Water can be pumped without using a pump!.  See the attachment for details.

What is new about this? This works the same way as a medical syringe. Put it in water and pull the rubber piston. Voila, water is going into the syringe.
You face a sub pressure when trying to pull out the inner cylinder due to the weight of the water you want to rise. It works, but it will require energy input to work.

Vidar

looked, understood, and it is a piston pump and air pressure dose act on V2.
when you bend the hose to pull on V2 you create a surface for air pressure to push it back in to
equalize the pressure.
thankyou for playing please play again.
fritznien

I think the flaw in the reasoning is in the statement "A very interesting point here to note is that work input to the system in only energy supplied to overcome friction between tube V2 and rubber ring . . ."
This is simply not the case.  The real energy input to the system is the energy required to overcome the friction of the o-ring plus the energy to overcome the air pressure differential between the surface of the water that is open to atmosphere and the air pressure in the tubes.
Consider figure 2.  If you were to try to reinsert V2 into V1 again, you would have to push the water back down the riser tube that runs from the open to atmosphere reservoir to the black box.
Nice try though.

I think the flaw in the reasoning is in the statement "A very interesting point here to note is that work input to the system in only energy supplied to overcome friction between tube V2 and rubber ring . . ."
This is simply not the case.  The real energy input to the system is the energy required to overcome the friction of the o-ring plus the energy to overcome the air pressure differential between the surface of the water that is open to atmosphere and the air pressure in the tubes.
Consider figure 2.  If you were to try to reinsert V2 into V1 again, you would have to push the water back down the riser tube that runs from the open to atmosphere reservoir to the black box.
Nice try though.

It is not the case.   Even when you reinsert V2 into V1 again, it pushes air (or water) from reservoir to jar without experiencing force since both ends of tube are open, making work input independent of work output.  Try it practically.

I have practically conducted this experiment.

Thanks.

you know that atmospheric pressure is equal in all directions and pressure at both ends of tube is also same. you won't pull the pipe against any air pressure which requires work

Kindly do the experiment practically.

No more arguments please.

Hello vineet_kiran,

let us assume the tank with water, the container the water is sucked up to and V1 is fixed to ground:

Why do you assume that there is no air-pressure on the flexible hose attached to V2 ?

According to what I can see, airpressure is acting from all sides on the flexible hose pushing it back in and the work necessary will be the the new potential energy Epot of the pumped up watermass equals m * g * h.


Regards

Kator01

Hello vineet_kiran,

let us assume the tank with water, the container the water is sucked up to and V1 is fixed to ground:

Why do you assume that there is no air-pressure on the flexible hose attached to V2 ?

According to what I can see, airpressure is acting from all sides on the flexible hose pushing it back in and the work necessary will be the the new potential energy Epot of the pumped up watermass equals m * g * h.


Regards

Kator01

There will be air pressure on the hose all round its circumference which will only try to flatten the hose and force due to this pressure will not be in a direction opposite to direction in which the hose is pulled out. (which is same as fixing a clip on the hose and pulling it out of the container V1.  The force of clip acting on both sides of the tube in no way can affect pulling of tube out of container).    In case of piston pump force due to atmospheric pressure acts exactly in the direction opposite to movement of piston opposing the movement of piston.

Regards,

Vineet.K.