# Talk:Dynamic pressure

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## R = 1716 what?

I can't figure out any units in which R=1716. I get 25.54 mbar/(°R·lb/ft3), for what it's worth. —Ben FrantzDale 16:00, 31 January 2007 (UTC)

It's 1716 ft·lbf/(slug·°R). I'll add it in. 164.107.197.169 06:47, 9 February 2007 (UTC)

## Usefulness of section: Alternative forms

What is the purpose of the formulas given in the section alternative forms? What use do they have? -- Crowsnest (talk) 14:54, 7 November 2008 (UTC)

Being unreferenced, and of no significant use (as far as I can see), I removed the following alternative form from the article:

 Alternatively, considering for example a spacecraft during launch and using the formula for the air density as a function of altitude (only valid below the tropopause), the dynamic pressure associated with the spacecraft can be expressed as:[citation needed] ${\displaystyle q\ =\ {\frac {v^{2}\cdot p_{0}}{2\cdot R\cdot \left(T_{0}+L\cdot h\right)}}\cdot \left({\frac {T_{0}}{T_{0}+L\cdot h}}\right)^{\frac {g\cdot M}{-R^{*}\cdot L}}}$ where (using SI units): ${\displaystyle q}$ = dynamic pressure in pascals ${\displaystyle v}$ = spacecraft's velocity in m/s ${\displaystyle p_{0}}$ = atmospheric pressure at sea level in pascals ${\displaystyle M}$ = molecular weight of dry air (28.9644 g/mol) ${\displaystyle R^{*}}$ = universal gas constant (8.31432×10³ N·m/(kmol·K) ) ${\displaystyle T_{0}}$ = absolute air temperature at sea level in kelvin ${\displaystyle L}$ = atmospheric temperature lapse rate (−0.0065 K/m) ${\displaystyle h}$ = altitude above sea level in m ${\displaystyle g}$ = acceleration due to gravity at the Earth's surface in (9.80665 m/s2)

You may put it back if you can provide reliable references and think it useful. Crowsnest (talk) 11:44, 13 November 2008 (UTC)

## Velocity Pressure and Flow

I looked up Velocity Pressure and was directed to this page on Dynamic Pressure. I find these formulas a little confusing. I was expecting to find the formula for determining air velocity from velocity pressure. The formula I expected to see was V = 1096.7 * sqrt(Pv / density). I'm further confused with the use of Q or q to represent Velocity Pressure since everywhere else that I've looked Q or q is used to represent flow. I also expected to see how Velocity Pressure is measured (Total Pressure - Static Pressure = Velocity Pressure).

I don't know enough to know if these are mistakes or if it's a matter of mixed nomenclatures. In the latter case, I would expect to see seperate pages for Velocity Pressure and Dynamic Pressure. Steveandaugie (talk) 19:30, 18 March 2009 (UTC)

Hi Steveandaugie. In fluid dynamics, and particularly aerodynamics, dynamic pressure is a very important concept. It is always abbreviated q but I concede that in hydrodynamics q is often used for volume flow rate.
In hydrodynamics, and in low-speed aerodynamics, q is always defined as ${\displaystyle {\frac {1}{2}}\rho V^{2}}$. You can safely use this formula to calculate air velocity when you know air density and dynamic pressure (or velocity pressure), providing the air velocity is no greater than about 30% of the speed of sound.
You are correct in writing:
dynamic pressure = q = total pressure - static pressure
(In high speed aerodynamics, when Mach number becomes significant, it becomes a little more complicated. In the USA and elsewhere, the term impact pressure is introduced. This is often abbreviated qc.)
Different abbreviations are often encountered in different countries, different fields, and among different authors. We all learn to live with that and use our favourite character, Greek or otherwise! Dolphin51 (talk) 22:42, 18 March 2009 (UTC)

## Unit balance

Care should be taken in balancing units. I believe ${\displaystyle {\tfrac {1}{2}}\rho V^{2}}$ mentioned in the top of the article should not this as to get the units to come out a 9.81m/s2 would need to be muliplied on the right hand side. —Preceding unsigned comment added by 208.250.9.26 (talk) 17:38, 22 April 2009 (UTC)

Dynamic pressure has the units of pressure. If ${\displaystyle {\tfrac {1}{2}}\rho V^{2}}$ is evaluated using density units of kg/m3 and speed units of m/s the result will be kg/m/s2 or kg.m/s2/m2 which is the same as N/m2 or Pa. These are units of pressure.
If ${\displaystyle {\tfrac {1}{2}}\rho V^{2}}$ is evaluated using density units of pounds (mass) per cubic foot and speed units of feet per second, it is necessary to multiply by g0 of 32.17 ft/sec2 in order to get a result in pounds (force) per square foot. That is because the lbm and lbf are not consistent units with respect to the formula ${\displaystyle {\tfrac {1}{2}}\rho V^{2}}$. Dolphin51 (talk) 23:29, 22 April 2009 (UTC)

Thank you! you are correct. —Preceding unsigned comment added by 208.250.9.26 (talk) 21:19, 23 April 2009 (UTC)

## Dynamic pressure equal to difference between stagnation and static pressures

The statement (referenced) on the page, "[t]he dynamic pressure is equal to the difference between the stagnation pressure and the static pressure" is not, s.s., true. It is a good approximation (how accurate does one wish to be?) but density variation becomes larger than about 5% after about Mach 0.3. The statement should be modified to read something akin to, "the dynamic pressure approximates the difference between the stagnation pressure and static pressure" (lobbing in the usual assumptions - time invariant, inviscid, irrotational, flows with gravity neglected, etc). See discussion page under Impact pressure, and/or talk page under Dolphin51's talk page on impact pressure. (Weirpwoer (talk) 16:07, 25 November 2009 (UTC))

I think the statement can be made completely accurate by changing it to: in incompressible flows, dynamic pressure ${\displaystyle {\tfrac {1}{2}}\rho v^{2}}$ is equal to the difference between the stagnation pressure and the static pressure. In compressible flows, dynamic pressure ${\displaystyle {\tfrac {1}{2}}\rho v^{2}}$ is a good approximation of the difference between stagnation and static pressures at low speeds, and is usually acceptable up to about thirty percent of the speed of sound. Dolphin51 (talk) 21:54, 25 November 2009 (UTC)
I wouldn't go so far as saying "completely accurate" as there is really no such thing as an incompressible flow which begs the question as to whether the words "equal to" should be used at all. I see where you're coming from, but it should be pointed out that this relationship is a convention and does not have a physical basis. Perhaps saying "is considered to be equal to" would be a safer position (noting the conditions). (Weirpwoer (talk) 15:44, 26 November 2009 (UTC))
Saying ${\displaystyle {\tfrac {1}{2}}\rho v^{2}}$ is equal to the difference between stagnation and static pressures is a statement of Bernoulli's equation. I am not aware of Bernoulli's equation ever being described as total pressure is considered to be equal to static pressure plus dynamic pressure.
Similarly, Bernoulli's equation is analogous to the Law of Conservation of Energy. This Law is always expressed in terms of energy remaining constant, or the energy before an event being equal to the energy after the event. The Law is not expressed in terms of considered to remain constant or considered to be equal.
I think the notion that liquids are compressible (or are not incompressible) is unnecessarily esoteric for Wikipedia articles on dynamic pressure, Bernoulli's principle, and others in this series. Bernoulli's principle still commands the greatest respect from scientists even though it relies on the idea that liquids are incompressible and inviscid. Dolphin51 (talk) 21:43, 26 November 2009 (UTC)
It is a statement of the Euler-Bernoulli equation inasmuch as density is held constant along a common streamline. This is easily shown by integrating Euler's equation with density held constant. The statement must be made with that condition expressed. Whilst I ought to have included at the end of the first sentence of my last reply above the words "where the fluid in question is a gas", it should be reasonably understood (given the use of the Mach number and the ratio of specific heats elsewhere in the text) that the Euler-Bernoulli equation finds widespread use in aerodynamics. Your remark regarding liquids is quite acceptable concerning the assumption that density is held constant. However, NACA 837 (Langley, 1949?) does not state that dynamic pressure equals the difference between total and static pressures - this definition is reserved for Impact pressure as previously discussed - and given the dissimilar domains for the application of dynamic pressure it seems a safer bet to stress the associated condition.
As for the amendment to the article, if you're content with the following then I would propose it be inserted under the section on 'Physical meaning': [Dynamic pressure:] In incompressible flow, dynamic pressure is the difference between total pressure and static pressure. (Footnote: Glover, D. R., Jr.. (1965). NASA Aerospace Dictionary. Cleveland: NASA GRC.)
I would further propose the removal of the remark on controversy (not limited to British authors, in my experience!) under "Compressible flow," renaming "Compressible flow" to "Alternative forms of the equation" and modifying the introductory sentence from "...sometimes called velocity pressure or impact pressure..." to "...sometimes called velocity pressure or kinetic pressure..." leaving the term "impact pressure" to be mentioned under the "See also" section. (If including the text you've proposed, mentioning "...thirty percent of the speed of sound..." it might be prudent to specify also "in a perfect gas.") (Weirpwoer (talk) 03:13, 28 November 2009 (UTC))

## Compressible flow

Putting the dynamic pressure equation in terms of the Mach number, ${\displaystyle \gamma }$ and static pressure under a Compressible flow heading is misleading. Acknowledging that the equation for dynamic pressure has limited usefulness at higher subsonic Mach numbers, it might be more appropriate to retitle this section as something like 'alternative forms of the [dynamic pressure] equation' and addressing any confusion between dynamic and impact pressures separately. (Weirpwoer (talk) 15:37, 26 November 2009 (UTC))

I agree. As I promised here, I will think about this over the next couple of days, and suggest a suitable way to rectify the errors I introduced by assuming ${\displaystyle {\tfrac {1}{2}}p_{s}\gamma M^{2}}$ is significant for all Mach numbers. Dolphin51 (talk) 21:55, 26 November 2009 (UTC)