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6

There's no magic behind it. It was done by non-dimensionalizing the momentum equation in the Navier-Stokes equations. Starting with: $$\frac{\partial u_i}{\partial t} + u_j\frac{\partial u_i}{\partial x_i} = -\frac{1}{\rho}\frac{\partial P}{\partial x_i} + \nu \frac{\partial^2 u_i}{\partial x_i x_j}$$ which is the momentum equation for an incompressible ...


5

The intuitive way to think about this is to consider a gas inside a glass container (that cannot expand). If the gas expands, then what must happen as a result? The gas leaks out of the container. Similarly, if try I put more gas into the container, then the gas compresses. The vector field $\mathbf F$ is what we use to describe the flow of a fluid. The ...


5

The front of the train compresses air which can blow you away, while at the back of the train air rushes back in after the train has displaced it. This backdraft is especially troublesome in closed areas such as subways, where a train exits a small tunnel near a platform and the displaced air rushes back into the vacated tunnel. Next time you see a big truck ...


4

Rising bubbles of air in a liquid oftentimes are anything but spherical. These bubbles have haphazard shapes because they are rising and because they are interacting with other nearby bubbles. The combination of drag, turbulence, and mutual interactions prevents those bubbles from taking on a nice, simple spherical shape. Here's a rather non-spherical ...


4

The key is the Reynolds number, $$ Re=\frac{\rho LV}{\mu}=\frac{LV}{\nu}\tag{1} $$ where $L$ and $V$ are characteristic lengths and velocities of the particular problem and $\mu$ & $\nu$ are the dynamic & kinematic viscosities, respectively. If you multiply (1) by $\rho LV/\rho LV$, you get $$ Re=\frac{\rho L^2V^2}{\mu LV} $$ The numerator is the ...


4

This technique is called supercavitation. So far it has only been applied to objects no larger than, say, a torpedo, because it not easy to produce a stable bubble the size of a ship when moving at high velocities.


4

The nonlinear term, $\left( \mathbf{V} \cdot \nabla \right) \mathbf{V}$, determines the steepening of a wave. This can be balanced/offset by loss terms like dispersion, diffusion, viscosity, resistivity, friction, etc. If the loss term dominates over the nonlinear term, then the wave cannot steepen as there is too much damping. If the loss term balances ...


4

The fact that the Bernoulli equation contains the kinetic and potential energy density (energies per unit volume) terms should strongly suggest to you that the conserved quantity is actually energy. Seeing the pressure as an energy term is a little bit trickier, but really is easiest to see from your units: $$ \left[\,{p}\,\right]=\frac{\rm N}{\rm ...


3

It forms a cone because it depends on a shock wave, and the region enclosed by the shock wave appears conical in shape. See, for example, the apparent cones here: They are also visible here: Wikipedia appears to be fairly clear on why vapor cones are related to shock waves. From the introduction to the article about vapor cones: Atmospheric water ...


3

Pressure is a scalar and does not have a direction. This is discussed in some detail in the answers to Define Pressure at A point. Why is it a Scalar?, though this might be a bit technical. When you measure a pressure you are actually measuring the force applied to a surface. For some small bit of surface $\delta {\bf A}$, the force produced on that surface ...


3

I would say pressure is better defined by $$ \vec{F} = P \vec{A}. $$ Yes, we are defining a quantity without having it all alone on the left-hand side. And yes, area is a vector. And as you guessed trying to divide one vector by another leads to trouble, so we won't do it. Let me explain where this comes from and what it is shorthand for. In continuum ...


2

I didn't entirely understand your question, aside from the last bit "what makes a solution of these equations unique?" And you are absolutely right, it is the boundary and initial conditions that entirely determine the solution (which may or may not be unique, that's a very open problem for the Navier-Stokes equations). You say that the situation is ...


2

Yes, it does protect against G forces because it spreads the pressure on the support surfaces of the body evenly. For example, and interesting article here


2

I can't answer the question as posed - but I can point you to something related. If you have an (infinite) cylinder perpendicular to a flow, you can calculate the flow around it (see for example this interesting page where these images and equations come from). The velocity profile would look like this: and the pressure profile is similar: They give ...


2

As an another attempt, I calculated the coefficients of the cubic regressions that describe the NIST data. I first calculated the cubic regression coefficients as a function of pressure for viscosity as a function of temperature. In other words, I calculated the coefficients for each isobar. In equation form that is, $$\mu ...


2

It depends on the fluid. Consider, for example, an ideal gas at fixed temperature near the surface of the Earth. Does the density vary in such a column? Yes. Let's investigate as follows. Imagine that the column is in the $z$-direction and has cross-sectional area $A$. Let $z=0$ at the ground. Consider a small, vertical "piece" of the column between ...


2

The parameter $h$ is the maximum distance that two smoothed particles, $a$ and $b$, can be before the distance between them is negligible for SPH purposes. If the distance, $\vert r_a-r_b\vert>h$ then the weight is zero. For any kernel, the integral over the particular region, e.g. $r\in(-h,\,h)$, is necessarily 1. Since $h$ is an unknown parameter, then ...


2

For the sake of the explanation I will assume you mean a gas bubble in a liquid*. David Hammen names a few conditions for a bubble to be spherical, in fact you could summarize these all as: for a bubble to be spherical the surface tension has to dominate over other forces (per unit length). If surface tension is indeed dominant than the pressure in the ...


2

Sorry for the enormous delay, I was caught with more work than I thought, then, here it is. @DanielSank, relativity is not necessary, it would help with what you said, since it would pinpoint exactly you are calling momentum in your system. My answer would be an extension of DanielSank comment. When there is the conservation of a continuous quantity, the ...


2

The way it was explained to me: you start by thinking of all the possible factors that could play in drag (size, velocity, density, viscosity, ...); then you do dimensional analysis and find dimensionless combinations - these tend to be "special" since they remain constant over different scales of time and space. Reynolds number is one such combination. The ...


2

Like any object moving through a fluid, a high-speed train distorts the air as it moves through it. Broadly speaking, there are three main regions of flow structure around a high-speed train: the upstream distortion, boundary layer and wake. These can be collectively referred to as the slipstream. The effects of the slipstream on a static observer (e.g. a ...


2

Your answer is correct on the basis of your model, which neglects the resistance to flow through the funnel (Bernoulli's equation assumes inviscid flow). If that resistance were included you'd find that $P_t < P_s$. [Addendum in response to comments:] As a result of the balance of forces and momentum change expressed by Bernoulli's equation, the body of ...


2

The rotation is part of the key to the storm itself. Primarily the pressure and temperature differences are what causes these systems to take the shape and forms that they do. Once a tropical depression starts to form you can already see rotation in the moisture around the low pressure zone, even through it typically looks nothing like a hurricane. Not ...


1

Unless the two containers are separate, i.e. have a wall sealing them off completely, the right set of tools for this question is fluid-dynamics rather than thermodynamics. for the sealed off problem, assuming ideal gasses, the end state for the coupled baths will be that of equal temperature. in that case it is essential you have the right number of ...


1

The answer is due to the area-Mach number relation for hydrodynamic shocks. G.B. Whitham has a great book (check out Chapter 8) on all sorts of various waves and has a good discussion of this topic. The idea is that one can define the Mach number as a function of the cross-sectional area of a ray tube. The simple form is: $$ \frac{ 1 }{ A } \frac{ d A }{ ...


1

You can get nothing out of equilibrium thermodynamic considerations for the rate at which pressure will equalize. What will matter is the speed of sound in the gas, as that is the rate at which density fluctuations travel in a fluid and assuming an equation of state, say $p(\rho)=\rho^{\gamma}$, the pressure is then enslaved to the density. So the sound ...


1

The thing you'll notice about a sphere is that it's symmetrical. very symmetrical. No matter how you rotate it, it looks the same. the surface tension pulls the surface of the bubble into a shape that has even surface tension over the entire bubble. The shape with even surface tension is a sphere. a sphere has the smallest possible surface area for an ...


1

I think expecting "fluid" behaviour in terms of a material that does not support shear, is not useful in the context of the various systems you have listed in the question. Instead, I believe you are intuitively connecting ideas and concepts pertaining to conservation laws. So the idea that in specific systems, conserved charges (in the sense of Noether) ...


1

How do slight changes in these properties result in a large change in pressure, microscopically? Slight change of volume is not so easy to accomplish for solids - it takes a great force to achieve it. Considerable external force applied by different body (wall) needs to be maintained. The pressure is a measure of this force per unit area and since the ...


1

Here we assume that OP is mostly interested in the Eulerian fluid picture (as opposed to the Lagrangian fluid picture). Both fluid pictures are discussed in great detail in Ref. 1. Note however that in the methods of Ref. 1, the mass density $\rho$ is a dynamical variable. The variation of $\rho$ is important in order to obtain a full set of eoms. But OP ...



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