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Originally Posted by Robert Tulip
I have not suggested any scenario in which the floating bag would change its depth.
Yes, you have. If you exert a vertical force on the float, it will change the depth to at which it is submerged to displace a volume of water with weight equal to the value of the force. This is called displacement; it is why you can tell the whether a cargo ship is loaded by looking at how low it rides in the water.

You are describing a system that would produce massive quantities of water at very, very low pressure - even ignoring changes in density due to the colder water deep below the surface, your pump could only lift that much water less than a meter above sea level. Why not just use land-based generation with undersea cables to electrically powered pumps submerged below your rig? That would be a much more functionally flexible system.

If you only want the nutrients, it might be more energy efficient to extract the nutrients at depth and just pump up the broth rather than huge volumes of seawater.

2. Originally Posted by VQkr
If you exert a vertical force on the float, it will change the depth to which it is submerged to displace a volume of water with weight equal to the value of the force. This is called displacement; it is why you can tell whether a cargo ship is loaded by looking at how low it rides in the water.
My reason for suggesting the depth would not change (and I may be wrong about the extent this is practical) was that a float with enough buoyancy to stay at the surface even with a massive lift weight would maximise efficiency. If the float is so weak that it just sinks a metre below the waves on a high tide it will not do any pumping. It needs to be powerful enough to lift the weight by all or nearly all the whole tidal range. A laboratory model in a fish tank could test what size float is needed to tidally lift a weight of 1 kg without sinking, for example.
Originally Posted by VQkr
You are describing a system that would produce massive quantities of water at very, very low pressure - even ignoring changes in density due to the colder water deep below the surface, your pump could only lift that much water less than a meter above sea level. Why not just use land-based generation with undersea cables to electrically powered pumps submerged below your rig? That would be a much more functionally flexible system.
It might be possible to lift a smaller amount of water to a higher level (eg lifting fresh water brought in a bag to a land based container) or use this method to shift water around within the ocean. My reason for suggesting this tidal method rather than electrical pumping is to minimise operating costs at scale. I would like to compare the unit capex for this tidal method to an equivalent solar or wind array for electrical pumping.
Originally Posted by VQkr

If you only want the nutrients, it might be more energy efficient to extract the nutrients at depth and just pump up the broth rather than huge volumes of seawater.
My interest is in obtaining nutrient for algae production, and I suspect that would best be done by pumping up water. I will check with CQ moderators though before going into that since I have raised it in other threads before, and would prefer to just restrict this thread to the physics of tidal pumping.

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Originally Posted by Robert Tulip
My reason for suggesting the depth would not change (and I may be wrong about the extent this is practical) was that a float with enough buoyancy to stay at the surface even with a massive lift weight would maximise efficiency... A laboratory model in a fish tank could test what size float is needed to tidally lift a weight of 1 kg without sinking, for example.
Archimedes already did this experiment; see the WP article on displacement.

Originally Posted by Robert Tulip
My reason for suggesting this tidal method rather than electrical pumping is to minimise operating costs at scale.
Minimizing the operating costs will mean minimizing the amount of infrastructure that is far offshore or deep underwater. Electric pumps are scalable, controllable, and quite efficient. You could start a tidal power plant along the coastline where the magnitude of the tides is 5x higher, and all your generation infrastructure will be within 20m of the surface. Easier to maintain. Lifting 1E10 liters of water 1 m requires 9.8E10 J of work (call it 1.4E11 J of electrical input at 70% efficiency). Do this four times per day for 5.6E11 J/day, or an average of 6.5 MW continuous electrical input (~8800 HP). This is not a whole lot of electricity on an industrial scale - at my local industrial power cost of \$0.07 USD/kWh it is \$450/h (same cost as somewhere between 5-10 well-trained staff; 1 if they are an underwater maintenance technician). At 12 kV, the three conductors serving the pump house would each be about 20mm diameter.

By contrast, that submarine we were talking about: a Los Angeles class SSN outputs 35,000 HP to the prop shaft. Its propeller could move enough water to do this job four times over.

Originally Posted by Robert Tulip
I ... would prefer to just restrict this thread to the physics of tidal pumping.
Fair enough. Moving fluids is in my wheelhouse, so I have no argument against staying on the topic.

4. The most effective tidal pump IMO would be a tethered float with a deep anchor and you would use both the vertical and horizontal movements, the latter being much larger from tidal flows. A 600m cable to the bottom and therefore to your pump is not impossible. However the net ocean current may not reverse, so you need to choose a point where there is a major tidal reversal.

5. Originally Posted by VQkr
Minimizing the operating costs will mean minimizing the amount of infrastructure that is far offshore or deep underwater. Electric pumps are scalable, controllable, and quite efficient. You could start a tidal power plant along the coastline where the magnitude of the tides is 5x higher, and all your generation infrastructure will be within 20m of the surface. Easier to maintain.
Such as this lagoon tidal power plant planned for the Welsh coastline:

http://www.bbc.com/news/science-environment-31682529

6. Originally Posted by geonuc
Such as this lagoon tidal power plant planned for the Welsh coastline:

http://www.bbc.com/news/science-environment-31682529
and please note this uses the horizontal motion of the tides to fill the lagoon.

7. I think it is clear that a surface vessel rising and falling with the tide would be able to exert vastly more force on an underlying bladder than would a submerged submarine of the same displacement in neutral buoyance. What is supposed to be an advantage of using the sub?

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Originally Posted by profloater
and please note this uses the horizontal motion of the tides to fill the lagoon.
In open-channel flow (such as a lagoon or the ocean), you can't have horizontal motion without vertical (and vice versa).

9. Originally Posted by VQkr
In open-channel flow (such as a lagoon or the ocean), you can't have horizontal motion without vertical (and vice versa).
yes indeed Bernoulli teaches that but it is still true that the tides are basically sideways motions of water, not pressure driven as in Bernoulli. Tides are a gravity effect the water is "falling" towards the moon and sun and the lifting of the tides is due to momentum when the water hits land such as the continental shelf. If the oceans were all the same depth, hypothetically, the tides would flow sideways with only vertical ridges where counterflows from momentum met. Of course we have all sorts of slopes and land in the way of the free movement of water. Atmospheric pressure changes also cause an effect too but for tidal power the ideal is to find a land shape where the water is funnelled into an estuary giving higher velocities and higher vertical movement too. The UK severn estuary is a good example.

10. Originally Posted by Hornblower
I think it is clear that a surface vessel rising and falling with the tide would be able to exert vastly more force on an underlying bladder than would a submerged submarine of the same displacement in neutral buoyance. What is supposed to be an advantage of using the sub?
I was just trying to get my head around the problem of how you would use tidal energy for pumping in deep water. My initial thought was that neutral buoyancy would provide tidal energy, but as the discussion has explained that is incorrect. A tanker hooked to an anchor designed to rise and fall with the tide, for example a large millstone held in place by vertical poles sunk into the ocean floor, can derive pumping energy from a bladder under the anchor. The maximum size and power of the weight is a function of the ship’s Plimsoll Line.
Last edited by Robert Tulip; 2015-Mar-03 at 08:22 PM.

11. Here is a diagram I have sketched of a tidal pump.Tidal Pump.png

12. The tidal pump described here is the theme of a proposal I have submitted to the MIT Climate Collaboration Energy-Water Nexus Challenge.

The judges have described my proposal as "technically very interesting indeed", and have selected it as a semi-finalist.

13. I I am still concerned that in deep ocean tidal movement is sideways and not vertically. If in deep ocean you propose to anchor something to the bottom then the sideways motion could be used to pump water but I fear your vertical motion proposal will only work in coastal waters.

14. Originally Posted by profloater
I I am still concerned that in deep ocean tidal movement is sideways and not vertically. If in deep ocean you propose to anchor something to the bottom then the sideways motion could be used to pump water but I fear your vertical motion proposal will only work in coastal waters.
Good point. This system is relevant to continental shelf locations, not to the abyssal plains. The length of the anchor line would not be worth it for very deep waters and might not work due to stretching.

A map of world tides at https://en.wikipedia.org/wiki/Tide#/...onstituent.jpg shows that there are substantial coastal areas with high tidal range where use of vertical water motion as described in my proposal should be a cost-effective pumping method. Horizontal currents obviously have a lot of energy, but it looks to me that a turbine to harness this work would be more expensive and complicated. Such a turbine could well be a useful way to augment the main vertical motion system though.

15. Originally Posted by Robert Tulip
Horizontal currents obviously have a lot of energy, but it looks to me that a turbine to harness this work would be more expensive and complicated. Such a turbine could well be a useful way to augment the main vertical motion system though.
I'm not sure what you're saying here. Wouldn't there be a turbine in your system? I can see the advantage of not having an open turbine, but not much more than that.

If I did my math right, the displacement of a float the size of a supertanker will generate a bit more than 50kWh per foot of tide. Or about 10kW of power spread out for rising and dropping tides. (140M lbs displacement converted from foot pounds to watts)
That doesn't sound like much to me.

Wave power might be more advantageous. Scale down the apparatus, but link a lot more of them together so they can each catch individual waves.
But, wave power is already being done, and it looks somewhat successful.
A [url=https://en.wikipedia.org/wiki/Pelamis_Wave_Energy_Converter]Pelamis[/quote] wave generator puts out 750kW with a float about a 3rd of the length of a supertanker. (I'm ignoring width, because they would have to be spaced apart anyway) That's still a magnitude more power even with a 7 foot tide.

16. Originally Posted by NEOWatcher
I'm not sure what you're saying here. Wouldn't there be a turbine in your system? I can see the advantage of not having an open turbine, but not much more than that.
No, there is no turbine. It is straight displacement of liquid into a pipe.
Originally Posted by NEOWatcher
If I did my math right, the displacement of a float the size of a supertanker will generate a bit more than 50kWh per foot of tide. Or about 10kW of power spread out for rising and dropping tides. (140M lbs displacement converted from foot pounds to watts)
That doesn't sound like much to me.
Lets check your calculation. Assume a supertanker is about 3 hectares (six acres) in size. Per foot of tide (or 0.3 metres) the displacement of the bladder in this example is about ten megalitres (~=10 acre feet) per tide.

To lift one ML one metre requires 9.81 MJ of energy (source),, or 2.7 kwH.
Therefore your figure here suggests this tidal pump would only lift the 10 ML by one inch, which is obviously trivial, and seems to me to be understating the physics by orders of magnitude.

My modelling rather suggests the entire 10 ML could be pushed into pipes which would fill whether they were horizontal or vertical, since the pressure inside and outside the pipe under the sea is the same. The distance of pumping would only be a function of the diameter of the pipe, friction etc, and the ability of the float to lift the weight.

Assuming the weight in my model is ten thousand tonnes, in your tanker-size example it would fully displace the water in the bladder beneath it on each falling tide, less friction. The question then is what size of buoyant pontoon is needed to lift a weight of 10,000 tonnes with a tide without sinking at all. The model I imagine is a floating bag of fresh water containing large sealed tanks of air.

http://www.ausmarinescience.com/mari.../oceanography/ has a map of Australian tides showing that the proposed deployment location on the North West Shelf has a tidal range about 15 times greater than this one foot example.
Last edited by Robert Tulip; 2015-Jul-08 at 03:39 AM. Reason: correct for KWH

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Originally Posted by Robert Tulip
Assuming the weight in my model is ten thousand tonnes, in your tanker-size example it would fully displace the water in the bladder beneath it on each falling tide, less friction. The question then is what size of buoyant pontoon is needed to lift a weight of 10,000 tonnes with a tide without sinking at all.
This is impossible. A massless pontoon would still need to displace 10,000 tonnes of water to lift the same amount of mass, and any real pontoon will have its own mass to lift as well. Using the limiting case of a massless pontoon and assuming fresh water for simplicity (density=1 tonne/m3), you would need to displace 10,000 m3 of water. You could do this with a 1 meter thick, 10,000 m2 pontoon or a 1 cm thick, 1x106 m2 pontoon.

18. Originally Posted by VQkr
A massless pontoon would still need to displace 10,000 tonnes of water to lift the same amount of mass
Consider the pontoon as a fabric spherical balloon filled with air, with radius 13.4m, displacing 10,000 m3 of water. How much weight could it carry without sinking?

A comparable question is how much weight would you have to tie to a basketball to make it sink?

19. Originally Posted by Robert Tulip
how much weight would you have to tie to a basketball to make it sink?
A two litre plastic bottle filled with air has neutral buoyancy in a fresh water pool holding weight of 2.6 kg (from an experiment I just did). But this contraption would not have lifting power to fill a bladder fixed between the pool floor and the weight driven only by changing surface water level. To make it work with the same weight, the bottle would need to be bigger, and it would need a winch to release the weight at the top of the tide and then tighten the anchor line at the bottom of the tide.

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Originally Posted by Robert Tulip
Consider the pontoon as a fabric spherical balloon filled with air, with radius 13.4m, displacing 10,000 m3 of water. How much weight could it carry without sinking?

A comparable question is how much weight would you have to tie to a basketball to make it sink?
Originally Posted by Robert Tulip
A two litre plastic bottle filled with air has neutral buoyancy in a fresh water pool holding weight of 2.6 kg (from an experiment I just did). But this contraption would not have lifting power to fill a bladder fixed between the pool floor and the weight driven only by changing surface water level. To make it work with the same weight, the bottle would need to be bigger, and it would need a winch to release the weight at the top of the tide and then tighten the anchor line at the bottom of the tide.
The spherical pontoon would be halfway submerged if pontoon+weight had a mass of 5,000 tonnes, and fully submerged if pontoon+weight had a mass of 10,000 tonnes (again, assuming water density of 1 m/tonne).

If the basketball has a volume of 7.5L and a mass of 0.625kg, then an additional mass of 6.875 kg would need to be added to the basketball's volume to make it sink in water with a density of 1kg/L.

By definition, a neutrally buoyant object has the same mass as the water it displaces. If the 2L bottle and weight assembly has a mass of 2.6 kg and is neutrally buoyant at a given depth, then the two together displace 2.6 L of 1 kg/L water at that depth.

21. Originally Posted by VQkr
The spherical pontoon would be halfway submerged if pontoon+weight had a mass of 5,000 tonnes, and fully submerged if pontoon+weight had a mass of 10,000 tonnes (again, assuming water density of 1 m/tonne). If the basketball has a volume of 7.5L and a mass of 0.625kg, then an additional mass of 6.875 kg would need to be added to the basketball's volume to make it sink in water with a density of 1kg/L. By definition, a neutrally buoyant object has the same mass as the water it displaces. If the 2L bottle and weight assembly has a mass of 2.6 kg and is neutrally buoyant at a given depth, then the two together displace 2.6 L of 1 kg/L water at that depth.
Many thanks, immensely helpful, that all makes very good sense.

Rather than talking about a pontoon as the lifting agent, a standard hot air balloon (filled with normal air) looks better.

The regular size of a hot air balloon is 2,800 m3, with some up to 17,000 m3 https://en.wikipedia.org/wiki/Hot_ai...s_and_capacity

At neutral buoyancy, a fully submerged balloon of 2.8 KT size (less reduction in size due to water pressure) would support a weight of 2.8 kilotonnes. As a sphere it would have radius of 8.75 metres.

As a first approximation, this float and weight apparatus should be able to fill and empty a connected bladder on the ocean floor containing 2.8 ML of water per metre of tide. In a one metre nearly twice a day tide, that would mean 5 ML per day.

I would like to work out how this will compare against diesel or electric pumping for cost, and start to estimate possible economies of scale.

22. Originally Posted by NEOWatcher
No, that is about electricity generation whereas the discussion here is about tidal pumping.

23. Originally Posted by Robert Tulip
No, that is about electricity generation whereas the discussion here is about tidal pumping.
That may be their end goal, but according to the diagram, it is pumping water up to an on shore power facility.

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Originally Posted by Robert Tulip
As a first approximation, this float and weight apparatus should be able to fill and empty a connected bladder on the ocean floor containing 2.8 ML of water per metre of tide. In a one metre nearly twice a day tide, that would mean 5 ML per day.

I would like to work out how this will compare against diesel or electric pumping for cost, and start to estimate possible economies of scale.
5*106 L of water weighs 9.81 N/kg*5*106 = 4.91*107 N. Work (energy) is FxD, so for a lift of 1 meter the daily energy output is 4.91*107 J. There are 86,400 s in a day, so 4.91*107/86,400=568 J/s=568 W.

The same mechanical work could be done by a 60% efficient motor/pump train with 1 kW of electricity (as you scale the system larger, combined motor/pump efficiency will rise asymptotically to about 80%). For an industrial consumer on the west coast of the USA electricity has a marginal cost of about \$0.06/kWh, so the continuous operation of the 1 kW motor would cost \$0.06/kWh*1kW*8760h/year=\$526 per year.

25. Originally Posted by VQkr
5*106 L of water weighs 9.81 N/kg*5*106 = 4.91*107 N. Work (energy) is FxD, so for a lift of 1 meter the daily energy output is 4.91*107 J. There are 86,400 s in a day, so 4.91*107/86,400=568 J/s=568 W.

The same mechanical work could be done by a 60% efficient motor/pump train with 1 kW of electricity (as you scale the system larger, combined motor/pump efficiency will rise asymptotically to about 80%). For an industrial consumer on the west coast of the USA electricity has a marginal cost of about \$0.06/kWh, so the continuous operation of the 1 kW motor would cost \$0.06/kWh*1kW*8760h/year=\$526 per year.
Thanks. I am not sure the cost of electric pumping is so low. This Irrigation Guide (p52) gives a table in Australian Dollars (AUD=USD*0.75) with price range for pumping which indicates that pumping 1825 ML per year (the quantity in our tidal pumping example) would require electicity cost in the range of \$50,000 to \$80,000, one hundred times your calculation. This is at flow rate of 2.6 ML per day so this sugar irrigation system would need to be doubled in the example here which is pumping 5 ML per day.

The linked example has parameters of electricity at AUD 0.21 / KwH (more realistic for remote operation than USD 0.06) and suction lift of 2m. One megalitre (ML) = 106 L of water.

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You are confusing suction head (how far below ambient the pressure is at the inlet) with discharge pressure. Those numbers are for pumps increasing the pressure by hundreds of kPa; in your question and my response you were asking about lifts of 1 meter (9.81 kPa).

Completing the two order of magnitude discrepancy is the cost of power; that publication lists \$0.20 AUD (\$0.15 USD) per kWh - 2.5x the US price I quoted. Reasonable assumptions on cost of power depend on country of manufacture; many costs of a remote installation (ie labor) will increase with remoteness more rapidly than the cost of electricity.
Last edited by VQkr; 2015-Jul-10 at 06:38 AM. Reason: add

27. Imagine four bladders joined together with a platform linking them, a weight to sink them, a vertical pole lever and a chain to a surface vessel. The tides will drag the vessel sideways and whatever orientation the sunken group is at, it will pump given a set of one way valves akin to a diode bridge. If the position is not good you haul it up and try elsewhere. The sideways tidal movement will give orders of magnitude more energy than vertical lift.

28. Quite a lively exchange on this topic is at http://climatecolab.org/web/guest/pl...essage_1349453