The pressure on each side of the pipe is given by $P = \rho g h$ where $\rho$ is the density, $g$ is the acceleration due to gravity and $h$ is the depth of the pipe.

From the Euler equations of motion in 1D and steady state, we have:

$$ \frac{1}{2}\frac{d u^2}{d x} = - \frac{1}{\rho} \frac{dP}{dx}$$

If we make some more assumptions, namely that the pipe is full of seawater (which makes sense in the steady state because the water will flow from the sea to the lake) and that the pressure at the fresh water end is constant (so we ignore dilution/mixing, the sea water just instantly drops under the fresh), and we integrate from $x = 0$ at the sea end to $x = L$ at the fresh end:

$$ u^2 |_0^L = - 2\frac{1}{\rho} P|_0^L$$

and taking the velocity to be zero at $x = 0$ gives:

$$ u(L)^2 = -2\frac{1}{\rho_s}(P(L)-P(0))$$
$$ u(L)^2 = -2\frac{1}{\rho_s}(\rho_f g h - \rho_s g h)$$

Taking $\rho_s = 1020 \text{kg}/\text{m}^3$ and $\rho_f = 1000 \text{kg}/\text{m}^3$ with $g = 9.8 \text{m}/\text{s}^2$ yields:

$$ u(L)^2 = 0.0961h $$

or

$$ u(L) \approx 0.31h^{1/2}$$

Obviously this makes some pretty big assumptions. No viscosity, which is probably not that bad of an assumption unless your pipe is really deep, and the pressure on the fresh water end is constant implying the salt water just "disappears" by dropping very quickly out of the pipe under the fresh water.

**Why does the length of the pipe matter**

It doesn't actually. You'll notice $L$ doesn't appear anywhere in the expression. The velocity at the end of a mile long pipe or a 1 inch long pipe is the same and given by that expression.

**What is the significance of $h^{1/2}$**

Again, there really isn't any significance. The units of pressure/density are $\text{m}^2/\text{s}^2$ which is what RHS of $ u(L)^2 = 0.0961h $ is. So the units on the 0.0961 are $\text{m}/\text{s}^2$ and $h$ is $\text{m}$. So when you take the square root of both sides to get into $\text{m}/\text{s}$, you end up with the $h^{1/2}$.

So the significance is really just that there is a non-linear relationship between velocity and the depth of the pipe. If you put your pipe four times deeper, you'll only get twice the velocity.

Let's do some rough estimation. Considering:

- CuriousOne's very good and fitting comment
- with regards to geometrical estimation given in question
- and the last but not least the human physiology ($10^{-12} \ W$ can hear only a full healthy man, with perception much less than a whisper and only on a frequency range improbable due to the diffraction)

we would need ca. $300 \ \mathrm{dB}$ at the source area to produce fine audible sound all around the globe.

Now, since the medium is air only, we can use $I \propto p^2$ for conversion from intensity to sound pressure. Therefore:

$$
20 \log \frac{p}{p_0} = 300 \ \mathrm{dB}
$$

With usual reference value $p_0 = 2\cdot10^{-5} \ \mathrm{Pa}$ we easily get:

$$
p \approx 10 \ \mathrm{GPa}
$$

that's a lot even for a static pressure in solid bodies and here it should be time dependent changing pressure in a gas. It would immediately result in state changes of the air.

*"Disussion":*

- We haven't used any reflections here (from the solid body of the earth and from "open end of the atmosphere" as well). It should work for us in the meaning of reconcentration of spreaded energy. The final value should therefore be lower. On the other hand, we would need to take at least big mountain ridges into account.
- The given estimation is of course invalid, because it is based on linear acoustics. It could provide only the
*very* rough guesstimate at its best. The nonlinear effects will be very pronounced.

*"Conclusion":*
"Audible all around the world" is moreless a figure of speech. In my opinion, the trick would not be in a focused sound source area of insane strength but in a large atmospheric breakdown (i.e. large and spreading source area).

## Best Answer

There's a good article about this at http://climate.nasa.gov/blogs/index.cfm?FuseAction=ShowBlog&NewsID=239. Since it's a NASA blog I assume it's reasonably trustworthy. The key points are:

I must admit I'm not sure about the last of these as it seems to me it would be a very small effect. Finally, another longer term effect I've seen mentioned in some discussions is that land rises as it's freed from the weight of the ice. The north of the UK is still rising, and the south still sinking, after the ice melted at the end of the last ice age.