# Tag Info

## Hot answers tagged viscosity

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 ...

10

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, ...

6

Imagine two two trains side by side - one going faster than the other. Frictionless rails. Start shoveling coal from the slow train to the fast one, and from the fast to the slow one. Every shovel of coal results in a transfer of momentum - until the two trains move at the same speed. In the same way, when layers of liquid move past one another at different ...

5

Yes there is. Let's focus on the kinematic viscosity ($\nu$), which is defined as the diffusion constant for momentum in the fluid. That is, it tells us how quickly a momentum disturbance would diffuse through the rest of the fluid. Or, in particular, it gives us the linear dependence on the mean square propagation of the momentum as a function of time, ...

5

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 ...

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

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 ...

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

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 ...

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 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: $$\tau_{ij} = \sum_k \sum_l ... 3 To put it in simple terms, at slow speed the drag is just due to the viscosity of the fluid. At high speed, the momentum you're imparting to each parcel of air is proportional to the speed, and the number of parcels of air per second you're doing it to is also proportional to speed. Since force is momentum/second, that's why it's proportional to ... 3 Your professor is correct, but I agree with you that the statement “vorticity can’t be destroyed or created” seems jarring - I would prefer to think of this as “vorticity is conserved” because the conservation of vorticity derives from the Navier-Stokes Eq and the conservation of angular momentum. I confess this is splitting terminology hairs (don’t push it ... 3 Friction is caused by two physical processes, both of which dissipate energy. It's the dissipation of energy that means work is required to slide over the surface, and this work is why we feel a frictional force. Anyhow, the first factor is the surface energy of the material because this affects the adhesion between the sliding object and the substrate. A ... 3 Your assumptions are correct (but r is often defined as the distance from the pipe centerline). However, this is a very specific case: laminar pipe flow. In general, the stress will be a tensiorial quantity, defined as$$ \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 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 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 ... 3 Yoghurt is a flocculated suspension of casein micelles. The acid secreted by the bacteria growing in the milk destabilises the casein particles and they aggregate together. Because the volume fraction of the casein particles is high in milk the aggregated particles form a gel. This image shamelessly stolen from Wikipedia gives a schematic illustration of how ... 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 ... 2 1D Burger's equation is not meant to model a physical phenomenon. Rather, it is a simplification of homogeneous incompressible Navier-Stokes equations that preserves (some of) its mathematical structure: the non-linear convection term and the second order derivative of viscous forces. It was initially intended as a useful simplification to try to ... 2 I think the reason is that the honey is significantly denser than the surrounding water, so it's merely falling and breaking up/dissolving at the same time. Because of this instability the honey droplets form odd shapes and which then cause them to fall in paths which aren't straight (picture a leaf falling in air). Here is an article about fluid thread ... 2 Let's take your last question first. Let the stress tensor at a point (x,y,z) in the fluid be given as \sigma. You can pick a Cartesian basis \{ e_1, e_2, e_3 \} and express the components of the tensor in that basis$$ \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & ...

2

Yes they would fall differently. For example, according to Stokes' Law the drag force is proportional to dynamic viscosity and the terminal velocity of a falling sphere is: $$\frac{2(\rho_s -\rho_f)}{9\mu}gR^2$$ where $\rho_s$ and $\rho_f$ are the densities of the sphere, $\mu$ is the dynamic viscosity, $g$ is the acceleration due to gravity, and $R$ is ...

2

The dynamics in case of relative motion will be different: e.g., the acceleration and terminal velocity when falling/sinking. But the equilibrium will be unaffected (e.g., floating and waterline). This is because viscosity is at the origin of a frictional force.

2

For a damped simple harmonic oscillator, the damping term is linear with the velocity - and the constant of proportionality includes the viscosity of the damping medium. If the equation of motion is $$m \ddot{x} + 2 \zeta \omega_0 \dot{x} + \omega^2x = 0$$ In this equation, $\zeta$ is proportional to viscosity (and contains other terms related to the ...

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