0
$\begingroup$

My primary school physics textbook trys to brainwash me with the idea that a region in a fluid with fast velocity (often) has lower pressure than a region that flows slowly. Examples of this include how a plane gains lift due to the different velocity of air above and below the wings.

In my attempt to formalize this statement, let the velocity of a fluid element (volume $=dx_1dx_2dx_3$) be $\mathbf v=(v_1,v_2,v_3)$. Consider newton's 2nd law in derection $x_1$: $$ \rho\times (dx_1dx_2dx_3) \frac{dv_1}{dt}=-Adp=-p'_1 dx (dydz)\\ \Rightarrow \rho\frac{dv_1}{dt}=-p_1'=-\frac{\partial p}{\partial x_1}, $$ where $p$ is the pressure, $\rho$ is the density and $A$ is the cross sectional area of the fluid element.

Combining similar results for directions $2,3$, we get $$ \rho \mathbf{\dot{v}}=-\nabla p. $$ I don't know if there is a name for the equation I've just derived.

So here's what that tells me

  1. the textbook is mostly right, in that the fluid accelerates down a pressure gradient, so when it reaches a region with lower pressure, it should be relatively faster.
  2. However, the book can be wrong. Counterexamples include a bomb explosion. There is a region with both high velocity and high pressure.

Is my understanding correct? Is there a better equation to explain this?

$\endgroup$
  • $\begingroup$ Your differential equation doesn't necessarily cause $\rho$ and $p$ to be anticorrelated. What your book was probably referring to was Bernoulli's principle, which is a result in steady state flows ($\partial_t \vec{v}=0$). $\endgroup$ – jacob1729 May 19 at 16:42
  • $\begingroup$ @jacob1729 Yes that's my point: my equation only partially supports the claim. Could you write an answer? $\endgroup$ – Ma Joad May 19 at 21:53
0
$\begingroup$

trys to brainwash me

that would be malicious intent, whch I don't think is the case here.

I don't know if there is a name for the equation I've just derived.

Of course. It's the "continuity equation" and it's a step to derive the acoustic wave equation which describes a lot of what you may be talking about. https://en.wikipedia.org/wiki/Acoustic_wave_equation

a region in a fluid with fast velocity (often) has lower pressure than a region that flows slowly.

Blanket statements like this are always problematic especially if they don't specify the what exactly you mean and what's the scope of applicability is. It would help if you could properly quote the exact wording that you are referring to.

Examples of this include how a plane gains lift due to the different velocity of air above and below the wings.

That is governed Bernoulli's principle which applies to fluids in more less constant motion. Sound on the other hand applies (mostly) to longitudinal compression waves that move through a fluid at rest (or at least relative to the bulk motion of the fluid).

Hence my comment about the "scope of applicability".

Proper definition is also really important. Let's look, for example, at spherical sound wave in the far field of the source. Both velocity and pressure decay with $1/r$ where $r$ is the distance from source. So on the macroscopic level, the statement would be wrong. However,if you zoom in to the level of local air compression, you can see that at the areas of high compression the AC pressure is high and the particle velocity is 0. The velocity is highest where the AC pressure is 0 since the particle want to move as quickly as possible from high to low pressure. So on the microscopic level the statement would still be correct. So it depends on how you define the term "region".

$\endgroup$
  • 1
    $\begingroup$ The equation derived in the question isn't the continuity equation -- it's the simplified momentum equation, where convection and viscous terms are ignored. (Unless the dot notation means substantial derivative, in which case maybe the convection term is there but another term is missing) $\endgroup$ – tpg2114 May 19 at 15:55
  • $\begingroup$ It is hard to specify the scope of applicability: the book says nothing about it. My example of aircraft shows what the book means. Could you also comment on my equation? Is it correct? $\endgroup$ – Ma Joad May 19 at 21:57
  • $\begingroup$ @tpg2114 Could you write down the proper version for the equation in the question? I can't find the simplified momentum equation online. $\endgroup$ – Ma Joad May 19 at 22:00
0
$\begingroup$

Bernoulli's theorem states that along streamlines the following is constant along streamlines of a fluid:

$H = \frac{u^2}{2} + \frac{p}{\rho} + \phi$

where $\phi$ is a potential generating a body force via $\vec{f}=-\rho \vec{\nabla}\phi$ (such as for example, the gravitational potential $gh$). The first term here is the kinetic energy per unit mass of fluid, whilst the last two terms can be thought of as the potential energy per unit mass. As such Bernoulli's principle is often said to be a consequence of energy conservation.

The requirements for this to be true are that the fluid be incompressible and in steady state. We also need to neglect viscocity. If the fluid is additionally irrotational, then not only is $H$ constant on streamlines, it is constant everywhere.

One typical illustration of this is that as a fluid passes through a constriction in a pipe, it must speed up in order for mass conservation to hold. Bernoulli's principle then implies that the pressure is lower in the higher velocity region in order to keep $H$ a constant.

$\endgroup$
-1
$\begingroup$

Higher the pressure ,higher will be the velocity.Because consider a tank containing water and you put a hole at the bottom and top of the tank.Pressure will be high at the bottom of the tank and so will be the velocity of the water coming out .But at the top the pressure is low and the velocity will be low.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.