PROBLEM SET #3 SOLUTIONS 3B.11 Radial flow between two coaxial cylinders
Consider an incompressible fluid, at constant temperature flowing radially between two porous cylindrical shells with inner and outer radii κR and R.
(a) Show that the equation of continuity leads to vr = C/r, where C is a constant. Solution
Assume that vz =vθ =0and that the flow is steady. Continuity equation in cylindrical coordinates:
(
)
1( )
( )
0 1 = ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ z r v z v r rv r r t ρ θ ρ ρ ρ θEliminate time, v and θ v terms to obtain: z
(
)
0 1 = ∂ ∂ r rv r r ρ Integrate to obtain: = r rv constant(b) Simplify the components of the equation of motion to obtain the following expressions for the modified pressure distribution:
0 , 0 , = = − = dz dP d dP dr dv v dr dP r r θ ρ Solution BSL page 848 equation B.6-4.
( )
r r r r r z r r r r v g r z v v r rv r r r r P r v z v v v r v r v v t v ρ θ θ µ θ ρ θ θ θ+ ∂ ∂ − ∂ ∂ + ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ − = − ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ 2 2 2 2 2 2 2 2 1 1Assume steady flow with vθ =vz =0, no viscous flow as in part (a):
r P r v v r r ∂ ∂ − = ∂ ∂ ρ
(c) Integrate the expression for dP/dr above to get
[
]
− = − 2 2 1 ) ( 2 1 ) ( ) ( r R R v R P r P ρ r Solution r P r v v r r ∂ ∂ − = ∂ ∂ ρ
Integrate between R and r:
(
)
− − = − − = − − = − = − 2 2 22 2 2 2 2 2 1 2 1 2 2 2 2 R r r v v v r v v v v P P r R r r r r R r r r r R R R ρ ρ ρ ρ ρNote: from (a) rvr =constant, so r2vr2 =C2 and 2 2 2 2 R r v v R R r r =
Direction of flow r
R mR
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.
(a) Show by use of the equation of continuity that r2vr = C, where C is a constant.
Solution
Begin with the equation of Continuity in Spherical Coordinates:
(
ρ)
θ θ(
ρ θ θ)
θ φ( )
ρ φ ρ v r v r v r r r t r ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ sin 1 sin sin 1 1 2 2 =0Assume radial flow, so that vθ =0=vφ.
( )
0 1 2 2 ∂ = ∂ r v r r rSo that after integration:
=
r
v r2
Constant
(b) Show by the use of the equations of motion that the pressure distribution in this
system is described by the equations:
φ ρ φ θ g p r ∂ = ∂ sin 1 θ ρ θ g p r ∂ = ∂ 1
( )
r r r gr r v v v r r r r p ρ ρ µ + ∂ ∂ − ∂ ∂ = ∂ ∂ 2 2 2 2Solution
Begin with the equation of motion in spherical coordinates (BSL equation B.6-7 page 848). r-component
( )
r r r r r r r r r g v r v r v r r r r p r v v v r v v r v r v v t v ρ φ θ θ θ θ θ µ φ θ θ ρ θ φ θ φ + ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ + ∂ ∂ − = − − ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ 2 2 2 2 2 2 2 2 2 2 2 sin 1 sin sin 1 1 sinAssuming steady state and that vφ =0=vθ, then the above equation reduces to:
( )
r r r r r v g r r r p r v v µ ρ ρ + ∂ ∂ + ∂ ∂ − = ∂ ∂ 2 2 2 2 1 Rearrange:( )
r r r r g r v v v r r r r p ρ ρ µ + ∂ ∂ − ∂ ∂ = ∂ ∂ 2 2 2 2 θ-component θ φ θ θ θ φ θ θ φ θ θ θ θ ρ φ θ θ θ φ θ θ θ θ θ µ θ θ φ θ θ ρ g v r v r v r v r r v r r r p r r v r v v v r v v r v r v v t v r r r + ∂ ∂ − ∂ ∂ + ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ − = − + ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ 2 2 2 2 2 2 2 2 2 2 2 sin cos 2 2 sin 1 sin ( sin 1 1 1 1 cot sin Since vφ =0=vθ, θ ρ θ g p r ∂ + ∂ − = 1 0 Rearranging, θ ρ θ g p r∂ = ∂ 1φ-component θ θ φ φ φ φ θ φ φ φ φ θ φ φ ρ φ θ θ φ θ φ θ θ θ θ θ µ φ θ θ φ θ θ ρ g v r v r v r v r r v r r r p r r v v r v v v r v v r v r v v t v r r r + ∂ ∂ − ∂ ∂ + ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ − = + − ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ 2 2 2 2 2 2 2 2 2 2 sin cos 2 sin 2 sin 1 sin ( sin 1 1 1 sin 1 cot sin Since vφ =0=vθ, φ ρ φ θ g p r ∂ + ∂ − = sin 1 0 Rearranging, θ
ρ
θ
g
p
rs
∂
=
∂
1
(c) Show that the radial pressure distribution may be expressed in terms of the quantity
P as: − = − = 4 2 1 2 1 r R v P P R r r R ρ Solution
Begin with equations from results of part (b),
( )
r r r r g r v v v r r r r p µ ρ +ρ ∂ ∂ − ∂ ∂ = ∂ ∂ 2 2 2 2 (1) using, θ ρ φ gφrsin p = ∂ ∂ (2) r g p θ ρ θ = ∂ ∂ (3) gr p gr p P= +ρ cosθ = +ρ θ ρ cosg r p r P + ∂ ∂ = ∂ ∂ θ ρ θ θ grsin p P − ∂ ∂ = ∂ ∂ Substitute (3) 0 sin = − = ∂ ∂ ρ ρ θ θ gθr gr P so P≠ P(θ)( )
ρ ρ µ( )
ρ ρ θ ρ θ µθ
ρ cos 2 2 cos cos
2 2 2 2 2 2 g g r v v v r r r g r v v v r r r g r p r P r r r r r r r ∂ + − ∂ − ∂ ∂ = + ∂ ∂ − ∂ ∂ = + ∂ ∂ = ∂ ∂
( )
2 2 2 2 2 2 r v r v v v r r r r P r r r r ρ µ µ − ∂ ∂ − ∂ ∂ = ∂ ∂ Since r vr = 2 Constant, then( )
0 2 = ∂ ∂ r v r r( )
( )
5 2 2 2 2 2 2 2 2 2 2 r C r v v v r v r r P r r r r ρ ρ ρ µ µ − + = = = ∂∂ where C is the constant from part (a)
dr r C dP 5 2 2ρ =
∫
∫
= r R P P r dr C dP r 5 2 2ρ − = − 2 14 14 4 2 R r C P P r ρ Substitute C=R2vr and C2=R4vr2( )
( )
− = − = − 4 4 2 4 4 4 2 4 1 1 2 1 1 1 2 1 r R R v R R r v R P P r ρ r ρ r And thus, − = − = 4 2 1 2 1 r R v P P R r r R ρ(d) Please write expressions for τrr, τθθ, τϕϕ, τrθ, τrϕ, and τθϕ for this system.
Solution
(
)
− ∂∂ ∂ ∂ − = − ∇• ∂ ∂ − = r v r v v r vr r r rr 3 2 2 3 2 2 µ µ τNext, utilize the result of (a) to obtain:
( )
∂ ∂ − ∂ ∂ − r v r r r vr r rr 2 2 3 2 2 µ τ Since r2vr = Constant, − − = r vr rr 2µ 2 τ r vr rr µ τ =4(
)
− ∂∂ + ∂ ∂ − = − ∇• + ∂ ∂ − = r v r v v r v r v v r r r r 3 2 1 2 3 2 1 2 θ µ θ µ τ θ θ θθUsing the same strategy as above,
( )
( )
∂
∂
−
+
∂
∂
−
=
∂
∂
−
+
∂
∂
−
=
r
v
r
r
r
v
v
r
r
v
r
r
r
v
v
r
r r r r 2 2 2 23
2
2
2
3
2
1
2
θ
µ
θ
µ
τ
θ θ θθ r vr µ τθθ =−2(
)
( )
∂ ∂ − − = • ∇ − + + ∂ ∂ − = r v r r r v v r v r v v r r r r 2 2 3 2 2 3 2 cot sin 1 2 θ µ φ θ µ τ φ θ φφ r vr µ τφφ =−2Since we have symmetric flow,
0 = = = φ θφ θ τ τ τr r
