While preparing to drain an "above ground" pool with a regular garden hose, I was thinking about how the hose actually generated a "suction" force. Place one end at the bottom of the pool, connect the other end to a water flow (water spigot), disconnect from spigot, and then place that end at an equal or slightly lower elevation to generate suction.

At this point, the hose is draining the pool of all its water until air stops the suction force at the submerged end. In this scenario, the hose is acting as an extension (increases the volume) of the source of water and the suction force should be just a partial vacuum (relative negative pressure).

Scaling this idea up, use the setup except now the ocean is the source of water. The atmosphere of the earth is pushing down on the entire 2-dimensional surface of the ocean. The water itself has mass that is pushing on all sides of the submerged hose opening. Once the flow has been reversed from adding to the ocean to removing, would it be possible to generate electricity with only the "negative flow"?

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    $\begingroup$ Just curious: what will you drain the ocean into? If you provide a drawing it will be easier to explain what's missing in your reasoning. $\endgroup$ – S. McGrew Apr 19 '20 at 13:49
  • $\begingroup$ It's not what the ocean is draining into which could be itself at different elevation, but the flow rate itself can be used to generate electricity. The flow rate is being provided by the "ocean" sink. $\endgroup$ – Doug Apr 19 '20 at 13:51
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    $\begingroup$ The flow rate is provided by the difference in gravitational potential energy at the top of the ocean and at the bottom of the siphon tube. There will be no flow if the bottom of the siphon tube is below the surface of a second reservoir whose surface is the same height as the ocean's surface. $\endgroup$ – S. McGrew Apr 19 '20 at 14:29
  • $\begingroup$ I agree with your statement 100%. If you don't care about where the water drains (since all water leads back to the ocean or the nearby source), aren't there elevations near the coast where it is below sea level? (Think about New Orleans or the Netherlands). $\endgroup$ – Doug Apr 19 '20 at 14:39
  • $\begingroup$ It kinda sounds like you're basically just describing hydro-power, which is very much used. $\endgroup$ – JMac Apr 19 '20 at 15:34

The first part of your question essentially refers to a so-called siphon, schematised below:

Siphon schematic

This is a 'machine' that converts potential (gravity) energy into kinetic energy. It is well described by Bernouilli's Principle:

$$P_1+\frac{\rho v_1^2}{2}+\rho gz_1=P_2+\frac{\rho v_2^2}{2}+\rho gz_2\tag{1}$$ where: $$P_1=P_0+\rho gd$$ where $P_0$ is the atmospheric pressure. If $d\ll z$, then $P_1\approx P_2=P_0$.

And if the higher reservoir is large, then $v_2 \gg v_1$.

In that case $\text{Eq. (1)}$ reduces to:

$$\frac{\rho v_2^2}{2}\approx \rho gz$$ or: $$v_2\approx \sqrt{2 gz}$$

In principle some of the kinetic energy could be converted to (for example) electrical energy by means of a suitably sized turbine.

Scaling this idea up, use the setup except now the ocean is the source of water.

But this raises the question: where will the siphoned flow flow into?

  • $\begingroup$ I can't deny your math, you are right to ask where the flow will flow into. In my example, I was trying to imagine that hose pouring back into the ocean. Now using the entire earth as the source, if the hose was long enough, wouldn't the curvature of the earth come into effect? $\endgroup$ – Doug Apr 19 '20 at 15:11
  • $\begingroup$ @Doug Gravity is part of a central force field. Its magnitude is dependant ONLY on a point's distance to the CoG of the Earth (causing small variations in $g$ across the Earth). So you 'can't beat the system' with curvature. Have look at the Hoover dam to see how that 'machine' works. $\endgroup$ – Gert Apr 19 '20 at 15:18
  • $\begingroup$ I'm not trying to beat gravity with curvature tricks, but with potential gravitational energy. As long as the hose mouths are at different relative elevations, doesn't that still count? $\endgroup$ – Doug Apr 19 '20 at 15:21
  • $\begingroup$ @Doug As long as the hose mouths are at different relative elevations, doesn't that still count? Of course that works, as I've shown. But how would it work siphoning off an ocean? Where would the lower outlet be? And even if you used a disused mine shaft (or such like) it would quickly fill up. And to empty it again would cost the same energy as was liberated filling it up! $\endgroup$ – Gert Apr 19 '20 at 15:26
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    $\begingroup$ Your best bet to exploit surface water remains to create a dam. As the rising water gains potential energy it can be used to drive turbines. $\endgroup$ – Gert Apr 19 '20 at 15:52

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