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Pressure question

by A.J.710
Tags: pressure
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A.J.710
#1
May1-14, 11:14 PM
P: 28
I am sitting in the library just thinking and I have a general question about pressure. Basically think of the problem as a 5 gallon bucket filled with water and a water tight circle of some sort sitting on top. Now say I stand on that circle and the force is exerted downward compressing the water. Now for simplistic reasons if there is a hole in the side wall at the bottom of the bucket with a 1" surface area, would the pressure of the water coming out of that whole be the sum of the force exerted by me standing on top and the water inside the bucket? The water pressure inside is minimal. I just basically want to know if my body weight of 200 pounds divided by the surface area of the hole 1" would make the pressure coming out of the hole 200 PSI plus the minimal water pressure inside. Thats what I am trying to get at. Is the pressure directly transferred from my weight on top to the hole on the bucket or is there some sort of loss somewhere in the system where there won't be that type of drastic build up in pressure when I stand on top of the system.

Thanks
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paisiello2
#2
May1-14, 11:24 PM
P: 558
Quote Quote by A.J.710 View Post
I am sitting in the library just thinking .
Libraries can be a dangerous place for thinking. Be careful.

Quote Quote by A.J.710 View Post
...would the pressure of the water coming out of that whole be the sum of the force exerted by me standing on top and the water inside the bucket? .
No.

Quote Quote by A.J.710 View Post
The water pressure inside is minimal. .
Actually it has enough pressure to support your weight and the weight of the water itself.

Quote Quote by A.J.710 View Post
I just basically want to know if my body weight of 200 pounds divided by the surface area of the hole 1" would make the pressure coming out of the hole 200 PSI plus the minimal water pressure inside.
No, the water flowing out would effectively have the same pressure as the atmosphere.

Quote Quote by A.J.710 View Post
Is the pressure directly transferred from my weight on top to the hole on the bucket or is there some sort of loss somewhere in the system where there won't be that type of drastic build up in pressure when I stand on top of the system.
There will be a build up of pressure in the water to support your weight which will continue to exist until all the water has left the bucket.
A.J.710
#3
May1-14, 11:26 PM
P: 28
Quote Quote by paisiello2 View Post
Libraries can be a dangerous place for thinking. Be careful.


No.


Actually it has enough pressure to support your weight and the weight of the water itself.


No, the water flowing out would effectively have the same pressure as the atmosphere.


There will be a build up of pressure in the water to support your weight which will continue to exist until all the water has left the bucket.
Is there some way to calculate the water pressure coming out of that 1" hole then?

paisiello2
#4
May1-14, 11:48 PM
P: 558
Pressure question

I already stated above it is effectively at atmospheric pressure.
A.J.710
#5
May1-14, 11:57 PM
P: 28
But with my weight on top of it wouldn't more water be forced out? Similar to a big balloon filled with water. If it is opened and the water is freely flowing out, when someone squeezes it it starts to shoot the water harder. Thats what I mean here. If the water is constantly being compressed down and the only source of escape is through that hole why wouldn't the pressure build up? It doesn't have to be just my weight. If there is a bucket of water and a 1" hole near the base and there is a huge hydraulic system pushing it down more water is going to come out as the force increases, correct?


Quote Quote by paisiello2 View Post
I already stated above it is effectively at atmospheric pressure.
256bits
#6
May2-14, 12:16 AM
P: 1,484
The pressure of the water at the bottom of the bucket is ρgh without you standing on it. Once you stand on the platform, the extra pressure at the bottom of the bucket will be your weight divided by the area of the platform, in addition to that from the height of the water column.

That is the pressure at the bottom of the bucket. If you open the hole, the water will exit to atmospheric pressure.

One cannot physically speak of the pressure of the water coming out of the hole. One would ask what is the extra velocity of the water coming out of the hole due to increased pressure, or what is the increase in water flow coming out of the hole due to increased pressure.
A.J.710
#7
May2-14, 12:20 AM
P: 28
I don't understand what you guys mean by atmospheric pressure. I am thinking of head pressure. For example if a tube was connected to the hole and extended upwards. The water will travel up the tube to the height of the rest of the water. Now if I stand on the bucket that water will be forced even farther upwards. Isn't that a reading of PSI of the water in the tube or is it something different?

Quote Quote by 256bits View Post
The pressure of the water at the bottom of the bucket is ρgh without you standing on it. Once you stand on the platform, the extra pressure at the bottom of the bucket will be your weight divided by the area of the platform, in addition to that from the height of the water column.

That is the pressure at the bottom of the bucket. If you open the hole, the water will exit to atmospheric pressure.
paisiello2
#8
May2-14, 02:03 AM
P: 558
Atmospheric pressure = 101.3kPA or 14.7psi.
http://en.wikipedia.org/wiki/Atmospheric_pressure

There's a subtle but important difference between the point just inside the bucket before it exits the hole and the point just outside the hole as the water flows out. Inside the bucket, the water velocity is negligibly small but the pressure is the total of the weight of the water and of you standing on it. Outside the bucket, the water is flowing very fast but the pressure has now dropped to atmospheric pressure.

If you attach a tube to the hole and extend it upwards, all you have done is effectively move the hole to the top of the tube. The hole and the top of the tube are both at atmospheric pressure. The only difference is that once the weight of the column of water matches the weight of the water remaining in the bucket plus your own weight, then the water column will come to rest.
crador
#9
May2-14, 09:24 AM
P: 25
A.J.710, I think the physical picture here is most intuitive of looked at in terms of energy (Disclaimer: I am not expert but have had dealings with problems like these in my own inventions).

The water in your bucket has a certain lot of energy that can be ascribed to it that is divided between kinetic energy (gross translation or movement -- speed), pressure-volume energy (due to its state and form -- takes energy to compress water vapour to liquid, energy consumed to evaporate etc.), and potential energy due to the water's elevation. These latter two entities make up the piezometric head -- which is reflected as gauge pressure (take a look at Bernoulli's equation, to make things clearer).

Now, standing on your circle will compress the water underneath you by your weight divided by the area of the circle, i.e. if you have a 200 square inch circle then you are only contributing 1 PSI to the pressure of the water. At the exiting outlet the pressure inside the bucket will then be that due to you standing on it, the atmospheric pressure pressing down upon it from above, and due to the weight of the water above the outlet. Outside the outlet the pressure is only that supplied by the atmosphere. Now, assuming the outlet is at a set elevation, there clearly is a difference in internal and external pressure. Water will thus be accelerated out of the bucket, and adopt the pressure outside. The surplus energy due to overpressure will be converted into kinetic energy as the water exits, which answers your questions as to why pressing harder causes a more vigorous flow of water to be expelled (as far as I know we can generally say and increase in velocity is accompanied by a drop in pressure if we are not adding energy to a fluid -- this is the idea behind the venturi tube and airplane wind, in the latter case we guide the air above the wing over a longer course than that below the wing, so it must be moving at a higher velocity, and hence we create a depression above the wing that keeps the airplane aloft).

Really what is happening is the water is transferring the energy stored in your elevated mass into kinetic energy of the water, with the pressure-volume energy of the water being the intermediate.

I hope my common-man's approach has been helpful, and am open to questions if you have any more. Also open to additions from our better informed peers, as I may be wrong.

Cheers
Khashishi
#10
May2-14, 04:12 PM
P: 887
Quote Quote by paisiello2 View Post
I already stated above it is effectively at atmospheric pressure.
No, that's wrong.

By standing on the bucket lid, you increase the pressure in the water by your weight divided by the area of the lid. The size of the hole doesn't really factor into the pressure.
paisiello2
#11
May2-14, 05:44 PM
P: 558
Are we taking about a point inside the bucket where the velocity is negligible or are we talking about a point outside the bucket where the velocity is significant?


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