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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)

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Hi Hanno, and welcome to physics.SE! We support Latex markup delimited with $, which is generally preferred to backticks for formulas and variables. A more in-depth list of supported Latex can be found here –  Chris White Dec 19 '12 at 23:02
    
@chris - yeah, I was already peeking at other people's source because I was jealous of their pretty formulas. Thanks for the pointer! –  Hanno Fietz Dec 20 '12 at 7:19

2 Answers 2

up vote 2 down vote accepted

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.

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I'm sorry, there's two things I don't seem to understand, probably because I'm kind of a physics noob: a) why is the length of the pipe important? and b) why do we get a square root of a length in the result and what does that mean? –  Hanno Fietz Dec 19 '12 at 14:38
    
@HannoFietz I edited to include those points –  tpg2114 Dec 19 '12 at 14:58
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About the $h^{1/2}$, the formula tpg is deducing is an ellaborate version of $v=\sqrt{2gh}$ for an object's velocity after free falling a distance $h$. –  Jaime Dec 19 '12 at 15:12
    
@jaime - Interesting. I can see that this would be related to free fall, as the driving force is gravity, but it seems clear to me that it's not the same, either. Accordingly, the Euler deduced version has a factor in it, which comes from the density values, and is dimesnionless, since the density units cancel each other out. Does that factor happen to have a particular name or represent something particular? –  Hanno Fietz Dec 19 '12 at 19:15
    
The math behind has you equating a KE to a gravity PE. And even though the densities units do cancel out, you still get a dimensionless factor multiplying $g$, so it is as if your salty water was flowing onto open air in a moon with reduced gravity... –  Jaime Dec 19 '12 at 19:26

at the same depth $h$ they should be at the same pressure, so no flow would occur. Diffusion yes, but that would be excruciatingly slow (except for the fish, who would probably prefer no salt at all in their lake)

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The densities of salt and fresh water are different, so the pressures at a given depth are different too. –  tpg2114 Dec 19 '12 at 14:09

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