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I pressure constant over time?

by victor.raum
Tags: constant, pressure, time
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Apr11-12, 11:54 AM
P: 71
I'm at the end of my vector calculus textbook, and there is a section on how vector calculus can be used to derive Euler's for fluid flow. The equations they give are:

[itex]\rho\left(\frac{dV}{dt} + \nabla V \cdot V\,\right) + \nabla p = 0[/itex]


[itex]\frac{d\rho}{dt} + \nabla \cdot (\rho V) = 0[/itex]

Where p stands for pressure, ρ stands for mass density, and V stands for velocity.

The derivations they give are totally understandable. It took me a few days to fully wrap my head around all of the derivations' nooks and crannies, but I now totally understand them.

Yet, now that I have my head out of the derivations I find myself looking at the equations and noticing that they don't describe how pressure changes with time. It looks to me as if the equations assume some preexisting spatial pressure gradient, and they describe how this gradient causes a fluid to move, but they don't describe how the fluid movement changes the pressure gradient.

Does that imply that this sort of fluid, in a perfect world, would be able to sustain a static pressure gradient? Or does the pressure change with time in some way that my book just has not bothered to mention?

Can a moderator change this thread's title to "Is pressure..." instead of "I pressure..." Silly type-o, heh.
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Apr11-12, 01:13 PM
P: 409
I'm not sure if I understood your question but in the equation the gradient of p is , in general, a function of x,y,z and t. If V and rho are not changing with t, then we have a static pressure gradient too.
Apr11-12, 04:05 PM
P: 71
I would think intuitively that the pressure function p is a function of time, yes. The problem is that I need to know [itex]\frac{dp}{dt}[/itex] in order to integrate, be it either numerically or symbolically, so that I can learn how p actually evolves with time. Yet the derivative of p with respect to t is not found in the equations. This confuses me. And so it leaves me wondering if perhaps p is *always* just constant over time in perfect fluids?

Apr11-12, 04:49 PM
P: 1,008
I pressure constant over time?

That's interesting. Your continuity equation (the second one) allows for compressibility effects, but your momentum equation (the first one) is in the incompressible, inviscid form (it also happens to neglect gravitational effects, but that's less important). For an incompressible, inviscid fluid, you could use those equations in combination with an equation of state to determine the pressure at any location in the flow. As for why the change in pressure doesn't show up explicitly? It's simply because those equations are simply continuity (conservation of mass) and conservation of momentum, neither of which has anything to do with the rate of change of pressure with time. The fluid flow is dependent only on the pressure at a given time, not the rate of change of pressure. One way to look at it is that you have 6 variables (P, T, density, 3-components of velocity), but only 4 equations at the moment (the continuity equation is one equation, but the momentum equation is 3 equations, as it is a vector equation). So, you still need 2 more equations to be able to solve the system. For solving the full Navier-Stokes equations, usually, you would add in an energy conservation equation and an equation of state, which then allows you to solve the system. For an incompressible, inviscid flow, you could neglect the energy equation, as the flow will be isothermal (but you still need one more equation than you have there).
Apr11-12, 08:54 PM
P: 71
<reply deleted because I accidentally hit "submit" too soon>
Apr11-12, 11:25 PM
P: 71
Thanks cjl for the reply.

That's correct, I do neglect gravity (and I also assume the fluid in question has no viscosity). I definitely haven't gotten to the full Navier-Stokes equations yet. But you say that incompressible inviscid flow can ignore energy and be represented by the system I already have if only I added in one more equation, and I'd be happy with that for the moment, so that's encouraging. But what is that equation exactly? Would that need to be this "equation of state" that you spoke of? All of this stuff so far is coming from a brief aside in my vector calc book, and I don't quite feel like tearing into any fluid mechanics books yet (though I have some). Is there anywhere I might look for a quick explanation, or is the explanation simply not quick?

Also, I'm confused when you say that the momentum equation I gave is for an incompressible fluid. To my understanding an incompressible fluid is defined as one for which [itex]\nabla \cdot V = 0[/itex], and this is by way of some interesting logic using the transport theorem which I don't feel the need to repeat here.

Yet, the way I understand the derivation of the momentum equation doesn't require using [itex]\nabla \cdot V = 0[/itex], or if it does then I don't see how. Nor does the resulting momentum equation imply [itex]\nabla \cdot V = 0[/itex], or if it does then (to repeat) lack of experience prevents me from seeing how. I'll give the derivation for that momentum equation here, and maybe someone can help me see why and how the result implies or requires incompressibility.

Firstly, here is the transport theorem just for reference. [itex]V[/itex] is the velocity field of a fluid, [itex]\phi[/itex] is the flow line function over that velocity field, [itex]W[/itex] is some region of space, [itex]d^3V[/itex] is a differential volume, and [itex]f(x, y, z, t)[/itex] is some scalar function. The transport theorem is then

\frac{d}{dt}\int_{\phi(W)} f \; d^3V =
\int_{\phi(W)}\frac{\partial f}{\partial t} + \nabla \cdot (fV) \; d^3V

If we assume conservation of mass and apply the transport theorem with [itex]f = \rho[/itex], where [itex]\rho[/itex] is the mass density function, then the continuity equation pops out. Specifically that is

[tex]\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho V) = 0[/tex]

If the fluid is incompressible then this simplifies to

[tex]\frac{\partial \rho}{\partial t} + \nabla\rho \cdot V = 0[/tex]

but we need not necessarily make that assumption for the rest of the derivations here in this post.

Now, to derive the momentum equation we first consider a volume of space which we will call [itex]W[/itex], the boundary surface of which we will call [itex]\partial W[/itex]. If we approximate [itex]W[/itex] as a polyhedron then [itex]\partial W[/itex] will simply be a collection of nice flat surfaces which are very easy to work with. The force on a given surface caused by the fluid pressure will be approximately

[tex]\vec{F} = - \, p \cdot \vec{n} \cdot A[/tex]

Where [itex]\vec{n}[/itex] is the normal vector to the surface, [itex]A[/itex] is the surface's area, and [itex]p[/itex] is the pressure sampled at some arbitrary point on the surface and then assumed to be constant across the entire surface.

The total force on the whole volume as the polyhedron's sides increase in number to form a better approximation of the volume's surface is then found by the vector-integral equation

[tex]\vec{F} = -\int_{\partial W} p \cdot \vec{n} \cdot d^2A[/tex]

If we momentarily consider only this equation's [itex]x[/itex] component then we can do some interesting manipulations. We will fiddle things into a form amenable to the application of the divergence theorem, and to do this we will use a vector [itex]\vec{e}[/itex] defined as [itex]\vec{e} = (1, 0, 0)[/itex]

F_x =
&-\int_{\partial W} p \cdot n_x \cdot d^2A =
-\int_{\partial W} p (\vec{e} \cdot \vec{n}) \cdot d^2A =
-\int_{\partial W} p \vec{e} \cdot d^2\vec{A} =
-\int_{W} \nabla \cdot (p \vec{e}) \; d^3V = \\
&-\int_{W} \nabla p \cdot \vec{e} + p(\nabla \cdot \vec{e}) \; d^3V =
-\int_{W} \nabla p \cdot \vec{e} \; d^3V =
-\int_{W} \frac{\partial p}{\partial x} \; d^3V

Similar arguments for the [itex]y[/itex] and [itex]z[/itex] dimensions using [itex]\vec{e} = (0, 1, 0)[/itex] and [itex]\vec{e} = (0, 0, 1)[/itex] then show that overall we have

[tex]\vec{F} = -\int_W \nabla p \; d^3V[/tex]

Yet, the force on a moving region is also defined as the rate of change of its momentum over time, which is yet another vector-integral, which in this case is one we can apply the transport theorem to. We will write out the case for the [itex]x[/itex] dimension, and other dimensions are similar.

F_x =
\frac{d}{dt}\int_{\phi(W)} \rho V_x \; d^3V =
\int_{\phi(W)} \frac{\partial}{\partial t}(\rho V_x) +
\nabla \cdot (\rho V_xV) \; d^3V

And so we have

\frac{\partial}{\partial t}(\rho V_x) + \nabla \cdot (\rho V_xV) =
-\frac{\partial p}{\partial x}

If we expand out the [itex]\frac{\partial}{\partial t}(\rho V_x)[/itex] then we can use the resulting [itex]\frac{\partial \rho}{\partial t}[/itex] term to combine this result with the continuity equation, and after the combination we have

[tex] \rho\left(\frac{\partial V_x}{\partial t} + \nabla V_x \cdot
V\right) = -\frac{\partial p}{\partial x} [/tex]

The more general vector form then reads as

\rho\left(\frac{\partial V}{\partial t} + \nabla V \cdot V\right) = - \nabla p

Now, to my (perhaps untrained) eyes it seems that [itex]\nabla \cdot V = 0[/itex] is not a required assumption nor an implication for this momentum equation.

Over and out, and still a bit confused.
But wow, this stuff is fun.

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