# Tag Info

0

One good resource for these kinds of questions which involves looking up some kind of quantity is wolfram alpha. As you can see you can get the answer for any temperature. As far as interpolating the values, it is fine if your interpolation error is small. That depends on the nature of the data you are trying to interpolate (is it linear/nonlinear) and the ...

0

The pump can apply only a fixed pressure/force from downwards . From the open end the atmosphere will apply a fixed force/pressure . It is only the height of the column which is variable here . If you need height $>10$m , you just need a bigger pump . If you can't get that , the water will rise upto a fixed height say $8$m,after which there will be ...

5

The premise of the question is not correct, but there is a general shape to rivers. From Leopold and Langbein, writing in Scientific American: A sample of 50 typical meanders on many different rivers and streams has yielded an average value for this ratio of ahout 4.7 to one. The ratio they use in that article is different from the definition you ...

4

which will go negative if $h>10m$. How is it possible? It is not possible that $P_{up}$ will go negative. It will experience cavitation. This is one of those cases where you have to realize that the equation you're using is a special case of a larger theory, step back, and employ the larger theory. If you step back from the concept of water as an ...

0

This realationship: $$P_{down}-P_{up}=\rho g h$$ will only hold for a standing water column. Definitely not for a free-falling one.

3

I am not convinced that this is true. However, if it is, one heuristic and simple argument might be to assume that a meander makes approximate semi-circles around the true path. IF we assume that intersections with a straight line are spaced by $r_1, \ldots, r_n$, $\sum_k r_k = l$ then the total path is $\sum \pi r_k = \pi l$. Since $l$ is the length of a ...

-1

that wavelength like structure approximates to circumfrence of a circle from crest to crest or trough to trough and the straight line is the diameter, so pi=C/d. Also if $\sqrt g$ is approximated to $\pi$, will this mean that $\pi$ and $g$ are related?

3

The subject you are asking about is flow through an orifice. The basic equation for inviscid flow is $Q = AVK$ where $Q$ is flow rate in mass/time, $A$ is orifice area, and $V$ is flow velocity. $K$ is a factor that depends on the shape of the orifice. Velocity $V$ generally follows $V = \sqrt{2gh}$ where $g$ is acceleration due to gravity, and $h$ is the ...

0

Actually, ducted fan propeller are much less efficient than open blade propeller soif efficiency is your priority, open propeller is your option

0

Ordinary velocity is just the velocity of an individual particle; a velocity field $\vec{u}(\vec{r}, t)$ is a function whose first argument lets you pick out the particle whose velocity you want to measure. If you follow an individual particle, you can write its position as $\vec{x}(t)$, a vector function of time. Its velocity is just the (time) derivative ...

12

The more interesting question is, "What does air feel like when it is moving away from me?" The answer is that there is really no sensation at all. You feel all the air coming into the vehicle, because it has a bulk momentum with respect to your frame of reference. However, air being sucked out of those same windows and doors is being pulled from a large ...

2

The answer to the valve-question,1 according Pilotfriend, seems to be the "floppy walled Eustachian tubes". During ascent the gas (air) in the middle ear cavity expands and a small amount of pressure builds up against the ear drum causing them to bulge outwards ever so slightly (that ‘fullness’ you feel in your ears just before they ‘Pop’). This pressure ...

3

If the cylinder is stationary, it is according to the hydrostatic pressure equation: $\vec{\nabla} p = \rho \vec{g}$ You can derive this equation by eliminating all velocities in the Navier-Stokes equations. If gravity is oriented in the negative $y$ direction, the equation becomes: $\frac{dp}{dy} = -\rho g$ EDIT: If you integrate this equation you get ...

0

Yes, provided you wait an infinite length of time :) First, forget the tube. It has no effect on the problem, except maybe to minimize splashing. Assuming the rate of flow out of the top bucket is proportional to the pressure at the hole, you can write a 1-term linear differential equation for the amount of fluid in that bucket, for which the solution is ...

2

Yes, at least if you ignore the little droplets that leave the bottom of the upper bucket a little damp. There will also be some water in the tube, but it won't be much higher than the level of the water in the lower bucket. (There might be some capillary action that raises it a little, but probably not much.) I suspect the reason you ask this is because ...

3

What distance can a cannonball traverse thru water without losing too much kinetic energy? For a back-of-the-envelope calculation we start from the observation that this distance scales with the ratio of the kinetic energy of the cannonball and the drag force exerted on the cannonball. Let's denote the ball's radius by $R$, its speed by $v$, and its mass ...

2

Consider an object that is floating and stationary in a liquid of density $\rho$ such that a volume $V_s$ of the objects is submerged. Suppose that we orient a cartesian coordinate system such that the positive $z$-axis points upward, orthogonal to the surface of the liquid. When the object is floating at rest, the net force it experiences is zero. If we ...

6

From the Wikipedia article for Reynolds number: In fluid mechanics, the Reynolds number (Re) is a dimensionless number that gives a measure of the ratio of inertial forces to viscous forces and consequently quantifies the relative importance of these two types of forces for given flow conditions. In addition to measuring the ratio of inertial to ...

0

Waves in fluids carry momentum not mass. When we see ripples diverging from a point, there is no radial (horizontal) movement of mass. Rather the mass moves in the vertical direction, some what sinusoidally. If you have observed birds floating on water surface, you see them moving up and down but not along the waves.

3

Turbulence is not one of the great unsolved problems in physics. Physics tells us exactly how turbulence emerges as a direct consequence of local mass and momentum conservation. We can create multiparticle computer models such as lattice gas automata that generate turbulence at large length and time scales. We can write down the equations that govern ...

1

I happen to have attended the ATW (Aerospace Thematic Workshop) twice in 3 years, so I can give you some first insight. I do not know precisely when plasma flow control was first experimented on, however this field gathers a quickly increasing amount of researchers around the world. The reason for that is that, before, people could use continuous DC ...

1

Jim is basically correct in what he has said. Consider a vertical cylindrical pipe which has an air-tight plunger at some distance from the top. Now you fill a section of that pipe with a small volume of incompressible fluid (water will do for this analogy as it is [for all intents and purposes] incompressible) resting on the plunger. If you now seal the top ...

1

Your statement: Assume density of fluid is same throughout. conflicts with your actual question But why is that at a greater depth pressure is higher when molecules are the same? If we imposed the very strict and non-physical constraint that the density of the fluid was uniform and isotropic, then we would have not variance in pressure what so ...

1

In short: because the weight does it so. Imagine a situation where several people are walking on each other in a small room: those that are at the top don't feel any discomfort whereas those at the bottom are crunched by the weight of the one above them. It's quite the same for the molecules.

0

The input stream does not only have a thermal energy - it also has a mechanical one. Mechanical energy can be used for work, and the gas temperature is easily changed by work - in adiabatic processes it rises when gas is pressed, and falls when gas is able to expand. This gives a general idea why this tube could work and at the same time not be a Maxwell's ...

0

Discretize the equations for conservation of matter, momentum, energy, and any other relevant physical quantity, and solve them on a computer. Increase the resolution of the simulation until the results converge. (Then think really hard about the results, supplement the detailed calculations with analytic treatments of simplified analogous cases, etc. But ...

0

Riblets can improve fluid flow in a pipe by 10-15%. Parallel riblets improve flow 10%.* Converging riblets speed flow 15%. Diverging riblets slowed flow 15%. The parallel riblets are believe to reduce wall-bound turbulence, but what do the converging riblets do? *Note the converging and diverging riblets were using air while the parallel ones I ...

0

I've made a small document featuring fluid dynamics equations in terms of vector-valued differential forms. The document with information on any further developments can be found on my page.

0

It is important to distinguish between liquids and gasses since liquids may be considered incompressible, while gases are compressible and therefore the density of gases cannot be treated as constant at high velocity. It is also important to distinguish unidirectional flow from bi-directional flow. There are four basic forms of wavy patterns known as ...

3

A perfect fluid is defined by the property that, in the local rest frame, it allows no energy fluxes and no anisotropic stresses. Thus, at a given space-time point, in the local rest frame [in which the components of the 4-velocity are $u^{\alpha} = (1, 0, 0, 0)^{\mathsf{T}}$], the energy momentum tensor components are $T^{\alpha\beta} = \mathrm{diag}(e, p, ... 1 Let$\mathbf v(\mathbf x,t)$denote the velocity of the fluid at position$\mathbf x$and time$t$. Suppose that we imagine traveling on a path$\mathbf x(t)$through the fluid, then we can ask ourselves If we travel along the curve$\mathbf x(t)$through the fluid, then what will be the rate of change of the flow velocity of each point in the fluid ... 2 Its a method called co-ordinate grid transformations that is used to transform an arbitrarily shaped geometry into a square; in the computational domain. Grid transformations work as parametric transformations do, in co-ordinate geometry. what these transformations basically do is, they map the complex geometry (viz. a curve) into a simpler geometry (a line; ... 0 You move with the plane because kinetically you are part of the plane's body. When you sit or walk in the plane gravity attaches you to the floor/body. you are going with the same velocity as the plane as everything in it. If you throw a ball in the air it will have the velocity of the plane already so no effect will be seen. We see effects in the plane when ... 3 Objects don't accelerate because they're inside other objects. Objects accelerate because other objects make forces on them. The chain of cause and effect here is that the box can affect the air, then the air can affect the helicopter. The answer to the question depends on how rapidly the box is accelerated. To pick an extreme case, suppose that the box is ... 4 I belive you have it pretty much settled already. If I was to change anything, I would shrink instead of adding more items: Identify the relevant quantities of your system: Energy, Momentum, entropy, electric charge, mass ... Which may or may not be conserved. If you have boundary conditions, most probably you don't have energy and/or momentum ... 3 First off, please don't use units with$c\ne 1$in GR. It makes everything horribly messy. What we normally think of as a ruler or clock measurement is represented in GR by an upper index quantity like$\Delta x^\mu$. Therefore in a Cartesian coordinate system in the fluid's rest frame, we are guaranteed that$u^\mu=(1,0,0,0)$, not$(-1,0,0,0)$. This is ... 1 There are (at least) three popular explanations for lift on an airfoil: Faster air on the top has lower static pressure than slow moving air on the bottom. The resulting pressure difference multiplied by the area is equal to the lift. The airfoil deflects air downward and by Newton's 3rd law an equal and opposite force (lift) is applied to the wing. Bound ... 2 To think of Bernoulli's principle as opposed to something else is not right. Bernoulli's principle can be thought of as the reason for the something else. Take a look at John Denker's explanation. It's the best I've seen. 1 It is absolutely true that the Bernoulli effect is not necessary in order for a wing to produce lift. Ultimately a wing produces lift by directing air flowing over the wings downward. The can be achieved by ramming air downward through the wing's "angle of attack" with respect to the air flow. This is why an airplane can fly upside down: the Bernoulli effect ... 0 The$i$th component of the integral is$\oint_S \epsilon_{ijk} x_j \sigma_{kl} n_l\, dS$We see that$\epsilon_{ijk} x_j \sigma_{kl}$has its$l$index contract with$\hat{n}$. Thus the divergence theorem allows us to convert this integral to$\int_V \partial_l \epsilon_{ijk} x_j \sigma_{kl} \, dV = \int_V \epsilon_{ijk} (\partial_l x_j) \sigma_{kl}\, dV + ...

9

Your assumption that there is a significant pressure differential due to fluid dynamics is correct. The assumption that it is a lifting force is not. An airplane generates lift because it has been engineered with lift in mind. An F1 car actually generates a powerful down force to push it against the track, allowing it to get better traction than it ...

3

Are you referring to the exact relativistic equivalent to Navier-Stokes equation or a more general Dissipative Relativistic Hydrodynamics Equation? The "relativistic equivalent to Navier-Stokes equation" would be something like this: There would be an energy momentum tensor with the following form: $T_{\mu\nu} = (e+p)u_\mu u_\nu - p g_{\mu\mu} + ... 0 First of all I am going to assume from your question that the viscosity is large enough, so the flow would be laminar (if it is not the case then the best answer would be numerical simulation or estimate based on the law of the wall velocity profile). Furthermore we will also neglect surface tension effects. Since we have a stationary problem the boundary ... 3 I remember attending a seminar by Unruh a few months ago and the same question arised. As far as I remember, he enfasized that in these hydrodynamic analogs of black holes, the flow is not quantized, it is a classical fluid, and everything is classical and that the dumb hole behaves like a quantum amplifier emitting quantum noise from the Horizon. ... 0 The surface area affects evaporation because if more area is exposed to air, allowing water molecules acquire more heat energy from the surroundings. Due to the increased heat energy (kinetic energy), there is more rapid movement of the water molecules which helps them to overcome the force of attraction and evaporate. The progression is essentially more ... 4 Overblowing is a phenomenon that exists in all wind instruments. The details of the physics are different from one instrument to the next, but there is a broad similarity, which is that it's the result of a nonlinear interaction between the air column and whatever is driving the air column. The recorder is in fact one of the simpler examples to understand. ... 4 This answer is a bit of a long story, but I have split it up for the different statements for your convenience. Having thought about it a bit more after the discussion with @Mephisto I actually believe that Bernoulli's equation is not applicable in points B and C, because it is based on conservation of energy and therefore only applies if wall friction is ... 2 The operating regime of a flue pipe such as a recorder is governed by the Ising equation (reference at end): $$I^2 = WST * 2 * P / \rho / F^2 / H^3$$ or equivalently $$I^2 = WST * v^2 / F^2 / H^3$$ where$WST$is the thickness of the flue (windsheet)$ P$is the blowing pressure$ F$is the frequency$H$is the cutup (mouth height)$v\$ is the ...

2

Warning: Another user has given a better answer. This one was chosen as the best one, before the other answer was written. First, a simplified approach based on Bernouilli's equation for incompressible fluids: Points B and C are directly in contact with the surface of the tube, thus they are nearly at zero speed with respect to it. But the fluid in A and D ...

2

Vorticity can certainly be destroyed, this is the basis of the energy cascade in 3D turbulence where energy is channeled across wavenumber space from large to small scales all the way down to the Kolmogorov scale at which point it dissipates into heat. Of course in order to do that you need to be looking at the complete equations... this link should answer ...

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