Talk:Pressure

fix the functions

I don't have the experience to fix what is wrong with this page... The math functions are all messed up.

How so? --Selket ^Talk 23:04, 17 May 2007 (UTC)

The helpme template has been removed for now. Please put it back when you answer, so we know to check back. Thanks--Werdan7^{T @} 00:00, 18 May 2007 (UTC)

What are "Imperial Units"?

(Trying to settle a small revert war between me and another user.)

People from Commonwealth countries often refer to U.S. units as "Imperial Units". That is not correct. U.S. units are a simplification of the customary systems that the colonies inherited from England. These were not Imperial units, which were not defined until 1824, long after the U.S. became independent.

Some units are the same in both systems, including weight. Others have the same name but are not the same. An Imperial gallon, for example, is larger than a U.S. gallon.

Thus it does not make sense to refer to any measure based on the customary definition of the pound as "US" (the Imperial system uses it too) or "Imperial" (the unit is used in at least one other system). --Isaac R (talk) 23:03, 26 July 2008 (UTC)

Uh, not a revert war. Customary units is a vague term and I'm not sure a proper one. This article only lists the unit names, so standards is not relevant. For maodern usage they should be labeled US Customary units. -Fnlayson (talk) 00:04, 27 July 2008 (UTC)

A respectful revert conflict? ¶ You're right, "customary units" is vague, probably not the right term. But I don't know what the right term is. I just know that "Imperial" and "U.S." are both wrong and misleading, since neither system owns the International pound. (See pound (mass).) Perhaps "International" is the correct word, but I'm not sure that adjective applies to units derived from the International pound. Isaac R (talk) 00:40, 27 July 2008 (UTC)

It'll be OK as "customary units" unless someone has a better, more correct name. -Fnlayson (talk) 00:49, 27 July 2008 (UTC)

Consider English units.Headbomb {ταλκ – WP Physics: PotW} 05:21, 27 July 2008 (UTC)

"English units" refers to the 18th century hodgepodge that U.S. units are based on. Which did not include a unit of pressure. ¶ Blatant editorialization: the fact that we're even having this discussion should be profoundly embarrassing to every American who works in any scientific or technical field. Our inability to make the change to the metric system is a symptom of our cultural arrogance and shortsightedness. Isaac R (talk) 17:05, 18 August 2008 (UTC)

If it's the units of mass we're talking about, we can call them avoirdupois, otherwise how about imperial/US units? We certainly wouldn't call them international this would seem to imply SI units. JIMp _talk·cont 20:21, 7 July 2010 (UTC)

About the table

I find these tables that compare various units interesting but

1) they are also confusing to the novice because they do not explicitly explain how to convert 2) they are necessarily limited to about a half dozen columns, that is, only relations among a few units can be shown.

Instead I suggest basing a new table on the standard international unit with all the other units listed with their explicit conversion factors to and from the SIU. Joseph Grcar (talk) 16:18, 29 July 2008 (UTC)

Pressure of real gas

Given that a real gas has an unknown fundamental thermodynamic relation, its equation of state must necessarily also be unknown, therefore I will remove the reference to an equation for a real gas. LeBofSportif (talk) 22:39, 29 July 2008 (UTC)

Pa:atm conversion in table is weird

• The order of magnitude seems out by a factor of 1000 in the relationship between Pa and atm. Nesbit (talk) 18:53, 19 September 2008 (UTC)
  • My calculations show the same value as listed in the table: 1 atm = 101.325 kPa = 101,325 Pa so 1 Pa = (1/101,325) atm = 9.8692 x 10^-6 atm. -Fnlayson (talk) 22:16, 19 September 2008 (UTC)

Statistical definition

Shouldn't a statistical definition to be included on this page be proper? —Preceding unsigned comment added by Mppf (talk • contribs) 01:07, 11 August 2010 (UTC)

Mistake in chart?

1 Pa should equal 0.00001 Bar, not 0.0001 Bar. — Preceding unsigned comment added by 166.147.64.27 (talk) 16:39, 11 September 2011 (UTC)

Question

A well must be perforated at 8500 ft.  The casing is full of 9.5 ppg mud.  The well has 32 degree API gravity oil in the well and the fluid level is at 4400 ft. with a gas surface pressure of 820 psi and a gas gravity of .8.  What is the pressure differential between the casing and tubing? How do I figure this out??

                                                                       32 degree API gravity oil is = 7.22 lbs/gal    
                                                                        the gas gravity of .8 is = 1.2894  — Preceding unsigned comment added by 12.230.15.34 (talk) 16:44, 14 November 2012 (UTC) 

Pascal's Vases

I expected to find a mention of Pascal's vases in this article, but there isn't one; nor do they seem to have an article of their own. Are Pascal's vases just considered to be a historical curiosity and not important any more? Dezaxa (talk) 13:52, 4 January 2013 (UTC)

Negative pressure

I would really contest the phrase "Negative pressure must exist at the top of any tree taller than 10 m". It doesn't sound scientific at all... You imagine a 50 meters tall ladder with minions that pass a bucket of water from hand to hand to the top... There is no negative pressure on the ladder, even if the minions are soaked wet and touch each other... In fact, the whole ascent of sap article (linked at the negative pressure section) is kinda gibberish...(if it would be true, then the sap will raise to the top of a dead tree too, because there is no structural difference between green wood and dead wood, they have the same capillarity, etc). — Preceding unsigned comment added by 223.27.210.130 (talk) 10:53, 25 October 2013 (UTC)

I agree. I have never been convinced by the arguments put forward in favor of negative pressures existing in fluids. Dolphin (t) 11:54, 25 October 2013 (UTC)

I think what is meant is an apparent negative pressure acting on water molecules only. A real negative pressure would imply a vacuum that whooshes everything in - air molecules, leaves outside the cell, etc., which is nonsense. However intermolecular forces are specific to the molecules interacting, and there are attractive forces on the water molecules which specifically pull water molecules up the tall tree. The article Ascent of sap in fact refers to the apparent existence of large negative pressures in some living plants. I will change this article to express this more accurately. Dirac66 (talk) 22:32, 6 August 2014 (UTC)

Note that the negative pressure acts on the wall of the capillary as well, which has to be quite strong or it would implode. Petr Matas 15:37, 7 August 2014 (UTC)

It is not nonsense. I've cited the existence of negative pressures in the article now. Oreo Priest ^talk 13:24, 31 January 2015 (UTC)

Introductory example of a "dollar bill" is parochial

The introductory section of this article contains the following sentence:

A pressure of 1 Pa is small; it approximately equals the pressure exerted by a dollar bill resting flat on a table.

This sentence is parochial: it would be an unhelpful example to anyone not familiar with US currency. Such an example is unhelpful. I suggest removing it unless a replacement can be found which is less culturally specific.

Chrislaing (talk) 21:20, 9 December 2013 (UTC)

I agree. We should just get rid of it. This is about pressure not the pascal anyway. Jimp 15:36, 24 February 2015 (UTC)

Wikilinks to units

Hello Trackteur, I think that in lists and comparisons of pressure units, each unit should wikilink to the article about the unit, even though it has been wikilinked already. I think that at such places the wikilink usefulness outweighs the repeated link ugliness. Petr Matas 20:11, 17 August 2014 (UTC)

Liquid pressure as energy equation

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. 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)

References

Pressure as relates to heating/cooling or fluids

I was surprised that I can't seem to find an article or section anywhere on Wikipedia where it actually discusses compression heating and decompression cooling, except for a brief bit in Compression ignition engine. Probably there is some discussion in pages on HVAC and refrigeration as well, but what I'm looking for is a technical page I can link to so people can read about why fluids (i.e. air) heat up when pressurized (well, compressed, anyway). I've looked on the page on Compressed air, and it says nothing. I believe it mentioned it on the page on Bleed air as well (it's been a while), but the point is, that none of these are really relevant to link to if I wanted to link to a page explaining why air heats as it's compressed. For example, I was on the page on Gas compressor(or maybe it was the page on Rotary screw compressors), and it mentions the elements of the system dedicated to "cooling the compressed air", but it doesn't say anything else. I thought it might be useful to link to a page explaining why the air would NEED cooling, but I can't find anything, unless I wanted to link to the page on diesel engines and let them read between the lines. Which doesn't seem very encyclopedic, somehow..45Colt 07:16, 19 August 2015 (UTC)

You could link to Adiabatic process#Adiabatic heating and cooling. NebY (talk) 20:00, 19 August 2015 (UTC)

Pressure on ears?

Pressure on ears is the same as pressure on other parts of the body at the same depth. Why specify ears? • • • Peter (Southwood) ^(talk): 05:59, 5 August 2016 (UTC)

Force over volume??

Is there a variant of pressure that is used for force over 3-dimensional volume?? If not, please explain why. Georgia guy (talk) 14:59, 1 May 2017 (UTC)

Please explain

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:

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

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)