Tag Info

Hot answers tagged

14

One's naive expectation would be that as the object moves through the medium, it collides with molecules at a rate proportional to $v$. The volume swept out in time $t$ is $A v t$, where $A$ is the cross-sectional area, so the mass with which it collides is $\rho A v t$. The impulse in each collision is proportional to $v$, and therefore the drag force ...


9

Actually, there are two different viscosity coefficients. You can see this from the stress tensor $$ \sigma_{ij} = -p_0 \delta_{ij} + \eta \left( \frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i} - \frac{2}{3} \delta_{ij} \frac{\partial v_k}{\partial x_k} \right) + \zeta \delta_{ij} \frac{\partial v_k}{\partial x_k} $$ which has the two ...


6

The article's preprint Mayer H. C., Krechetnikov R. "Walking with coffee: why does it spill?," Phys. Rev. E 85, 046117 (2012). is available from the UCSB site. From a glance of the article the phenomenon is not specific only to coffee. The authors make use of the next formula: The natural frequencies of oscillations of a frictionless, ...


5

Do we have viscous force acting between two layers even if there is no relative motion? No. From the Wikipedia article on viscosity: In general, in any flow, layers move at different velocities and the fluid's viscosity arises from the shear stress between the layers that ultimately opposes any applied force When the fluid is stationary, there's ...


4

A version of your proof without a stream function: The Laplace acting on the velocity may be expressed via the curl of the curl identity and aside from the $\nabla\times (\nabla\times \vec u)$ which vanishes, you also get another term $\nabla\cdot (\nabla\cdot \vec u)$ which vanishes (only) if one assumes incompressibility (it's the conservation of the ...


4

The high speed expression, proportional to $v^2$ is the ram pressure, which is wholly a momentum transfer effect and has nothing to do with viscosity - in contrast with the low flow speed Stokes law you cite above. To understand the ram pressure, which arises particularly for supersonic objects, witness the object is just shoving fluid out of its way, and ...


3

The general form of Newton's Law for constant mass is $$ m \frac{dv}{dt} = F $$ so in your case, $F = mg - \gamma v$ is the provided force law. In your case your force happens to depend on the velocity; the greater the velocity, the more negative the force, so it is a kind of friction or drag. $\gamma$ is just the proportionality constant between the ...


3

this will happen if the cup material's density is very close to that of water itself. completely submerged, the buoyant force on the cup is very similar to its own weight, so it appears to neither sink nor float underwater. Polystyrene cups are about 1g/cm³. these will do. heavier duty plastic cups will definitely sink faster, although not as fast as a ...


3

Your professor is correct, but I agree with you that the statement “vorticity can’t be destroyed or created” seems jarring - I would prefer to think of this as “vorticity is conserved” because the conservation of vorticity derives from the Navier-Stokes Eq and the conservation of angular momentum. I confess this is splitting terminology hairs (don’t push it ...


3

To put it in simple terms, at slow speed the drag is just due to the viscosity of the fluid. At high speed, the momentum you're imparting to each parcel of air is proportional to the speed, and the number of parcels of air per second you're doing it to is also proportional to speed. Since force is momentum/second, that's why it's proportional to ...


3

Your assumptions are correct (but $r$ is often defined as the distance from the pipe centerline). However, this is a very specific case: laminar pipe flow. In general, the stress will be a tensiorial quantity, defined as $$ \tau_{ij}= \eta \frac{\partial u_i}{\partial x_j}$$ which is true for turbulent flow, in arbitrary geometries. Where $i,j$ are in the ...


3

Friction is caused by two physical processes, both of which dissipate energy. It's the dissipation of energy that means work is required to slide over the surface, and this work is why we feel a frictional force. Anyhow, the first factor is the surface energy of the material because this affects the adhesion between the sliding object and the substrate. A ...


3

If "not much less than water" means "not an order of magnitude lower than water at room temperature" this is probably correct. However there are substances like http://en.wikipedia.org/wiki/Pentane with a viscosity 4 times less than that of water or http://en.wikipedia.org/wiki/Acetone with a viscosity 3 times less than water (at 20 degrees celcius). At ...


3

As far as I remember viscosity of the gases, unlike liquids, increases with increasing temperature. So in order to decrease viscosity, You would have to cool the air (which is already cold at some 10 km). When it comes to it's effect on planes, viscosity is responsible for creating pressure gradient between top and bottom side of an airfoil or wing. That ...


2

Your viscometer measures the viscosity of your Ficoll-70 solution. This viscosity is made up partly from the viscosity of the water and partly from the viscosity of the Ficoll-70 dissolved within it, and obviously the viscosity you measure increases with the concentration of Ficoll-70. Intrinsic viscosity is a rather different concept. It's really a measure ...


2

No it cannot in the general case. The formulas giving the pressure loss in a duct are always given by a form P = K.geometry.rho.V² where K is an empirical friction coefficient, geometry contains geometrical parameters (diameter, length etc), rho is the density and V the velocity. Now K depends typically on the Reynolds number and on the roughness of the ...


2

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


2

I think the answer has to do with the velocity gradient in the radial direction being zero in the rotating frame (i.e. consider a Lagrangian fluid element). Conceptually, it should be clear that two radially separated parcels of fluid have no relative motion in the cylindrical case, while they do in the planar case. In other words, there is no gradient in ...


2

You have lots of questions here, so I'll try to address them all but I may miss/gloss over some. A fluid that is considered incompressible is divergence-free. This comes from the conservation of mass equation: $$\frac{\partial \rho}{\partial t} + \frac{\partial \rho u_i}{\partial x_i} = 0$$ where if we say the density does not change, the time derivative ...


2

This is an excellent question and requires more discussion. Therefore, my answer will also have questions in it for others to weigh in. Bird and Stewart explain this very well in their Transport Phenomena book. In its general form, the viscous stresses may be linear combinations of all the velocity gradients in the fluid: $$\tau_{ij} = \sum_k \sum_l ...


2

Yes. Yes. Yes. See below. The Falkenhagen relation (NB: paywall, but (a) it's on the first page of the "Look Inside" option and (b) your University's library might have a copy) suggests that $$\frac{\eta_s}{\eta_0}=1+A\sqrt{c}$$ where $\eta_s$ is the solution viscosity, $\eta_0$ the solvent viscosity, $A$ a constant that depends on the electrostatic ...


2

As you say in your question, the advantage of a cone and plate viscometer is that the shear rate is constant and therefore well defined. If you use parallel plates the shear rate increases lineraly with distance from the centre. The reason this matters is that many fluids are shear thinning, or to use the more precise term non-Newtonian. This means the ...


1

Yes: up to a point The Darcy Weisbach equation accounts for frictional losses in a pipe: $f L D \frac{v^2}{2 g}$ $f$ typically comes from a Moody Diagram. Looking at one you'll see that the friction factor for a pipe decreases with increasing Reynolds number until the flow is deemed "fully turbulent" at which point the friction factor levels off. The ...


1

The classic Poisseuille flow approach is a fine approximate solution for situations that satisfy its assumptions. The effect of gravity can be accounted for well by including it in the pressure-drop term. It should work fine for water being sucked through a soda-straw. Surface tension forces won't be large for water or milkshakes sucked through an ...


1

I would say that the friction coeficient would make sense for liquid as well if the internal cohesion forces in the object are comparable with the friction superficial contact forces. Imagine a drop of mercury on a flat glass surface; it stays almost spherical and it slides very easily - the friction to glass is comparable to the internal cohesion forces. ...


1

I think that deep inside it is simply turning the 'gradient halves' concept into a pretty picture, but the classical intuitive image for viscosity is a stack of paper... The force you apply tangentially is applied by each sheet to the next one. For a given speed of the top sheet, the more sheets you have, the lower the speed difference between adjacent ...



Only top voted, non community-wiki answers of a minimum length are eligible