# Thread: Help in calculating kilowatts needed to pump water that is already in motion.

1. I am designing a project where solar power will be used to store potential energy, basically I will have two tanks to store the water, one tank will be 25 meters higher than the other. The tank that will store potential energy will be air tight so as to create suction to pull the water that has dropped to the lower tank.

Before the water reaches the lower tank I will have a turbine.

There will be pipe that will connect the lower tank with the upper tank with an inline pump to assist, this pipe will take the water that is being sucked back up.

How much power will I need to pump water that has the suction force to the tank that is 25 meters higher

Flow rate 2000Gpm

Friction lose standard pipe work will be used with minimal bends, two bends of 45° at most.

2.

3. Are you sure about that 2000 gallons per minute figure? That's a buttload of water....it would take 2 pumper fire trucks to produce that. You also need to specify pipe diameter.

Remember, in any system...you can't get more energy out than you put in. There will always be loses due to friction etc. What is the purpose of this system? Is it to provide backup power in case of an outage, or something provide continuous power?

4. I don't get the idea of the vacuum. Yes, it would help pull the water up from the lower level to the upper, but it will retard the water from flowing down through your turbine. It will also cause a lot of practical problems such as in keeping the system airtight. Your head loss calculations will need to account for the pipe diameter, which you haven't specified.

5. Apologies I meant litters, I will use solar to pump water back up to the higher tank. I need help to find out if save significant energy can be saved if I use gravity lift water back with the help of a pump.
The system is to store solar energy in the form of potential energy for use when there is no sunshine.

As you know the batteries in the market to store enough energy for domestic use are expensive and have a recharge cycle limit.

I can store the potential energy using only solar, but if I can save energy by use of suction force generated by gravity it would make the whole system more efficient and the cost of setting it up and running the system would reduce.

If the gains that will be made by using gravity for suction are eliminated by friction, will there be any energy saving as the pump will have less work to do because it will be pumping in the same direction as the suction force?

Originally Posted by MacGyver1968
Are you sure about that 2000 gallons per minute figure? That's a buttload of water....it would take 2 pumper fire trucks to produce that. You also need to specify pipe diameter.

Remember, in any system...you can't get more energy out than you put in. There will always be loses due to friction etc. What is the purpose of this system? Is it to provide backup power in case of an outage, or something provide continuous power?

6. Originally Posted by strep
If the gains that will be made by using gravity for suction are eliminated by friction, will there be any energy saving as the pump will have less work to do because it will be pumping in the same direction as the suction force?
As Harold said, the vacuum will reduce, and eventually stop, the flow of water out. Which rather defeats the object.

7. Originally Posted by strep
Apologies I meant litters, I will use solar to pump water back up to the higher tank. I need help to find out if save significant energy can be saved if I use gravity lift water back with the help of a pump.
The system is to store solar energy in the form of potential energy for use when there is no sunshine.

As you know the batteries in the market to store enough energy for domestic use are expensive and have a recharge cycle limit.

I can store the potential energy using only solar, but if I can save energy by use of suction force generated by gravity it would make the whole system more efficient and the cost of setting it up and running the system would reduce.

If the gains that will be made by using gravity for suction are eliminated by friction, will there be any energy saving as the pump will have less work to do because it will be pumping in the same direction as the suction force?
You can't get something from nothing. As my more knowledgeable comrades have pointed out, a sealed system would impede the downward flow of the water. You can either use the gravitational energy of the water to turn your turbine, or use it to pull the water back up, but not both (at full efficiency). The amount of energy the pump would use will always be greater than the output of your turbine. That's just the law of thermodynamics. It would be much more efficient just to use the energy from the solar panels to get the power.

Now, I could see a non-sealed system like this used as a back-up to temporarily generate power for a few hours during an emergency. Plus..you have the benefit of having a couple thousand gallons of fresh water if it's needed.

8. Thanks for taking a look at this.
Gravity will most probably take care of the flow rate as all I have to do is to increase the size of outlet. The turbine will be in the open only the higher tank will be sealed. Keeping the system airtight is no problem, although it has to be vented periodically.
Suction force would help reduce the number of the solar panels because I will use less force to pump the water back up, the question is how much power would such a setup save?

Originally Posted by Harold14370
I don't get the idea of the vacuum. Yes, it would help pull the water up from the lower level to the upper, but it will retard the water from flowing down through your turbine. It will also cause a lot of practical problems such as in keeping the system airtight. Your head loss calculations will need to account for the pipe diameter, which you haven't specified.

9. Originally Posted by strep
Gravity will most probably take care of the flow rate as all I have to do is to increase the size of outlet.
You seem to be missing the point the vacuum will act against gravity, reducing the flow of water through the turbine.

10. Originally Posted by strep
Thanks for taking a look at this.
Gravity will most probably take care of the flow rate as all I have to do is to increase the size of outlet. The turbine will be in the open only the higher tank will be sealed. Keeping the system airtight is no problem, although it has to be vented periodically.
Suction force would help reduce the number of the solar panels because I will use less force to pump the water back up, the question is how much power would such a setup save?

For every Joule of energy that you save from the "suction force" (BTW, suction is not really a scientific term) is one less joule the turbine will produce. The potential energy of the elevated water is finite.

11. The system will not be a sealed loop, only the top tank will be sealed.

Here is a diagram; sorry I had to do a quick draw using touchpad.

I fully understand that there is no perpetual motion. The suction energy is the onc I want to use but I need to know if it is worth the effort at any scale.

Originally Posted by Strange
Originally Posted by strep
Gravity will most probably take care of the flow rate as all I have to do is to increase the size of outlet.
You seem to be missing the point the vacuum will act against gravity, reducing the flow of water through the turbine.
thescience.JPG

12. Originally Posted by strep
The system will not be a sealed loop, only the top tank will be sealed.
And that is what will slow/stop the water flowing out. Look up Torricelli; he proved this will not work 400 years ago.

I fully understand that there is perpetual motion.
I hope that does not mean what it says.

The suction energy is the once I want to use but I need to know if it is worth the effort at any scale.
IT WILL PREVENT THE WATER COMING OUT.

13. The entire system must be sealed to achieve what you want. As the water drains out of the upper tank, it creates a vacuum. Air pressure from the bottom tank then "pushes" the water back up. If there's a break anywhere in the system, you lose the vacuum.

Think about a drinking straw. Your mouth creates a small vacuum at the top of the straw that is lower pressure than the surrounding air. The surrounding air "pushes" down on the top of the liquid in the cup, and "pushes" it up the straw. If you poke a hole in the straw...air rushes in from the hole to fill the vacuum, and the liquid stops flowing upward.

14. Please have a look at this video

As you can see there is suction and the system is not sealed.

Originally Posted by MacGyver1968
The entire system must be sealed to achieve what you want. As the water drains out of the upper tank, it creates a vacuum. Air pressure from the bottom tank then "pushes" the water back up. If there's a break anywhere in the system, you lose the vacuum.

Think about a drinking straw. Your mouth creates a small vacuum at the top of the straw that is lower pressure than the surrounding air. The surrounding air "pushes" down on the top of the liquid in the cup, and "pushes" it up the straw. If you poke a hole in the straw...air rushes in from the hole to fill the vacuum, and the liquid stops flowing upward.

15. Strange, could you field this one?...I got to go rake leaves if I want to have dinner tonight.

Funny enough...my leaf vacuum uses the same principles we are discussing. Hurray for science!

16. The system is being assisted by a pump powered by solar panels. Plus the wieght of water of the upper storage tank will be more that the water in the pipe that is taking the water to it.

Originally Posted by Strange
Originally Posted by strep
The system will not be a sealed loop, only the top tank will be sealed.
And that is what will slow/stop the water flowing out. Look up Torricelli; he proved this will not work 400 years ago.

I fully understand that there is perpetual motion.
I hope that does not mean what it says.

The suction energy is the once I want to use but I need to know if it is worth the effort at any scale.
IT WILL PREVENT THE WATER COMING OUT.

17. Originally Posted by strep
The system is being assisted by a pump powered by solar panels. Plus the wieght of water of the upper storage tank will be more that the water in the pipe that is taking the water to it.
In that video, the upper cup is open to the air. If it was a sealed container, the water flow would stop.

18. To apply the same physics in my query, the storage tank would be enclosed in the pipe (for arguments sake) situated just past the middle of the pipe, in the direction of the flow, take the two cups as the lower tank. That suction power is the one I want to harness.

Originally Posted by Strange
Originally Posted by strep
The system is being assisted by a pump powered by solar panels. Plus the wieght of water of the upper storage tank will be more that the water in the pipe that is taking the water to it.
In that video, the upper cup is open to the air. If it was a sealed container, the water flow would stop.

19. Here is a video that shows what I am looking to do, I intend to add solar energy to help pump the water back up.

Originally Posted by Strange
Originally Posted by strep
The system is being assisted by a pump powered by solar panels. Plus the wieght of water of the upper storage tank will be more that the water in the pipe that is taking the water to it.
In that video, the upper cup is open to the air. If it was a sealed container, the water flow would stop.

20. Originally Posted by strep
Here is a video that shows what I am looking to do
I assume that is meant as a joke?

21. I read that a few times, and still couldn't tell what you were trying to say. Do you have one of those pet watering dishes? I mean the kind where the water stays in the bottle until the pet drinks out of it, which then allows a bubble of air up inside, so some more water can come down into the dish. That's what would happen. Except for the fact that atmospheric pressure can only hold up a column of water about 10 meters high.

22.

23. As I wrote up there I hope to stick a pump on system. There is no perpetual motion, I will use solar to compensate for friction and gain the 9.8m/s² force provided by gravity.

The video was for illustration only, the inlet to the storage tank would be on top.

Originally Posted by Strange
Originally Posted by strep
Here is a video that shows what I am looking to do
I assume that is meant as a joke?

24. OK. So you plan is: an elevated tank; pump water up into it using solar power; take water from the tank to drive a turbine?

The tank smooths out the difference between demand and available electricity.

No problem. That will work.

But you will not make it any more efficient by stopping the water coming out of the tank by not letting air in.

25. What I want to do is to reduce the energy needed to store the potential energy. The system in the link you have provided uses pumps powered by access energy from the grid, while I will use solar.

Originally Posted by Strange

26. I am trying to use the force of gravity to reduce the energy needed to store potential energy.

Originally Posted by Harold14370
I read that a few times, and still couldn't tell what you were trying to say. Do you have one of those pet watering dishes? I mean the kind where the water stays in the bottle until the pet drinks out of it, which then allows a bubble of air up inside, so some more water can come down into the dish. That's what would happen. Except for the fact that atmospheric pressure can only hold up a column of water about 10 meters high.

27. Originally Posted by strep
What I want to do is to reduce the energy needed to store the potential energy.
Which will, therefore, reduce the energy you can get out of it. How hard is that to understand?

Conservation of energy - Wikipedia, the free encyclopedia

The system in the link you have provided uses pumps powered by access energy from the grid, while I will use solar.
That is not relevant (and not true either). In both cases, I assume, the idea is to smooth out the difference in supply and demand. Otherwise you would just use the solar power immediately.

28. Originally Posted by strep
I am trying to use the force of gravity to reduce the energy needed to store potential energy.
It doesn't take any energy to store potential energy. Once it's up at the higher elevation, you can just close a valve or something and it will stay there indefinitely. The problem is how much energy it takes to get it there and how much you get back out of it when it runs back down to the lower elevation. It's always going to take more to get it there than you get back out. The vacuum doesn't help that situation at all.

29. Originally Posted by Strange
OK. So you plan is: an elevated tank; pump water up into it using solar power; take water from the tank to drive a turbine?

The tank smooths out the difference between demand and available electricity.

No problem. That will work.

But you will not make it any more efficient by stopping the water coming out of the tank by not letting air in.

Now that I'm through raking...allow me to demonstrate using the cause and solution to all life's problems....beer. Strep...I'm not sure how old you are...but have you ever "shotgunned" a beer? If you take a beer and open it, and turn it upside down, the beer will flow out in an uneven gulg..gulg..gulg. This is because as the beer leaves the can, it creates a vacuum, and air rushes in from the only opening to fill this vacuum and disrupts the beer flowing out. When you shotgun a beer, you poke a hole in the side of the beer can. This allows air to enter in the hole, instead of the regular opening. This produces an even, larger flow of beer coming out, and you can drink a whole beer in just a few seconds.

30. I want to design a solar power system that gives power even when the sun is not shinning, I am planning to store solar power as potential energy. The plan is to raise water to a reasonable height using minimal energy.

The less energy I use, the less solar panels I will have and the pump will be smaller, the system will be made cheaper to install and run.

Is it possible to reduce the energy needed to pump the water by exploiting the suction effect?

Originally Posted by Harold14370
Originally Posted by strep
I am trying to use the force of gravity to reduce the energy needed to store potential energy.
It doesn't take any energy to store potential energy. Once it's up at the higher elevation, you can just close a valve or something and it will stay there indefinitely. The problem is how much energy it takes to get it there and how much you get back out of it when it runs back down to the lower elevation. It's always going to take more to get it there than you get back out. The vacuum doesn't help that situation at all.

31. Originally Posted by strep
Is it possible to reduce the energy needed to pump the water by exploiting the suction effect?
Possibly (although I don't think it is practical).

BUT (and this is the bit you don't seem to get): If you put less energy in, you will get less energy out. "Suction" is not some magic source of free energy.

If you use energy E to pump the water up, the most energy you can get out is slightly less than E.

So, in conclusion: if you reduce the energy needed to pump the water by exploiting the suction effect you will reduce the energy you can recover from that water by (at least) the same amount.

TANSTAAFL.

32. There is no such thing as "the suction effect". "Suction" is just a layman term. There is only difference of pressure. Pressure pushes..it doesn't pull. If you have an area of lower pressure next to an area of higher pressure...there will be a flow towards the low pressure until the pressure equalizes. You're idea of storing solar energy as potential energy will definitely work....however attempting to use a vacuum to assist the filling of the top tank would only result in a loss of flow. Keep your idea, just lose the vacuum idea. I'm not sure it would be as effiecient as storing the energy in batteries...but in addition to extra electrical power, you would have a supply of pressurized fresh water, in case something ever happened to your main water supply.

33. I have seen the beer thing work on barrels where they have two openings, a small opening sits at the top to let air in for a faster drain.

I am beginning to doubt this, I have been building a small unit for the last couple of hours, but have not been successful in even finding out the force needed to move the water round. I will try and get a miniature pump and see what happens. Off course I will get two small power meters and a small electric motor.

Originally Posted by MacGyver1968
Originally Posted by Strange
OK. So you plan is: an elevated tank; pump water up into it using solar power; take water from the tank to drive a turbine?

The tank smooths out the difference between demand and available electricity.

No problem. That will work.

But you will not make it any more efficient by stopping the water coming out of the tank by not letting air in.

Now that I'm through raking...allow me to demonstrate using the cause and solution to all life's problems....beer. Strep...I'm not sure how old you are...but have you ever "shotgunned" a beer? If you take a beer and open it, and turn it upside down, the beer will flow out in an uneven gulg..gulg..gulg. This is because as the beer leaves the can, it creates a vacuum, and air rushes in from the only opening to fill this vacuum and disrupts the beer flowing out. When you shotgun a beer, you poke a hole in the side of the beer can. This allows air to enter in the hole, instead of the regular opening. This produces an even, larger flow of beer coming out, and you can drink a whole beer in just a few seconds.

34. Thanks guys, I have lost the argument and will now try to build a model and see what happens.

Originally Posted by Strange
Originally Posted by strep
Is it possible to reduce the energy needed to pump the water by exploiting the suction effect?
Possibly (although I don't think it is practical).

BUT (and this is the bit you don't seem to get): If you put less energy in, you will get less energy out. "Suction" is not some magic source of free energy.

If you use energy E to pump the water up, the most energy you can get out is slightly less than E.

So, in conclusion: if you reduce the energy needed to pump the water by exploiting the suction effect you will reduce the energy you can recover from that water by (at least) the same amount.

TANSTAAFL.

35. You've lost nothing...just gained a little knowledge. If people didn't come up with new ideas, then we'd all still be sitting around a fire saying "Ung!" I love coming up with new ideas...maybe only 1 in 100 would actually work...but hey...it's a start!

If you need any help with your model...let us know.

36. Hello Strep,

As my distinguished colleagues pointed out, you cannot improve the energetic efficiency using any "succion" effect.
According to me, the only interresting calculation we can do, is to compare the quantity of energy you can store using a water tank (potential energy) compared with an electric battery.

Let us consider an ordinary car battery : 12V with a capacity of 70Ah than can be find for a cost around 100 USD.
The energy you can store in it is E = 12 x 70 x 3600 = 3 024 000 J = 3 024 KJ.

Now, what will be the mass of water M, in the upper tank, at an elevation h from the lower tank that can accumulate the same potential energy ?

Potential energy is given by the formula E = M x g x h. Where g is the acceleration of gravity (9.81 m/s2).
So M = E / g x h. Your hypothesis was h = 25m, => M = 3 024 000 / 9.81 x 25 = 12 330 Kg. That is 12 cubic meters of water.

What will be your decision ? a 100 USD battery ? A tower of 25m able to support 12 000Kg, a tank, a pump, a turbine and a couple of electro-valves ?

37. Batteries are the most convenient form of storing solar energy but may not the cheapest, in the long run. Just recently we have started to see batteries that can hold even 20kwh enough to power 3 homes for several hours. Problem is that, the batteries are expensive as it is new technology and their performance deteriorates rapidly. A concrete water reservoir could last for a very long time.

If I was to set up a 5kw system, I would only need to switch water between two swimming pools
If I am lucky to have a neighbour that has a pool that is lower or higher than mine, I could be in business without investing a lot.

Basically a 5 kw system would need about 40,000 litters of water to get power throughout the day.

I was looking into using gravity to assist the pump to lift the water to a reasonable height.

Near free energy is not far away, Professor Sadoway gives us hope, listen to him.

Originally Posted by caKus
Hello Strep,

As my distinguished colleagues pointed out, you cannot improve the energetic efficiency using any "succion" effect.
According to me, the only interresting calculation we can do, is to compare the quantity of energy you can store using a water tank (potential energy) compared with an electric battery.

Let us consider an ordinary car battery : 12V with a capacity of 70Ah than can be find for a cost around 100 USD.
The energy you can store in it is E = 12 x 70 x 3600 = 3 024 000 J = 3 024 KJ.

Now, what will be the mass of water M, in the upper tank, at an elevation h from the lower tank that can accumulate the same potential energy ?

Potential energy is given by the formula E = M x g x h. Where g is the acceleration of gravity (9.81 m/s2).
So M = E / g x h. Your hypothesis was h = 25m, => M = 3 024 000 / 9.81 x 25 = 12 330 Kg. That is 12 cubic meters of water.

What will be your decision ? a 100 USD battery ? A tower of 25m able to support 12 000Kg, a tank, a pump, a turbine and a couple of electro-valves ?

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