# Tag Info

## New answers tagged fluid-dynamics

0

You can think about it in analogy with density and current. So let's review the situation with density and current. Let's suppose I have a fluid whose spatially varying density is described by $\rho =\rho(\vec{x})$. Since the fluid is allowed to flow, $\rho$ may change with time. However, the change of $\rho$ with time is not arbitrary, since the mass ...

2

The momentum flux tensor comes from the momentum equation of Navier-Stokes equations: $$\frac{\partial\left(\rho\mathbf{u}\right)}{\partial t}+\nabla\cdot\mathbf{P}=0$$ Or, using indices (where it is easier to see that $\mathbf{P}$ is a rank-2 tensor): $$\frac{\partial\left(\rho u_i\right)}{\partial t}+\frac{\partial\Pi_{ij}}{\partial x_j}=0$$ We can ...

0

What you need to do is take into account the thermodynamic properties of freezing water into ice and the elastic properties of the ice. You will likely not be forming different types of ice, but at high (for ice) temperatures you will probably have a two phase system of ice and water where the strain that would form from freezing prevents some of the water ...

1

If you force your liquid in a container you will increase the pressure during the freezing process. This will lower the freezing temperature and thus will indeed "stop the water from freezing". However, if you cool down further, probably at some point a different crystal structure (or the same with lower inter-atomic distances) will form and thus the water ...

1

The simple answer is that a wing moves through the air generally at a non-zero angle of attack. The air flow below the wing sort of impacts the wing surface, compressing and slowing down as it is deflected. See this Drawing:

0

This is a variation on the semi-well-known helium balloon experiment. Sit in a car while holding a He balloon so it floats freely. Accelerate hard & observe which way the balloon moves.

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This is really a continuation of my comment, but it got a bit long for a comment. As I mentioned in my original comment, if the acceleration is constant then we get a static pressure gradient just like we get a pressure gradient in the atmosphere. Incidentally, the Earth's gravitational field is approximately constant up to say the stratopause, so it's a ...

6

Courtesy of the book Carl found we have an answer! Consider the element of the liquid helium at a height $h$ above the fluid surface and distance $y$ from the wall. To raise that element above the fluid surface costs an energy $mgh$, but because there is a Van der Waals attraction between the helium atoms and the wall you get back an energy $E_{VdW}$. ...

1

To make a less apetizing point: The argument leading to the water levels being equal starts from an ideal fluid. Now a sewer backing up is far from ideal. I would expect solid pieces like toilet paper and feces clogging the pipe leading up to the shower (especially if the shower drain has a siphon), while the toilet drain is made for larger pieces fitting ...

2

Possibly helpful: http://arxiv.org/ftp/arxiv/papers/1103/1103.0517.pdf www.paper.edu.cn/download/downPaper/200812-856‎ The bizarre behaviour of superfluids! Climbing up walls and geting out of glass beakers EDIT: A googlebooks excerpt seems more useful:

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The critical point for that phenomenon is the superfluid's viscosity being zero. For that, the deposit formed on the wall doesn't prevent the fluid from flowing out as it does for the normal fluid case.

0

the same kind of phenomenon appears for normal liquid but the explanation is different. for normal liquid, this phenomenon is called capillarity. In short, capillarity can be explained because the liquid like more to be in contact with the wall of the container than with the air. for suprafluidity, heat creates the phenomenon. As a superfluid can't transmit ...

1

It took me quite some time to clearly understand the experiment you're describing. Actually, pouring a full bottle in a container is a quite intriguing thing. Consider the following starting configuration : This of course is an unstable situation, as the pressure $P_0$ cannot be at the same time the pressure of the air in the bottle, and the atmospheric ...

1

Assume that the liquid has a uniform density, $\rho$, and that the diameter of the U-tube is large enough to preclude capillary effects. Tha acceleration of gravity, $g$, is the same for both arms of the U-tube. Pick a reference point at the bottom of the U-tube, where the absolute pressure is defined as $P_0 \text{ }$. Move from this point up each arm ...

2

I believe this decomposition has a specific name "Cauchy–Helmholtz theorem (regarding the decomposition of the velocity of a point within an infinitesimal continuum particle)". You can read about this decomposition in the wikipedia page on Strain rate tensor, sections about symmetric and antisymmetric parts and shear rate and compression rate (though this ...

3

Matrices It's a general property of square matrices (and 2-tensors). Any matrix $M_{ij}$ can be decomposed into a part containing the trace, and a part that is traceless. So we begin $$M_{ij} = \frac{1}{N} (\operatorname{tr} M) \delta_{ij} + \left( M_{ij} - \frac{1}{N} (\operatorname{tr} M) \delta_{ij}\right)$$ I hope that much is evident. The first part ...

3

I am not an expert, but the thing I would do would be to use conformal invariance of potential flow. You would find a conformal transformation take a circle into a square, then take your potential function for the circular cylinder, and put this function through the conformal transformation. Derivatives give you the velocities for the square geometry. I am ...

2

The flow around a square is dominated entirely by viscous effects and the vortex shedding due to the boundary layer. Additionally, at very high Reynolds numbers such that viscous effects are minimal, the square has considerable separation which cannot be solved with the potential equations. Because potential flow requires both irrotational and inviscid ...

4

Flow around a square cylinder Similar solution to a cylinder but more violent von-karman vortex street because separation happens at an edge.. Notation of variables for a flow around a square cylinder The Reynolds number is defined as $Re = Ud/\nu$ and stands for a ratio between the inertial and viscous forces. a) Re<55 Re=30, alpha=0 Re=30, ...

-2

Suppose the cylinder is very wide. Then certainly centrifugal force would cause the fluid pressure to be higher at the perimeter than it is at the center. So if the hole is near the periphery, there is a greater "head" there, so fluid should be ejected at higher velocity. Ignoring viscosity, the velocity should be proportional to square root of pressure. ...

2

Indeed, in Kundu & Cohen, Fluid mechanics, the derivation of Lorenz system is rather sketchy. So let us turn to the source. Original paper by E.N. Lorenz is: Lorenz, E. N. (1963). Deterministic nonperiodic flow. Journal of the atmospheric sciences, 20(2), 130-141. doi, Free pdf. (12000+ citations!) The text is quite accessible, so for all the ...

0

The incomparable Chris Hadfield did a related experiment on the ISS using water on a cloth. You can see that the water does not fly off the cloth. To simplify the experiment consider water on a flat surface: The air/water interface has an energy per unit area, so increasing the area of the air/water interface takes energy. This is also true for the ...

1

If you only consider viscous dissipation within the droplet, this should indeed go to zero in the vanishing velocity limit: the (local) dissipation rate is quadratic in the velocity, so that decreasing the velocity by a factor of $\lambda$ reduces the (local and global) dissipation rate by $\lambda^2$. Of course, the process takes $\lambda$ times longer, ...

1

Imagine if all the astronauts and cosmonauts inside the ISS started bouncing off the walls, would this impact the trajectory of the ISS. The physics says no. The ISS actually had a problem like this, but it does not result in orbital trajectory change. The center of mass of an object in space will move along its path regardless of motion within or about the ...

2

Any reshaping of the droplet will require flow of water inside the droplet and there will be viscous losses. Presumably the energy would come from an increased torque on whatever motor was moving the droplet and substrate.

1

Let us start with 1D continuity and Euler equations written in terms of $p$ and $u$: \begin{gather} \partial_t p + u \partial_x p + \rho a^2 \partial _ x u=0,\\ \partial_t u + u \partial_x u + \frac1\rho \partial _ x p=0.\\ \end{gather} Here we used an equation $d \rho = a^{-2} d p$, derived from definition of speed of sound. Dividing the first equation by ...

3

Bernoulli to the rescue! Does this answer the question? Keep in mind, hurricane speeds are often twice small aircraft stall speeds, and typical aircraft wing loading is in the range of $50 kg/m^2$, so a roof could see 4 times that. Roofing material would have to be really heavy not to be lifted by that.

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Refer Bernoulli's Theorem. Watch this video for demonstration http://dornsife.usc.edu/labs/lecture-support-lab/wind-storm/. Brief explanation: When the velocity of the wind is great enough, the air pressure above the surface is lower compared to that underneath. This cause the roof to blow off. The aeroplane work in the same principle (lower pressure on top ...

6

Pressure is force per unit area, yes, but it also represents the difference in kinetic energy density across a surface - only the energy of random motion of particles, though not large-scale coherent motion like wind. Accordingly, the faster a fluid moves, the more of its kinetic energy goes into large-scale motion, and the less is left for random motion of ...

1

Axisymmetry implies that there is no change in anything in the $\theta$ direction, i.e. $$\frac{\partial}{\partial\theta}(\text{anything}) = 0$$ Which would mean \begin{align} \frac{\partial p}{\partial\theta} &= 0 \\ \frac{\partial \vec{V}}{\partial\theta} &= 0 \\ \implies &\frac{\partial v_r}{\partial\theta} = 0 \\ \implies ...

0

If its a semicircular arc and we aren't taking into account gravity or viscosity, then the pipe must exert a centripetal force on the water to make it move in a circular arc. Now, if you know the radius of curvature of the pipe, $r$, by dividing the pipe length by the subtended angle in radians, then you only need the mass flow rate ($\dot m$) and the ...

4

The formula for capillary rise that most people know is easily derived through a pressure balance between the capillary pressure and the hydrostatic pressure. The hydrostatic pressure equals $$\Delta P_h=\rho g h$$ whereas the capillary pressure is $$\Delta P_c=\frac{2\gamma}{R}=\frac{2\gamma \cos \theta}{r}$$ So balancing these we get our 'famous' ...

0

The answer is A. Think of what capillary action really is. It does not pull on the bulk of the water. It is a edge effect that pulls on the meniscus.

1

No, it won't overflow. That should be obvious since doing so would create a constant flow, constantly using energy, but without any energy input. Put another way, that would be a perpetual motion machine, one you could actually extract free power from. The same force that pulls the water along the inside of the capillary tube also holds it there when it ...

1

Fluid dynamics problems such as this are generally best approached by control volume analysis. Consideration of conservation of mass, momentum, energy, and sometimes angular momentum for an isolated control volume system generally provide an engineering answer. To figure out the force exerted on the pipe by the fluid it would seem appealing to isolate the ...

0

Find the net momentum of a cross-section of fluid entering the pipe. Write this as a vector, like $\vec{p_{in}} = \dot{m_1}(u_1\hat{i} + u_2\hat{j} + u_3\hat{k})$. Next, find the net momentum of the fluid leaving the pipe. Write this as $\vec{p_{out}} = \dot{m_2}(v_1\hat{i} + v_2\hat{j} + v_3\hat{k})$. Suppose the fluid element you're considering takes a ...

2

Yes. Yes. Yes. See below. The Falkenhagen relation (NB: paywall, but (a) it's on the first page of the "Look Inside" option and (b) your University's library might have a copy) suggests that $$\frac{\eta_s}{\eta_0}=1+A\sqrt{c}$$ where $\eta_s$ is the solution viscosity, $\eta_0$ the solvent viscosity, $A$ a constant that depends on the electrostatic ...

5

The motivation comes from applying the no-slip boundary condition on a fluid flow. This is probably easier to understand pictorially, The fluid at the top travels at $u$ while the fluid at the bottom does not move, hence the gradient $\partial u/\partial y>0$. In order to properly model fluid flows, this needs to be accounted for in the Navier-Stokes ...

0

It's a tautology. The viscosity is the proportionality constant for Newtonian fluids. We can imagine a small column of fluid that undergoes a shear strain $\frac{\Delta x}{\Delta y}$ in a time $\Delta t$, so the rate of shear strain is $\frac{\Delta x}{\Delta t \Delta y} \to \frac{\partial u}{\partial y}$ . In general, one would expect a layer of fluid ...

1

The stretching of the air-liquid meniscus matches the pressure exerted by the liquid, not the mass of liquid in the tube. In a closed tube of water the pressure at the top of the tube is $P = P_0 - \rho g h$ where $h$ is the vertical distance (P_0 is atmospheric pressure), so in the diagram above $P_1 = P_2$. So if we replace the closed end of the tubes ...

3

Look on the water from the point of view of the accelerated reference frame oriented in such way that the surface of the water is parallel to plane $x'y'$ and depth below the water surface is measured by $z'$. In this frame, the total gravity (due to Earth's gravity and due to inertial force of acceleration) is directed perpendicular to the water surface and ...

0

you will have to do this in two parts first of all take into account the horizontal difference and then the vertical distance

2

The problem with the phase space flow in Hamiltonian mechanics is that the flow itself is non-dynamical, that is, the flow is immediately defined for a given Hamiltonian, so there is no independent equation governing its evolution. Thus, Liouville equation is simply a transport of a scalar variable in a given flow. So, dimensional analysis of the flow ...

0

I will give you mathematical background that explains where sound waves, as well as shear, and other waves come from in continuum approximations and why viscosity has no influence on such waves: In Eulerian form, the vector equations of motion for a fluid or solid continuum can be written as $$\frac{\partial \mathbf{q}}{\partial t} + ... 4 Your mistake is to assume that the water will stop "no matter how long a bottle you take". It will not - you just need a longer bottle than you expect. To be precise, you need a column of water 10 meters high to counteract atmospheric pressure. -1 I think it is a question how hard you can suck the water in. The force F you need to accelerate the water column depends on the mass of water:$$F=mg. And the mass depends on the density $\rho$ and the volume $V=hA$ with the length $h$ and the surface area inside the straw $A$. So, the force you need to accelerate the water column is proportional to ...

1

It looks like there can be a difference of up to 5 m/s between cloud velocity and wind velocity (http://journals.ametsoc.org/doi/abs/10.1175/1520-0450%281976%29015%3C0010%3AWEFCMP%3E2.0.CO%3B2 ).

4

Clouds move with the wind, so the cloud velocity is just the wind velocity. The recent storm in the Philipines reached wind velocities of 200 mph, though the higest speed reported is apparently 253 mph. The fastest moving clouds known are on Neptune, where the winds reach 1340 mph.

6

You will have two forces that act on an elementary mass element $dm$ on the surface. The force in the $x$-direction will be $dF_{x}=\omega^{2}xdm$ and in the $y$-direction $dF_{y}=gdm$. Also, we know that the slope of a curve is $\tan{\alpha}=dy/dx$. However, the tangent is equal also to $\tan{\alpha}=dF_{x}/dF_{y}$. So from this you have that ...

-1

Fans have a unique quality of balancing air. They can bring in air surrounding the fan and push-out the air existing there, where it is running, this will take time.Its a low pressure high pressure game.

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