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So the assumption when linearizing is that the deviations are very small compared to the reference (averaged) values. Perhaps starting from the Euler equations does not give a good representation, lets instead start more general with the Navier-Stokes equations (neglecting viscosity): $$\partial_{t}\rho+\nabla\cdot\rho\boldsymbol{v}=0$$ ...

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https://www.youtube.com/watch?v=YfYPJZCSI-E&index=9&list=PLq1e8D3Q-o-W-QrxTK4UIiJGyMKEsaBjo this should help you, as the cavity collapse the 2 fronts collides and form a jet. depending on the length of the cavity form, they might be 2 jets formed. The only question now is how does the speed relates to the height of the jet?

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As Thomas has commented, the trick is that we only assume first order terms and this convective acceleration would be small of the second order. In fact, that is one of the first assumptions to drop when you consider more general cases. See e.g. Burger's equation for first generalizations and/or Lighthill's equation for source terms arising in the wave ...

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To derive the Bernoulli equation for inviscid fluids, the plan is to rewrite the Euler equation in such a way that we have gradients. I'll write the Euler equation with gravity here $$\frac{\partial \vec{u}}{\partial t} + \vec{u} \cdot \vec{\nabla} \vec{u} = -\frac{1}{\rho} \vec{\nabla} p + \vec{g}.$$ Recall $g = - \vec{\nabla} \Psi$, and $\vec{u} \cdot ... 0 It completely depends on how the maximum flow rates are enforced. For one extreme, I could imagine a sensor that losslessly watches the flow and, if the max flow is exceeded, closes a valve to limit the flow. In that case, there'd be no difference between the two pipes up to the smaller pipe's 10LPM limit, so the flow would be equally split. On the other ... 0 Torricelli's formula can be derived for water jet running out from a tank, where water does not move much and obeys laws of hydrostatics. Inside a hose, all the water moves roughly with the same velocity and the Toricelli's formula does not apply, because moving water experiences considerable friction as it moves along the hose and behaves in a more ... 1 Torricelli's law is just a restatement of the conservation of energy of a non-viscous, non-turbulent and incompressible liquid flow. Thus, the maximum speed that can be obtained by making water flow through a hose (purely by the force of gravity, ie. a tank on the roof) is the formula that is given. The water flows slower when it moves through a hose with ... 1 If you severely reduce the cross-section of the hose by putting your finger into it, you increase the pressure drop across the restriction and thus the flow rate decreases (you can block flow altogether too, of course, if your finger tightly fits or covers the hose's open end, or by 'kinking' the hose). This causes the volumetric flow rate$\dot{Q}$... 0 Let's use the balloon as an example. When you inflate it the contents are under pressure because the latex wants to shrink back to its natural size. When you pop the balloon that pressure is released. It rushes outward temporarily creating a void. The void then sucks in the surrounding air. The popping sound is the air molecules slamming into each otherwhen ... 1 Smoothed particle hydrodynamics (SPH) is a fluid simulation approach that has been initially developed for astrophysics fluids (galaxies, nebula, exploding stars...), showing huge range of possible "densities", and embedded in ambient vacuum. see https://en.wikipedia.org/wiki/Smoothed-particle_hydrodynamics 3 With hydrodynamics we normally find that at low shear rates the flow is limited by the viscosity of the liquid while at high shear rates it's limited by inertial forces and the viscosity doesn't matter. This is the case for flow in a pipe. At low flow rates the pressure drop$\Delta P$is related to the flow rate$Q$by the Hagen-Poiseuille equation: $$... 0 Without more details, we can't find a specific case that will match what you are doing. However, I did find a decent example to show you what to look for to answer your question. Data for a nearly-sonic round jet can be found in this paper. If you look at Figure 5(a), you'll see how the normalized centerline velocity from several experiments collapses ... 0 I am trying to understand the question, so I will make a few assumptions. A container, under pressure, holds water. This pressure could presumably be measured, and may result in some slight deformity of the container. (I am imagining a metal container, under pressure, containing water). You open a cap, and the pressure is reduced. But the only change in ... 1 In Newtonian physics, fluids, like anything else, obey conservation of momentum, or F=ma. The Bernoulli principle is just a re-statement of conservation of momentum. The only thing that can change a parcel of fluid's speed (i.e. accelerate it) is a force. One kind of force is a pressure gradient. Another kind of force is gravity. If the fluid is ... 1 from the figure: liquid that cannot be compressed... so forget compression of springs. Molecularly, particles in a liquid have much less space to move around (short mean free paths) compared to particles in a gas. The result is that any applied pressure to these particles is instantaneously transferred through the fluid. The mechanism by which this happens ... -1 No, water moving out of a bottle rocket is not nearly fast enough to generate the energy sapping shock waves and mach which the converging-diverging nozzle seeks to eliminate. 2 One way to think about the Reynolds number is to note that is measures the relative magnitude of viscous and ideal (inertial) stresses in the fluid$$ {\it Re}^{-1} = \frac{\mu\nabla u}{\rho u^2} = \frac{\mu}{\rho Lu}\, . $$However, the stress tensor of an ideal fluid has two terms, one related to pressure, and one related to inertial stresses$$ ... 0 In any dimensionless formulation of the Navier-Stokes equations you will never find a dimensionless number as a coefficient infront of the dimensionless pressure gradient. This is because the pressure gradient must always be of the same order as one of the other viscous or inertial terms: $$\boldsymbol{v}\cdot\boldsymbol{\nabla}\boldsymbol{v} = ... 3 I am unaware of any dimensionless quantity that this represents, and Wikipedia seems to agree. However, what you have defined there is the ratio of the pressure stress to the viscous stress, and this is in some sense similar to the Bingham Number (yield stress to viscous stress). So what would your number mean? Let's go ahead and call it the Toliveira ... 1 According to these lecture notes, the Coriolis parameter at mid-latitudes is on the order of f_c = 1\times10^-4 \text{s}^{-1} and this needs to be multiplied by a wind speed to get a force. This is the first important note -- Coriolis forces do not create wind/motion, they merely change the direction of it. For a pressure force, let's look at a ... 0 when in doubt, use math. t in above is the time required to empty the water from height h1 to h2 in the container 0 The flow rate is dependent on the pressure, and the pressure is dependent on the height of water above the hole. Over a short enough period of time, the change in water level can be ignored and the flow rate can be considered constant. At an infinitesimally short scale we get the concept of instantaneous flow, which again can be applied over a short time ... 0 The hydraulic radius is half of the geometrical radius. This is odd, no? The hydraulic diameter (in general) is defined to be 4A/P (A being cross-sectional area; P being the wetted perimeter). The hydraulic radius (mostly in thermoacoustics) is defined to be A/P (check out Swift's book). Subbing in, you get \frac{\pi r^2}{2 \pi r} = ... 1 First Part. No. The vortex is caused by some internal forces of the fluid. It can't end inside the fluid. (Helmholtz's theorem) Some rotation can, and will be transferred like in any surface contact, but this will never create any vortex. Theoretically it might be possible if you have some thin layer of less dense fluid above the vortex fluid. But in case ... 2 Good question. My Problem is that I can't give you a "Mainstream answer". The reason is that Turbulence is considered happening In three space dimensions and time, given an initial velocity field, there exists a vector velocity and a scalar pressure field, which are both smooth and globally defined, that solve the Navier–Stokes equations. But with these ... 1 Personally, i think understanding the fundamentals of that equation is beyond that of a 11th grader but i will give it a go. I'm going to start with something which seems completely unrelated; a warm house losing heat to its colder surroundings. Assuming no wind is blowing outside, the difference between the temperature inside and outside drives the heat ... 1 Interesting question which even made me laugh. As already pointed out the Power of an hurricane is too high to be connected to any grid. My laugh came about this; "Giant heaters out to sea which dump heat into sea water?" -Why? Because it's basically the heat of the sea which feeds the energy to the Hurricane, and thus this kind of system would not do ... 1 What is the relation between pressure and velocity of a fluid in a closed pipe flow? Bernoulli's equation: By continuity equation velocity at all points is the same. Then shouldn't the pressure be same at both points? No the pressure won't be the same at all points in the pipe. Considering Pascal's Law, a change in pressure at any point in an ... 1 Assuming A_h is small, and the pressure at the orifice is P_2, you have a situation of orifice flow. The flow rate strongly depends on the shape of the orifice. It also depends on the thickness of the pipe wall, because if it is thick, the leak will take the form of a channel dominated by viscous flow. So it will take some experimentation. You can't ... 0 The bernoulli equation is derived for streamline flow of inviscid fluid. So in practical situations, where flow is often turbulent, the equation can give incorrect results. Consider green and red point. Let green point be 1 and red point be 2. Then, we have$$P_1 + \rho g h + \dfrac{\rho v_1^2}{2} = P_2 + \rho g h + \dfrac{\rho v_2^2}{2} $$The presure ... 1 A stagnation looks like this: Fluid does not accumulate at the stagnation point; it flows away one way or the other. Close to the stagnation point, it flows very slowly and the closer you get, the slower it flows. The diagram for the stagnation point over a pitot probe looks very similar: Again, all the fluid is moving. It is that the closer you get ... 3 You seem to only have a blurry idea of the hydrodynamic approach, so I will add a tad more about the whole idea, mainly to give you a better intuition. Hopefully this will be a useful addition to Samuel Weir's wonderful answer. A hydrodynamic state is described by the variables: mass density field, energy density field and momentum density field. These are ... 6 I think that the key requirement is that the material be in local thermodynamic equilibrium. Even if it is a dynamic situation with mass flow or shock waves running through the material under consideration, if the material is in local thermodynamic equilibrium at every instant in time, then equilibrium thermodynamic concepts such as temperature, pressure, ... 2 The Finite Volume Method, and Method of Characteristics are both ways to go about solving partial differential equations. The Finite Volume Method is a way to take complicated geometry that would be impossible or difficult to model analytically and break it into regions that adhere to geometry that is easy to model analytically such as tetrahedrons. The ... 0 The reference values come from the physical system you are modeling, i.e. oil in a pipeline will have the density, viscosity of oil, diameter of the pipe and the cross-sectional velocity as reference values. Often the exact physical situation doesn't matter as long as you keep your dimensionless numbers constant (geometric and dynamic similarity) See also ... 2 The amplitude does not depend on lake depth, and the celerity depends on deph only when the depth is less than a few wavelenghts ("shallow water" case). Since the largest triggered wavelenght is the order of magnitude of the size of your stone, you should throw stone having about (e.g. 25-400%) the size of the expecte lake depth. 2 You are correct that your chest muscles are in fact pulling the lungs "open," which creates a pressure differential and draws air into the lungs. When the muscles relax, the chest cavity collapses to its original state, expelling the air (not 100% of it!). You may have heard of a "collapsed lung" injury. What happens there is that the lung is ripped loose ... 3 From what I've gathered, I think my initial guess is correct. Air tries to maintain a constant pressure. According to the ideal gas law, there are two ways to maintain the same amount of pressure with an increasing volume: 1) increase the amount of gas, and 2) increase the temperature of the gas. 0 Just like when you "enlarge" an accordion the air rush inside. In both cases, muscles pull the air bag, in order to suck air. 0 The continuity equation is \frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{\rho \mathbf{v}} = 0 , Now you can substitute directly for \nabla \cdot \mathbf{\rho \mathbf{v}} with the expression for divergence in spherical co-ordinates \nabla \cdot \mathbf{A} = {1 \over r^2}{\partial \left( r^2 A_r \right) \over \partial r} + {1 \over ... 1 On the scale of galactic spiral arms, the central black hole is gravitationally utterly insignificant. I'll illustrate with an example, NGC 524. Of spiral galaxies (this is technically an S0, but there's still spiral structure) with measured black hole masses, NGC 524 has one of the most massive. Here's a picture of the galaxy: The visible disk has a ... 1 The following (approximate) solution relates the flag's angle to wind speed. Three forces act on the flag: Weight, Drag, and the Reaction Forces (from the flag pole). Because the flag is flexible, it is modeled as a parallelogram (ie. the corner angles are not fixed) to capture the general shape and center of mass of the flag. This is illustrated below. ... 1 The size of drops of a Newtonian liquid ejected from an aperture in a laminar regime without simultaneous air intake at that aperture are governed by two non-dimensional numbers, that combine the inertial forces, viscous forces and capillary forces (which have different dependences on the length scale, i.e. size of aperture). There are several ways to ... 0 \vec{v}=\vec{w}\times \vec{r} \nabla \times \vec{v}=\nabla \times (\vec{w}\times \vec{r}) =\vec{w}(\nabla .\vec{r})-(\vec{w}.\nabla)\vec{r} \nabla .\vec{r} = 3 and (\vec{w}.\nabla)\vec{r}=w.(\nabla\vec{r})=w therefore \nabla \times \vec{v}=2w or see:$$\nabla \times \left( {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} ... 1 You seem to understand the$\rho g h$term, so I will explain the pressure energy in terms of that. Really the$P$term and the$\rho g h$term are very similar. For now lets just ignore the$P$term and focus on the$\rho gh + \frac{1}{2} \rho v^2 = \textrm{constant}\$. This is basically just conservation of energy. It says that if you throw an object up ...

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The technical answer First I need to explain entropy. Suppose you have any system which you can only see at a certain granularity: we say that you see its "macrostate" but this could be any set of "microstates" which all "look alike". Since particle-interactions tend to multiply and distribute our uncertainty about a system, we could imagine under certain ...

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Pressure "energy" P is simply the measured pressure anywhere in the system. For example, if we have a venturi whose entry and exit cross section are 1m^2 and velocity 1m/s, but it has a throat of cross section 0.1m/s, the gas will speed up through the throat, reaching a velocity of 10m/s. The measured pressure on a pressure gauge with an inlet ...

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When a fluid is squeezed, as in a cylinder by a piston, work is done on the fluid. This work 1) elevates the pressure (pressure energy), and 2) the temperature (heat energy). (If the cylinder is insulated, this is called "adiabatic".) In an ideal gas, these are all related by the ideal gas law, which says roughly that volume times pressure equals heat.

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The vessel do matters. But you can only loose. If you look the capillarity from wikipedia, you can notice that the contact angle has an influence in the height of a liquid column. But the equation reveals you, that the influence is COS(contact angle), and it's a plain multiplier. Cos (0) is the maximum. You will have an effect if you use various materials; ...

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