Flow between infinite moving inner cylinder and outer cylinder. http://sites.google.com/site/peeterjoot2/math2012/twoCylinders.pdf Peeter Joot — [email protected] Revision https://github.com/peeterjoot/physicsplay commit 269612195ac31041e194e8124aa83e7359a29fe0 Apr/15/2012 twoCylinders.tex Keywords: Navier-Stokes, PHY454H1S, PHY454H1

Contents 1

Motivation.

1

2

Statement.

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3

Solution. 3.1 Part 1,3. Velocity and pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Part 2. Friction force per unit length. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Part 4. With external pressure gradient. . . . . . . . . . . . . . . . . . . . . . . . . . .

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1. Motivation. A problem from the 2009 phy1530 final that appears appropriate for phy454 exam prep. 2. Statement. An infinite cylinder of radius R1 is moving with velocity v parallel to its axis. It is places inside another cylinder of radius R2 . The axes of the two cylinders coincide. The fluid is incompressible, with viscosity µ and density ρ, the flow is assumed to be stationary, and no external pressure gradient is applied. 1. Find and sketch the velocity field of the fluid between the cylinders. 2. Find the friction force per unit length acting on each cylinder. 3. Find and sketch the pressure field of the liquid. 4. If an external pressure gradient is present, how do you think your answer will change? Sketch your expectation for the velocity and pressure in this case. 3. Solution. 3.1. Part 1,3. Velocity and pressure Let’s start with the illustration of figure (1) to fix coordinates. We’ll assume that we can find a solution of the following form u = w(r )zˆ 1

(1a)

Figure 1: Coordinates for flow between two cylinders. p = p (r ).

(1b)

We’ll also work in cylindrical coordinates where our gradient is ˆ φ ∂φ + zˆ ∂z . r Let’s look at the various terms of the Navier-Stokes equation. Our non-linear term is

∇ = rˆ ∂r +

u · ∇u = w∂z (w(r )zˆ ) = 0,

(2)

(3)

Our Laplacian term is µ ∇2 u = µ

=



 1 1 ∂r (r∂r ) + 2 ∂φφ + ∂zz w(r )zˆ r r

µ (rw0 )0 zˆ . r

Putting the pieces together we have µ (rw0 )0 zˆ . r Decomposing these into one equation for each component we have 0 = −rˆ p0 +

(4)

p0 = 0,

(5)

(rw0 )0 = 0.

(6)

p(r ) = constant,

(7)

rw0 = A,

(8)

and

The pressure can be trivially solved

and for our velocity equation we get

Short of satisfying our boundary value constraints our velocity is 2

w = A ln r + B.

(9)

Our boundary value conditions are given by w ( R2 ) = 0

(10)

w( R1 ) = v,

(11)

so our integration constants are given by 0 = A ln R2 + B

(12)

v = A ln R1 + B.

(13)

v = A ln( R1 /R2 ).

(14)

Taking differences we’ve got

So our constants are v ln( R1 /R2 )

(15a)

B=−

v ln R2 , ln( R1 /R2 )

(15b)

w (r ) =

v ln(r/R2 ) . ln( R1 /R2 )

(16)

A=

and

A plot of this function can be found in figure (2), and the Mathematica notebook that generated this plot can be found in phy454/twoCylinders.cdf. That notebook has some slider controls that can be used interactively.

Figure 2: Velocity plot due to inner cylinder dragging fluid along with it.

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3.2. Part 2. Friction force per unit length. For the frictional force per unit area on the fluid by the inner cylinder we have 1 (σ · rˆ ) · zˆ = − pˆr · zˆ + 2µ 2 ln r = µv ln( R1 /R2 )



 ∂uz ∂ur +  ∂r ∂z

So the forces on the inner and outer cylinders for a strip of width ∆z is frictional force on inner cylinder = −2πR1 ∆zµvzˆ frictional force on inner cylinder = 2πR2 ∆zµvzˆ

ln R1 ln( R1 /R2 )

ln R2 ln( R1 /R2 )

(17a) (17b)

3.3. Part 4. With external pressure gradient. With an external pressure gradient imposed we expect a superposition of a parabolic flow profile with what we’ve calculated above. With dp , dz our Navier-Stokes equation will now take the form G=−

0 = −rˆ p0 − (− G zˆ ) +

(18)

µ (rw0 )0 zˆ . r

(19)

We want to solve the LDE

−

Gr = (rw0 )0 = rw00 + w0 µ

(20)

The homogeneous portion of this equation

(rw0 )0 = 0,

(21)

we have already solved finding w = C ln r + D. It looks reasonable to try a polynomial solution for the specific solution. Let’s try a second order polynomial w = Ar2 + Br

(22a)

w0 = 2Ar + B

(22b)

w00 = 2A.

(22c)

We need

−

Gr = 2Ar + 2Ar + B, µ

So B = 0 and 4A = − G/µ, and our general solution has the form 4

(23)

w=−

G 2 r + C ln r + D. 4µ

(24)

requiring just the boundary condition fitting. Let’s tweak the constants slightly, writing w=

G 2 ( R − r2 ) + C ln r/R2 + D, 4µ 2

(25)

so that D = 0 falls out of the w( R2 ) = 0 constraint. Our last integration constant is then determined by the solution of G 2 ( R − R21 ) + C ln R1 /R2 . 4µ 2

(26)

  G 2 G ln r/R2 ( R2 − r2 ) + v − ( R22 − R21 ) . 4µ 4µ ln R1 /R2

(27)

v= Or w=

A plot of this, with a pressure gradient small enough that we still see the logarithmic profile is shown in figure (3). An animation of this with different values for R1 , v, and G/4µ is available on http://youtu.be/BNgpnYeRpLo, but the Mathematica notebook above can also be used.

Figure 3: Pressure gradient added. Even cooler is to look at some plots of the velocity profiles in 3D

Figure 4: 3D plot 1

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Figure 5: 3D plot 2

Figure 6: 3D plot 3

Figure 7: 3D plot 4

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An animation of this can be found at http://youtu.be/OiJTopWx7L8, and the Mathematica notebook that generated it at phy454/twoCylinders3D.cdf.

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