Chapter 5 Navier-Stokes Equationsstokeslet.ucsd.edu/mae210adocs/ProblemSet5.pdf · Chapter 5...

Chapter 5Navier-Stokes Equations

Problem 5.1. Stokes second problemConsider an infinite flat plate y = 0 subject to oscillations with velocity Uw cos !t in the x-direction.The fluid in the half-space y > 0 is Newtonian, homogeneous, and incompressible. Assume that bodyforces are negligible, that the pressure is uniform and constant, and that the flow driven by the plate isunidirectional along x.

(a) Write down the equation and boundary conditions satisfied by vx(y, t) (x-component of the velocityfield).

(b) Assume that the solution can be written

vx(y, t) = Re[U(y)e

i!t], (5.1)

where Re denotes the real part. Find an equation and boundary conditions for U(y), and solve that equa-tion. Infer the solution for vx(y, t).

(c) Sketch the velocity profile at t = 0. What is the characteristic thickness � of the layer of fluid that ismoving next to the wall?

Problem 5.2. Startup of shear flowA Newtonian incompressible fluid with constant density ⇢ and kinematic viscosity ⌫ is placed betweentwo parallel infinite flat plates separated by a distance h. Initially, both plates are at rest. At t = 0, thebottom plate (at y = 0) starts to translate with a constant velocity U = Uex. Assume that the flow isunidirectional in the x-direction, that body forces can be neglected, and that the pressure is constant anduniform everywhere in the fluid.

(a) Determine the velocity field u = u(y, t)ex between the two plates.

(b) Sketch the velocity profile for different values of the parameter ⇡2⌫t/h2. What flow do you recoverwhen ⇡2⌫t/h2 ! 1?

Problem 5.3. Circular Couette flowConsider the steady flow of a Newtonian fluid between two infinite concentric cylinders of radii Ri (inner)and Ro (outer) that are rotating around their common axis at angular velocities !i and !o. The density⇢ and viscosity µ of the fluid are both constant and uniform. In cylindrical coordinates (r, ✓, z), assumethat the flow is uniform in the z- and ✓-directions, and that its direction is azimuthal in the (r, ✓) plane:u = u✓(r)e✓.

(a) Write down the Navier-Stokes equations and boundary conditions for this problem using cylindricalcoordinates. Simplify as much as you can using the stated assumptions.

(b) Solve for the velocity profile u✓(r), and determine all the integration constants. Sketch the velocityprofile.

(c) Determine the pressure p(r) up to an additive integration constant.

Problem 5.4. Flow down an inclined planeAn incompressible liquid (constant density ⇢ and viscosity µ) is flowing under the influence of gravity g

17

18 5 Navier-Stokes Equations

down a very long plane inclined at an angle ✓ to the horizontal, forming a film of constant thickness h(see figure). Assume that the flow is steady, unidirectional along x, and uniform in the x and z directions:v = vx(y)ex, and that the pressure only depends on y: p = p(y).



!

h!

y = h

!

y = 0

!

x

!

y

!

g

!

v(y)

liquidatmosphere

!

"

!

n

atmosphereliquidn

v(y)

g

hy = h✓xy

(a) Justifying every step in the derivation, show that the momentum equation, projected along the x andy directions, leads to the two equations

⇢g sin ✓ + µd2vx

dy2

= 0, (6.2)

⇢g cos ✓ +

dp

dy= 0. (6.3)

(b) Knowing that the pressure at the interface between the liquid and the atmosphere is given by theconstant atmospheric pressure pa, solve for p(y).

(c) The viscous traction t = n·⌧ (where ⌧ is the viscous stress tensor) is zero at the interface between theliquid and the atmosphere. Using this fact, derive a boundary condition for the velocity vx(y) at y = h.What is the boundary condition at y = 0?

(d) Solve for the velocity vx(y). Sketch the velocity profile.

(e) Calculate the volumetric flow rate Q (per unit length in the z direction), defined as

Q =

Z h

0

vx(y) dy. (6.4)

Problem 6.5. Wind-driven flow inside a lakeConsider a large lake, over which wind is blowing and exerts a constant shear stress S in the x-direction(see figure below). The goal of this problem is to determine the steady flow field established in the lakeby the wind. Assume that the lake has a constant depth h

0

before the wind starts blowing. The wind will



!

h!

y = h

!

y = 0

!

x

!

y

!

g

!

v(y)

liquidatmosphere

!

"

!

n

atmosphereliquidn

v(y)

g

hy = h✓xy



dy2

= 0, (6.2)

⇢g cos ✓ +

dp

dy= 0. (6.3)





Q =

Z h

0

vx(y) dy. (6.4)


0




!

h!

y = h

!

y = 0

!

x

!

y

!

g

!

v(y)

liquidatmosphere

!

"

!

n

atmosphereliquidn

v(y)

g

hy = h✓xy



dy2

= 0, (6.2)

⇢g cos ✓ +

dp

dy= 0. (6.3)





Q =

Z h

0

vx(y) dy. (6.4)


0




!

h!

y = h

!

y = 0

!

x

!

y

!

g

!

v(y)

liquidatmosphere

!

"

!

n

atmosphereliquidn

v(y)

g

hy = h✓xy



dy2

= 0, (6.2)

⇢g cos ✓ +

dp

dy= 0. (6.3)





Q =

Z h

0

vx(y) dy. (6.4)


0




!

h!

y = h

!

y = 0

!

x

!

y

!

g

!

v(y)

liquidatmosphere

!

"

!

n

atmosphereliquidn

v(y)

g

hy = h✓xy



dy2

= 0, (6.2)

⇢g cos ✓ +

dp

dy= 0. (6.3)





Q =

Z h

0

vx(y) dy. (6.4)


0




!

h!

y = h

!

y = 0

!

x

!

y

!

g

!

v(y)

liquidatmosphere

!

"

!

n

atmosphereliquidn

v(y)

g

hy = h✓xy



dy2

= 0, (6.2)

⇢g cos ✓ +

dp

dy= 0. (6.3)





Q =

Z h

0

vx(y) dy. (6.4)


0




!

h!

y = h

!

y = 0

!

x

!

y

!

g

!

v(y)

liquidatmosphere

!

"

!

n

atmosphereliquidn

v(y)

g

hy = h✓xy



dy2

= 0, (6.2)

⇢g cos ✓ +

dp

dy= 0. (6.3)





Q =

Z h

0

vx(y) dy. (6.4)


0




!

h!

y = h

!

y = 0

!

x

!

y

!

g

!

v(y)

liquidatmosphere

!

"

!

n

atmosphereliquidn

v(y)

g

hy = h✓xy



dy2

= 0, (6.2)

⇢g cos ✓ +

dp

dy= 0. (6.3)





Q =

Z h

0

vx(y) dy. (6.4)


0




!

h!

y = h

!

y = 0

!

x

!

y

!

g

!

v(y)

liquidatmosphere

!

"

!

n

atmosphereliquidn

v(y)

g

hy = h✓xy



dy2

= 0, (6.2)

⇢g cos ✓ +

dp

dy= 0. (6.3)





Q =

Z h

0

vx(y) dy. (6.4)


0




!

h!

y = h

!

y = 0

!

x

!

y

!

g

!

v(y)

liquidatmosphere

!

"

!

n

atmosphereliquidn

v(y)

g

hy = h✓xy



dy2

= 0, (6.2)

⇢g cos ✓ +

dp

dy= 0. (6.3)





Q =

Z h

0

vx(y) dy. (6.4)


0




!

h!

y = h

!

y = 0

!

x

!

y

!

g

!

v(y)

liquidatmosphere

!

"

!

n

atmosphereliquidn

v(y)

g

hy = h✓xyy = 0



dy2

= 0, (6.2)

⇢g cos ✓ +

dp

dy= 0. (6.3)





Q =

Z h

0

vx(y) dy. (6.4)

Problem 6.5. Wind-driven flow inside a lakeConsider a large lake, over which wind is blowing and exerts a constant shear stress S in the x-direction(see figure below). The goal of this problem is to determine the steady flow field established in the lake


⇢g sin ✓ + µd

2vx

dy2

= 0, (5.2)

⇢g cos ✓ +

dp

dy= 0. (5.3)





Q =

Z h

0

vx(y) dy. (5.4)

Problem 5.5. Hydrodynamic slipExperiments in microfluidic devices have shown that the no-slip boundary condition can sometimes be in-accurate, especially when the channel walls are made of hydrophobic surfaces. A more accurate boundarycondition in this case is the following:

vt = bn · rvt at the wall, (5.5)

where vt is the tangential component of the velocity vector, n is a unit normal vector pointing into thefluid, and b is a given constant.

(a) What are the dimensions of b?

(b) Consider the pressure-driven flow of an incompressible homogeneous Newtonian fluid in a cylin-drical microchannel of radius a (cylindrical Poiseuille flow). Solve for the velocity vx(r) in cylindricalcoordinates, using the boundary condition (5.5).

5 Navier-Stokes Equations 19

(c) What is the value vx(a) of the velocity at the wall? Sketch the velocity profile, and give an interpreta-tion for the constant b.

Problem 5.6. Flow in a porous channel with injection/suctionAn incompressible Newtonian fluid of density ⇢ and viscosity µ flows beween two parallel flat plates oflength L, separation 2H ⌧ L, and infinite width. The flow is induced by a pressure difference (p

0

� pL)

over the length of the plates. The two plates are porous: by injecting more of the same fluid through one ofthe plates and removing it through the other, a uniform crossflow is generated. Use Cartesian coordinates(x, y, z) with origin on the centerplane of the entrance to the plates, x-direction aligned with the lengthof the plates, and y-direction aligned normal to the plates. Uniform crossflow means that the transversevelocity component uy is constant, say U . Show that the axial velocity component ux is given by

ux(y) =

(p0

� pL)H2

µL

1

Re

y

H� 1 +

e

Re � e

Rey/H

sinh Re

�(5.6)

where Re = ⇢UH/µ is the crossflow Reynolds number, assuming that there is no slip at the porousplates. Sketch the axial velocity profile for Re ⌧ 1, Re ⇠ 1, and Re � 1.

Problem 5.7. Pipe flow of two immiscible liquidsConsider the incompressible flow of two immiscible homogeneous Newtonian viscous liquids in a cylin-drical pipe of radius R, driven by a constant pressure gradient dp/dz = G along the axis of the pipe(see figure below). Liquid 1 (with viscosity µ

1

) occupies the center of the pipe (0 < r < R(1 � "),where 0 < " < 1 is a given parameter), while liquid 2 (with viscosity µ

2

) occupies the periphery. Weassume that the interface between the two liquids is also cylindrical, and that the flow in both liquids isaxisymmetric and unidirectional in the z direction:

v

1

(x) = v1

(r)ez and v

2

(x) = v2

(r)ez. (5.7)


Q =

8<

:

0 when ↵ � 1

⇢gH3

3µ(1 � ↵)

2

⇣1 +

↵

2

⌘when ↵ < 1

(6.27)

where ↵ = ⌧0

/⇢gH .


0

� pL)


ux(y) =

(p0

� pL)H2

µL

1

Re

y

H� 1 +

e

Re � e

Rey/H

sinh Re

�(6.28)



1

) occupies the center of the pipe (0 < r < R(1 � ✏),where 0 < ✏ < 1 is a given parameter), while liquid 2 (with viscosity µ

2


v

1

(x) = v1

(r)ez and v

2

(x) = v2

(r)ez. (6.29)

liquid 1 liquid 2

z

view of a cross-section

R

R(1-e)

liquid 1liquid 2view of a cross-sectionzRR(1 � ")


Q =

8<

:

0 when ↵ � 1

⇢gH3

3µ(1 � ↵)

2

⇣1 +

↵

2

⌘when ↵ < 1

(6.27)

where ↵ = ⌧0

/⇢gH .


0

� pL)


ux(y) =

(p0

� pL)H2

µL

1

Re

y

H� 1 +

e

Re � e

Rey/H

sinh Re

�(6.28)



1


2


v

1

(x) = v1

(r)ez and v

2

(x) = v2

(r)ez. (6.29)

liquid 1 liquid 2

z


R

R(1-e)



Q =

8<

:

0 when ↵ � 1

⇢gH3

3µ(1 � ↵)

2

⇣1 +

↵

2

⌘when ↵ < 1

(6.27)

where ↵ = ⌧0

/⇢gH .


0

� pL)


ux(y) =

(p0

� pL)H2

µL

1

Re

y

H� 1 +

e

Re � e

Rey/H

sinh Re

�(6.28)



1


2


v

1

(x) = v1

(r)ez and v

2

(x) = v2

(r)ez. (6.29)

liquid 1 liquid 2

z


R

R(1-e)



Q =

8<

:

0 when ↵ � 1

⇢gH3

3µ(1 � ↵)

2

⇣1 +

↵

2

⌘when ↵ < 1

(6.27)

where ↵ = ⌧0

/⇢gH .


0

� pL)


ux(y) =

(p0

� pL)H2

µL

1

Re

y

H� 1 +

e

Re � e

Rey/H

sinh Re

�(6.28)



1


2


v

1

(x) = v1

(r)ez and v

2

(x) = v2

(r)ez. (6.29)

liquid 1 liquid 2

z


R

R(1-e)



Q =

8<

:

0 when ↵ � 1

⇢gH3

3µ(1 � ↵)

2

⇣1 +

↵

2

⌘when ↵ < 1

(6.27)

where ↵ = ⌧0

/⇢gH .


0

� pL)


ux(y) =

(p0

� pL)H2

µL

1

Re

y

H� 1 +

e

Re � e

Rey/H

sinh Re

�(6.28)



1


2


v

1

(x) = v1

(r)ez and v

2

(x) = v2

(r)ez. (6.29)

liquid 1 liquid 2

z


R

R(1-e)



Q =

8<

:

0 when ↵ � 1

⇢gH3

3µ(1 � ↵)

2

⇣1 +

↵

2

⌘when ↵ < 1

(6.27)

where ↵ = ⌧0

/⇢gH .


0

� pL)


ux(y) =

(p0

� pL)H2

µL

1

Re

y

H� 1 +

e

Re � e

Rey/H

sinh Re

�(6.28)



1


2


v

1

(x) = v1

(r)ez and v

2

(x) = v2

(r)ez. (6.29)

liquid 1 liquid 2

z


R

R(1-e)


(a) Starting from the incompressible Navier-Stokes equations, write down the differential equations sat-isfied by v

1

(r) and v2

(r). Integrate these equations to show that the velocity in each liquid can be writtenas

v1

(r) =

G

4µ1

r2

+ A1

ln r + B1

, v2

(r) =

G

4µ2

r2

+ A2

ln r + B2

. (5.8)

(b) What is the boundary condition on v2

(r) on the pipe wall at r = R?

(c) At the interface between the two liquids r = R(1 � "), both the fluid velocity and shear stress mustbe continuous:

v1

(r) = v2

(r), and ⌧1

rz(r) = ⌧2

rz(r) at r = R(1 � "). (5.9)


Using these two conditions and the boundary condition of (b), solve for the four integration constants A1

,B

1

, A2

, and B2

.

(d) Assuming that the liquid layer on the periphery of the pipe is very thin (" ⌧ 1), show by neglectingterms of order "2 that the velocity fields v

1

(r) and v2

(r) can be expressed as

v1

(r) =

G

4µ1

r2 � R2 � 2R2"

✓µ

1

µ2

� 1

◆�, v

2

(r) =

G

4µ2

(r2 � R2

). (5.10)

(e) Sketch the velocity profile qualitatively in the cases where µ1

> µ2

and µ2

> µ1

. In which case doesthe presence of the thin liquid layer on the pipe wall help to enhance the flow rate of liquid 1 in the centerof the pipe?

Problem 5.8. Flow of paint down a vertical wallThe surface of a large, flat, vertical wall is to be protected from the atmosphere by being covered with athin film of paint of uniform thickness H . When it is wet, the paint is a non-Newtonian fluid that can beregarded as a Bingham fluid, for which the shear stress ⌧ = ⌧xy in a parallel flow u = ux(y)ex (wherethe x-axis points in the negative vertical direction and the y-axis points in the direction normal to thewall) is related to the shear rate � = dux/dy by

⌧ = µ� + ⌧0

if ⌧ > ⌧0

(5.11)� = 0 if ⌧ ⌧

0

(5.12)

where µ denotes the viscosity and ⌧0

the yield stress. Because of gravity, there is a volumetric flow rateQ of paint down the wall per unit width of the wall. Let g denote the magnitude of the gravitationalacceleration and ⇢ the density of the paint. The boundary condition for the flow is no slip at the wall(ux(0) = 0), and zero shear stress at the free surface (⌧(H) = 0). Assume that the pressure is constantand uniform.

(a) Assuming steady flow, show using the momentum conservation equation that the shear stress ⌧(y)

satisfiesd⌧

dy+ ⇢g = 0. (5.13)

Solve for ⌧(y) and sketch it.

(b) Treating the cases ⌧0

� ⇢gH and ⌧0

< ⇢gH separately, solve for the velocity profile ux(y) across thepaint film. Sketch the velocity profile in each case.

(c) Show that the volumetric flow rate is given by

Q =

8<

:

0 when ↵ � 1

⇢gH3

3µ(1 � ↵)

2

⇣1 +

↵

2

⌘when ↵ < 1

(5.14)

where ↵ = ⌧0

/⇢gH .

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