Pressure...Help I'm confussed

No, it is not correct. If you have a cylinder of any thickness with an internal pressure of 3 bars while the external pressure is 1 bar, then the difference in pressure is 2 bars. That means that the net force on the bottom of the cylinder is equal to Area*2 bars. That force is constant without any opening in the cylinder.

If you now cut an opening in the side of the cylinder measuring 100 mm², say 10mm x 10mm or 1mm x 100mm or a circular opening of diameter 11.3mm, the air will start to leak out. If the cylinder is very large, the pressure will equalize at a slow rate. Say the cylinder had the same volume as an arena! A small hole like that will take a long time to allow the air to escape. If the cylinder is small, the pressure will equalize much more quickly. Eventually, the interior pressure will be 1 bar, same as the exterior pressure.

So, the pressure difference at the beginning of leakage is 2 bars. The pressure difference at the end of leakage is 0 bars. The time taken to leak completely is dependent on the original volume of the cylinder, which we do not know. At any given time, the force on the bottom of the cylinder is equal to the area of the bottom times the difference between external and internal pressure, or the pressure differential.

I hope that answers your question, but, if not please let us know.

What an Answer ba/ael. Now I'm confussed and need time to read and understand the technalities. But good job

If the pressure is due to air or gas in an enclosed sealed cylinder then the pressure, whatever it is before and after, will be the same throughout at all levels.

Or is the pressure due to the weight of liquid? in which case, height matters, and so will the length of the slot especially if vertical.

The slot will weaken the cylinder. Thus a large diameter cylinder might split lengthwise, and if filled with air, then perhaps with dramatic effect.

Either way, the force downward on the end of the cylinder will be 'pressure X area'.

You are assuming facts that weren't given. First, if the intenal pressure is 3 Bars (43.5 psi) that already assumes an area. If you are looking at it as psi, then the pressure would be 43.5 pounds on every square inch of the internal surface of the cylinder regardless of the position of the cylinder in the case of a gas-filled cylinder. If, however, you are considering a liquid-filled cylinder, then you are correct in that the pressure would be a function of the depth of the liquid insde. For instance, if the cylinder is filled with water, it would have to be just over 100 feet tall to exert 43.5 psi (3Bars) at the bottom regardless of the area of the bottom. Your statement that "either way, the downward force on the end of the cylinder will be 'pressure X area' is wrong. If the cylinder is air-filled, the pressure at any point on the inside would be the same. Liquid would be different as stated above. Further, if the liquid is other than water, then the specific gravity of the liquid would enter into the equation whether it is heavier or lighter. There are other factors that impact this scenario as well depending on what kind of accuracy the Guest is requiring. These factors include such as the assumption that one bar is equal to one atmosphere is erroneous. One bar equals 0.9869 atmosphere. If there is air compressed on the top of the liquid in the cylinder, the pressure would be whatever the pressure of the compressed air above the liquid is, plus the weight of the remaining depth of the cylinder and then with the added dimension of the specific gravity of the liquid, and, for that matter, the gas above the liquid if it is not air. Not an answerable question without having the total picture. Talk about confusing!!!

Miguel

In a closed volume the pressure exerts the same force in all directions and against all surfaces. If the exit path is up, down or around it will not affect the flow rate.

I work with inches instead of mm and to save the time will convert 100 mm^2 to .155 in^2.

Above 1 bar (1 bar (14.5 psig) or 2 bar absolute (29 psia)) this formula will work if you can estimate the coefficient of discharge (Cd) of the exit passage. An orifice with well rounded entrance may be .9. A square, rectangular, or ragged gash with burrs and sharp edges may have a Coefficient of discharge below .5 caused by turbulence.

Flow, Qscfm = 18.46 x .155in^2 x P (Delta P psig + 14.5 psi) x Cd

You can find the Cd by starting at your initial pressure of 2 Bars and time the discharge from 29 psig down to 11 psig (37%).

If the Cd were 100% your time would be 2 (compression factor) x Cylinder Volume ft^3 /18.46 x .155 x 29 = time in minutes. If the actual time is longer then divide the calculated (perfect time) by the calculated perfect time. This is the coefficient of discharge (Cd). 10V/Q = Time (minutes) to vacate the volume to .067 psig.

I think the pressure measured as 3 bar is a relative pressure not absolute pressure, so the value of pressure to be taken into consideration is 3 bar. And the net force is equal to the pressure 3 bar multiplied by the area.

Note. It was used to read the measured pressure as a relative pressure or gauge pressure PSIG (sometimes written as PSI without G), and the absolute pressure must be mentioned as it is an absolute and must be written as PSIA to differentiate between the both.

In accordance with ASME code, the internal pressure (as a relative pressure or a gauge pressure) must be taken in design whatever the pressure of the atmospheric. But in case of requiring designing/checking for full vacuum FV, the atmospheric pressure as 15 PSI shall be used, and considering the vacuum case will create zero gauge pressure inside the vessel.

Note. The gauge design pressure inside a vessel is that pressure measured at the top of that vessel. Therefore, the pressure at the bottom of the vessel must be increased by the additional pressure due to the weight of column of liquid above the point under consideration.