[Jayan2007]

It's area of cross section is 1 metre squire.The pipe is vertically placed.It's bottom end is closed.Inside the pipe, at the closed bottom end, there is 1 metre cube of air,closed with a disc. This disc is movable, like a piston in the cyllinder.(The cyllinder is the pipe.)
Now, I'm pressing on the disc downwards and compressing the air to 5 bar of pressure.(the volume of the air decreased.).
Now I'm pouring water on to the disc, 1000kg of water.
Now in the pipe,there is the compressed air at the very bottom, then the piston on it, and on the piston, filled with water of 1 metre hight(1 metre cube,i.e. 1 tone of weight).).
Now, if I release the pressure of the compressed air at the bottom SUDDENLY, to how much height the water colunm
Will go up(will be shooted up) ?
Consider the atmospheric pressure also,which blocking downwards.
PLEASE HELP..

[JudyKayTee]

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[samcreed]

"Simple" logical problem? I have a 2 Bachelors degrees, and this really doesn't look simple to me. You might ask your professor or teacher to give you some help with this one. Good luck.

[ebaines]

It's actually not too difficult. Consider the work done on the compressed air as the piston is compressed from 1 meter down to 0.2 meters, giving the increase from 1 bar to 5 bars of pressure. That work is the pressure times piston area integrated over the distance compressed, and a reasonable approximation of pressure is 1 Bar divided by the height of the piston from the bottom (P = 1 Bar/x). Set this amount of work to the kinetic energy imparted to the water and piston when the mechanism fires. That will give you the velocity at the moment when the piston is at x = 1 meter, which is the moment of its greatest velocity (at x>1 there is actually a vacuum in the piston and so it slows down), and from that you can determine the height that the water shoots.

Note that this method ignores the fact that as the air expands it will naturally cool a bit (under the ideal gas law), and consequently the pressure is not quite equal to 1 Bar/x. Also in reality the upward water column will be affected quite substantially by the drag of air resistance, which this method ignores. And as for considering the pressure of the atmosphere, which as you say opposes the piston - just use relative pressures (i.e. the volume of the chamber is compresed from 0 to 4 Bar, not from 1 to 5).

Post back with what you calculate and we'll compare notes.

[Jayan2007]

I'm equating 500000 Newton to 1000kg of mass and unknown accelaration.(F=ma)
will 500000 equal to 1000 into accelaration? So,a=500000/1000=500m/s^2 ? Am I corect?

[ebaines]

The initial acceleration while the pressure is 5 Bar is indeed 500m/s^2. But as the piston moves upward the air pressure behind it is reduced, and consequently the acceleration is reduced as well. Hence acceleration is not constant but rather is a function of the position of the piston. To find the resulting velocity you need to use the following:



where 'a(x)' = acceleration. From the data you provided we know that a(x)=100p(x), where 'p' = pressure in the cylinder in bar and is a function of displacement 'x.' If you use a simple equation such as [math] p(x)=6.25-6.25x[math] you can then put it into the above integral to find the final velocity of the piston after it expands from x = 0.2 meter to x = 1 meter.

Post back with what you get for the final velocity.

[joshfella]

"Simple" logical problem? I have a 2 Bachelors degrees, and this really doesn't look simple to me. You might ask your professor or teacher to give you some help with this one. Good luck.

Unless, yours is Bachelor of Science, then you will find this logical problem is simple logical. If in just regular Physics, maybe normal student needs to think a little hard to get this through. But for someone who are taking Fluids Mechanics or Applied Differential Equation, then the problem will lay out clearer.