Solving the Dilemma of Moving Water Without External Energy

  • Thread starter Thread starter CIVILian
  • Start date Start date
  • Tags Tags
    Water
AI Thread Summary
Moving water from a lower tank to a higher tank without external energy is challenging due to gravitational constraints. A siphon cannot elevate water above the level of the source tank, and while a gravity pump could theoretically move some water higher, it is inefficient and requires sacrificing some water to achieve this. The discussion emphasizes that without external power, achieving a higher water level in the second tank is impossible. Suggestions include using a gravity pump or a powered pump, but the original challenge is to avoid external energy sources. Ultimately, the most viable solution appears to be constructing a crude gravity pump.
  • #51
brewnog said:
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.
 
Engineering news on Phys.org
  • #52
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.
 
  • #53
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.. :)
 
  • #55
What is the significance of the "20 secondish" time frame? There are easy solutions if you take your tiem.
Freddie
 
  • #56
Do you rule out a really tall but skinny supply tank and a really, squat but fat receiving tank?
 
  • #57
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.
 
  • #58
Where do you intend to get the energy to power your compressor/bike pump?
 
Back
Top