Hydro Differential pressure exchange over unity system.

[fletcher]

Hi .. I think you'll find your answers in the attached pic I made - it is a representation, & I used 'g' at 10 m/s^2 just for ease of calculations readability.

In essence it's a tortoise & hare race story - lifting slowly & surely with very little fall gradient uses least/minimum Work Done Joules by you to give the system PE J's - quick is costly because you have to lift higher to get fast transfer.

NOTE that the system cannot gain PE without a minimum of the SAME expenditure of Work energy it took to raise the system PE - SLOW is the most optimal & efficient method to achieve this efficiency, IMO.

[fletcher]

Also, a quick question on this.

I thought, thanks to wikipedia and other sites I've found to help me understand (a forum on physics is pretty good), that buoyancy was based on volume of the displaced water. I know we've emptied 50kg of water into the tank, but that's not what's displaced.... is it?

I thought that the initial, soon to be rapidly decreasing, buoyancy force would have been the water density * volume of the cylinder (as it's totally submerged at this point) * g would mean 1000 * .44426 * .44426 * .95 * 9.81 = 1839.36 N over 0.2m = 367.87 J. Even if we took an average between the start and stop point (0.2m), it would be: 329.15 J.

If we link in MT's idea of having a bit more water in the blue tank to start with, and as we release the pod lift the blue water tank a little further to empty the remaining contents and keep the keep the buoyancy constant, are we getting close to the OU tipping point?

I've been trying to do more drawings but it's frustratingly slow. They should be worth more than 1000 words for the effort it takes

Amo

Now, that's why we do experiments to find these things out.

Buoyancy is a factor of water volume displacement [weight of water] & the Work the NET Upthrust force can do is dependent on the stroke length [ WD = f x d ] - for an Archimedes system this is easy to visualize - in this case we are using 'virtual water displacement' due to the pod taking up volume - so the pressure gradient is still created on which buoyancy force is predicated but it is not supported by 'true water volume' to complete the stroke length otherwise - IOW's, it is very powerful for a very very short time & length of which both drop of exponentially IINM [high acceleration potential for low inertia systems] - IMO, the total Work Done capability being identical or thereabouts.

Sounds like its time to get your hands wet.

[MT]

Hi guys,
attached is updated spreadsheet v2.1. Before opening it, rename file extension to xlsx so it opens correctly. Checked but could not find further errors in it. Still getting COP>1, see for yourself. COP>1 is not just for a specific dimensions of cylinder and pod, dims can be in certain range and still be over 100%. How is the work calculated is described in my previous post.

Interesting (also a bit ironic after spending quite some hours in Excel) was that when I finished checking the sheet I realized we can compute work done on pump water much easier.
Using the dimensions of cylinder and pod as in the sheet:
Precharge step needs 28 liters raised up to 0.9m
Stroke step needs 78 liters raised up to 0.1m + we need to lift the precharge 28 volume 0.1m


precharge work = m*g*h = 28 * 10* 0.9 = 252
stroke part 1 = 78 * 10 * 0.1 = 78
stroke part 2 = 28 * 10 * 0.1 = 28

total work 252 + 78 +28 = 358
Guess what that 358 is the same work that is relatively complicatedly computed in spreadsheet, cell F84.

respect,
Marcel

[neptune]

Hi Marcel. I tried to work this out in my own way. You say that precharge work is m*g*h and your answer is in what units? I got it to 24.93 kilogram metres.

Work for stroke. See my last post . We need to raise 75.5 litres a distance of 0.2 metres which comes to 15.1 Kg metres.
Total input is 24.93 + 15.1 =40.03 Kg metres 


Volume of pod is 678 Litres. Upthrust is 678 Kg


Stroke is 0.1 metres . Output is  67.8 Kg Metres.


COP is 67.8/ 40.03 = 1.693.

[neptune]

Here is my set up for the above calculations. We have a main tank [A] , containing a pod, same dimensions as your . This tank has two connections fitted into the sidewall. |One is level with the top of the pod, and one os near the bottom. We have a second tank [C] this tank is 0.9 metres tall, and contains enough water for precharge. It is connected to the bottom connector of tank A by a flexible hose . To precharge we lift this tank 0.9 meters. This tank very shallow, and large in diameter.
To stroke, we have a Third tank , one meter diameter and 0.1 metres deep, connected by a flexible hose to the top connector of tank A. To stroke we need to raise it 0.2 metres , cso the bottom of tank B is level with the top of tank A.


I just realised my initial mistake. I assumed that we needed to raise tank C by 0.9 metres . If tank C is 0.9 metres tall we would need to raise it 1.8 meters.  But if we made it only 10 one   centimetre tall, we would only need to raise it one meter to precharge .


So now energy to precharge is 27.7 Kg metres.


So now total input27.7 +15.1 =42.8 Kg metres


So COP is now 67.8 /42.8 = 1.584.


If I am wrong, please tell me where.