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A fountain head pumps water out of the main tank into a 'pond' reservoir. Can the water recirculate back into the main tank without the help of another pump?

I'm sorry if this a dumb question. I'm guessing it would not function as the diagram shows, as the pressure of the water in the main tank would not let any water in at the bottom,right? Any solutions? (not requiring additional pumps)

enter image description here

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    $\begingroup$ You don't need to actively pump from the main tank to the reservoir, the hydrostatic pressure to achieve that is already there. All you have to do is to pump from the reservoir back into the tank. $\endgroup$ – CuriousOne Jun 15 '16 at 23:54
  • $\begingroup$ the pump in the main tank is used to output water through a fountainhead (it needs a certain flow and psi to work), this leads the water to the reservoir and i need to be able to get it back into the tank $\endgroup$ – psedro Jun 16 '16 at 0:04
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    $\begingroup$ What is the height of the reservoir? can you make it so that the water level in the reservoir will equal or greater to the water level of the tank(like in my answer)? $\endgroup$ – philip_0008 Jun 16 '16 at 0:07
  • $\begingroup$ If you need more pressure than is available hydrostatically, then you will need two pumps, I am afraid. $\endgroup$ – CuriousOne Jun 16 '16 at 0:10
  • $\begingroup$ the height of the reservoir will always be less then the tank's (water level in tank might get lower then reservoir, but I need it work always not just when the conditions are such) and it will always hold less water in it too.. guess 2 pumps it is then? $\endgroup$ – psedro Jun 16 '16 at 0:20
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The set up shown will work just fine, though to be safe I'd put the check valve on the pump so as to prevent water and or air from flowing back into the tank from above.

I also made a few assumptions:-

  1. The free space in the top of the tanks is small enough
  2. The tank is not too tall (less than about $9~\text{m}$ should suffice)
  3. The volume of the pond is large enough
  4. Water cannot flow back through the pump
  5. The diameter of the outflow pipe is small compared to the tank diameter.

At first, the pressure of the air gap in the tank is at atmospheric pressure: this means it has a gauge pressure (pressure above atmospheric) of 0.

Check valve must be placed on the outlet pipe as close to the pump as possible.
Before the pump is turned on, the weight of the water will cause it to flow out into the pond through the bottom pipe connecting the two. As this does, the volume of the air gap will increase, and thus it's pressure will drop. Eventually the pressure will reach a point such that the force it exerts on the water will cancel out the gravitational force pulling the water out. (Negative gauge pressure)

When the pump turns on, it will remove water from the main tank again increasing the air gap volume. This will mean there is an overall negative gauge pressure at the tank side of the inlet pipe, thus causing water to be drawn into the tank to replace the water removed by the pump.

The reason this all works is due to the difference of the external pressure and the internal pressure, and works on a similar principle this.

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The water level in the pond must be the same as in the tank, so:
enter image description here

:-)

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  • $\begingroup$ @psedro If the reservoir is lower than the tank, then you don't need to pump it to transfer water to the reservoir. It will readily flow through the valve. What you only need is to pump the water from the reservoir to the tank, as CuriousOne said. $\endgroup$ – philip_0008 Jun 16 '16 at 0:21
  • $\begingroup$ philip, i know it seems counter-intuitive to pump water out of the tank that way (from top) just to get it to the reservoir, but that is not the goal, just the result). The sole purpose of this is to be a water-feature, like a fountain, so that flow out of the top is the point of the whole system. $\endgroup$ – psedro Jun 16 '16 at 0:30
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How about:
enter image description here
open valve 2 to transfer water from the reseroir to the tank. Close valve 2 to create the 'fountain' feature. Open valve 3 to transfer water from tank to reservoir.

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  • $\begingroup$ yesss!!! this isn't the exact solution but you gave me exactly i needed.. will post potential solution diagram as another answer! $\endgroup$ – psedro Jun 16 '16 at 1:24
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enter image description here

@phillip_0008 Pretty sure this one would work! Basically, no need for valve 3. And wouldn't be able to access valve 2. So split pump inlet between bottom of tank and bottom of reservoir, directly outlet to fountain. Water coming out of fountain sits in reservoir until it gets sucked back in thru pump. makes sense in my mind but let me know if you see anything i didn't!

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  • $\begingroup$ You must have a valve at the pipe that comes from the tank also, because the pressure at the tank might be greater, it would make the reservoir water level rise significantly. Also, you don't have a way to pump water from reservoir to tank. I think you should split the outlet of pump also, one to the fountain, and one to the tank, and also add valve to the one going to the tank. $\endgroup$ – philip_0008 Jun 16 '16 at 1:55
  • $\begingroup$ Oh, I didn't see, there's a check valve from the tank already.. $\endgroup$ – philip_0008 Jun 16 '16 at 1:57
  • $\begingroup$ in this drawing how is the main tank filled? only out going arrows $\endgroup$ – anna v Feb 7 '17 at 7:06
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The answer is no, simply because the pressure on the check valve on the side of the open pond must be greater than that within the main tank before the valve can open and let water flow from the open pond and the main tank.

This will only happen if either:

  1. The open pond's water level is higher than that of the open tank; as drawn, the system will simply overflow the open pond and keep draining the main tank until the two tank levels are equal;

  2. A pump raises the pressure of the water from the open tank to force the valve to open and push water back into the main tank.

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Another suggestion:
enter image description here
close valve 1 and 2 and turn on pump to make 'fountain' effect. Close valve 1 and open valve 2 and turn on pump to transfer water from reservoir to tank. Close valve 2, open valve 1 and turn off pump, to transfer water from tank to reservoir.

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  • $\begingroup$ Oh, so the check valve is not an ordinary valve, How about replace the valves with ordinary ones instead? $\endgroup$ – philip_0008 Jun 16 '16 at 2:25
  • $\begingroup$ Just having fun:) $\endgroup$ – philip_0008 Jun 16 '16 at 2:30
  • $\begingroup$ But I'm serious with my answers though.. $\endgroup$ – philip_0008 Jun 16 '16 at 2:47
  • $\begingroup$ That's great! This way you can really control the water and direct it anywhere. I might just have a check valve instead of an ordinary at valve #2, just in case... But hey I really appreciate your time and help! Thanks a ton! $\endgroup$ – psedro Jun 16 '16 at 16:35
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Huh, people? :-)

He says the main tank is closed!

That means it can be pressurised, and the pressure will ofc push out water, depending on pump efficiency.

Nothing else needed than a valve that hinders the water from flowing back to the open reservoir when the pump is off.

In the pic below a self-adjusting system. - If there is too little water in the main tanh, it pushes out air until the water level rises to touch the pipe, after which it pushes out water. - The amount of air defines the functionality, and need be matched against the pump efficiency. Some component dimensioning needed, after which it must work :-).

enter image description here

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