[Mootle (OP)]

Hi folks - I've made a bit of an animation for a new type of hydro power renewable energy system. Hopefully, this will show how it works but it would be good to know if I've managed to get the idea across. Please let me have any comments (good or bad,) relating to my Buoyancy Engine idea so I can focus emphasis in the right places for future videos in the series.

http://www.youtube.com/watch?v=q2pUlbusnXo

Many thanks

Mootle

[Geezer]

Nice annimation! It's not clear to me what all the bits do. I understand the pontoon rising and falling in the tide, but I don't understand what the other tank thing is doing. Maybe a bit of a description might help?

BTW, assuming the energy is derived from the rise and fall of the pontoon, have you worked out a size to power ratio? Something along the lines of kilowatts per cubic meter or 1000 kg of displacement might be helpful. I think you should be able to get a good idea of it from the mass of water displaced by the pontoon, and the distance it moves in time.

[RD]

Please let me have any comments (good or bad,) relating to my Buoyancy Engine idea

 [ Invalid Attachment ]

What is preventing horizontal and rotary motion by the “storage vessel" and “pontoon" due to ocean currents, causing the cables to become twisted/tangled which would prevention operation ?.

If the answer is “they are enclosed in an almost watertight lift-shaft built on the seabed” that’s gonna cost a fortune to construct to a standard which will survive being in the ocean.

BTW I suspect it would only be a matter of time before marine fouling would clog-up the pulleys.

[Geezer]

I couldn't resist trying to work out the power.

Let's say the pontoon displaces 1000 metric tonnes and it moves 2 meters up and 2 meters down every 12 hours.

The force produced by the pontoon is 9.81*1000 = 9810 kN (kilo-newtons)

The average speed of the pontoon is 8/60/60 = 0.0022 m/s

Power is force times distance in time, so the average power (assuming the device produces power in both directions) is

0.0022*9810 = 21.6 kW (before you account for any parasitic losses)

(Did I get that right?)

[Geezer]

(Did I get that right?)

No, I didn't!

I mucked up the speed. It should be 8/24/60/60 = 0.000093 m/s (93 micrcometers per second)

The average power would be 0.9 kW, so, if you are lucky, you might get about 1kW for every 1000t displacement. If I have that right, it means a very large supertanker could generate about 500kW.

[Geezer]

Mootle,

I looked your idea and wondered if you have thought of using another method that might use compressed air instead of the cables and pontoons.

Imagine you have a (very large) thing like a diving bell that is anchored to the sea floor. The top of the bell does not need to be far below sea level. It just needs to be low enough that it remains fully submerged at the lowest tide.

There is a (very big) hose connected to the top of the bell. At low tide, we force air into the bell to dispel all the water from the bell. We then disconnect the compressor and seal the hose. The only thing that remains attached to the hose is a pressure gauge.

As the tide rises, the pressure in the hose will also rise as water is forced into the bell by the increased head of water relative to the bell. As the tide falls, the opposite happens.

The changing air pressure in the hose is a potential source of useful work. This may not be the most elegant way of tapping tidal energy, but in certain situations it might overcome a lot of the problems associated with corrosion and marine fouling. Also, it can be very robust, is largely invisible, and it has no impact on the flow of water through estuaries and the like.

The downside is that you either need one gigantic bell, or an enormous number of smaller bells connected in parallel, which is probably the only way to do it. I think it would be best if they are made of reinforced concrete.

The slowly fluctuating compressed air could be made to do a lot of different things, but I think the best thing to do with it would be to use it to pump water into an elevated reservoir that could produce electric power from water driven turbines when required.


 

[Bored chemist]

It seems much simpler to hang a big heavy float in the sea and let it rise and fall with the tide. Tie a rope to it and then connect that rope to a pulley. Have the other end of the pulley connected to a spring (or a counterweight).
When the tide falls it pulls the rope and turns the pulley. When the tide falls the spring or counterweight pulls the rope and turns the pulley the other way.
With this system there is less rope, it's all above water (and so is everything else apart from some sort of frame to hold it in place.)
Your system looks unduly complicated. Why have 6 ropes + pulleys when you can just put a gearbox on the generator shaft?

[Mootle (OP)]

Nice annimation! It's not clear to me what all the bits do. I understand the pontoon rising and falling in the tide, but I don't understand what the other tank thing is doing. Maybe a bit of a description might help?

Thanks. The Storage Vessel is submerged to a desired depth toward the bottom of the ocean by virtue of the pulley system which connects it to the pontoon via a cable. In the animation I've shown a 6:1 gearing ratio so for the 2m tidal range the Storage Vessel with descend by 2 x 6 to give 12m. The pulley is ratchetted so multiple cycles can be used to achieve the desired depth. Once the desired depth is achieved a valve can be opened to allow an inrush of water to a turbine. The turbine is linked to a generator. The greater to depth and flow the greater the potential for power generation.

BTW, assuming the energy is derived from the rise and fall of the pontoon, have you worked out a size to power ratio? Something along the lines of kilowatts per cubic meter or 1000 kg of displacement might be helpful. I think you should be able to get a good idea of it from the mass of water displaced by the pontoon, and the distance it moves in time.

A turbine and generator rated at 1.99MW with a working head of 50m and a Storage Vessel volume of 67,000m3 would provide the following revenue:

RO @ £100/MWh   £434,637.01/yr
FIT @ 11p/kWh   £478,100.71/yr
Export @ 3p/kWh   £130,391.10/yr
         £1,043,128.82/yr
   
£20,862,576.34/20yr based on 25:1 gearing or £208,625,763.39/20yr for 10 systems based on simple scaling.

[Geezer]

Ah, but if the storage vessel is 67,000m3, the pontoon will have to displace six times that volume, or about 400,000 cubic meters, otherwise it will sink. A cubic meter of water weighs about 1t, so your pontoon displacement is about 400,000t which is about 80% of the supertanker I mentioned, and that is only capable of generating 0.5MW, so you are only going to get 0.4MW, and that's without allowing for the inevitable losses in the system

It doesn't matter how you cut it, but ultimately, the thing that is doing the work is the tide lifting the mass of water displaced by the pontoon against gravity. If your calculation says that you are doing more work than that, you are trying to get something for nothing, which has never worked thus far.

[Mootle (OP)]

Ah, but if the storage vessel is 67,000m3, the pontoon will have to displace six times that volume, or about 400,000 cubic meters, otherwise it will sink. A cubic meter of water weighs about 1t, so your pontoon displacement is about 400,000t which is about 80% of the supertanker I mentioned, and that is only capable of generating 0.5MW, so you are only going to get 0.4MW, and that's without allowing for the inevitable losses in the system

It doesn't matter how you cut it, but ultimately, the thing that is doing the work is the tide lifting the mass of water displaced by the pontoon against gravity. If your calculation says that you are doing more work than that, you are trying to get something for nothing, which has never worked thus far.

The power generation calculation depends on the depth and the size of the pipe. I've applied typical turbine and motor efficiency and calculated the energy loss due to pumping out. These are standard calculations which I'm happy to post if necessary or you could post your calculation for review so I can see where I've managed to lose you?

Think of the idea as a portable dam. With a working head of 50m a much higher power output can be achieved by opting for a larger pipe or choosing a greater working head for the generation phase. There is always downside though and this idea is no exception, i.e., the storage tank would fill much faster and a higher rated turbine / generator set would cost more. This would only be worth investing in if the selling price of high instantaneous electricity upon demand was valued higher than a lower base load provision. At the moment the feed in tariff and renewable obligation are geared toward providing a base load provision. The pontoon sizing is another example where we can't get something for nothing, i.e., the greater the gearing ratio the greater the volume of the pontoon will need to be for a given storage vessel. The optimum gearing ratio for a 50m depth and 2m range would be 25:1. At this size we would need to make full use of the pontoon. Using it as a low cost platform for wind turbines or solar PV would be an obvious way to improve the return on investment. 

[Geezer]

The power generation calculation depends on the depth and the size of the pipe.

Alas, it does not. To generate power you need to do work, and the only thing that's doing any work is the tide elevating the pontoon against the force exerted by the cable. It's quite simple to determine how much work the tide does, or the power, which is the rate of doing work. I calculated that in my previous posts.

You can do anything you like with systems of gears and cables, but you will never be able to use them to multiply the amount of work done by the tide acting on the pontoon. The second law of thermodynamics has yet to be broken.

[imatfaal]

seems to me that Geezer is correct (damn) - I will note also that 400k metric tonnes deadweight (available displacement for cargo - or in your case provide buoyancy force) is bigger than all but a dozen or ships in the world today.  Large cargo ships (no one says supertankers apart from the press) are not designed to be pulled from underneath by cables but to have cargo spread evenly over about 15000m2 of bottom.  For your guidance - to construct a tanker of 400k dwt you use about us$25-30million worth of steel

[Geezer]

seems to me that Geezer is correct (damn) -

Actually, he's not! I cocked up the distance by counting both the rising and falling distance. That's not right; work is only done on the pontoon when the tide is rising. I overestimated the power by a factor of two.

A half million tonne pontoon can only generate 250kW (average) which, coincidentally, is the same as the maximum power output of the engine in my truck  [] 

[Mootle (OP)]

The power generation calculation depends on the depth and the size of the pipe.

Alas, it does not. To generate power you need to do work, and the only thing that's doing any work is the tide elevating the pontoon against the force exerted by the cable. It's quite simple to determine how much work the tide does, or the power, which is the rate of doing work. I calculated that in my previous posts.

You can do anything you like with systems of gears and cables, but you will never be able to use them to multiply the amount of work done by the tide acting on the pontoon. The second law of thermodynamics has yet to be broken.

I offered to review your calculations or show you the calculations I've used. You've chosen a third option. Unfortunately, your way takes the debate no further forward and denies us a learning opportunity, where either you learn something new or I do.

However, strongly you assert something it means nothing without proof!

[Mootle (OP)]

What is preventing horizontal and rotary motion by the “storage vessel" and “pontoon" due to ocean currents, causing the cables to become twisted/tangled which would prevention operation ?.

Hi, thanks for your feedback. The schematic animation is a little misleading, it is more clear in the scaled model that I'm developing. However, I will try to describe in words. The Storage Vessel would be shaped like a submarine. The Pontoon would be shaped such that the Storage Vessel rises up into a hollowed out centre (a bit like an oval shaped polo mint). The Pontoon would include various overhead gantries spanning the hollowed centre (shortways,) each with a pair of compound pulleys. Each set of cables would pass through guides mounted onto the outer shell of the Storage Vessel. The cables would be highly tensioned at all times. I would anticipate that this would be sufficient To minimise drift but I would agree that some simulation modelling will be needed to verify the effects and account for variability. The storage vessel would be turned into any prevailing current but ordinarily the site would be selected to avoid strong current systems.

If the answer is “they are enclosed in an almost watertight lift-shaft built on the seabed” that’s gonna cost a fortune to construct to a standard which will survive being in the ocean.

Again, I can see why you might get that impression. The idea is intended for the open ocean. I had to place the animation in a box to avoid strange effects in the video as it rescaled with the movement of the tide.

BTW I suspect it would only be a matter of time before marine fouling would clog-up the pulleys.

This is a real engineering challenge but I'm sure it's not a new problem. I would seek consultation with marine engineers but anticipate a few practical mitigation's as follows;

1. Keep things moving.
2. Select large diameter pulleys.
3. Don't compromise on material selections.
4. Fit scraping devices.
5. Include planned maintenance tasks.

[Bored chemist]

OK, let's have another look at the maths.
"A turbine and generator  with a working head of 50m and a Storage Vessel volume of 67,000m3 would provide the "

In the right units, pressure times volume = energy
50m of depth in water with a density of 1000 kg/m^3 will exert a pressure of
(rho) g h =
1000 X 10 * 50 =500,000 pascal (that's about right: it's 5 bar which is what you would expect with 10m head of water being about 1 bar)
The volume is 67000 so the stored energy is
500,000*67,000
33.5 GJ
That sounds good, but it's only roughly the energy stored in 1000 litres of cooking oil or gasoline.
That energy is available twice a day so that's once every 12 hour
720 minutes
43200 seconds
So the power is 33,500,000,000 /43,200 which gives you
0.78MW average power. That's not a lot.

To do that you would need 67000 tons moving up and down by 50 metres or roughly 500,000 tons moving up and down 8.3 metres.

But it only moves by about 4 metres so the energy is only half what I calculated.
So the power output is about 400KW for a roughly supertanker sized pontoon.
That's the same ballpark as Geezer's figure. (Possibly the reason it's bigger is because I haven't included the energy needed to blow the tank clear each time)

Do you have any real evidence that the output will be better?
If not, perhaps you would do better to study some thermodynamics rather than say things like "However, strongly you assert something it means nothing without proof!"


Incidentally, you might like to answer the questions I asked earlier.
"Why have 6 ropes + pulleys when you can just put a gearbox on the generator shaft?"

[Mootle (OP)]

seems to me that Geezer is correct (damn) - I will note also that 400k metric tonnes deadweight (available displacement for cargo - or in your case provide buoyancy force) is bigger than all but a dozen or ships in the world today.  Large cargo ships (no one says supertankers apart from the press) are not designed to be pulled from underneath by cables but to have cargo spread evenly over about 15000m2 of bottom.  For your guidance - to construct a tanker of 400k dwt you use about us$25-30million worth of steel

Thanks for this. Accurate costing information is vitally important and I would be most grateful if you can provide any source information on this. I'm second guessing that the Pontoon sizing that has been used above is a false assumption carried forward from the schematic animation. As stated, the scaled animation is based on a 25:1 gearing ratio and would thus require a larger pontoon.

[Mootle (OP)]

OK, let's have another look at the maths.
"A turbine and generator  with a working head of 50m and a Storage Vessel volume of 67,000m3 would provide the "

In the right units, pressure times volume = energy
50m of depth in water with a density of 1000 kg/m^3 will exert a pressure of
(rho) g h =
1000 X 10 * 50 =500,000 pascal (that's about right: it's 5 bar which is what you would expect with 10m head of water being about 1 bar)
The volume is 67000 so the stored energy is
500,000*67,000
33.5 GJ
That sounds good, but it's only roughly the energy stored in 1000 litres of cooking oil or gasoline.
That energy is available twice a day so that's once every 12 hour
720 minutes
43200 seconds
So the power is 33,500,000,000 /43,200 which gives you
0.78MW average power. That's not a lot.

To do that you would need 67000 tons moving up and down by 50 metres or roughly 500,000 tons moving up and down 8.3 metres.

But it only moves by about 4 metres so the energy is only half what I calculated.
So the power output is about 400KW for a roughly supertanker sized pontoon.
That's the same ballpark as Geezer's figure. (Possibly the reason it's bigger is because I haven't included the energy needed to blow the tank clear each time)

Do you have any real evidence that the output will be better?
If not, perhaps you would do better to study some thermodynamics rather than say things like "However, strongly you assert something it means nothing without proof!"

Hi, thanks for this, happy to show how I've come to my generator rating. fyi I have studied thermodynamics during my Masters in BSE & Sustainable Energy but I didn't mean anything by the comment other than trying to get to the bottom of the descrepancy. If I'm wrong I want to know why and I cannot do that without knowing how the figures are reached but I've suggested a possible reason in my previous email.

The equation used for generation and pumping is the same:

Pgen = (gen eff*rLLw*g*dLL*Agen*(2*9.81*dLL)^0.5)/1000000
2MW = (0.85 * 1,025 * 9.81 * 50 *0.15 * (2 * 9.81 * 50)^0.5)/1000000
Where
gen eff, generator efficiency = 85%
rLLw, ocean density at depth = 1025kg/m3
g, gravity = 9.81m/s2
dLL, depth at LL = 50m
Agen, Gen Supply Pipe Area = 0.15m2

Following your calculation I think the issue is one of time. The difference seems to be in the time frame applied. For the optimum operation, i.e., generation would only be possible for a shorter amount of time than the 12hr period indicated.

Incidentally, you might like to answer the questions I asked earlier.
"Why have 6 ropes + pulleys when you can just put a gearbox on the generator shaft?"

I hadn't considered the gearbox idea and can see a number of benefits which are well worth exploring. Thank you.

[Geezer]

The power generation calculation depends on the depth and the size of the pipe.

Alas, it does not. To generate power you need to do work, and the only thing that's doing any work is the tide elevating the pontoon against the force exerted by the cable. It's quite simple to determine how much work the tide does, or the power, which is the rate of doing work. I calculated that in my previous posts.

You can do anything you like with systems of gears and cables, but you will never be able to use them to multiply the amount of work done by the tide acting on the pontoon. The second law of thermodynamics has yet to be broken.

I offered to review your calculations or show you the calculations I've used. You've chosen a third option. Unfortunately, your way takes the debate no further forward and denies us a learning opportunity, where either you learn something new or I do.

However, strongly you assert something it means nothing without proof!

Apparently you didn't like my calculations  []

OK, here we go again, from the top!

To generate power something has to do work. We are trying to derive power from tidal energy, so it must be the tide that's doing the work - so far so good, (I hope).

According to Wikipedia (Work) - "The SI unit of work is the joule (J), which is defined as the work done by a force of one newton acting over a distance of one meter."

In this case it's the vertical pull on the cable caused by the pontoon rising in the tide that is doing all the work (if you don't believe me, please tell me what other source of energy you are tapping into.)

So, if we know how far the pontoon was raised by the tide and the force exerted by the cable, we can multiply one by the other to determine the amount of work done (in joules).

The force in the cable equals the force produced by the buoyancy of the pontoon acting against gravity, which is 9.81 times the mass of water displaced - (don't blame me, blame Archimedes.)

If the pontoon displaces 400,000t or 400,000,000kg (which it must in order to submerge a 67,000 cubic meter storage vessel with a 6:1 mechanical ratio) the force in the cable is 9.81 times 400,000,000 = 3,924,000,000N. (Could be quite a thick cable!)

We know the tide lifts the pontoon 2 meters twice a day, or 4 meters in 24 hours.

Therefore, the work done per day (force times distance) is 9.81 x 400,000,000 x 4 = 15,700 MJ (megajoules)
The average work done in one second must be that value divided by 86,400.

Therefore, average work per second is 15,700,000,000/86,400 = 182,000 J/s = 182kJ/s

A watt is a rate of doing work at 1 J/s (joules per second), so the average power is 182kW. It looks to me that you are assuming continuous generation in your calculations, in which case you better be using average power, not peak power.

BTW, 1 megawatt-hour equals 3,600 MJ (megajoules), so 15,700 MJ per day can generate a maximum of 4.36 MWh per day.

[Mootle (OP)]

Apparently you didn't like my calculations  []

OK, here we go again, from the top!

Thanks for this. I think our last posts crossed in the ether but my last post identifies the descrepancy as the generation time. There would not be a full 12hrs available to generate. I have allowed a smaller amount of time, i.e., your equation would balance.