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The ancients undoubtedly discovered the ideal shape of a water clock by trial and error. In examining some ancient water clocks I notice the shape is different depending on the size. For example, a 9-inch water clock will have a different shape than a 5-inch water clock. Obviously the size of the orifice is very important also, and for that reason water clocks were made from metal, even gold, or from hardened ceramic, so that the orifice could be sized very exactly.

Should a modern person wish to make a water clock without going through the agony of many hundreds of hours of experimentation, what theory could be used to determine the ideal shape using the principles of physics alone?

The type of clepsydra I am envision is one that would provide the height of the water as a linear function of time.

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There are (at least) two types of water clock: constant flow per unit time, and constant drop in height per unit time.

If you want constant flow, you need a mechanism to keep the pressure constant - this was the subject of this question

If you want constant change of height with time, you need to change the area as a function of height above the orifice. It is easy to show (Bernoulli) that the velocity of the flow goes as the square root of the pressure (height). The area at a height $h$ needs to be such that the level drops at a constant rate. If the diameter at $h$ is $r(h)$, then $r(h)^2 \propto \sqrt{h}$. It follows that the shape of the side of the wall of the clock is of the form $$r \propto \sqrt[4]{h}$$

Note that the above assumes non-viscous flow: that is, the additional pressure difference due to the viscous drag across the aperture is neglected in the Bernoulli equation. This is a good assumption when the aperture is quite short and the Reynolds number of the flow is high - see this article which shows that the discharge coefficient changes by about 1% for a Reynolds number from 10$^4$ to 10$^7$. However, at lower flow rates the effect can be significant, and this tends to make flow meters less accurate at low flow rates, as expressed by the turndown ratio.

In principle you can take this into account in the construction of your water clock: to do so, you need an accurate determination of the pressure drop across the nozzle due to the viscous forces.

Assuming that you have a 24 hour water clock with a total height of 1.20 m, and you let the water run down from 120 cm to 24 cm (1 cm per 15 minutes) with a diameter of 25 cm at a height of 100 cm, then the flow rate at that height would be calculated as follows:

$$\frac{dh}{dt}=\frac{0.01~\rm{m}}{15\cdot 60~\rm{s}}=11.1~\rm{µm / s}\\ \frac{dV}{dt} = A\frac{dh}{dt} = \frac{\pi}{4}\cdot 0.25^2 \cdot 11.1\cdot 10^{-6} = 0.545~\rm{ml/s}$$

According to the Bernoulli equation, the velocity of the liquid that dropped 100 cm is $\sqrt{2*g*h}=4.42~\rm{m/s}$. The "pure Bernoulli" aperture needed with a pressure difference of 9.81 kPa would be just 0.198 mm radius. If we assume that the wall of the vessel is 2 mm thick, we have a "nozzle" that's about 0.4 mm diameter and 2 mm long. If liquid flows through such a nozzle at a volume flow rate of 0.545 ml/s, what would be the (additional) pressure drop?

For pure Poisseuille flow, flow rate and pressure are (linearly) related by

$$Q = \frac{\pi r^4}{8\mu L}P$$

So for the given dimensions and flow rate, and using $\mu= 0.001 kg/m/s$, we find P = 1.8 kPa - that is significant.

If we want to take account of this additional linear term, then it follows that the equation of our shape needs to be modified. For a given height $h$, the flow rate is found by solving for $v$, noting that $Q = A v$ and that $\Delta P = \frac{8\mu L}{\pi r^4} Q = \alpha Q$ . The pressure available to accelerate the water is then $\rho g h - \Delta p$ so that

$$\rho g h - \alpha A v = \frac12 \rho v^2\\ v^2 + \frac{\alpha A}{\rho} v - 2gh = 0$$

This is a quadratic equation in $v$, and the roots are

$$v = -\frac{\alpha A}{2\rho} ± \sqrt{\left(\frac{\alpha A}{2\rho}\right)^2+2gh}$$

We need the positive root (to get positive velocity) and can simplify the expression to

$$v = \frac{\alpha A}{2\rho}\left(\sqrt{1+\frac{8gh\rho^2}{(\alpha A)^2}}-1\right)$$

As a sanity check, the second term under the square root will be large when viscous forces can be ignored; in that case

$$v = \frac{\alpha A}{2\rho}\frac{\sqrt{8gh\rho^2}}{\alpha A}\\ =\sqrt{2gh}$$ as before.

Now we can simplify the expression so we can determine the shape of the vessel. Put

$$v = a\left(\sqrt{1+bh}-1\right)$$

Once again, we need to make the area as a function of height such that $\frac{dh}{dt}=\rm{const}$. We can write

$$\frac{dh}{dt} = \frac{Q}{\pi R^2}$$

Where $Q$ is the volume flow rate, and $R$ is the radius of the vessel. Since $Q \propto v$,

$$\pi R^2 \propto v\\ R \propto \sqrt{\frac{\alpha A}{2\rho}\left(\sqrt{1+\frac{8gh\rho^2}{(\alpha A)^2}}-1\right)}$$

When the viscosity is very small, this reduces to the equation we had before; when it is very large, it tells us that the radius is proportional to $\sqrt{h}$ instead of $\sqrt[4]{h}$. In between - it's something in between. Obviously, if viscous terms matter, this clock will lose accuracy as the viscosity changes - and that's a pretty big problem. From 10 to 30 C, viscosity changes a lot:

T(C)  mu (mPa s)
 10    1.308
 20    1.002
 30    0.7978

On cold days, time will slow by 30%... and on warm days it will speed up by 20%. There are actually techniques for mitigating this - it involves a more complex clock design. See this interesting analysis

UPDATE

I found an interesting analysis that is quite critical of some literature regarding water clocks, and that reminds us that for a small orifice, surface tension will modify the above considerably - especially when the water level gets close to the bottom. One might consider having a long (and sufficiently wide not to restrict flow) vertical pipe at the bottom of the clock (before the nozzle) to ensure that pressure at the nozzle is always "high" - not only will the shape of the clock be more uniform, but the viscous forces will be less important and the clock will be less sensitive to temperature.

Interesting but irrelevant tidbit: water clocks were apparently used in the brothels of Athens to time the customers' visits; if these clocks were operated in the viscous regime, I suppose customers could stay longer when it was cold. How considerate.

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  • $\begingroup$ It may not be that simple because the friction at the orifice will vary with pressure. Also, concerning previous question, it was closed, so it is of no use. I have updated my question to specify constant drop in height per unit time. $\endgroup$ – Ambrose Swasey Jan 11 '16 at 22:11
  • $\begingroup$ "Closed" does not mean "of no use" - I had posted what I thought was a useful answer... Why do you expect the friction to vary with pressure? Water is fairly incompressible at the accuracy of a water clock... $\endgroup$ – Floris Jan 11 '16 at 22:13
  • $\begingroup$ The formulas I have seen for the friction at a vena contracta always involve pressure as a variable. $\endgroup$ – Ambrose Swasey Jan 11 '16 at 22:16
  • $\begingroup$ Ah yes - if the pressure difference across the orifice is a function of flow rate then you do get an additional dependence. This happens when viscous flow effects are significant compared to Bernoulli effects. That becomes more significant when the opening is not a simple hole in a thin wall, but a "pipe" with a length that is a significant fraction of the diameter. If I have some time I will update. $\endgroup$ – Floris Jan 11 '16 at 22:44
  • $\begingroup$ I have added the analysis for the case when viscosity is significant. $\endgroup$ – Floris Jan 12 '16 at 5:20

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