Hot answers tagged fluid-dynamics
12
The more interesting question is, "What does air feel like when it is moving away from me?" The answer is that there is really no sensation at all. You feel all the air coming into the vehicle, because it has a bulk momentum with respect to your frame of reference. However, air being sucked out of those same windows and doors is being pulled from a large ...
9
Your assumption that there is a significant pressure differential due to fluid dynamics is correct. The assumption that it is a lifting force is not. An airplane generates lift because it has been engineered with lift in mind. An F1 car actually generates a powerful down force to push it against the track, allowing it to get better traction than it ...
7
There is one famous example in which there is no solution for the Stokes' flow case: Stokes flow around a cylinder, which is approriately named Stokes' paradox. In this case it is impossible to match the boundary conditions
both at infinity (uniform flow) and at the cylinder surface (no-slip) with Stokes flow dynamics. See e.g. paragraph 6.4 of the Fluid ...
6
From the Wikipedia article for Reynolds number:
In fluid mechanics, the Reynolds number (Re) is a dimensionless number that gives a measure of the ratio of inertial forces to viscous forces and consequently quantifies the relative importance of these two types of forces for given flow conditions.
In addition to measuring the ratio of inertial to ...
4
Overblowing is a phenomenon that exists in all wind instruments. The details of the physics are different from one instrument to the next, but there is a broad similarity, which is that it's the result of a nonlinear interaction between the air column and whatever is driving the air column.
The recorder is in fact one of the simpler examples to understand. ...
4
excellent discussion, Im currently researching small scale water falls in the UK. In my research so far using 1/3 octaves I have found the spectra of various shaped waterfalls to be similar (and a combination of Pink and White). Special features, rocks, curvature, head height seem to have a considerable effect as does width and angle of inclination with ...
4
I belive you have it pretty much settled already. If I was to change anything, I would shrink instead of adding more items:
Identify the relevant quantities of your system: Energy, Momentum, entropy, electric charge, mass ...
Which may or may not be conserved. If you have boundary conditions, most probably you don't have energy and/or momentum ...
4
This answer is a bit of a long story, but I have split it up for the different statements for your convenience. Having thought about it a bit more after the discussion with @Mephisto I actually believe that Bernoulli's equation is not applicable in points B and C, because it is based on conservation of energy and therefore only applies if wall friction is ...
3
Turbulence is not one of the great unsolved problems in physics. Physics tells us exactly how turbulence emerges as a direct consequence of local mass and momentum conservation. We can create multiparticle computer models such as lattice gas automata that generate turbulence at large length and time scales. We can write down the equations that govern ...
3
A perfect fluid is defined by the property that, in the local rest frame, it allows no energy fluxes and no anisotropic stresses. Thus, at a given space-time point, in the local rest frame [in which the components of the 4-velocity are $u^{\alpha} = (1, 0, 0, 0)^{\mathsf{T}}$], the energy momentum tensor components are $T^{\alpha\beta} = \mathrm{diag}(e, p, ...
3
Objects don't accelerate because they're inside other objects. Objects accelerate because other objects make forces on them. The chain of cause and effect here is that the box can affect the air, then the air can affect the helicopter.
The answer to the question depends on how rapidly the box is accelerated.
To pick an extreme case, suppose that the box is ...
3
First off, please don't use units with $c\ne 1$ in GR. It makes everything horribly messy.
What we normally think of as a ruler or clock measurement is represented in GR by an upper index quantity like $\Delta x^\mu$. Therefore in a Cartesian coordinate system in the fluid's rest frame, we are guaranteed that $u^\mu=(1,0,0,0)$, not $(-1,0,0,0)$. This is ...
3
Are you referring to the exact relativistic equivalent to Navier-Stokes equation or a more general Dissipative Relativistic Hydrodynamics Equation?
The "relativistic equivalent to Navier-Stokes equation" would be something like this:
There would be an energy momentum tensor with the following form:
$T_{\mu\nu} = (e+p)u_\mu u_\nu - p g_{\mu\mu} + ...
3
What distance can a cannonball traverse thru water without losing too much kinetic energy? For a back-of-the-envelope calculation we start from the observation that this distance scales with the ratio of the kinetic energy of the cannonball and the drag force exerted on the cannonball.
Let's denote the ball's radius by $R$, its speed by $v$, and its mass ...
3
I remember attending a seminar by Unruh a few months ago and the same question arised. As far as I remember, he enfasized that in these hydrodynamic analogs of black holes, the flow is not quantized, it is a classical fluid, and everything is classical and that the dumb hole behaves like a quantum amplifier emitting quantum noise from the Horizon. ...
3
Your professor is correct, but I agree with you that the statement “vorticity can’t be destroyed or created” seems jarring - I would prefer to think of this as “vorticity is conserved” because the conservation of vorticity derives from the Navier-Stokes Eq and the conservation of angular momentum. I confess this is splitting terminology hairs (don’t push it ...
3
If the cylinder is stationary, it is according to the hydrostatic pressure equation:
$\vec{\nabla} p = \rho \vec{g}$
You can derive this equation by eliminating all velocities in the Navier-Stokes equations. If gravity is oriented in the negative $y$ direction, the equation becomes:
$\frac{dp}{dy} = -\rho g$
EDIT:
If you integrate this equation you get ...
2
The Wikipedia article on the equation of state discusses this.
Matter
Consider non-relativistic matter first, because this is easy. We write the pressure as:
$$ p = \rho_m \space RT = \rho_m \space C^2 $$
where $\rho_m$ is the mass density, $RT$ is the thermal energy of the matter (or whatever) in your fluid, and $C$ is some sort of characteristic ...
2
If we assume the flow is laminar, the flow in the straw is best described by Poiseuille flow. The linked website will tell you, that the pressures drop scales with the flow-rate $Q$ and diameter $d$ as
$$ \Delta p \propto \frac{Q}{d^4} $$
The proportionality further depends on the length of the straw and the viscosity of the fluid. You can directly derive ...
2
Yes.
You need to solve two problems.
First, given the natural radius of the spherical membrane $R_0$ (the radius with no tension in the membrane) and the current membrane radius $R$, membrane's modulus of elasticity $E$ and Poisson's ratio $\mu$, calculate tension stress in the membrane. If you consider an infinitesimally small square (with the side of ...
2
The operating regime of a flue pipe such as a recorder is governed by the Ising equation (reference at end):
$$
I^2 = WST * 2 * P / \rho / F^2 / H^3
$$
or equivalently
$$I^2 = WST * v^2 / F^2 / H^3
$$
where
$WST$ is the thickness of the flue (windsheet)
$
P$ is the blowing pressure
$
F$ is the frequency
$H$ is the cutup (mouth height)
$v$ is the ...
2
Warning: Another user has given a better answer. This one was chosen as the best one, before the other answer was written.
First, a simplified approach based on Bernouilli's equation for incompressible fluids:
Points B and C are directly in contact with the surface of the tube, thus they are nearly at zero speed with respect to it. But the fluid in A and D ...
2
Vorticity can certainly be destroyed, this is the basis of the energy cascade in 3D turbulence where energy is channeled across wavenumber space from large to small scales all the way down to the Kolmogorov scale at which point it dissipates into heat. Of course in order to do that you need to be looking at the complete equations... this link should answer ...
2
To think of Bernoulli's principle as opposed to something else is not right. Bernoulli's principle can be thought of as the reason for the something else.
Take a look at John Denker's explanation. It's the best I've seen.
2
Its a method called co-ordinate grid transformations that is used to transform an arbitrarily shaped geometry into a square; in the computational domain. Grid transformations work as parametric transformations do, in co-ordinate geometry. what these transformations basically do is, they map the complex geometry (viz. a curve) into a simpler geometry (a line; ...
2
Yes, at least if you ignore the little droplets that leave the bottom of the upper bucket a little damp. There will also be some water in the tube, but it won't be much higher than the level of the water in the lower bucket. (There might be some capillary action that raises it a little, but probably not much.)
I suspect the reason you ask this is because ...
2
Consider an object that is floating and stationary in a liquid of density $\rho$ such that a volume $V_s$ of the objects is submerged. Suppose that we orient a cartesian coordinate system such that the positive $z$-axis points upward, orthogonal to the surface of the liquid.
When the object is floating at rest, the net force it experiences is zero. If we ...
2
The answer to the valve-question,1 according Pilotfriend, seems to be the "floppy walled Eustachian tubes".
During ascent the gas (air) in the middle ear cavity expands and a small amount of pressure builds up against the ear drum causing them to bulge outwards ever so slightly (that ‘fullness’ you feel in your ears just before they ‘Pop’). This pressure ...
1
Let $\mathbf v(\mathbf x,t)$ denote the velocity of the fluid at position $\mathbf x$ and time $t$.
Suppose that we imagine traveling on a path $\mathbf x(t)$ through the fluid, then we can ask ourselves
If we travel along the curve $\mathbf x(t)$ through the fluid, then what will be the rate of change of the flow velocity of each point in the fluid ...
1
Your statement:
Assume density of fluid is same throughout.
conflicts with your actual question
But why is that at a greater depth pressure is higher when molecules are the same?
If we imposed the very strict and non-physical constraint that the density of the fluid was uniform and isotropic, then we would have not variance in pressure what so ...
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