# Tag Info

10

If the metal pan was cool then you would expect to see water droplets staying in the same place once any original movement had dissipated. You would have a combination of cohesive forces within each water droplet and adhesive forces between the water and metal surfaces. With the metal having a temperature well above the boiling point of water, the water ...

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 ...

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 ...

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 $E_{VdW}$. ...

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 ...

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' ...

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.

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, ...

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.

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 ...

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 ...

3

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 ...

3

The definition of the stress tensor is (in Einstein summation notation): $$\tau_{ij} = \mu \left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i} \right)$$ So, if you look at $\tau_{ii}$, you get $2\frac{\partial u_i}{\partial x_i}$. That's really where the factors come from, not from "averaging" over any fluid element or anything like ...

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 ...

2

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 ...

2

Let's start with a few issues considering Amperè's law. Amperè's law describes the magnetic field generated by current. The current can be localized at a certain point in space, but its magnetic field is spread everywhere. Applying Amperè's law to a certain point where current is zero does not generate a magnetic field. However, it doesn't necessarily mean ...

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 ...

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 ...

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 ...

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 ...

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:

2

Suppose you want to use real measurements in this equation, then the density is actually the mass $m_0$ contained in a mesh unit volume (or voxel) divided by the volume of the voxel, $a^3$ such that $\rho=m_0/a^3$. As far as I know in the lattice Boltzmann method there is a finite number of velocities $\vec c_i$ and $f_i$ is the mass $m_i$ of matter moving ...

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 ...

1

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 ...

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 ...

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 ...

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 ...

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 ...

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