Tidal Power Gen Multiplier? [closed]

Regarding tidal power generation, the following scheme uses an outer tank and an inner tank that floats on the water of the inner tank. You fill the inner tank first (while it's at high-tide level) and then fill the second tank which raises the innner tank doubling it's potential energy.

Question: Would this work?

UPDATE: I think it may be important to consider that water pouring into the tank looses energy. Meaning the water at the bottom of the tank has less energy than at the top. So it could be that this mechanism captures some of that energy that would otherwise be lost.

Also, the issue does not appear to be buoyancy since force of 1L of water = force of 1L of air in water = 9.8N. So that's not the problem.

closed as off-topic by StephenG, Kyle Kanos, sammy gerbil, Jon Custer, YashasJul 27 '18 at 12:47

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• The center of gravity o the 2nd tank only moves the same vertical distance as the first = same potential energy. ps tidal effect tag, doesn't mean tides in this sense – Martin Beckett Jul 23 '18 at 18:29
• You should explain how you think it will work. Note the volume of air you would need to "raise" the water, and what that volume represents... – Jon Custer Jul 23 '18 at 18:29
• So the inner tank would not be raised as high as depicted in the graphic? The weight of the water displaced equals the weight of the water raised. So you're saying the upward force of water displaced is not equal to the downward force of water in inner tank and so it would simply not be raised as high? – squarewav Jul 23 '18 at 19:14
• Why bother with 2 tanks? What is the point? If you can generate energy by draining the outer tank, why not fill it completely during high tide then drain it during low tide? I don't see any multiplier effect here. Where are your calculations to show multiplication? – sammy gerbil Jul 23 '18 at 19:46
• Sounds like you're trying to harvest energy, $2e$, from a sequence of two tide cycles where $e$ is the energy that you could have harvested from a single cycle by using a simpler machine. – Solomon Slow Jul 23 '18 at 20:04

Much to my surprise I think that this does work. However you only get a finite possible benefit from it, even when iterated of course. I also may be wrong somewhere: I would welcome someone pointing out why the following is wrong!

So, here is what I think the situation you are describing. There are two tanks which we assume fit closely inside each other (this is just to simplify the maths), and they each have height $H$ and cross-sectional area $A$ (so the inner one has area $A-\Delta A$ if you like so it fits inside the outer one, and $\Delta A$ is small). The inner tank has a partition half way up it, with a tap. Both tanks have taps in their bases (which is how you get the water out to get power) and both tanks have open tops to allow water to flood in. The outer tank is attached to the ground (ie it does not float away), the inner tank isn't. The tanks are light (so I can neglect their mass which will reduce the efficiency of the thing as you have to lift them).

At high tide the water level is at or above the top of the tanks, and at low tide it is at or below the bottoms of them. There is lots of water, the density of the water is $\rho$.

So, OK, let's just consider the outer tank first of all, with no inner tank. At high tide it fills with water, and at low tide the water is allowed to drain out and do work. How much work does it do? Well, each little disk of water of thickness $\delta h$ at height $h$ does work $(\rho A\delta h) \times g h)$, so the total work we can get from the thing is

\begin{align} W_1 &= \int\limits_0^H \rho A g h\,dh\\ &= \frac{\rho A g H^2}{2} \end{align}

So, now consider the trick you suggest:

1. place the tanks one inside the other, with all the taps closed;
2. fill the top of the inner tank from the tide (so the inner tank is now half full);
3. now fill the outer tank, raising the inner tank up by half its height;
4. wait for the tide to go out.

Now how much work can we get from this? Well to get work out of it:

1. open the tap in the partition of the inner tank, allowing the water to flow into the bottom half of it -- trivially this does work $\rho AgH^2/4$ (the mass of water is $\rho AH/2$, and it falls a distance $H/2$);
2. now open all the taps and let the water run out as before, doing $\rho AgH^2/2$ again.

So, now

\begin{align} W_2 &= \frac{\rho AgH^2}{2} + \frac{\rho agH^2}{4}\\ &= \frac{3\rho AgH^2}{4}\\ &= \frac{3}{2}W_1 \end{align}

And this is more than $W_1$. Well, we can iterate this with a third tank. But the third tank can only be half as high as the second one as it needs to float in the second tank. And we can repeat this, of course:

\begin{align} W_1 &= \rho A g H^2\frac{1}{2}\\ W_2 &= \rho A g H^2\frac{1}{2} + \frac{1}{4}\\ W_3 &= \rho A g H^2\frac{1}{2} + \frac{1}{4} + \frac{1}{8}\\ &\cdots\\ W_\infty &= \rho A g H^2\sum\limits_{n=1}^\infty \frac{1}{2^n}\\ &= \rho A g H^2 \end{align}

This is the ideal case: in real life the tanks must get smaller in radius so they fit inside each other and will not be massless.

As I said: I'm not convinced by this answer: I feel I must have made a mistake somewhere.

• I think it actually does work as well. The important bit which I mentioned in my comments above is that without the inner tank the water falls to the bottom and in the process wastes energy. But with the inner tank, the outer tank fills immediately and no energy is wasted filling the upper half of the outer tank. That energy instead applied to float the inner tank. However, note that if one simply made a tank with twice the area an half the depth (such that the volume of water were the same), it would also have more energy being elevated above low tide. So this all equates to a space saving op – squarewav Jul 24 '18 at 19:24