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Friction can be recognized in a statistical description of a microscopic view, not only in macroscopic view. It is when energy (or any other conserved quantity) that was previously organized becomes disorganized. In this case, when the atoms that were flowing along the surface, after interaction with that surface, lose that flow, and the energy associated ...

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The first term derives the torque for the curved surfaces and the second term derives the torque for the top and bottom of the cylindrical surface. I don't really know why but apparently the cross sectional area at the top keep changing as $r\rightarrow R$ where $R=Dc/2$. so $dA=\pi(r+dr)^2-\pi r^2$ $$dA=2\pi r dr+\pi.dr^2$$ I also don't understand why ...

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This is known as Helmholtz resonance. Essentially, the volume of air in the cavity acts as a spring where the spring constant is dependent on the volume of the air, and damping is dependent on the inertia of air in the neck of the bottle or container. The frequency is: or: frequency = speed of sound / 2 pi * sqrt (opening area / cavity volume * length ...

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I assume you are referring to what happens when such a plane wave is propagating parallel to a solid surface. You are correct that strictly speaking, the acoustic particle velocity, like the mean flow velocity, must obey the no-slip condition at the wall. Viscous effects become important very close to the wall only, such that the acoustic motions are no ...

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From my point of view, when a raindrop falls, it has some initial torque, which rotates the rainfall as it falls, but due to the drag force of the air, there is a frictional torque exerted by this force which cancels out the net torque, because of which the raindrop rotate with a constant angular velocity.

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This is a standard problem in flow through porous media. The equation you are looking for is the Ergun Equation. This is found at the following site: https://en.wikipedia.org/wiki/Ergun_equation. The Ergun equation is used extensively in modeling filtration also. It takes into account the viscous drag in the bed. There is also an extended version of ...

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Beside my previous answer to use a check valve, another simple way is to extend the vertical line from the pump several inches above the top of the tank and at the top of that line turn the line 180 degrees downward. Install a tee connection in that vertical line at the point where you branch off to direct water to the plant pots and connect the line to the ...

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Install a one way check valve in the vertical line at the exit of the pump discharge. Check valve will open in direction of the water flow due to pump pressure. When the pump is shut off, the valve will close either due to its weight or by a small spring when energy to pump is terminated.

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static pressure at B can be larger than at A and can be lower. This is not used for estimating stagnation pressure change. If there is no frictional loss, the fact, that the two total pressures are the same, means the energy is conserved. When this is loss, of course energy at A is larger than that at B in order to conserve the energy.

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A few micrometres is general gap between cotton shirt threads. Even as slight as 0.0000001Pa pressure should be enough for gas to pass, with atomic size being in nanometers. If cone had water instead of air, surface tension would have prevented it. But because you fill cone with air, vanderwall forces here are very weak. So, surface tension is weakest or ...

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There are two primary things that draw the ball into the waterfall (or push the ball upstream), and both of these are much stronger than the two effects mentioned in the question. As a physicist and a white water kayaker, I have a lot of experience with the relevant forces, and they can both be very strong and sometimes life threatening. The first force: ...

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Bernouli does not explain wing lift. You can measure an older light plane with a "plank" wing, factor in the wing area, distance over the upper and lower surfaces, cruise speed, and air density, and come up with a total lift figure of about 25% of the aircraft weight. Bernouli equations were published in an aviation text decades ago and the error propagated ...

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UPDATE IN RESPONSE TO YOUR COMMENT I apologise : as you suggest, there might be a lift force on the sphere if there is a shear flow in the fluid (see Discussion in 1st Link). However, this force is likely to be much smaller than the drag on the particle. ...

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The other way around is more intuitive; if the pressure is lower on the right, the fluid would feel a net positive force in that direction and accelerate toward right. hence it will have higher velocity there. So, lower pressure will result in higher velocity. you can rephrase the above in a way that it sounds as what you may like but is not scientifically ...

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There are many ways to think about the entropy of a vacuum (assuming there is no radiation and thus T=0), but all give the same result, the entropy is zero. One easy way is to notice that the walls are made of something (it doesn't matter what) that cannot change its state, so the number of microstates, $\Omega$, is equal to 1. Then $S=k\ln\Omega=0$.

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Don't let the size and length of the tube confuse you. What you described can be viewed as a big nozzle with a diameter of 20 meters with a water pressure of 500 atm. entering the back side of the nozzle. The formulas that can be used are: V = sq. root of 2gh where: V = velocity in ft/s. g = acceleration const = 32.2 ft/s^2 ...

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The frictional forces try and reduce the relative motion between the water and the spheres. If the water is travelling faster than the spheres then the water exerts a frictional force on the spheres to try to make the spheres move faster and the spheres exert a frictional force on the water to try and make the water move slower.

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Drag force opposes the motion of a body relative to the surrounding fluid. In this case the surrounding fluid moves to the right and relative to that the solids move to the left. The drag force is opposing the motion to the left, hence it is towards the right. The solids are being swept away by the fluid.

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Motion is a very diffuse concept :) you have to add a frame of reference to make it meaningfull. In the frame of reference of the surrounding water the force definitely tries to stop the particle. So if you have a stone rolled along the ground by a swift stream, the force goes in the direction of motion (in the usual, external, frame of reference), since ...

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The idea is all based on scaling laws, which is a very common theme in turbulence. As I mentioned in a comment, this assumes that the energy produced is equal to the energy dissipated. The energy produced is proportional to $u^2$. The turbulence is assumed to dissipate according to some time scale. A reasonable time scale is the eddy-turnover time, or ...

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Even though capillary forces may be the dominant force in this situation, hydrostatic (gravitational body) forces are still there. And that results in the equilibrium seen in the left most figure as you start this experiment. So as you bend the tube over, the head (force due to height of the fluid) is reduced and that would allow the surface tension of the ...

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My guess is that you take the oil composition, maybe its octane, C4H8, look at its ionization energy which is 10.01 eV and calculate the probability of spontaneous emission using Einstein coefficients. (Last formula in the document)

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Concerning your wording "force is transmitted (and maybe decreases because of loss of energy)" - no, no, the decrease of force is not easily connected to the loss of energy. Force can be decreased because there is friction, but this does not imply a loss of energy (not if nothing moves). And also energy can be lost (plastic deformation of the rope) without a ...

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You can use the same explanation : "The first molecule pushes its nearest neighbours (which also push back), the nearest neighbours push their nearest neighbours, and so on until the force is transmitted throughout the fluid." The total force transmitted to or by a surface is in proportion to the number of molecules pushing, which is proportional to the ...

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Pascal's law states that when there is an increase in pressure at any point in a confined fluid, there is an equal increase at every other point in the container. To explain this, the hydraulic jack is a better example. Car jack works on mechanical forces. A mechanical jack employs a screw thread for lifting heavy equipment. But hydraulic jacks use force ...

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I've talken your diagram and attempted to draw on the grid lines for $Q = 2e$, $Q=3e$, etc: The problem is that the vertical spacing between the points is much smaller than $e$. If you were picking up ambient charge that charge would still have to increase in steps of $e$ and it isn't doing so. I would guess that your experimental errors are larger than ...

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The continuity equation is described by: $$\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho{\vec u}) = 0$$ For incompressible and steady flow, it reduces to: $$\nabla\cdot{\vec u} = 0$$ The incompressible continuity equation in spherical coordinates is: $$\nabla\cdot{\vec u} = \frac{1}{r^2}\frac{\partial}{\partial ... 1 This is a really good question. What you are wondering about is why, when the cross section is constant, the Bernoulli equation doesn't predict free fall. It doesn't have anything to do with viscosity if the fluid is considered inviscid. The reason that the Bernoulli equation cannot be extended to the case of constant cross section is that the usual form ... 1 Yes, the viscosity of a dilatant suspension will decrease if the viscosity of the solvent decreases. Surprisingly I struggled to find experimental data to back this up. Perhaps everyone thinks it's too obvious to be worth publishing. The best I could do is this school exeriment report. The authors timed the fall of a ball through the suspension, so lower ... 2 I think you are asking two questions. First : Can falling water create suction? The answer is Yes, but the effect is quite weak. It makes use of the siphon effect and is limited by atmospheric pressure, so the maximum column of water you could lift is about 10m. https://en.wikipedia.org/wiki/Siphon Second : Is it is possible to use this effect to ... 2 While the answer of wbeaty is very interesting in showing points relevant in practice, I think all the answers are still missing an important and simple theoretical point, which you should consider to understand the process. vapour pressure does mean two different things as used above. First, the pressure, the existing water vapour would have (if it were ... 2 When the vapor pressure is equal to the external pressure, there will form a bubble. Not true. Instead, when the vapor pressure is equal to the external pressure, then any existing bubbles will begin growing continuously. And, if no bubbles are already present, then the water will superheat far above the boiling temperature, yet no bubbles will ... 2 You have the right starting point with energy basically, but I'm finding your homework hint more useful than where you go from energy of a differential unit. It says "The energy density should be easy to identify." The energy density is:$$ \frac{\text{energy}}{\text{volume}} = \frac{\text{mass}}{\text{volume}} \frac{\text{energy}}{\text{mass}} = \rho ...

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This got me thinking though, what would happen if the cross section does not change, and you have vertical flow of fluid in some pipe with an obvious change in elevation, but then had both the entrance and exit at atmospheric pressure. The energy equation, which is supposed to remain constant for a continuous fluid without losses, would have a differing ...

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The following interpretations are taken from Thorne [2014]. Chapter 17, entitled Miller's Planet, discusses the issue of the large waves on the water planet in the movie Interstellar. There Kip mentions that the waves are due to tidal bore waves with height of ~1.2 km. In the appendix entitled Some Technical Notes, Kip estimates the density of Miller's ...

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Nondimensional groups such as the Reynolds number do not generally characterise the flow as a whole, but a feature that you choose in the flow. If the flow considered is not an academic problem, you will have several such features, which have different lengths, velocities... So there may be several different relevant Reynolds and Nusselt numbers in your ...

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If the bottom had tiny holes like the ones on the air hockey games and the floor was smooth, it may slide quite easily just off the floor on the cushion of air.

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Heat transfer coefficients are a bit tricky, for good reason. You need to know something about the conditions that exist on both sides of the pipe, and you only stated that you have water in a stainless steel pipe. For heat transfer to occur, you need a second fluid on the outside of the pipe that you can transfer heat with. In addition to the above, ...

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As mentioned by @Chester, Bernoulli isn't a good approximation for viscous flows which blood flow is. Instead you should use the Hagen-Poiseuille law which relates the average volumetric flowrate and the pressure gradient in the pipe. From it we find that the flowrate $Q$ is proportional to: $$Q \propto R^4 \Gamma$$ where $R$ is the radius of the pipe and ...

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The Bernoulli equation is a good approximation only if viscous flow resistance is not important. In blood flow through arteries, veins and (particularly) capillaries, viscous flow resistance is very important.

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From a multiscale analysis it is found that the lattice Boltzmann equation solves the continuity and Navier-Stokes equations in the incompressible limit: $$\partial_t\rho+\boldsymbol{\nabla}\cdot\rho\boldsymbol{u}=0$$ \partial_t\rho\boldsymbol{u}+\boldsymbol{\nabla}\cdot\rho\boldsymbol{u}\boldsymbol{u}=-\boldsymbol{\nabla}p + ...

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Your many questions shows remarkable effort on questioning. You seem to note the point. Does anydody could explain me what is the physical reason that the jet entrains air? "Diffusion" is now given as an answer. But If that would be true, a lot of air would be diffused also on the diesel stored in the tank. But there this doesn't happend in 10 years. ...

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I am not sure if the same laws apply to the heart as that of mechanical pump, but for a given flow rate, say X gallons per minute, the mechanical pump must develop a pressure P to overcome pipe friction and any other force trying to retard flow. If the pipe in a system is reduced in size, to pump the same flow rate a higher pressure will be required. The ...

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Sorry, didn't complete my answer above. The area A of the 20 m exit hole you mentioned is 313 m^2. To determine that calculate the area of a circle with a 20 m diameter. Example: A = .785(20^2) = 313. If you want to calculate the vol. of water mult. The V by the A as follows: Vol(m^3) = 26(m/s) * 313(m^2) = 8,112 (m^3/s)

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So we have a tower (tube) 5000 m (16500 ft) tall. The bottom of the tower has a hole 20 m ( 66 ft ) in diameter. To determine the pressure the column of water (assuming specific gravity = 1.0) exerts on the bottom of the tower at point of water exit is: P = (H * SG)/2.31 where: H = height of the liquid in feet. SG = 1.0 ...

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You are going down the right path. First decide upon the temperature change you want for your key stream. This will establish the heat load. For the counter flow, then calculate the temperature change, based on the flow rate you intend to use and the heat load. This will give you your four inlet and exit temperatures. After that, the rest is easy.

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As far as I know in counter flow heat exchangers the temperature difference between hot and cold fluid remains roughly the same for any distance from either end.

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$Pressure = \dfrac{Force}{Area}$ Suppose your initial pressure is: $P_1=\dfrac{F_1}{A_1}$ Now you make the cross sectional area of the tube have $\dfrac{1}{4}$ the initial area. And this makes the total volume, hence total mass and total force $\dfrac{1}{4}$ what it was. The two $\dfrac{1}{4}$'s cancel. \$P_2=\dfrac{(\dfrac{1}{4}F_1)}{(\dfrac{1}{4}A_1)} ...

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There is a pressure differential. When the falling water hits the partial obstruction in the middle of the straw, the pressure around the edges of the straw increases. This increased pressure accelerates the water into the smaller passage below (and also tries to decelerate the water coming from above). This effect is not consistent over the period in ...

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As the problem is initially described, the nozzle is located on the left bottom side of the tank with the nozzle exit facing downward. if this is the case, there will be no horizontal force to act as a thrust to start the tank in a horizontal motion. Any thrust that may be developed by the water exiting the nozzle will be in the opposite direction of the ...

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