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BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.4

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Problem 3B.4 Creeping flow between two concentric spheres (Fig. 3B.4). A very viscous Newtonian fluid flows in the space between two concentric spheres, as shown in the figure. It is desired to find the rate of flow in the system as a function of the imposed pressure difference. Neglect end effects and postulate that vθ depends only on r and θ with the other velocity components zero.

(a) Using the equation of continuity, show that vθ sin θ = u(r), where u(r) is a function of r to be determined. (b) Write the θ-component of the equation of motion for this system, assuming the flow to be slow enough that the [v · ∇v] term is negligible. Show that this gives    1 ∂P 1 1 d 2 du 0=− +µ r (3B.4-1) r ∂θ sin θ r2 dr dr (c) Separate this into two equations ∂P sin θ = B; ∂θ

µ d r dr

 r

2 du

dr

 =B

(3B.4-2, 3)

where B is the separation constant, and solve the two equations to get P2 − P1 2 ln cot 12 ε    (P1 − P2 )R  r R u(r) = 1− +κ 1− 4µ ln cot(ε/2) R r B=

(3B.4-4) (3B.4-5)

where P1 and P2 are the values of the modified pressure at θ = ε and θ = π − ε, respectively. (d) Use the results above to get the mass rate of flow w=

Solution

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π(P1 − P2 )R3 (1 − κ)3 ρ 12µ ln cot(ε/2)

(3B.4-6)

BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.4

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Part (a) For two concentric spheres a spherical coordinate system (r, θ, φ) is used, where θ represents the angle from the polar axis. We assume that the fluid flows only in the θ-direction and that the velocity varies as a function of r and θ. v = vθ (r, θ)θˆ If we assume the fluid does not slip on the walls, then it has the wall’s velocity at r = κR and r = R. Boundary Condition 1:

vθ (κR, θ) = 0

Boundary Condition 2:

vθ (R, θ) = 0

The equation of continuity results by considering a mass balance over a volume element that the fluid is flowing through. Assuming the fluid density ρ is constant, the equation simplifies to ∇ · v = 0. From Appendix B.4 on page 846, the continuity equation in spherical coordinates becomes 1 ∂ 2 1 ∂vφ 1 ∂ (r vr ) + (vθ sin θ) + = 0. 2 r sin θ ∂φ |r ∂r{z } r sin θ ∂θ | {z } =0

=0

Multiply both sides by r sin θ. ∂ (vθ sin θ) = 0 ∂θ Integrate both sides partially with respect to θ, to obtain vθ sin θ = u(r), where u(r) is an arbitrary function of r. Divide both sides by sin θ to solve for vθ . vθ (r, θ) =

u(r) sin θ

Use the boundary conditions for vθ to obtain those for u. u(κR) =0 sin θ u(R) vθ (R, θ) = =0 sin θ

vθ (κR, θ) =

→

u(κR) = 0

→

u(R) = 0

Part (b) The equation of motion results by considering a momentum balance over a volume element that the fluid is flowing through. Assuming the fluid viscosity µ is constant in addition to ρ, the equation simplifies to the Navier-Stokes equation. ∂ ρv + ∇ · ρvv = −∇p + µ∇2 v + ρg ∂t www.stemjock.com

BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.4

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Creeping flow is assumed, so the acceleration terms on the left side are zero. 0 = −∇p + µ∇2 v + ρg As this is a vector equation, it actually represents three scalar equations—one for each variable in the chosen coordinate system. From Appendix B.6 on page 848, the Navier-Stokes equation yields the following three scalar equations in spherical coordinates.     1 ∂ ∂vr 1 ∂ 2 vr ∂p 1 ∂2 2 + ρgr sin θ + 2 2 0=− + µ 2 2 (r vr ) + 2 ∂r ∂θ r sin θ ∂φ2 } |r sin θ ∂θ{z |r ∂r{z {z } | } =0 =0 =0      1 ∂p ∂v 1 1 ∂ 1 ∂ ∂ θ 0=− r2 + 2 +µ 2 (vθ sin θ) r ∂θ r ∂r ∂r r ∂θ sin θ ∂θ  2 ∂vr 2 cot θ ∂vφ 1 ∂ 2 vθ + − + ρgθ + 2 2 r sin θ ∂φ2 |r2{z∂θ} r2 sin θ ∂φ {z } | {z } | =0 =0 =0      ∂v 1 ∂p 1 1 ∂ ∂ 1 ∂ φ 0=− + 2 +µ 2 r2 (vφ sin θ) r sin θ ∂φ r ∂r ∂r r ∂θ sin θ ∂θ | {z } {z } | {z } | =0 =0 =0  ∂ 2 vφ 1 2 ∂vr 2 cot θ ∂vθ + 2 2 + 2 + + ρgφ r sin θ ∂φ r2 sin θ ∂φ |{z} r sin θ ∂φ2 {z } | {z } | {z } | =0 =0

=0

=0

The relevant equation for the velocity is the θ-equation, which has simplified considerably from ˆ the assumption that v = vθ (r, θ)θ.      1 ∂ 1 ∂ 1 ∂ 1 ∂p 2 ∂vθ 0=− +µ 2 r + 2 (vθ sin θ) + ρgθ r ∂θ r ∂r ∂r r ∂θ sin θ ∂θ It was found from part (a) that vθ sin θ is only a function of r, so the second term in square brackets is zero.    1 ∂p 1 ∂ 2 ∂vθ 0=− +µ 2 r + ρgθ r ∂θ r ∂r ∂r Replace vθ with u(r)/ sin θ. Also, assuming that gravity points downward, the vector is ˆ in spherical coordinates, so gθ = g sin θ. g = −gˆ z = −g[(cos θ)ˆ r + (− sin θ)θ]    1 ∂p 1 ∂ 2 ∂ u(r) 0=− + ρg sin θ + µ 2 r r ∂θ r ∂r ∂r sin θ 1 0=− r



    ∂p 1 1 d 2 du − ρgr sin θ + µ r ∂θ sin θ r2 dr dr

Therefore,    1 ∂P 1 1 d 2 du 0=− +µ r , r ∂θ sin θ r2 dr dr where P = P(r, θ) = p(r, θ) + ρgr cos θ is the modified pressure.

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BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.4

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Part (c) Bring the term with P to the left side. d 1 ∂P µ = 2 r ∂θ r sin θ dr

 r

2 du



dr

Multiply both sides by r sin θ. ∂P µ d sin θ = ∂θ r dr

 r

2 du



dr

If we assume that the space between the spheres is small, then the modified pressure is approximately constant in r.   dP µ d 2 du sin θ r = dθ r dr dr The only way a function of θ can be equal to a function of r is if both are equal to a constant B.   dP µ d 2 du sin θ = r =B dθ r dr dr Solve the first equation for P. sin θ

dP =B dθ

Multiply both sides by dθ/ sin θ. dP = Integrate both sides.

ˆ

B dθ sin θ ˆ

P2

θ2

dP = P1

θ1

B dθ, sin θ

where θ1 = ε, θ2 = π − ε, and P1 and P2 are the modified pressures at these angles, respectively. θ π−ε P2 − P1 = B ln tan 2 ε   π−ε ε = B ln tan − ln tan 2 2 π−ε tan 2 = B ln ε tan 2 ε cot 2 = B ln ε tan 2  ε 2 = B ln cot 2 ε = 2B ln cot 2 Dividing both sides by 2 ln cot(ε/2), therefore, B=

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P2 − P1 . 2 ln cot(ε/2)

BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.4

Now solve the second equation for u. µ d r dr

  2 du r =B dr

Multiply both sides by r/µ. d dr

  Br 2 du r = dr µ

Integrate both sides with respect to r. r2

du Br2 = + C1 dr 2µ

Divide both sides by r2 . du B C1 = + 2 dr 2µ r Integrate both sides with respect to r once more. u(r) =

C1 B r− + C2 2µ r

Apply the boundary conditions here to determine C1 and C2 . B C1 κR − + C2 = 0 2µ κR B C1 u(R) = R− + C2 = 0 2µ R

u(κR) =

Solving this system of equations yields C1 = −

BκR2 2µ

and C2 = −BR

κ+1 . 2µ

So then B BκR2 1 κ+1 r+ − BR 2µ 2µ r 2µ   BR r κR = + −κ−1 2µ R r    BR  r R =− 1− +κ 1− 2µ R r     P2 − P1 R r R =− 1− +κ 1− 2 ln cot(ε/2) 2µ R r

u(r) =

Therefore,    (P1 − P2 )R  r R u(r) = 1− +κ 1− . 4µ ln cot(ε/2) R r

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BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.4

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Part (d) The volumetric flow rate is obtained by integrating the velocity distribution over the area the fluid is flowing through. ¨ dV = vθ dA dt To get the mass flow rate, multiply both sides by the density ρ. ¨ dV =ρ vθ dA ρ dt Bring ρ inside the derivative. d(ρV ) =ρ dt

¨ vθ dA

Density times volume is mass. ¨ dm =ρ vθ (r, θ) dA dt ˆ 2π ˆ R =ρ vθ (r, θ)(dr)(r sin θ dφ) 0 κR ˆ 2π ˆ R u(r) =ρ (r sin θ dr dφ) 0 κR sin θ ˆ 2π ˆ R =ρ ru(r) dr dφ 0 κR ˆ 2π  ˆ R =ρ dφ ru(r) dr 0

ˆ

κR R

= 2πρ

ru(r) dr κR ˆ R

   (P1 − P2 )R  r R = 2πρ r 1− +κ 1− dr R r κR 4µ ln cot(ε/2)  ˆ  π(P1 − P2 )Rρ R r2 = r− + κr − κR dr 2µ ln cot(ε/2) κR R π(P1 − P2 )Rρ R2 · (1 − κ)3 = 2µ ln cot(ε/2) 6 Therefore, letting w = dm/dt, the mass flow rate is w=

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π(P1 − P2 )R3 (1 − κ)3 ρ . 12µ ln cot(ε/2)