3. Slow Viscous Flow

Question 1

Stokes Flow

Definition

Answer 1

-for flows where the Reynold’s number:
Re = ρUD/µ &laquo_space;1
-we can neglect the term ρDu_/Dt in the Navier-Stokes equation so that the equations governing fluid flow become the Stokes equations

Question 2

Stokes Flow

Stokes Equations

Answer 2

ρ∇²u_ = p
∇.u_ = 0

Question 3

Stokes Flow

Stokes Equations for a Gravity Driven Flow

Answer 3

ρ∇²u_ = ∇P - ρg_
∇.u_ = 0

Question 4

What are the differences between the Navier-Stokes equations and the Stokes equations?

Answer 4

• the stokes equations are linear so can be solved using the principle of linear superposition
• also since there are no time derivatives in the Stokes equations, the solutions are time reversible under the reflection, (u_,t)->-(u_,t)

Question 5

Translating Sphere

Description

Answer 5

-consider the Stokes flow generated by a sphere of radius a moving at speed U
-using spherical polar coordinates centred on the sphere, the velocity of the sphere is given by:
(Ucosθ, -Usinθ, 0)
-where θ represents the angle between the first vector of the coordinate frame and the direction of motion

Question 6

Translating Sphere

Potential Flow Solution

Answer 6

-the potential flow solution for a sphere of radius a moving at speed U is given in spherical polar coordinates by:
u_ = ∇ϕ = (U a³/r³ cosθ, -U a³/2r³ sinθ, 0)
-this satisfies the boundary condition for the r component of velocity, u but does not satisfy the condition for v
-this solution also has the unrealistic property that the drag force on the sphere is 0, the D’Alembert paradox

Question 7

Translating Sphere

Slow Viscous Solution

Answer 7

-2D flow, so assume form:
u_ = (u(r,θ), v(r,θ), 0)
-substitute into the Stokes equations and the incompressibility condition (3 equations)
-given boundary conditions, seek solutions of the form:
u = f(r) cosθ
v = g(r) sinθ
-where f(r) and g(r) satisfy f(a)=U and g(a)=-U, and f and g both tend to zero as r tends to infinity
=>
f(r) = 3Ua/2r - a³U/2r³
g(r) = -3Ua/4r - a³U/4r³

Question 8

Translating Sphere

Pressure

Answer 8

-all terms coming from the fluid velocity are proportional to cosθ, suggesting:
p(r,θ) = p0 + h(r)cosθ
-where p0 is a constant
=>
p = p0 + 3µUa/2r² cosθ

Question 9

Translating Sphere

Stokes Drag Force

Answer 9

-total force on the sphere is found by integrating the force density over the surface:
F_ = - ∫ f_ dS = - ∫n_.τ_ dS
-the outward pointing normal vector is -er^ from the perspective of the fluid
-the drag force is the ex^ direction, since this is the direction of U_
=>
Fd = -6πµUa
-this is the drag force around a sphere for flow with small Re and with viscosity taken into account

Question 10

Translating Sphere

Sphere Falling Due to Gravity

Answer 10

-dropping a sphere into viscous fluid, an equilibrium between the weight and drag forces is reached
-the sphere asymptotically approaches this terminal velocity, found by:
||Fw|| = ||Fd||
g 4/3 πa³ Δρ = 6πµUa
-where Δρ is the difference in density between the sphere and the fluid it is falling through
U = 2a²Δρg/9µ

Question 11

Stokes Paradox

Answer 11

• if you try to attempt a similar solution for a translating cylinder as for the translating sphere, you reach a general solution for which it is not possible to apply the boundary conditions r=a and r->∞ simultaneously
• the result is known as Stokes paradox

Question 12

Translating Sphere

Determining Viscosity

Answer 12

-given the terminal velocity of a sphere falling through viscous fluid:
U = 2a²Δρg/9µ
-if a and Δρ are known, you can measure U and use these to determine the viscosity of the fluid
µ = 2a²Δρg/9U

Question 13

Slider Bearing Flow

Description

Answer 13

-wall on the x axis
-slider at an angle, length L
-height d2 above the x axis on the left and height d1 above the x axis on the right
-wall translating at speed U to the right
-general height of fluid above the x axis given by:
h(x) = d1 + (d2-d1)/L * x
-where h’ = (d2-d1)/L
-two cases to consider, h’=0 and |h’|«1

Question 14

Slider Bearing Flow

Case 1: h’=0

Answer 14

-incompressibility condition 
=> 0 + 0 = 0
-Stoke's Equation:
μ*∂y²*u = ∂x*p
0 = ∂y*p
-boundary conditions:
--u(y=0) = U
--u(y=h) = 0
--v(y=0) = 0
--v(y=h) = 0
-gives velocity:
u = U * (h-y)/h

Question 15

Slider Bearing Flow

Case 2: |h’|«1

Answer 15

-incompressibility:
∂x*u + ∂y*v = 0
-Stokes
μ(∂x²*u+∂y²*u) = ∂x*p
μ(∂x²*v+∂y²*v) = ∂y*p
-boundary conditions:
--u(y=0) = U
--u(y=h) = 0
--v(y=0) = 0
--v(y=h) = 0
-non-dimensionalise

Question 16

Cylinder Approaching a Wall

Outline

Answer 16

-wall at the x axis
-cylinder, radius a, axis of the cylinder in the z direction
-cylinder falling downwards at speed V
-minimum distance from the cylinder to the wall, d
-general distance from the cylinder to the wall is h(x)
-total distance from the wall to the centre of the cylinder given by:
a + d = h + √[a²-x²]
=> h = a + d - √[a²-x²]
h = a + d - √[a²(1-x²/a²)]
-focus on |x|«a></a>

Question 17

Cylinder Approaching a Wall

Steps

Answer 17

-incompressibility
∂x*u + ∂y*v = 0
-Stokes Equation
μ(∂x²*u+∂y²*u) = ∂x*p
μ(∂x²*v+∂y²*v) = ∂y*p
-boundary conditions
--u(y=0) = v(y=0) = 0
--u(y=h) = 0
--v(y=h) = -V
-non-dimensionalise
-introduce ε = √[d/a] << 1

Question 18

How to choose the non-dimensionalisation for pressure?

Answer 18

• in 2D, the Stokes equation gives two possible expression for non-dimensionalisation of P
• always take the largest one as it then the range of values covered encompasses the smaller value as well