Pumping water with atmospheric pressure

[JamesBe1]

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.

Before I build something like this, I would like to understand a little more about it.
Are you saying that when you insert V2 into V1, that it takes no energy to move the air out of the system?
Also, what do you mean when you say that "both ends of the tube are open"?  Which tube are you referring to?  The end of tube V1 is closed in your diagram, and the other end of tube V2 is connected to a hose that is connected to a box.  This doesn't appear to make sense to me.
Thanx.

[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

[Low-Q]

Medical syringe works like a reciprocating pump. When you pull the rubber piston, external atmospheric pressure straight away acts on the piston.  The idea of keeping both ends of tube V2 open is to prevent atmospheric pressure acting on it.

Please understand the concept before making comments.

It's hard to understand a concept which doesn't work. You cannot pump water with a static pressure.

Vidar

[vineet_kiran]

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.

[vineet_kiran]

It's hard to understand a concept which doesn't work. You cannot pump water with a static pressure.

Vidar

Mr.Vidar,


When you suck the air from a tube immersed in water,  the water will be pumped into the tube by static atmospheric pressure only.   Here the working principle is difference of pressure or drop of pressure and not static pressure.

When this experiment has worked for me, it should work for you also.  Certain things are difficult to understand unless you experience it practically.   Only theoritical genious like Einstein the GREAT can understand everything theoritically.


Thanks and regards

Vineet.K.