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25

Yes it works. But let's not use it on a massive scale, lest we damage the ecosystem (tip of the hat to @phi1123). A hint to the mechanism can be found in Behroozi et al (Am J Phys, 2007) They state in the abstract: From the attenuation data at frequencies between 251 and 551Hz, we conclude that the calming effect of oil on surface waves is principally ...


14

To work out the force needed to move a ship there are two considerations: the mass of the ship the hydrodynamic drag due to the ship's motion through the water At low velocities the force is likely to be dominated by the mass of the ship because the drag is roughly proportional to velocity. Newton's second law tells us that the acceleration of the ship ...


9

Before everyone freaks out, no, you don't use petroleum oil. You use vegetable, fish or animal oil. In earlier times, whale oil would be used. The OP's picture looks like a fuel oil leak, not an attempt at wave calming. I have seen references of this technique being used since at least the early 1800s, probably much earlier. Ernest Shackleton made use of ...


7

The Reynolds number, with $\rho$ the density, $u$ the velocity magnitude, $\mu$ the viscosity and $L$ some characteristic length scale (e.g. channel height or pipe diameter) is given by $$\text{Re}=\frac{\rho~u~L}{\mu}.$$ This is a dimensionless relation of the ratio of inertial forces ($\rho u u$) to viscous forces ($\mu\frac{u}{L}$). It therefore signifies ...


5

The general requirement you are looking for is that the particular function be of class $C^1$, where ...if all order $p$ partial derivatives evaluated at a point $\mathbf a$: $$\frac{\partial^p}{\partial x_1^{p1}\partial x_1^{p2}\cdots\partial x_n^{pn}}f\left(\mathbf x\right)\vert_{\mathbf x=\mathbf a}$$ exist and are continuous, where $p1,\,p2, ..., ...


5

First, remove the flaps and replace them a single flap (or valve) where the balls enter the water on the bottom right which prevents the water to flow out. In order to get the red ball into the water, you'll then have to overcome the excess pressure corresponding to the water column, which will cost an energy exactly equal to the energy gained by the balls ...


5

The places in physics where commutation of partial derivatives tends to be important are in the identities of vector calculus. The situations where these identities might seem to break down is when there is some kind of topological winding. Then the partial derivatives commute at almost all points except some small set where they are undefined but still can ...


4

The question you ask is actually the central question of a huge sub-discipline of fluid dynamics. Some have even referred to it as "the last great unsolved problem in classical physics." If you get a complete answer, please let me know! (And don't tell anyone else. Just keep between us, eh?) Generally, there are always small fluctuations in any flow, ...


3

The physics of a mushroom cloud due to an explosion are described in this Wikipedia page. To summarize: the explosion causes a massive fireball above the ground which rapidly floats upwards like a hot air balloon. The internal motion of the fireball causes a vacuum effect which draws dust up from below causing the stalk of the mushroom. As to the second ...


3

This is the pressure-gradient term integrated over all volume, converted to a surface integral and using Gauss' theorem. Note that physicists prefer the differential form of such equations (see also this Wikipedia article), when the corresponding equation becomes $$ \frac{\text{d}\boldsymbol{u}}{\text{d} t} = \frac{\partial\boldsymbol{u}}{\partial t} + ...


3

The only thing I can think of is that this is done in cylindrical coordinates: $$\partial_t \rho + \partial_x \rho u_x + \frac{1}{r}\partial_r r \rho u_r=0$$ Using the product rule the last term can be written as: $$\frac{1}{r}\partial_r r \rho u_r=\partial_r \rho u_r + \frac{1}{r}\rho u_r$$ Update: The reference in vol. ii indicates that my feeling about ...


3

The chemical kinetics of air depend on both how fast you are flying and your altitude. Fortunately, NASA has studied these issues. The figures below are from NASA Report NACA-TN-4359. The predominant chemistry in the stagnation region of an airfoil as a function of flight speed and altitude are shown below: You say $M=7$. If your vehicle is near sea ...


3

The fastest point of sail depends on the boat (both its hull shape and its sail plan), the wind strength, and the sea state. In general, a beam reach is not the fastest point of sail. For instance, in very light wind some boats will go fastest on a close reach due to the increased apparent wind from going toward the wind. For boats that sail faster than ...


3

There are two different kinds of wave here. The ones that you see on the surface are the (IMO) badly named Gravity Wave (not to be confused with a Gravitational Wave), which is the familiar phenomenon of waves on the ocean surface and arise at the interfaces between dissimilar fluids. Their phase velocity is $\sqrt{\frac{g}{k}}$ and the group velocity ...


2

For a given volume (for raindrop - a given amount of water translates to volume with the relevant density value) - the shape with the least surface area is a sphere. This is important because there is an energetic difference between molecules inside the drop and on it's surface - molecules inside the sphere have more connections to other molecules , which ...


2

Take a 1-cm square tube and place it vertically in the container from top to bottom, touching the bottom so that the bottom of the container is the bottom of the tube. The pressure at the bottom of the tube is nothing but the weight of water it is supporting - the water in the tube. Supporting means to keep from falling. (Forget the air pressure - that's ...


2

A discontinuity in the flow of water could be a wall, or a clog that water is still getting around, but not flowing directly through. In finite electric current it could be a substance with a different conductivity, notably zero or ∞. In theory, the mixed partial second derivatives would not be generally equal, just on the cusp of a boundary such as these. ...


2

Deliberately introducing turbulence can often reduce overall drag, counterintuitive as it seems. On the wing of this aircraft, you can see vortex generators fitted along it's length. From Wikipedia Turbulent Flow and Drag In turbulent flow, unsteady vortices appear on many scales and interact with each other. Drag due to boundary layer skin friction ...


2

The situation is exactly the same as if the water was stationary and the body was moving (well, assuming you're far enough from the walls for edge effects not to matter). In that case the equation of motion for the body will be: $$ \frac{dv}{dt} = A(v) $$ where $A$ will be given by something like the quadratic drag equation: $$ mA = \frac{1}{2} \rho ...


2

If you want to be physical, you'd have to have a physical interpretation of the derivatives. If you've already taken two derivatives you can ask yourself whether it is possible to take the gradient of those second derivatives. If so, then the second derivatives commuted, if not then the second derivatives are weird (if something wasn't weird you could take ...


1

The enthalpy flux balance equation is in essence a transport equation. Its use over the more familiar enthalpy balance equation has to do with a breakdown of the diffusive approximation for transport phenomena in cases where the mean free-path of the fluid particles becomes comparable to the typical length-scale of fluid. That is, transport phenomena can be ...


1

The conditions you describe are the idealized ones when studying hydraulic jumps. What you will have is a radial flow along the plate: pressure forces will convert the momentum in the $z$ direction into momentum in the $r$ direction. This may of course be unstable. See e.g. http://web.mit.edu/lienhard/www/hydraulic_jump.pdf There are some inconsistencies in ...


1

This is a nice question on $\color{blue}{\text{Hydraulic Turbine}}\ \color{red}{ \text{(FLUID MECHANICS)}}$ Given $$\text{rated power}, P_{\text{rated}}=650 \ H.P.=650\times 746=484900 \ W$$ $$\text{discharge}, Q=0.85\ m^3/sec$$ $$\text{efficiency }, \eta=84\ \text{%}$$ Let, $H$ be head under which the turbine is working Now, the efficiency of the turbine ...


1

My absolute favorite book on the subject is the one that we used in our Gas Dynamics class: Introduction to Physical Gas Dynamics by Vincenti and Kruger. I had never had an introduction to statistical mechanics prior to this book and it does a great job developing the requirements as they are needed and providing motivation for the path it takes. I also ...


1

T1: $w_r/r$ is just $Q_o/2\pi r^2$ which cancels the first term. T2: If $w_z$ is not given, one should assume that the flow is perpendicular to the $z$-axis i.e. that $w_z=0$. In that case, the curl only has a $z$-component $$\frac{1}{\rho}(\frac{\partial(\rho w_\phi)}{\partial\rho}-\frac{\partial w_\rho}{\partial \phi}) $$ which turns out to be $0$.


1

Task 1 Find a calculus textbook and look for cylindrical coordinates, you will find that the divergence of a vector in those coordinates is given by: $$\vec{\nabla}\cdot\vec{\omega} = \frac{1}{r}\frac{\partial}{\partial r}(r\omega_r) + \frac{1}{r}\frac{\partial}{\partial \phi}\omega_{\phi}+\frac{\partial\omega_z}{\partial z}$$ Now apply the product rule to ...


1

prerequisite Thrust produced by a nozzle can be given by $$F_T = \dot{m} V_e + (p_e - p_0)A_e$$ Thrust component in a nozzle can be split into two component that is pressure thrust ($(p_e - p_0)A_e$) and momentum thrust ($\dot{m} V_e$). In most of the nozzles we try to achieve exit pressure equal to ambient pressure, this phenomena is called fully expanded ...


1

Make sure you've normalized everything correctly in both the analytical and numerical solutions so you're comparing apples to apples. Is $n_x$ the wavelength? If so, then the factor of $\cos\left(k_x\frac{n_x}{4} \right)$ is just 0. That seems right, since $u'_x$ is then $\pi/2$ out of phase with $\Delta \rho$, and the velocity perturbation is ...


1

It's simply a definition of the properties of a fluctuating component. We require that, on large enough time scales, the fluctuating component averages out to zero: $$ \frac{1}{T}\int_t^{t+T}u'(t)\,dt=0 $$ so that you are left with just the bulk flow term, $U$, that contributes to the mean velocity, $\bar{u}$. I think this is more easily seen visually than ...


1

The term $(u \cdot \nabla)u $ describes non-linear advective acceleration through a fixed point in the stationary frame of reference (Eulerian frame of reference). In your momentum equation, you need to multiply by this term by $\rho$ to have rate of change of momentum per unit volume. If $(u \cdot \nabla)u = 0$ in your momentum equation, then you have ...



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