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fluid dynamics
|
Why is pressure gradient assumed to be constant with respect to radius in the derivation of Poiseuille's Law?
|
https://physics.stackexchange.com/questions/67398/why-is-pressure-gradient-assumed-to-be-constant-with-respect-to-radius-in-the-de
|
<p>Poiseuille's Law relies on the fact that velocity is not constant throughout a cross-section of the pipe (it is zero at the boundary due to the no-slip condition and maximum in the center). By Bernoulli's Law, this means that pressure is maximum at the boundary and minimum at the center. But in the book I have it is assumed that the pressure gradient is independent of radius (distance from the center of the pipe), and the pressure gradient is thus extricated from a radius-integral. Can anyone justify this?</p>
|
<p>First of all, Bernoulli's law is applicable only to inviscid flow, while Poiseuille's flow is for the viscous fluid. The fact that pressure is constant along the orthogonal cross section of the pipe could be derived from the assumption that the flow is parallel, that is everywhere inside the pipe the velocity field has only z-component (assuming the cylindrical coordinate system, with pipe oriented along the z-axis). Then the r-component of Navier-Stokes equation is then reduced to $0 =- \frac1{\rho}\frac{\partial p}{\partial r}$ (all terms containing velocity components are equal to zero here), which gives the pressure independent of radial coordinate.</p>
<p>The assumption that the flow in the pipe would be parallel in derivations of Poiseuille's flow is just an ansatz, compatible with the symmetries of the problem, that is later justified by producing correct solution of Navier-Stokes equation. However one should remember that such flow would be stable only for relatively large viscosity of the fluid (that is for Reynolds number not exceeding the certain critical value). </p>
| 800
|
fluid dynamics
|
Boundary layer equation
|
https://physics.stackexchange.com/questions/74853/boundary-layer-equation
|
<p>I was trying to understand the derivation of the boundary layer equations at p.145 of <a href="http://www.unimasr.net/ums/upload/files/2012/Sep/UniMasr.com_919e27ecea47b46d74dd7e268097b653.pdf" rel="nofollow">http://www.unimasr.net/ums/upload/files/2012/Sep/UniMasr.com_919e27ecea47b46d74dd7e268097b653.pdf</a>. : the derivation is completely given at that page. </p>
<p>I almost figured it all out, but I don't understand (12-8). With the approximations they made, I should get (12-8), but not the last term $\mu\frac{\partial^2v_y}{\partial x^2}$. I don't see how they get this term, when approximating $\tau_{xy} = \mu \frac{\partial^2 v_y}{\partial x^2}$ and $\sigma_{yy} = -P$ does not give the result shown. I wonder if I'm wrong, or the book is just mistaken? I hope someone can help me out here.</p>
|
<p>Let's remove this from the list of unanswered questions.</p>
<p>The derivation in the book is a bit odd. I favor the derivation in Schlichting's book "Boundary-Layer Theory", because it's cleaner.</p>
<p>Usually the derivatives of $\sigma_{xx}$, $\sigma_{yy}$, $\tau_{xy}$ and $\tau_{yx}$ in equations (12-5) and (12-6) are treated in combination. For (12-6) the right hand side becomes
$$
\frac{\partial \tau_{xy}}{\partial x} + \frac{\partial \sigma_{yy}}{\partial y} = \mu \frac{\partial^2 v_x}{\partial x \, \partial y} + \mu \frac{\partial^2 v_y}{\partial x^2} - \frac{\partial P}{\partial y} + 2 \mu \frac{\partial^2 v_y}{\partial y^2}\\ = - \frac{\partial P}{\partial y} + \mu \frac{\partial^2 v_y}{\partial x^2} + \mu \frac{\partial^2 v_y}{\partial y^2} + \mu \left( \frac{\partial^2 v_x}{\partial x \, \partial y} + \frac{\partial^2 v_y}{\partial y^2} \right)
$$
Then
$$
\frac{\partial^2 v_x}{\partial x \, \partial y} = \frac{\partial^2 v_x}{\partial y \, \partial x} \quad\Rightarrow\quad
\frac{\partial^2 v_x}{\partial x \, \partial y} + \frac{\partial^2 v_y}{\partial y^2} = \frac{\partial}{\partial y} \left( \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} \right)
$$
which is zero due to the continuity equation for incompressible fluids (12-10). Therefore,
$$\frac{\partial \tau_{xy}}{\partial x} + \frac{\partial \sigma_{yy}}{\partial y} = - \frac{\partial P}{\partial y} + \mu \left( \frac{\partial^2 v_y}{\partial x^2} + \frac{\partial^2 v_y}{\partial y^2} \right) \quad\mbox{,}
$$
which is the usual right hand side for the $y$ component of the Navier-Stokes equations.</p>
<p>The argument from Prantl is that the boundary layer thickness is small compared to the characteristic length of the object. For derivatives this implies that derivatives with respect to $x$ are smaller than derivatives with respect to $y$, and that second derivatives with respect to $x$ are even smaller than second derivatives with respect to $y$.</p>
<p>With this argument, the last term in (12-8) should read
$$
\mu \frac{\partial^2 v_y}{\partial y^2}
$$
and not
$$
\mu \frac{\partial^2 v_y}{\partial x^2} \quad\mbox{.}
$$
It's a typo as user23660 pointed out.</p>
| 801
|
fluid dynamics
|
Blasius boundary layer solutions
|
https://physics.stackexchange.com/questions/74906/blasius-boundary-layer-solutions
|
<p>I'm trying to understand the Blasius boundary layer solution, but I'm having some difficulties. Using wikipedia, I wonder how they get the first formula: <a href="http://en.wikipedia.org/wiki/Blasius_boundary_layer" rel="nofollow">http://en.wikipedia.org/wiki/Blasius_boundary_layer</a>. $$\frac{U^2}{L}\approx\nu \frac{U^2}{\delta^2}$$
And how they get to the fourth formula: $$\delta(x) \approx \sqrt{\frac{\nu x}{U}}$$
I feel that those formulas are correlated somehow, but I don't really see how they derive those. I hope someone can help.</p>
|
<p>The first formula (scaling argument)</p>
<p>$$ \frac{U^2}{L} \sim \nu \frac{U}{\delta^2}, \tag{*} $$</p>
<p>comes directly from equations for boundary layer equations (and not specifically for Blasius boundary layer). We have continuity equation:
$$\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}=0 $$
and x-component of momentum equation:
$$u\dfrac{\partial u}{\partial x}+v\dfrac{\partial u}{\partial y}=-\dfrac{1}{\rho}\dfrac{d p}{dx}+{\nu}\dfrac{\partial^2 u}{\partial y^2}.$$
The general assumptions behind this equations is that along the x-axis quantities vary slower than along y-axis. Let us denote $L$ the characteristic length scale along the x- axis, $\delta$ the scale along the y-axis and $U$ is the characteristic velocity of the fluid. ($L$ could be, for instance, the length of the body and $\delta$ the boundary layer thickness.) We thus have $L \gg \delta$ and $u \sim U$. Various derivatives could be estimated by
$$\dfrac{\partial }{\partial x} \sim \frac{1}{L},\qquad \dfrac{\partial }{\partial y} \sim \frac{1}{\delta}.$$</p>
<p>Now applying this approximations to continuity equation we get for instance the scale of y-component of the velocity: $$v \sim \frac \delta L U.$$</p>
<p>Now for the x-momentum equation both terms of the left side and the term with pressure on the right have all the following estimate:
$$ u\dfrac{\partial u}{\partial x}\sim v\dfrac{\partial u}{\partial y} \sim \dfrac{1}{\rho}\dfrac{d p}{dx} \sim \frac{U^2}L ,$$
(the term with pressure could be rewritten using Bernoulli's equation as $u_\infty \dfrac{ d u_\infty}{dx}$ which would lead to the same estimate).</p>
<p>The single remaining term of x-momentum equation will have the following approximation:
$$ {\nu}\dfrac{\partial^2 u}{\partial y^2} \sim \nu \frac {U}{\delta^2}.$$</p>
<p>Combining this estimates we get $(*)$. Note, that I wrote it using the $\sim$ and not $\approx$ as in wiki page. </p>
<p>Now we further assume that the problem at hand is the Blasius boundary layer, that is semi-infinite plate in uniform flow parallel to it. In this problem there is <strong>no</strong> intrinsic <strong>length scale</strong> (because the plate is infinite), so role of scale along the x-axis would be played by the current value of the x-coordinate $L = L (x) = x$, and the traversal scale $\delta$ also has to be dependent on $x$. Then we simply find from $(*)$
$$\delta (x) \sim \sqrt{\frac{x \nu}{U}},$$ which is yours second equation.</p>
| 802
|
fluid dynamics
|
Non-uniform field pressure on fluid
|
https://physics.stackexchange.com/questions/76127/non-uniform-field-pressure-on-fluid
|
<p>I have been studying fluid mechanics and currently I was trying to understand Euler's equation for fluid flow. For that purpose, I was following <a href="http://www.av8n.com/physics/euler-flow.htm" rel="nofollow">this</a> webpage.</p>
<p>My problem is in understanding Eq.(12). The author states that that is the contribution due to an external $\underline{\text{uniform}}$ (gravitational) field, $g$. What I don't understand is why does that definition only applies to an uniform field. What would the contribution due to a, say, central field be?</p>
<p>Can anyone help me? </p>
|
<p>In general, any external body force would give a contribution $\int \mathbf{f} \ dV$, where $\mathbf{f}$ is the force per unit volume resulting from the external force. For those who didn't read the link, the integral is over a small volume at the point in consideration. Here external means that the source of the force is anything but the fluid itself.</p>
<p>The specific case, which is common in fluids, is that the external force is gravity, and $\mathbf{f} = \rho \mathbf{g}$. </p>
<p>If you didn't want to make a uniform gravity approximation and instead took the Newton's gravitational law for the gravitational force you would get $\mathbf{f} = \frac{G \rho m}{ r^2}$, where $m$ is the mass gravitational field source, and $r$ is the distance to that body (assuming for simplicity a spherically symmetric object).</p>
| 803
|
fluid dynamics
|
What does the Froude number represent?
|
https://physics.stackexchange.com/questions/75585/what-does-the-froude-number-represent
|
<p>While reading on <a href="http://en.wikipedia.org/wiki/Froude_number" rel="noreferrer">Wikipedia</a>, I read the following</p>
<blockquote>
<p><em>The Froude number is defined as:</em></p>
<p>$$\mathrm{Fr} = \frac{v}{c}$$</p>
<p><em>where $v$ is a characteristic velocity, and $c$ is a characteristic water wave propagation >velocity. The Froude number is thus analogous to the Mach number. The greater the Froude >number, the greater the resistance.</em></p>
</blockquote>
<p>While reading the following <a href="http://isites.harvard.edu/fs/docs/icb.topic1191136.files/solitons-intro.pdf" rel="noreferrer">paper</a> on shallow water waves, I could not understand how/why the Froude number was given to be </p>
<p>$$\mathrm{F} = \frac{gt_0^2}{L}$$</p>
<p>Where $t_0$ is the time scale, and $L$ is the length scale along the X and Z axes. Note that the scale for the amplitude of the surface above the mean depth is some other $A$ and not $L$.</p>
<p>Please explain how the author has defined the Froude number. Is it the same as the one given on Wikipedia? If so, then please provide a step-by-step reasoning to show that they are indeed the same. If not, then please explain what the Froude number mentioned in the paper represents.</p>
<p>EDIT: I'd like to add another minor question as it is related to the abovementioned <a href="http://isites.harvard.edu/fs/docs/icb.topic1191136.files/solitons-intro.pdf" rel="noreferrer">paper</a>. It mentions that the surface tension is accounted for but I do not understand how that is possible given that there is no term containing the surface tension $\gamma$ explicitly.</p>
|
<p><strong>To answer the question in the title:</strong> <em>What does the Froude number represent?</em></p>
<p>The Froude (<span class="math-container">$Fr$</span>) number is a non-dimensional value that (typically) is used to quantify the degree of linearity/nonlinearity of a gravity wave through the ratio of a characteristic fluid velocity (<span class="math-container">$v$</span>) and the gravity wave speed (<span class="math-container">$c$</span>). As shown below, the measure is analogous to wave steepness, <span class="math-container">$s = ak$</span>, where <span class="math-container">$a$</span> is the wave amplitude and <span class="math-container">$k$</span> is the wavenumber. Consider a fluid particle embedded in the wave, its orbital (characteristic) velocity, <span class="math-container">$v$</span>, is given by <span class="math-container">$\omega a$</span>, where <span class="math-container">$\omega$</span> is the wave frequency. This makes sense because as the wave advances fluid particles (at first order) move around in circles with a spatial scale given by the wave amplitude <span class="math-container">$a$</span> and a temporal scale given by <span class="math-container">$\omega^{-1}$</span>. Moreover, we know that the wave speed is given by, <span class="math-container">$c = \omega/k$</span>, so:</p>
<p><span class="math-container">$Fr = (v)/(c) = (\omega a)/(\omega/k) = ak = s$</span>. Hence, the Froude number is analogous to wave steepness.</p>
<p>The importance of this result is that, when you nondimensionalize the Navier-Stokes equations, wave steepness (or <span class="math-container">$Fr$</span>) ends up multiplying the convective acceleration term <span class="math-container">$\mathbf{u} \cdot \nabla \mathbf{u}$</span>. As you know, this term is responsible for much of the beautiful fluid dynamics arising from nonlinear wave breaking such as turbulence. You can read a bit more in this <a href="https://physics.stackexchange.com/questions/60370/navier-stokes-system/60377#60377">other post</a>.</p>
<p>So in essence, the <span class="math-container">$Fr$</span> number tells you if a wave is big or not. In theory, if it is much less than 1 then the wave is linear (small), if it is larger than 1 then the wave will break because the fluid speed <span class="math-container">$v>c$</span> and the orbital fluid particle velocity exceeds the wave speed. That is, the wave topples over and breaks.</p>
<p>The Froude number is also used in other contexts but this is strictly for surface gravity waves.</p>
<p>Also, what the author uses in the text is fine, as long as you are consistent with the scaling throughout your analysis you can use any sensible nondimensional number. It may not the most conventional way of representing it but it is not wrong.</p>
| 804
|
fluid dynamics
|
how to reduce pressure drop in a 5/16 ss tube
|
https://physics.stackexchange.com/questions/76019/how-to-reduce-pressure-drop-in-a-5-16-ss-tube
|
<p>I have a fluid line which passes high density fluid at $60 \text{psi}$. At the end there is a valve. I have a $30 \text{psi}$ dynamic pressure drop in the line valve itself creates a back pressure of $10 \text{psi}$. How do I reduce my pressure drop without changing tube parameters?</p>
|
<p>I couldn’t get any answer for this question and so ended up in performing experiments in the lab to find answer by myself. To reduce my dynamic pressure drops in the SS tube.</p>
<p>The factors externally (Without changing tube parameters) which I can vary to reduce dynamic pressure drop in the line is maintaining temperature of the liquid. We know that viscosity reduces by increasing the temperature. I increased the temperature of the liquid in which the liquid starts to flow easier (Reducing Shear stress). The pressure drop is mainly created by major and minor losses and major is ofcourse friction loss.</p>
<p>Increase temperature of the liquid flowing.
Creating a back pressure in the system. We can add back pressure by adding a valve (which we can control the flow rate by reducing orifice diameter)
Using a booster within the line to generate pressure.
Reducing the flow velocity of the liquid as equation rightly says-ΔP=fD x L/D x ρV²/2.</p>
<p>If we can change the parameters of SS tubes, then we can
Increase the diameter of the tube (helps reduce flow velocity inside the line)
Reduce length of the tube-Less friction losses.</p>
<p>The application of this was to design a heat exchanger. Thought to share these leanings with all of you. i have lot of laboratory tested data regarding effect of temperature Vs viscosity, shear stress, kinematic viscosity, and many more</p>
| 805
|
fluid dynamics
|
Aerofoil Theory Project
|
https://physics.stackexchange.com/questions/76765/aerofoil-theory-project
|
<p>I'm doing a project (dissertation) on the mathematics of Aerofoil Theory. I wonder if I could get some advice on a possible structure. I'm new to fluid dynamics, so it's quite hard to know where to start. </p>
<p>So far I've looked at different types of flows; streamlines and velocity potentials. I've also covered a fair amount of complex variable theory - and can (loosely) see how conformal mappings are used to map for instance a cylinder to an aerofoil shape using Joukowski's transform. I would guess from here, I would want to look at Bernoulli's equation for lift and start looking at modelling an example. </p>
<p>It would be useful to know if this is the right kind of structure to follow or if I'm approaching this the wrong way! So any help would be appreciated. Thanks! </p>
|
<p>To use the Joukowski concept of an airfoil I have learnt from the professors Claes Johnson and his student professor Johan Hoffman is the "old" and wrong theory of flight because it is totally based on 2D calculation. It is a beautiful theory in complex variables but completely unphysical it is by no means based on physical laws (complete mumbo-jumbo, so to say) according to these two experts on Numerical analysis from the calculation center of (NADA/KTH). Their road to the truth is stony but correct. Remember marquis Pierre Simon de Laplace's word of wisdom: <em>La verite seul est belle</em> (only the truth is beautiful).</p>
<p>So instead of trying to develop the wrong 2D (or even worse) 3D ideas of Prandtl I advise You to study the "New theory of flight" of Johnson-Hoffman. This theory is based on FEM-analysis which has it's origin in the famous Russian mathematician and engineer Galerkin. This is the only way to solve this problem of fluid dynamics to start with Navier-Stokes equation with a slip-boundary-condition which is totally different to the no-slip condition of Dirichlet type. With this You can reach a more profound understanding of the fluid dynamics of airfoils. The Görtler vortices then goes from secondary to primary phenomenons and important in the understanding of stall and flow separation for instance.</p>
| 806
|
fluid dynamics
|
(Why) is a solid easier to suspend in a thick slurry than in a liquid?
|
https://physics.stackexchange.com/questions/80023/why-is-a-solid-easier-to-suspend-in-a-thick-slurry-than-in-a-liquid
|
<p>When making cacao (the non-instant type), one usually mixes the powder with a small amount of milk or water before mixing this slurry into the rest of the milk/water. Some manufacturers of slurry feeding systems for biogas plants claim the same principle for their product: One mixes the solid feedstock into a small amount of slurry from the fermenter (in the hopper of a pump <a href="http://www.netzsch-pumpen.de/en/products-solutions/nemo-progressing-cavity-pumps/nemo-bmax-mixing-pump.html" rel="nofollow">like this</a>), the mixture is then pumped into the reactor vessel and mixed into the slurry there with conventional agitators. The manufactureeres of these feeding systems claim, and we have observed this (but did not do a real experiment), that less energy is needed for agitators if a liquid feeding system is in use (compared to dumping the solid feedstocks into the vessel via an auger or similiar).</p>
<p>Now, the only reason this should be the case is that with the smaller volume it is easier to exert shear forces on the conglomerates of solids and to thus disperse them. But this assumes that the solids cling to each other - evident for cocoa powder, less so for the substrates our plants handle.</p>
<p>On the other hand, I would assume that thicker liquids are always harder to mix, as there is less shear (if all else is equal) and particles or conglomerates are somewhat protected by the liquid.</p>
<p>So I assume that one major factor is that the real factor here is the size of the mixing implement relative to the volume (think spoon in cup vs. spoon in pot), so the forces acting on the particle conglomerates don't depend on flow.</p>
<p>So my question is ... </p>
<ul>
<li>is this even the general case (it's easier to mix solids into a thicker slurry) </li>
<li>if so why, is the reasoning given above correct?</li>
<li>Can one translate my reasoning above from prose into proper physics?</li>
</ul>
|
<p>If you're trying to mix anything with starch grains (cornflour is the classic example) a problem you'll run into is that concentrated suspensions are <a href="http://en.wikipedia.org/wiki/Dilatant" rel="nofollow">dilatant</a>. If you add the powder slowly this may not be a problem, but if you throw it in as a lump you get a ball of dry powder with high viscosity dilatant suspension insulating it from the rest of the mix. This ball can be very difficult to break up even with high shear because it responds to the shear by becoming more viscous.</p>
<p>Mixing the powder into a concentrated slurry still requires care, but at least the high viscosity of the concentrated slurry makes it easier to break up any chunks of dilatant goo. Once you've got a slurry of less that say 30% solids by volume, this can be mixed into a low viscosity mixture without any special precautions.</p>
| 807
|
fluid dynamics
|
Energy of liquid and gas at same pressure is different?
|
https://physics.stackexchange.com/questions/81156/energy-of-liquid-and-gas-at-same-pressure-is-different
|
<p>Air compresses (change in volume) and it creates pressure. It's the internal energy (momentum of each molecules) creates that pressure. But in liquids as it can't be compressed (may be by .3% if I am sure about water) it can't be pressurised, right? This is only possible by an external force on water itself. Does it means that work done to generate same pressure on liquid is lower than gas?</p>
<p>In this case pressurized water holds less energy that pressured gas of same pressure.</p>
<p>This is my thought, could someone tell me my thought is right or wrong?</p>
|
<p><em>All</em> molecular forms of matter can be compressed and consequently pressurized. Generally speaking you can pressurize fluids by adding energy to them. This includes heat transfer as well as external forces performing work. The energy needed to achieve a given pressure change in a liquid will in general be less than it would be for a gas at the same initial temperature and pressure. The energy added to a fluid by work done is simply the force applied multiplied by the distance acted over. It is then easy to understand why it requires less energy to modify a liquid's pressure since the distance required for the force to act over will inevitably be smaller in comparison to that for a gas in order to achieve the same pressure change.</p>
| 808
|
fluid dynamics
|
Pressure difference in water tanks at different heights
|
https://physics.stackexchange.com/questions/82132/pressure-difference-in-water-tanks-at-different-heights
|
<p>Suppose there are two tanks(airtight) fully filled with water.one is kept on top of a ten storey building and one on top of a twenty storeyed one.They are NOT interconnected.Will there be a pressure difference?</p>
|
<p>Forget the airtight part. That just confuses the issue by making you deal with the air pressure (or lack of it) at the top and bottom.</p>
<p>The pressure depends on how far below the surface of the water you measure it.</p>
<p>The pressure on a unit of area is nothing but the weight of the water in a slender column from that unit of area all the way to the surface.</p>
<p>So if you measure at the surface, it is nothing.
If you measure at the bottom of each tank, it just depends on the height of the surface of the water above the bottom.</p>
| 809
|
fluid dynamics
|
What is the motivation behind $\tau=\mu \frac{\mathrm{d}u}{\mathrm{d}y}$ for Newtonian fluid?
|
https://physics.stackexchange.com/questions/88897/what-is-the-motivation-behind-tau-mu-frac-mathrmdu-mathrmdy-for-new
|
<p>Where does the motivation come from? $\tau$ is the shear stress, $u$ is the velocity and $\mu$ is the shear viscosity. </p>
<p>EDIT: Since I wrote the question on phone I wasn't clear enough about what I was actually wondering about (see comments below); thus let me clarify. </p>
<p>I was simply wondering where did the derivative term come from and not why the expression is linear in the derivative term or why it is the first derivative and not e.g. the second and so on. </p>
<p>The diagram in the answer I accepted shows exactly how the gradient comes into the picture. </p>
|
<p>The motivation comes from applying the <a href="http://en.wikipedia.org/wiki/No-slip_condition" rel="noreferrer">no-slip boundary condition</a> on a fluid flow. This is probably easier to understand pictorially,</p>
<p><img src="https://i.sstatic.net/F20QX.png" alt="enter image description here"></p>
<p>The fluid at the top travels at $u$ while the fluid at the bottom does not move, hence the gradient $\partial u/\partial y>0$. In order to properly model fluid flows, this needs to be accounted for in the Navier-Stokes equations; it is called $\tau$ for "simplifying" the equations.</p>
| 810
|
fluid dynamics
|
Streamlines in Boundary Layer
|
https://physics.stackexchange.com/questions/90456/streamlines-in-boundary-layer
|
<p>Suppose that a fluid is flowing parallel to and over a flat plate. Obviously, a boundary layer develops in which the velocity ranges from 0 to 99% of the upstream velocity U. Could somebody please show me how the streamlines would look in this boundary layer? What do they typically look like in boundary layers?</p>
| 811
|
|
fluid dynamics
|
Festive physics: gold flake vodka
|
https://physics.stackexchange.com/questions/91243/festive-physics-gold-flake-vodka
|
<p>I have a bottle of vodka that has a load of gold flakes suspended in it. It has been sat still for over 24 hours and the flakes are all still suspended within the liquid: they have not risen to the surface or sunk to the bottom. Any ideas as to the physics behind this? </p>
|
<p>The viscosity of water ethanol mixtures isn't especially high, though the wetting properties of vodka may make it seem oily. Actually water ethanol mixtures are highly non-ideal: both water and ethanol have a viscosity of about 1 mPa.s at room temperature, but a mixture can achieve a viscosity of over 3 mPa.s. See <a href="http://link.springer.com/article/10.1007/s11814-011-0239-6#page-1">this paper</a> or Google for many such tables.</p>
<p>The real reason gold leaf will stay suspended for so long is that it is extraordinarily light. The thickness of gold leaf is <a href="http://www.britannica.com/EBchecked/topic/237354/gold-leaf">around 100nm</a>, and since the density of gold is 19300 kg/m$^3$ a flake of gold leaf 1mm by 1mm weighs just 2 $\mu g$ so the downward force due to gravity is 20 nano-Newtons. At such low forces water is viscous enough to keep the gold suspended for long periods.</p>
| 812
|
fluid dynamics
|
Approximating density from pressure and flow
|
https://physics.stackexchange.com/questions/92554/approximating-density-from-pressure-and-flow
|
<p>I want to get the density of a fluid going through a pipe. I can measure the flow and pressure with a flowmeter and the temperature using a thermometer. With this information, I want to calculate (or approximate) the instantaneous density of the fluid passing through my instruments.</p>
<p>How would I go about doing this?</p>
<p>I'm looking at a liquid (not a gas), mostly incompressible. Think water with varying amounts of solute (salt, sugar) in it.</p>
| 813
|
|
fluid dynamics
|
Efflux speed of ideal fluid dependent on diameter?
|
https://physics.stackexchange.com/questions/94217/efflux-speed-of-ideal-fluid-dependent-on-diameter
|
<p>I have a cylinder full of water with diameter $D$ with a round opening on the bottom with diameter $d$. The water is friction-free and incompressible. Now I need a relationship for the efflux speed $v$ with which water exits the cylinder and I shouldn't use the approximation $d \ll D$, but formulate a general relationship.</p>
<p>Ok. So what I thought is to equate the Bernoulli law on the top of the cylinder with that on the bottom of the cylinder which gives me $v=\sqrt{2gh}$. Solved. How does the speed of efflux depend on the diameter of the efflux hole? I googled quite a lot and all I could find was the above relationship...</p>
|
<p>While calculating velocity of efflux you use bernoulli's theorem as follows : </p>
<p>At the top of container :$ P + \frac{\rho {v_1}^2}{2} + \rho gh_1 = k$<br>
At the efflux : $P_a + \frac{\rho {v_2}^2}{2} + \rho g h_2 = k$ </p>
<p>$P - P_a + \rho g (h_1 - h_2) = \frac{\rho ({v_2}^2 -{v_1}^2)}{2} $</p>
<p>Now, in accordance with equation of continuity,<br>
$ v_1 A_1 = v_2 A_2$<br>
$ v_1 (\pi D^2) = v_2 (\pi d^2) $<br>
$ v_1 = v_2 \frac{d^2}{D^2}$<br>
If $ d<<D$ , then $v_1 = 0$.</p>
<p>This gives us :</p>
<blockquote>
<p>$v_2 = \sqrt{2gh + \frac{2(P-P_a)}{\rho}}$</p>
</blockquote>
<p>Also if top of container is open, or pressure inside container is $P = P_a$ , then $v_2 = \sqrt{2gh}$</p>
<p>If the condition $d<<D$ was not given, we would have to put value of $v_1$ along with $v_2$ in bernoulli's equation.</p>
| 814
|
fluid dynamics
|
Can ship/boat propellers be placed, with adequate protection, alongside the fuselage instead of at the back?
|
https://physics.stackexchange.com/questions/32532/can-ship-boat-propellers-be-placed-with-adequate-protection-alongside-the-fuse
|
<p>On small private aircraft the engine is placed at the front of the fuselage, or on the wings
E.g. Stationair, Otter </p>
<p>On combat aircraft the engine/s is placed at the end of the fuselage
E.g. Raptor, Typhoon </p>
<p>On larger aircraft constructed to carry cargo/passengers the engines are placed on the wings
E.g. DC3, A380</p>
<p>The positioning probably has to do with the amount of load to be lifted into the air, and manouverability desired... which seems sensible.</p>
<p>Yet most water-going craft (boats, power-boats, merchantmen, naval craft) have their propellers/push-pull placed at the back of the fuselage. </p>
<p>What is the advantage in placing the driving component for boats in this position in addition to protecting them against damage? Can the propellers on a boat, with adequate protection say- a tubular enclosure, placed alongside the fuselage not drive the watercraft with equal efficiency and power? </p>
|
<p>The main reason is mechanical: the size of the the engine, shaft and propeller require them to be aligned. However many modern ships do have auxiliary engines located in 360° oriantable pod and located at different places under the boat hull. They are used to rotate the boat.</p>
<p><a href="https://www.google.com/search?num=10&hl=en&site=imghp&tbm=isch&source=hp&biw=1366&bih=643&q=pod+propeller&oq=pod+propeller&gs_l=img.3..0i24j0i5i24.1161.6141.0.6328.17.15.2.0.0.0.182.1472.6j9.15.0...0.0...1ac.DntMyneXvoU" rel="nofollow">https://www.google.com/search?num=10&hl=en&site=imghp&tbm=isch&source=hp&biw=1366&bih=643&q=pod+propeller&oq=pod+propeller&gs_l=img.3..0i24j0i5i24.1161.6141.0.6328.17.15.2.0.0.0.182.1472.6j9.15.0...0.0...1ac.DntMyneXvoU</a></p>
| 815
|
fluid dynamics
|
If there is significant temperature difference between indoor air and outdoor air, will that significantly increase the rate of air exchange?
|
https://physics.stackexchange.com/questions/43236/if-there-is-significant-temperature-difference-between-indoor-air-and-outdoor-ai
|
<p>There would be heat diffusion, of course, but heat diffusion occurs even without the exchange of fluid parcels between each environment.</p>
<p>We do know that cold air tends to be denser than warm air, and that pressure differences drive the exchange of fluid parcels. But let's assume that there is no pressure difference for the sake of this question.</p>
|
<p>I would like you to precise some point : you want to know the rate of air exchange without any air displacement due to pressure difference?
If you want to know what happens during a cold winter and the diffusion of heat through walls (which block any air flow), the heat flux will be proportional to the temperature gradient (temperature difference divided by wall thickness).
An interesting phenomena occurs when you suddenly open your door to the cold outdoor air : as you said, cold air is denser than warm air, so two layers of warm and cold air separated by a vertical boundary is not a stable configuration. The two layers will mix dynamically through a <a href="http://en.wikipedia.org/wiki/Rayleigh%E2%80%93Taylor_instability" rel="nofollow">Rayleigh-Taylor instability</a>.</p>
| 816
|
fluid dynamics
|
Why does central jet velocity of a flute air stream not include aperture size?
|
https://physics.stackexchange.com/questions/43355/why-does-central-jet-velocity-of-a-flute-air-stream-not-include-aperture-size
|
<p>Some fellow flutists and I are pretending to be experts at fluid dynamics and reading <a href="https://ccrma.stanford.edu/~pdelac/research/MyPublishedPapers/Thesis.pdf" rel="nofollow">Patricio Cuadra's thesis</a> on the topic:</p>
<p>On page 17, it says:</p>
<blockquote>
<p>The central speed of the jet $U_j$ at the channel exit can be estimated assuming atmospheric pressure outside the channel and using Bernoulli’s equation:</p>
<p>$$U_j = \sqrt{ \displaystyle\frac{2P_f}{\rho_1} }$$</p>
<p>Where $P_f$ = pressure in the cavity before the channel, and $\rho_1$ = gas density</p>
</blockquote>
<p>Our question, why does this omit the channel size, how big the lip-hole is, which intuitively seems to have a large affect on speed (e.g. a hose with the nozzle half open vs. open).</p>
<p>What is "central velocity"? Is it what we are thinking of when we say, "jet velocity"?</p>
<p>Or maybe the key phrase is "at the channel exit" (e.g. after the constraints of the channel size no longer apply?)</p>
|
<p><a href="http://en.wikipedia.org/wiki/Bernoulli%27s_principle" rel="nofollow">Bernoulli's principle</a> states that</p>
<p>$$ \frac{1}{2}U^2+gz+\frac{P}{\rho} = \text{constant} $$</p>
<p>A very important remark here, is that the quantity is only constant a long a streamline. That means that the constant is different from streamline to streamline (e.g., between the center of your flute and the side). Be careful that the theorem does not hold at the side, because viscosity will come into play at some point.</p>
<p>There is a streamline, from the cavity with pressure $P_f$ (pressure on top over atmospheric pressure) and zero velocity, to the center of your flute, with atmospheric pressure (this is an approximation) and velocity $U_j$.</p>
<p>This reduced Bernoulli's equation to</p>
<p>$$ \frac{P_f}{\rho} = \frac{1}{2}U_j^2 $$</p>
<p>which directly reduces to the equation you gave in the question.</p>
<p>As you're looking at streamlines, you don't <em>see</em> how large the system is. The size of the channel is secretly hidden in the measured variable $P_f$. You can imagine, that for a wider channel to blow the same amount of air, you need to blow a lot harder, thus need a higher pressure.</p>
<p>To address <em>jet velocity</em> versus <em>central velocity</em>. In texts like you're referring to, people rarely use two expressions for the same thing, as this is confusing. They are always defined somewhere, but I would assume that the central velocity is the maximum velocity at the centerline, whereas jet velocity is the mean velocity across the whole jet.</p>
| 817
|
fluid dynamics
|
When playing the flute, does lip-hole aperture size affect air pressure in the mouth?
|
https://physics.stackexchange.com/questions/43364/when-playing-the-flute-does-lip-hole-aperture-size-affect-air-pressure-in-the-m
|
<p>I apologize if this is too basic of a question for this forum, but when having arm-chair discussions about the fluid dynamics of flute playing, a friend has said:</p>
<blockquote>
<p>The lip-hole area together with the lung pressure is the cause of the pressure built in the mouth.</p>
<p>When measuring the mouth pressure you are already taking into account the lip-hole area and the lungs pressure.</p>
</blockquote>
<p>My intuition is that this is incorrect--mouth pressure is caused by the "upstream" pressure in the lungs, and not any "downstream" container/flow size changes.</p>
<p>But intuition is often wrong in these situations--can anyone clarify the situation for us?</p>
|
<p>Unless the air moving through the lips goes sonic, the downstream does influence the pressure in the chamber. </p>
<p>So long as the fluid, any fluid, is moving less than the speed of sound, the downstream influences the upstream because pressure waves can move against the flow direction. </p>
<p>The smaller the opening in your lips, the higher the pressure in your mouth (and the harder you have to push with your lungs to make the air leave). This is a simple test to validate -- make a tiny opening with your lips and push as hard as you can. You'll notice your cheeks have to work hard to hold the air in your mouth and you have to push quite hard. Open your mouth all the way and repeat, you'll notice the pressure in your mouth is much lower and you don't have to push as hard.</p>
| 818
|
fluid dynamics
|
How can these fluid dynamical smoke-ring phenomena be explained?
|
https://physics.stackexchange.com/questions/46506/how-can-these-fluid-dynamical-smoke-ring-phenomena-be-explained
|
<p>The Navier-Stokes fluid dynamics equations, said that, as Sir William Thomson (or Lord Kelvin) predicted:</p>
<ol>
<li><p>When two smoke-rings are moving in the same direction, with the same speed, one behind the other, the 'leading' ring will slow down and enlarge, while the 'following' one will get smaller and speed up, passing through the ring in front, and this will keep on happening until they fade away. It was experimentally proven by Kelvin.</p></li>
<li><p>When two smoke-rings move towards each other, rather than colliding and annihilating in a smokey mess, they actually slow down and enlarge, never meeting, just getting larger until they fade away. It was experimentally proven by Kelvin. </p></li>
</ol>
<p>My questions:</p>
<ol>
<li>How does phenomena number 1 happen?</li>
<li>How does phenomena number 2 happen?</li>
</ol>
|
<p>I currently don't have access to the paper, but this should meet your needs:</p>
<p><strong>Vortex Ring Interactions</strong> (by Wakelin-Riley)<br />
<a href="https://doi.org/10.1093/qjmam/49.2.287" rel="nofollow noreferrer">https://doi.org/10.1093/qjmam/49.2.287</a></p>
<p>Here is my interpretation: Consider two rings moving co-axially in the positive x-direction with their rings in the y-planes (ring A in front of ring B). The particles in each circle of the (fattened) ring circulate and force spreading of the air in front of it (see <a href="http://en.wikipedia.org/wiki/File:Vortex_ring.gif" rel="nofollow noreferrer">http://en.wikipedia.org/wiki/File:Vortex_ring.gif</a>). Then</p>
<ol>
<li>A's circulations will force B to spread and enlarge (visually look at the vectors in the vicinity of A). Then A moves through and this repeats, because B's circulation will now spread A.</li>
<li>Here ring A is moving in positive x-direction and B in negative x-direction. Their respective circulations will stretch each other and slow them down until they meet at rest and fade.</li>
</ol>
<p>Here are more direct papers:<br />
<strong>A Note on the Leapfroggeing Between Coaxial Vortex Rings at Low Reynolds Numbers</strong> (by Lim)
<a href="https://doi.org/10.1063/1.869160" rel="nofollow noreferrer">https://doi.org/10.1063/1.869160</a><br />
<strong>Interaction of Two Vortex Rings Moving along a Common Axis of Symmetry</strong> (by Oshima)<br />
<a href="https://doi.org/10.1143/JPSJ.38.1159" rel="nofollow noreferrer">https://doi.org/10.1143/JPSJ.38.1159</a></p>
| 819
|
fluid dynamics
|
Could the Bernoulli effect be causing my ceiling to come down?
|
https://physics.stackexchange.com/questions/47925/could-the-bernoulli-effect-be-causing-my-ceiling-to-come-down
|
<p>Very 'applied' question, but I have nowhere else to turn, so I'm asking the physics experts here: I have a carport whose ceiling is made of very lightweight paneling. I've had several times now that those panels have fallen down, without an evident reason. So I'm wondering: the wind (because of the placement of other buildings in the area) blows very strong underneath the carport. Could it be that strong wind gusts, through the Bernoulli principle, are causing 'pull' on the panels and tearing them from the staples they are attached with? Stated more generally, does a fluid flowing fast beneath a surface cause a force on that surface? Thanks.</p>
|
<p>If a downward force was applied to the panels to 'pull' them down it would be the result of a pressure difference on either side of the panel. Consider a flat plate in a wind tunnel. On each side of the plate the freestream velocity is equal, thus the static pressures are equal (assuming the flow is homogeneous and the plate is at zero angle of attack). Because there is no pressure difference, there is no lift on the plate.</p>
<p>If the plate is at an angle of attack, there is lift present. If the wind is impinging on your carport's roof at an angle, this could be the cause. If this is the case there isn't much you can do about it other than improving the structure.</p>
<p>Now instead consider the walls of the wind tunnel. The freestream velocity is the same as in the flat plate case, so the static pressure in the tunnel is the same as before. However, the outside of the tunnel is at approximately stagnation pressure (assuming open-return wind tunnel). In this case there can be a huge pressure difference applied over a large area, resulting in a huge force.</p>
<p>If there is a open sealed cavity above the panels in your carport, a strong gust of wind could setup this pressure difference and cause a large downward force. If possible, maybe try drilling some ventilation holes in the panel to reduce the pressure difference.</p>
| 820
|
fluid dynamics
|
How to calculate the Darcy-Weissbach friction factor for shear thinning laminar flow in a pipe?
|
https://physics.stackexchange.com/questions/48587/how-to-calculate-the-darcy-weissbach-friction-factor-for-shear-thinning-laminar
|
<p>The <a href="http://en.wikipedia.org/wiki/Darcy%E2%80%93Weisbach_equation" rel="nofollow">Darcy-Weissbach</a> friction factor for laminar flow would be $\frac{64}{Re}$</p>
<p>Now, having a shear thinning (non-newtonian) fluid where the viscosity is not constant how do I arrive at $Re$?<br>
To know an apparaent viscosity, I'd need to know the shear rate, but that is not constant over the diameter of the pipes.
Obviously I need to make allowances anyway (like assuming that my fluid obeys a power law over the relevant shear rates), so the method doesn't neet to be uber-exact. Bu I will want to know where I'm off.</p>
<p>Googling this, I only ound numerical/CFD solutions to far more complex problems an I couldn'T draw my answer from there.</p>
|
<p>In my view, the objective of knowing the friction factor, is for one to be able to calculate what is the pressure drop needed to push a given flow $Q$ through a given pipe diameter. This kind of relations exists for several models of non Newtonian fluid, take for example the power law model:</p>
<p>$\tau=K\gamma^n$</p>
<p>In this case the solution gives:</p>
<p>$Q=\pi(\frac{\Delta P}{2KL})^{1/n}(\frac{n}{1+3n})R^{(1+3n)/n}$</p>
<p>where $R$ is the pipe radius and $L$ is the pipe length. You can rearrange an expression of this type to obtain an effective viscosity, depending of your definition of "effective viscosity". For example "the value of viscosity that plugged into the Newtonian pressure drop-flow relation will give the correct value of pressure drop for given $Q$".</p>
<p>You can review the solutions for Bingham plastic or other type of models also.
A more general approach is the one used in the Rabinowitsch-Mooney relations, where you determine experimentally a relation between flow and pressure drop, which allows you to find the shear rate at the wall, and deduce a shear rate- shear stress curve for the fluid. There are also definitions of "generalized Re", for non Newtonian flows.</p>
<p>Fluid dynamics books treat this topics in an accessible(algebraic, not CFD) manner, search for chapters on "non Newtonian fluids", or review Perry's Chemical engineerss handbook.
My experience is from chemical engineering though.</p>
<p>Hope this helps.</p>
| 821
|
fluid dynamics
|
Swimming and forces
|
https://physics.stackexchange.com/questions/52187/swimming-and-forces
|
<p>I was told that the total integral of the stress over the surface of a swimmer (i.e. the total force exerted by the swimmer on the fluid) always vanishes, because there are no external forces applied on it. That seems fair by Newton’s third Law.</p>
<p>But, how does it take into account the effects of the Newton's second Law? If, for example the swimmer starts from a stationary state, and at the time <span class="math-container">$t_1$</span> reaches a velocity <span class="math-container">$v_1$</span>, where does the force generating the acceleration <span class="math-container">$v_1/t_1$</span> come from?</p>
<p>Could you help me to clarify all this? </p>
|
<p>Let's imagine a swimmer in a swimming pool and approximate the Earth as an inertial frame. The swimmer can certainly accelerate relative to the Earth frame (in the direction parallel to the Earth's surface); we see this happen all the time in real life. It follows from Newton's second Law, as you point out, that the net external force on the swimmer is nonzero. The only object that can possibly exert a <em>horizontal</em> force on the swimmer is the water in the pool (the swimmer is making physical contact with nothing else, and the force due to gravity cannot affect his/her horizontal acceleration). It follows that the net horizontal force of the water on the swimmer is nonzero.</p>
<p>If you had a blob of water floating around in outer space, and a swimmer inside, and if the blob of water and swimmer began stationary relative to an inertial frame, then the swimmer attempting to swim would cause some of the water to move backward, and this would propel the swimmer forward relative to the inertial frame. However, in this case the total external force on the blob+swimmer system would be zero, so the center of mass of the water+blob would remain stationary, but even in this case, the swimmer could accelerate relative to the inertial frame.</p>
<p>Essentially, the swimmer is doing the same thing that a torpedo would do; he/she expels water backward, and this propels him/her forward.</p>
| 822
|
fluid dynamics
|
How does a hinge affect the amount of a submerged material?
|
https://physics.stackexchange.com/questions/51275/how-does-a-hinge-affect-the-amount-of-a-submerged-material
|
<p>Suppose I have a rod that has a density of $X <1$. If I were to submerge that rod in water (density 1), I would expect $X$ of the rod to be below water and $1-X$ of it to be above water (simple buoyancy).</p>
<p>But here's the catch. Suppose I attach one corner of the rod to a frictionless hinge, suspended close to the water. How will that affect what percent of the rod is above water in equilibrium now? My teacher told me it is NOT simply $1-X$. He also didn't specify how high above the water the hinge was, so I suppose that doesn't matter. Please tell me if more info is needed.</p>
|
<p>Here's a solution assuming a "thin rod"; i.e. assuming that the free end of the rod is fully submerged and "the stretch where the rod begins to dive and is only partially submerged" is negligibly small compared to the stretch (from the end attached to the hinge) that is lifted completely above the water as well as compared to the fully submerged stretch.
(Without this assumption or approximation the problem may be considered and even solved as well, but the solution seems too unwieldy to be presentable.)</p>
<p>Let's call $H$ the height above water level of the end with the hinge. According to the stated assumptions, the "stretch above water" therefore has a length of $H / \text{Sin[} \, \phi \, \text{]}$, where $\phi$ is the angle the rod makes with the horizontal water surface, and the remaining fully submerge stretch has a length of $L - H / \text{Sin[} \, \phi \, \text{]}$.</p>
<p>This configuration is stable </p>
<ul>
<li><p>if the hinge is sufficiently supported to hold the rod end in height $H$. (That's supposed to be satisfied by construction. See the calculation of the required supporting force below.) And:</p></li>
<li><p>if the torque ("around the hinge") due to the weight of the rod, which acts to turn/pull the rod down towards the vertical (i.e. towards increasing $\phi$) is balanced by the torque due to the force/buoyancy/lift acting on the submerged stretch of the rod:</p></li>
</ul>
<p>$0 = \int_0^L \, \mathbf{r \cdot dF}_{\text{weight}} - \int_{ H / \text{Sin[} \, \phi \, \text{]} }^L \, \mathbf{r \cdot dF}_{\text{lift}}$,</p>
<p>$0 = \int_0^L \, \mathbf{r \cdot g} \, \rho_{\text{rod}} \, A \, dl - \int_{ H / \text{Sin[} \, \phi \, \text{]} }^L \, \mathbf{r \cdot g} \, \rho_{\text{water}} \, A \, dl$,</p>
<p>$0 = \int_0^L \, l \, \text{Cos[} \, \phi \, \text{]} \, g \, \rho_{\text{rod}} \, A \, dl - \int_{ H / \text{Sin[} \, \phi \, \text{]} }^L \, l \, \text{Cos[} \, \phi \, \text{]} \, g \, \rho_{\text{water}} \, A \, dl$,</p>
<p>where $A$ is the (otherwise considered negligible) cross-sectional area of the rod. Let's assume that it is constant along the length of the rod; therefore it cancels. Also assume that the rod is of uniform density. This leads to:</p>
<p>$0 = \rho_{\text{rod}} \, \int_0^L \, l \, dl - \rho_{\text{water}} \, \int_{ H / \text{Sin[} \, \phi \, \text{]} }^L \, l \, dl$,</p>
<p>$\rho_{\text{rod}} \, L^2 \, / 2 = \rho_{\text{water}} \, (L^2 \, / 2 - (H / \text{Sin[} \, \phi \, \text{]})^2 \, / 2).$</p>
<p>Solving for the length of the stretch of rod above water:</p>
<p>$H / \text{Sin[} \, \phi \, \text{]} = L \, \sqrt{ 1 - \rho_{\text{rod}} / \rho_{\text{water}} }$,</p>
<p>or following the terminology given in the question:</p>
<p>$H / \text{Sin[} \, \phi \, \text{]} = L \, \sqrt{ 1 - X }$.</p>
<p>If the height of the hinge reaches or exceeds the value $L \, \sqrt{ 1 - X }$ then the rod simply hangs vertically (dipping in the water as long as the height of the hinge doesn't exceed $L$, of course). If the height of the hinge approaches $0$ it also approaches the (physically inevitable) actual thickness of the rod and the initially made assumption breaks down; the rod then swims more or less horizontally, with $1 - X$ of its volume above water.</p>
<p>Finally, the force supported by the hinge has to compensate the weigth and buoyancy of the rod:</p>
<p>$\mathbf{F}_{\text{hinge}} = $<br>
$\int_0^L \, \mathbf{dF}_{\text{weight}} - \int_{ H / \text{Sin[} \, \phi \, \text{]} }^L \, \mathbf{dF}_{\text{lift}} = $<br>
$\int_0^L \, \mathbf{g} \, \rho_{\text{rod}} \, A \, dl - \int_{ H / \text{Sin[} \, \phi \, \text{]} }^L \, \mathbf{g} \, \rho_{\text{water}} \, A \, dl = $<br>
$\mathbf{g} \, \left( \rho_{\text{rod}} \, A \, L - \rho_{\text{water}} \, A \, (L - H / \text{Sin[} \, \phi \, \text{]}) \right)$.</p>
<p>Inserting the result from above:</p>
<p>$\mathbf{F}_{\text{hinge}} = $<br>
$\mathbf{g} \, \left( \rho_{\text{rod}} \, A \, L - \rho_{\text{water}} \, A \, (L - L \, \sqrt{ 1 - \rho_{\text{rod}} / \rho_{\text{water}} }) \right) = $<br>
$\mathbf{g} \, \left( m_{\text{rod}} - \rho_{\text{water}} \, A \, L \, (1 - \sqrt{ 1 - \rho_{\text{rod}} / \rho_{\text{water}} }) \right) = $<br>
$\mathbf{g} \, \left( m_{\text{rod}} - m_{\text{rod}} \, \rho_{\text{water}} / \rho_{\text{rod}} \, (1 - \sqrt{ 1 - \rho_{\text{rod}} / \rho_{\text{water}} }) \right) = $<br>
$\mathbf{g} \, m_{\text{rod}} \, \left( 1 - \rho_{\text{water}} / \rho_{\text{rod}} (1 - \sqrt{ 1 - \rho_{\text{rod}} / \rho_{\text{water}} }) \right) \equiv $<br>
$\mathbf{g} \, m_{\text{rod}} \, \left( 1 - 1/X (1 - \sqrt{ 1 - X }) \right) $, </p>
<p>which is less than $\mathbf{g} \, m_{\text{rod}}$ for $X \lt 1$.</p>
| 823
|
fluid dynamics
|
Query about Bernoulli's principle
|
https://physics.stackexchange.com/questions/55820/query-about-bernoullis-principle
|
<p>We know that the lower atmosphere has high pressure and as we go up, the pressure decreases, if it's so then why doesn't all gases fly up into the upper atmosphere from the lower following <a href="http://en.wikipedia.org/wiki/Bernoulli%27s_principle" rel="nofollow">Bernoulli's theorem</a>? I do expect that gravitational effect on gases isn't that worth notable. Do correct me for my mistake if it exists! </p>
|
<p>If you have ever swum to the bottom of a swimming pool you'll know that in water the pressure increases as you go deeper. At a depth of about 10 metres the pressure is twice what it is at the surface, but the water 10 metres down doesn't burst up to the surface because it is held down by the weight of water above it. In fact the increase of pressure with depth is exactly the weight of water above.</p>
<p>Exactly the same is true of the atmosphere. The pressure at ground level is 101,325 Pa because each square metre of the ground has about 10,329 kg of air above it (10329 kg times the acceleration due to gravity 9.81 m/sec$^2$ = 101325 Pa). If you could magically remove the 100 km or so of atmosphere that's above some patch of air at ground level that air would indeed immediately expand upwards.</p>
<p>Incidentally, <a href="http://en.wikipedia.org/wiki/Bernoulli%27s_principle" rel="nofollow">Bernoulli's principle</a> is unrelated to the problem.</p>
| 824
|
fluid dynamics
|
Water, how do you make ripples
|
https://physics.stackexchange.com/questions/56032/water-how-do-you-make-ripples
|
<p>How do I make a ripple effect in a bowl of water. I have tried throwing small pebble sin but it just splashes and sinks. I have read that I need to make the water a thicker liquid ie add glycerine, will this work and why? </p>
|
<p>Making water ripples in the bowel of water by throwing a pebble in it would be difficult an it will also be little bit bigger and faster to see it in a water bowel. Rather you can try vibrating the bowel at it's centre at the bottom of the bowel with something like electronic raser or a massage machine or you can just hit the bottom with a spoon the vibration will case small amplitude ripples to form on the surface of water. Yes if you add some thing to the water to make it make it a little thicker then according to me it will require a little bit more energy to form ripple.</p>
| 825
|
fluid dynamics
|
Membrane that allows liquid to pass only when forced under pressure
|
https://physics.stackexchange.com/questions/56046/membrane-that-allows-liquid-to-pass-only-when-forced-under-pressure
|
<p>Is it possible to have a membrane that will not let a liquid through it at normal pressures due to gravity, but pass that liquid when substantially pressurised?</p>
<p>For instance, a few inches of water (say 0.1psi) would be blocked, but 100psi would pass through. </p>
|
<p>Such membranes do exist, like in <a href="http://en.wikipedia.org/wiki/Reverse_osmosis" rel="nofollow">reversed osmosis</a> plants creating fresh water from sea water. Such plants use huge pressure to press fresh water through the membrane, and without pressure very little water will come through. If there are air on the other side of the membrane surface tension of the water in the pores might hold back all the water, if its not highly pressurized.</p>
<p>Membranes where pressure will expand the pores is well suited for your typical application. Puncture a balloon with a needle numerous times and attach it to the tap, this experiment might give you a clearer picture.</p>
| 826
|
fluid dynamics
|
Flow in parallel paths after pressure regulator
|
https://physics.stackexchange.com/questions/55954/flow-in-parallel-paths-after-pressure-regulator
|
<p>Consider the following:</p>
<pre><code> /‾‾ Valve --- Watering hose
Pump --- Pressure regulator ------ Valve --- Watering hose
\__ Valve --- Watering hose
</code></pre>
<p>A pump generating an arbitrary pressure and a pressure regulator
to curb it at 1.5 bar. After the regulator we have three valves in parallel.</p>
<p>If I close one of the valves partially or completely will the volume
flowing through the other two valves change? I hope that the pressure
regulator would compensate closing/opening valves by keeping the pressure
constant but I'm not sure. </p>
<p>Background is a watering solution in gardening. I aim to decouple several
strands of watering hose so they can be adjusted independently.</p>
|
<blockquote>
<p>I hope that the pressure regulator would compensate closing/opening valves by keeping the pressure constant but I'm not sure. </p>
</blockquote>
<p>In the absence of confounding factors, this is the correct viewpoint.</p>
<p>In order for the pressure regulator to be "perfect", you would ideally have a large volume between the pressure regulator and the valves, and the valves would be spaced a good distance away from each other. In the current picture, the pressure regulator controls the pressure by controlling the flow from the pump. At least this is my understanding.</p>
<p>Non-ideal conditions will be seen when the interacting kinetic forces between the different valves are significant. In that case, the pressure that pressure-regulator sees can't exactly be called the back pressure behind each valve. If we treat the region up to the valve as a stagnant, constant pressure, region then there is no problem with your view. We would also require that the pressure-regulator performs its job perfectly, obviously.</p>
<p><em>note: I'm assuming the pressure-regulator controls the pressure with its own valve. The representation could be trying to imply that the pressure-regulator is just a sensor and the controller opens and closes one or more of the hose valves, but I don't think this is the system because the OP implies that the hose valves are manually controlled</em></p>
| 827
|
fluid dynamics
|
What does Euler equation mean?
|
https://physics.stackexchange.com/questions/56116/what-does-euler-equation-mean
|
<p>In order to prove the Bernoulli’s principle ($\frac{p}{\rho} + \frac{1}{2}u^2+\phi = constant$ ), I have to use the <a href="http://en.wikipedia.org/wiki/Euler_equations_%28fluid_dynamics%29" rel="nofollow">Euler equation</a>: $\frac{Du}{Dt} = -\frac{1}{\rho}\nabla p + g$.</p>
<p>I know how to prove it, but I didn't understand what does it mean and say (Euler equation)?</p>
<p>please explain me.</p>
<p>any help appreciated!</p>
|
<p>Euler equation is the special case of the Navier-Stokes equation that describes fluid motion. Euler equation stands for inviscid flow, i.e. flow with zero viscosity.</p>
<p>To get it a bit more explicit please visit
<a href="http://en.wikipedia.org/wiki/Euler_equations_%28fluid_dynamics%29" rel="nofollow">wikipedia page</a>, and for further reference I would suggest you to search for some lecture notes online and select those you would like the most. This is quite popular topic, so there should not be any problem. As a suggestion you may want to have a look <a href="http://www.jpoffline.com/mphys_y4.php" rel="nofollow">here</a>.</p>
| 828
|
fluid dynamics
|
Bernoulli's theorem: $\frac{p}{\rho}+\frac{1}{2}u^2+\phi$ is constant along a streamline
|
https://physics.stackexchange.com/questions/56323/bernoullis-theorem-fracp-rho-frac12u2-phi-is-constant-along-a-st
|
<p>I am trying to understand <strong>the Bernoulli's theorem:</strong></p>
<ul>
<li><strong>$\frac{p}{\rho}+\frac{1}{2}u^2+\phi$ is a constant along a streamline</strong></li>
</ul>
<p>I got that:</p>
<p>$\frac{\partial u}{\partial t}$ + ($\nabla \times u)\times u$ = $-\nabla(\frac{p}{\rho}+\frac{1}{2}u^2+\phi)$</p>
<p>For a steady flow: $\frac{\partial}{\partial t}$ = 0 and then:</p>
<p>($\nabla \times u)\times u$ = -$\nabla H$ with the scalar: $H$ = $\frac{p}{\rho}+\frac{1}{2}u^2+\phi$</p>
<p><strong>now, I didn't understand the next steps:</strong></p>
<p>Taking the “dot product” of ($\nabla \times u)\times u$ = -$\nabla H$ the left hand side vanishes, as ($\nabla \times u)\times u$ is perpendicular to $u$ and we get:</p>
<p>$(u \cdot \nabla)H = 0$</p>
<p>This implies that $H$ is constant along a streamline</p>
<p><strong>Can someone explain me this thing in other words please?</strong></p>
|
<p>Let $\mathbf u(t, \mathbf x)$ represent the velocity vector field of the fluid. Let $\mathbf x(t)$ denote the position of a particle moving with the fluid, then the velocity $\dot{\mathbf x}(t)$ of the particle at a time $t$ will be equal to the velocity of the fluid flow at the point $(t, \mathbf x(t))$, namely
$$
\mathbf u(t, \mathbf x(t)) = \dot{\mathbf x}(t)
$$
Now suppose that $\mathbf u(t, \mathbf x)\cdot\nabla H(t, \mathbf x) = 0$. He want to show that this implies that $H$ is constant along the path of a particle moving with he fluid. Notice that for <em>any</em> path $\mathbf x(t)$ we have
$$
\frac{d}{dt}H(t, \mathbf x(t)) = \frac{\partial H}{\partial t}(t, \mathbf x(t))+\dot{\mathbf x}(t)\cdot\nabla H(t, \mathbf x(t))
$$
Assuming then that $\partial_t H = 0$, and assuming that the path $\mathbf x(t)$ is that of a particle moving with he fluid, the equations written above imply
$$
\frac{d}{dt}H(t, \mathbf x(t)) = \mathbf u(t, \mathbf x(t))\cdot\nabla H(t, \mathbf x(t)) = 0
$$
so the quantity $H$ is constant along a flow line, as desired!</p>
| 829
|
fluid dynamics
|
Wall pressure of a fluid flow in a pipe of variable radius
|
https://physics.stackexchange.com/questions/57254/wall-pressure-of-a-fluid-flow-in-a-pipe-of-variable-radius
|
<p>Using all cylindrical coordinates, pipe with z-axis vertically upward and radius of $r = G(z)$, flow is incompressible, inviscid and steady, Using appropriate boundary conditions I want to find the wall pressure.Flow is axis symmetric and irrotational, and of the form $(a(r,z), 0, c(r,z))$ in cylindrical coordinates.</p>
<p>The boundary conditions I have are the the fluid is stationary at the wall.</p>
<p>$a(G(z), z) = c(G(z), z) = 0, $</p>
<p>Using Bernoilli's equation for steady flow </p>
<p>$\frac{p}{\rho} + \frac{v^2}{2} + \Omega = G$</p>
<p>$\frac{p}{\rho} + \frac{c^2}{2} = G - \Omega$</p>
<p>$p_w = (G - \Omega - \frac{c^2}{2})\rho$</p>
<p>But I don't think this is correct for some reason.</p>
| 830
|
|
fluid dynamics
|
blood fluidics - can you determine the force exerted on a bound red blood cell under shear stress?
|
https://physics.stackexchange.com/questions/59304/blood-fluidics-can-you-determine-the-force-exerted-on-a-bound-red-blood-cell-u
|
<p>Is it possible to calculate the force exerted on a bound, infected red blood cell under various shear stresses? The strength of an adhesive interaction is normally measured by SPR or AFM, however using microfluidics, you can increase the shear stress until the cell dislodges, therefore, can you use this data to calculate force?</p>
| 831
|
|
fluid dynamics
|
Spinning liquid to create a centrifuge effect
|
https://physics.stackexchange.com/questions/59629/spinning-liquid-to-create-a-centrifuge-effect
|
<p>I'm in the business of purifying used cooking oil. Normally, I heat the oil up and let it settle for a couple of days. Water and solids settle to the bottom and cleaner oil remains on top. I'm trying to accelerate this process. </p>
<p>So, the idea is to take a vertical tank 5ft diameter, about 10ft tall and put a paddle mixer on top with vertical blades that would extend to about 3" from the wall. If I turn on the mixer, it will spin at about 60 rpm and spin the liquid with it. According to my calculations this should give me a force of 3.5g and thus decrease settling time by about 3.5 times. </p>
<p>Please let me know if this can work or if I'm missing something. Your input will be highly appreciated.</p>
|
<p>This approach will not work; since it will result in mixing of the oil and particles, not a settling/separation. This is called a "stir-tank reactor" in chemical-engineering jargon. If you look into this properly, you will realize that this configuration will REDUCE THE MIXING TIME, and in fact, deter separation. Refer:
<a href="http://en.wikipedia.org/wiki/Continuous_stirred-tank_reactor" rel="nofollow">http://en.wikipedia.org/wiki/Continuous_stirred-tank_reactor</a>
for a more detailed explanation. It is a model for PERFECT MIXING.</p>
<blockquote>
<p>Spinning the whole thing is not really an option. The tank is not well balanced and when full weighs about 15000 lbs. Potential for a huge mess is tremendous.</p>
</blockquote>
<p>I suggest thinking about a Couette flow instead (<a href="http://en.wikipedia.org/wiki/Couette_flow" rel="nofollow">http://en.wikipedia.org/wiki/Couette_flow</a>) since centrifugation is not possible.
The RPM of the tank should be low enough, so that turbulence isn't generated, (Reynold's number: <a href="http://en.wikipedia.org/wiki/Reynolds_number" rel="nofollow">http://en.wikipedia.org/wiki/Reynolds_number</a>) since turbulence causes increased mixing.</p>
| 832
|
fluid dynamics
|
how to explain the upright force for the plane?
|
https://physics.stackexchange.com/questions/59977/how-to-explain-the-upright-force-for-the-plane
|
<p>I remember in the high school physics, my teacher told us that the design of the plane wing is because we want the air above the wing flowing faster than the air flowing below so the pressure above and below will be different so the net force is pointing upright. I am thinking if it is possible to explain this by bernoulli equation? But I soon stuck there because I don't know if bernoulli equation applicable to the case of the air flow in the open region? If so, what cross section we should choose?</p>
| 833
|
|
fluid dynamics
|
What is a good reynolds number for this process?
|
https://physics.stackexchange.com/questions/62786/what-is-a-good-reynolds-number-for-this-process
|
<p>I’m trying to convince my boss that the mixers we are using are too much. I’m trying to prove that we are over-mixing our product. Our product is ink…just your basic ink found in your printer at home. We mix in a 23 inch diameter 50 gallon vessel using a 2.75 inch diameter axial flow impeller going at 1050 rpm with a 3 inch rotor stator 5 inches below it. The inks have a viscosity of around 3. What Reynolds number should be sufficient for this process or can you give me any further input? Any help would be GREATLY appreciated. Thanks.</p>
|
<p>At least in the industry I work in, sizing mixers is more an art than a science. If you are convinced the process works at too high a Reynolds Number, try if you can make an experiment:</p>
<ul>
<li>find out why it is done the way it is done and how the mixer was sized</li>
<li>when you still think the Re is too high, convince your boss to make an experiment with one batch</li>
<li>mix a batch or two with lower speed (rpm) and see if the ink still has the required quality, maybe also try different mixing times</li>
<li>If it does not work, you should be able to achieve the required quality by continuing mixing, so no material is lost in your experiment</li>
</ul>
<p>In general, you want to achieve two things with mixing: A global flow in the container, to homogenize the mixture, and high shear forces (associated with high turbulence) to disperse small droplets. If the ink is destined for a printer, I assume that small pieces of pigments and the like may pose a huge problem - so the process may designed like this to guarantee a minimum size of particle.</p>
| 834
|
fluid dynamics
|
Water ripples and nonzero divergence
|
https://physics.stackexchange.com/questions/64038/water-ripples-and-nonzero-divergence
|
<p>In 2-d, one ripple would mean the velocity of water particles move out radially forming a circular wavefront. The Navier Stokes equations say the divergence of velocity has to be zero, but this circular radiation pattern has nonzero divergence. What am I missing here, since water ripples clearly exist?</p>
|
<p>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.</p>
<p>If you have observed birds floating on water surface, you see them moving up and down but not along the waves.</p>
| 835
|
fluid dynamics
|
What keeps the sugar suspended in the tea?
|
https://physics.stackexchange.com/questions/5562/what-keeps-the-sugar-suspended-in-the-tea
|
<p>At room temperature....</p>
<p>How long will the sugar stay suspended once dissolved ?</p>
<p>What governs the rate of settling ? </p>
<p>What part does Brownian motion play ?</p>
<p><em>Sugar might be a bad example...please substitute as needed.</em></p>
|
<p>Anything dissolved is kept "suspended" by brownian motion.
Theoreticly there will be a small gradient of concentration
due to gravity, but this is unmeasurable for sugar.
(Barometric formula). For much bigger particles (ãbout µm) , this
gradient can be determined. </p>
<p>The sugar will stay dissolved until the water evaporates,
then the sugar will form a solid residue. (if You are lucky,
as crystals) Settling does not occur with molecules.</p>
| 836
|
fluid dynamics
|
Pressure vs wind speed, on a rectangular surface
|
https://physics.stackexchange.com/questions/5850/pressure-vs-wind-speed-on-a-rectangular-surface
|
<p>How do I go about finding the pressure exerted on a rectangular surface in a free flowing air stream?</p>
<p>I wouldn't imagine that this is directly related to the airspeed / surface area, but have no idea where to start. Is there even an equation, or does one need to do some kind of FEA?</p>
<p>For instance a 1.2m x 2.4m metal sheet suspended some distance above ground level, if I have a gust of wind at 8m/s (directly perpendicular to the sheet), what is the average pressure across the face of sheet?</p>
|
<p><a href="http://www.reasonablepower.com/nice_to_know/an_introduction_to_wind_loads.htm" rel="nofollow">Wind Load Formula:</a></p>
<p>$F_d = \frac{1}{2} \rho v^2 A C_d$</p>
<p>where<br>
$F_d$ is the force of drag (or in this case Force Against the flat plate)<br>
$\rho$ is the density of the air<br>
$v$ is the speed of the air against the object<br>
$A$ is the area of the object which the air is blowing against<br>
$C_d$ is the drag coefficient</p>
| 837
|
fluid dynamics
|
Reynolds number, turbulence regime, and drag force
|
https://physics.stackexchange.com/questions/6098/reynolds-number-turbulence-regime-and-drag-force
|
<p>I am trying to model a system in which cubes of about 2 cm in size are floating in a circular water thank of about 30 cm in diameter. The cubes move around under the influence of the fluid flow induced by four inlets that point toward the center of the tank, and are located at the positions $0$, $\pi/2$, $\pi$, and $3\pi/2$. The flow velocity ranges from 0 to 10 cm/s, with an average velocity around 6 cm/s.</p>
<p>My questions are the following:</p>
<ul>
<li>What would be the Reynolds number of the system? In particular, should I take as characteristic length the size of the cubes, or that of the tank?</li>
<li>For such a system, what is the limit Reynolds number for the turbulent regime?</li>
<li>What would be the correct form of the drag force, and do you intuitively think that the orientation of the blocks is negligible from a drag coefficient point of view?</li>
</ul>
<p>Thanks for your help!</p>
|
<p>I'd say that you have several regimes that are well defined:</p>
<ul>
<li>The behavior of the fluid as it exits in inlet jets and enters the bulk without interference from the cubes. [Length scale set by the exit aperture?]</li>
<li>Flow of the fluid around isolated cubes when far from the edges of the tank (far being several times the characteristic size <em>of the cube</em>). [Length scale set by the side of the cube.]</li>
<li>Flow of the fluid toward, along and away from the sides of the tank away from the jets and without interference from the cubes. [Length scale set by the boundary behavior?]</li>
</ul>
<p>which is the good news, unfortunately you also have all the cases that mix and match the various length scales:</p>
<ul>
<li>case with cubes interacting with the jet near the aperture</li>
<li>case with cubes in motion near the walls</li>
<li>case with cubes in close proximity to one another</li>
</ul>
<p>You can probably find existing treatments for all the former cases, but the latter ones are going to be tricky, and you'll note that <em>they</em> feature at least two length scales.</p>
<p>Yuck.</p>
<p>This must be part of why they say CFD is hard.</p>
| 838
|
fluid dynamics
|
Sutherland's constant of argon
|
https://physics.stackexchange.com/questions/9995/sutherlands-constant-of-argon
|
<p>I want to compute the viscosity of argon at different temperatures. What is <a href="http://en.wikipedia.org/wiki/Viscosity#Gases" rel="nofollow">Sutherland's constant</a> of argon?</p>
|
<p>I meanwhile found the constant by intensive googling (for example in <a href="http://www.cse.scitech.ac.uk/ceg/papers/Gu_et_al_Nanoscale_Microscale_Thermophysical_Eng.pdf" rel="nofollow">this article</a>). It is (depending on the source) around 144 K.</p>
| 839
|
fluid dynamics
|
Entropy decrease using Stoke's flow?
|
https://physics.stackexchange.com/questions/11065/entropy-decrease-using-stokes-flow
|
<p>There is a <a href="http://io9.com/5811236/this-is-the-coolest-science-experiment-youll-see-all-week" rel="noreferrer">video</a> of an experiment from University of Mexico using corn syrup (highly viscous) and water. They are "mixed together" in a container by turning a crank but when the crank is turned in the opposite direction they neatly un-mix.</p>
<p>I understand that the two liquids actually do not blend, but how can the syrup droplets be returned to very close to their initial state. It looks as if the entropy is greatly reduced. Is the increase in temperate of the liquids (through the turning of the crank) the answer?</p>
|
<p>Entropy never increases during this "mixing process" as it normally would, because it is <em>not</em> really a mixing process. The only thing that happens is a huge distortion of the distribution of the colors, but no information about it is lost. Compare it to performing a Fourier transform on some function: this also results in something that looks completely different, sometimes chaotic and high-entropy-like, but it's really just an invertible transformation.</p>
| 840
|
fluid dynamics
|
communicating vessels formula
|
https://physics.stackexchange.com/questions/10778/communicating-vessels-formula
|
<p>I having trouble with this formula (asked first in math.Se, I didn't know the existence of physics.se)</p>
<p><span class="math-container">$$
Z_1(t) = Z_e+(\sqrt{Z_1-Z_e}-\frac{2S_0}{S_1}\sqrt{2g\left(1+\frac{S_1}{S_2}\right)}.t^2
$$</span></p>
<p>where <span class="math-container">$Z_1(t)$</span> and <span class="math-container">$Z_2(t)$</span> are the levels of liquid in the vessels at time <span class="math-container">$t$</span>, <span class="math-container">$S_1$</span> and <span class="math-container">$S_2$</span> are the cross-section areas, <span class="math-container">$S_0$</span> is the cross-section area of the tube that connects them, and <span class="math-container">$Z_e$</span> is the final height of the two vessels.</p>
<p>but I only get incorrect values and don't know if the formula is wrong of if its me.</p>
<p>For testing is use <span class="math-container">$Z_1 = 45$</span>, <span class="math-container">$Z_2 = 5$</span>, <span class="math-container">$S_0=2\pi{}\cdot{}0.3=1.884$</span>, <span class="math-container">$S_1=2\pi{}\cdot{}10=62.8$</span> and <span class="math-container">$S_2=S_1$</span> and</p>
<p><span class="math-container">$Z_e = \frac{S_1 Z_1 + S_2 Z_2} {S_1+S_2}=\frac {62.8{}\cdot{}45 + 62.8{}\cdot{}5}{62.8+62.8}=25$</span></p>
|
<p>like the comment aheadd said , i was doing it wrong , using perimeter instead of area ($2\pi R$ instead of $\pi R^2$).</p>
<p>if you wan to watch the result look <a href="http://www.wolframalpha.com/input/?i=plot%2025%2b%28Sqrt%5B45-25%5D-%282Pi0.3%C2%B2%29/%28Pi10%C2%B2%29Sqrt%5B2%2a980%281%2b%28Pi10%C2%B2%29/%28Pi10%C2%B2%29%29%5Dt%29%C2%B2%20from%200%20to%2038" rel="nofollow">here</a></p>
| 841
|
fluid dynamics
|
Pressing a fluid through leaks
|
https://physics.stackexchange.com/questions/13242/pressing-a-fluid-through-leaks
|
<p>Say I have a cylindrical container of some volume $V$, that I fill with water or some other fluid with dynamic viscosity $\mu$. Here, the bottom of the container is milled such that some fraction, $p$, of its surface area, $A$, is removed to allow passage of fluid. One could, for example, drill pores until this criterion is met.</p>
<p>After positioning a plate at the top of the container, I press down on the plate with some force $F$ to compress the fluid. As a function of $F$, how quickly will I drain the fluid through the bottom of the container? </p>
<p>To calculate this, what additional measurements/parameters, such as surface hydrophobicity/etc., will I require, and how does this change with the minimum cross-sectional area of the pores I mill in the bottom of the container?</p>
|
<p>Your answer will depend greatly upon the size of the holes you drill. If I remember correctly, fluid flow through an orifice is proportional to the fourth power of the orifice diameter. So, for a given force, you would get twice as much fluid out of a single hole of diameter 1 mm as you would out of eight holes, each of 1/2 mm diameter. </p>
| 842
|
fluid dynamics
|
Can I measure the Reynolds number or the heat transfer coefficient
|
https://physics.stackexchange.com/questions/13321/can-i-measure-the-reynolds-number-or-the-heat-transfer-coefficient
|
<p>Suppose I have a flow of hot air around a cold and unevenly shaped object with holes and tunnels (think about it as a bed packed with some objects). I would like to know the Reynolds number of this flow and its convective heat transfer coefficient. The definition of the Reynolds number contains a "characteristic length" that is somehow mysterious to me and that I do not have at hand. And I am a bit reluctant to use the formula for the packed bed Reynolds number. Is it possible to measure it? How would I design an experiment for this?</p>
<p>I would like to avoid temperature readings of the object. In <a href="http://www.sciencedirect.com/science/article/pii/096014819190107Z" rel="nofollow">this</a> article I found a relation between the Reynolds number of a packed bed and the heat transfer coefficient, so measuring only the Reynolds number would be a good start, even though I do not have a real packed bed. Could this be done by simple pressure drop readings?</p>
|
<p>From what you describe there are two different length scales associated with this problem. The one ascociated with flow through the porous medium (the 'bed packed with some objects'), and the second ascociated with the flow around the 'bed'. I will assume you are talking about the flow through the porous 'bed'. </p>
<p>Normally the characteristic dimension or length scale for internal flows is taken to be the hydraulic diameter. This is defined to be four times the cross-sectional area (of the fluid), divided by the wetted perimeter. However, for such things as 'pebble beds' etc. the Reynolds number is defined differently.</p>
<p>For flow of fluid through a bed of approximately spherical particles of diameter D in contact, if the voidage (fraction of the bed not filled with particles) is ε and the superficial velocity V (that is, the fluid velocity through the bed as if the spheres/objects were not present) then a Reynolds number can be defined as:</p>
<p>$$Re = \frac{\rho V D}{\mu(1 - \epsilon)}$$</p>
<p>Laminar conditions apply up to Re = 10, fully turbulent from 2000 (Wikipedia). There are more advanced formulas for this, and they work in a variety of regimes; from not-so-packed beds, to very packed-beds, also with a variaty of pebble/object shapes.</p>
<p>Many experiments have been done on convective and radiative heat transfer in pebble bed nuclear reactors and other such heat exchangers. I am sure you should be able to find some journal papers on this stuff along with the standard correlations you need for your particular flow.</p>
<p>For the convective heat transfer coefficient for this flow however, you should be using the Nusselt Number which is a measure of the ratio of convective to conductive heat transfer a solid-fluid boundary.</p>
<p>I hope this helps.</p>
| 843
|
fluid dynamics
|
Behaviour of liquid in vaccume
|
https://physics.stackexchange.com/questions/14722/behaviour-of-liquid-in-vaccume
|
<p>Is it possible for a liquid to exist in a high quality vacuume? For example, a few Torr.</p>
<p>If so what are the methods for doing this?</p>
|
<p>It's absolutely possible to have stable liquids at low pressures. It all depends on the equilibrium phase diagram of the liquid in question.</p>
<p>Looking at the <a href="http://en.wikipedia.org/wiki/Triple_point" rel="nofollow">triple points</a> (a good estimate of the lowest pressure at which the liquid is stable) for a variety of liquids, you can see that <a href="http://www-d0.fnal.gov/hardware/cal/lvps_info/engineering/elements.pdf" rel="nofollow">mercury</a> for example (it's in this pdf, I promise) has its triple point at -38 C and ~$10^{-6}$ Torr. In other words, liquid mercury is thermodynamically stable near this pressure and temperature and can thus certainly be stable at a pressure of a few Torr.</p>
| 844
|
fluid dynamics
|
Clarification of equilibrium expression for upside down can in water
|
https://physics.stackexchange.com/questions/16147/clarification-of-equilibrium-expression-for-upside-down-can-in-water
|
<p>Let's say I initially have an open, empty, soda can. I then turn it over and lower it into a bowl of water, and then release it. Obviously water rises to some level in the cup and then there is air at the top of the can, which was initially the bottom.</p>
<p>Are there only three forces on the can once it reaches equilibrium?</p>
<ol>
<li><p>Buoyant force from the water (up)</p></li>
<li><p>Weight of the can itself (down)</p></li>
<li><p>Air pressure inside the can acting on the water surface (down)</p></li>
</ol>
<p>Let me know if all of these are right, otherwise my equation for total forces is...</p>
<p>F_b = mg + P where m is the mass of the can, and P is the air pressure in the cup</p>
|
<p>Maybe you can be a bit more clear in your formulation. Unless the soda can changes into something else, do not use the word 'initially'. Are cup and bowl the same in your experiment?</p>
<p>In small scale experiments like this(unlike the earths atmosphere) air does not really fall anywhere, and the pressure will be the same everywhere inside the can giving a net force of zero. You do not need it here, but remember that the force from a pressure is calculated by multiplying with the area. F=P*A</p>
| 845
|
fluid dynamics
|
What formulae for calculating fluid pressure change involves these parameters?
|
https://physics.stackexchange.com/questions/16403/what-formulae-for-calculating-fluid-pressure-change-involves-these-parameters
|
<p>Okay... this is a bit desperate...</p>
<p>I am reverse engineering an excel program that is used for calculating outputs for a pressure vessel used for industrial process fluid heating.</p>
<p>There is a part in the excel sheet where a pressure drop is being calculated. But there are too many unit conversions and other magic constants in intermediate steps i cannot regonise. I have some physics background but not enough in fluid mechanics to decipher.</p>
<p>For starters, I have intermediate step formulas like:</p>
<pre><code>(fluid flow rate[lbs/h]]) * (1545) * ((inlet plus outlet temp[deg c] / 2)*9/5+460) /
((molecular weight[??])*((pressure[psig])+14.5)*144)
</code></pre>
<h2>The formula looks like this:</h2>
<p>$\frac{\dot{m} \times 1545 \underbrace{\left[\frac{(T_\text{in} + T_\text{out})}{2}\times(\frac{9}{5}) + 460 \right]}_\text{Degrees C to Deg F to Rankine}}{M_w \times (P_g + 14.5) \times 144}$</p>
<p>I am hoping someone who deals with fluid mechanics will be familiar with many formulae to tell what it might be.</p>
<p>The inputs to the entire calculation steps consists of flow rate, inlet and outlet temperatures, molecular weight, viscosity and thermal conductivity; and also dimensions of the vessel.</p>
<p>Greatly appreciate everyones help! Thanks.</p>
<p><strong>EDIT</strong>: The actual problem is here <a href="http://www.jakesee.com/se/physics_se_16403.png" rel="nofollow noreferrer">http://www.jakesee.com/se/physics_se_16403.png</a> and the birthplace of the image is here <a href="https://stackoverflow.com/questions/7647478/generate-a-flat-list-of-all-excel-cell-formulas">https://stackoverflow.com/questions/7647478/generate-a-flat-list-of-all-excel-cell-formulas</a></p>
| 846
|
|
fluid dynamics
|
Timber floating in the river
|
https://physics.stackexchange.com/questions/16696/timber-floating-in-the-river
|
<p>There's an old puzzle (but the physical it is):</p>
<p>Lumberjack chop wood and float it down the river. He noticed an interesting feature: in the spring, when the water is comming, timber is nailed to the shore. In the hot summer, when the water lowers, timber, by contrast, floats exactly in the middle of the river. Why is that?</p>
|
<p>I believe the answer has to do with currents perpendicular to the main flow of the river. In temperate climates, lakes freeze over winter; in the spring, the lake "turns" because the melting ice water is colder than the insulated under layers. In a river, you might not get such a dramatic movement all at once, but you might still get some of the same effect.</p>
<p>In the winter, the water is low and only the central channel is covered by water. It may be cold, but this land is warmer than the low banks that are frozen solid. When the snow melt comes, it fills the river so that the lower banks now form the edges of the riverbed. Since the central channel is warmer than the edges, the water naturally wells up from the center and sinks by the edges. This current would be enough to drive a floating object to the edges.</p>
<p>In the summer, the opposite occurs. The shallow portions of the river on the edges are warmer than the deep central channel. The perpendicular currents reverse and the log moves to the middle.</p>
<p>I've never studied this, but it sounds pretty darned reasonable to me. :)</p>
| 847
|
fluid dynamics
|
Rotating fluid under gravity, fluid dynamics question
|
https://physics.stackexchange.com/questions/19880/rotating-fluid-under-gravity-fluid-dynamics-question
|
<p>An incompressible inviscid fluid is rotating under gravity g with constant angular
velocity $\Omega$ about the z-axis, which is vertical, so that $u = (−\Omega y, \Omega x, 0)$ relative
to fixed Cartesian axes. We wish to find the surfaces of constant pressure, and
hence surface of a uniformly rotating bucket of water (which will be at atmospheric
pressure).
Bernoulli's equation suggests that $$p/\rho+|u|^2/2+gz=\text{constant. So,}$$</p>
<p>$$z=\text{constant}-\frac{\Omega^2}{2g}(x^2+y^2)$$</p>
<p>But this suggests that the surface of a rotating bucket of water is at its highest in
the middle, where is this going wrong?</p>
<p>Many thanks</p>
|
<p>In the Physics Forums you should find the answer to your exact problem <a href="http://www.physicsforums.com/showthread.php?t=476134" rel="nofollow">'Fluid dynamics - finding pressure for a rotating fluid'</a>. Another answer is <a href="http://www.physicsforums.com/showthread.php?t=38112" rel="nofollow">here</a> (the point is that Bernoulli's law is applicable only along a streamline so that we must use a rotating frame and add the centripetal acceleration).</p>
<p>From another point of view see Newton's <a href="http://en.wikipedia.org/wiki/Bucket_argument" rel="nofollow">'Bucket argument'</a>.</p>
| 848
|
fluid dynamics
|
Speed of a dynamic hydraulic system
|
https://physics.stackexchange.com/questions/21184/speed-of-a-dynamic-hydraulic-system
|
<p>We all have noticed that changing the temperature of the water in the shower is not instantaneous, rather the result is felt when the water that was in the tap works its way up to the showerhead. However, changing the pressure does feel instantaneous. I wonder at what speed the change in pressure propagates in the direction of flow in a flowing fluid. My guess would be at the speed of sound in the medium, based on the fact that it is faster than the speed of the flow of the water (like the temperature is), yet obviously at or below C. I am obviously missing the correct keywords to google for as I cannot find any references to this phenomenon. What is the answer to the query, or better yet, how could I have found this information (short of empirically)?</p>
<p>Thanks.</p>
|
<p>The pressure wave does indeed travel with the speed of sound of the media. This is much higher for water then for air with a speed of $\approx$ 1500 m/s. </p>
<p>This effect is well known and feared under the name <a href="http://en.wikipedia.org/wiki/Water_hammer" rel="nofollow">water hammer</a>. Basically by rapidly closing a valve you create a longitudinal pressure wave that can be powerful enough to damage to destroy metal pipe joints. The kinetic energy of the flowing liquid causes this phenomenon.</p>
| 849
|
fluid dynamics
|
Why tea dust in a cup of tea seems to concentrate in the bottom center?
|
https://physics.stackexchange.com/questions/21041/why-tea-dust-in-a-cup-of-tea-seems-to-concentrate-in-the-bottom-center
|
<blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://physics.stackexchange.com/questions/3244/vortex-in-liquid-collects-particles-in-center">Vortex in liquid collects particles in center</a> </p>
</blockquote>
<p>Sorry for my bad English. :-) Hello !</p>
<p>Here is my question :
If you enjoy some cup of tea, you may see that there are some tea dust in suspension in your cup. If you turn your spoon in the liquid, the tea will rotate, and the dust will concentrate in the bottom of the cup (because the dust is denser than the water) but too in the center !</p>
<p>This is conter intuitive for me. The dust have to spread near the sides of the cup, because it's more denser than the water ! No ?</p>
<p>Edit : We can think to the centrifugation of the blood. When the blood is collected from a people, the pocket pass into a centrifuge (? do you see what I mean ? top of the pocket in the center, bottom of the pocket near the walls) and the diverse parts of the blood finish separated in function of their density. But the denser, the closer of the bottom of the pocket...</p>
| 850
|
|
fluid dynamics
|
Is it possible to calcualte the yield-stress of a fluid by measuring the smalles bubble that rises through it?
|
https://physics.stackexchange.com/questions/29153/is-it-possible-to-calcualte-the-yield-stress-of-a-fluid-by-measuring-the-smalles
|
<p>Consider a yield-stress liquid in gravity. I assume that the buyouncy of a small gas-bubble will not be enough to overcome the yield stress (so the liquid doesn't behave liquid), thus leaving the bubble trapped. Is this so in theory, or is there a flaw in my thinking?</p>
<p>Sme argue that there are no yield-stress liquids in sense of one stress, below wich there is no yield, but rather a very high apparent viscosity. Would this imply that even miniscule bubbles will rise, but very slowly?</p>
<p>Lastly, have there been tries to measure viscosity or apparent viscosity or some measure for yield stress from rising bubbles?</p>
|
<p>Yes, your method is a perfectly good way of measuring viscosities. It's a bit limited because it's hard to measure the bubble diameter exactly, but assuming you can do this you can calculate the viscosity of the fluid using <a href="http://en.wikipedia.org/wiki/Stokes%27_law" rel="nofollow">Stokes' law</a>.</p>
<p>However note that I haven't mentioned yield stress. That's because yield stress is a slippery thing to define. Indeed a rheologist friend of mine, <a href="http://books.google.co.uk/books/about/An_Introduction_to_Rheology.html?id=B1e0uxFg4oYC" rel="nofollow">Howard Barnes</a>, maintains that there is no such thing as a yield stress. The problem is that for non-Newtonian fluids the viscosity often rises very rapidly as the shear decreases and it may get too high for you to measure it. So how do you you tell the difference between a true yield stress and a viscosity too high for your equipment to measure?</p>
<p>If you shake up your non-Newtonian liquid to get some bubbles in it then come back in an hour or so you'll see some apparently stationary bubbles, and you can use these to calculate a "yield stress". But if you come back a bit later you may find the bubbles that you thought were stationary have moved a bit. You could wait a bit longer, but you can't run the experiment for very long because <a href="http://en.wikipedia.org/wiki/Ostwald_ripening" rel="nofollow">Ostwald ripening</a> will be changing the bubble sizes as you watch. That means there's inevitably a limit to how high a viscosity you can measure. Unless the yield stress is below this limit you won't be able to measure it.</p>
| 851
|
fluid dynamics
|
How do Kolmogorov scales work in shear thinning fluis?
|
https://physics.stackexchange.com/questions/30125/how-do-kolmogorov-scales-work-in-shear-thinning-fluis
|
<p>My understanding of Kolmogorov scales doesn't really go beyond this poem:</p>
<blockquote>
<p>Big whirls have little whirls that feed on their velocity,<br>
and little whirls have lesser whirls and so on to viscosity.</p>
</blockquote>
<p>The smallest scale according to wikipedia* would be
$\eta = (\frac{\nu^3}{\epsilon})^\frac{1}{4}$</p>
<p>But can I assume the same shear across all scales, and hence (for a shear thinning liquid) the same apparent viscosity?<br>
Are there practical observations about this?</p>
<ul>
<li><a href="http://en.wikipedia.org/wiki/Kolmogorov_microscales" rel="nofollow">http://en.wikipedia.org/wiki/Kolmogorov_microscales</a></li>
</ul>
<p>Update: Maybe I need to clarify my question. I'm not so much interested in the theory as in one real physical phenomenon this theory describes: That there is a lower limit to the size of a vortex for a given flow, and this size can at least be estimated using above equation. Now, a lot of real fluids are non-Newtonian in one way or the other, I'm asking about shear because the apparant viscosity is (also) shear dependent.<br>
While the theory of Kolmogorv may be hard to translate for non-Newton flow, the actual physical phenomenon of an observable (or evenmeasureable) lower limit for vortex size should still hold - are there any measurements or observations? </p>
|
<p>Yes it is not clear why you are mentionning shear. Hence it is not clear whether your are interested by very exotic cases or on the contrary by the classical Kolmogorov turbulence theory.</p>
<p>I will give an answer assuming the latter.
The Kolmogorov theory starts by analogy with statistical mechanics by assuming an isotropic and homogeneous distribution of vortices in a Newtonian fluid.
Viscosity is constant for a Newtonian fluid. Then assuming self similarity, Kolmogorov established the energy spectrum in relation to the wave number.
Because the wave number <b>is</b> the scale, he established how the energy is distributed on the different scales.</p>
<p>The originality of this theory is that there is a minimum scale, called Kolmogorov length where the dissipation of energy by viscosity happens. Of course this length is not constant but depends on the flow and the largest scales (inertial scale) L are related to the Kolmogorov scale l by L/l = Re^(3/4) where Re is the Reynolds number. The scale invariance of viscosity is simply given by the fact that we deal with a Newtonian fluid.</p>
<p>Now to shear thinning. Shear thinning (or thickening) fluids are <b>non Newtonian</b>. Viscosity depends on stress and even on time.
The conditions of isotropy, homogeneity and self similarity are not given so the Kolmogorov theory and its lengths have nothing to say about these exotical substances.
I add that you won't observe turbulent vortex cascades in them either.</p>
<p>These substances behave at the same time like solids (plastics) and fluids so that Navier Stokes equations alone are not the best way to study them. The previous sentence is an understatement.</p>
| 852
|
fluid dynamics
|
Concentration of fluid pumped through a tube?
|
https://physics.stackexchange.com/questions/30907/concentration-of-fluid-pumped-through-a-tube
|
<p>I have a bottle filled with a fluid A, and a tube of volume $V_0$ filled with a fluid B leading away from it.
When I now start pumping fluid A out of the bottle, I am interested in the concentration $c(V)$ of fluid A in fluid B at the end of the tube after the volume $V$ exited the tube.</p>
<p>In my second situation (I am afraid there are two), after having pumped $V_0$, a valve at position $V_1$ switches and the pump is reversed. The mixed fluids are now pumped into a second tube of volume $V_2$. Again, I'd like to know what drips out of there.</p>
<p><strong>Situation 1</strong></p>
<pre><code>| A |
| | V0
| ==================== <- B
| |
| |
|_________|
</code></pre>
<p><strong>Situation 2</strong></p>
<pre><code>| A |
| | V0
| =========++========== <- B
| | V1 ||
| | || V2
|_________| ||
^- B
</code></pre>
<p>I've tried using the <a href="http://en.wikipedia.org/wiki/Hagen%E2%80%93Poiseuille_equation" rel="nofollow">Hagen-Poiseuille equation</a> (or its proof as given in the Wikipedia), but that grows ugly fast. Is there a simple(r) way to approach this?</p>
| 853
|
|
fluid dynamics
|
Behaviour of fluids in thin spaces
|
https://physics.stackexchange.com/questions/427610/behaviour-of-fluids-in-thin-spaces
|
<p>Suppose you have a simple setting: a thin (scale of hundreds of nanometer to single digit micrometer) but in relation to that very wide (up to millimeters) box with an input and an output hole in it.</p>
<p>Given how thin the space is, would a fluid even go in there and fill out the box? How long would it take? </p>
<p>I had a bit of a hard time to find literature on this, so hints in that direction would also be appreciated.</p>
|
<p>Your setup is known as a Hele Shaw cell, in which the bulk of the flow is a balance between pressure and viscosity, and around the edge of a region if fluid the balance is between pressure and surface tension.</p>
<p>Also look at the <a href="https://en.m.wikipedia.org/wiki/Hele-Shaw_flow" rel="nofollow noreferrer">Wikipedia page about Hele-Shaw flow</a></p>
| 854
|
fluid dynamics
|
Blood as Newtonian fluid
|
https://physics.stackexchange.com/questions/428813/blood-as-newtonian-fluid
|
<p>In some of the literature I read that blood can be considered as Newtonian fluid when a larger vesses with high shear stress is considered... How is the shear stress calculated for aorta and how do they claim that shear stress is more for larger vessel when compared to the smaller ones . What is the relationship between the shear stress and the size of the vessel</p>
|
<p>welcome. As a general rule, when you read something provide a reference.</p>
<p>Laminar blood flow in cylindrical blood vessels is <a href="https://en.wikipedia.org/wiki/Hagen%E2%80%93Poiseuille_equation" rel="nofollow noreferrer">Poiseuille</a> flow. In Poiseuille flow the velocity profile is parabolic
$$
u_z = \frac{dP}{dz} \frac{R^2-r^2}{4\mu}
$$
where $R$ is the radius of the blood vessel, $dP/dz$ is the pressure gradient driving the flow, and $\mu$ is viscosity. The shear stress only has a $zr$ component
$$
\tau_{zr} \sim \frac{\partial u_z}{\partial r} \sim
\frac{dP}{dz} r
$$
and the maximum stress occurs on the wall, $r=R$. For fixed pressure gradient this increases with radius, but pressure gradient is not fixed in a branching network. </p>
<p>One way to look at this is using the flow rate
$$
Q \sim \frac{dP}{dz}\frac{R^4}{\mu}.
$$
We can re-express shear stress using flow rate
$$
\tau_{rz}(R) \sim \frac{Q}{R^3}.
$$
In vascular branching the total flow rate of an incompressible fluid must be conserved (in binary branching each branch has flow $Q/2$). The magic question is then how vessel radius scales in binary branching. </p>
<p>In principle this could follow any relation (as an engineer, you can just decide), but nature presumably tries to optimize things in some way. There is an empirical relation, called <a href="https://en.wikipedia.org/wiki/Murray%27s_law" rel="nofollow noreferrer">Murray's law</a> (together with somewaht hand-waving derivations) that states that the sum of the cubes of $R$ is conserved. This would imply that the shear stress is constant across the network.</p>
| 855
|
fluid dynamics
|
Justification of an assumption in fluid mechanics (free surface)
|
https://physics.stackexchange.com/questions/445776/justification-of-an-assumption-in-fluid-mechanics-free-surface
|
<p>What justifies this assumption on the free surface of the sea, taken from Faltinson's book (page 15), for a potential flow?</p>
<blockquote>
<p>A fluid particle on the free-surface is assumed to stay on the free-surface</p>
</blockquote>
| 856
|
|
fluid dynamics
|
Is gas flow always compressible?
|
https://physics.stackexchange.com/questions/449903/is-gas-flow-always-compressible
|
<p>From Franz Durst's <em>Fluid Mechanics: An Introduction to the Theory of Fluid Flows:</em></p>
<blockquote>
<p>When a fluid element reacts to pressure changes by adjusting its volume and consequently its density, the fluid is called compressible. When no volume or
density changes occur with pressure or temperature, the fluid is regarded as
incompressible although, strictly, incompressible fluids do not exist.</p>
</blockquote>
<p>So, strictly speaking, although the <strong>fluid</strong> is always compressible is there a case where gas fluid <strong>flow</strong> maintains constant density?</p>
|
<p>At the steady state, the density of a flow will be constant. That's tautological.</p>
<p>Fluids like water can be treated as incompressible because its response to a pressure change is negligible for most practical calculations.</p>
<p>For more information see <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/permot3.html" rel="nofollow noreferrer">here</a>.</p>
| 857
|
fluid dynamics
|
Does pressure decreases in accelerating fluids?
|
https://physics.stackexchange.com/questions/461861/does-pressure-decreases-in-accelerating-fluids
|
<p>Bernuolli says that in a pipe when speed <strong>increases</strong> pressure also <strong>decreases</strong> and that happen when pipe become <strong>narrower</strong>. </p>
<ol>
<li>So,if a fluid <strong>accelerate</strong> (increases speed) due to pressure difference in a pipe that <strong>DO NOT</strong> changes area and <strong>do not</strong> become narrower, WILL THE PRESSURE STILL BE DECREASING DURING THIS ACCELERATION?</li>
</ol>
|
<p>Assuming the area doesn't change and do not become narrower, and that the pipe is oriented horizontally, the Bernoulli equation becomes: <span class="math-container">$P_2 - P_1 = \frac{\rho}{2}(v_1^2 - v_2^2)$</span>. If a fluid accelerates, then <span class="math-container">$v_1^2 - v_2^2$</span> will be lowered, so the difference of pressure will be lowered. You are turning piezo (pressure) energy into kinetic energy.
However, by conservation of mass, since the flow rate is constant because we are in a stationary flow, then speed cannot change. So the Bernoulli equation doesn't hold. We need to introduce the extended Bernoulli equation that takes into account the head loss. The head loss is responsible for the "pressure decreasing" along the axis of the pipe.</p>
<p>The effect you described typically what is used in nozzle in turboreactors with two differences: in nozzles, the area changes, and in nozzles, you use a tiny bit of the internal energy in addition with the piezo energy.</p>
| 858
|
fluid dynamics
|
Diver's air torus rising from 10m underwater (edited)
|
https://physics.stackexchange.com/questions/490190/divers-air-torus-rising-from-10m-underwater-edited
|
<p>[Edit: earlier version was even more messy]
I just watched a video of a jellyfish caught in a diver's 'air ring' - a torus blown for the sake of watching it rise. The jellyfish gets drawn into the torus and spun like a top as the bubble rises.
I want to know how fast, in rpm, the jellyfish is spinning at a given distance from the surface.
I think I need to know a formula for the circumference of the torus section at any given point as it rises , and a formula for the speed of rise. </p>
<p>My assumptions:
- air torus is released at 10m, 2 atmospheres.
- the exhaled volume at 10m is 500mls, so that'll double to 1000mls at sea-level
- the jellyfish is isobuoyant, so is the same mass as the seawater it displaces - so can be treated as water.
-the radial rotation of the torus is arbitrary (ie due to the mechanics of how the diver blew the bubble and is at a normal to the axial rotation), so can be ignored.</p>
<p><a href="https://youtu.be/JXkWSgU-CL0" rel="nofollow noreferrer">https://youtu.be/JXkWSgU-CL0</a></p>
<p>[Apologies if this is completely the wrong forum for these questions - in which case signposting to other places appreciated]</p>
|
<p>Bernoulli's Principle provides a sufficient explanation. In the toroidal flow, water moves faster near the long circular axis. Water is not flowing <em>toward</em> the axis; but flows faster <em>around</em> the axis as you get closer. Bernoulli's Principle indicates that the pressure, then, decreases as the distance to the axis decreases. An extended body will experience greater pressure on the side away from the axis and lesser pressure on the side toward the axis, so will be pulled toward the axis. When the extended body is <strong>on</strong> the axis, the pressure is the same on all sides so it gets trapped there, spinning with the fluid flow.</p>
| 859
|
fluid dynamics
|
Can you create a pressure differential with a material that only allows flow in one direction?
|
https://physics.stackexchange.com/questions/498348/can-you-create-a-pressure-differential-with-a-material-that-only-allows-flow-in
|
<p>(Theoretically) suppose you had a material that only allowed water through one way but not the other.</p>
<p>Then you have a box of water and then put this material at the halfway point dividing the box in two. Eventually one side of the box would have less water than the other?</p>
| 860
|
|
fluid dynamics
|
What is a boundary layer, exactly?
|
https://physics.stackexchange.com/questions/504467/what-is-a-boundary-layer-exactly
|
<p>I've been struggling with this concept for a very long time. I know about boundary layers in a very informal sense, such as in boundary layer separations at the trailing edge of airfoils and other objects immersed in flowing fluid. I also know about the boundary layer development in a pipe, as a plug flow develops into the parabolic distribution expected of a Newtonian fluid.</p>
<p>I've referred to <a href="https://en.wikipedia.org/wiki/Boundary_layer" rel="nofollow noreferrer">Wikipedia</a>, but I could not find a proper definition of what a boundary layer is:</p>
<blockquote>
<p>In physics and fluid mechanics, a boundary layer is an important concept and refers to the layer of fluid in the immediate vicinity of a bounding surface where the effects of viscosity are significant.</p>
</blockquote>
<p>I can expect the boundary layer connects points that have some common parameter, but what is it?</p>
|
<p>The boundary layer is the region of a flow close to a surface, where there is a gradient of velocity between zero at the wall and the free-stream velocity (<span class="math-container">$v_\infty$</span>), which is caused by viscosity and the no-slip condition. It is also possible to have boundary layers for other flow variables, e.g. temperature or scalar concentration.</p>
<p>From your question, it seems that what you are looking for is a firm definition for where the <em>edge</em> of the boundary layer is:</p>
<blockquote>
<p>I can expect the boundary layer connects points that have some common
parameter, but what is it?</p>
</blockquote>
<p>Typically, the edge of the boundary layer is defined as some percentage of the free-stream velocity. From the same Wikipedia article:</p>
<blockquote>
<p>Paul Richard Heinrich Blasius derived an exact solution to the above
laminar boundary layer equations. The thickness of the boundary
layer <span class="math-container">$\delta$</span> is a function of the Reynolds
number for laminar flow.</p>
<p><span class="math-container">$$\delta \approx 5.0 \frac{x}{\sqrt{Re}}$$</span></p>
<p><span class="math-container">$\delta$</span> = the thickness of the boundary layer: the region of flow where the velocity is less than 99% of the far field velocity <span class="math-container">$v_\infty$</span>; <span class="math-container">$x$</span> is position along the semi-infinite plate, and <span class="math-container">$Re$</span> is the Reynolds Number given by ...</p>
</blockquote>
| 861
|
fluid dynamics
|
Can pitot tubes be used for unsteady flows?
|
https://physics.stackexchange.com/questions/512675/can-pitot-tubes-be-used-for-unsteady-flows
|
<p>The classical derivation of the Pitot tube equation assumes that the fluid is in a steady flow to use the Bernoulli equation version for steady flows. But in practice, does Pitot tube can be used also for unsteady flows? </p>
|
<p>Certainly. Unsteady flows (i.e., where the flow velocity changes with time, sometimes quickly) happens all the time in aviation, and pitot tubes are used almost universally to monitor airspeed in planes. </p>
| 862
|
fluid dynamics
|
What does the Laplacian of a fluid velocity potential mean?
|
https://physics.stackexchange.com/questions/523906/what-does-the-laplacian-of-a-fluid-velocity-potential-mean
|
<p>I was studying something related to fluid mechanics and then I found that <span class="math-container">$\nabla^2 \Phi = 0$</span> where <span class="math-container">$\Phi$</span> is the fluid velocity potential (<span class="math-container">$\vec{V}=\nabla \Phi$</span>). So I was wondering what does it mean that the laplacian of the fluid velocity potential is equal to zero (everywhere) and also in general. </p>
<p>I know the gradient of a scalar function tells us about the direction in which the function increases and the divergence of a field about how it spreads out in space. Then I could guess that the Laplacian is equal to zero because the fluid could be uniform and thus the velocity field doesn't spread out.</p>
<p>Is this a correct assumption? </p>
|
<p>One of the consequences of <span class="math-container">$\nabla\cdot V = 0$</span> is the incompressibility of the fluid. It is perfectly possible for air, for example, that some portions of space are temporary being depleted <span class="math-container">$\nabla\cdot V > 0$</span> or filled <span class="math-container">$\nabla\cdot V < 0$</span>. </p>
<p>edit (Jan,<span class="math-container">$15^{th}$</span>, 2020):
if <strong>V</strong> can be expressed as a gradient of some scalar, its rotational (only for x component) is:</p>
<p><span class="math-container">$\nabla$</span> X <span class="math-container">$V$</span> = <span class="math-container">$\nabla$</span> X <span class="math-container">$ (\nabla \phi) =$</span> </p>
<p><span class="math-container">$\frac{\partial (\nabla \phi)_z}{\partial y} - \frac{\partial (\nabla \phi)_y}{\partial z} = \frac{\partial (\partial \phi/\partial z)}{\partial y} - \frac{\partial (\partial \phi/\partial y)}{\partial z} = \frac{\partial^2\phi}{\partial z \partial y} - \frac{\partial^2\phi}{\partial y \partial z} = 0$</span></p>
<p>The same for the other components.</p>
<p>Resuming: <span class="math-container">$\nabla\cdot V = 0$</span> means incompressibility, but the additional condition of <span class="math-container">$\nabla^2 \phi = 0\;$</span>, (what implies <span class="math-container">$V = \nabla \phi)\;$</span>means the fluid has rotational = 0 everywhere.</p>
| 863
|
fluid dynamics
|
Molecular explanation of floating and sinking
|
https://physics.stackexchange.com/questions/595362/molecular-explanation-of-floating-and-sinking
|
<p>If we have an object submerged in water with a density greater than it then it will sink while objects with less density will float. This is found in many scientific articles and videos.. but why?</p>
<p>Particularly speaking, How do we understand this in a molecular picture? If we think of it using a free body diagram it makes sense but I can't imagine why it is so if we look at it from a microscopic scale.</p>
|
<p>At a macroscopic level, <a href="https://en.wikipedia.org/wiki/Buoyancy" rel="nofollow noreferrer">buoyancy</a> is the result of the increase of pressure with depth. This creates a net force on a submerged object which is equal and opposite to the weight of the volume of liquid that it displaces. Hence if a fully submerged object displaces an amount of liquid with a weight that is greater than itself then it will float; if not then it will sink.</p>
<p>You can also think of buoyancy in terms of energy. If you start with an object is partially submerged and push it down slightly so that an additional volume <span class="math-container">$\delta V$</span> is submerged, then the same volume <span class="math-container">$\delta V$</span> of liquid is displaced upwards. The object loses potential energy but the displaced liquid gains potential energy. If the potential energy gained by the displaced liquid is greater than the potential energy lost by the object then it must require a force to push the object down (otherwise we would have a perpetual motion machine). This force is the net buoyancy force on the object (buoyancy force minus object's weight).</p>
<p>One interesting consequence of this is that the level at which an object floats is independent of the force of gravity. If you put a rubber duck in a bathtub on the moon (and put the bathtub in a sealed environment at a pressure of one atmosphere), the duck will float with exactly the same proportion out of the water as on earth. The smaller weight of the duck and the smaller weight of displaced water cancel out.</p>
<p>At a microscopic level <a href="https://en.wikipedia.org/wiki/Pressure" rel="nofollow noreferrer">pressure</a> is the result of collisions between the atoms or molecules of the liquid and the atoms or molecules of the submerged object. Gravity causes an increase of liquid density with depth; in other words there are more atoms per unit volume in the liquid. This results in more collisions per unit time with the submerged object as you go deeper in the liquid, which at a macroscopic level means greater pressure.</p>
| 864
|
fluid dynamics
|
Does the bottom of a rotating cylinder influence the shape of the fluid surface?
|
https://physics.stackexchange.com/questions/604116/does-the-bottom-of-a-rotating-cylinder-influence-the-shape-of-the-fluid-surface
|
<p>Given a cylinder of fluid that is rotating the surface is of a parabolic shape. From what I can tell when deriving this the shape of the bottom of the cylinder should not influence the shape of the surface (as long as it is bellow the fluid surface). Is this true in practice?
Consider a parabolic bottom very close to the shape of the fluid surface, does surface tension and other such effects distort the surface in any significant way?</p>
| 865
|
|
fluid dynamics
|
Derive the Momentum Equation from a Lagrangian (Fluid Dynamics)
|
https://physics.stackexchange.com/questions/616297/derive-the-momentum-equation-from-a-lagrangian-fluid-dynamics
|
<p>I'm trying to derive the fully compressible Euler-Momentum equation for the given Lagrangian.</p>
<p>We wish to derive <span class="math-container">$$\rho\frac{D\boldsymbol{u}}{Dt} + \nabla P + \rho\nabla\phi = 0 $$</span> from the Lagrangian <span class="math-container">$$ \mathcal{L} = \rho \left( \frac{|\boldsymbol{u}|^{2}}{2} -e(\rho,s) - \phi \right) $$</span></p>
<p>I've gotten as far as</p>
<p><span class="math-container">$$\frac{\partial}{\partial t}(\rho\boldsymbol{u}) + \nabla\cdot(\rho\boldsymbol{u}\boldsymbol{u}) + \rho\nabla\frac{|\boldsymbol{u}|^{2}}{2} -\rho\nabla \left(\frac{|\boldsymbol{u}|^{2}}{2}-e-\frac{P}{\rho}-\phi \right) - \rho T\nabla s = 0 $$</span>
which has been verified as correct.
I've recognized the following relationships, <span class="math-container">$$\frac{\partial}{\partial t}(\rho\boldsymbol{u}) + \nabla\cdot(\rho\boldsymbol{u}\boldsymbol{u}) = -\nabla P $$</span> <span class="math-container">$$\frac{D\boldsymbol{u}}{Dt} = -\nabla \frac{P}{\rho} $$</span>
<span class="math-container">$$ \rho T\nabla s = \rho\nabla e - \frac{P}{\rho}\nabla\rho $$</span> which are the Euler-Momentum equation in conservative form, the Euler-Momentum equation, and the 2nd Law of Thermodynamics, respectively.</p>
<p>but after implementing these I arrive at:</p>
<p><span class="math-container">$$\frac{\partial}{\partial t}(\rho\boldsymbol{u}) + \nabla\cdot(\rho\boldsymbol{u}\boldsymbol{u}) + \rho\nabla\frac{|\boldsymbol{u}|^{2}}{2} - \rho\nabla\frac{|\boldsymbol{u}|^{2}}{2} + \rho\nabla e + \rho\nabla\frac{P}{\rho} + \rho\nabla\phi - \rho\nabla e + \frac{P}{\rho}\nabla\rho = 0 $$</span>
<span class="math-container">$$ \Rightarrow -\nabla P - \rho \frac{D\boldsymbol{u}}{Dt} + \rho\nabla\phi + \frac{P}{\rho}\nabla\rho = 0 $$</span>
which is "close" but incorrect. Where am I going wrong here?</p>
|
<p>It works out just fine</p>
<p>Starting from your equation:
<span class="math-container">$$\frac{\partial}{\partial t}(\rho\boldsymbol{u})+\nabla\cdot(\rho\boldsymbol{u}\boldsymbol{u})+\rho\nabla\frac{|\boldsymbol{u}|^{2}}{2}-\rho\nabla\left(\frac{|\boldsymbol{u}|^{2}}{2}-e-\frac{P}{\rho}-\phi\right)-\rho T\nabla s=0$$</span>
it can be simplified:
<span class="math-container">$$\frac{\partial}{\partial t}(\rho\boldsymbol{u})+\nabla\cdot(\rho\boldsymbol{u}\boldsymbol{u})+\rho\nabla\left(e+\frac{P}{\rho}+\phi\right)-\rho T\nabla s=0$$</span>
by canceling kinetic energy terms.</p>
<p>Then applying the 2nd law of thermodynamics:
<span class="math-container">$$\rho T\nabla s=\rho\nabla e-\frac{P}{\rho}\nabla\rho$$</span>
substituting, cancelling and rearranging we get:
<span class="math-container">$$\frac{\partial}{\partial t}(\rho\boldsymbol{u})+\nabla\cdot(\rho\boldsymbol{u}\boldsymbol{u})+\rho\nabla\frac{P}{\rho}+\frac{P}{\rho}\nabla\rho+\rho\nabla\phi=0$$</span></p>
<p>The pressure terms can be combined using the product rule:
<span class="math-container">$$\nabla P = \nabla \left[\rho\frac{P}{\rho}\right]=\rho\nabla\frac{P}{\rho}+\frac{P}{\rho}\nabla\rho$$</span>
to yield:
<span class="math-container">$$\frac{\partial}{\partial t}(\rho\boldsymbol{u})+\nabla\cdot(\rho\boldsymbol{u}\boldsymbol{u})+\nabla P+\rho\nabla\phi=0$$</span></p>
<p>Then using the continuity equation:
<span class="math-container">$$\frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\boldsymbol{u})=0$$</span>
we simplify the first two terms on the left:
<span class="math-container">$$\frac{\partial}{\partial t}(\rho\boldsymbol{u})+\nabla\cdot(\rho\boldsymbol{u}\boldsymbol{u})=\rho\left[\frac{\partial\boldsymbol{u}}{\partial t}+\boldsymbol{u}\nabla\cdot\boldsymbol{u}\right]+\left[\frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\boldsymbol{u})\right]\boldsymbol{u}=\rho\frac{D\boldsymbol{u}}{Dt}$$</span></p>
<p>Finally we get to what you wished to derive:
<span class="math-container">$$\rho\frac{D\boldsymbol{u}}{Dt}+\nabla P+\rho\nabla\phi=0$$</span></p>
| 866
|
fluid dynamics
|
Mokka pot coffee homogeneity
|
https://physics.stackexchange.com/questions/620048/mokka-pot-coffee-homogeneity
|
<p>I've noticed some people pouring coffee from a mokka pot alternates between two cups instead of directly filling one after the other. The reason argued is related to the beverage homogeneity, as the initial vapor will go through totally fresh coffee, in opposition to the very last drops. Neglecting the possible solid residuum, I intuitively see no reason supporting this, as for the few minutes while the coffee is raising to the upper recipient, it will fall along the external walls of the central chimney, even inducing a ring-vortex-like circulation around it, depending on the geometry of the chimney aperture.</p>
<p>I've thought a simple transparent model of the pot to see this, but I would like to ask for possible Comsol(/alternative software) model results that may already exist and I don't manage to find, as I have no active licenses. I would also thank further arguments supporting what may actually happen.</p>
|
<p>I think the explanation about the stages of the percolation process is incorrect, for the reasons that the OP describes. What seems more plausible is that more dense coffee layers (regardless of when they originate) descend while the pot is boiling, so the coffee at the top is more diluted (another possibility is that the coffee at the top is less polluted with small cofee partciles, i.e., the sediment). For other beverages this is usually fixed by shaking the bottle/container, but this method is impractical with a mokka pot.</p>
| 867
|
fluid dynamics
|
Why liquid in a duct flows when there is a pressure difference?
|
https://physics.stackexchange.com/questions/650028/why-liquid-in-a-duct-flows-when-there-is-a-pressure-difference
|
<p>Liquid flows in a pipe when a pressure difference is applied on the ends. But can a liquid flow even if there is no pressure difference?
Like in case the liquid (ideal) is flowing with uniform velocity through a horizontal pipe, the pressure along the horizontal direction remains same.</p>
<p>Please explain this in a detailed way considering the liquid to be ideal.</p>
|
<p>No. When a pressure difference exists between two points, the fluid will start to move from the higher pressure point to the low pressure point. This happens to minimize the pressure difference.</p>
<p>Without such a pressure differential, the fluid is stationary, and the system is absent of any flow.</p>
<p>But if we consider a long horizontal pipe, and consider a small segment of this pipe, Bernoulli's equation tells us that since <span class="math-container">$$p_1 +\rho gh_1 + \frac{1}{2}\rho v_1^2 = p_2 +\rho gh_2 + \frac{1}{2}\rho v_2^2$$</span> then since <span class="math-container">$h_1=h_2$</span> <span class="math-container">$$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$$</span> Now since we are considering a small section, we get that <span class="math-container">$p_1-p_2\approx 0$</span> which will yield <span class="math-container">$$v_1=v_2$$</span> meaning that we can have fluid flow over a small pressure differential.</p>
<p>But again, for large sections, the pressure at one end of the pipe must be larger than the pressure at the other end for there to be flow.</p>
| 868
|
fluid dynamics
|
Euler equation of fluid dynamics
|
https://physics.stackexchange.com/questions/45648/euler-equation-of-fluid-dynamics
|
<p>I'm trying to obtain <a href="http://en.wikipedia.org/wiki/Euler_equations_%28fluid_dynamics%29" rel="nofollow">Euler equation</a> for a perfect fluid in laminar or stationary flow. A particle fluid is submitted at volume forces and surface force. The fist, in my case, is giving only by gravity and the second by pressure. By Newton's second law I obtain:</p>
<p>$$\vec{F}_V + \vec{F}_s = m\frac{d\vec{v}}{dt}.$$</p>
<p>An element of volume force is given by
$$d\vec{F}_V = dm\vec{g}=\rho d\omega\vec{g}$$
and an element of surface force is given by
$$d\vec{F}_S = -pd\vec{S}.$$</p>
<p>Integrating I obtain</p>
<p>$$ \int_V \rho \,d\omega\vec{g} - \int_S p\,d\vec{S} = \frac{d\vec{v}}{dt}\int_V \rho\, d\omega$$.</p>
<p>Now Euler equation is written in local form as
$$\rho\vec{g} - \nabla p = \rho \frac{d\vec{v}}{dt}.$$</p>
<p>My question is this: where the gradient of $p$ comes from? I must have the following identity
$$-\int_S pd\vec{S} = -\int_V \nabla p\,d\omega.$$</p>
<p>Why the transformation from a surface integral to a volume integral is given by the gradient and not by the divergence? I'm doing something wrong in the previous calculations?</p>
|
<p>This confusion is caused by vector calculus. You should treat each component separately, and then it is obvious. For example, for the x component:</p>
<p>$$ \int_V \partial_x p dx dy dz = \int_{\partial V} p(x) dy dz = \int_{\partial V} p dS_x $$</p>
<p>by the fundamental theorem of calculus (do the x integral first). Likewise for the other components. You can make up a proof for this from the divergence theorem by introducing the fictitious vector field</p>
<p>$$ Q = (p,0,0) $$</p>
<p>And then the divergence of Q is the left hand side, while the right hand side is $Q\cdot dS$. But it's really just the fundamental theorem of calculus.</p>
| 869
|
fluid dynamics
|
Water coming out of dam gates
|
https://physics.stackexchange.com/questions/343249/water-coming-out-of-dam-gates
|
<p>I recently went to a dam and when I saw the water coming out of its gates with huge pressure. The water instantly got converted to cloudy fog and there was fog all around. Why and how this water gets converted to fog? Does pressure has to do something with this?</p>
|
<p>Sudden decrease in pressure causes some part of water to vaporise or split into small droplets
But if the dam is high on the ground it can also happen as water falling hits air and spreads as seen in waterfalls</p>
| 870
|
fluid dynamics
|
Centripetal forces acting on fluid flow in a bent pipe?
|
https://physics.stackexchange.com/questions/208430/centripetal-forces-acting-on-fluid-flow-in-a-bent-pipe
|
<p>Yesterday I answered <a href="https://physics.stackexchange.com/questions/208353/force-applied-to-a-90-degree-elbow-by-hydraulic-flow/208362#208362">this question</a>, using analysis of forces <a href="http://www-mdp.eng.cam.ac.uk/web/library/enginfo/aerothermal_dvd_only/aero/fprops/cvanalysis/node49.html" rel="nofollow noreferrer">from this web page</a>.</p>
<p>But today I started having doubts regarding the <em>completeness</em> of this analysis. It occurred to me this approach does not take into account any centripetal forces caused by the changing direction of the fluid flow. Simply put:</p>
<p>$\vec{F}=m\frac{d\vec{v}}{dt}$.</p>
<p>To analyse this, consider the following geometry for a $90^0$ bend in a pipe lying in a horizontal $(x,y)$ plane:</p>
<p><a href="https://i.sstatic.net/j8wSu.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/j8wSu.png" alt="Bent pipe."></a></p>
<p><strong>Assumptions:</strong></p>
<ul>
<li>turbulent plug flow with constant fluid speed $v$, fluid density $\rho$, cross-section of pipe $A$, bend radius $R$.</li>
<li>$\frac{R}{A}\gg 1$</li>
</ul>
<p>Right now I am only concerned with centripetal forces needed to keep the flow on its rotational trajectory.</p>
<p>Take an infinitesimal element at $\theta$:</p>
<p>$dm=\rho A Rd\theta$.</p>
<p>The centripetal forces are:</p>
<p>$dF=-\frac{v^2 dm}{R}=-\rho A v^2 d\theta$</p>
<p>$dF_x=-\rho A v^2\cos\theta d\theta$.</p>
<p>$dF_y=-\rho A v^2\sin\theta d\theta$.</p>
<p>So these would be the forces the pipe needs to exert on the fluid:</p>
<p>$F_x=-\rho A v^2\int_0^{\frac{\pi}{2}}\cos\theta d\theta=-\rho A v^2$.</p>
<p>$F_y=-\rho A v^2\int_0^{\frac{\pi}{2}}\sin\theta d\theta=-\rho A v^2$.</p>
<p>If this is correct then I obviously need to amend my answer. <strong>So my question is: is this correct?</strong></p>
|
<p>I got the same answer using momentum considerations.</p>
<p>In time $dt$ the amount of mass hitting the side of the pipe is: </p>
<p>$$dm=\rho A v dt$$</p>
<p>Now, we know that the force must satisfy:</p>
<p>$$F_{x}dt=dp=0-vdm=-\rho A v^2 dt $$</p>
<p>Dividing both sides by $dt$ gives:</p>
<p>$$F_{x}=-\rho A v^2$$</p>
<p>The force in the $y$ direction is the same as here. </p>
| 871
|
fluid dynamics
|
Angular velocity of a fluid element
|
https://physics.stackexchange.com/questions/210072/angular-velocity-of-a-fluid-element
|
<p>If we have a fluid element that is subjected to:</p>
<ul>
<li>Translation</li>
<li>Rotation</li>
<li>Extensional strain (dilatation)</li>
<li>Shear strain</li>
</ul>
<p>As in this picture starting from time <span class="math-container">$t$</span>
<a href="https://i.sstatic.net/eIXgz.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/eIXgz.png" alt="enter image description here" /></a></p>
<p>it can be shown that angles <span class="math-container">$d\alpha$</span> and <span class="math-container">$d\beta$</span> equal <span class="math-container">$\frac{\partial v}{\partial x} dt$</span> and <span class="math-container">$\frac{\partial u}{\partial y} dt$</span> respectively, most textbooks I've encountered defines the rate of rotation (angular velocity) of this fluid element as the average of the rate of rotation of the two angles (the minus sign due to the difference in rotation directions):
<span class="math-container">$$\omega = \frac{1}{2}\left (\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}\right )$$</span></p>
<p>My question is about the accuracy of this definition, how do we know that the rate of rotation of the fluid element is simply the rate of the arithmetic mean of <span class="math-container">$d\alpha$</span> and <span class="math-container">$d\beta$</span>? What if one side is deforming at very much high speed than the other?</p>
|
<p>It's important to note that these displacements are all infinitesimal, so while one may be thousands of times the other, during an infinitesimal time slice the displacement is still infinitesimal. If one displacement is much much greater than the other, then that displacement will effectively be half rotation and half distortion, if you rotated the fluid back, then the two axes would displace equally. The reason there's no square term or sines or cosines is that this is a linear approximation, that works because all of the displacements are infinitesimal.</p>
| 872
|
fluid dynamics
|
Why jets entrain mass?
|
https://physics.stackexchange.com/questions/244864/why-jets-entrain-mass
|
<p><a href="https://i.sstatic.net/FztW0.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/FztW0.jpg" alt="enter image description here"></a></p>
<p>This very beautiful picture shows a jet of fuel in the combustion chamber of a diesel engine. Does anydody could explain me what is the physical reason that the jet entrains air? Why jets and plumes entrain mass from the environment?
Edit. I believe it has to do with the conservation of momentum and i always wait for the happy day when people will stop mentioning the Bernoulli equation.</p>
|
<p>Your many questions shows remarkable effort on questioning. You seem to note the point.</p>
<blockquote>
<p>Does anydody could explain me what is the physical reason that the jet entrains <strong>air</strong>? </p>
</blockquote>
<p><strong>"Diffusion"</strong> 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. <a href="https://www.youtube.com/watch?v=1_oyqLOqwnI" rel="nofollow">This video on 14:00-></a></p>
<p><strong>"Celerity"</strong> is nothing else <a href="https://en.wikipedia.org/wiki/Celerity" rel="nofollow">than saying it happends fast</a>. But why should it happen fast?</p>
<p>I bolded "air" on your question. You are seeking the solution for Turbulence. I mostly avoid linking <a href="https://www.researchgate.net/publication/283451471_Turbulence" rel="nofollow">this paper of mine here</a>, as it's not Mainstream Physics; Mainstream Physics can't explain turbulence. I would also like to link you these <a href="https://www.youtube.com/playlist?list=PLgUc9kJnDMMExJivT2dWh9dAjdYYUgOFE" rel="nofollow">videos of mine</a>. As it seems that you are able to ask you through this, up to the solution.</p>
<p>Turbulence is fluid Split in parts. Surfaces inside a fluid. Cracks which aloud the air to be Entrained. These cracks are caused by the <strong>"friction"</strong> this part of the first answer is correct. I rather use word collision. Look the Prince Rubert drops video in my playlist. And you sure find a certain <strong>celerity.</strong></p>
| 873
|
fluid dynamics
|
I want to increase pressure of water in a closed vessel
|
https://physics.stackexchange.com/questions/249734/i-want-to-increase-pressure-of-water-in-a-closed-vessel
|
<p>I am trying to make a high pressure water jet using DIY hand pump which pushes water into a sealed vessel with non-return valve at the inlet. I am planning to do it by first closing the valve to the nozzle until i pump in enough water into the vessel so that pressure is created.then after open the nozzle valve to create a jet at very high pressure. I have never done it before, Is it possible? Please I need guidance...What is the maximum pressure I can raise? I need to raise upto 100 PSI. Thank you. I have attached a picture to illustrate the idea.<a href="https://i.sstatic.net/pNBd3.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/pNBd3.png" alt="enter image description here"></a></p>
|
<p>I recommend that you add some air in your system. Water is mostly incompressible - so while you can raise the pressure to 100 psi quite easily, the moment you open the nozzle and a little bit of water leaves, the pressure will drop right down. As you add air, the air will expand and maintain pressure in the vessel (when the air volume doubles, the pressure will drop in half, roughly).</p>
<p>However, such air volume gives the system certain ballistic properties: if the pressure container fails, there will be more oomph behind the fragments flying around. Making good pressure vessels is not an amateur pastime. </p>
<p>The kind of thing you are looking for is commonly used with irrigation pumps - between the pump and the sprinkler is a vessel with an air bladder in it. As you pump water in, the air is compressed. This does exactly what you are talking about.</p>
<p><a href="http://www.sprinklerwarehouse.com/DIY-Pressure-tanks-s/7024.htm" rel="nofollow">Here is a link that describes this in more detail</a></p>
| 874
|
fluid dynamics
|
Efficient way to move water pumping vs circulation
|
https://physics.stackexchange.com/questions/249742/efficient-way-to-move-water-pumping-vs-circulation
|
<p>I have a basic question about efficiency of pumping. </p>
<p>If I have a well of 100 meters depth, but water filled till top, and I want to circulate the water. What will be most energy efficient. A) Pumping the water to 100 meters head and then discharging the water at top of that well but below water level or B) having a discharge pipe run all the way down to the same level as intake pipe.</p>
<p><a href="https://i.sstatic.net/381nb.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/381nb.png" alt="enter image description here"></a></p>
<p>Questions</p>
<p>1) with option A, If the well is filled to the top, and discharge point is below water level. Then will the Total Dynamic Head be 10m or 110m and what will be energy required pump 1KG of water.</p>
<p>2) with option B, what will be the energy required ?</p>
<p>3) any other major influences on the pumping energy ? like pressure if the intake depth increases but water level remains same ?</p>
|
<p>Both will be of the same efficiency. It only matters how far you raise water above the surface. Where you take it from within the bulk of the water is irrelevant.</p>
| 875
|
fluid dynamics
|
How does a fluid force behave at different angles?
|
https://physics.stackexchange.com/questions/249805/how-does-a-fluid-force-behave-at-different-angles
|
<p>I have a mock exam problem that deals with fluid forces acting at different angles. Here are the scans:
<a href="https://i.sstatic.net/PE9AW.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/PE9AW.png" alt="enter image description here"></a>
<a href="https://i.sstatic.net/vAu0n.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/vAu0n.png" alt="enter image description here"></a>
Based on the description of the systems, I know that in (a) and (c), F is going to equal the sum of the weight of the piston and the force exerted by the water on the surface of the piston. In (b), the weight of the piston is negligible so F is equal to just the force exerted by the water on the piston.</p>
<p>This is what I got for F in (a):</p>
<p>$$
F = mg + \rho_{water}XY(H-d)g
$$</p>
<p>The solution to (b) is essentially the same, except for the weight of the piston. My question is about (c)... I think it's a trick question. The fluid force acts at a 45 degree angle to the surface, so the y-component will be smaller than in (a). However, there is more surface area.</p>
<p>To me it seems intuitive that the y-component of fluid force will be weaker in (c) than it is in (a), but in my rough calculations they wind up being essentially the same. So how does it work? Are they actually the same?</p>
<p>Thank you!</p>
|
<p>The solution to part (B) is not same as part (a) because the fluid pressure along YZ plane would vary as $$P=P_o+\rho(gh)$$ <br><br>Thus<br><br>$force_{yz}=\int_{y}^{z}(P_o+\rho(gy))Zdy$<br><br>$$force_{yz}=P_o(Z)(Z-Y)+\frac{\rho(g)(Z)(Z^2-Y^2)}{2}$$
Therefore force on YZ plane would have to found by integration.<br><br> On similar grounds, force in part(c) would have to found by integration keeping in mind the fact that only Y component of force is taken into account(i.e $cos45$). </p>
| 876
|
fluid dynamics
|
How much load does an aqueduct support?
|
https://physics.stackexchange.com/questions/250104/how-much-load-does-an-aqueduct-support
|
<p>Recently, I found out about /r/InfrastructurePorn, and I found a particularly interesting photo of the Gouwe Aqueduct in Gouda, NE: </p>
<p><a href="https://i.sstatic.net/HG7Wi.jpg" rel="noreferrer"><img src="https://i.sstatic.net/HG7Wi.jpg" alt="enter image description here"></a></p>
<p>It seems like the bridge that is supporting the boat wouldn't be able to do it. Is the weight of the actual boat being supported by the aqueduct? </p>
|
<p>The ship is floating and so is supported by the upthrust due to the water and so the weight of water displaced by the ship.<br>
If the ship travels very slowly so that the level of water does not rise but rather flows away then the weight supported by the aqueduct does not change between the ship present and no ship present situation.<br>
In practice I would imagine the water level does rise but probably by only a little.</p>
| 877
|
fluid dynamics
|
A question about droplets formation
|
https://physics.stackexchange.com/questions/251934/a-question-about-droplets-formation
|
<p><a href="https://i.sstatic.net/Eptst.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Eptst.jpg" alt="enter image description here"></a></p>
<p>This picture shows a fuel ejector of a diesel engine that propels ships.</p>
<p>When the fuel line pressure exceeds 380 bars, the tension of the spring is overcomed and fuel is ejected into the combustion space.</p>
<p>It is common knowledge among engineers and i have read it on manuals that tension of the spring could be adjusted to lift in higher pressures but above this pressure the jet will not penetrate further; instead it will break into smaller droplets.</p>
<p>My question is; Why after a critical value, the surplus energy/pressure we give to a flow does not translate to kinetic energy increase but creates smaller droplets, in other words surfaces?</p>
<p>The flow is sonic, if that is of any help and the speed of the flow is about 500m/sec</p>
| 878
|
|
fluid dynamics
|
Pressure drop through a flat vs. pleated filter
|
https://physics.stackexchange.com/questions/254754/pressure-drop-through-a-flat-vs-pleated-filter
|
<p>Consider two identical pipes where two identical air flows take place. The pipes are "obstructed" by two filters made of the same fabric, one flat, the other pleated as shown in the following picture.</p>
<p><a href="https://i.sstatic.net/u7WKp.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/u7WKp.png" alt="enter image description here"></a></p>
<h2>Qualitatively</h2>
<p>In which pipe will the pressure drop be larger? I see two opposing factors:</p>
<ul>
<li>The folded filter has more "holes" overall, so its hydrodynamic resistance is lower.</li>
<li>If you imagine the filter fabric to be made of small channels normal to its surface, air undergoes more shear to cross the pleated filter than to cross the flat one.</li>
</ul>
<p>Mechanical ventilation systems commonly use pleated filters for fine fabrics: is it only to increase their lifespan (larger area means it takes longer to seal them up), or does it also reduce pressure drop?</p>
<h2>Quantitatively</h2>
<p>What section should a flat filter have to yield the same pressure drop as a pleated filter? This question boils down to: Are there formulas or charts to compute the pressure drop as a function of dimensions $a$ and $t$?</p>
|
<p>In a first approximation, and if you have a laminar flow in the pipes, the flux will be $Q=\frac{-\kappa A}{\mu} \frac{(p_b - p_a)}{L}$, from <a href="https://en.wikipedia.org/wiki/Fluid_flow_through_porous_media" rel="nofollow">Darcy's law</a>, see notations there. So increasing the area $A$ will increase the flux in proportion. Conversely, as you can see from the equation, the pressure drop is inversely proportional to $A$, if you impose flux. You can easily calculate $A$ from $a$ and $t$.</p>
| 879
|
fluid dynamics
|
Why is the velocity of fluid on top a hydrofoil higher than that on the bottom? What keeps enlarging the difference in these 2 velocities?
|
https://physics.stackexchange.com/questions/258858/why-is-the-velocity-of-fluid-on-top-a-hydrofoil-higher-than-that-on-the-bottom
|
<p>I am trying to learn how hydrofoil works. A big part of it is Bernoulli's principles.</p>
<p>I found this <a href="http://web.mit.edu/2.972/www/reports/hydrofoil/hydrofoil.html" rel="nofollow noreferrer">article</a> on MIT website.</p>
<p>There are two explanation for how it works:</p>
<ol>
<li>The conservation of linear momentum or Newton's third law.If water is pushed down, the water pushes back on the hydrofoil.
<a href="https://i.sstatic.net/uqKlr.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/uqKlr.png" alt="enter image description here"></a></li>
</ol>
<p>For the change for momentum of water, their is a equal but opposite change in momentum of the hydrofoil, directed upward.</p>
<ol start="2">
<li>Bernoulli'principles: $$P_o=P+1/2dv^2+dgy$$</li>
</ol>
<p>$P_o$ is called the stagnation of pressure and is a constant. P is pressure. d is density. v is velocity of fluid. and g is gravitational acceleration. So if v increases on top, the top pressure will decreases, and as v of fluid decreases on the bottom, the pressure with increase. The hydrofoil then starts to rise when the bottom pressure overcomes the top pressure. It rises till there is no water on top.</p>
<p><strong>QUESTION</strong></p>
<ol>
<li><p>Why the velocity of fluid on top is greater than that of the fluid on bottom? The link says "this is due in part to viscous effects which lead to formation of vertices at the end of foil." But I am not sure what those "effect" and "vertices" are.</p>
<ol start="2">
<li>What keeps the top velocity increasing, and the bottom velocity decreasing?
I guess it's this enlarging difference in velocities that produces larger and larger pressure difference, and thus larger net force.</li>
</ol></li>
</ol>
| 880
|
|
fluid dynamics
|
Attractive force between hose and floor when a pool is filled
|
https://physics.stackexchange.com/questions/262495/attractive-force-between-hose-and-floor-when-a-pool-is-filled
|
<p>I was filling a rubber pool for my baby using a hose. But I noticed that if I put the hose outlet directly with the floor there is an attractive force!
This is strange, because the hose is throwing water, and by action/reaction it should experiment a repulsive force upward (...I noticed this repulsive force when the hose was a little separated from the floor).
How can this phenomenon be explained?</p>
<p><a href="https://i.sstatic.net/gE1qH.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/gE1qH.png" alt="enter image description here"></a></p>
|
<p>An apparently simple situation can be extremely complex (effects from nozzle on tube, possible helical flow inside tube, ram pump effect, cavitation...), but the most common attractive force in this situation is the <strong>Venturi effect</strong>.
Please check <a href="https://en.wikipedia.org/wiki/Venturi_effect" rel="nofollow">the Wikipedia article</a> and <a href="https://www.youtube.com/watch?v=Na9ORhYjvJU" rel="nofollow">this vidéo</a>.
or if this video is not available from where you are, [via a search engine] (<a href="https://duckduckgo.com/?q=venturi+effect&ia=videos" rel="nofollow">https://duckduckgo.com/?q=venturi+effect&ia=videos</a>)</p>
| 881
|
fluid dynamics
|
How does an hydraulic ram pump work?
|
https://physics.stackexchange.com/questions/264398/how-does-an-hydraulic-ram-pump-work
|
<p>Hydraulic ram (<a href="https://en.wikipedia.org/wiki/Hydraulic_ram" rel="nofollow noreferrer">wikipedia</a>)(<a href="https://www.youtube.com/results?search_query=ram%20pump" rel="nofollow noreferrer">youtube</a>) uses the water hammer (<a href="https://en.wikipedia.org/wiki/Water_hammer" rel="nofollow noreferrer">wikipedia</a>) to pump water.</p>
<p>Please help me understand how it works.<br>
I understand that the check valve catches the pressure from the water hammer (since when the pressure below the valve is higher then above it, it opens)<br>
What i do not understand is why does the waste valve <strong>open</strong> (and close)? (and how does it sync with the check valve?)<br>
I guess it has to do with the air chamber (I do not exactly understand what is it for), but I do not understand why does it open?</p>
<p>Images are from <a href="http://www.rampumps.net/face/20120410150106.html" rel="nofollow noreferrer">here</a><br>
<a href="https://i.sstatic.net/ZWuwi.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ZWuwi.jpg" alt="enter image description here"></a> </p>
<p><a href="https://i.sstatic.net/iLnms.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/iLnms.gif" alt="enter image description here"></a></p>
|
<p>I doubt if the person who made that animated gif knew how the pump works. It's not at all clear why the valves open and close, and the regular movement of the water is nothing like an actual ram pump. It's more likely to confuse than enlighten people imo. </p>
<p><a href="https://i.sstatic.net/PSWtD.gif" rel="noreferrer"><img src="https://i.sstatic.net/PSWtD.gif" alt="enter image description here"></a>
This one (found on <a href="http://www.meribah-ram-pump.com" rel="noreferrer">http://www.meribah-ram-pump.com</a>) does a better job showing the strong waterhammer pulse. </p>
<p>The waste valve can be a spring-operated normally-open valve that closes when the flow velocity generates sufficient drag. The water in the drive pipe, which is long and rigid (steel pipes will generate higher pressure than pvc), has gathered significant momentum (p=m*v) by that time.
When the valve closes, that momentum generates pressure, slamming open the check valve and compressing the air in the reservoir. The check valve closes again, the compressed air expands, pushing water up the delivery pipe. The waste valve reopens and the process repeats.</p>
<p>The air reservoir stores the energy, absorbs the pressure pulses, and provides continuous flow at the delivery pipe. It's comparable to a smoothing capacitor: can handle any incoming current, absorbs voltage (pressure) peaks, and delivers the energy during the whole cycle. Without the air, every time the waste valve closes, the two water columns are in direct contact, one at full speed, the other at rest. That will cause either large pressure peaks (when they meet head on), or large movement of the pump. At 30,000 times a day, they may not last very long. </p>
<p><strong>But the fundamental difference between a pump with air reservoir and one without is the efficiency</strong>: without reservoir, energy is transferred by a <a href="https://en.wikipedia.org/wiki/Inelastic_collision#Perfectly_inelastic_collision" rel="noreferrer">perfectly inelastic collision</a>: the water in the drive pipe and the water in the delivery pipe have the same velocity after the collision. Not only is that the collision where most energy is lost, the mass in the drive pipe is also the largest mass, and will keep most of its energy after the collision. </p>
<p>With a reservoir, most of the energy of the drive mass is transferred to the reservoir, and all that energy is used to pump the water in the delivery pipe. </p>
| 882
|
fluid dynamics
|
How much input fluid is needed to fully replace the contents of my inline chamber
|
https://physics.stackexchange.com/questions/272878/how-much-input-fluid-is-needed-to-fully-replace-the-contents-of-my-inline-chambe
|
<p>I have built a simple inline chamber for my irrigation system. I have made it so I can add powdered fertilizer to the chamber which will then be pumped through the system to the drip feed sprinklers. What I need to know now is how many liters of water must enter the chamber to replace the original contents. The inside diameter of the chamber is 88mm and the length is 596mm. That gives me approximately 3.6 liters (ignoring the small volume for the thread cap in the center). Now both the input and output hoses are 13mm (outside diameter) which connect to two taps both are 28mm (outside diameter) as they pass through the chamber ends. You can see it below
<a href="https://i.sstatic.net/17S9a.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/17S9a.jpg" alt="enter image description here"></a>
I expect the input (on the right) to mix with the fluid in the chamber which means I will need to push more than 3.6 liters through the system to get the bulk of the fertilizer out of the chamber. I also expect the concentration of the fertilizer to drop relatively quickly at first but slow down as the mixing process continues to dilute the mix in the chamber. What I need is a rough estimate of how many liters I need to push through the chamber to get the majority of the fertilizer through.</p>
|
<p>The answer will really depend on the degree of mixing that happens. If there is "no" mixing in the chamber, then the entire volume $V$ will be replaced by pushing just $V$ additional fluid through. But the question is more interesting when you have perfect mixing - so let's set up that problem and do the math:</p>
<p>A chamber of volume $V$ has a certain initial concentration $C_0$ of fertilizer. We add $q$ water per unit time, and remove an equal volume of water-plus-fertilizer. What is the concentration of fertilizer $C(t)$ as a function of time?</p>
<p>In a short time interval $dt$ during which we assume the concentration is constant, the quantity of fertilizer removed is $C(t)~q~dt$. This means the equation for concentration with time is</p>
<p>$$C(t+dt) - C(t) = -\frac{C(t) q}{V} dt\\
\frac{dC}{dt} = -\frac{Cq}{V}$$</p>
<p>We can integrate this to get</p>
<p>$$C(t) = C_0 e^{-qt/V}$$</p>
<p>That means that every time you get a volume $V$ of liquid through your mixing chamber, the concentration drops by a factor $1/e$ or roughly 37%.</p>
<p>To get down to 1% of the initial concentration, you would need 4.6 V of water, since $e^{-4.6} = 0.01$</p>
<p>I hope you can work with this. What exactly consists of "majority of fertilizer" is of course something you need to decide for yourself. The table below may help:</p>
<pre><code>conc volume
0.001 6.91 x V
0.003 5.76 x V
0.010 4.61 x V
0.032 3.45 x V
0.100 2.30 x V
0.316 1.15 x V
1.000 0.00 x V
</code></pre>
| 883
|
fluid dynamics
|
Question about a fountain
|
https://physics.stackexchange.com/questions/299136/question-about-a-fountain
|
<p><img src="https://i.sstatic.net/GnUeT.jpg" alt="enter image description here"></p>
<p>There's this fountain where I work. I haven't really studied fluid dynamics, so my question is about how it works. </p>
<p>The way the fountain is designed, is that there is one long big pipe that runs down the middle of the fountain. There are holes drilled into the pipe about every 8 inches or so. Into this holes goes a little clear plastic piece of pipe making a jet for the water to go straight up.</p>
<p>My question is: how are all the "jets" pushing water to about the same height? Intuition would tell me that the water would be higher at one end or the other..the end that the Lon main pipe is connected to the pump. I would assume the water height of the jets to taper off the further away from the pump.</p>
<p>I hope that makes sense.</p>
<p>Thanks,
Kyle</p>
|
<p>There would be a slight difference in height, but obviously they design these to be negligible.</p>
<p>The key issue is the drag on the water from the inside surface of the pipe. If you were to explore this system in a static situation (no water movement), you would find the pressure at the outlets of each fountain was <em>exactly</em> the same. However, as we have water movement, we see drag. This drag will decrease the pressure the final fountain outlet sees.</p>
<p>One solution to this is to use a large enough pipe to simply ignore this effect entirely. Another is to counteract the effect by making the pipe more narrow as you go to convert some of the water's dynamic pressure due to velocity into static pressure (which is what you need to spray out of the outlet).</p>
<p>This effect can be observed in several places. One is a children's toy for the pool, which shoots a large stream of water using this effect.</p>
<p><a href="https://i.sstatic.net/ajOAU.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ajOAU.jpg" alt="Water Cannon Toy"></a></p>
<p>And another place you can see it is in air conditioning vents. Next time you are in a large commercial building where the air conditioning vents are exposed, take a good look at them. The vents are very large near the air conditioner itself, and at every outlet (or nearly every outlet), they decrease the diameter of the venting. This keeps the output of each vent equal. You can see the vent go from right to left in this picture.</p>
<p><a href="https://i.sstatic.net/3vFnHm.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/3vFnHm.jpg" alt="Air conditioning"></a></p>
| 884
|
fluid dynamics
|
Why the excess pressures are equated in hydraulic press?
|
https://physics.stackexchange.com/questions/302454/why-the-excess-pressures-are-equated-in-hydraulic-press
|
<p>Let's say we have a closed chamber filled with incompressible liquid and there are two area's(smaller(a) and larger(A)) where we can apply force or keep objects.We keep an object of weight W on larger area.Due to this,pressure at every point in a liquid will increase by W/A ,according to pascal's law.The situation is depicted in picture below.</p>
<p><a href="https://i.sstatic.net/0TJZT.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/0TJZT.png" alt="enter image description here"></a>
Now,we apply force f on smaller area .Due to this the pressure at every point will even further increase by f/a.Now the situation looks something like the picture below.</p>
<p>Now from what I have read in books
W/A=f/a</p>
<p>Why the pressure exerted by the box W has to be equal to the pressure exerted by force f?</p>
<p><a href="https://i.sstatic.net/uUl9D.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/uUl9D.png" alt="enter image description here"></a></p>
|
<p>This equality is a equilibrium condition, i.e. a condition for the fluid to be in rest. In general, inertia force terms must be added if the fluid is not in equilibrium.</p>
| 885
|
fluid dynamics
|
Manometer Reading from Fluid Dynamics
|
https://physics.stackexchange.com/questions/305140/manometer-reading-from-fluid-dynamics
|
<p>so I have been working on this problem :</p>
<ol>
<li>Diameter1 at wide end: 8cm || V1 = 1.56m/s</li>
<li>Diameter2 at narrow end: 3cm || V2 = 11.094m/s</li>
</ol>
<p><strong><em>Find the manometer reading</em></strong></p>
<p><a href="https://i.sstatic.net/ZKpT0.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ZKpT0.jpg" alt="enter image description here"></a></p>
<p>I know that due to Bernoulli's Equation the pressure where velocity is at 11.094m/s is much lower than at the other end. This can also be seen from the Mercury(Hg) level. </p>
<p><strong>I have take P2 as being 0 because it is open to the atmosphere -- I am unsure whether this should be taken as 0 or 1.</strong></p>
<p>I have used the following equation to work out P1 :</p>
<pre><code>P1+(1/2 ρv1^2)=P2+(1/2 ρv2^2) and I end up with P1 = 60.321kPa
</code></pre>
<p>Can I use : 𝑃1 − 𝑃2 = (𝜌2 −𝜌1)𝑔ℎ in order to find the height ? Or is thus equation only valid for a differential manometer that has a constant cross sectional area ?</p>
<p>I am still unsure how to proceed in this case. Any help or tips would be appreciated.</p>
<p><strong>EDIT</strong> This question is not related to finding any sort of velocity as asked in the possible related question. I am interested in the manometer reading and if the equation 𝑃1 − 𝑃2 = (𝜌2 −𝜌1)𝑔ℎ is relevant in this case.</p>
|
<p>For a problem like this, you first calculate the velocity in the two sections - the ratio of the velocities is the inverse of the ratio of the areas for an incompressible fluid.</p>
<p>The manometer will read a pressure that depends on the difference in density between mercury and water, so</p>
<p>$$\Delta p = \Delta \rho g h$$</p>
<p>The absolute pressure doesn't matter here, just the difference. So you can rewrite this as</p>
<p>$$\frac12 \rho_w (v_2^2 - v_1^2) = (\rho_m - \rho_w) g h$$</p>
<p>You asked:</p>
<blockquote>
<p>Can I use : 𝑃1 − 𝑃2 = (𝜌2 −𝜌1)𝑔ℎ in order to find the height ?</p>
</blockquote>
<p>The short answer is "yes", assuming you are careful with your signs. $P_1<P_2$, but $\rho_1 < \rho_2$ so in your expression you would get a negative $h$ but the way you drew it, the value looks positive. But that's easy enough to get right.</p>
| 886
|
fluid dynamics
|
Simplest model of simplest organism swimming in fluid
|
https://physics.stackexchange.com/questions/313395/simplest-model-of-simplest-organism-swimming-in-fluid
|
<p>(Hopefully this question isn't too elementary for this StackExchange.)</p>
<p>Given a micro-organism we'll define as two "cells" (which are the
atomic things in this model):</p>
<ol>
<li>A "main" larger cell: <code>O</code></li>
<li>A "tail" smaller cell: <code>o</code></li>
</ol>
<p>The <code>o</code> is connected by a locus which has free movement around the
circumference of the <code>O</code>. It looks like this:</p>
<pre><code>O
o
</code></pre>
<p>Swimming action in fluid might look like this (think of a spermatozoa,
but with an even simpler tail):</p>
<pre><code> O
o
O
o
O
o
</code></pre>
<p>For the purpose of modeling this swimming motion in a computer
program, I'm trying to figure out the physics of it and the minimum
necessary to simulate realistic movement. Assumptions:</p>
<ol>
<li>A 2D plane.</li>
<li>A certain level of friction (the fluid is e.g. water).</li>
<li>No buoyancy; the organisms are fully submerged and are at
equilibrium with the surrounding water.</li>
<li>The body as a whole has a momentum and direction at all times for
calculating its next position. The mass of <code>O</code> is 2 and the mass of
<code>o</code> is 1.</li>
</ol>
<p>It's my intuitive understanding that, for example, when the tail moves
by n°, the fluid which was surrounding the tail is displaced on both
sides:</p>
<pre><code>Clockwise rotation
O
Fluid here moves away, -> o <- Fluid moves in to fill the
but creates resistance. gap created by the tail.
</code></pre>
<p>That resistance lets the <code>O</code> cell rotate. So in order to rotate itself
anti-clockwise, it moves the tail clockwise.</p>
<p>But that just lets the organism rotate left and right. Here is where
I'm stuck. How does an organism like this create momentum other than
rotation? Is there a natural movement that comes from rotation? Do we
need a third cell to extend the tail? If so, why?</p>
<p>Supposing that there is a natural movement coming from this action in
a fluid, what's the formula for that? Supposing <code>O</code> is at <code>(x1,y1)</code>,
and <code>o</code> is at <code>(x2,y2)</code>, <code>O</code> has radius <code>r1</code> and and <code>o</code> has <code>r2</code>, and
we do n° of rotation with mass <code>1</code> for <code>o</code> and mass <code>2</code> for <code>O</code>, at
velocity <code>v</code>. I would imagine there's something from fluid dynamics
that deals with how to calculate the whole body's new momentum and
direction?</p>
<p>I've been looking at this page on Bacterial motility and behavior,</p>
<p><a href="http://www.rowland.harvard.edu/labs/bacteria/movies/showmovie.php?mov=tethered_strep" rel="nofollow noreferrer">http://www.rowland.harvard.edu/labs/bacteria/movies/showmovie.php?mov=tethered_strep</a></p>
<p>which states that:</p>
<blockquote>
<p>If a single flagellar filament, hook or polyhook is attached to the
substratum, the cell body will spin (Silverman and Simon, 1974).</p>
</blockquote>
<p><a href="https://i.sstatic.net/KHYGD.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/KHYGD.png" alt="enter image description here"></a></p>
<p>So it seems that, indeed, tethering the <code>O</code> to a point would make it
spin, but that being unpinned would automatically make it move in some
direction.</p>
<p>For your entertainment, E coli. wiggle to move around, as you can see
in this funny video:</p>
<p><a href="http://www.rowland.harvard.edu/labs/bacteria/movies/showmovie.php?mov=laser_trap" rel="nofollow noreferrer">http://www.rowland.harvard.edu/labs/bacteria/movies/showmovie.php?mov=laser_trap</a></p>
<p>They use a laser to capture one E coli and move it around, finally
after releasing it, it wiggles away. I believe that wiggling is, in
the model defined here, a series of well-timed rotations at n loci
of m cells; so we come back to the base question.</p>
<p>My hope is to model organisms with arbitrary connected cells which
perform rotation actions, and for their movement in the fluid to be
realistic, if not in truth accurate by skimping on details.</p>
| 887
|
|
fluid dynamics
|
Expressions for Analytical Solutions of the Sedov blast wave problem?
|
https://physics.stackexchange.com/questions/601690/expressions-for-analytical-solutions-of-the-sedov-blast-wave-problem
|
<p>There is a large amount of expository material out there about the self similar solution to the Sedov Blast wave problem, and I am interested in obtaining the analytical expression so I can code it up and compare it to my numerical results.</p>
<p>From Chapter 17 of Shu's "The Physics of Astrophysics (Vol 2)", we consider a point release of a large amount of energy <span class="math-container">$E$</span> into a static medium of homogenous density <span class="math-container">$\rho_1$</span>. By self similarity arguments we can obtain a variable <span class="math-container">$\xi = r\left(\frac{\rho_1}{Et^2}\right)^{1/5}$</span> (17.1 of Shu) and ultimately define explicit expressions for the primitive variables of the governing equations</p>
<p><span class="math-container">\begin{align}
\rho(r, t) & = \frac{\gamma + 1}{\gamma - 1}\rho_1\alpha(\xi) \\
u(r, t) & = \frac{4}{5(\gamma+1)}\frac{r}{t}v(\xi) \\
P(r, t) & = \frac{8}{25(\gamma + 1)}\rho_1\left(\frac{r}{t}\right)^2p(\xi)
\end{align}</span></p>
<p>where <span class="math-container">$\gamma = 1.4$</span> (17.10-17.12 of Shu). He then goes on to say that Sedov (miraculously!) derived analytical expressions for <span class="math-container">$\alpha(\xi), v(\xi), p(\xi)$</span>.</p>
<p>There are multiple writeups that give that present the same equations with slightly different notation, and then pass the buck to Sedov's original monograph for these expressions. Such as equations (5.23)-(5.25) of <a href="http://www.astronomy.ohio-state.edu/%7Eryden/ast825/ch5-6.pdf" rel="nofollow noreferrer">http://www.astronomy.ohio-state.edu/~ryden/ast825/ch5-6.pdf</a>, or (10.17) - (10.20) of <a href="https://www.ita.uni-heidelberg.de/%7Edullemond/lectures/num_fluid_2011/Chapter_10.pdf" rel="nofollow noreferrer">https://www.ita.uni-heidelberg.de/~dullemond/lectures/num_fluid_2011/Chapter_10.pdf</a>. The notes <a href="http://www.pmaweb.caltech.edu/Courses/ph136/yr2012/1217.2.K.pdf" rel="nofollow noreferrer">http://www.pmaweb.caltech.edu/Courses/ph136/yr2012/1217.2.K.pdf</a> mention that these expressions can be found in Section 99 of Landau and Lifshitz. I looked there and in Sedov's Monograph and am not able to easily tell what they should be.</p>
<p>Can somebody help me make this more explicit? My background is in numerics so this literature is all quite unfamiliar to me.</p>
| 888
|
|
fluid dynamics
|
Is the pressure of an incompressible fluid discontinuous or continuous across an aperture, like the nozzle of a hose?
|
https://physics.stackexchange.com/questions/607513/is-the-pressure-of-an-incompressible-fluid-discontinuous-or-continuous-across-an
|
<p>Is the pressure of an incompressible fluid discontinuous or continuous across an aperture? E.g. inside the fire hose water is at high pressure, outside of it it's at 1atm, but what does the transition look like at the aperture of the nozzle?</p>
<p>Since water is only <em>mostly</em> incompressible I'm sure there's at least continuity to the extent that it was very slightly compressed in the hose and returned to a very slightly greater volume once outside. But what about an idealized incompressible fluid?</p>
<p>If it's not discontinuous, are there formulae for how it looks for different flow types?</p>
|
<p>The flow in and around a fire hose can be modelled in a similar way to a jet in the vicinity of a wing - a configuration studied by Carl Shollenberger and Peter Lissamon in 1971.</p>
<p><a href="https://thesis.library.caltech.edu/6018/" rel="nofollow noreferrer">Shollenberger_PhDThesis</a></p>
<p>In this idealization, there is a jump in total pressure across a thin actuator disk at the inlet, and there are velocity discontinuities (modelled by vortex sheets) across the wake boundaries, but the static pressure is continuous everywhere.</p>
<p><a href="https://i.sstatic.net/Z32Er.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Z32Er.jpg" alt="enter image description here" /></a></p>
| 889
|
fluid dynamics
|
Is dropping water from a cup in a dry bucket an incompressible flow?
|
https://physics.stackexchange.com/questions/608793/is-dropping-water-from-a-cup-in-a-dry-bucket-an-incompressible-flow
|
<p>I'm learning hydrodynamics by googling stuff so forgive me if the question is super simple.</p>
<p>Imagine that I'm holding a cup of water, which I proceed to drop in a bucket laying next to my feet. The water will splash, and after some movement it will spread across the bottom of the bucket forming a thin layer.</p>
<p>From my poor understanding, this flow should be a <em>compressible flow</em>. After all, the water is spreading across the bucket floor so the divergence should be greater than zero. I assume this would also be true in all scales, so "small packets" of water will also spread themselves across the space, each occupying more space than they did a few seconds before.</p>
<p>Nonetheless, every reference I found seem to state that water can be treated as an incompressible flow with the exception of very extreme conditions (under extreme pressure, next to the speed of sound, etc.).</p>
<p>What am I missing? How come water continues to behave as incompressible fluid, when it is clearly changing density and it's molecules are spreading through a larger area (which I understand means diverging)?</p>
|
<blockquote>
<p>[...] when it is clearly changing density [...]</p>
</blockquote>
<p>The shape the water is in has nothing to do with its density. It is changing shape, not density. Fill a cube of 1 m side length with water. Now put the same water in a cube with 10 m side length. The water will fill the larger cube only 1 cm high and is spread out over a much larger area. However, the volume is the same as before, the weight is the same as before, hence the density did not change, meaning it is incompressible.</p>
| 890
|
fluid dynamics
|
Would birds be able to fly if the air had no viscosity?
|
https://physics.stackexchange.com/questions/610126/would-birds-be-able-to-fly-if-the-air-had-no-viscosity
|
<p>There are already many answers to the general question of why birds or planes can fly. But my question is more specific: I would like to understand the relative importance of the viscosity. The air has a very low viscosity, so I am wondering whether it really makes a difference that the viscosity is non-zero. Or could a bird fly in a non-viscous fluid?</p>
| 891
|
|
fluid dynamics
|
If a fluid element defined by the fluid particle it contains, it becomes useless for turbulent flow, doesn't it?
|
https://physics.stackexchange.com/questions/610408/if-a-fluid-element-defined-by-the-fluid-particle-it-contains-it-becomes-useless
|
<p>In the book of Batchelor, <em>Introduction to Fluid Mechanics</em>, at page 71 the author defines the fluid/material elements as a volume that contains the same fluid particles all the time. In other words, the fluid elements defined by the fluid particles that they contain.</p>
<blockquote>
<p>Nevertheless, the notion of material volumes, surfaces and lines which consist always of the same fluid particles and move with them is indispensable, ...</p>
</blockquote>
<p>Assuming an incompressible fluid, the volume of such a volume elements shouldn't change, but this doesn't mean that its shape will not change either. However, due to diffusion and especially in the case of turbulent flow, this implies that fluid particles can easily separate from each other in such a way that after some time t it would be hard to call our fluid volume a <em>volume</em>.</p>
<p>If so, isn't this causing any problem in our formulation of fluid mechanics? Or am I totally missing something?</p>
| 892
|
|
fluid dynamics
|
Why does measured pressure change over time in closed hose with temperature gradient
|
https://physics.stackexchange.com/questions/2158/why-does-measured-pressure-change-over-time-in-closed-hose-with-temperature-grad
|
<p>I have a 4' hose that is closed at one end and connected to a Airdata Test Set (precise control of pressure) and a high accuracy pressure monitor on the other end with a T and valve. The valve allows the Airdata Test Set connection to be closed off resulting in a hose connected to the pressure monitor and closed at the other end. The valve is a high quality needle valve. The pressure monitor is a a Druck DPI 142. Half the length of the hose is in a temperature chamber controlled to 70 C. The Druck connected end is outside the chamber at roughly 22 C. When the airdata test set is commanded to a pressure of 1300 mb and allowed to settle for 30 seconds or so, then the valve closed, the pressure reported by the Druck drops over 20 minutes or so with a decreasing rate of change. The airdata test set draws air from the room when operating. The hose is ~0.190" ID neoprene, Saint-Gobain P/N 06404-15. The temp chamber, hose, etc, are given 1 hour to thermally stabilize prior to commanding the pressure to 1300 mb. The difference between initial pressure and stable pressure is ~18 mb. Why does the pressure take 20 minutes to stabilize?</p>
|
<p>Firstly, and forgive me for asking the obvious, are you <em>certain</em> that there are no leaks anywhere in your setup? I'd suggest getting someone else to check it over in person - it may be something obvious you've overlooked that a fresh pair of eyes would see. Again, apologies if you've already tried this :-)</p>
<p>Assuming that's not it:</p>
<p>As you increase the pressure in the hose, the air temp in the hose will also increase (work is done on the gas to increase the pressure). As the temp settles back to equilibrium, the pressure decreases slightly according to $pV = NRT$. (I think you say you allow 1 hour for equilibrium <em>prior</em> to increasing the pressure). </p>
<ul>
<li><p>If this is true, then increasing the pressure again should result in a much smaller pressure drop (since an increase of 18mb will result in much less heating than 1300mb). </p></li>
<li><p>Are you able to monitor temperatures inside the hose?</p></li>
</ul>
<p>Given that the pressure does seem to settle, this is where I'd put my money.</p>
| 893
|
fluid dynamics
|
Mixing of fluid in a rotating barrel
|
https://physics.stackexchange.com/questions/4038/mixing-of-fluid-in-a-rotating-barrel
|
<p>A barrel/drum with a diameter of 60cm is rotating at 20RPM to get a good mixing of the fluid contained (type thick oil). </p>
<p>At what RPM should a barrel of 30cm rotate to get the same mixing efficiency with the same fluid? To get the same peripheral velocity, the RPM should be 40RPM, but will this give the same mixing of the fluid? I guess gravitational and centripetal forces play a role her...</p>
<p>EDIT:
More information to my otherwise incomplete question :)</p>
<p>The barrel is rotating around the length axis and the degree of fill is 90%.</p>
|
<p>Knowing a bit about process engineers' attitude to science I would say you just need to have same Reynolds number. $U\approx\Omega R$, $L\approx R$, then $Re\approx\frac{\Omega R^2}{\nu}$ so 80RPM.</p>
| 894
|
fluid dynamics
|
How to pump liquid aerosol?
|
https://physics.stackexchange.com/questions/577913/how-to-pump-liquid-aerosol
|
<p>I have a use case where I want to produce fog (an aerosol) using an industrial grade high-volume fog generator and then I would like to direct the resulting fog to different locations i.e. small boxes in the same room. This way, I can have concentrated fog where I need and keep rest of the room space clear.</p>
<p>Being a physics noob, I'm running the thought experiment in my head, and I believe since fog too is a fluid, I can just 'pump' it somehow. What kind of pump would one use in case one had to pump fog from one place to another?</p>
| 895
|
|
fluid dynamics
|
How do fluids behave in a vacuum?
|
https://physics.stackexchange.com/questions/566883/how-do-fluids-behave-in-a-vacuum
|
<p>My textbook says that the volume of liquids is assumed to be such under atmospheric pressure. What if the atmospheric pressure is reduced? Will the liquid stop exerting pressure according to Pascal's law? Will there be an intrinsic force that the liquid will exert?</p>
|
<p>If you pull a vacuum on a liquid, when the ambient pressure exerted on the liquid falls to the vapor pressure of the liquid at that temperature, the liquid starts to boil. Pulling such a vacuum very suddenly on a sample of water will cause the water to explode violently into vapor by boiling all at once.</p>
| 896
|
fluid dynamics
|
How would air flow inside a toroidal space station?
|
https://physics.stackexchange.com/questions/309124/how-would-air-flow-inside-a-toroidal-space-station
|
<p>The torus-shaped rotating space station, familiar from science fiction, is a way to produce artificial gravity in space. Would the fluid dynamics of gas in a rotating torus cause a person standing inside to notice any airflow?</p>
<p>It's noted (for example <a href="https://physics.stackexchange.com/questions/112354/artificial-gravity-on-rotating-spaceship">Artificial gravity on rotating spaceship?</a>) that inside such a space station the air would rotate with the structure. It's clear that this rotation would be the main motion of the air. however <em>would the variation in centrifugal force or other properties of fluid flow inside a torus cause a secondary motion</em> So that a person standing inside the torus would experience this secondary motion as wind.</p>
<p>Several websites mention fluid flow, for example, Wikipedia notes that fluids can freely move in a <a href="https://en.wikipedia.org/wiki/Vortex#Irrotational_vortices" rel="nofollow noreferrer">vortex</a>. And other sites deal with this <a href="https://www.gamedev.net/topic/680801-flying-inside-a-rotating-torus-space-station-artificial-gravity/" rel="nofollow noreferrer">https://www.gamedev.net/topic/680801-flying-inside-a-rotating-torus-space-station-artificial-gravity/</a> mentions "There will be airflow caused by the rotating torus. Laminar flow, oh the joy. That means fastest flow near the floor with quadratic or something falloff". I have also unsuccessfully tried googling for "gas flow inside a torus" </p>
|
<p>No, the atmosphere will rotate with the space station, just like Earth's atmosphere.</p>
| 897
|
fluid dynamics
|
Bernoulli's principle in a closed loop?
|
https://physics.stackexchange.com/questions/588481/bernoullis-principle-in-a-closed-loop
|
<p>In most simple terms, Bernoulli's principle dictates that</p>
<ul>
<li>a fluid or gas moving in one direction exposes less pressure in orthogonal directions.</li>
<li>the sum of the pressures, in the direction of movement and otherwise, is equal the pressure as measured in rest.</li>
</ul>
<p>Essentially, we loose static pressure and gain dynamic pressure. At least, that's the version we are taught in aerodynamics.</p>
<p>Imagine a bicycle tire being filled with air. Imagine it being freely suspended, and within a vacuum. Imagine spinning it up from rest. During acceleration, the tire is constantly somewhat faster than the air within. Considering both reference frames equal, the air moves through the tire, so the tire should lose some diameter.</p>
<p>So my question is: If the "total pressure" remains unchanged, where is the missing pressure applied? Or if that's paradoxical, what’s wrong with my initial description of Bernoulli's principle?</p>
| 898
|
|
fluid dynamics
|
Evaluating the gradient of pressure at the boundary of container
|
https://physics.stackexchange.com/questions/595136/evaluating-the-gradient-of-pressure-at-the-boundary-of-container
|
<p>In the hydrostatic case with no viscosity, we can write the gradient of the pressure inside a container of water within the gravitational field as
<span class="math-container">$$ \nabla P = g \rho$$</span></p>
<p>Moving close to a wall vertical wall, We encounter a normal force which should reduce our pressure, however, due to the isotropic nature of pressure, the pressure downwards must be the same as total pressure to the wall and hence the normal shouldn't affect the gradient.</p>
<p>What exactly is the intuition behind the forces at the boundary not affecting the pressure at all even though the fluid is clearly being 'pushed'?</p>
|
<p>I'm not sure why you think the normal force should reduce the pressure. The normal force on the container is a <em>reaction</em> force. The normal force on the ground when you step on it doesn't cause you to not weigh anything, it just prevents your weight from continuing to accelerate you downwards. It's similar with pressure. The force that the walls are feeling does not mean that it takes away from the pressure in the fluid, it is just the force that the walls of the container apply in response when pushed against by pressure.</p>
<p>You ask about:</p>
<blockquote>
<p>"the forces at the boundary not affecting the pressure at all"</p>
</blockquote>
<p>But that is not correct. The forces of the boundary are <em>necessary</em> for this pressure gradient to develop, so they have a very significant effect.</p>
<p>Without a solid boundary of the container to "push" at the fluid (or something else to keep it pushed together, like gravity on a larger scale), the force of the fluid pushing against itself would cause it to spread apart. It would not be able to develop a pressure gradient because the water above pushing down on the water below would just cause the water below to push away due to the force of the water above, with nothing to stop it from moving away.</p>
<p>The normal force of the container is a <em>requirement</em> to contain the water pressure in the container. Without that normal force pushing back, static pressure would not be able to develop. Instead, the water would just spread out into as thin of a surface as it can, so that it is essentially all at rest at atmospheric pressure. Even the normal force of the atmosphere helps keep the water contained, but that's more because that without the atmospheric pressure, the water would evaporate to fill the vacuum as much as possible.</p>
| 899
|
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