New answers tagged fluid-dynamics
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analytical solution for navier stokes where non-linear term is important
In cylindrical Couette flow the non-linear term is non-zero: in fact it is the centripetal acceleration $$\mathbf{ u}\cdot\nabla\mathbf{u}=
-\frac{u^2}{r}\mathbf{\hat r}.$$
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Incompressible fluid flow in tilted pipe setup - Continuity Equation, Flow Rate, Mass Conservation
The water does not necessarily accelerate going through a downwards stretch of pipe. It is prevented from doing so by the other water ahead of and behind it, and by the walls of the pipe. Just as you ...
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Derivation of flow rate and pressure drop
That was my question. Friction losses are usually characterized in terms of dynamic pressure:
$$ \Delta P = k \frac{1}{2} \rho u^2
$$
($ \Delta P$ pressure drop, $ \rho$ density, $u$ average flow ...
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Incompressible fluid flow in tilted pipe setup - Continuity Equation, Flow Rate, Mass Conservation
Very good question!
I myself would love to have a good explanation. For now, though, I can give you an explanation which is not good but may satisfy you.
To describe fluid motion, we use the field ...
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Why flapping rudder produce net thrust if one half-stroke produce thrust and second half-stroke drag?
The key point may be that stern of the boat moves laterally. During the first half of the stroke, the force exerted on the rudder by the water is forward and to the side. The sideways component causes ...
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Additional Pressure/Force Impact on a solid body as it moves through a fluid
The amount of force exerted by a fluid on an immersed moving object is just the opposite of the amount of force applied by the object on its surrounding. As far as I know, there is no global formula ...
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Why is the wake of boats leaving a long lasting print on the sea?
A ship's wake is just very fine air bubbles churned into the water by the propeller.
It is not oil because boat engines do not leak that much oil. You'll also find other type of surface watercraft, ...
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Why flapping rudder produce net thrust if one half-stroke produce thrust and second half-stroke drag?
I guess that during the forward thrust portion of the stroke the skipper pushes harder and faster creating more turbulence and drag and thus more thrust. During the reverse thrust part they slow down ...
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Can Mach number be greater than 1 if mass flow rate is decreasing?
The vertical axis is labeled Normalised Mass Flow Rate ( and the brackets [-] indicates no units). It is the value calculated on the right side of the equation divided by the value of the right side ...
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Why flapping rudder produce net thrust if one half-stroke produce thrust and second half-stroke drag?
Below the horizontal line is my original answer, submitted 5 hours ago, but there is a better explanation that I overlooked.
In a comment to another answer Gordon McDonald points out that since the ...
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Why flapping rudder produce net thrust if one half-stroke produce thrust and second half-stroke drag?
I suspect it has nothing to do with regions of higher or lower pressure (if those even exist). When you pump the rudder you are pushing water backward and by Newton's Third Law that water exerts an ...
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$\Delta P=QR $ in fluids
Its just a matter of convention which direction voltage/pressure difference is calculated. When using Ohms law in a circuit, youre supposed to follow the so called "passive sign convention". ...
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Why does a stream of falling water get narrower at the bottom?
There will be an additional contraction term due to the fact that the water entering the stream sideways carries with it a velocity component in the x and y directions, but I do not know how to ...
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Comparison of 2 fans
Poor design of the blade shape will cause more useless churning of the air and greater noise generation. Better design will cause less churn, less noise, and better power transfer from the motor to ...
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Why does a stream of falling water get narrower at the bottom?
Is my line of reasoning correct? Is there any incorrect assumptions that I made? Is this resoult coherent or even correct?
Seems reasonable to me, at least to zeroth order.
You are basically using:
$$...
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What is the $P_o$ or $P_i$ in Bernouilli's equation?
Here is the hyper-physics link for Bernoulli's principle
The Bernoulli Equation can be considered to be a statement of the conservation of energy principle appropriate for flowing fluids.
You ask ...
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What is the $P_o$ or $P_i$ in Bernouilli's equation?
In a static situation, each volume element has its weight balanced by the forces of the surroundings. The sideways pressure (say directions x and y) is equal at both sides of the element by symmetry. ...
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What is the $P_o$ or $P_i$ in Bernouilli's equation?
$P_i$ can be thought of as an initial pressure measurement at some point in a system that contains a flowing fluid. $P_0$ can be thought of as another pressure measurement somewhat downstream of the ...
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Propagation velocity of circular gravity-capillary surface waves in shallow water
The first answer is the dispersion relation for plane or one dimensional surface waves in shallow water; $h$ is the depth. If the surface tension term is negligible (low-$k$ regime) it also applies to ...
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What is the speed of the information transmission between water molecules?
Water molecules in liquid form "communicate" between their nearest neighbors via electrostatic forces, which propagate at the speed of light. But because each water molecule is electrically ...
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Conflicting result for velocity and radius in physiology
When you conclude from $V=Q / A$ that $V$ is proportional to $r^{-2}$ you are implicitly assuming that $Q$ is independent of $A$ and $r$, which is incorrect.
Poiseuille’s law tells us that $Q$ is ...
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Would water flow from the higher container to the lower one?
Shouldn't the pressure be the same in both buckets, meaning that the
water does not flow?
This is from a comment, and is the crux of the reasoning for the question. The answer below explains why ...
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Would water flow from the higher container to the lower one?
I like the willingness to experiment!
The result of the experiment is indeed expected. Basically, because there is a connection this is all one body of water. If the surface of a body of water is ...
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How far should I spread my fingers when swimming?
fitness.stackexchange link
Scholarly literature quotes:
The optimum finger spacing in human swimming (2009) by Minettia / Machtsiras / Masters, 10.1016/j.jbiomech.2009.06.012
optimal finger spacing (...
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How far should I spread my fingers when swimming?
For swimming, your goal is to maximize the surface area of your hand to allow for more pulling. So you don't want to "shrink" the area by squeezing your hand into a smaller cup shape. But at ...
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Navier-Stokes Equations in Einstein Notation and its relation to Poisson's Equation
You are taking the inner product of $\nabla$ and $\mathbf v$, so you need to make sure they both have the same index:
$$\partial_i\left[\partial_tv_i+v_j\partial_jv_i\right]=\partial_i\left[-\...
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Galilei transformation of mass flux
You need five-vectors to take the Galilei limit of special relativity. Check out Davison Soper Classical Field Theory. He uses the Galilei limit of the NS equation as an example if I remember ...
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Navier-Stokes Equations in Einstein Notation and its relation to Poisson's Equation
You can only repeat an index twice: that means sum over it. Your equation with three js is meaningless. You need to take a curl in order to eliminate the grad p term, not a div. That will involve the ...
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Mass flux in a multicomponent mixture with a concentration-dependent density
First, recognize that because the densities are always uniform, the diffusive mass fluxes are always zero. Therefore, the continuity equation for species $j$ in this special case is:
$$
\frac{\partial ...
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Change in appearance of liquid drop due to gravity
It is not an answer but an unfinished draft of how to get the shape :
Let $y(x)$ be the profile function of this axisymmetric shape. Supposing a laminar regime friction implies the total counterforce ...
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If water is nearly as incompressible as ground, why don't divers get injured when they plunge into it?
Try this in a pool or bucket of water or whatever:
Slap the water with your open palm. OUCH ! BELLY FLOP
Stab the water with your outstretched fingers, blade-like
they go right in like an Acapulco ...
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Reducing yawing in a kayak with a rudder
I want to know how much the kayak will yaw $~\psi(t)~$ and where the center of rotation is.
I) the equation of motion
$${\frac {d}{d\tau}}\beta \left( \tau \right) -{\frac {{}{\it Fy
}-m \left( {\...
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Bernoulli's equations on a falling (not freefall) bucket of water
The air pressure above and external to the spout do not change. But, if the bucket is accelerating downwards, the water in the bucket experiences a (fictitious) upward force that counters gravity and ...
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Does the Continuity Equation Imply it's Zero for Incompressible Flow?
Since the dot product is commutative, doesn't this imply that $(\mathbf u\cdot\nabla)\mathbf u=0$?
No because $\nabla$ is an operator that acts on the object to its right. The term in the parenthesis ...
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Velocity at a place for fluid with a given pressure gradient?
I think the answer should be
$$v \propto \frac{1}{\sqrt r}.$$
Indeed, if we assume a purely radial, steady, incompressible, inviscid flow, then the Navier-Stokes equation along the radial coordinate ...
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If water is nearly as incompressible as ground, why don't divers get injured when they plunge into it?
It can also be viewed in acceleration terms, a sufficiently accelerated water still will pass through a concrete wall, neatly
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Submerged Landau jet
The same total momentum flux (the "momentum of the jet") must pass through any closed surface surrounding the origin (in particular, through an infinitely distant surface). For this to be so,...
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Submerged Landau jet
Since the problem has an axial symmetry then he rid off the $\phi$ angle, i.e. no dependence on the azimuthal angle $\phi$ hints to the absence of this variable in the solution in the function $F$ or $...
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Navier-Stokes Equation Derivation in "A Mathematical Introduction to Fluid Mechanics" (Stress Tensor)
Assumption 1) in the book means that each component $\sigma_{ij}$ is a linear and homogeneous function of each $\frac{\partial u_k}{\partial x_l}$.
To express that mathematically you need $3^4$ ...
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Navier-Stokes Equation Derivation in "A Mathematical Introduction to Fluid Mechanics" (Stress Tensor)
No, not a matrix, but a rank 4 tensor. In index notation it means:
$\sigma_{ij}=A_{ijkl}(\nabla u)_{kl}$
You can think of it this way: A linear transformation taking one vector (1-index object) to ...
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If water is nearly as incompressible as ground, why don't divers get injured when they plunge into it?
Basically, you do experience an impact when you hit the free surface of a liquid (which could be very substantial), but the water moves out of the way so it's not like hitting concrete.
When a solid ...
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If water is nearly as incompressible as ground, why don't divers get injured when they plunge into it?
As others have pointed out, water isnt very different from solids in terms of its resistance to changes in volume; but the defining difference between a liquid and a solid, lies in its resistance to '...
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If water is nearly as incompressible as ground, why don't divers get injured when they plunge into it?
I don't quite understand the question, but isn't it:
Because water is a fluid and ground is not.
Falling with 100m/s on water might be much the same as falling on ground, but when diving, then your ...
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If water is nearly as incompressible as ground, why don't divers get injured when they plunge into it?
Well, depends on the way someone dives in and the altitude.
You might have wondered at some point, why does it hurt when you fall into the water with your belly, but when you fall in with your feet ...
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If water is nearly as incompressible as ground, why don't divers get injured when they plunge into it?
If water is treated as incompressible it has constant volume, but being a fluid it can change shape. The less the viscosity the easier it is to change its shape.
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If water is nearly as incompressible as ground, why don't divers get injured when they plunge into it?
For very skilled divers, they will also use a technique where they enter the water hands first, with a particular shape that pulls air in with them. They then spread the hands out, creating a 'hole' ...
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If water is nearly as incompressible as ground, why don't divers get injured when they plunge into it?
Concrete is rigid to a good approximation: it will keep its shape and its volume.
Water is just incompressible: it will keep its volume but not its shape. Water will move out of your way on impact.
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If water is nearly as incompressible as ground, why don't divers get injured when they plunge into it?
Adding another perspective to the existing answers:
In your usual diving scenario, water is not confined to the points in space it occupied before, while a slab of ground is – on account of water ...
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If water is nearly as incompressible as ground, why don't divers get injured when they plunge into it?
In simple terms, water (or any fluid) will move out of the way; concrete won’t (unless it is hit very hard). The important properties are viscosity and elasticity rather than compressibility.
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If water is nearly as incompressible as ground, why don't divers get injured when they plunge into it?
Incompressible doesn't mean that it has to keep the same shape.
But, due to viscosity, water can be "slow" to change its shape under external influence. So when a diver arrives too fast, ...
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