hole at the bottom) is exponential of the form :

$$Ae^{kx}$$

I know that Torricelli's law is:

$$\sqrt{2gh}$$

But this only proves a square root relationship. I have data points every 10 seconds and graphed it suggests a decay function. I know the distance between the pipe is 1.5M and the internal diameter is 5cm. The hole diameter is 0.25cm, if this helps. I need to prove that the water leaving the pipe is exponentially decaying.

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  • $\begingroup$ Please show the graph. $\endgroup$ – Deep Aug 20 '16 at 11:11
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We don't give solutions to homework problems, but we can explore the physics a bit.

You need to set up a differential equation for the change in height with time. The way to do this is to note that if the speed of the water flowing out through the hole is $v$, and the area of the hole is $a$, then the volume flowing out per second is just $av$. The outflow per second is just the change in the volume in the pipe, so we get the differential equation:

$$ \frac{dV}{dt} = -av $$

where $V$ is the volume in the pipe. The next step is to note that if the area of the pipe is $A$ then the volume in the pipe is $V = Ah$ and because $A$ is constant:

$$ \frac{dV}{dt} = A\frac{dh}{dt} $$

So our equation for $dV/dt$ turns into:

$$ \frac{dh}{dt} = -\frac{a}{A}v $$

You mention Torricelli's law. If you use this to substitute for $v$ it's going to give you:

$$ \frac{dh}{dt} = -\frac{a\sqrt{2g}}{A}\,h^{1/2} \tag{1} $$

We solve differential equations like this by rearranging to give:

$$ \frac{dh}{h^{1/2}} = -\frac{a\sqrt{2g}}{A}\,dt $$

and then just integrate:

$$ \int\frac{dh}{h^{1/2}} = -\frac{a\sqrt{2g}}{A}\int dt $$

But this isn't going to give you an exponential decay of $h$ with time. If you measure an exponential decay (you don't say whether this is what you measure in the experiment) then Torricelli's law cannot apply.

Torricelli's law is based on inertial forces i.e. it balances the potential energy lost as the liquid falls with the kinetic energy gained at the outflow. However in fluid dynamics there are always two effects to consider - inertial forces and viscous forces. Torricelli's law considers only inertial forces and ignores viscosity of the liquid, so it applies only when inertial forces dominate. When viscous forces dominate the flow velocity will be determined by the Hagen-Poiseuille equation and that gives a different dependence of the velocity on the pressure:

$$ v \propto P \propto h $$

This is going to change our differential equation to:

$$ \frac{dh}{dt} \propto -C\,h \tag{2} $$

for some constant $C$ that will depend on the geometry of the hole and the viscosity of the fluid. Without going into the details, this will give you an exponential decay of $h$ with time.

So how $h$ varies with time in your experiment is going to depend on the details i.e. whether inertial or viscous forces dominate. I think you need to graph $\ln(h)$ against time and $\sqrt{h}$ against time and see which gives you a straight line.

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  • $\begingroup$ I understand your concern in it being a homework question. Perhaps my phrasing suggested so but I can assure you it is not a homework question. I am looking for an explanation so I can justify data modelling. Thankyou for your help however, It proved quite useful. $\endgroup$ – Zano Aug 20 '16 at 11:25
  • $\begingroup$ @AndrewAnon: if/when you try graphing the data I'd be interested to know which graph gives a straight line. $\endgroup$ – John Rennie Aug 20 '16 at 12:15
  • $\begingroup$ @AndrewAnon: If I can restate what John is saying, if the hole at the bottom consists of a long narrow horizontal tube, so that viscosity dominates, then pressure and velocity will be linearly related, and the decay will be exponential. If the hole is simply a round hole, nothing more, velocity will go as the square root of pressure, so the decay will not be exponential. Then doubling the flow rate requires four times the height, not twice. $\endgroup$ – Mike Dunlavey Aug 20 '16 at 22:12
  • $\begingroup$ @AndrewAnon - perhaps what is a squared decay (the result from the above analysis) looks like an exponential decay to you? It would help if you could show the graph. $\endgroup$ – nluigi Aug 21 '16 at 7:44

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