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Suppose you have a spherical particle being pushed up a slope of angle $\theta$ by the current: Assume that the system is dominated by inertial forces not viscous forces, in which case the force on the particle is equal to the momentum change per second of the fluid striking it. If the flow velocity is $v$ then the amount of water hitting the particle per ...

6

Imagine two two trains side by side - one going faster than the other. Frictionless rails. Start shoveling coal from the slow train to the fast one, and from the fast to the slow one. Every shovel of coal results in a transfer of momentum - until the two trains move at the same speed. In the same way, when layers of liquid move past one another at different ...

5

If you consider a boat moving through water, or alternatively a stationary boat, with water flowing past it, then the water has a momentum relative to the boat. The blue circle is supposed to show some small volume of the water of mass $m$. If the velocity of the water relative to the boat is $v$ then the momentum of this bit of water is $p = mv$. Now ...

5

It can be shown via simple dimensional analysis. We know that $[u]=m/s$, so just multiply by 1 in terms of an area: $$[u]=\frac{m}{s}\cdot\frac{m^2}{m^2}=\frac{m^3}{m^2\cdot s}=\color{red}{\frac{1}{m^2}}\cdot\color{blue}{\frac{m^3}{s}}$$ The blue term is the volumetric flow rate while the red term is the area, thus we have a volumetric flow rate per unit ...

5

Yes there is. Let's focus on the kinematic viscosity ($\nu$), which is defined as the diffusion constant for momentum in the fluid. That is, it tells us how quickly a momentum disturbance would diffuse through the rest of the fluid. Or, in particular, it gives us the linear dependence on the mean square propagation of the momentum as a function of time, ...

3

Edit: The original explanation focused on the vortex production. For a description of what's happening when the vortex breaks off into child vortices, I've added this new section. The breakup of a single vortex ring into many smaller rings looks sounds like it should be driven by an azimuthal instability. Indeed, under certain conditions, azimuthal waves ...

3

The latter is definitely more standard because suppose you have a collection of different particles (e.g., $H_2O$ and $H_2O_2$). The masses of the individual molecules are different, but over an infinitesimal volume, the density could be taken as an average value. I suppose, since $dV$ is an infinitesimal, then $\rho dV=m$ and the two pictures are ...

3

You have to use the differentials properly: If $V = \frac{m}{\rho}$, then $$\mathrm{d}V = \frac{\partial V}{\partial \rho}\mathrm{d}\rho = -\frac{m}{\rho^2}\mathrm{d}\rho$$

3

If $V = \frac{m}{\rho}$, $dV = \frac{m}{d\rho}$ This is the source of your error. You can re-write the above as $\rho V = m$, and this yields $\rho dV + Vd\rho = 0$, or $V \frac d {dV} = -\rho\frac d {d\rho}\,$ as a differential operator. This leads directly to the alternate form for the bulk modulus $B = \rho \frac{dP}{d\rho}$. Being a bit more ...

3

I'll answer the easy question first -- you are looking at it the right way. Now for the other question... it's really impossible to say that "most" flows are locally compressible. Although that's also a lie because every flow is compressible! Incompressibility is an approximation that makes the math easier, but even the slowest flows of air are technically ...

3

Before the discussion of a sphere, I would like to mention how the flow across a long cylinder (i.e. a circle in 2 dimensions) progresses (and why so) with an increase in Reynolds number (Re). Consider a flow across the cylinder in the creeping flow regime ($Re\leq 1$). This means that the inertial forces are low compared to the viscous forces. Consider ...

3

You could express flow rate as a velocity. But if you want to have a quick measure of how much material flows through (for example) a pipe, you need to know both the velocity and the area - a quick diagram shows you that velocity x area = volume that passes through the area per unit time. So if $$v \cdot A = Vol/time$$ Then it follows that $$v = ... 3 If you neglect viscosity, Bernoulli's equation (just Navier-Stokes without frictional or stress terms) will get you into the ballpark:$$P_g + \frac{1}{2}\rho_g v_g^2 = P_a$$Where the g subscripts pertain to the gas and the a subscript to the ambient. The gas density \rho_g \equiv M / V is the ratio of the mass of gas (M) in the tank to the volume ... 3 You've pretty much answered your own question; you've motivated it and derived the expression, just with a different notation. To rephrase what you've said: the material derivative is the derivative along the path defined by the integral curves of \vec{u}. This is useful because this is the path taken by a small element of the fluid. As a motivating ... 2 The explanation is using an energy argument. That for the normal case of a submerged piece of wood, you can assume that if the wood and a parcel of water above it switch places, then the water (which is heavier/more massive) drops in the gravitational field releasing potential energy. This release is not offset by the rising wood since it is not as ... 2 First one can get killed even by coming in contact (with speed from high altitude) with the water surface, which at this speed and momentum it appears as a "block of cement" (or more correctly, develop high enough forces to break your bones as per @dmckee's comment). This depends what wil be the impact surface (that is why seals and olympic divers fall into ... 2 Bernoulli's equation does not require that the flow be irrotational, just inviscid. Let's consider a vortex filament, and denote its surface by S and volume by V. Using the identity you mentioned above, Euler's equation can be written as:$$ \rho\frac{\partial \boldsymbol{u}}{\partial t} + ...

1

The usual integral for the divergence of the velocity field is over a volume. Since $u$ does not depend on $y$ and $v$ does not depend on $x$, we have  \begin{align} \int_V \left(\nabla\cdot \vec{U}\right) \mathrm{d}V & = \iint \left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}\right) \mathrm{d} x \mathrm{d} y \\ & = \iint ...

1

Everything depends on the size of the air pocket, since you can treat the water as incompressible. As water is lost, the air pocket expands, lowering the pressure. If the air pocket is large, it takes a lot of water loss to lower the pressure a certain amount. If the air pocket is small, the pressure will be very sensitive to loss of water. Check out ...

1

Assumptions: The hole is in the region below the air pocket (so water, not air, is leaking) Air pocket volume is $V_p$ when the pressure is $P$ Isothermal process (slow expansion: temperature constant) Volume of container doesn't change with pressure (probably not true… - this will underestimate the leakage rate) you can write the rate of change of the ...

1

This is a wild-ass guess. You understand why the lemon drops form into a "smoke ring". OK so far. My guess is that after a while, this coherent structure falls apart, just like nice laminar smoke going up from a candle does after some distance. This is based on the Reynolds number. Roughly when that distance is exceeded, the smoke ring structure ...

1

Yes, forces are exerted at the molecular level that have macroscopic effects. This is what thermodynamics is all about. If you hold a sheet of tissue paper in the middle of a gas-filled box, molecules of gas on one side are constantly bouncing off it, creating a force (pressure). Molecules on the other side are also doing so, so the tissue doesn't move. ...

1

You could solve numerically the first problem that you postulate, but the solution will not converge to the steady state solution you solve later. In the initial condition that you propose and in the steady state solution, in both cases the energy of the fluid is purely gravitational, since it is not moving. But the homogeneous column has more potential ...

1

I diluted some vodka with flakes and found that they sank. Also if you spin the bottle and let go the liquid inside rotates forward and then backwards. Must be a non newtonian fluid. I suggest it has long chain molecules formed by an organic addative. These may be trapping the gold in suspension. My bottle has been standing for over a month and there is no ...

1

Short answer: I think the notation is the main problem here. In your second equation, the LHS $\rho\mathbf{u}$ is a function of $\mathbf{x}_0$ and $t$, while your RHS $\rho\mathbf{u}$ is a function of $\mathbf{x}$ and $t$. The subtle difference is that $\mathbf{x}_0$ should be treated as a particle label, not an actual position. As you suspected, the ...

1

The area of the manometer tube makes no difference. All that matters is the difference in the heights of the two ends (labelled $x$ in your diagram). That's why pressure units like the torr exist that are (or rather were) defined as the pressure difference when the difference in height of a mercury manometer is 1mm. All that matters is the height difference. ...

1

I think that it makes little difference... but anyway... You should pour the room-temperature water first, and then the cold water until you get the target temperature. Why? Because if you pour the cold water first, it will immediately start to warm. If you leave it enough time it will actually reach equilibrium at room temperature. So the amount of cold ...

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I suggest the same answer for a different reason out of personal experience, as I am rather fussy: If you pour cold water first, when you pour warm water (unless the jet is powerful) it will stay above and the drink will have an unpleasant difference of temperature. If you pour warm water first, when you pour cold water it will sink to the bottom, creating ...

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Interesting question. I suppose one should compare several scenarios Lie still Go forward - either straight, or hard to port, or hard to starboard Go in reverse The rate at which water enters the ship is (to first order) proportional to the pressure differential - lower the pressure and live longer. Maybe even long enough for the Carpathia to come and ...

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I looked briefly at the Blandford and Thorne notes. The analogy between $\omega$ and B appears to be mainly illustrative and not to be extended too far. (I don't see the reference to the analogy between E and $\omega \times \bf{u}$ there at first glance, but it does not seem to be apt at all.) It seems to be intended to draw upon any previous intuition ...

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