Calculating electrical generation from tidal piston

  • #1
I have been kicking around an idea for a while about generating electricity from ocean tides. The traditional approach is to use the flow of water through a channel to turn a turbine, which essentially blocks the channel for sea life and water craft.

It seems more practical to use the rise and fall of the tide like a slow moving, yet powerful piston. The way I envision it, there would be a barge attached to posts set in the ocean floor. As the tide moved the barge, gears on the posts would turn a turbine on the barge.

Say two complete turns of the piston occur every 24 hours. The height of the tide is 3 Meters. The weight of the barge is 1,500,000 kg. Assuming 100% of the energy could be converted into electricity, how much electricity could this device put into the grid?

Thanks in advance for the help.
 

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  • #2
berkeman
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I have been kicking around an idea for a while about generating electricity from ocean tides. The traditional approach is to use the flow of water through a channel to turn a turbine, which essentially blocks the channel for sea life and water craft.

It seems more practical to use the rise and fall of the tide like a slow moving, yet powerful piston. The way I envision it, there would be a barge attached to posts set in the ocean floor. As the tide moved the barge, gears on the posts would turn a turbine on the barge.

Say two complete turns of the piston occur every 24 hours. The height of the tide is 3 Meters. The weight of the barge is 1,500,000 kg. Assuming 100% of the energy could be converted into electricity, how much electricity could this device put into the grid?

Thanks in advance for the help.
Welcome to the PF.

The energy from lifting a mass m to a height h is E=mgh, where g=9.8m/s^2 is the acceleration due to gravity at the Earth's surface.

So for each raising of the barge, the water would do the following amount of work:

E=(1,500,000kg) * (9.8m/s^2) * (3m) = 44.1*10^6 Joules.

If you can extract 10% of that on the rising stroke, then hold off lowering it until the water is at low tide so you can extract energy on the downward stroke, you could get 40% of that energy per day. So E = 17.6*10^6 Joules.

A Watt is a Joule per second, so the average power generated is Pave = (17.6*10^6 J) / [(3600s/hour) * (24 hours/day)] = 204 Watts.

Assuming I didn't make any math errors, that's a pretty low average power... :smile:
 
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  • #3
Thank you Berkman for the info! I appreciate you taking the time to help me with this. If you don't mind, I have couple additional questions based on your response.

You calculated Pave (what is Pave?) as the E / (3600 * 24) - but the piston changes every 6 hours, so would the 24 be replaced with 6 as in E / (3600 * 6)?

Also, to help me put this in perspective, would the opposite also be true, i.e. that you could lift the barge with 510 Watts (adding back the the estimated 40% loss)?

Thanks again for your help. It's been a long time since I've worked with physics, but I still love the fact that you can conceptually figure something out without having to build anything. This is a great forum.
 
  • #4
berkeman
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You calculated Pave (what is Pave?) as the E / (3600 * 24) - but the piston changes every 6 hours, so would the 24 be replaced with 6 as in E / (3600 * 6)?
I used the symbol Pave for the average power. You said 2 tidal cycles per day, at 10% conversion efficiency, so that's why I took 40% of the total energy per rise of the barge.
Also, to help me put this in perspective, would the opposite also be true, i.e. that you could lift the barge with 510 Watts (adding back the the estimated 40% loss)?
That's a good question. Assuming very little friction, I suppose it would be something like that. But as a practical matter, there would be a lot of friction in the gear reduction chain from whatever motor you used to lift such a heavy object so slowly. I'm not sure what the real numbers would turn out to be.

Have you done much reading yet on such power extraction? There probably are some good references. I have seen articles on extracting power from ocean waves, but of course waves move much more quickly than the tides. :smile:
 
  • #5
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I have seen articles on extracting power from ocean waves, but of course waves move much more quickly than the tides. :smile:
A big advantage of extracting power from ocean waves is a large float can be directly connected to a small high pressure piston which pumps seawater to a relatively high storage location.
 
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  • #7
sophiecentaur
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The weight of the barge is 1,500,000 kg
Here is the basic problem with your system. The total mass being moved is very limited so mgh, which is the change in potential energy as the barge floats between two levels of water, will be (surprisingly) small. (See the calculations in the above posts)
If, instead of your barge, you use a vast amount of water behind a tidal dam, you could be dealing with many thousands of times the total mass involved and the available energy from a similar change in height of the mass of water itself is much greater. It is true that the conversion could be more lossy, when turbines are involved but that would be offset by the vast difference in masses involved.
I once considered a system for charging the batteries on my boat, using the changing height between tides and a tether to the bottom but, for a moderate (four ton) size of craft, the energy is peanuts and I chose wind and solar as the better options. Consider the power requirement for a simple electric motor, driving a winch with very efficient gearing if you had six hours to raise a sail boat / car / lump of concrete. You could probably get away with a scalextric motor.
 
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  • #8
jbriggs444
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One additional quibble -- depending on the draft of the boat, you may or may not be able to make effective use of its entire 1,500,000 kg displacement. If, for instance, the boat had a 40 foot draft and the tides were 10 feet high to low then you would only have roughly 25% of the 1,500,000 kg force available to drive your generator at most. The average through the stroke would be roughly half of that.
 
  • #9
sophiecentaur
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Unless the boat takes the ground near low tide, it's draught is not relevant. It's CM moves up and down by the 10m, whatever if it is always floating.
 
  • #10
jbriggs444
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Unless the boat takes the ground near low tide, it's draught is not relevant. It's CM moves up and down by the 10m, whatever if it is always floating.
The fact that its CM moves up or down by 10m is not relevant. The force with which it does so relates to the craft's horizontal cross-section in the water and that is inversely proportional to its draft.

Consider, for instance, the extreme case of a craft shaped like a vertical pencil with a draft of several miles. How much more buoyant will it be at high tide than at low?
 
  • #11
I had this very problem presented to me in real life. A neighbor was keeping his 15'x30' barge moored in his salt water lagoon near Alaska's Cook Inlet. After having watched that barge go up and down twice a day practically forever, he told me he was going to install a system of cables and pulleys to drive a generator t provide electricity to his cabin. This area has a tidal range of about 30'. However, no water could enter this lagoon until the tide reached +13' and an average high tide was around 20', yielding an average up/down movement of 7' twice a day. AT an estimated 20 tons for the barge, the average daily potential energy would have been about 1.12 million foot pounds (40,000# x 7' x 4 movements/day) which is about 0.42 KW hour. After friction losses, the likely net energy would have yielded perhaps 10% of that or enough juice to light a 40w bulb for an hour. Though intelligent, his education ended with high school and his science probably ended at 8th grade and there was simply no way to convince him this scheme wouldn't work. "It just stands to reason that something that heavy could make a lot of power", he insisted. The moral of the story is that all too many folks think that math and science are plots devised to keep them from getting what they want. We need look no further than climate change deniers for ample evidence of this.
 
  • #12
sophiecentaur
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The fact that its CM moves up or down by 10m is not relevant. The force with which it does so relates to the craft's horizontal cross-section in the water and that is inversely proportional to its draft.

Consider, for instance, the extreme case of a craft shaped like a vertical pencil with a draft of several miles. How much more buoyant will it be at high tide than at low?
I think we must have different models in mind. The suggested model is just a floating barge (?). There is no energy to be had from the moving volume of the water below the barge because the process is so slow that the rising water would just be deflected by the hull - that's basically Archimedes' Principle. A wide, flat bottomed barge would displace exactly the same amount of water as a narrow V shaped barge and so the Upthrust would be the same.
Otoh - and this may be the basis of your idea. A large, heavy piston in a sealed cylinder would, have work done on it that is equal to the actual volume of water flowing in (and out) times the pressure. The larger the area of the cylinder, the more work would be involved because you would be dealing with displaced water volume, rather than displaced weight (as with a boat). I haven't put that very well but you may get the idea.

Why would there be a difference, if it floats all the time?
It's Energy, not Force that counts, when you are generating power.
 
  • #13
jbriggs444
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I think we must have different models in mind. The suggested model is just a floating barge (?). There is no energy to be had from the moving volume of the water below the barge because the process is so slow that the rising water would just be deflected by the hull - that's basically Archimedes' Principle. A wide, flat bottomed barge would displace exactly the same amount of water as a narrow V shaped barge and so the Upthrust would be the same.
If you are pushing down on a barge during a rising tide, you do not get the full tidal stroke. The barge rides lower in the water due to your push. If you are lifting up on a barge during an ebbing tide, you do not get the full tidal stroke. The barge rides higher in the water due to your push. This effect is smallest for a boat with a large area and a shallow draft. It is largest for a boat with a small area and a deep draft.
 
  • #14
anorlunda
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If you Google "bay of fundy hydropower", you'll find a wealth of information on this topic. This idea has been investigated many times around the world. It is undoubtedly possible, but not necessarily attractive.
 
  • #15
To add another twist to the discussion, consider if the barge were not a floating vessel at all, but rather an empty cylinder which filled and emptied with the tide, and used turbines to harness energy from the water moving in and out. Theoretically, the energy would be identical to a barge rising and lowering in the tide. The total volume of water moving into and out of the empty cylinder is equivalent to the volume of water that raises the vessel. Therefore, if the vessel weighted the same as the volume of water in the column, the piston could attain identical power in the up and down stroke. The appeal of using a flat-bottomed barge, is that they are much cheaper to produce than creating a self-enclosed bay.

I love this discussion. Thanks all for participating. It is an inviting idea, even if not practical. I especially appreciate that story about the neighbor building this device, I love that!! It is sometimes difficult to satisfy an itch theoretically, but so much faster!

Also, if anyone has a link they could share on this topic I would be most grateful. I will post any if I find them as well.

Thanks again all, I'm loving Physics Forums!!
 
  • #16
sophiecentaur
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If you are pushing down on a barge during a rising tide, you do not get the full tidal stroke. The barge rides lower in the water due to your push. If you are lifting up on a barge during an ebbing tide, you do not get the full tidal stroke. The barge rides higher in the water due to your push. This effect is smallest for a boat with a large area and a shallow draft. It is largest for a boat with a small area and a deep draft.
Yes, pulling or pushing the hull lower in the water will increase the force (i just realised that). The limit is, of course, the weight of water displaced when the barge is submerged. And, as you say, the effect is maximised with a flat bottom and as big an area as possible. There are several variations on the idea. For instance if you use a very light hull, you will get all the energy on a rising tide and very little as the tide falls. But it is easier to engineer it that way with a tether on the sea bed than using the downward force on the falling tide.
However, the volume involved is limited by the size of hull you use, which is a tiny fraction of what's available if you use the total volume of water behind a tidal dam. I suppose the environmental impact would be proportionally less, though.
 

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