Average shear rate of of particle sureface

[hizen_14]

Homework Statement

Diffusion coefficient (D) of 100 nm particles ( density = 1050 kg/m³) in a viscous solvent at 25 C is 2.2 x 10^-13 m³/s

What is the average shear rate at the particle surface?

Homework Equations

D = kT/ 6πηa (1)
where k is Boltzmann constant, T = temperature, η_susp = viscosity and a = radius

Pe = a²γ /D (2)

where γ = shear rate

The Attempt at a Solution

Assumed particles are in water

To find shear rate of particle surface I am not too sure that eq2 is the right equation to use or not. Can we assume that Pe = 1 ??

Pe = 1 means the solvent in the viscous state ?

Thank you

Zen

 

[Chestermiller]

This can't be the complete problem statement. Please provide the complete problem statement, with its exact wording.

Chet

 

[hizen_14]

Hi Chet, This is the full question

Thanks

 

[Chestermiller]

If I understand correctly, you have completed parts a, b, and c, and now you want want to determine the answer to part d, correct? If so, what was your answer to part c.

Chet

 

[hizen_14]

Hi Chet !

Yes you are right I have done part a-c. My answer on C was 0.05 m/s

Thanks

Zen

 

[Chestermiller]

Hi Chet !

Yes you are right I have done part a-c. My answer on C was 0.05 m/s

Thanks

Zen

Are you sure about that average velocity of the particles? It seems awfully large.

Chet

 

[hizen_14]

Thanks Chet

I have changed the radius of particle from 100µm to 50µm but the velocity was larger than 50 µm. it doesn't make sense. :(

V_av = ((3kT)/m)^1/2

Volume of particle = 4/3 πr³ = 4/3 π (50x10^-9)³ = 5.236 x10^-22 m³
Mass of particle = ρV = 1050 x 5.236x 10^-22 = 5.498 x 10^-19 kg

V_av = ((3x1.38 x 10^-23 x 298 )/ 5.498 x 10^-19 )^1/2 = 0.149 m/s

 

[hizen_14]

I think this one is more reasonable ?

v = (2r² Δρg) / 9η
= ( 2x( 50x10^-9 )² x (1050-1000) x 9.8) / (9 x 8.9x10^-4)
= 3.059 x 10^-10 m/s

THanks

 

[Chestermiller]

I think this one is more reasonable ?

v = (2r² Δρg) / 9η
= ( 2x( 50x10^-9 )² x (1050-1000) x 9.8) / (9 x 8.9x10^-4)
= 3.059 x 10^-10 m/s

THanks

This is much better, if analyzed correctly. I can't vouch for that part of the calculation (because my experience is not in that area), but I can help you get the average shear rate at the particle surface once the particle velocity is established. Sometimes the shear rate is taken as the radial velocity gradient at the particle surface, and some times it is taken as the square root of the second invariant of the rate of deformation tensor. How is it defined in your situation?

Do you know the equations for the components in spherical coordinates of the velocity vector in Stokes flow past a moving sphere? If so, please write them down.

Chet

 

[hizen_14]

Hi Chet.

Sometimes the shear rate is taken as the radial velocity gradient at the particle surface, and some times it is taken as the square root of the second invariant of the rate of deformation tensor. How is it defined in your situation?

I am not really sure about but I can see as the radial velocity gradient at the particle surface (dv/dx)

Do you know the equations for the components in spherical coordinates of the velocity vector in Stokes flow past a moving sphere? If so, please write them down.

Fg-Fb-Fd = 0 at Terminal Velocity
where Fg = gravity force, Fb = buoyancy force and Fd = drag force.

Fg = (Vρ_pg)
Fb = (Vρ_fg)
Fd = 9πηrv

v = (2r² Δρg) / 9η

I am not really sure this is what you asked for.

Thanks

 

[Chestermiller]

Hi Chet.
I am not really sure about but I can see as the radial velocity gradient at the particle surface (dv/dx)
Fg-Fb-Fd = 0 at Terminal Velocity
where Fg = gravity force, Fb = buoyancy force and Fd = drag force.

Fg = (Vρ_pg)
Fb = (Vρ_fg)
Fd = 9πηrv

v = (2r² Δρg) / 9η

I am not really sure this is what you asked for.

Thanks

I'm looking the equations for the components as a function of r and θ.

Chet

 

[hizen_14]

?

 

[Chestermiller]

View attachment 82555

View attachment 82554 ?

I'm more interested in the tangential velocity.

Chet