Flow Between Coaxial Cylinders And Concentric Spheres..pdf
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BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.1
Page 1 of 8
Problem 3B.1 Flow between coaxial cylinders and concentric spheres. (a) The space between two coaxial cylinders is filled with an incompressible fluid at constant temperature. The radii of the inner and outer wetted surfaces are κR and R, respectively. The angular velocities of rotation of the inner and outer cylinders are Ωi and Ωo . Determine the velocity distribution in the fluid and the torques exerted by the fluid on the two cylinders needed to maintain the motion. (b) Repeat part (a) for two concentric spheres. Answers: r κR κR 2 (a) vθ = (Ωo − Ωi κ ) + (Ωi − Ωo ) 1 − κ2 κR r " # r κR κR 2 3 (Ωo − Ωi κ ) sin θ (b) vφ = + (Ωi − Ωo ) 1 − κ3 κR r Solution Part (a) We assume that the fluid flows only in the θ-direction and that the velocity varies as a function of radius only. v = vθ (r)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. The tangential velocity is obtained by multiplying the angular velocity by the distance from the axis of rotation (the moment arm). Boundary Condition 1:
vθ (κR) = Ωi κR
Boundary Condition 2:
vθ (R) = Ωo R
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.
(1)
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 also constant, the equation simplifies to the Navier-Stokes equation. ∂ ρv + ∇ · ρvv = −∇p + µ∇2 v + ρg ∂t
(2)
As this is a vector equation, it actually represents three scalar equations—one for each variable in the chosen coordinate system. Using cylindrical coordinates is the appropriate choice for this problem, so equations (1) and (2) will be used in (r, θ, z). From Appendix B.4 on page 846, the continuity equation becomes 1 ∂ 1 ∂vθ ∂vz (rvr ) + + = 0, ∂θ} |{z} ∂z |r ∂r{z } |r {z =0
www.stemjock.com
=0
=0
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.1
Page 2 of 8
which doesn’t tell us anything. From Appendix B.6 on page 848, the Navier-Stokes equation yields the following three scalar equations in cylindrical coordinates. ∂vr vθ ∂vr ∂vr vθ2 1 ∂ 2 vr ∂ 2 vr ∂vr ∂p ∂ 1 ∂ 2 ∂vθ ρ + vr + + vz − =− +µ (rvr ) + 2 + − 2 + ρgr 2 |{z} ∂t ∂r} |r {z∂θ} | {z ∂z} r ∂r ∂r r ∂r r {z ∂θ2} |∂z r {z∂θ} |{z} | {z | {z } | {z } | =0 =0
=0
=0
=0
=0
=0
=0
=0
∂vθ ∂vθ vθ ∂vθ ∂vθ vr vθ 1 ∂p ∂ 1 ∂ 2 ∂vr 1 ∂ 2 vθ ∂ 2 vθ ρ + vr + + vz + =− +µ + + 2 + ρgθ (rvθ ) + 2 2 |{z} ∂t ∂r} |r {z∂θ} | {z ∂z} | {z r} ∂r r ∂r ∂θ2} |∂z |{z} | {z | r{z∂θ} {z } |r {z∂θ} |r {z =0 =0 =0 =0 =0 =0 =0 =0 =0 =0 ∂vz vθ ∂vz ∂vz ∂vz ∂p 1 ∂ 1 ∂ 2 vz ∂ 2 vz ∂vz + vr r + + vz =− +µ + 2 + + ρgz ρ ∂t ∂r} |r {z∂θ} | {z ∂z} ∂z r ∂r ∂r r ∂θ2 ∂z 2 |{z} | {z | {z } | {z } | {z } =0
=0
=0
=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. d 1 d 0=µ (rvθ ) dr r dr Divide both sides by µ. d 1 d (rvθ ) = 0 dr r dr Integrate both sides with respect to r. 1 d (rvθ ) = C1 r dr Multiply both sides by r. d (rvθ ) = C1 r dr Integrate both sides with respect to r once more. rvθ = C1
r2 + C2 2
Divide both sides by r. r C2 vθ (r) = C1 + 2 r Apply the two boundary conditions here to determine C1 and C2 . κR C2 + = Ωi κR 2 κR R C2 vθ (R) = C1 + = Ωo R 2 R
vθ (κR) = C1
Solving the system of equations yields C1 =
2(Ωo − κ2 Ωi ) 1 − κ2
We then have vθ (r) = www.stemjock.com
and C2 =
κ2 R2 (Ωi − Ωo ) . 1 − κ2
2(Ωo − κ2 Ωi ) r κ2 R2 (Ωi − Ωo ) 1 + . 1 − κ2 2 1 − κ2 r
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.1
Page 3 of 8
Therefore, upon factoring κR/(1 − κ2 ), r κR κR 2 (Ωo − Ωi κ ) . + (Ωi − Ωo ) vθ (r) = 1 − κ2 κR r The torque on the inner cylinder is obtained by multiplying (−τrθ )|r=κR , the viscous force per unit area in the θ-direction on a plane perpendicular to the r-direction, by the lateral surface area of the cylinder by the moment arm. The minus sign in front of τrθ indicates that the fluid is at a higher radius than the cylinder it is acting upon. The expression for τrθ is given in Table B.1 on page 844. Ti = (−τrθ )|r=κR · 2π(κR)L · κR ∂ vθ = µr · 2πκ2 R2 L ∂r r r=κR 2µκ2 R2 (Ωo − Ωi ) · 2πκ2 R2 L = r2 (1 − κ2 ) r=κR Thus, the torque exerted by the fluid on the inner cylinder is Ti =
4πµLκ2 R2 (Ωo − Ωi ) . 1 − κ2
The torque that needs to be applied to the inner cylinder to maintain the motion is therefore Tai = −
4πµLκ2 R2 (Ωo − Ωi ) . 1 − κ2
Similarly, the torque on the outer cylinder is To = (τrθ )|r=R · 2πRL · R. The sign in front of τrθ is positive here because the fluid is acting on the outer cylinder, which is at a higher radius than the fluid. ∂ vθ = −µr · 2πR2 L ∂r r r=R 2µκ2 R2 (Ωo − Ωi ) =− · 2πR2 L r2 (1 − κ2 ) r=R
Thus, the torque exerted by the fluid on the outer cylinder is To = −
4πµLκ2 R2 (Ωo − Ωi ) . 1 − κ2
The torque that needs to be applied to the outer cylinder to maintain the motion is therefore Tao =
www.stemjock.com
4πµLκ2 R2 (Ωo − Ωi ) . 1 − κ2
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.1
Page 4 of 8
Part (b) 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 φ-direction and that the velocity varies with the distance from the axis of rotation (that is, as a function of r and θ). ˆ v = vφ (r, θ)φ If we assume the fluid does not slip on the walls, then it has the wall’s velocity at r = κR and r = R. The tangential velocity is obtained by multiplying the angular velocity by the distance from the axis of rotation (the moment arm). Boundary Condition 1:
vφ (κR, θ) = Ωi κR sin θ
Boundary Condition 2:
vφ (R, θ) = Ωo R sin θ
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.
(3)
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 also constant, the equation simplifies to the Navier-Stokes equation. ∂ ρv + ∇ · ρvv = −∇p + µ∇2 v + ρg ∂t
(4)
As this is a vector equation, it actually represents three scalar equations—one for each variable in the chosen coordinate system. From Appendix B.4 on page 846, the continuity equation in spherical coordinates is 1 ∂ 1 ∂vφ 1 ∂ 2 (r vr ) + (vθ sin θ) + = 0, 2 θ ∂φ |r ∂r{z } |r sin θ ∂θ{z } |r sin{z } =0
=0
=0
which doesn’t tell us anything. From Appendix B.6 on page 848, the Navier-Stokes equation yields the following three scalar equations in spherical coordinates. =0
z}|{ vθ2 +vφ2 vφ ∂vr ∂vr vθ ∂vr ∂p ∂vr + vr + + − ρ =− ∂t ∂r} |r {z∂θ} r sin θ ∂φ r ∂r |{z} | {z | {z } =0
=0
=0
=0
1 ∂2 2 1 ∂ ∂vr 1 ∂ 2 vr + µ 2 2 (r vr ) + 2 sin θ + 2 2 + ρgr ∂θ r sin θ ∂φ2 |r ∂r{z } |r sin θ ∂θ{z {z } } |
=0
www.stemjock.com
=0
=0
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.1
Page 5 of 8
=0
z}|{ 2 vφ ∂vθ vr vθ −vφ cot θ ∂vθ ∂vθ vθ ∂vθ 1 ∂p + + ρ + vr + =− ∂t ∂r} |r {z∂θ} r sin θ ∂φ r r ∂θ |{z} | {z | {z } =0
=0
=0
=0
1 ∂ 1 ∂ 1 ∂ 2 ∂vθ +µ 2 r + 2 (vθ sin θ) r ∂r ∂r r ∂θ sin θ ∂θ | {z } | {z } =0
=0
∂ 2 vθ 2 ∂vr 2 cot θ ∂vφ 1 + − + ρgθ r2 sin2 θ ∂φ2 |r2{z∂θ} r2 sin θ ∂φ {z } | {z } |
+
=0
=0
=0
vφ ∂vφ vφ vr + vθ vφ cot θ ∂vφ vθ ∂vφ ∂vφ 1 ∂p + + =− ρ + vr + ∂t ∂r} |r {z∂θ} r sin θ ∂φ | r r sin θ ∂φ {z } |{z} | {z | {z } | {z } =0
=0
=0
=0
=0
∂vφ 1 ∂ +µ 2 r2 r ∂r ∂r
=0
1 ∂ 1 ∂ + 2 (vφ sin θ) r ∂θ sin θ ∂θ ∂ 2 vφ 2 ∂vr 2 cot θ ∂vθ 1 + 2 + + ρgφ + 2 2 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 φ-equation, which has simplified considerably from ˆ the assumption that v = vφ (r, θ)φ. 1 ∂ 1 ∂ 1 ∂ 2 ∂vφ 0=µ 2 r + 2 (vφ sin θ) r ∂r ∂r r ∂θ sin θ ∂θ Multiply both sides by r2 /µ. ∂vφ ∂ ∂ 1 ∂ r2 + (vφ sin θ) = 0 ∂r ∂r ∂θ sin θ ∂θ From the boundary conditions, we hypothesize that the solution is of the form vφ (r, θ) = f (r) sin θ. Substitute this into the PDE ∂ ∂ 1 ∂ 2 ∂ 2 r [f (r) sin θ] + [f (r) sin θ] = 0 ∂r ∂r ∂θ sin θ ∂θ and the boundary conditions. vφ (κR, θ) = f (κR) sin θ = Ωi κR sin θ
→
f (κR) = Ωi κR
vφ (R, θ) = f (R) sin θ = Ωo R sin θ
→
f (R) = Ωo R
Expand the left side of the PDE. (2rf 0 + r2 f 00 ) sin θ + (−2 sin θ)f = 0 Divide both sides by sin θ. r2 f 00 + 2rf 0 − 2f = 0 www.stemjock.com
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.1
Page 6 of 8
Because θ no longer appears in the equation, the hypothesis made about the form of the solution is legitimate. What we have now is an equidimensional ODE for f , so it has solutions of the form f (r) = rn . Find the derivatives of it f (r) = rn
→
f 0 (r) = nrn−1
→
f 00 (r) = n(n − 1)rn−2
and substitute these expressions into the ODE to determine n. n(n − 1)rn + 2nrn − 2rn = 0 n(n − 1) + 2n − 2 = 0 n2 + n − 2 = 0 (n + 2)(n − 1) = 0 n = {−2, 1} Consequently, C4 . r2 Use the boundary conditions for f to determine C3 and C4 . f (r) = C3 r +
C4 = Ωi κR κ2 R2 C4 f (R) = C3 R + 2 = Ωo R R
f (κR) = C3 κR +
Solving this system of equations gives C3 =
Ω o − κ3 Ω i 1 − κ3
and C4 =
κ3 R3 (Ωi − Ωo ) . 1 − κ3
We then have Ω o − κ3 Ω i κ3 R3 (Ωi − Ωo ) 1 r + 1 − κ"3 1 − κ3 r2 2 # r κR κR = (Ωo − κ3 Ωi ) + (Ωi − Ωo ) . 1 − κ3 κR r
f (r) =
Therefore, " # r κR 2 κR 3 sin θ. vφ (r, θ) = (Ωo − Ωi κ ) + (Ωi − Ωo ) 1 − κ3 κR r Since vφ depends on θ, the torque on the inner sphere is obtained by integrating (−τrφ )|r=κR · κR sin θ, the product of the moment arm and the viscous force per unit area in the φ-direction on a plane perpendicular to the r-direction, over the sphere’s surface area. The minus sign in front of τrφ indicates that the fluid is at a higher radius than the sphere it is acting upon.
www.stemjock.com
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.1
Page 7 of 8
The expression for τrφ is given in Table B.1 on page 844. ¨ Ti = (−τrφ )|r=κR · κR sin θ dS ˆ
Si 2π
ˆ
π
= 0
ˆ
0
2π
ˆ
2π
ˆ
2π
ˆ
π
= 0
ˆ
0 π
= 0
ˆ
0
(−τrφ )|r=κR · κR sin θ[(κR)2 sin θ dθ dφ] ∂ vφ µr · κ3 R3 sin2 θ dθ dφ ∂r r r=κR 3µκ3 R3 (Ωo − Ωi ) sin θ · κ3 R3 sin2 θ dθ dφ 3 3 r (1 − κ ) r=κR
π
3µ(Ωo − Ωi ) sin θ · κ3 R3 sin2 θ dθ dφ 3 1 − κ 0 0 ˆ 2π ˆ π 3µκ3 R3 (Ωo − Ωi ) 3 dφ sin θ dθ = 1 − κ3 0 0 3µκ3 R3 (Ωo − Ωi ) 4 = (2π) 1 − κ3 3 =
Thus, the torque exerted by the fluid on the inner sphere is Ti =
8πµκ3 R3 (Ωo − Ωi ) . 1 − κ3
The torque that needs to be applied to the inner sphere to maintain the motion is therefore Tai = −
8πµκ3 R3 (Ωo − Ωi ) . 1 − κ3
Similarly, the torque on the outer sphere is ¨ To = (τrφ )|r=R · R sin θ dS So
The sign in front of τrφ is positive here because the fluid is acting on the outer sphere, which is at a higher radius than the fluid. ˆ 2π ˆ π = (τrφ )|r=R · R sin θ(R2 sin θ dθ dφ) 0 0 ˆ 2π ˆ π ∂ vφ = −µr · R3 sin2 θ dθ dφ ∂r r r=R 0 0 ˆ 2π ˆ π 3µκ3 R3 (Ωo − Ωi ) = − sin θ · R3 sin2 θ dθ dφ 3 (1 − κ3 ) r 0 0 r=R ˆ 2π ˆ π 3 3µκ (Ωo − Ωi ) = − sin θ · R3 sin2 θ dθ dφ 3 1 − κ 0 0 ˆ 2π ˆ π 3µκ3 R3 (Ωo − Ωi ) 3 =− dφ sin θ dθ 1 − κ3 0 0 3µκ3 R3 (Ωo − Ωi ) 4 =− (2π) 3 1−κ 3 www.stemjock.com
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.1
Page 8 of 8
Thus, the torque exerted by the fluid on the outer sphere is To = −
8πµκ3 R3 (Ωo − Ωi ) . 1 − κ3
The torque that needs to be applied to the outer sphere to maintain the motion is therefore Tao =
www.stemjock.com
8πµκ3 R3 (Ωo − Ωi ) . 1 − κ3
Page 1 of 8
Problem 3B.1 Flow between coaxial cylinders and concentric spheres. (a) The space between two coaxial cylinders is filled with an incompressible fluid at constant temperature. The radii of the inner and outer wetted surfaces are κR and R, respectively. The angular velocities of rotation of the inner and outer cylinders are Ωi and Ωo . Determine the velocity distribution in the fluid and the torques exerted by the fluid on the two cylinders needed to maintain the motion. (b) Repeat part (a) for two concentric spheres. Answers: r κR κR 2 (a) vθ = (Ωo − Ωi κ ) + (Ωi − Ωo ) 1 − κ2 κR r " # r κR κR 2 3 (Ωo − Ωi κ ) sin θ (b) vφ = + (Ωi − Ωo ) 1 − κ3 κR r Solution Part (a) We assume that the fluid flows only in the θ-direction and that the velocity varies as a function of radius only. v = vθ (r)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. The tangential velocity is obtained by multiplying the angular velocity by the distance from the axis of rotation (the moment arm). Boundary Condition 1:
vθ (κR) = Ωi κR
Boundary Condition 2:
vθ (R) = Ωo R
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.
(1)
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 also constant, the equation simplifies to the Navier-Stokes equation. ∂ ρv + ∇ · ρvv = −∇p + µ∇2 v + ρg ∂t
(2)
As this is a vector equation, it actually represents three scalar equations—one for each variable in the chosen coordinate system. Using cylindrical coordinates is the appropriate choice for this problem, so equations (1) and (2) will be used in (r, θ, z). From Appendix B.4 on page 846, the continuity equation becomes 1 ∂ 1 ∂vθ ∂vz (rvr ) + + = 0, ∂θ} |{z} ∂z |r ∂r{z } |r {z =0
www.stemjock.com
=0
=0
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.1
Page 2 of 8
which doesn’t tell us anything. From Appendix B.6 on page 848, the Navier-Stokes equation yields the following three scalar equations in cylindrical coordinates. ∂vr vθ ∂vr ∂vr vθ2 1 ∂ 2 vr ∂ 2 vr ∂vr ∂p ∂ 1 ∂ 2 ∂vθ ρ + vr + + vz − =− +µ (rvr ) + 2 + − 2 + ρgr 2 |{z} ∂t ∂r} |r {z∂θ} | {z ∂z} r ∂r ∂r r ∂r r {z ∂θ2} |∂z r {z∂θ} |{z} | {z | {z } | {z } | =0 =0
=0
=0
=0
=0
=0
=0
=0
∂vθ ∂vθ vθ ∂vθ ∂vθ vr vθ 1 ∂p ∂ 1 ∂ 2 ∂vr 1 ∂ 2 vθ ∂ 2 vθ ρ + vr + + vz + =− +µ + + 2 + ρgθ (rvθ ) + 2 2 |{z} ∂t ∂r} |r {z∂θ} | {z ∂z} | {z r} ∂r r ∂r ∂θ2} |∂z |{z} | {z | r{z∂θ} {z } |r {z∂θ} |r {z =0 =0 =0 =0 =0 =0 =0 =0 =0 =0 ∂vz vθ ∂vz ∂vz ∂vz ∂p 1 ∂ 1 ∂ 2 vz ∂ 2 vz ∂vz + vr r + + vz =− +µ + 2 + + ρgz ρ ∂t ∂r} |r {z∂θ} | {z ∂z} ∂z r ∂r ∂r r ∂θ2 ∂z 2 |{z} | {z | {z } | {z } | {z } =0
=0
=0
=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. d 1 d 0=µ (rvθ ) dr r dr Divide both sides by µ. d 1 d (rvθ ) = 0 dr r dr Integrate both sides with respect to r. 1 d (rvθ ) = C1 r dr Multiply both sides by r. d (rvθ ) = C1 r dr Integrate both sides with respect to r once more. rvθ = C1
r2 + C2 2
Divide both sides by r. r C2 vθ (r) = C1 + 2 r Apply the two boundary conditions here to determine C1 and C2 . κR C2 + = Ωi κR 2 κR R C2 vθ (R) = C1 + = Ωo R 2 R
vθ (κR) = C1
Solving the system of equations yields C1 =
2(Ωo − κ2 Ωi ) 1 − κ2
We then have vθ (r) = www.stemjock.com
and C2 =
κ2 R2 (Ωi − Ωo ) . 1 − κ2
2(Ωo − κ2 Ωi ) r κ2 R2 (Ωi − Ωo ) 1 + . 1 − κ2 2 1 − κ2 r
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.1
Page 3 of 8
Therefore, upon factoring κR/(1 − κ2 ), r κR κR 2 (Ωo − Ωi κ ) . + (Ωi − Ωo ) vθ (r) = 1 − κ2 κR r The torque on the inner cylinder is obtained by multiplying (−τrθ )|r=κR , the viscous force per unit area in the θ-direction on a plane perpendicular to the r-direction, by the lateral surface area of the cylinder by the moment arm. The minus sign in front of τrθ indicates that the fluid is at a higher radius than the cylinder it is acting upon. The expression for τrθ is given in Table B.1 on page 844. Ti = (−τrθ )|r=κR · 2π(κR)L · κR ∂ vθ = µr · 2πκ2 R2 L ∂r r r=κR 2µκ2 R2 (Ωo − Ωi ) · 2πκ2 R2 L = r2 (1 − κ2 ) r=κR Thus, the torque exerted by the fluid on the inner cylinder is Ti =
4πµLκ2 R2 (Ωo − Ωi ) . 1 − κ2
The torque that needs to be applied to the inner cylinder to maintain the motion is therefore Tai = −
4πµLκ2 R2 (Ωo − Ωi ) . 1 − κ2
Similarly, the torque on the outer cylinder is To = (τrθ )|r=R · 2πRL · R. The sign in front of τrθ is positive here because the fluid is acting on the outer cylinder, which is at a higher radius than the fluid. ∂ vθ = −µr · 2πR2 L ∂r r r=R 2µκ2 R2 (Ωo − Ωi ) =− · 2πR2 L r2 (1 − κ2 ) r=R
Thus, the torque exerted by the fluid on the outer cylinder is To = −
4πµLκ2 R2 (Ωo − Ωi ) . 1 − κ2
The torque that needs to be applied to the outer cylinder to maintain the motion is therefore Tao =
www.stemjock.com
4πµLκ2 R2 (Ωo − Ωi ) . 1 − κ2
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.1
Page 4 of 8
Part (b) 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 φ-direction and that the velocity varies with the distance from the axis of rotation (that is, as a function of r and θ). ˆ v = vφ (r, θ)φ If we assume the fluid does not slip on the walls, then it has the wall’s velocity at r = κR and r = R. The tangential velocity is obtained by multiplying the angular velocity by the distance from the axis of rotation (the moment arm). Boundary Condition 1:
vφ (κR, θ) = Ωi κR sin θ
Boundary Condition 2:
vφ (R, θ) = Ωo R sin θ
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.
(3)
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 also constant, the equation simplifies to the Navier-Stokes equation. ∂ ρv + ∇ · ρvv = −∇p + µ∇2 v + ρg ∂t
(4)
As this is a vector equation, it actually represents three scalar equations—one for each variable in the chosen coordinate system. From Appendix B.4 on page 846, the continuity equation in spherical coordinates is 1 ∂ 1 ∂vφ 1 ∂ 2 (r vr ) + (vθ sin θ) + = 0, 2 θ ∂φ |r ∂r{z } |r sin θ ∂θ{z } |r sin{z } =0
=0
=0
which doesn’t tell us anything. From Appendix B.6 on page 848, the Navier-Stokes equation yields the following three scalar equations in spherical coordinates. =0
z}|{ vθ2 +vφ2 vφ ∂vr ∂vr vθ ∂vr ∂p ∂vr + vr + + − ρ =− ∂t ∂r} |r {z∂θ} r sin θ ∂φ r ∂r |{z} | {z | {z } =0
=0
=0
=0
1 ∂2 2 1 ∂ ∂vr 1 ∂ 2 vr + µ 2 2 (r vr ) + 2 sin θ + 2 2 + ρgr ∂θ r sin θ ∂φ2 |r ∂r{z } |r sin θ ∂θ{z {z } } |
=0
www.stemjock.com
=0
=0
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.1
Page 5 of 8
=0
z}|{ 2 vφ ∂vθ vr vθ −vφ cot θ ∂vθ ∂vθ vθ ∂vθ 1 ∂p + + ρ + vr + =− ∂t ∂r} |r {z∂θ} r sin θ ∂φ r r ∂θ |{z} | {z | {z } =0
=0
=0
=0
1 ∂ 1 ∂ 1 ∂ 2 ∂vθ +µ 2 r + 2 (vθ sin θ) r ∂r ∂r r ∂θ sin θ ∂θ | {z } | {z } =0
=0
∂ 2 vθ 2 ∂vr 2 cot θ ∂vφ 1 + − + ρgθ r2 sin2 θ ∂φ2 |r2{z∂θ} r2 sin θ ∂φ {z } | {z } |
+
=0
=0
=0
vφ ∂vφ vφ vr + vθ vφ cot θ ∂vφ vθ ∂vφ ∂vφ 1 ∂p + + =− ρ + vr + ∂t ∂r} |r {z∂θ} r sin θ ∂φ | r r sin θ ∂φ {z } |{z} | {z | {z } | {z } =0
=0
=0
=0
=0
∂vφ 1 ∂ +µ 2 r2 r ∂r ∂r
=0
1 ∂ 1 ∂ + 2 (vφ sin θ) r ∂θ sin θ ∂θ ∂ 2 vφ 2 ∂vr 2 cot θ ∂vθ 1 + 2 + + ρgφ + 2 2 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 φ-equation, which has simplified considerably from ˆ the assumption that v = vφ (r, θ)φ. 1 ∂ 1 ∂ 1 ∂ 2 ∂vφ 0=µ 2 r + 2 (vφ sin θ) r ∂r ∂r r ∂θ sin θ ∂θ Multiply both sides by r2 /µ. ∂vφ ∂ ∂ 1 ∂ r2 + (vφ sin θ) = 0 ∂r ∂r ∂θ sin θ ∂θ From the boundary conditions, we hypothesize that the solution is of the form vφ (r, θ) = f (r) sin θ. Substitute this into the PDE ∂ ∂ 1 ∂ 2 ∂ 2 r [f (r) sin θ] + [f (r) sin θ] = 0 ∂r ∂r ∂θ sin θ ∂θ and the boundary conditions. vφ (κR, θ) = f (κR) sin θ = Ωi κR sin θ
→
f (κR) = Ωi κR
vφ (R, θ) = f (R) sin θ = Ωo R sin θ
→
f (R) = Ωo R
Expand the left side of the PDE. (2rf 0 + r2 f 00 ) sin θ + (−2 sin θ)f = 0 Divide both sides by sin θ. r2 f 00 + 2rf 0 − 2f = 0 www.stemjock.com
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.1
Page 6 of 8
Because θ no longer appears in the equation, the hypothesis made about the form of the solution is legitimate. What we have now is an equidimensional ODE for f , so it has solutions of the form f (r) = rn . Find the derivatives of it f (r) = rn
→
f 0 (r) = nrn−1
→
f 00 (r) = n(n − 1)rn−2
and substitute these expressions into the ODE to determine n. n(n − 1)rn + 2nrn − 2rn = 0 n(n − 1) + 2n − 2 = 0 n2 + n − 2 = 0 (n + 2)(n − 1) = 0 n = {−2, 1} Consequently, C4 . r2 Use the boundary conditions for f to determine C3 and C4 . f (r) = C3 r +
C4 = Ωi κR κ2 R2 C4 f (R) = C3 R + 2 = Ωo R R
f (κR) = C3 κR +
Solving this system of equations gives C3 =
Ω o − κ3 Ω i 1 − κ3
and C4 =
κ3 R3 (Ωi − Ωo ) . 1 − κ3
We then have Ω o − κ3 Ω i κ3 R3 (Ωi − Ωo ) 1 r + 1 − κ"3 1 − κ3 r2 2 # r κR κR = (Ωo − κ3 Ωi ) + (Ωi − Ωo ) . 1 − κ3 κR r
f (r) =
Therefore, " # r κR 2 κR 3 sin θ. vφ (r, θ) = (Ωo − Ωi κ ) + (Ωi − Ωo ) 1 − κ3 κR r Since vφ depends on θ, the torque on the inner sphere is obtained by integrating (−τrφ )|r=κR · κR sin θ, the product of the moment arm and the viscous force per unit area in the φ-direction on a plane perpendicular to the r-direction, over the sphere’s surface area. The minus sign in front of τrφ indicates that the fluid is at a higher radius than the sphere it is acting upon.
www.stemjock.com
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.1
Page 7 of 8
The expression for τrφ is given in Table B.1 on page 844. ¨ Ti = (−τrφ )|r=κR · κR sin θ dS ˆ
Si 2π
ˆ
π
= 0
ˆ
0
2π
ˆ
2π
ˆ
2π
ˆ
π
= 0
ˆ
0 π
= 0
ˆ
0
(−τrφ )|r=κR · κR sin θ[(κR)2 sin θ dθ dφ] ∂ vφ µr · κ3 R3 sin2 θ dθ dφ ∂r r r=κR 3µκ3 R3 (Ωo − Ωi ) sin θ · κ3 R3 sin2 θ dθ dφ 3 3 r (1 − κ ) r=κR
π
3µ(Ωo − Ωi ) sin θ · κ3 R3 sin2 θ dθ dφ 3 1 − κ 0 0 ˆ 2π ˆ π 3µκ3 R3 (Ωo − Ωi ) 3 dφ sin θ dθ = 1 − κ3 0 0 3µκ3 R3 (Ωo − Ωi ) 4 = (2π) 1 − κ3 3 =
Thus, the torque exerted by the fluid on the inner sphere is Ti =
8πµκ3 R3 (Ωo − Ωi ) . 1 − κ3
The torque that needs to be applied to the inner sphere to maintain the motion is therefore Tai = −
8πµκ3 R3 (Ωo − Ωi ) . 1 − κ3
Similarly, the torque on the outer sphere is ¨ To = (τrφ )|r=R · R sin θ dS So
The sign in front of τrφ is positive here because the fluid is acting on the outer sphere, which is at a higher radius than the fluid. ˆ 2π ˆ π = (τrφ )|r=R · R sin θ(R2 sin θ dθ dφ) 0 0 ˆ 2π ˆ π ∂ vφ = −µr · R3 sin2 θ dθ dφ ∂r r r=R 0 0 ˆ 2π ˆ π 3µκ3 R3 (Ωo − Ωi ) = − sin θ · R3 sin2 θ dθ dφ 3 (1 − κ3 ) r 0 0 r=R ˆ 2π ˆ π 3 3µκ (Ωo − Ωi ) = − sin θ · R3 sin2 θ dθ dφ 3 1 − κ 0 0 ˆ 2π ˆ π 3µκ3 R3 (Ωo − Ωi ) 3 =− dφ sin θ dθ 1 − κ3 0 0 3µκ3 R3 (Ωo − Ωi ) 4 =− (2π) 3 1−κ 3 www.stemjock.com
BSL Transport Phenomena 2e Revised: Chapter 3 - Problem 3B.1
Page 8 of 8
Thus, the torque exerted by the fluid on the outer sphere is To = −
8πµκ3 R3 (Ωo − Ωi ) . 1 − κ3
The torque that needs to be applied to the outer sphere to maintain the motion is therefore Tao =
www.stemjock.com
8πµκ3 R3 (Ωo − Ωi ) . 1 − κ3
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