A 2 -mm-diameter sand particle having a density of 2.40 Mg / m^3 is released from rest at the su

A 2 -mm-diameter sand particle having a density of 2.40 Mg /...

A 2 -mm-diameter sand particle having a density of $2.40 \mathrm{Mg} / \mathrm{m}^{3}$ is released from rest at the surface of oil that is contained in the tube. As the particle falls downward, "creeping flow" will be established around it. Determine the velocity of the particle and the time at which Stokes' law becomes invalid, at about $\mathrm{Re}=1 .$ The oil has a density of $\rho_{o}=900 \mathrm{~kg} / \mathrm{m}^{3}$ and a viscosity of $\mu_{o}=30.2\left(10^{-3}\right) \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}$. Assume the particle is a sphere, where its volume is $\forall=\frac{4}{3} \pi r^{3}$.

A 2 -mm-diameter sand particle having a density of $2.40 \mathrm{Mg} / \mathrm{m}^{3}$ is released from rest at the surface of oil that is contained in the tube. As the particle falls downward, "creeping flow" will be established around it. Determine the velocity of the particle and the time at which Stokes' law becomes invalid, at about $\mathrm{Re}=1 .$ The oil has a density of $\rho_{o}=900 \mathrm{~kg} / \mathrm{m}^{3}$ and a viscosity of $\mu_{o}=30.2\left(10^{-3}\right) \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}$. Assume the particle is a sphere,
where its volume is $\forall=\frac{4}{3} \pi r^{3}$.

Key Concepts

Buoyancy Effect

The buoyancy effect is the upward force exerted by a fluid on an immersed object, equal to the weight of the fluid displaced. In the context of a settling particle, buoyancy reduces the effective gravitational force acting on the particle, thereby influencing the terminal velocity and overall settling behavior.

Reynolds Number

The Reynolds number is a dimensionless parameter that compares inertial forces to viscous forces within a fluid flow. It is crucial for determining the flow regime, with low values (typically Re < 1) indicating creeping (laminar) flow and higher values indicating the onset of inertial effects. In this context, the transition point where Re reaches about one marks the limit of validity for Stokes' law.

Terminal Velocity

Terminal velocity is the constant speed that a particle eventually reaches when the net forces acting on it, such as gravitational, buoyancy, and drag forces, balance each other. In a viscous medium under creeping flow conditions, terminal velocity can be determined using Stokes' law, and it represents the steady state of the particle's motion.

Stokes' Law

Stokes' Law provides an expression to calculate the drag force experienced by a small spherical particle moving at low Reynolds numbers in a viscous fluid. It assumes that the flow is steady, incompressible, and that inertial effects are negligible, which is often applicable in creeping flow conditions. This law is central to analyzing the terminal velocity of particles in viscous media.

Creeping Flow

Creeping flow refers to fluid motion where viscous forces are dominant over inertial forces, commonly characterized by a very low Reynolds number. Under these conditions, the flow around objects such as small particles is smooth and steady, and linear approximations like Stokes' law can be applied to predict fluid behavior accurately.

Transcript

00:02 We're considering a spherical sam particle that's released from the surface of oil and starts falling.

00:09 We're given the diameter of the particle, which we converted to meters, the density of the particle, which i converted to kilograms per meters cubed, the density of the oil, the viscosity of the oil, and we're given the volume of a sphere.

00:23 We are asked to consider or to determine the velocity in time at which stokes law becomes invalid, which is approximately equal when the reynolds is about one.

00:41 What else? okay.

00:45 So let's figure the volume of our sphere will be equal to four thirds times pi times, and this will be r will be 2 times 10 to the minus third over 1 ,000.

01:24 No, i got this.

01:25 I don't need to do that.

01:33 Cued.

01:33 Let me see what this equals.

01:46 Math x cubed i'm going to take that times four divided by three enter times second pi and i get 4 .188 8 times 10 to the minus 9 cubic meters okay and then our forces acting on our y will be buoyancy and our drag and our mg and that will equal zero.

02:35 Okay, and the mass of the particle will equal the density times my volume, and this will equal a one times 10 to the minus fifth kilograms.

03:12 My buoyant force will equal 900 kilograms per meter cubed, my oil times 9 .81 meters per second squared times 4 .1 888 times 10 to the minus 9th meters cubed.

03:36 So my buoyant force will be equal to 3 .698 times 10 to the minus 5th newtons.

03:47 Okay.

03:49 So the characteristic length for spheres down.

03:51 My reynolds number.

04:09 And this will be 900 times you times.

04:21 0 .002 meters divided by 30 .2 times 10 to the minus third n s over m squared.

04:40 This will equal 59 .6 u...

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