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84

Alright, let's start with your direct question. Since $d = vt$ the time it takes to travel a certain distance is inversely proportional to your speed $ t \propto v^{-1} $, and so the fractional change in time is proportional to the negative fractional change in your speed. $$ \frac{dt}{t} = - \frac{dv}{v} $$ So, if we consider typical a typical highway ...


11

Intuitive start of an answer: If you have counter rotating vortices they have zero net angular momentum (to first order). If they merged they would have to have no motion -> where did the energy go. In between the two axes of rotation the fluid moves in the same direction and has no mechanism for dissipation. By contrast for two vortices with the same ...


9

Because where they come close together the air in between circulates in such a way as to join them in a single path. Floris is right, but maybe this picture helps.


5

For the case that you have drawn, the behavior of the drop is actually the exact opposite of what you mention: it will move from right to left. This is caused by surface tension and the curvature of the droplet caps which creates a larger pressure in the drop at side B than at side A. To make it more quantitative. Let's assume that the funnel is ...


5

If the speed limit is 60 mph, it would take 60 min to go 60 miles. To go the same 60 miles at 65 mph, it takes 55.4 minutes. A time savings of 4.6 minutes. Is it really worthwhile to speed if all you are going to save is a few minutes (even less for shorter distances).


4

When a cup of coffee is hot, the air molecules directly above it get hot as well. After some time, they reach equilibrium and no heat transfer (or maybe very little transfer) occurs. By blowing, you disturb that equilibrium and replace the hot air molecules directly above the cup with colder air and therefore create once again a steeper temperature gradient. ...


4

The main problem with continuum equations is that it is just a "lucky guess" of macroscopic dynamics otherwise governed by microscopic ones. You have encountered the moment where this "guessing" isn't entirely consistent. Before delving into the technical derivation, the non-$D_t \rho$ in the Newton's equation is just a consequence of the fact that the ...


4

Newtonian Aspect So the force generated by the sail will perpendicular to the sail. This is because only the momentum of the air particles perpendicular to the sail changes the component changes the component along the sail is unchanged. However, most sailing boats have a daggerboard or keel under the boat. Essentially this is a big plank under the boat in ...


4

In hydrodynamics, conservation means that what flows into the control volume is equivalent to the flow out of the control volume. With respect to momentum, we mean precisely that any change in momentum of the fluid within a control volume is due to the net flow of fluid into the volume and the action of external forces on the fluid within the volume ...


4

Fluid dynamics models might have a practical value in heavily congested areas, but then you can't speed, save for some random short bits after traffic signals, rendering it irrelevant. Guess they're more of a traffic distribution models, smaller roads will attract more traffic if the counterpressure (congestion) gets higher on the main pipe (road). It ...


4

Aside from the effect on one driver you might consider the effect on traffic in general, that is what happens when everyone breaks the speed limit. As far as fluid dynamics is concerned, the side-effects of your speeding (if any) are felt by the people behind you. Reaction distance increases linearly with speed, but stopping distance must include a term ...


4

This is a great question, and the answer relates intimately to why turbofan engines equipped with afterburners require variable geometry exhaust nozzles. Without increasing the throat area to accommodate the larger volumetric flow rate, lighting the afterburner would back-pressure the fan and very possibly lead to a compressor stall. Similarly, mechanically ...


4

No-slip condition requires that the velocity is $0$ at the boundaries, thus you can see intuitively that the velocity profile has to behave in $R^2-r^2$ (at least as leading terms in its expansion). To get the hydraulic resistance, you need the mass flux: integrating $R^2$ and $r^2$ on a section (circle) you'll get terms in $R^4$ : the mass flux is in ...


3

A stationary state, usually referred to as the steady state, is the particular case when the partial time-derivative of the variable in question is zero. For your question, $$ \frac{\partial v}{\partial t}=0 $$ Or, in macroscopic terms, $$ \frac{\Delta v}{\Delta t}=0\to v_{in}=v_{out} $$ which is what jhobbie said in the comments. The water level will ...


3

Actually, in your rotating torus your are mimicking gravity, which is point outward. I.e. the outside of the torus acts as a floor. The centrifugal acceleration would be $g\approx\omega^2 R$. This $g$ plays the same role as the gravitational acceleration on liquid or gas pressures under normal gravity conditions, so you can say $\Delta p = \rho g h$, or in ...


3

Not a complete answer, but this is a classic "sloshing" problem. The interaction between the fluid and the container wall, under the influence of the external (periodic) force sets up a (self-reinforcing) and harmful resonance. This is of immense practical interest: jet-fuel sloshing inside airplane tanks, for instance.


3

Reynold's number is defined to be: $$ \text{Re} = \frac{ v D }{ \nu } $$ where $v$ is the characteristic velocity for the flow, $D$ is a characteristic size and $\nu$ is the kinematic viscosity. Now, why should we care? Why is Reynold's number important? Well, the first thing to realize is that the Reynolds number is a dimensionless number. This means ...


2

This is low-Reynolds number particle sedimentation. It turns out that the problem is strikingly difficult, despite the simplicity of the setup and even of the equations (Stokes plus dynamics of pointwise solid particles). Check the webpage of E Guazzelli who's been working a lot on this. However, I believe you can get a fair rendering with simply a ...


2

PhotonicBoom is correct in saying that the airflow created by blowing across the top of the coffee will replace the coffee-heated air with cooler air that will absorb more heat from the coffee. It also allows more of the coffee to evaporate (which might seem like a bad thing, but evaporation is simply the hottest molecules becoming gaseous and leaving, so it ...


2

Well, I can't say anything about the microscopic details, but in day to day language, the fast stream is adding extra material to the slow stream and it has to go somewhere. Since we are talking about flow that is confined it has to be up. Let me try a goofy metaphor that should not be taken too seriously but does relate to everyday experience. You may ...


2

EDIT: replaced "fluid" with "liquid", thanks to Kyle. I am not aware of any material with a liquid phase in near-vacuum. Probably, the liquid would evaporate and maybe a part of it freezes solid due to evaporation cooling. EDIT: NeuroFuzzy pointed to a youtube video containing an ionic liquid, which is able to retain liquidity in very near vacuum. What ...


2

1D Burger's equation is not meant to model a physical phenomenon. Rather, it is a simplification of homogeneous incompressible Navier-Stokes equations that preserves (some of) its mathematical structure: the non-linear convection term and the second order derivative of viscous forces. It was initially intended as a useful simplification to try to ...


2

If the flow is laminar, i.e. not turbulent, then the relationship between flow rate and pressure is given by the Hagen–Poiseuille equation: $$\text{Flow rate} = \frac{\pi r^4 (P - P_0)}{8 \eta l}$$ where $r$ is the radius of the pipe or tube, $P_0$ is the fluid pressure at one end of the pipe, $P$ is the fluid pressure at the other end of the pipe, ...


2

Impact on water is a very complex topic. Your simple calculation just figures out the velocity of a free-falling body after a 50 m drop. That just tells you the initial relative velocity of body and water surface. It doesn't tell you much about the force at impact, or whether the person survived. There are two things that might kill on impact: high local ...


2

If one instead uses $F=\dot p$ so that $$f = D_t\left(\rho v\right)$$ The formulation of the second law by the equation $$ \mathbf F=\frac{d\mathbf p}{dt} $$ where $\mathbf p$ is total momentum of the system is valid only provided the system considered does not lose parts. This is usually the case in hydrodynamics, because we think of a fluid element ...


2

For a damped simple harmonic oscillator, the damping term is linear with the velocity - and the constant of proportionality includes the viscosity of the damping medium. If the equation of motion is $$m \ddot{x} + 2 \zeta \omega_0 \dot{x} + \omega^2x = 0 $$ In this equation, $\zeta$ is proportional to viscosity (and contains other terms related to the ...


2

A fluid is modelled as a vector field and therefore we use vorticity to describe its spinning motion. Angular momentum is more often used for a single object or particle, but not so often for a vector field (even though it is still applicable in principle). For a fluid in general, vorticity is twice the mean angular velocity and this fact to me makes it less ...


2

Vella and Mahadevan explain the effect as follows: For simplicity, we consider the latter case schematically illustrated in Fig. 2, although the explanation of the clustering of many bubbles is similar. Here, the air–water interface is significantly distorted by the presence of the wall the well-known meniscus effect, and because the bubble is ...


2

It is not a force exerted by the building that causes the wind, it is just the way the air current is manipulated. The air currents have to go somewhere. Due to the building being tall, they are unable to flow over it. Thus, they travel up along it, down along it, and around it. Moreover, if there is a high density of tall buildings in an area, you get the ...


2

Let's take your last question first. Let the stress tensor at a point (x,y,z) in the fluid be given as $\sigma$. You can pick a Cartesian basis $\{ e_1, e_2, e_3 \}$ and express the components of the tensor in that basis $$ \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & ...



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