[Mootle (OP)]

I understand perfectly well how it works, and I also understand that you have magically made the pontoon more than four time larger to let you reverse into your original, incorrect, power calculation.

The pontoon is now 3.5 times larger than the World's biggest supertanker (er, or whatever Matt wants us to call it.)

This isn't the Tommmy Cooper Show ewe know  []

"The 'reality' is that Mootle put the wrong figures into the calculation - a point that I've made several times. If anything, I find it amusing that people who obviously have an understanding of the principles cannot concede a point when the error has been clearly identified."

So, do you now agree that when the correct volume figures are used for the Pontoon the energy calculation balances and that the power rating is also correct?

I find it hard to see how anyone following the thread would draw the conclusion that I have used the wrong figures since my calculation has not changed throughout. I cannot understand why you are reacting the way you are, as it is easy to see why the misunderstanding took place. However, I would once again apologise for my part in not making the schematic video as clear as it could have been.

Fyi, I have stated several times throughout the course of this thread that the schematic animation is to demonstrate the principle only. I'm working on the scaled animation and this thread has taught me the need to emphasis certain key aspects. Furthermore, the 25:1 gearing ratio is mentioned toward the end of the scaled animation and I also mention the 25:1 gearing ratio when setting out the revenue calculation for the scaled version, earlier in this thread.

Since you correctly identified that increasing the gearing would result in a larger Pontoon I did not feel the need to show all the intermediate steps, although in hindsight this would have been beneficial.

[Geezer]

What I agree is that you never even bothered to look at what BC and I posted (despite repeated requests), because, if you had, you would have had ample opportunity to object to the numbers we were using.

If you were proposing a displacement of 1.7Gt, why didn't you simply point that out five or more days ago? Did I hide that information in my calculation? Was the pontoon displacement never discussed? In fact, up until now, you have not provided ANY value for the displacement of the pontoon, despite the fact that it is critical in determining the amount of work done. 

I don't suppose that's because you only just worked it out based on the information we gave you?

EDIT: Correction - that should be 1.7Mt (not 1.7Gt)

[Mootle (OP)]

What I agree is that you never even bothered to look at what BC and I posted (despite repeated requests), because, if you had, you would have had ample opportunity to object to the numbers we were using.

If you were proposing a displacement of 1.7Gt, why didn't you simply point that out five or more days ago? Did I hide that information in my calculation? Was the pontoon displacement never discussed? In fact, up until now, you have not provided ANY value for the displacement of the pontoon, despite the fact that it is critical in determining the amount of work done. 

I don't suppose that's because you only just worked it out based on the information we gave you?

EDIT: Correction - that should be 1.7Mt (not 1.7Gt)

The truth of the matter is, those were your calculations and I pointed out (albeit indirectly,) that you had used incorrect values, stating that the figures balanced on several occasions in the knowledge that you could self-correct. I try to avoid redoing other peoples calculations, preferring to guide toward an answer - I find this more respectful. Having reviewed the thread I could have spelled things out to you in a more step by step way but in fairness I did make a real effort to reply fully to all the posts, even though some of yours came with a distinct sarcastic undertone.

Sadly, it looks as though you will never concede the points on capacity or total energy despite the evidence supporting this prognosis, preferring instead to making unsubstantiated speculation as to my understanding of my own idea despite ready acknowledgement of any contributions that I hadn't already considered.

I really don't understand why it is so hard for you to concede these points, nor why you are so quick to be disrespectful but eh oh - each to their own!

[Geezer]

Mootle, old bean, if you look back down the thread you will see I said this;

"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."

I also asked you several times to point out any errors in my calculation.

Now you are saying you knew all along that I should have been using a displacement of more than four times that amount and a ratio of 25:1?

I really hope you didn't know that all along, because if you did, and you didn't bring it up, you were simply being a troll. I will give you the benefit of the doubt and assume that you had no idea what the displacement was until you recently figured it out.

[Johann Mahne]

Mootle,
What's the reason for not building a scale model?

[Mootle (OP)]

Mootle, old bean, if you look back down the thread you will see I said this;

"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."

I also asked you several times to point out any errors in my calculation.

Now you are saying you knew all along that I should have been using a displacement of more than four times that amount and a ratio of 25:1?

I really hope you didn't know that all along, because if you did, and you didn't bring it up, you were simply being a troll. I will give you the benefit of the doubt and assume that you had no idea what the displacement was until you recently figured it out.

More insults, whilst it would be amusing to reply in kind I refuse to lower myself to that level. Most people reading the thread back will conclude that your comments are unjust and uncalled for. Furthermore, your refusal to acknowledge that the power and energy ratings that were stated were infact correct is just a measure of the Geezer. It would also be jolly decent of you to retract the false accusations regarding violations of the laws of thermodynamics.


A possible explanation for my not realising I needed to spell every step out to you sooner could be that you were reluctant to give your calculation or that I'm not rude or perhaps we're speaking a different language - this video might offer a possible explanation  [].

http://www.youtube.com/watch?v=6D9Kt0sTWy8

Tallyho old bean!

[Mootle (OP)]

Mootle,
What's the reason for not building a scale model?

I have worked up the sketch designs, component selections and costs for a small system (3kW,) which came out at ca. £50k in materials and special pool rental plus my time to build and transport. Having already invested in software and time I think my wife would think me quite selfish to spend this kind of money on my idea rather than the kids college fund. Since the fundamentals of the system are well established I was hoping to get investment for the pilot.

But maybe I should research a design for a micro-scaled model to say power an LED lamp (3W) just to demonstrate the principles. The problem is I would probably have to develop my own turbine and generator set as I couldn't find anything that small on the market.

[imatfaal]

Peter - could you explain the equation you use to get the figure of 2MW - I don't recognize it - but then I probably wouldn't.  It is dimensionally correct - but I cannot quite see the logic behind it.  I would have thought power needed for a pump (or vice vers) would be along the lines of (ignoring efficiency)

power = Δheight * density * flow * gravity -> m * kg.m^-3 * m^3.s^-1 * m.s^-2 -> kg.m^2s^-3

your equation dissects flow into two components cross sectional area and sqrt(2.Δh.g)

[imatfaal]

If you reverse the equation

power = Δheight * density * flow * gravity and put 2MW in as power generated you find you need to move 345000 cubic metres of water by 50 metres every twenty four hours

2,000,000 = 50m * 1000 * 10 * flow
flow = 4m^3 per second

4m^3 * 60 * 60 * 24 = 345600 m^3 per day

[Mootle (OP)]

Imatfaal - I agree, the calculation is less well known as those used by Bored chemist and Geezer but the principles are very similar. I didn't derive the equation, it is something that is commonly used in the field of hydropower. Ignoring efficiency, I would break down the equation as follows: the first element (density * gravity * head) accounts for the static head or pressure.

The balance of the equation Agen*(2*9.81*dLL)^0.5 deals with the dynamic flow, where (2*9.81*dLL)^0.5 is the jet velocity and the product of the pipe area and the velocity is the volumentric flow rate.

Using the figures I gave earlier, you should get a flow rate of:
4.65 m3/s

The product of the two is the theoretical power hence the need for an efficiency. Fortunately, large Pelton Wheel type turbines are efficient energy converters. As a designer you have a choice (limited by the size of the Storage Vessel,) you can have a high power output for a short duration (this might be useful to meet power surges that are predictable,) or a low power output over a longer duration (to meet base load,) or something in between. I opted for the later since this enables the optimum revenue from the Feed in Tariff, a cost effective turbine / generator selection, and it gives time for the ascent and purge phases to occur. For this idea, it is desirable to synchronise the Buoyancy Engine cycle with the tidal pattern. There are a number of key time frames that must be achieved: the Storage Vessel must be purged and ready for the descent phase at the each low tide to allow the maximum energy to be extracted from the tidal rise, the Generation phase must be completed during the tidal fall and give time for the Storage Vessel ascent and purge phases.

For a volume of 67,000m3 this gives ca. 14,400s of flow or 4hr worth of 2MW power generation available twice per day.

[imatfaal]

Peter - OK I worked it out - its just a replacement of the flow by the cross-sectional area multiplied by a rearrangement of the old suvat equation v^2=u^2+2as.

[imatfaal]

Just recap on the economics -
1. your tank that sinks must be able to withstand 5 atmospheres pressure and hold 67000m3 of water - being very generous this tank itself will have to be constructed of about 10k mt of steel and will total about 80k m3
2. to drag this to the bottom with a ratio of 25 to 1 your pontoon which is moving by 2m will have to displace around 2M m3
3. you will need miles of steel wire - the breaking strength of a 300m 42mm steel wire rope is about 900k N but the safe working load is around 200k N

1.  your tank will cost 5 million bucks in steel alone
2.  easiest way to get 2M m3 of floating pontoon is to buy 7 vlccs (large tankers not supertankers) - scrap price of about 20 million bucks each
3.  i seem to remember that each SWR costs us about 5,000 bucks and weighs over a tonne- you will be needing lots.  and these are designed for use above sea not underwater. 

Set up costs are well over 150 million bucks lets say £100M - there is just a lot of steel involved

Wholesale electricity prices last month in the uk were £47 per MWh

Your set up generates 16 MWh per day (on your calcs 2MW for 4 hours twice daily) thats an income of £750 per day

To break even at today's prices and not including interest, labour, maintenance, insurance, and downtime you would need to generate 365 days a year for 365 years

[Bored chemist]

This is going nowhere.
Mootle,
Please tell us the numbers you are working from, in particular,
The volume of the floating pontoon,
Its rise and fall distance.
The volume of the moving tank
The number of pulleys or the vertical range of the moving tank.
The number of tides each day.

That is (more than) enough information to work out how much energy is stored each day.
For the sake of this  bit of work we can assume that water is incompressible and has a density of 1 tonne per m^3. We can also assume, to make the maths easy, that the efficiency of the turbine and generator are 100%.

Then we can have a sensible look at
(1) are you actually ignoring the rules of physics and
(2) are you ignoring the rules of economics.

Incidentally, I think for the record, that your equation is correct, provided that you are calculating the right quantity. My best guess is that somewhere or other we are at crossed purposes.
If you can give us the information above then we can all get a better look at the problem.
Imatfaal,
Your dissection of his equation is right when you say.
"your equation dissects flow into two components cross sectional area and sqrt(2.Δh.g)"

the root 2gh factor is the speed at which water would fall if it dropped down a pipe with no viscous losses.
Multiply that by an area and you have cubic metres per second.
Multiply by density and you get mass per second. Multiply by acceleration and you get force per second (an odd unit, but it's legitimate)
Multiply by distance and you get force times distance divided by time; which is work done/ time which is power.

The formula is OK. I think the values put in as the volume etc need clarification.

[Mootle (OP)]

Just recap on the economics -
1. your tank that sinks must be able to withstand 5 atmospheres pressure and hold 67000m3 of water - being very generous this tank itself will have to be constructed of about 10k mt of steel and will total about 80k m3

2. to drag this to the bottom with a ratio of 25 to 1 your pontoon which is moving by 2m will have to displace around 2M m3
3. you will need miles of steel wire - the breaking strength of a 300m 42mm steel wire rope is about 900k N but the safe working load is around 200k N

1.  your tank will cost 5 million bucks in steel alone
2.  easiest way to get 2M m3 of floating pontoon is to buy 7 vlccs (large tankers not supertankers) - scrap price of about 20 million bucks each
3.  i seem to remember that each SWR costs us about 5,000 bucks and weighs over a tonne- you will be needing lots.  and these are designed for use above sea not underwater. 

Set up costs are well over 150 million bucks lets say £100M - there is just a lot of steel involved

Wholesale electricity prices last month in the uk were £47 per MWh

Your set up generates 16 MWh per day (on your calcs 2MW for 4 hours twice daily) thats an income of £750 per day

To break even at today's prices and not including interest, labour, maintenance, insurance, and downtime you would need to generate 365 days a year for 365 years

Imatfaal - thanks for this, I really need to get some accurate costings.

First of all I would agree that the idea has little scope for value for money if wholesale prices alone are used. Fortunately, government incentives are available for technologies such as this so we can substitute your revenue figure with the ones I gave, i.e., ca. £1m/yr or £2,8k/d including the FIT / RO and Export Values to which I can give references if required. Even so, based on your figure, the platform would need to serve additional functions such as forming a platform for wind turbines, in order to provide a return on investment within a reasonable period of time.

Secondly, whilst I like the idea of using the nearest comparison, I have no reference for the scrap value of a tanker or the Storage Vessel, so will need to research this for myself to gauge the accuracy of the information, but any references your have will be gratefully received. However, there is a lot of scope to value engineer the Pontoon, which in comparison to a tanker would be a far simpler design.

[Mootle (OP)]

This is going nowhere.
Mootle,
Please tell us the numbers you are working from, in particular,
The volume of the floating pontoon,
Its rise and fall distance.
The volume of the moving tank
The number of pulleys or the vertical range of the moving tank.
The number of tides each day.

I have already given all of this information but once more:
The volume of the floating pontoon = 1,675,000m3
Its rise and fall distance = 2m
The volume of the moving tank = 67,000m3
The number of pulleys or the vertical range of the moving tank = 50m
The number of tides each day = 2.

That is (more than) enough information to work out how much energy is stored each day.
For the sake of this  bit of work we can assume that water is incompressible and has a density of 1 tonne per m^3. We can also assume, to make the maths easy, that the efficiency of the turbine and generator are 100%.

Sea water density is 1,025kg/m3
Generator efficiency is taken as 85%
Refer to calculation given earlier.

Then we can have a sensible look at
(1) are you actually ignoring the rules of physics and
(2) are you ignoring the rules of economics.

Incidentally, I think for the record, that your equation is correct, provided that you are calculating the right quantity. My best guess is that somewhere or other we are at crossed purposes.
If you can give us the information above then we can all get a better look at the problem.

That would be very useful - thank you.

[Geezer]

This is going nowhere.
Mootle,
Please tell us the numbers you are working from, in particular,
The volume of the floating pontoon,
Its rise and fall distance.
The volume of the moving tank
The number of pulleys or the vertical range of the moving tank.
The number of tides each day.

That is (more than) enough information to work out how much energy is stored each day.
For the sake of this  bit of work we can assume that water is incompressible and has a density of 1 tonne per m^3. We can also assume, to make the maths easy, that the efficiency of the turbine and generator are 100%.

Then we can have a sensible look at
(1) are you actually ignoring the rules of physics and
(2) are you ignoring the rules of economics.

Incidentally, I think for the record, that your equation is correct, provided that you are calculating the right quantity. My best guess is that somewhere or other we are at crossed purposes.
If you can give us the information above then we can all get a better look at the problem.
Imatfaal,
Your dissection of his equation is right when you say.
"your equation dissects flow into two components cross sectional area and sqrt(2.Δh.g)"

the root 2gh factor is the speed at which water would fall if it dropped down a pipe with no viscous losses.
Multiply that by an area and you have cubic metres per second.
Multiply by density and you get mass per second. Multiply by acceleration and you get force per second (an odd unit, but it's legitimate)
Multiply by distance and you get force times distance divided by time; which is work done/ time which is power.

The formula is OK. I think the values put in as the volume etc need clarification.

It should always come back to work done by the tide against the pontoon. The displacement of the pontoon and the elevation of the tide defines the maximum energy that can be produced.

All the other stuff about pressures and flows is in regard to the extraction of the energy, and there are a bunch of ways of doing that, some of which may be more efficient than others. Whether we use Mootle's "Turbine in a giant submarine" approach, pump water up a hill and let it drive a turbine on the way back down, or use BC's pontoon pulling a string idea, might have some effect on the efficiency, but it won't alter the available energy one bit.

A tidal dam works the same way, except the head is the tidal variation. It has the advantage that it can have an enormous "displacement" with a very simple mechanism. Ignoring ecological implications, the downside is that the head is small which means the turbines are rather chunky.

Not having a good handle on the energy input is a bit like trying to design a coal fired power station without having any idea about how much coal it's going to burn. With any tidal system, the displacement (or water elevated) dominates the design and cost. After that, it's "simply" a question of extracting the energy as efficiently as possible.

A water turbine that's designed for the available flow and head should operate with an efficiency of at least 90%, so, to get a reasonable estimate of the energy output we can avoid a huge amount of futzing around by simply multiplying the work done by the tide on the displacement by 0.9

Any calculation that produces a significantly different result is suspect, not to mention mucked up  []

[Geezer]

I should have added that Mootle really only needs to specify the pontoon's displacement and the tide height.

EDIT: I forgot to factor in the efficiency of the generator! That might knock off another 10% perhaps? 80% efficiency to get to kWh might be more realistic.

[Bored chemist]

Right, now we have an agreed set of data.
The weight of the pontoon is g * rho * V
9.81 m/s/s * 1025 kg/m^3 *  1,675,000m3
So the weight is 1.68 E 10 Newton
It travels up 2 metres on each tide so that's
3.36 E 10 Joules

Lets all agree on something here.
That, if I have got the arithmetic correct, is all the energy that the tide provides to the system, so that's all the energy that could ever hope to get into the turbines and thus to the generator and eventually as electricity. In fact it will be less than that but, since it's not my field, I can't come up with a realistic guess for the efficiency so I will pretend it's 100%. We can always allow for that later.
It does that every 12 hours
i.e. every 43200 seconds

So the mean power is 0.78 MW.

The air conditioning system where I work takes more power than that.
Now you can get that power from this system, but it needs, at the very least a big metal box as the float.
That box has to displace nigh 1.7 million tonnes of water. So it's at least comparable with building some large tankers or buying them as scrap (probably not a great move- they will have been scrapped for a reason). You will need about six of them.
That's roughly $120M

It's never going to pay.

0.78MW
6833 MW Hr per year.
At £47/ MW Hr
£300,000
Lets assume that a bizzare government subsidy pays 5 times the going rate (that's rather more than all your figures added together)
If you  invest $120M or £75M  you could- with no other building costs and no allowance for the other infrastructure like the pulleys , not maintaining it nor building labour.... and with absurdly generous subsidy and assuming 100% efficiency get a return of £1.5M

2%

LOL

[Geezer]

Yes, as I said about a week ago,

A half million tonne pontoon can only generate 250kW (average)

So a 1.7Mt pontoon will only generate 0.85 MW (average) on that basis. I think I was a bit generous with my rounding. 

[Geezer]

Mootle,
What's the reason for not building a scale model?

I have worked up the sketch designs, component selections and costs for a small system (3kW,) which came out at ca. £50k in materials and special pool rental plus my time to build and transport. Having already invested in software and time I think my wife would think me quite selfish to spend this kind of money on my idea rather than the kids college fund. Since the fundamentals of the system are well established I was hoping to get investment for the pilot.

But maybe I should research a design for a micro-scaled model to say power an LED lamp (3W) just to demonstrate the principles. The problem is I would probably have to develop my own turbine and generator set as I couldn't find anything that small on the market.

I actually agree with Mootle! (Watch for the carefully concealed insults.) There isn't much point in producing a scale model of the turbine end because it's already well known what a water turbine will do at various heads and flow rates, so why bother? Also, turbines don't scale well either, so the results could be very misleading. The pulley system should work too. Why wouldn't it?

The questions are to do with the sheer size of all the working parts, their ability to survive and function in a hostile marine environment, and their cost relative to the power generated. Some of those can only be answered by building a model that is large enough to operate in the hostile marine environment.

I don't think there is any question that the "turbine in a submarine" (subturbine? turbmarine? soupterrine?) could not be made to work, so there isn't much point in modelling that part of the system. What could be done is to model the system with a dummy turbmarine that doesn't produce any power. That would help to keep the cost down, a lot.