Vorticity Equation Return to Viscous Incompressible ﬂow

13.021 – Marine Hydrodynamics, Fall 2004 Lecture 9

13.021 - Marine Hydrodynamics Lecture 9

Vorticity Equation Return to viscous incompressible ﬂow. ³ ´ * ∂ v * * p 2* N-S equation: ∂t + v · ∇v = −∇ ρ + gy + ν∇ v

* ¡ ¢ ∂ ω * * 2* ∇ × () −→ ∂t + ∇ × v · ∇v = ν∇ ω since ∇ × ∇φ = 0 for any φ (conservative forces)

Now:

1 ¡ ¢ ¡ ¢ (*v · ∇)*v = ∇ *v · *v − *v × ∇ × *v 2 µ 2 ¶ v ¯ ¯2 = ∇ − *v × ω* where v2 ≡ ¯*v¯ = *v · *v 2 µ ¶ ¡ ¢ v2 ¡ ¢ ¡ ¢ ∇ × *v · ∇ *v = ∇ × ∇ − ∇ × *v × ω* = ∇ × ω* × *v 2 ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ = *v · ∇ ω* − ω* · ∇ *v + ω* ∇ · *v + *v ∇ · ω* | {z } | {z } = 0 = 0 since incompressible ∇ · (∇ × ~v) = 0 ﬂuid

1 Therefore,

∂ω* ¡ ¢ ¡ ¢ + *v · ∇ ω* = ω* · ∇ *v + ν∇2ω* ∂t or Dω* ¡ ¢ = ω* · ∇ *v + ν∇2ω* Dt | {z } diﬀusion • Kelvin’s Theorem revisited.

* ¡ ¢ D ω * * * * If ν ≡ 0, then Dt = ω · ∇ v, so if ω ≡ 0 everywhere at one time, ω ≡ 0 always. • ν can be thought of as diﬀusivity of (momentum) and vorticity, i.e., ω* once generated (on boundaries only) will spread/diﬀuse in space if ν is present.

G G ω ω G G Dv 2 G Dω G = υ∇ v +... = υ∇2ω+ ... Dt Dt

∂T 2 • Diffusion of vorticity is analogous to the heat equation: ∂t = K∇ T , where K is the heat diffusivity ¡√ ¢ Also since ν ∼ 1 or 2 mm2/s, in 1 second, diffusion distance ∼ O νt ∼ O (mm), whereas diffusion time ∼ O (L2/ν). So for a diffusion distance of L = 1cm, the necessary diffusion time needed is O(10)sec.

2 * ∂ * * * • For 2D, v = (u, v, 0) and ∂z ≡ 0. So, ω = ∇ × v is ⊥ to v (parallel to z-axis). Then,  

¡* ¢ *  ∂ ∂ ∂  * ω · ∇ v = ω + ω + ω v ≡ 0, |{z}x y z  ∂x |{z} ∂y |{z}∂z 0 0 0 so in 2D we have

Dω* = ν∇2ω.* Dt

D*ω If ν = 0, Dt = 0, i.e. in 2D, following a particle, the angular velocity is conserved. Reason: in 2D, the length of a vortex tube cannot change due to continuity. • For 3D,

Dω ∂v ∂2ω i = ω i + ν i Dt j ∂x ∂x ∂x | {z j} | {zj j} vortex turning and stretching diffusion e.g.

Dω ∂u ∂u ∂u 2 = ω 2 + ω 2 + ω 2 +diﬀusion Dt 1 ∂x 2 ∂x 3 ∂x | {z 1} | {z 2} | {z 3} vortex turning vortex stretching vortex turning

1 ∂ u2 3 ∂ x1

3 Example: Pile on a River

Scouring

What really happens as length of the vortex tube L increases?

IFCF is no longer a valid assumption.

Why? Ideal ﬂow assumption implies that the inertia forces are much larger than the viscous eﬀects (Reynolds number). UL Re ∼ ν

Length increases ⇒ diameter becomes really small ⇒ Re is not that big after all. Therefore IFCF is no longer valid.

4 3.3 Potential Flow - ideal (inviscid and incompressible) and irrotational ﬂow If ω* ≡ 0 at some time t, then ω* ≡ 0 always for ideal ﬂow under conservative body forces by Kelvin’s theorem.

Given a vector ﬁeld *v for which ω* = ∇ × *v ≡ 0, then there exists a potential function (scalar) - the velocity potential - denoted as φ, for which

*v = ∇φ Note that ω* = ∇ × *v = ∇ × ∇φ ≡ 0 for any φ, so irrotational ﬂow guaranteed automatically. At a point ~x and time t, the velocity vector ~v(~x,t) in cartesian coordinates in terms of the potential function φ(~x,t) is given by µ ¶ ¡ ¢ ¡ ¢ ∂φ ∂φ ∂φ *v *x, t = ∇φ *x, t = , , ∂x ∂y ∂z

φ(x)

u → ←u

x u=0

The velocity vector ~v is the gradient of the potential function φ, so it always points towards higher values of the potential function.

5 Governing Equations:

Continuity:

∇ · *v = 0 = ∇ · ∇φ ⇒ ∇2φ = 0 Number of unknowns → φ

Number of equations → ∇2φ = 0

Therefore the problem is closed. φ and p (pressure) are decoupled. φ can be solved independently ﬁrst, and after it is obtained, the pressure p is evaluated. ¡ ¢ p = f *v = f (∇φ) → Solve for φ, then ﬁnd pressure.

3.4 Bernoulli equation for potential flow (steady or unsteady) Euler eq: µ ¶ µ ¶ ∂*v v2 p + ∇ − *v × ω* = −∇ + gy ∂t 2 ρ Substitute *v = ∇φ into the Euler’s equation above, which gives: µ ¶ µ ¶ µ ¶ ∂φ 1 p ∇ + ∇ |∇φ|2 = −∇ + gy ∂t 2 ρ or ½ ¾ ∂φ 1 p ∇ + |∇φ|2 + + gy = 0, ∂t 2 ρ which implies that ∂φ 1 p + |∇φ|2 + + gy = f(t) ∂t 2 ρ everywhere in the fluid for unsteady, potential flow. The equation above can be written as · ¸ ∂φ 1 p = −ρ + |∇φ|2 + gy + F (t) ∂t 2

6 which is the Bernoulli equation for unsteady or steady potential ﬂow.

Summary: Bernoulli equation for ideal flow. • Steady rotational or irrotational flow along streamline. µ ¶ 1 p = −ρ v2 + gy + C(ψ) 2 • Unsteady or steady irrotational flow everywhere in the fluid. µ ¶ ∂φ 1 p = −ρ + |∇φ|2 + gy + F (t) ∂t 2

* ∂ • For hydrostatics,v ≡ 0, ∂t = 0.

p = −ρgy + c ← hydrostatic pressure (Archimedes’ principle)

∂ • Steady and no gravity eﬀect ( ∂t = 0, g ≡ 0)

ρv2 ρ p = − + c = − |∇φ|2 + c ← Venturi pressure (created by velocity) 2 2 • Inertial, acceleration eﬀect

∂φ p ∼ −ρ ∂t + ··· ∂ * ∇p ∼ −ρ ∂t v + ···

∂p p p + δx ∂x δx

7 3.5 - Boundary Conditions • KBC on an impervious boundary

∂φ *v · nˆ = *u · nˆ no ﬂux across boundary ⇒ = U given |{z} |{z} ∂n n nˆ·∇φ Un given

• DBC: specify pressure at the boundary, i.e., µ ¶ ∂φ 1 −ρ + |∇φ|2 + gy = given ∂t 2

Note: On a free-surface p = patm.

3.6 - Stream function • continuity: ∇ · *v = 0; irrotationality: ∇ × *v = ω* = 0

• velocity potential: *v = ∇φ, then ∇ × *v = ∇ × (∇φ) ≡ 0for any φ, i.e. irrotationality is satisﬁed automatically. Required for continuity:

∇ · *v = ∇2φ = 0

* • Stream function ψ deﬁned by

* *v = ∇ × ψ

³ *´ * Then ∇ · *v = ∇ · ∇ × ψ ≡ 0 for any ψ, i.e. satisﬁes continuity automatically. Required for irrotationality:

³ *´ ³ *´ * ∇ × *v = 0 ⇒ ∇ × ∇ × ψ = ∇ ∇ · ψ − ∇2ψ = 0 (1) | {z } still 3 unknowns * *v ←→ ψ

8 * • For 2D and axisymmetric ﬂows, ψ is a scalar ψ (so stream functions are more useful for 2D and axisymmetric ﬂows).

* ∂ For 2D ﬂow: v = (u, v, 0) and ∂z ≡ 0. ¯ ¯ ¯ ˆi ˆj kˆ ¯ µ ¶ µ ¶ µ ¶ * ¯ ¯ * ¯ ∂ ∂ ∂ ¯ ∂ ˆ ∂ ˆ ∂ ∂ ˆ v = ∇ × ψ = ¯ ∂x ∂y ∂z ¯ = ψz i + − ψz j + ψy − ψx k ¯ ¯ ∂y ∂x ∂x ∂y ψx ψy ψz

∂ψ ∂ψ Set ψx = ψy ≡ 0 and ψz = ψ, then u = ∂y ; v = − ∂x

So, for 2D:

* ∂ ∂ ∂ ∇ · ψ = ψ + ψ + ψ ≡ 0 ∂x x ∂y y ∂z z

Then, from the irrotationality (see (1)) ⇒ ∇2ψ = 0 and ψ satisﬁes Laplace’s equation.

* ∂ • 2D polar coordinates: v = (vr, vθ ) and ∂z ≡ 0.

y ê êr r x

v v ¯ ¯ z }|vr { z }|θ { z }|z { ¯ eˆr reˆθ eˆz ¯ µ ¶ * ¯ ¯ * 1 ¯ ∂ ∂ ∂ ¯ 1 ∂ψz ∂ψz 1 ∂ ∂ v = ∇ × ψ = ¯ ∂r ∂θ ∂z ¯ = eˆr − eˆθ + rψθ − ψr eˆz r ¯ ¯ r ∂θ ∂r r ∂r ∂θ ψr ψθ ψz

9 1 ∂ψ ∂ψ Again let ψr = ψθ ≡ 0 and ψz = ψ, then vr = r ∂θ and vθ = − ∂r .

* • For 3D but axisymmetric ﬂows, ψ also reduces to ψ (read JNN 4.6 for details). • Physical Meaning of ψ. ∂ψ ∂ψ In 2D: u = ∂y and v = − ∂x .

We deﬁne

Z ~x Z ~x ψ(~x,t) = ψ(~x0, t) + ~v · ndlˆ = ψ(~x0, t) + (udy − vdx) | ~x0 {z } ~x0 total volume ﬂux from left to right accross a curve C between~x and ~x0 G x

C’ G t

G C xo nˆ

R For ψ to be single-valued, must be path independent. Z Z Z Z I ZZ * * = or − = 0 −→ v · nˆ dl = |∇{z · v} ds = 0 C C0 C C0 C−C0 S continuity Therefore, ψ is unique because of continuity.

10 * * * Let x1, x2 be two points on a given streamline (v · nˆ = 0 on streamline)

streamline

*x Z 2 ¡ ¢ ¡ ¢ ψ *x = ψ *x + *v · nˆ dl | {z2} | {z1} |{z} ψ ψ * 2 1 x 1 = 0 along streamline

Therefore, ψ1 = ψ2,i.e., ψ is a constant along any streamline. For example, on an impervious stationary body *v · nˆ = 0, so ψ = constant on the body is the appropriate boundary condition. If * the body is moving v · nˆ = Un Z ψ = ψ + U dl on the boddy 0 |{z}n given

11 ∂φ ψ=constant≡ = 0 ∂ u=0 n

ψ=given

ψ o

∂ψ ∂ψ Flux ∆ψ = −v∆x = u∆y. Therefore, u = ∂y and v = − ∂x

u streamline -v streamline (x+ ∆x,y) (x,y)

ψ+ ∆ψ ψ

12 Summary: Potential formulation vs. Stream-function formulation for ideal ﬂows

potential stream-function * definition *v = ∇φ *v = ∇ × ψ continuity ∇ · *v = 0 ∇2φ = 0 automatically satisfied ³ *´ ³ *´ * irrotationality ∇ × *v = 0 automatically satisfied ∇ × ∇ × ψ = ∇ ∇ · ψ − ∇2ψ = 0

∂ In 2D : w = 0, ∂z = 0 * 2 2 ∇ φ = 0 for continuity ψ ≡ ψz : ∇ ψ = 0 for irrotationality

Cauchy-Riemann equations for (φ, ψ) = (real, imaginary) part of an analytic complex function of z = x +iy

u = ∂φ u = ∂ψ Cartesian (x, y) ∂x ∂y v = ∂φ ∂ψ ∂y v = − ∂x

∂φ 1 ∂ψ vr = ∂r vr = r ∂θ Polar (r,θ) 1 ∂φ ∂ψ vθ = r ∂θ vθ = − ∂r

For irrotational flow use φ For incompressible flow use ψ For both flows use φ or ψ

Given φ or ψ for 2D ﬂow, use Cauchy-Riemann equations to ﬁnd the other: e.g. φ = xy ψ = ? ) ∂φ ∂ψ 1 2 ∂x = y = ∂y −→ ψ = 2 y + f1(x) 1 2 2 ∂φ ∂ψ 1 2 ψ = (y − x ) + const ∂y = x = − ∂x −→ ψ = − 2 x + f2(y) 2