# Thread: Your ideas on energy-efficient communication between vessels

1. In the sketch below i propose a challenge to find solutions, brainstorm and evaluate upon.

One purpose is to see how much better we can do than 500 Liters from A to B.
Another purpose later on, is to check if the best solution could also be verifiably energy-efficient,in for instance a concept to feed a turbine.

(No intention on my part here to blabla on perpetuum mobile stuff, or to defy science,
just a friendly piece of infotainment that may or may not become relevant as we elaborate it.)

Feel free also to correct any mistakes or incompleteness in the setup.

List of added info and remarks due to questions asked below :
-The containers must stay in place
-The legs must stay in place

2.

3. Waving.

4. Siphons?

5. Join container A and B with a pipe fitted to the sides (at the bottom).
Fit the top 55% of container B with an absorbant material.
That should transfer more than 50% of the liquid to container B.

Or does that count as "making compartments in container B"?

6. Remove the legs from container B and connect a tube from the bottom of container A to container B.

Or does removing the legs count as "using energy"?

7. Originally Posted by RedPanda
Remove the legs from container B and connect a tube from the bottom of container A to container B.

Or does removing the legs count as "using energy"?
Removing the legs is not allowed, the containers must stay in place.

The legs are merely intended to let the area under the containers be used.

8. Originally Posted by RedPanda
Join container A and B with a pipe fitted to the sides (at the bottom).
Fit the top 55% of container B with an absorbant material.
That should transfer more than 50% of the liquid to container B.

Or does that count as "making compartments in container B"?
Nothing is to be inserted on the inside of container B , other than the water of A using no extra energy.

But absorbtion is in interesting idea to keep in mind i think.

9. I,ll post my suggested solution om monday, under the form of a small sketch with info, to be assessed.

10. Ok, how about this then:

Connect a tube between container A and Container B.
Get a plastic rectangular cylinder which is 100m tall and fits tightly into Container A - but has a very narrow hole throught its centre.
Put the rectangular cylinder into container A.

11. segment A as much as possible, with a seperate valve fed from the bottom of each segment feeding a manifold and then to a single pipe into bottom of B with a check valve stopping flowback from B.

Drain each segment of A one at a time, then shut off the valve for that drained segment
when done
there will be more fluid in b than in a

12. Join container A to Container B with a tube.
As the water drains out of Container A, add pebbles to Container A.

13. Originally Posted by sculptor
segment A as much as possible, with a seperate valve fed from the bottom of each segment feeding a manifold and then to a single pipe into bottom of B with a check valvestopping flowback from B.

Drain each segment of A one at a time, then shut off the valve for that drained segment
when done
there will be more fluid in b than in a
Yes Sculptor, that is my line of thought also. You can get upto 80-90% of the water to container B with a construction along a similar idea,
yet not bringing the water to the bottom of B.

14. Originally Posted by Noa Drake
Originally Posted by sculptor
segment A as much as possible, with a seperate valve fed from the bottom of each segment feeding a manifold and then to a single pipe into bottom of B with a check valvestopping flowback from B.

Drain each segment of A one at a time, then shut off the valve for that drained segment
when done
there will be more fluid in b than in a
Yes Sculptor, that is my line of thought also. You can get upto 80-90% of the water to container B with a construction along a similar idea,
yet not bringing the water to the bottom of B.
Really?
As soon as you have more water in B than the compartment in A why would it flow from A to B?
The pressure of water in B would prevent any flow from A to B, no?
Especially as B, not being compartmented, would have a larger surface area = more air pressure.
I would have thought (after giving it not much thought) that the maximum you could transfer just using this way be around 50% (and that'd probably be without compartments).
As soon as you get matching quantities there's nothing to cause flow either way.

15. Originally Posted by Dywyddyr
Really?
As soon as you have more water in B than the compartment in A why would it flow from A to B?
The pressure of water in B would prevent any flow from A to B, no?
Especially as B, not being compartmented, would have a larger surface area = more air pressure.
I would have thought (after giving it not much thought) that the maximum you could transfer just using this way be around 50% (and that'd probably be without compartments).
As soon as you get matching quantities there's nothing to cause flow either way.
But if there were multiple vertically stacked compartments (with tubes and valves, etc.) and you released (and then sealed) each valve in turn (going from bottom to top) then you shouldn't you end up with more water in B than A?

To keep the maths simple (for my sake, not yours) let's just imagine 2 vertically stacked compartments - let's call them C1 and C2 - and between them they contain all the water in Container A.

Opening the valve in C1 would cause 50% of the water in C1 to transfer to Container B. 50% of 500 litres = 250 litres.
Now close the C1 valve.
Opening the valve in C2 would cause 75% of the water in C2 to transfer to Container B. 75% of 500 litres = 375 litres.
In total, Container B will have 625 litres.

Or have I missed something important?

16. Once the compartment setup has run it's course:

Boil the rest of the water in the top container of A with a large conical reflector and sunlight by connecting a small diameter tube to the bottom of the top and let it run through a radiator on it's way to container B. You can also do this from the start and you'll purify all the water in the process as well.

17. Originally Posted by KALSTER
Boil the rest of the water in the top container of A with a large conical reflector and sunlight by connecting a small diameter tube to the bottom of the top and let it run through a radiator on it's way to container B.
I expect that breaks the "No energy except gravity" rule.

18. Originally Posted by RedPanda
Originally Posted by KALSTER
Boil the rest of the water in the top container of A with a large conical reflector and sunlight by connecting a small diameter tube to the bottom of the top and let it run through a radiator on it's way to container B.
I expect that breaks the "No energy except gravity" rule.
I expect it does.

19. Originally Posted by sculptor
segment A as much as possible, with a seperate valve fed from the bottom of each segment feeding a manifold and then to a single pipe into bottom of B with a check valvestopping flowback from B.

Drain each segment of A one at a time, then shut off the valve for that drained segment
when done
there will be more fluid in b than in a
You still can't transfer more than half the water. The first of your compartments will empty almost completely because the amount of water it contains will only cause a small rise in B. The next one will drain a little less. ... The last one will drain half way, but that will be a small amount of water because the segments are small.

Sum up the total amount of water and it will be <= 50%.

20. Originally Posted by RedPanda
Or have I missed something important?
I was about to point out the obvious reason that doesn't work. But then I realised it does!

I *think* this might (slowly) approach 100% with an infinite number of compartments. (But, of course, it is not reversible.)

21. strange
Did i forget to mention that A is segmented by vertical walls
when you open the last valve on the last segment of container A, the water in that compartment will still be at original height, and therefre have more pressure than the water height of container b

Think height of water column, and not volume.
And, ignore air pressure.

as you progress, each subsequent virtical segment of container A will also have a higher level than the preceeding segment(s).
RP got a good rough start on the math.

The difference in final volume of water in B vs A is directly related to the number of verticle segments in conatiner A

22. Using my 2 compartment example:
Originally Posted by Strange
The first of your compartments will empty almost completely because the amount of water it contains will only cause a small rise in B.
Won't the first compartment empty half of it's water?
i.e. 50% of 50% of 1000 litres?
250 litres?

Originally Posted by Strange
The next one will drain a little less. ...
The next compartment (2) is higher than both compartment 1 and the water level of Container B.
It would require 250 litres to bring the level of Container B up to meet the bottom of Compartment 2.

The remaining water in Compartment 2 would then distribute evenly between A and B.
(500 - 250) / 2 = 125 litres would transfer to Container B.

Remember, Compartment 1 has been sealed - water can't run back into it when you empty Compartment 2.

23. Originally Posted by RedPanda
Using my 2 compartment example:
Originally Posted by Strange
The first of your compartments will empty almost completely because the amount of water it contains will only cause a small rise in B.
Won't the first compartment empty half of it's water?
In your vertically stacked version, yes. The next will be 1/2 + 1/4.

If you have 4 compartments, then the amounts will be 1/2 + 3/4 + 7/8 + 15/16 (I think). So the total transferred is (1/2 + 3/4 + 7/8 + 15/16)*250 = 766 litres.

24. Here's my suggested solution to bring 80-90 percent of the water to B.

-Make in advance 10 horizontal 'bassins' in A, using horizontal compartiments
-Connect a tube from each bassin of A to B horizontally, except no tube at the lowest bassin.
(The tubes at the A bassins would have to be connected at the right down flank of each A bassin.)
-Start valving of each A bassin towards B, starting with the second bassin (! The lowest bassin is left untouched)
So the natural pressure in bassin A will make the water flow to the lowest level available, being container B, at a level just below the level of bassin A.
-Continu until you have transferred the water of 9 A bassins into B.

Note : At each stage you valve off the right tube connection at compartiment B after transferring the water.
Nota also that an air ventile would be necessary at the topfank of each A bassin, otherwise it would vacume itself while transferring the water.

Does this make sense ?

(Maybe you could even increase the result if you made even more bassins, but this would lead to a point of decreasing positive effect)

Images : START POSITION > TRANSFERRING > END POSITION

25. Originally Posted by sculptor
Did i forget to mention that A is segmented by vertical walls
No.

when you open the last valve on the last segment of container A, the water in that compartment will still be at original height, and therefre have more pressure than the water height of container b
But by this time, the water in B will be at (almost) 50%. So the water in this compartment of A is twice the height of B. So half the water will flow out of the compartment. But, because the compartments are small, this is a small amount of water.

Think height of water column, and not volume.
I am.

Unless I have misunderstood your scheme. I am assuming it is something like:
||||||||||||||||||||
||||||||||||||||||||
||||||||||||||||||||
||||||||||||||||||||
||||||||||||||||||||
--------------------

Whereas, Red's is more like:
|----------------------|
|----------------------|
|----------------------|
|----------------------|
|----------------------|
------------------------

26. Originally Posted by Noa Drake
Here's my suggested solution to bring 80-90 percent of the water to B.
Very good. That transfers slightly less(*) than RedPanda's method but is simpler to understand (always a good thing in my book!)

(*) With 8 divisions, his will transfer 87.55% while this one will only do 87.5%

27. Originally Posted by Strange
... but is simpler to understand (always a good thing in my book!)
Pish!
I claim victory!! :P

28. Interesting approach RedPanda and Sculptor

29. RP wins

30. Originally Posted by RedPanda
Originally Posted by KALSTER
Boil the rest of the water in the top container of A with a large conical reflector and sunlight by connecting a small diameter tube to the bottom of the top and let it run through a radiator on it's way to container B.
I expect that breaks the "No energy except gravity" rule.
Gravity is responsible for compressing the hydrogen in the centre of the sun to cause fusion that leads to the emission of the sunlight. So, you should be OK.

31. Ok, let's assume theoretically that we can transfer, by anyone's method, 95% of the water from A to B.

Imagine placing a lever between the vessels, vessel B being a harmonica structure or compressable bag of a sort.

On top of the lever above vessel B we put a weight of ? KG.

>Vessel B compresses and pumps up the water.

>Vessel A goes up

>> Can we refill vessel A to a certain level, so it will go back down and we can retransfer the water from A to B, and repeat the sequence ?
Baring in mind that we would have to 'help' it with extra water or extra torque.

The question would be : Just how much help would be needed to have an interesting energy-efficiency going on.

Many such levers placed in sequence along side an axel could provide torque, for any purpose.

>> My question is to see if we can create some sort of what i would call a Bicycle-effect :

Compare the performance and labour needed for a running man on foot, to a man using a bycicle, to obtain a certain steady velocity : it is a very big difference.
The man 'helps' the incomplete perpetuum mobile (the bycicle) to keep it going if you will (Damn, i wasn't going to use that word..,),)

In parallel just how much help would the 'lever-water-motor' need in comparison to it's usefull output ?

A quick very simple sketch (incomplete)

32. Of course the simplest answer is to have a cube of heavy material that fits almost exactly into container A.
And merely let gravity pull it to the bottom: it will displace all of the water (apart from a neglible film on all surfaces) as it sinks.

33. Originally Posted by Noa Drake
The man 'helps' the incomplete perpetuum mobile (the bycicle) to keep it going if you will (Damn, i wasn't going to use that word..,),)

In parallel just how much help would the 'lever-water-motor' need in comparison to it's usefull output ?
Give it up. You can't beat the conservation of energy principle. Many clever inventors have tried and failed.

34. Originally Posted by Noa Drake
The question would be : Just how much help would be needed to have an interesting energy-efficiency going on.
The tricks used for moving more water from A to B are, by definition, non-reversible. So it makes no difference to the amount of energy that can be extracted from the system.

Laws of thermodynamics - Wikipedia, the free encyclopedia

It's not just a good idea; it's the law.

35. No intent to defy the laws, i'm looking to use a small continuous extra force applied to keep it going, as illustrated with the bicycle.

A remark on reversing the transfer.

If you install compartments in conrtainer B and a set of 'B to A' tubes and valves,
then you can transfer 95% of 95% of the water back to A.

And if you first put the residual 5% water from A into B, then it's a 100% back transfer.

But what you imply is that the input needed for putting the 50 liters up into B, is equal to the gain from transferring 950 liters from A to B.
Minus even som losses due to friction etc.

Yes ?

36. Originally Posted by Noa Drake
And if you first put the residual 5% water from A into B, then it's a 100% back transfer.
Oh boy...

37. Originally Posted by sculptor
RP wins
no he doesn't!

Move one of my legs into container "B"

perfect transfer (running for the hills)

38. Originally Posted by John Galt
Siphons?
My thought also, just like kids transferring gas from one vehicle to another.
Is atmospheric pressure related to gravity?

39. Originally Posted by Dywyddyr
Originally Posted by Noa Drake
And if you first put the residual 5% water from A into B, then it's a 100% back transfer.
Oh boy...
Meaning 100% of 95% there, not transferring back 100% of the water from B to A of course

40. Originally Posted by Noa Drake
Originally Posted by Dywyddyr
Originally Posted by Noa Drake
And if you first put the residual 5% water from A into B, then it's a 100% back transfer.
Oh boy...
Meaning 100% of 95% there, not transferring back 100% of the water from B to A of course

41. Why would a siphon not be able to transfer all the water?

You take the water from a hole in the bottom of A and add it to the top of B, by means of vacuum tubing. The energy is supplied by downward air pressure (due to gravity) on the surface of the water in A.

Even if, after transfer, any water is left in the tubing, one can then lift the tubing to drain the remainder of the water into B. 100% transfer.

42. Originally Posted by Write4U
Why would a siphon not be able to transfer all the water?
Because, in a syphon, the outlet has to be below the level of the water in A.

43. Originally Posted by Strange
Originally Posted by Write4U
Why would a siphon not be able to transfer all the water?
Because, in a syphon, the outlet has to be below the level of the water in A.
I can see this in a purely gravitational transfer, but does that hold with a vacuum system as well?

From what I remember of transferring gasoline from one vehicle into another, all we did was suck on the hose until the hose was full with gasoline, then capping it with the thumb to keep the hose filled and sticking it into the gastank of the other vehicle. This never failed even if the other tank (B) was higher than the original.

It was my impression that by draining the hose a vacuum was created in the hose strong enough to "suck up" (due to atmospheric pressure), an equal amount from tank A. As soon as the vacuum in the hose was broken the transfer would stop, thus my assumption of a vacuum effect.

Similar to the sucking a hardboiled egg into a bottle from which air was removed by burning a little piece of paper. It makes no difference if you hold the bottle upside down, the egg always gets pushed into the bottle.

I visualized a similar arrangement might work in this problem as long as a vacuum can be maintained inside the tube.

Perhaps I am missing an important point here and frankly I never did pay attention to the relative depths of the hose ends although I seem to recall that we always had to keep the outlet above the gas level in tank B or the transfer process would slow down or stop.

44. Originally Posted by Write4U
Perhaps I am missing an important point here .
Yep. The outlet must be lower than the fluid level.

45. Originally Posted by Strange
Originally Posted by Write4U
Perhaps I am missing an important point here .
Yep. The outlet must be lower than the fluid level.

ok, thank you.

46. Originally Posted by Strange
Originally Posted by Write4U
Perhaps I am missing an important point here .
Yep. The outlet must be lower than the fluid level.
Pour me some coffee....I just got a chuckle out of this

47. I believe I have found a way to transfer 100% from container A to container B. We need to make a slight modification to the containers and instead of two square containers, we use one cylinder of equal volume as the original containers.
At the bottom of each we attach a valved tube as well as an additional valve at the bottom of container B.
If we just open the valves 50% of the water will flow from A to B (achieve equal levels). But now we lower a sealing piston (of sufficient weight) into cylinder A which will force all the water from A to B, except for the remaining water in the tube between the cylinders. We close the valve at the bottom of B and both valves at the tube ends. Then we detach the tube from both cylinders and tilt the tube downward into cylinder B, open the valves and the remaining water will flow into B. 100 % transfer, using the criteria of the problem except for the shape of one container (but of equal volume).

Theoretically, all we need do is force the water out of the cylinder A with the gravitational force of a heavy piston, at which time we can transfer the water to any container, at any reasonable height when sufficient gravitational pressure is applied to the surface of the water in A.

A very simple design, but 100% effective. IMO

48. Write4U,
You violated the rule of no external energy source.

49. Originally Posted by Harold14370
Write4U,
You violated the rule of no external energy source.
I understand your point, but how does my design violate the instruction that gravity can be the only source of energy? The instructions allow for modification. We can assemble tubes, valves, compartments to the set-up. And (as I stated) the piston is not powered by anything other than it's gravitational weight.
But I guess the intent was to use the water's gravity only and another gravitational source is not allowed.
But it would work, no?

50. Originally Posted by Write4U
Originally Posted by Harold14370
Write4U,
You violated the rule of no external energy source.
I understand your point, but how does my design violate the instruction that gravity can be the only source of energy? The instructions allow for modification. We can assemble tubes, valves, compartments to the set-up. And (as I stated) the piston is not powered by anything other than it's gravitational weight.
But I guess the intent was to use the water's gravity only and another gravitational source is not allowed.
But it would work, no?
If you want to insert a piston of sufficient weight, then you must first lift this weight above tank level, this requires external energy input.

51. Just reading along.....can't contribute.....just learning a little

52. Originally Posted by Noa Drake
Originally Posted by Write4U
Originally Posted by Harold14370
Write4U,
You violated the rule of no external energy source.
I understand your point, but how does my design violate the instruction that gravity can be the only source of energy? The instructions allow for modification. We can assemble tubes, valves, compartments to the set-up. And (as I stated) the piston is not powered by anything other than it's gravitational weight.
But I guess the intent was to use the water's gravity only and another gravitational source is not allowed.
But it would work, no?
If you want to insert a piston of sufficient weight, then you must first lift this weight above tank level, this requires external energy input.
Ok, I had not considered that additional aspect.

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