PROBLEM SET #3 SOLUTIONS 3B.11 Radial flow between two coaxial c oaxial cylinders Consider an incompressible fluid, at constant temperature flowing radi ally between two porous cylindrical shells with inner and outer radii κ R and R. (a) Show that the equation of continuity leads to vr = = C/r, where C is a constant. Solution Assume that v z = vθ
= 0 and that the flow is stead y.
Continuity equation in cylindrical coordinates: ∂ρ 1 ∂ ( ) 1 ∂ ( ) ∂ ( ) + ρrv + ρv + ρv = 0 ∂t r ∂r r r ∂θ θ ∂ z z Eliminate time, vθ and v z terms to obtain:
∂( ) ρrvr = 0 r ∂r
1 Integrate to obtain:
rvr = constant (b) Simplify the components of the equation of motion to obtain the following expressions for the modified mod ified pressure distribution: dP = − ρvr dvr , dP = 0, dP = 0 dr dr d θ dz Solution BSL page 848 equation B.6-4.
∂vr ∂v v ∂v ∂v v 2 ∂ P ∂ 1 ∂ ( ) 1 ∂ 2vr ∂ 2vr 2 ∂vθ rvr + 2 + vr r + θ r + v z r − θ + 2 − 2 + ρ g r 2 = − ∂r + µ ∂r t r r z r r r ∂ ∂ ∂ θ ∂ ∂ ∂ θ ∂ z r ∂θ r Assume steady flow with vθ = v z = 0 , no viscous flow as in part (a): ρ
ρvr
(c)
∂vr ∂ P =− ∂r ∂r
R 2 Integrate the expression for dP/dr above to get P (r ) − P ( R ) = ρ [vr ( R ) ] 1 − 2 r 1
2
Solution ρvr
∂vr = − ∂ P ∂r ∂r
Integrate between R and r: r ρvr 2 ρvr 2 − ρvr 2r vr 2 r − ρvr 2r r 2 − ρv R2 1 − 2 = ( P − P R ) = = − − = 1 − R 2 2 R 2 2 2 2 v R r r R
Note: from (a) rvr = constant, so r v
2 2 r
= C and 2
vr 2 v R2
r R
=
r 2 R 2
Radial Flow Between Concentric Spheres Consider an isothermal, incompressible fluid flowing radially between two concentric porous spherical shells. Assume steady laminar flow with vr = vr (r). Note that the velocity is not assumed zero at the solid surfaces. See the figure below.
R mR
r
Direction of flow
2
(a) Show by use of the equation of continuity that r vr = C, where C is a constant. Solution Begin with the equation of Continuity in Sphe rical Coordinates: ∂ρ 1 ∂ 2 ∂ ( ∂ 1 + 2 (ρr vr ) + 1 ρvθ sin θ ) + (ρvφ ) =0 ∂t r ∂r r sin θ ∂θ r sin θ ∂φ Assume radial flow, so that vθ
= 0 = vφ . 1 2
r
∂ 2 (r v ) = 0 ∂r r
So that after integration: r 2vr = Constant
(b) Show by the use of the equations of motion that the pressure distribution in this system is described by the equations: 1 ∂ p = ρ g φ r sin θ ∂φ 1 ∂ p r ∂θ
= ρ g θ
∂ p µ ∂ 2 ( 2 ) ∂v = 2 2 r vr − ρvr r + ρ g r ∂r r ∂r ∂r
Solution Begin with the equation of motion in spherical coordinates (BSL equation B.6-7 page 848). r-component
∂vr vφ ∂v v ∂v + vr r + θ r + ∂t ∂r r ∂θ r sin θ
ρ
1 ∂2 2 1 ∂ ∂vr ( ) r v sin + θ + r 2 ∂r 2 r r 2 sin θ ∂θ ∂ θ ∂vr vθ2 − vφ2 ∂ p + ρ g r − = − + µ 2 ∂φ ∂r r ∂ vr 1 2 2 2 r sin θ ∂φ
Assuming steady state and that vφ
= 0 = vθ , then the above equation reduces to:
∂vr ∂ p 1 ∂ 2 (r 2v ) g v ρ r = − + µ r + ρ r ∂r ∂r r 2 ∂r 2 Rearrange:
∂ p µ ∂ 2 ( 2 ) ∂v = 2 2 r vr − ρvr r + ρ g r ∂r r ∂r ∂r
θ-component
∂vθ vφ ∂vθ vr vθ ∂v v ∂v vr θ + θ θ + + + ∂t r r r sin r ∂ ∂ θ θ ∂ φ
ρ
Since vφ
1 ∂ 2 ∂vθ r r 2 ∂r ∂r 1 ∂ 1 ∂ + 2 (vθ sin θ 2 r ∂θ sin θ ∂θ v cot θ 1 ∂ p − φ = − + µ + ρ g θ 2 r r ∂ θ 1 v 2 v ∂ ∂ + r 2 sin 2 θ ∂φ 2θ + r 2 ∂θr 2 cosθ ∂vφ − 2 2 r sin θ ∂ φ
= 0 = vθ , 0=−
1 ∂ p r ∂θ
+ ρ g θ
Rearranging, 1 ∂ p r ∂θ
= ρ g θ
φ-component
∂v v ∂v v ∂vφ vr vφ vθ vφ cot θ ∂vφ 1 + vr φ + θ φ + φ + − =− ∂r r ∂θ r sin θ ∂φ r r ∂t r sin θ
ρ
Since vφ
1 ∂ 2 ∂vφ r 2 ∂r r ∂r 1 ∂ 1 ∂ (vφ sin θ + 2 ∂ p r ∂θ sin θ ∂θ + ρ g + µ θ 2 ∂φ ∂ vφ 1 2 ∂ v r + 2 2 + r sin θ ∂φ 2 r 2 sin θ ∂φ 2 cos θ ∂v θ − 2 2 r sin θ ∂φ
= 0 = vθ , 0=−
1 r sin θ
∂ p + ρ g φ ∂φ
Rearranging,
1 rs
∂ p = ∂θ
ρ g θ
(c) Show that the radial pressure distribution may be exp ressed in terms of the quantity P as: P − P R
=
1 2
2 ρ vr
R 4 1 − r = R r
Solution Begin with equations from results of part (b),
∂ p µ ∂ 2 ( 2 ) ∂v = 2 2 r vr − ρvr r + ρ g r ∂r r ∂r ∂r
(1)
using,
∂ p = ρ g φ r sin θ ∂φ ∂ p = ρ g θ r ∂θ P = p + ρ gr cosθ = p + ρ gr ∂ P = ∂ p + ρ g cosθ ∂r ∂r ∂ P = ∂ p − ρ gr sin θ ∂θ ∂θ Substitute (3)
∂ P = ρ g θ r − ρ gr sin θ = 0 so P ≠ P (θ ) ∂θ
(2)
(3)
∂ P ∂ p µ ∂2 ∂ µ ∂2 ∂ = + ρ g cosθ = 2 2 (r 2vr ) − ρvr vr + ρ g r = 2 2 (r 2vr ) − ρvr vr + ρ g cosθ − ρ g cosθ ∂r ∂r ∂r ∂r r ∂r r ∂r v ∂ P µ ∂ 2 2 ∂v = 2 2 (r vr ) − ρvr r − 2µ r 2 r ∂r r ∂r ∂r
Since r vr = Constant, then 2
∂ (r 2vr ) =0 ∂r
2 v 2 2 C 2 ∂ P µ ( ) µ ( ) = 2 2vr − 2 2vr + 2 ρvr 2 = ρ r = ρ 5 where C is the constant from part (a) r r r ∂r r
dP =
2ρC 2 r 5
P
∫
P r
dP = 2ρC 2
P − P r = 2
2
Substitute C=R vr and C
P − P r =
2
ρ ( R v
dr
∫ r R
5
1 4 − 4 r R
4
2 =R vr 4 2 r
r
2 ρC 2 1
4
1
dr
4 1 1 1 1 R 4 2 ) 4 − 4 = ρ ( R vr ) 4 1 − R r r R 2
And thus, P − P R
=
1 2
2 ρ vr
R 4 1 − r = R r
(d) Please write expressions for τrr , τθθ , τϕϕ , τr θ , τr ϕ , and τθϕ for this system. Solution
∂vr − 2 (∇ • v ) = − µ 2 ∂vr − 2 ∂vr ∂r 3 ∂r ∂r 3
τ rr = − µ 2
Next, utilize the result of (a) to obtain:
∂vr 2 ∂ (r 2vr ) τ rr − µ 2 − 2 r 3r ∂r ∂ Since r 2vr = Constant,
τ rr = −2µ − 2 τ rr = 4µ
τ θθ
vr r
vr r
1 ∂vθ vr 2 1 ∂vθ vr 2 ∂vr = − µ 2 + − (∇ • v ) = − µ 2 + − r ∂θ r 3 r ∂θ r 3 ∂r
Using the same strategy as above,
1 ∂vθ vr 2 ∂(r 2vr ) 2 ∂vθ 2vr 2 ∂(r 2vr ) τθθ = −µ2 + − 2 = −µ + − 2 ∂ θ ∂ ∂ θ ∂ r r 3 r r r r 3 r r =−2
τ θθ
µvr r
1 ∂vφ vr vθ cot θ 2 2vr 2 ∂ (r 2vr ) τ φφ = − µ 2 + + − 3 (∇ • v ) = − µ r − 3r 2 ∂r ∂ r sin r r θ φ τ φφ
= − 2µvr r
Since we have symmetric flow, τ r θ
= τ r φ = τ θφ = 0