Solving the Dilemma of Moving Water Without External Energy

[brewnog]

I think I understand what you're getting at, could you redo the diagram and show more clearly which pipes are connected to what, and which way your check valves operate?

 

[Artman]

@ lurch and Artman

Again this is external energy.
Being the weight of the lid and the stone (or whatever) on top of the lid.
The case is to NOT use external power.

I know it doesn't qualify. I still like it though.

 

[Marijn]

in that case the isdea of sealing it and blowing 2000 psi air on top is a helluve lot more fun :D

 

[BobG]

There is no way to do the job with no external energy. Do you mean no electrical/gasoline energy, perhaps?

Using weights to push the water through is a good idea, but I'd make a small improvement to it. Build two pistons, one on each side of a see-saw above the tanks. Hold the bowling balls (or cannon balls, or other incredibly heavy rolling objects since you're attempting to move 1000 kg of water) on the high side of the see-saw with a board. Remove the board, the balls roll to the low side, pushing the low piston and pulling the high piston (via torque) at the same time.

You hold one side of the see-saw high with a wedge. Lubricated rubber around a wooden disk should be sufficient for the pistons. You don't need a perfect vacuum, you just need the pressure to be very low in the high tank and very high in the low tank.

Oh, and be careful balancing that last bowling ball on the end of the see-saw. It can hurt your foot.

 

[Artman]

in that case the isdea of sealing it and blowing 2000 psi air on top is a helluve lot more fun :D

Absolutely! Of course it has to be done scientifically. Like counting the number of seconds it takes from the time the lid flies off till it strikes the ground (or a passerby) then calculating the height it achieved.

 

[Averagesupernova]

Your first post says something about not being able to use a siphon. But it has been discussed at length. Assuming you are permitted to use a siphon, your drawing is accurate, and contrary to popular discussion in this thread, you could move about half the water with a siphon. Just run the hose into both tanks all the way to the bottom. Assuming the hose was full of water when inserted it will level off the 2 tanks with half in each.

 

[BobG]

This is the problem in case my explanation hadn't been sufficient.

Yes, your explanation wasn't sufficient.

I have a question about the drawing in Post #27. Where does the water go after it goes into the inlet? If the eventual destination is below the first tank, then a siphon is the answer, provided you get a good seal between the pipe and inlet of the destination.

It makes no difference where the inlet is. The only thing that matters is where the water eventually winds up.

A siphon is a lot more boring than the other solutions proposed, however.

 

[Civilian2]

Averagesupernova and Bobg: No I can't use a syphon unfortunately.

Brewnog: Doing a better diagram is difficult because the forum only allows me to draw in a 400x400 canvas. The red arrows show the direction of the flow, the valves stop the water going back the other way- out of the system.
PVC pipe coming out of the water tank outlet would be fed into the inlet in my 'gizmo' via the bucket. This pipe would have a hole in the side so when the system is full, water will spew out the side which would then enter the bucket, weighing it down.
When the piston is pushed down by the bucket, water would be pushed through the lower of the two bottom horizontal pipes (I drew them above each other for diagram purposes but in reality they would both be on the ground) and up the vertical pipe that I have labelled outlet into the collecting inlet. (not in the gizmo diagram but in the original problem diagram).

Bobg: I'm not attempting to move 1000 kg of water. I'm trying to move as much water as possibly in 20-30 seconds. 10 mL would be enough! As long as I could move some! At the moment, I'm doubting whether I"m going to manage 1mL.
Using bowling balls, I think I'd be using the gravitational potential of the balls, rather than the water which would defeat the purpose of the exercise.

Does my system look feasible? I just want to move SOME water. I don't want to buy all the materials to find it doesn't work, and I don't really know how to work it out theoretically.

 

[BobG]

First, the drawing problem. I did mine in power point, and then copied the picture into Paint. Just about every computer has that program installed on their computer when they buy it.

If you have paint, once you copy your picture into it, select "Image", then "Attributes". Change your units to pixels. That shows you the size of your picture.

Mine was about 716x496. Not wishing to spend a lot of time on this, I just shrunk it by half. Select "Image", then "Stretch/Skew". 'Stretch' both the horizontal and the vertical by 50%. The result is a 358x248 picture, small enough to past on the board. The picture probably has to be saved as a GIF or the file size will be too big (this might mess up your colors, but that's life).

Second, I think your general idea with the bucket might be close (it's always hard to visualize your explanation). If you look at the bowling ball picture, you could reverse the angle of the see-saw (empty tank lower - full tank higher) and hang an empty bucket or series of buckets from the high side. You'd need two outlets from the full tank. One to fill the buckets - as the bucket became full and began to lower the see-saw, the next bucket would drop low enough to be filled, and so on. The other outlet is connected to a pipe or hose so the water can be drawn through by the lowering pressure on the empty tank.

If the water will flow out of the full tank into the bucket(s), it's going to lower the pressure, forcing the piston down (plus the force of the extra weight in the bucket from the water), but getting the water to start flowing out of the outlet in the first place might be a problem - something has to displace the volume of the water leaving the outlet. I'm thinking at least some air will run into the outlet to get the water started, and once the whole piston scheme has been set into motion, the water flow increases and you have no further problems.

It might just be better to leave the top of the full tank open and rely solely on the piston creating a low pressure in the empty tank.

Edit: Obviously, both tanks are going to have to be elevated so the buckets have room to lower.

 

[Civilian2]

Oh I see what you are getting at Bobg. My piston idea involved pistons entirely separate from the tanks, not turning the tank into a massive piston. I might find it difficult to get an airtight disc to fit the tank itself perfectly. But I like the idea. I've got a mate of mine calculating the water flow rate etc, and friction in the pipes etc which should make this easier to develop.

I'll have a hard think about this this evening.

As for my first piston idea though I am beginning to think it will be impossible for me to get the piston to first rise and suck water into the piston before the bucket fills and pumps it down so I am considering alternatives. I have started to consider that I don't need to fill the piston with water at all. If I have large volume in my piston and a very small volume in the pipe which is full of water, then perhaps even though the air into piston would compress, it would still be able to force water through the pipe and up into the higher tank.

I've also realized that I can create larger images if I don't upload them to this site, but upload them elsewhere and link them. So I'll get around to doing that.

Again I appreciate the help mate. Lol, I found it quite funny that you put a man in your bowling ball graphic haha. Are you sure that's where I'm meant to stand when it all happens?

 

[LURCH]

Something I've not seen mentioned yet, can you use the weight of the tank itself? Maybe the water could be drained into a narrow tube, and the weight of the (now empty) tank could press down on that tube, forcing the water up.

 

[Random]

Hydraulic Ram Pumps

I'm not attempting to move 1000 kg of water. I'm trying to move as much water as possibly in 20-30 seconds. 10 mL would be enough! As long as I could move some! At the moment, I'm doubting whether I"m going to manage 1mL.

I had thought this was some sort of logic impossi-puzzle until I read this. Your answer is a hydraulic ram (sold as a "Platypump" in Australia). These devices are not very efficient but are excellent for raising water well beyond the supplied head pressure. For every 10 volumes of water past the pump, 1 volume can be raised to approximately 100 times the supply head pressure. (Great for topping a tank from a lower creek etc.) Hope this helps. Google search the hydraulic ram & you will find plans etc enabling you to build your own for only a few dollars.

 

[brewnog]

Yup, that's a gravity pump!

 

[Artman]

I had thought this was some sort of logic impossi-puzzle until I read this. Your answer is a hydraulic ram (sold as a "Platypump" in Australia). These devices are not very efficient but are excellent for raising water well beyond the supplied head pressure. For every 10 volumes of water past the pump, 1 volume can be raised to approximately 100 times the supply head pressure. (Great for topping a tank from a lower creek etc.) Hope this helps. Google search the hydraulic ram & you will find plans etc enabling you to build your own for only a few dollars.

My only worry with either a Ram or other form of gravity pump is that there is not enough height to provide enough head pressure to produce adequate results. Some of the head pressure is lost to operating the pump mechanisms. If the suction side is large enough for a large mass to drop a short distance to move a small mass a longer distance perhaps it could work.

 

[brewnog]

But in this case, even getting a tiny amount of water raised would be considered an "adequate result" by Civilian, - see his earlier posts.

 

[Artman]

But in this case, even getting a tiny amount of water raised would be considered an "adequate result" by Civilian, - see his earlier posts.

I think it is worth a shot. It just depends on how it is done.

Say you were going to drop a full bucket of water on the raised handle of a spoon to catapult a spoonfull of water into the air. If the bucket is dropped on the spoon handle the water will certainly fly, however if the bucket is slowly lowered onto the spoon handle the water will just trickle down the handle of the spoon onto the ground.

 

[Marijn]

You can always make a small piston pump that is driven by a water wheel (in effect a gravaty pump).
The tricky bit would be building the piston and getting it to work (fiddle around with some valves).
This thing should be able to get quite some water up into the other container.
Even when the waterstream on the water wheel is getting weaker, it will still pump.
It will pump to the point where the stream isn't strong enough to turn the wheel anymore.

 

[Random]

Have calculated the efficiency of of Hydraulic Ram. As the question / challenge requires a very much smaller ultimate head than I had been requested.. had to re-evaluate..
Now Supose the initial head is h and the height to be delivered to is H. Where H/8h is not exceeded (certainly so here) efficiency is about 80%
So of the original 1000 litres supplied, I have 800 delivered in a single effort. The 200 lost can be recovered for a subsequent (80% efficient) effort.

{ ie: If W pounds (do you fellas think in CGI units mostly or not.. no matter.. convert till you're happy).. of water decend the supply pipe per second, the work available per second is W times h foot pounds. If e is the efficiency of the ram, the total weight of water that can be lifted is :Total = W times h / e times H.
Think upon it and .. move that water.. :)

 

[brewnog]

>You can always make a small piston pump that is driven by a water wheel <

No need. Take a look at this. It's far more simple than you imagine :-)

http://www.animatedsoftware.com/pumpglos/ram_pump.htm

 

[Freddie]

What is the significance of the "20 secondish" time frame? There are easy solutions if you take your tiem.
Freddie

 

[GENIERE]

Do you rule out a really tall but skinny supply tank and a really, squat but fat receiving tank?

 

[Evil Knievel]

Spraygun solution?

Hi all. Couldn't one possibly use the same principle as a spraygun, where you force an airstream past the mouth of the issuing tank and thus create a local area of low pressure? Of course, this pressure would have to be let off somewhere.

Another alternative may be to pressurise the issuing tank above the water level. This could probably be done with a tyre pump but do not expect much speed.

 

[brewnog]

Where do you intend to get the energy to power your compressor/bike pump?