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## New answers tagged fluid-dynamics

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The milk is a thick liquid (more viscous than water). When it boils the bubbles formed don't break as quickly as they do in water. This means more bubble form which take up room in the pot and this causes the milk to boil over. Water rises, too, just not as much since the bubble break more quickly.

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The direction of the Resultant force $R$ is always dependent on the direction of the $V\infty$ But however, the direction/orientation of the Normal force $N$ is dependent on the orientation of the body itself ($N$ is perpendicular to the body and axial force $A$ is parallel to the body.) In the above case, since there is not much of surface interaction, ...

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Let there be given an $n$-dimensional manifold $(M,\nabla)$ endowed with a connection $\nabla$. [In particular, we do not assume that the manifold $M$ is equipped with a metric tensor.] Let there be given a curve $\gamma:\mathbb{R}\to M$. Here the reader should think of $\mathbb{R}$ and $M$ as time and space, respectively. If $f: M\times \mathbb{R}\to ... 2 You correctly identify the residual velocity of the water after bouncing off the bucket as a critical parameter in the calculation. Where you go wrong is in assuming that you can assign any value you want to it. If your bucket's bottom was shaped in such a way as to "turn around" the water jet hitting it, then you would have the maximum possible momentum ... 3 You've pretty much answered your own question; you've motivated it and derived the expression, just with a different notation. To rephrase what you've said: the material derivative is the derivative along the path defined by the integral curves of$\vec{u}$. This is useful because this is the path taken by a small element of the fluid. As a motivating ... 0 You have to think according to the situation. Consider a tank with high pressure, if we open the tap means water flows with high velocity and eventually after sometime the velocity will start to reduce because of pressure is decreasing accordingly. Velocity and pressure is directly proportional. 0 I'm not quite sure if this correct, but I think you can solve this with an energy balance, the equation of state and the mass flow: Assuming$\dot{v}$is zero, no losses in the pipe: $$\dot{E}_{kin}=\frac{1}{2} \dot{m} v^2$$ $$\dot{m}=\rho A V$$ $$P = \rho R T$$ Substitute them, and evaluate them for both the incoming and outgoing location(s) On second ... 0 You are right, difference in the bounds of the integrals represents the difference between the Eulerian and Lagrangian perspectives. The Lagrangian frame of reference follows a volume of fluid, so Newton's laws can be directly applied. However, the bounding surfaces of the volume can change over time. This makes it difficult to actually apply the ... 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 ... 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 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 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 ... 14 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 ... 0 In a viscous fluid the shear stress is proportional to the velocity gradient.$\sigma=\eta \frac{dv}{dy}$where$\eta$is the viscosity, and$v$is the fluid speed at right angles to the$y$axis. Therefore as the small distance$dy$tends to zero, the change of fluid speed$dv$also tends to zero, for any non-zero viscosity. Let us now follow Navier ... 0 Non-turbulent flow through a pipe of radius$r$is proportional to the cross section$\pi r^2$times the average flow velocity. The latter is proportional to the time$t$it takes for the momentum in the fluid to diffuse to the wall:$t \approx (1/\nu) r^2$. Here$\nu$is the kinematic viscosity ($\eta/\rho$). It follows that the flux (area times velocity) ... 0 Usually we think of friction as something like this (not a formal definition, but I think it's close enough to be understood as "friction"): when two objects move past one another and are in contact, the differential velocity between them leads to a force we call "friction" At the boundary between a liquid and a solid, if we permit a different velocity ... 0 Laplace law will give you the curvature of the surface for a given pressure in the film, which you can integrate to get the shape... provided you know the volume of fluid that remains in the slit. I believe the volume of fluid itself will be dependent on the dynamics of the process of getting the frame out of the water, and also on the frame itself: its ... 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 ... 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 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 ... 0 I don't know much about fluid mechanics, but I think this can be answered by remembering that everything we can touch/see is three dimensional on some scale. A two dimensional object would be infinitely thin if we could see its side. I don't know about in the higher-level physics (When you talk about superstrings etc. Which can exist in several dimensions, ... 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 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 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$x, we have \begin{align} \int_V \left(\nabla\cdot \vec{U}\right) \mathrm{d}V & = \iint \left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}\right) \mathrm{d} x \mathrm{d} y \\ & = \iint ... 2 Bernoulli's equation does not require that the flow be irrotational, just inviscid. Let's consider a vortex filament, and denote its surface by S and volume by V. Using the identity you mentioned above, Euler's equation can be written as: \rho\frac{\partial \boldsymbol{u}}{\partial t} + ... 1 The area of the manometer tube makes no difference. All that matters is the difference in the heights of the two ends (labelledx$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. ... 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 ... 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 = ... 0 Because of gravitational forces you liquid is at the bottom of both phases. A pressure gradient, due to those forces exist in both phases; the pressure increases linearly through bot fluid, with a discontinuous ramp at the interface liquid/gaz. So if you push down the piston, by increasing the volume you will modify the equilibrium point and so decrease ... 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 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 ... 1 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 ... 1 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 ... 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 ... 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 ... 1 I propose the following order-of-magnitude, very rough line of reasoning. The rate of momentum escaping from the aperture is$$\frac{dp}{dt}=\rho\pi(D_2/2)^2 v^2,$$all in the horizontal direction. I assume that a fraction (1-\cos\theta) of the horizontal momentum will be lost by the fluid since the change of direction of its motion. This fraction of ... 1 I looked briefly at the Blandford and Thorne notes. The analogy between \omega and B appears to be mainly illustrative and not to be extended too far. (I don't see the reference to the analogy between E and \omega \times \bf{u} there at first glance, but it does not seem to be apt at all.) It seems to be intended to draw upon any previous intuition ... 2 The explanation is using an energy argument. That for the normal case of a submerged piece of wood, you can assume that if the wood and a parcel of water above it switch places, then the water (which is heavier/more massive) drops in the gravitational field releasing potential energy. This release is not offset by the rising wood since it is not as ... 3 If you neglect viscosity, Bernoulli's equation (just Navier-Stokes without frictional or stress terms) will get you into the ballpark:$$P_g + \frac{1}{2}\rho_g v_g^2 = P_a$$Where the g subscripts pertain to the gas and the a subscript to the ambient. The gas density \rho_g \equiv M / V is the ratio of the mass of gas (M) in the tank to the volume ... 1 When a pipe is bent the outside curve - what would be the longest path through the curve - has the highest pressure and the lowest speed. The inside curve - the shortest path through the curve - has the lowest pressure and the highest speed. In short, when the path of a fluid in steady-state flow bends, the pressure on the outside of the bend is always ... 0 Viscosity is the material's resistance to shearing stress. According to the Wikipedia entry, water has a viscosity that is exponential:$$ \mu\sim A\cdot10^{B/(T-C)} $$If we then look at the Navier-Stokes equations (which holds for Newtonian fluids, something grease is not (instead a non-Newtonian fluid)),$$ \rho\left(\color{blue}{\frac{\partial\mathbf ... 0 Indeed, "This of course goes a bit haywire at r=0". In fact the whole argument is just a trick to hide the fact that infinite vorticity has been inserted at the axis - see vortex. By stirring the tea you have introduced vorticity and hiding it in a singularity doesn't help. Nevertheless, "In the absence of external forces, a vortex usually evolves fairly ... 0 Ultimately, the Navier-Stokes equations explain this :) OK, that's not a useful answer: here's how they explain the phenomenon in some cases. Under steady state conditions for a fluid (inviscid, incompressible) that doesn't differ too much from a cup of tea, the Vorticity Transport Equation shows that the vorticity$\omega = \nabla \times \vec{v}(the curl ... 0 If I understand correctly, the cup itself doesn’t rotate. In normal conditions it implies that linear velocity of the liquid at the boundary (the cup’s wall) equals to 0 (search for “no-slip condition” in Web for explanations). This is the main reason why the liquid moves “quicker in the middle”. Also, if you stir your tea especially vigorously, you can ... 0 I am not sure the parameters you mentioned are enough. For example, in some viscous liquids, one can observe the Mössbauer effect, where gamma-ray absorption differs dramatically for the liquid and a single particle; therefore, even details of the nuclear spectra can be important. 1 how should I notate my bounds of integration? I guess you mean for this integral, passing to gauge pressure \begin{align} \int_{P_s(o,t)}^{P_s(L,t)}(P_s(x,t)+b) \ dP_s(x,t). \end{align} SinceP_s=P_g+P_a$, all you need to do is that change of variables. Besides,$dP_s=dP_g$because$P_a$is a constant. Also, there is no need to say$dP_s(x,t)$, since ... 1 The Jeans equations can be a bit tricky. Their simplest form (in cartesian coordinates, with no particular assumptions) is: $$\frac{\partial\nu}{\partial t}+\frac{\partial(\nu\bar{v_i})}{\partial x_i} = 0$$$$\nu\frac{\partial\bar{v_j}}{\partial t}+\nu\bar{v_i}\frac{\partial\bar{v_j}}{\partial x_i} = -\nu\frac{\partial\Phi}{\partial ... 0 In a fluid the molecules collide and interact randomly. As a consequence, local physical features typically dampen out over timescales of a few collisions. However, quantities that are conserved in collisions form an exception to this rule. Quantities such as mass and momentum can not be 'destroyed' and therefore do not dampen out over collision ... 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 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, ...

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