# Tag Info

## New answers tagged fluid-dynamics

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I think you are right. (Involute spiral: Wikipedia) If you take a string wound around a stationary ball and unwind it, its end traces an involute spiral. Similarly, if the ball is rotating and the string runs out in one direction, it is the same curve with respect to the ball. A drop of water leaving the surface of the ball should travel in a straight line ...

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You can calculate air velocity as v = f/A where f=volumetric air flow and A = cross-sectional area of the air passage. Also, you have m=p*f, where m=mass flow rate and p=air density. Here 1 it says that "By conservation of energy, the energy consumed in rotating the fan is the same as the energy required to deliver the air: Pfan ∝ υ"

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Generally you would have $p(t) = p_0 + \beta V(t)$. Yes, the velocities at both end are equal but you should conservation of mass to get a relation between volume of balloon and the velocity. As for your last question I would say not using the potential form of unsteady Bernoulli equation is more intuitive but the velocity potential is $\phi(x,t)=u(t)x$ so ...

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Yes, after a fashion, increase of entropy is always coupled to some kind of "friction". Though isentropic could be argued to be stronger than frictionless, as it does not only refer to mechanical friction. For example a Rayleigh flow with heat transfer will not be isentropic (credits to Azad for pointing out the example). Nearly, it is $v_2 = \sqrt{v_1^2 - ... 4 The heat flow (per unit area) through some thin layer, e.g. a boundary layer of water, is given by: $$\frac{dQ}{dt} = \frac{K\Delta T}{d}$$ where$K$is the thermal conductivity,$d$is the thickness of the layer and$\Delta T$is the temperature difference between the two sides of the layer. So a high thermal conductivity does indeed mean a high heat ... 0 Consider the one-dimensional flow of quantity$q(x,t)$. The advection equation is generalized to, $$\partial_tq+\partial_xF(q,x)=0\tag{1}$$ where$F(q,x)$is the flux of$q$in the flow--in the case of the advection equation$F(q,x)=qu$. However, we can use the chain rule to write the above as $$\partial_tq+\frac{\partial F}{\partial q}\frac{\partial ... 0 With this answer I'll try to note the mistake committed in the original formulation "Landau implies that he has not assumed incompressibility at this point so how is it that one can choose the first form for Newton's second law without loss of generality?" Well the problem is originated when you assert: "If one instead uses F=\dot p so that$$f = ... 1 First, it's not only drag that slows down the air, every force has to be matched by an equal and opposite force including lift. Second, friction is not the sole cause of drag, indeed there's drag on an airfoil in an inviscid fluid due to pressure. So why we need sophisticated airfoil design when traditional windmills can work and how modern wind turbines ... 1 Perhaps I could share some idea for further research. If we could make actual and correct pressure measurements in the cochlea to reveal wether the non-stationary Bernoulli effect is a good description of the actual physics-of-how-the-cochlea-isolates-frequencies-along-its-length? I would consider: I would propose to use a pitot tube, with sensor in the ... 1 How would you solve this for a single small hole? What happens if you now move that hole down by a small amount? What if you add up the contributions of all these holes? Congratulations, you just integrated the expression for the flow rate over the aperture. 4 The cochlea has a complex physical structure, with multiple membranes and fluid-filled chambers. Therefore to explain the separation of frequencies along the basilar membrane of the cochlea is complex to. Sure, there are a lot of very general descriptions (even the answer of theblackcat) and a lot never go into the actual physics of the system. This ... 0 The force that the seal has to keep water from escaping must have the same force as the balloon being pushed upwards at least, this will cause it to stop the balloon from going upwards and no motion will occur. 3 Don't do it that way. You have what is called a "Change Point". Run it up until the time when the change should occur. Then stop the solver. Perform the instantaneous state change. Then restart the solver. So much silliness happens when people try to run ODE solvers over discontinuities. 0 RK 4th order is a good numerical approach - but it is only accurate up to fourth order terms in the Taylor expansion of your series. As long as the fifth (and higher) order derivatives of the function are small, you are fine. But when you introduce a step function, or even a piecewise linear approximation, that assumption is violated. I would recommend, as ... 1 Yes, the flow rate is related to the pump pressure by the Darcy-Weisbach equation: $$\Delta P = f_d \frac{L}{D} \frac{\rho v^2}{2}$$ where$L$is the length of the pipe,$D$is the diameter,$\rho$is the water density,$v$is the flow velocity and$f_D$is a fudge factor called the Darcy friction factor.$f_D$varies with the pipe diameter, density and ... 0 I think I found what I would call a rigorous treatment of the subject : http://www.math.ubc.ca/~jfeng/CHBE553/Scan_Notes/Modeling_Similarity.pdf 1 Static pressure in a compressible flow depends on the density but not the speed (not directly). Speed and geometry may affect the density. For isentropic flow (neglecting gravitational potential): $${p \over \rho^\gamma} = constant, \gamma = {c_p \over c_v}$$ which could be turned into this: $${p \over p_0} = ({1 \over 1+{(\gamma-1) \over ... 0 Your equation applies to a perfect fluid that has no viscosity. N-body simulations are for dark matter only and the dark matter is generally assumed to only interact gravitationally. There are hydrodynamical simulations that include baryons but those are more computationally intensive and so generally done for smaller simulations. If you are just interested ... 7 Note that this is almost identical to this previous SE question; the answers on that question however, including the favourite answer with 13 upvotes, are somewhat erroneous (or rather oversimplified). This is a good question: the exact details of how such elevations, called 'central uplifts', form are not well known. What follows is a brief summary of ... 0 Terminal velocity of a free falling object is obtained at the moment its acceleration vanishes $$\Sigma\mathbf{F}=0.$$ The forces that act on the object while falling are the gravity force, $$F_g=-mg,$$ and the drag force $$F_d=\frac{1}{2}v_t^2dC_dA$$ where v_t is the ... 0 Hele-Shaw flow - the flow between two closely spaced parallel plates can be regarded as a spaciel case of 3D Poiseuille flow. It's governing equations are: \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0 \frac{1}{\mu}\frac{\partial p}{\partial x}=\frac{\partial^2 u}{\partial z^2} \frac{1}{\mu}\frac{\partial p}{\partial ... 0 I think your problem is that,according to what you said on April 26 at 11:15, you really don't understand what incompressible means. If you have a certein initial non uniform distribution of \rho in space it doesn't mean that the distribution will stay constant with respect to time, on the contrary, for the value of density of a certein particle(small ... 1 1/f spectra have the unique distinction of being "scale invariant" in the sense that the energy in an interval df is proportional to df. The 1/f spectra in fact have the property that the in an interval with width df available energy is proportional to df but not with f. There, namely "scale invariant" attribute for. It is not the energy, but the signal ... -1 It' simple. If the lip profile curls down-ward at the point of pour (or even down and back) the poured liquid cannot dribble, no matter how slowly you pour, because the liquid would have to travel up-wards after leaving the spout in order to do so, and even the most sticky liquid (imagine golden syrup) won't do that. Certain manufacturers of plastic electric ... 2 Azad provides a link to one of the many empirical models that can predict pressure drop, but if you are seriously planning and investing in plumbing a new home - take caution. A general comment to begin is that tubing or pipe in general will create a resistance to the flow of fluid going through the pipe for any given pressure, so the bigger the pipe, the ... 1 I'd say option 1. Doubling the diameter would reduce the pressure loss in that pipe by 32 times in a constant flow rate scenario. Overall pressure loss will also decrease but not that much. Surface area increases 4 times then if the flow rate is constant the flow velocity will decrease by 4 times according to continuity equation. Then we have this ... 0 If you raise 1.2 liter of water (mass 1.2 kg) by 55 cm (h) in 3 minutes (t) against a gravitational acceleration g of 9.81 m/s2, the average power needed is$$\frac{m\cdot g\cdot h}{t} = \frac{1.2\cdot 9.81 \cdot 0.55}{180} < 40 \mathrm{\;mW}$$. But if the depth of your water is 30 cm, then on average the water needs to be raised just 40 cm, not 55 ... 0 My educated guess is that a large block of ice, delivered to space somehow, would last quite a while. If we assume it is in Earth orbit, the side facing the sun would sublimate (go directly from solid to gas) and dissipate. The rate of sublimination would depend on the insolation (power per unit area), which is about 400 W/m^2 and the absorption ... 0 When the surface area of liquid is increased, more molecules of liquid can go or evaporate at the same time. Because it also follow the law of physical equilibrium. The vapor pressure of liquid remains constant, then if surface area increases then evaporation also increases. 1 Pretty often I notice that a small rain system or even a large one will congregate or follow a large interstate highway It is like urban pollution. A storm releases rain over cities: City pollution may also impact cloud formation and rainfall. “Water vapor doesn’t ordinarily spontaneously condense into drops to form clouds,” says climate scientist ... 1 For our three compartment hearing sense, from a physics point of view, there is a basilar membrane stimulation, from base to apex, in its pathway in the cochlea, to a place on the basilar membrane. By periodic movement of perilymph, non viscous fluid, backwards and forewards, in the cochlear duct meet the conditions of a potential flow. The basilar ... 1 The Navier-Stokes equations assume (assuming we are looking at a vector conservative form): The continuum hypothesis, which is applicable for Knudsen numbers of much less than unity. The Navier-Stokes equations must specify a form for the diffusive fluxes (e.g. otherwise you would have the Cauchy momentum equation not the Navier-Stokes momentum ... 1 Here is a sketch of where it comes from. First just consider the perfect fluid terms and note the thermodynamic relation$$ \rho + p = \mu n + T s, $$where T and s are temperature and entropy, \mu and n are a chemical potential and number density. We also have a relation for derivatives of p$$ dp = n d\mu + s dT. $$Now if you take the ... 0 The bubble will get bigger. This is because in a pipe, when you look at a cross-section the same amount of liquid will flow through in a certain amount of time, even if it widens or narrows. This means that current must speed up at narrow points. Bernoulli's principle states that when a fluid increases speed, its pressure decreases. This means that, when the ... 0 It's a good question, and I would be inclined toward B or C. You are worried about blockage, but I view that as an issue of aerodynamics, like the difference between a lifting airfoil and a stalled airfoil. If you are not asking the air to follow sharp turns, so the flow doesn't separate, you should be good to go. If you do need to make the air follow ... 1 Joule heating is typically associated with increases in random kinetic energy (i.e., heat) due to \mathbf{j} \cdot \mathbf{E}. Ohmic dissipation and resistive heating are similar in a sense to Joule heating, as all three result from fluctuating electric fields acting as an effective drag force on an otherwise free flowing charged particle. Ion drag is ... 0 The bubbles move upwards because of their density. Since there density is less than that of the liquid, the has inside displaces a weight of liquid greater than its own. This results in an upward buoyant force. 3 The idea (theory) behind the selfsimilarity Parameters like Reynolds- or Machnumber is: that fundamental flow features of a specific flow have a dimensionless number connected to it (Dimensional homogeneity). This means: not the dimensional units (like inch, meters, tons, horsepower) should be used to describe (in this case) flow but dimensionless numbers. ... 1 As a planetary science and aviation enthusiast I can offer these tidbits, although a bit late for the 2013 posted question.... http://www.wired.com/2010/05/gallery-clouds/ This shows mountain-induced Van Karman vortex street (Strouhal instability) in a cloud layer as viewed from space. and so does this: ... 0 The wiki page on Blasius boundary layers is a useful and thorough resource in this case. Blasius boundary layers arise in steady, laminar 2D flow over a semi-infinite plate oriented parallel to the flow. In this scenario, the Navier-Stokes equations are particularly simple and amount to a leading-order balance between inertia and viscous forces. The ... 1 Initially, I agreed with Olaf Chujko's answer to this question; however, on further reflection, I think the most accurate answer is 'it depends': Firstly, from the schematic that is given, when switch A is pressed, the cylinder volumes above the two cylinders will be vented to a reservoir at ambient pressure (trust me, I work in Oil & Gas and I look at ... 1 A fluid in motion possesses momentum, just like any massive body in motion. And if the momentum changes, for example if the fluid hits a plate as you suggest, then there will be a force defined by:$$ F = \frac{d\vec{p}}{dt} $$And remember that momentum is a vector, so the force we get by differentiating it is also a vector (obviously) and is dependant on ... 2 For the modeling of surface wave motion there are only two restoring forces to consider: surface tension and gravity. Compared to gravity, surface tension forces are very weak and therefore have a greater influence on the regime of the smaller, capillary waves. Waves in deep water carry away the energy dissipated by shear wind forces - perhaps from a storm ... 1 Probably you have your question answered already, however, let me point out that: You are incorrect. You can't think of a hydrostatic system as it was an electric circuit. In an electric circuit what you (or the source) supply(ies) is the voltage and the result of that voltage acting upon the resistor of given resistance is the current. In hydrostatic ... 0 A quick google research (link) gives some references. As far as I can understand, to premultiply allows to filter the spectrum and identify a result on a specific wave number range. For instance, by defining the energy spectra by E(k,t), the kinetic energy dissipation \varepsilon and the wave number k, you can try to find scalings in the inertial ... 0 It is a common misunderstanding to say that \hat{n} \times \mathbf{v} is the tangential or transverse velocity. This vector is orthogonal to both \mathbf{v} and \hat{n}, so it is not a component of either vector. The transverse-to-\hat{n} component of \mathbf{v} is defined as:$$ \mathbf{v}_{t} = \hat{n} \times \left( \mathbf{v} \times \hat{n} ... 0 This condition means that the tangential components of the velocities of the two fluids are the same at their interface. It is equivalent to saying that there is no slip between the two fluids. Equality of tangential components is not restricted to fluid dynamics alone. You can find them in electrodynamics as well, where the tangential component of the ... 3 The argument is that the air was flowing through the hole at around 700 mph, so the air inside the aircraft had a substantial velocity in the direction of the hole. The air velocity inside the plane would have been less than 700 mph because the flow was converging on the hole, but the speed of the air would still have been hundreds of mph. When the hole was ... 1 At low Reynolds number, as in a creeping flow, one can ignore the advective acceleration terms in the Navier-Stokes equation. If we also assume a steady state, the equation becomes $$0 = -\nabla p + \mu\nabla^2\vec{v},$$ where$p$is the hydrostatic pressure,$\mu$is the viscosity of the fluid and$\vec{v}\$ is the flow velocity. ...

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If you solve stokes flow (time-independent navier-stokes equation without the non-linear term) for a sphere, you will find a linear dependence between the drag and the velocity.

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