# Tag Info

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one side of the ball is rougher while the other side is softer .the rougher side brings the stream lines above the ball closer decreasing the cross sectional area.then the velocity at that side becomes greater decreasing the pressure at side. this doesnt happen on softer side causing a lesser velocity and greater pressure on that sidetherefore ball travels ...

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1) They are non-polar, which means that it does not have positive and negative ends and consequently do not attract each other. 2) Also the viscosity of these fluids is less and hence have low resistance to flow. Therefore, they are slippery

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This is essentially Pascal's law. I would like to add that it depends on the fluid being in equilibrium, as well as incompressible. Otherwise local pressure differences can build up and the fluid can behave rather differently. I like to look at this problem from a thermodynamics point of view. One way to think of this is that the pressure of a fluid is a ...

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It is not true that the pression of a fluid is due to gravity. Do you think fluids inside a bottle in the space station have no preassure? The right answer is the one you have quoted. The first link only refers to buoyancy, that does require gravity.

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The question is therefore : why doesn't a fluid flow out a bottle smoothly ? The "bottleneck" phenomenon is caused by the lack of pressure in the can/bottle. As the liquid flows out, the pressure inside decreases because the volume of the container is fixed. When the pressure inside the bottle reach a given threshold, the outside air tends to flow in the ...

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Because the bag deflates as milk leaves it, the volume of the bag decreases and the pressure remains constant so the milk pours smoothly. When pouring from a can, which does not deform like the bag, the pressure inside the can decreases as liquid leaves the can. The pressure differential creates a potential that pulls air into the can, interrupting the flow ...

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As @JánLalinský nicely explains, surface tension is measured between two fluids, while viscosity is measured within one. Say that you have a droplet of some liquid this means that if you change the surrounding medium the liquid-surrounding surface tension changes, while the viscosity of the droplet will not. That said, if you keep the surrounding medium ...

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I don't believe that this would be necessary. Most pumps (not constant volume ones) have a relationship between flow rate and pressure jump (often called a pump curve). Since the entrance diameter of the pump doesn't change, you've got (basically) the same momentum in the pipe at a given flow rate regardless of how it gets that way. Another way of looking ...

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Consider the following image, we have some fluid volume, $V$, having density $\rho$ and traveling at a velocity $v$ along a pipe with some cross-sectional area $A$. The rate at which the water flows through the pipe is called the volumetric flow rate. This is given by, $$\frac{dV}{dt}\equiv Q=\mathbf v\cdot\mathbf A$$ where $V$ is the volume of the ...

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Both viscosity and surface tension are connected theoretically to inter-molecular forces, but they are still very different concepts. Viscosity force is a force that acts only when the fluid is moving and acts to decrease the gradient of velocity in it. Viscosity is a characterization of the fluid itself. Roughly speaking, it says how fast momentum of ...

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The specific gravities would be the same if the levels of the two side were the same after liquid-II was added. When I try this, my logic seems to be flawed too. I don't get 1.12. The level on side II has not changed. The level on side I has risen 2 cm. So 2 cm of liquid-II were added. Consider the horizontal plane 2 cm below the top of side II. Below ...

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Pascal used a serynge that allowed leakage of fluid, to demonstrate that increasing pressure at one point would increase pressure at all points. Although water is leaking, as Babou said, we can still consider the fluid enclosed, where, instead of holes (that allow leakage) we could use sensors to mesure pressure. Watch a Wolfram demonstration of the ...

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Every time, when you deal with differential equations, the first step is to put it into dimensionless form. There are more reasons for that. First, "small" and "large" has no meaning in dimensional forms, since you can always change the system of units. Second, nature knows no units. Now, when there is no exact solution (this is often the case), you can ...

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The system you describe, a water tower with pumps and turbines, is a version of a pumped hydro storage system. The amount of energy stored in a pumped hydro storage system is defined by the elevation, i.e. the height that the water is lifted; multiplied by the volume of water that is lifted (assuming that the height the water will fall, is the same as the ...

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The fluid in the tube is not water as some might think but an organic solvent called Dichloromethane. The reason the bubbles form is due to the fact that the fluid is heated at the base of the tube to it's boiling point which is a low 103.3 F degrees. You can almost get it boiling by holding it in your hand. The bubble is actually the vapor form of the ...

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There was a recent publication addressing exactly this question. From the abstract: Here, we show that the overall foaming-over process can be divided into three stages where different physical phenomena take place in different time scales: namely, the bubble-collapse (or cavitation) stage, the diffusion-driven stage, and the buoyancy-driven stage.

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In your approach, drag force is a result of the calculation rather than additional effects: it is the surface integral of the normal stress vector $T n$ where $n$ is the normal to $B$. Buoyancy force will arise if you add gravity to your fluid. Of course in order to close the system you need to prescribe how the body moves: if you're making it move ...

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As is given in Jamie's answer I'll assume the surface is a revolution about $r=0$, that the mean curvature is proportional to the pressure difference, and that the radius of the cup is much larger than the inverse of this mean curvature. In this case the mean curvature can be specified as $$K_m = \frac{r''}{2(1+r'^2)^{\frac32}}$$ As in Jamie's answer the ...

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The friction between a solid and liquid is a function of viscosity. The best way to answer this is with a model setup called Couette flow where a fluid sandwiched between two plates is sped up by the movement of the top plate: Image source: University of Virginia, Physics 152 taught by Michael Fowler The friction force $F$ that the fluid exerts on ...

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Dust sticking to things is a complex process but can be broken down into several stages and analyzed. First though lets define our dust. Dust Size The aerodynamics of dust are most easily approximated by pretending all of the particles are spheres with a density equal to water ($1000 \frac{kg}{m^3}$). Each particle is assigned an aerodynamic diameter that ...

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The centrifugal force (said pseudo force) pushes the liquid towards the walls of the cylinder on its entire length. Since the walls restrict further motion of the liquid from the centre of rotation, and the liquid is incompresible, it cannot squish any further acting on side walls only, and thus the bulk of it will start pushing on the top and bottom of the ...

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Well, in this case the origin presents an unstable equilibrium point, in the sense that if you put a test particle exactly at the origin, and there is no noise/fluctuation, it will stay there, but if there is even a tiniest fluctuation it will drift off origin. The reason is that $u=(0,0)$ at $(x,y)=(0,0)$. Put it another way, the equations of motion for a ...

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Tribology (not the study of tribes!) is the study of what happens when things 'rub'. This involves friction and wear when solids rub against other solids (such as in mechanical bearings) and the effect of liquids (such as 'lubricants') and other fluids. Friction at a solid-liquid interface is still called friction. It is a 'damping' or 'dissipative' force, ...

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I see an error in the Lindeburg correction formula presented by you. Instead of 2.1r it should be 2.4r. I found in an article of Smoluchowski, that the coefficient (1+2.4r/R) was proved by Landenburg in the paper: R. Ladenburg, Ann. d. Phys. 23, p.447 (1907). The error is probably responsible for the non-constant value of viscosity.

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Hint: If you divide by $\sqrt {h(t)}$ the equation separates.

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For all practical purposes, in the Earth's atmosphere, terminal velocity is less than the speed of sound in water by a large margin ie around 15x. So water will cushion an impact. That Mythbusters episode left something to be desired - accuracy. The world record for a shallow dive into 30cm of water is about 11m in height, meaning an impact velocity of ...

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If anyone is looking for the same thing, here is the solution : $$m = \rho V = constant$$ $$\Leftrightarrow \rho dV + V d\rho = 0$$ $$\Leftrightarrow \frac{d\rho}{\rho} = - \frac{dV}{V}$$ $$\Leftrightarrow \chi_T = - \frac{1}{V} \left(\frac{\partial V}{\partial p}\right)_T = \frac{1}{\rho} \left(\frac{\partial \rho}{\partial p}\right)_T$$ It's simple but ...

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I would say no, it is not easier. The equation of state you mentioned will involve new variable(s) as the temperature for instance. Therefore you should add an equation for energy in the Navier-Stokes system to find the evolution of the temperature field. Dealing with compressibility means you have density variations. You may also have to solve the ...

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There is two types of drag: The viscous drag which is caused by shear of fluid on surface area, and The pressure drag, Viscous Drag is caused by shear caused by liquid flowing on object surface and is significant in laminar flow, Pressure Drag is caused by the difference in pressure at the front and rear-end of surface of object and it is significant in ...

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I'm not a fluid mechanics expert, but my mechanical systems knowledge suggests it might be simply a natural oscillatory behavior, which is always present but in this case is more noticeable due to the aggressive initial response (i.e fast influx of air) your chamber experiences. So what is causing this inexplicable pressure drop? Once the chamber has ...

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The working for that last solution was correct except for the final answer. It should be: $z = \frac{-m}{2 \pi U}+0i$

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The maximum rate at which the pressure "information" can be transferred in the pipe is the speed of sound for the liquid. This can be calculated using the Newton-Laplace equation: $$a = \sqrt{\frac{K}{\rho}}$$ where, a = speed of sound K = bulk modulus of elasticity $\rho$ = density So the time for the pressure pulse to reach the other end of ...

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To answer both parts of your question: In general the flow coefficient will not change with the head height of the system. The head height will only impact on the static pressure and hence the flow rate through the nozzle. This will be true provided the liquid level is not so low that it is influencing the flow pattern near the orifice, in which case the ...

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In the formula for capillary rise only θ is a variable. So when the capillary tube of insufficient length is kept then the θ will change accordingly so that capillary length h comes out to be height of the capillary tube. Limiting case is when θ = 90 degrees when height of capillary tube is 0. This is meaningless so this is not possible. Concluding thing is ...

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From an engineering standpoint, the propulsive efficiency (which is relevant for engines, rather than the aerodynamic effiency) is defined as: $$\eta_{prop}=\frac{\mathrm{Power Available}}{\mathrm{Jet Power}}=\frac{P_a}{P_j}$$ In this, the jet power is the amount of energy added to the flow: $$P_j=\frac{1}{2}m V_j^2-\frac{1}{2}m V_0^2=\frac{1}{2}m ... 2 If you use Newton's 3rd law of motion, which states that for every action, there is a equal and opposite reaction i.e. when the fan accelerate the air particles by a force, it also experiences a backward reaction force and according to 3rd law of motion these action and reaction force must be equal.Hence the net force acting of the box is zero . That's why ... 0 No! You have to make an opening in the front and in the back. The theory is that the sum of the momentum of all moving particles and things \displaystyle\sum_{k=1}^nm_i\mathbf{v}_i should be zero. If some particles are accelerated by the fan and moving backwards a force will be applied to the box in direction forward. If it wasn't a smooth stream of ... 1 It depends. The incompressibility helps you to get rid of sound waves. Often they are just a nuisance and limit your time-step. Especially in low Mach number flows the sound waves are much faster than the flow and can be very limitting. E.g., in meteorology they must be filtered even when solving the compressible equations. The incompressible Navier-Stokes ... 1 Assuming you do survive the impact with the water, according to this answer all you need is 4m. 3 The forces slowing you are (1) drag as you note and (2) buoyancy. The former, assuming ram drag is the main one, is given by:$$F_D = -\frac{1}{2}\,A\,\rho_W\,C_D\,v^2$$where \rho_W is the density of water, v the velocity of the dragged object, A the cross-sectional area presented to the water as you fall and C_D is a fudge factor called the drag ... 2 There are bound to be more detailed answers and analysis, but let's look at one of the oldest possible results in the field. Isaac Newton proposed that, for bodies of equal density, the penetration depth during ballistic impact is equal to the length of the penetrating body. Note -- this is independent of the speed (which means independent of the height) of ... 0 As mentioned in comments the dynamic pressure is the equivalent of the kinetic energy in fluid dynamics. It is dynamic since it can change with time (like kinetic energy) and also like kinetic energy it is an invariant under coordinate transformations (e.g rotations) as such it is a scalar and not a vector (just like energy which also depends on velocity in ... 1 Here the radius of the pump piston according to your question is 0.015 m. So the area of the pump piston is πr^2=π(〖0.015)〗^2= A(say) If x N force require then we create pressure on that piston is P= F/A=x/(π〖(0.015)〗^2 ). Hence, F/A=x/(π〖(0.015)〗^2 )=2500/(π〖(0.12)〗^2 ) or, x= 39.06 N. 0 Simplifying: Two vessels 1 and 2 with volumes V_1 < V_2 and pressure P_1 < P_2 contain methane at temperature T. A small valve is opened between them so gas starts to flow from 2 to 1. After time t the pressures are P_1' and P_2'. The rate of change of pressure at t=0 is estimated at R_1, and at t=t_1 it is estimated at ... 1 As some other materials in addition of those definitely helpful references proposed by Michiel, I found the following very useful resources: a full review of many concepts in droplet dynamics has been presented in this book: Ashgriz, N. Handbook of Atomization and Sprays. Vol. 11. Springer New York, 2010. for droplet collision: Rein, Martin. ... 1 Books For fundamentals I prefer books over papers, because they are typically more thorough and a little bit more 'slow' in the introduction of concepts. There are many books that will cover, some of, the topics that you mention. I will mention below the 3 books that where most useful to me in the past. 1) An excellent resource for a theoretical foundation ... 1 Interesting question - I have to speculate about the answer. First off, to generate an acoustic wave, you need to bend the diaphragm... so in that sense the question makes no sense. But I think you are really asking whether the situation should be thought of as static or dynamic. I am pretty sure that the advantage of the piezo transducer is that it can ... 0 In the equation \mathbf{j} = \mathbf{v}\rho, \mathbf{v} is the average velocity of charges and the equation describes the net flow of charge. A black body spectrum describes a system at thermodynamic equilibrium and at thermodynamic equilibrium there is, by definition, not net transfer of heat. This means that the net energy flux through a surface is ... 0 All the forces acting on the bead will accelerate the center of mass as though the force is acting there - so the mass will accelerate with a magnitude and direction given by the vector sum of$$\frac{\vec{F_{up}} + \vec{F_{spring}}}{m} . The bead will further undergo a rotation. The torque is entirely due to the spring - but we are missing the angle of ...

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