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During the derivation you are considering a small particle but not an infinitesimal one. In the final step, the integrand on LHS and RHS are equated because if it's true over a finite particle, it must be true at every infinitesimal point. But yeah - please try to rewrite using MathJax.


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In this derivation is it necessary to write the triple integral, as I thought that if we are dealing with one fluid particle it only contains one "point" and hence we do not have to take a sum? The fluid "particle" in this case is not a mathematical point, it is just a "small" region of the fluid. Indeed, you have explicitly drawn it as an extended ...


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The continuum hypothesis means the following: at each point of the region of the fluid it is possible to construct one volume small enough compared to the region of the fluid and still big enough compared to the molecular mean free path. Why is that important? Because of two things. First, since the volume you can build at each point is very small compared ...


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The derivation shown in the paper is a real mess. Not only he forgot the area, but he doesn't work well with the differentials. The first green underline should read $|\vec{dF}|=pdS$ and the second one $\vec{dF}=p\vec{n}dS$


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I don't really conceptually understand why the mass flux/mass flow rate into the arbitrary volume BCDE is given by [...] $u$ is speed. This is a distance per second. Multiply a distance (per second) with a cross-section area, and you get volume (per second). So $$\dot V=Au$$ where the dot in $\dot V$ simply means per second. If some particles of water ...


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I'm assuming $\vec u(\vec r, t)$ is the vector field describing the velocity of the fluid. The flux, generally, will be of the form $\rho \vec A \cdot \vec u$. $\rho(\vec r, t)$ gives us the amount of mass that is flowing at $(\vec r, t)$, $\vec u$ gives us its velocity, and $\vec A$ gives us the area through which it is flowing, so it makes sense that we ...


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Consider a control volume $\Sigma(t)$ of a fluid with density $\rho(\mathbf x,t)$. The mass inside $\Sigma(t)$ is clearly given by $$M(t):=\int_{\Sigma(t)}\rho(\mathbf x,t)\text d^3\mathbf x.$$ The way $\Sigma(t)$ is defined is that its mass content doesn't change with time, that is, a control volume is representing the time evolution of a certain amount of ...


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You are missing the bit under-braced here: $$ \dot{m}=\rho\underbrace{\mathbf A\cdot\mathbf v} $$ This is a dot product, which takes two vectors and returns a scalar: $$ c=\mathbf a\cdot\mathbf b=a_xb_y+a_yb_y+a_zb_z $$ So the mass flow rate is indeed a scalar value.


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You're calculating the rate of liquid flow through a tube under a specified pressure gradient. The rate of flow changes depending on whether the flow is laminar or turbulent. Laminar flow is described by the Hagen-Poiseuille equation: $$ \Delta P = \frac{8\mu\ell V}{\pi r^4} $$ where $\Delta P$ is the pressure drop, $\mu$ is the viscosity of your saline ...


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Place a glass rod against the cylinder at right angles. Observe the size of the contact patch - the length of the patch should be easy to measure through the glass rod. Make sure that the contact force (weight of the rod) is the same in both cases. You will be able to see that one cylinder deforms more (larger patch) than the other. That makes a objective ...


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First up thanks to all who took an interest especially @irishphysics who stuck with the question for some time. It turns out that the phenomena was analysed and solved by Lord Kelvin and is known as the Kelvin wave pattern. The pattern itself is the result of a spreading pressure wave which manifests itself as the curved diverging wave crests (the ones I ...


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Do you mean this equation $A_1v_1=A_2v_2$ In that case, we are supposing the liquid to be an ideal fluid i.e. it is incompressible fluid. That should mean that mass entering per second in the tube is equal to mass exiting per second.


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If your tube has a hole in which mass can flow out, then we can lose mass at a rate $Q$. Similarly, if your tube has an inlet, then the mass can accumulate at a rate $P$. Thus, the time rate of change of mass would be $$ \frac{dm}{dt}=P-Q $$ However, with a tube, we usually consider it closed, such that $P=Q=0$, leading directly to $$ \frac{dm}{dt}=0 $$ from ...


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Your fractal pattern will fail for reasons already given. However, given a large enough lake, you should be able to stand on a frame which is supported by a very long (perhaps circular) wire. All you need is for the force per meter (assuming uniformly applied) caused by your body weight and the structure itself to be less than the surface tension force ...


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You are referring to the equation: $$\frac{\partial \rho}{\partial t} + \frac{\partial \rho u_i}{\partial x_i} = 0$$ which is the conservative form of the continuity equation in Eulerian form (fixed domain, fluid moving through it). This can also be written as: $$\frac{D \rho}{D t} = 0$$ which is the Lagrangian form (the density of a moving region of ...


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The problem lies in your simplistic assumption that the perimeter is the only thing that matters. The actual force can be no greater that the weight of displaced water (see for example a capillary) and as the force you try to exert, so the amount of water displaced will increase. That doesn't mean you could not use surface tension to "walk on water" - just ...


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Since the force is based on the wetted perimeter, any configuration that would make the perimeter very large in a very small area would be overwhelmed by the surface tension of the water droplets connecting nearby perimeters. So the effective perimeter would be much lower. So you are sunk!


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In every boundary layer (except for exotic hypersonic cases), the speed at the wall is zero. At the trailing edge, the upper and lower layers meet, and if you imagine a plane which extends from the trailing edge backwards and follows the streamlines, the speed at the trailing edge is equally zero. The more you now move away from the trailing edge along this ...


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Inviscid flow doesn't exist. That's so important to understand, I'll say it again: Inviscid flow doesn't exist! However, we use it all the time. So what gives? It turns out that many of the effects we are interested in are viscous, but the viscous effects can be modeled various other ways. This is effectively the same type of question as Does a wing in a ...


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The mechanism is known as Bernoulli's law (see e.g. wikipedia or scienceworld): A moving fluid, regardless of viscosity, has a different apparent pressure along the direction of motion and perpendicular to it. Along perpendicular directions, the pressure is reduced. This is the situation seen by your cylinder, which (when rotating as in your diagram) will ...


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when you look at an airplane propeller, you will se that parts of the propeller blades are slanted. These slants, when rotating, push the air back, creating a low pressure zone, sucking more pressure in, while at the same time pushing more air back.


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This is a most excellent and astute question. Ultimately it comes down to experiment: the model below works pretty well for many fluids. What this must mean therefore is that the loss is small enough that each particle of fluid, in flowing past the region of disturbance, loses a fraction of its energy that is small enough that it doesn't upset the energy ...


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Wiki/google quote: Dynamic pressure is the kinetic energy per unit volume of a fluid particle. Dynamic pressure is in fact one of the terms of Bernoulli's equation, which can be derived from the conservation of energy for a fluid in motion. So Dynamic Pressure is just local impulse/energy of movement that is being passed from particles to other ...


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No,the velocity increases but the pressure decreases. This pressure that you are refering to,is the dynamic pressure,it is the pressure that drives the fluid.That keeps the momentum going. The fluid entering the narrower piece of pipe means that fluid tends to push back the fluid that tries to enter after it.That compresses the micro movements of the tiny ...


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As you've indicated in your title, the correct question is "why doesn't this work". The system, as described would continue to produce energy indefinitely without any being added to it (perpetual motion, violation of conservation of energy...) . So you can be sure there's a problem. I believe that part of the trouble here is in the assumption that the ...


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1) The $sin$ term appears to be resolving the velocity of the water relative to the face of the disk. $U$ is the just the speed, it's necessary to determine what part of that velocity will produce force on the object. It might help to visualize extreme cases, setting $\alpha$ and $\beta$ to 0, $\pi /2$ etc. For $\alpha = \beta =0$ you have a perfectly level ...


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You asked a similar question on worldbuilding for a story. Hit youtube for Operation Crossroads, Baker test. The US Navy detonated a 21kt device 90 feet underwater. Produced a nice fountain but no overly-destructive wave action after a few kilometers. A few years later, Castle Bravo (15Mt) also failed to produce any significant damage* outside the ...


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As if you will see that a capillary tube kept in a beaker filled with water,so the water level rises but if the length of the capillary tube is insufficient than the angle of cos theta will be of 90 it means that its just impossible,and the surface tension force will be just stopped.


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It has to do with the volume of the object, the height is irrelevant. If I put a tower that reaches to the bottom of the ocean but has negligible width and thickness then the tower displaces almost no water. The water height raise will be determined by the displaced water. An equation for the water rise: $h = \frac{V_{obj}}{Surface Area}$ Where the ...


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The reason why oil and water do not mix on Earth doesn't really have to do with their densities. Water is a polar molecule, which means it has distinct regions that are more positively or negatively charged. Oil is non-polar, which means that it does not have differences in charge across the molecule. As a general rule, "like dissolves like"; which means ...


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With no gravity and no impulse given to the blobs of fluid, my guess is that they would remain one inside the other and do nothing. If present, gravity would play on the blobs of fluid by making the less dense fluid (the oil) rise relative to the denser fluid, due to buoyancy. The oil would also "rise" if the spacecraft were rotating due to the greater ...


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Books generally teach sine or cosine waves because according to Fourier thereom any wave can be written as linear combination of sine or cosine waves. FOURIER THEOREM A mathematical theorem stating that a periodic function f(x) which is reasonably continuous may be expressed as the sum of a series of sine or cosine terms (called the Fourier series), each ...


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The sine function is just an idealized way to approximate wave motion, and indeed suitable for teaching the basic principles of how waves propagate, reflect and interfere with one another to create standing waves, but as with any real physical system, including the motion of waves, the closer you look the more you see non-ideal behavior. For surface waves ...


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From this link below " Refraction is the bending of waves because of varying water depths underneath. The part of a wave in shallow water moves slower than the part of a wave in deeper water. So when the depth under a wave crest varies along the crest, the wave bends. See "http://www.coastal.udel.edu/ngs/waves.html So my guess is its due ...


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Quite clearly time itself doesn't slow down, but the time scale over which the drops form and fall is strongly dependent on the viscosity of the fluid. The times cale for viscous flow through a constriction under the action of gravity is given through dimensional arguments as $$ \tau \sim \nu / (dg) $$ where $\nu$ is the kinematic viscosity of the fluid, ...


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JonT provided a nice answer, but I'd also like to add that heat is always generated whenever wind is. When turbulence is generated (and there is always some level of turbulence generated when one generates wind), the energy in each eddy cascades down to smaller eddies (with very little energy lost) and eventually the eddies reach a size where viscous forces ...


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There are also conservative and non-conservative forms of the Burgers' Equation. The non-conservative form you presented will only solve for smooth shocks. You need the conservative form if there are likely to be any sharp changes in the solution. This might also help: http://people.maths.ox.ac.uk/trefethen/pdectb/burgers2.pdf


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Your question is ultimately about scaling. Scaling is best explored by non-dimensionalizing equations. Let's look at a free-fall equation of a mass $m$ in the absence of air friction : $$ m \frac{d^2 z}{dt^2} = - m g $$ Obviously the time to fall of a given height will depend on gravity $g$. But your question is, will it be the same up to a scaling factor ...


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If I understand correctly what you're asking, you are referring to the vacuum created from an explosion. Think about it like this: An explosion is rapidly expanding gas from a central point (the explosive material). As that gas expands, it pushes all of the material (the gas of the atmosphere) away from that central point. The front of the gas (shockwave) is ...


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I have to disagree a little bit with @NoMorePen. In an incompressible fluid (the assumption usually made at low Mach numbers) there is no energy associated with a pressure change. The principle at work is conservation of momentum. You understand that the velocity of the flow changes. Well, the only way a parcel of fluid can change velocity is by falling ...


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Not necessarily, it depends on how different the viscosities are. @MonkeysUncle got it right. If the Reynolds number is < 2,000 the flow is laminar; if it's > 4,000 the flow is turbulent. Since the Reynolds number depends on viscosity, if the viscosity of the two fluids is different enough that it changes the flow from laminar to turbulent, then you ...


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I think it will be clear if you understand where Bernoulli's principle comes from. If you have a duct, and you decrease its diameter, the same amount of water you put in has to come out the other side. If furthermore, the fluid is incompressible you have that: $A_1V_1=A_2V_2$ where $V_{1,2}$ is the velocity of the fluid where the duct cross section area ...


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When you rock the espresso from side to side you make the surface larger and stretch the bubbles. When the surface comes back to its original size it is energetically more favorable for the bubbles to bunch (lowering their exposed surface area). Every time you rock the cup, the thinner region is stretched more (relatively) and so the bunches end up ...


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Yes you will create wind sound and heat. The figure you show is a simplification of the actual interactions that occur to allow students to learn visually. Consider a jet aircraft as an example of 'accelerated mass'. if you've ever stood near a runway as a plane is taking off you'll know how much 'wind' is generated by its passage. A supersonic jet in ...


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Terminal velocity is only achieved by object in free fall with air resistance balancing the weight. The main resistance to water flow on a river (at an angle to the surface so that the downward acceleration is much less than $$g$$ of course) is caused due to friction with the mountain surface. Air resistance (i.e. fluid friction/drag) has practically no ...


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The air in an idealized stream (no white-water rapids) would be at the top of the stream, so that the water would flow inhindered.


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I found a NASA technical memo for very accurate "Real-Time Aerodynamic Heating and Surface Temperature Calculations for Hypersonic Flight Simulation." The authors also discuss a bit how they obtained their expression. I have to say, though, Dave's dissertation and tgp2114's answer are wholly sufficient and more straightforward.


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For viscous hypersonic flows, the heating takes a form: $$ q_w = \rho_\infty^N V_\infty^M C $$ where the parameters $N$, $M$, and $C$ depend on the configuration and $q_w$ is the heating in $W/cm^2$ (this is all from Hypersonic and High Temperature Gas Dynamics and I highly recommend this book). For the stagnation point (like the leading edge of a body): ...


0

The force on the dam is the integral of the pressure over the area. $$F_{dam} = \int_{A}{p(z)dA}=\int_{0}^{Z}p(z)dz\int_{0}^{L}{dx}$$ The pressure is a function of mass density, gravitational acceleration, and depth. $$p(z)=\rho g z$$ Then the force on the dam is $F_{dam}=\frac{\rho g L Z^2}{2}$



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