Pressure in River Pump with No Moving Parts

[Leafhill]

Pump with no moving parts!

Hi!

I'm trying to build a "pump" pushing water a few meters up of the river bank, and wondered if I could do it without ANY moving parts (except water ).

My plan is to mount a tube (say 50cm in diameter) with one end sealed into a wide river. The open end of the tube will be facing up-river into the stream. The river floats at approximately 1 m/sec towards (and around) the tube.

Now, if I mount a small hose to the sealed end of the tube, how high above the water level can I expect the level of water to rise in the hose (IE how much pressure is in the tube)?

To simplify:
Hose cross-section = 10 cm^2 (3.57 cm diameter)
Tube diameter = 50cm (~2000 cm^2)
Tube length = 1 meter (don't think it matters)
River flow velocity = 1 meter / second

I guess we can simplify by disregarding water viscosity and vortexes. Right?

Any help is appreciated!

 

[Bob S]

You are describing the hydraulic equivalent of the pitot tube used on aircraft to measure velocity by static air pressure of onrushing air. Google pitot tube.
Bob S

 

[russ_watters]

...use Bernoulli's equation.

 

[Leafhill]

Thanks guys!
So I looked up Bernoulli and pitot-tube and found this piece of math:

Stagnation pressure = 0.5 * water density * velocity^2 + Pstatic
I'll assume I can ignore Pstatic (Correct me if I'm wrong) and fill in the rest with my original numbers:

Stagnation pressure = 0.5 * 1000 * 1^2 = 500 pa = 0.005 Bar

So this pressure should be able to push the water in my 10 cm^2 hose up 10 meter above the river.

Does this sound right?

 

[russ_watters]

Stagnation pressure = 0.5 * 1000 * 1^2 = 500 pa = 0.005 Bar

So this pressure should be able to push the water in my 10 cm^2 hose up 10 meter above the river.

Does this sound right?

I was with you until the last line. How do you get from 500 pa to thinking that you can push water up 10 meters?

The pressure of a column of water is the weight of that column of water divided by the area. Or p=rho*g*h. From 500 pa, you get

500 n/m^2= 1000 kg/m^3 * 10 * h
h= 0.05 m (5 cm)

 

[Leafhill]

Sorry, might have skipped a bit much :-) I agree with your equation. My reasoning was that since I got 500pa (0.05bar), this should actually mean 0.05kg/cm^2. With a 2000cm^2 entry-tube * 0.05kg = a total pressure of 10kg on the tube.

I was assuming this means I could lift a total of 10kg above the waterlevel of the river, and with a tiny hose (10cm^2) this would mean 10 meters height.

So I'm way of?

 

[russ_watters]

Bernoulli's principle says that the total pressure along a streamline is constant. In other words, you can't concentrate pressure with a funnel.

 

[gmax137]

Hi!

I'm trying to build a "pump" pushing water a few meters up of the river bank

Farmers use (or used to use before electrification) some clever little devices, I think they're called 'sling pumps' or 'coil pumps.' The stream flow rotates a tube that dips in a little water with each rotation; it is a positive displacement device that will pump the water uphill (say to a livestock trough). Low flowrate but it works 24/7.

and wondered if I could do it without ANY moving parts (except water ).

Oops, I guess the sling pumps don't qualify. But I always thought they were interesting.

 

[Leafhill]

Thanks for all the input! Both the ram pump and coil pump is definitely interesting. I'm just trying to understand all the aspects of water flow and pressure to construct the simplest pump that suits my needs.

I should have known my pitot-tube design was too easy, or they would have been used all over the place :-) It looks like I need a more conventional approach.

 

[Bob S]

For ram pumps, See
http://www.theramcompany.com/
I used to use one. It would pump water over a 100-ft hill with only 50 ft of head. My main problem was that I had to walk about a mile every morning to pull twigs out of the pump, or off the input screen.
Bob S

 

[Leafhill]

Hmmm... So I'm researching the possibilities of using a venturi. The calculator found here (http://www.pipeflowcalculations.com/venturi/index.htm ) is very helpful, but as it is meant to calculate flow through a pipe, so it doesn't allow me to enter velocity of the river. It also demands to know what pressures I have both at the inlet and the throat. I'd like to know that myself.

How can I calculate inlet pressure of a restricted tube? As shown above I can calculate the stagnation pressure, but in this case we're not talking about a full stop in the water flow. Instead we're trying to increase the speed as much as possible and insert a pitot-tube in the outlet stream of the throat. If I can reach 5-6 m\s through the throat I should be able to get sufficient pressure in the pivot to lift a few meters.

 

[michelcolman]

Sorry, might have skipped a bit much :-) I agree with your equation. My reasoning was that since I got 500pa (0.05bar), this should actually mean 0.05kg/cm^2. With a 2000cm^2 entry-tube * 0.05kg = a total pressure of 10kg on the tube.

I was assuming this means I could lift a total of 10kg above the waterlevel of the river, and with a tiny hose (10cm^2) this would mean 10 meters height.

So I'm way of?

Yes, quite a bit off :)

If the water inside the funnel and tube is stationary, the pressure in the entire funnel will be the same. It will not be higher at the narrow end. So you really have to multiply the pressure with the cross section of the narrow end, not the big end. The water will get to the same height no matter how thin the tube gets (excluding capillary effects, of course).

You seem to be confused about what happens to all the force (10 kg) that's being applied to the water at the entrance of the funnel. There is indeed a large total force there, but this is not just transferred to the water inside the narrow tube. Instead, the vast majority of this force is canceled out by the walls of the funnel itself. Since the pressure inside the funnel is higher than outside, the water pushes against the walls of the funnel and the funnel pushes back, action=reaction. This force has a horizontal component, and that's exactly where you are "losing" the vast majority of your 10 kg. The remaining force on the water at the narrow end will exactly match the pressure times the local cross section there.

 

[sophiecentaur]

I would be inclined to think in terms of pipes - like the School demo of the Bernoulli effect. A wide pipe, narrowing to a throat and then widening again, will produce a lowering of pressure in the narrowsection which could draw air in, through a narrow tube.
The tube could be taken to the top of a sealed container on the bank, lowering the pressure in the container. Another thin tube, taken upstream of the construction, could then draw water up and into the container. With enough flow and the appropriate 'waisting', you'd have a working system. It would rely on having a river with a fairly hefty flow, though and the use of large diameter piping.
Snag is that the water that has been pumped up would be in the sealed container and would need to be let out, occasionally.

 

[Leafhill]

The tube could be taken to the top of a sealed container on the bank, lowering the pressure in the container. Another thin tube, taken upstream of the construction, could then draw water up and into the container.

Yes, I've considered using the lower pressure (or suction) at the waist of a venturi to suck water from the river side up to a sealed container. This container should have a manual valve on the inlet- and outlet tubes to close it of, and a third valve to equalize the pressure and get the water out.

This should work up to about 7 meters before I get cavitation, but as you state, would require the appropriate venturi design and I don't know how to calculate the sizes. Any Ideas? This could be a beautiful solution with no moving parts and a very long lifetime.

I know there are other pump designs out there. I'm just curious why this particular one has not been used much. Or has it?

 

[sophiecentaur]

Dunno but I imagine the words BIG, WIDE and LONG would be involved! Not cheap.
I would have thought that a couple of metres of lift would be a more realistic target. Of course, as the 'suction tube would not carry water, there would be nothing to stop you from having several stages of lift as long as you can get the water that is inside at each stage to the outside for the next stage.
On the down side, the School demo only gives you a couple of cm of water of pressure difference! :-)

 

[Leafhill]

Well, I made a tiny venturi design (using Pro Engineer) for my kitchen tap. 23mm inlet and outlet and 3mm throat. I printed it using the Zcorp 3D-Printer (amazing machine) and mounted the finished bonafied venturi to my tap. I measured the total pressure drop between inlet and throat to 2.8 bars. The throat pressure was at 0.2 bars compared to the normal 1 bar of atmospheric pressure! Quite cool.

If I opened the tap more than half the water would start to cavitate and the pressure would rise again.

I was hoping to use something like this on a big scale, but won't have these kinds of pressures or water speeds available.Pictures below. Sorry about the picture quality on the "VenturiTest"-picture. It shows -0.78 bar on the dial.
http://home.online.no/~dan-l/Venturi1.jpg
http://home.online.no/~dan-l/Venturi2.jpg
http://home.online.no/~dan-l/VenturiTest1.jpg

 

[sophiecentaur]

Did you ever see this link?
http://home.earthlink.net/~mmc1919/venturi.html#animation

 

[Leafhill]

Did you ever see this link?
http://home.earthlink.net/~mmc1919/venturi.html#animation

Cool! No, I haven't seen that particular one before.

 

[sophiecentaur]

I just did some sums, using that formula and ignoring any actual drop in height of the river.
It seems to suggest that, if you can increase the water speed from 1m/s (typical river) to 3m/s you would only get 0.4m (head of water) of pressure reduction. Feel free to show I'm wrong but it may just be the reason this system has not been used in a serious way.

I'm fairly confident that my figure is in the right ballpark so you'd at least need several stages to get water up onto the river bank. It doesn't seem likely that you'd get much more than a threefold speed increase without using some sort of dam arrangement. (Dammed if you do and dammed if you don't?) But, as the pressure difference is proportional to the square of the speed, it could be worth looking in detail (some depth, perhaps - glug glug) at your river regarding its actual speed etc.
It also struck me that, if you were to draw up a bubbly mix of water by admitting some air into the sucked-up water, you could achieve a greater height.
Was your idea for an industrial scale or just for watering the veg or topping up the fish pond?.

 

[Leafhill]

Hehe, yeah the scale hasn't really been defined yet. I'm doing this with a friend and will keep it at a relaxing spare time project for now. The fish pond version as a start and scale it up later.

I've noticed as you said, that the pressure rises with the square of the speed:
0.5*density*velocity^2 = 500*velocity^2

Velocity (m\s) | Water height (m)
1 | 0,05
2 | 0,25
3 | 0,45
4 | 0,80
5 | 1,25
6 | 1,80
7 | 2,45
8 | 3,20

And so forth... Its not difficult, but could easily become impractical and unusable. I need to know the pressure and water flow through a venturi placed in a river to calculate this. Maybe do some field tests.

 

[sophiecentaur]

We seem to be agreeing about the sums. That's good. Now you have to find your river. . .

 

[Leafhill]

Wait, I just realized something. Isn't the Coefficient of Discharge actually just a value for the efficiancy, or if you will the loss in pressure through the venturi? This value tends to be really high (such as 0.95), so at the inlet I should be getting about 95% of the velocity of the water outside. This defines the flowrate through the venturi and I have the numbers I need.

http://home.online.no/~dan-l/VenturiTubeCalculator.JPG

Check out V2 (Throat). This is exciting :-)