If I took a fresh water lake* whose surface is exactly as sea level, and connect it to the sea with a pipe filled with sea water, with both ends of the pipe at exactly the same depth from the surface, $h$, what is the correct way to calculate the velocity with which the salt water leaves the pipe from the depth $h$, and the densities of fresh and sea water, $\rho_\text{fw}$ and $\rho_\text{sw}$?
I can't seem to apply Torricelli's law here, since that is not influenced by the fluid's density, yet that would be the only driving force here. Can I treat this as a hydraulic head of a fluid with a density of $\rho = \rho_\text{sw} – \rho_\text{fw}$? Should I treat this the same as a pipe of water on land that's pressurized with $(\rho_\text{sw} / \rho_\text{fw}) g \rho_\text{fw} h$ above ambient pressure ($g$ being gravitational acceleration)?
For the purpose of this question, let's assume both water reservoirs are very large in comparison to the volume exchanged through the pipe, and the salt dissipates very rapidly in the fresh water (or the water sinks to a far away bottom immediately), so we neither alter the surface level (which would happen in communicating vessels containing liquids with different density) of either body nor do the salinities level out.
*(no actual lakes will be ruined, promise)
Best Answer
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.