Talk:Pressure

I think that DHeyward's statement about energy equation is incorrect. Unlike pressure, the gravitational potential energy is not a function of position of measurement, but rather of position of the liquid body's center of gravity (and depends also on the choice of the zero height level), because it is an integral quantity. Therefore, the statement "energy is constant throughout the vessel" is a nonsense. Unless this is fixed somehow (I doubt that it is possible), I think that the paragraph should be removed. Petr Matas 06:46, 7 November 2014 (UTC)

• I think DHeyward's statement is correct. It is true that the total potential energy of the body of liquid is proportional to the height of the centre of mass of the body above the datum, but that isn't what Heyward's statement is about; it is about the total energy of any small parcel of liquid. A small parcel of liquid near the surface of the liquid has high potential energy per unit of mass because of it's large height above the datum. A small parcel of liquid near the bottom of the liquid has low potential energy per unit of mass because of it's small height above the datum.Dolphin (t) 13:18, 7 November 2014 (UTC)

• It's mentioned in a similar fashion in Bernouilli's principle. Thus an increase in the speed of the fluid – implying an increase in both its dynamic pressure and kinetic energy – occurs with a simultaneous decrease in (the sum of) its static pressure, potential energy and internal energy. If the fluid is flowing out of a reservoir, the sum of all forms of energy is the same on all streamlines because in a reservoir the energy per unit volume (the sum of pressure and gravitational potential ρ g h) is the same everywhere.^[1] It's the reason there is a static pressure drop when fluid changes from static to flowing in a horizontal pipe: pressure energy is converted to kinetic energy. I'll add the citation. --DHeyward (talk) 15:30, 7 November 2014 (UTC)
  • So it was possible to fix it. Thanks to you both. Petr Matas 19:38, 7 November 2014 (UTC)

• ??. It's more clear with unit volume but the concept is the same. A cubic-centimeter at the top of a fluid has the same total energy as a CC at the bottom. The top surface CC has more gravitational potential energy and less pressure energy while the CC at the bottom has more pressure energy and less gravity energy. The sum is constant. (gravity head + pressure head = constant) --DHeyward (talk) 22:43, 7 November 2014 (UTC)
  • I was confused by the expression "total energy" - I thought that you were speaking about the energy of the entire liquid body. The body's position is given, so the expression "throughout the body" is meaningless for it. When it's per volume element, it's ok. Petr Matas 16:12, 8 November 2014 (UTC)

• Ahh. It's a property of fluids. The center of mass potential energy is give but the tradeoff as the energy is calculated away from the center is the same at any other point. Per unit volume is easier to express and understand. The principle is the same as the volume can be made infinitely small such that a point calculation yields the same constant. It's the reason a deep ocean of 2 miles depth has the same level as the same ocean in two feet of water. Take an empty tube that's two miles long with a valve at the bottom, submerse it to 2 miles and open the valve - the water rises to the same level as the rest of the ocean because the energy is constant. --DHeyward (talk) 04:42, 9 November 2014 (UTC)

  @DHeyward how??????? 41.210.141.245 (talk) 11:15, 13 May 2024 (UTC)

Please explain to me this paragraph and the following formula after this paragraph- "Pressure is a scalar quantity. It relates the vector area element (a vector normal to the surface) with the normal force acting on it. The pressure is the scalar proportionality constant that relates these two normal vectors: {I am unable to copy the formula?}" WorldDiagram837 (talk) 15:26, 4 January 2025 (UTC)

The formula is:

$${\displaystyle d\mathbf {F} _{n}=-p\,d\mathbf {A} =-p\,\mathbf {n} \,dA.}$$

The formula F = pA is used to calculate the force F on a surface of area A when the pressure p is uniform across the whole of the surface. An example of this is a flat, horizontal surface of area A in a stationary fluid such as water or air. The surface is horizontal so the pressure is the same at all points on the surface, and the vertical force is determined by F = pA.

If the surface is not flat and horizontal the pressure varies according to the depth of the liquid or the gas. The force acting on the surface cannot be determined by a linear equation; it requires a differential equation such as the one posted above. This differential equation can then be integrated to calculate the total force acting on the surface. Dolphin (t) 06:39, 5 January 2025 (UTC)

alright, but why does the second part (the one with A bold) suddenly equal the third part (the one with bold n). Like where did that bold n come from? WorldDiagram837 (talk) 10:17, 6 January 2025 (UTC)

Good question. The force F_n is a vector but its direction is not determined by the pressure p because pressure is a scalar quantity. In the case of a flat surface, the direction of the force is determined by the orientation of the surface. The area of the surface and its orientation are denoted by the vector A which is normal to the surface.

In the case of a curved surface we consider a vanishingly small area, represented by the differential dA, that is a tiny part of the whole curved surface. This vanishingly small area experiences a vanishingly small force dF. Both dF and dA are vectors and are related by the differential equation:

dF = p dA

An alternative way to determine the differential force dF is to make use of a vector n that represents the direction that is normal to the vanishingly small area, but has a magnitude of exactly unity (one). It is the unit vector in the direction that is normal to the vanishingly small area dA. Now that the direction of the differential force dF is determined by the unit vector n, the magnitude of the differential area can be given by dA which is now a scalar quantity. This alternative way to determine dF is to use the alternative differential equation:

dF = p n dA

where dF and n are vectors, but p and dA are scalars. Dolphin (t) 12:02, 6 January 2025 (UTC)

ah, got it. thank you very much

just one thing- what do you mean by - In the case of a flat surface, the direction of the force is determined by the orientation of the surface.[ I suppose from your following statement that the orientation has to do something with the direction of the area vector]

Also, why are we talking about area as a vector here? WorldDiagram837 (talk) 13:10, 6 January 2025 (UTC)

I used the word orientation to avoid writing about the direction of a surface. It would be odd to talk about a surface having a direction, even though the force clearly has a direction.

When fluid pressure acts on a surface it causes a force to act normal to that surface. It is therefore necessary to treat the area (A or dA) of the surface as a vector. This also seems a little odd so you can see the attraction of defining the normal vector that exists normal to the surface, and then making use of the unit vector having a magnitude of unity (one). In that way we avoid having to explain the “direction of a surface of area A” (or dA).

In physics the concept of the unit vector is often used to avoid having to define things like the direction of a circle, or direction of a cube etc. Dolphin (t) 14:19, 6 January 2025 (UTC)

alright. thanks for the explanation ( seems like area being a vector is still going to itch my brain for a while) WorldDiagram837 (talk) 15:22, 6 January 2025 (UTC)

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Please add this template. The page is linked on the template but the template is missing from the page. 2405:6E00:657:CC5D:50AA:F668:14BE:7152 (talk) 21:38, 27 August 2025 (UTC)

 Not done: According to the page's protection level you should be able to edit the page yourself. If you seem to be unable to, please reopen the request with further details. Day Creature (talk) 21:57, 27 August 2025 (UTC)

"1 atm = 101,325 Pa / 33.066 = 3,064.326 Pa"
This doesn't make sense. Should it be:
1 lfs = 101,325 Pa / 33.066 lfs = 3,064.326 Pa
or conversion factor = 101,325 Pa / 33.066 lfs = 3,064.326 Pa/lfs
? Eaberry (talk) 17:08, 25 November 2025 (UTC)

Light or sound ~2026-43508-6 (talk) 20:25, 20 January 2026 (UTC)

There are some similarities (between light and sound) as well as many differences. What is your question? Dondervogel 2 (talk) 21:41, 20 January 2026 (UTC)