Marine Hydrodynamics (Lecture 8)

2.20 - Marine Hydrodynamics, Spring 2005 Lecture 8 2.20 - Marine Hydrodynamics Lecture 8 In Lecture 8, paragraph 3.3 we discuss some properties of vortex structures. In paragraph 3.4 we deduce the Bernoulli equation for ideal, steady flow. 3.3 Properties of Vortex Structures 3.3.1 Vortex Structures • A vortex line is a line everywhere tangent to ω . r vortex line ω2 r u2 Ω2 r ω1 r u1 Ω1 • A vortex tube (filament) is a bundle of vortex lines. vortex tube r u r ω vortex lines 1 • A vortex ring is a closed vortex tube. A sketch and two pictures of the production of vortex rings from orifices are shown in Figures 1, 2, and 3 below. (Figures 2,3: Van Dyke, An Album of Fluid Motion 1982 p.66, 71) v u side view U v u cross section U v u Γ v u ω v u ω Figure 1: Sketch of vortex ring production 2 v ω v v 3.3.2 No Net Flux of Vorticity Through a Closed Surface Calculus identity, for any vector v : ∇ · (∇ × v) = 0 ⇒   ω  ∇ ·  ∇ · ω V =0⇒ ω   = ⇑ ω  · n̂  dS = 0 Divergence S vorticity flux Theorem i.e. The net vorticity flux through a closed surface is zero. (a) No net vorticity flux through a vortex tube: (Vorticity Flux)in = (Vorticity Flux)out ⇒ (ω · n̂)in δAin = (ω  · n̂)out δAout (ωv ⋅ n̂ )out v ω ⋅ nˆ = 0 (ωv ⋅ n̂ )in (b) Vorticity cannot stop anywhere in the fluid. It either traverses the fluid begin­ ning or ending on a boundary or closes on itself (vortex ring). r r ω ω 3 3.3.3 Conservation of Vorticity Flux  0 = Γ3 =  v · dx = C3 n̂1 C3 Γ1 = C1 C2  v · dx = ω  · ndS ˆ =0 n̂2 C1  S3 S1  ω  · n̂1 dS = S2 ω  · n̂2 dS = Γ2 Therefore, circulation is the same in all circuits embracing the same vortex tube. For the special case of a vortex tube with ‘small’ area: Γ = ω1 A1 = ω2 A2 ω1 ω2 A1 A2 An application of the equation above is displayed in the figure below: ω2 ω1 A2 = 2A1 ω2 = ω1/2 A1 4 3.3.4 Vortex Structures are Material Structures Consider a material patch Am on a vortex tube at time t. Am ∂Am By definition, ω · n̂ = 0 on An Then,   v · dx = Γ∂Am = ω  · nds ˆ =0 Am ∂Am At time t + Δt, Am moves, and for an ideal fluid under the influence of conservative body forces, Kelvin’s theorem states that Γ∂Am = 0 So, ω · n̂ = 0 on Am still, i.e., Am still on the vortex tube. Therefore, the vortex tube is a material tube for an ideal fluid under the influence of conservative forces. In the same manner it can be shown that a vortex line is a material line, i.e., it moves with the fluid. 5 3.3.5 Vortex stretching Consider a small vortex filament of length L and radius R, where by definition ω is tangent to the tube. R Γ = ↑ Stokes Theorem A ωA = ↑ constant (in time) Kelvins’ Theorem But tube is material with volume = AL = πR2 L = constant in time (continuity) ∴ ω Γ ωA = = = constant Volume LA L As a vortex stretches, L increases, and since the volume is constant (from continuity), A and R decrease, and due to the conservation of the angular momentum, ω increases. In other words, Vortex stretching ⇔ L ↑ ⇒ ω ↑ (conservation of angular momentum) ⇒ A and R ↑ (continuity) 6 3.3.6 Summary on Vortex Structures r ω Γ Γ R r A ω r Vortex ring length L = 2πR [L] Cross sectional area A = πr2 [L2 ] Vortex ring volume ∀ = AL = const ⇑ [L3 ] continuity Vorticity ω = ∇ × v [T−1 ] Circulation Γ [L2 T−1 ] = const ⇑ Kelvin’s theorem Γ = ⇑ ωA = const [L2 T−1 ] vorticity flux through A Γ ∝ U r = const 7 [L2 T−1 ] Continuity relates length ratios ∀ = LA = const ⎧ ⎪ ⎪ ⎪ A ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ r ∀ L ∝ ∀ L ∝ ∴ as L ↑ A↓ ∴ as L ↑ r↓ Kelvin’s theorem + Continuity relate length ratios to Γ, ω, U U r ∝ Γ = const Γ → U∝ r ωA ∝ Γ = const → ω∝ Example 1: A∝∀/L → ω∝Γ U↑ ∴ as L ↑ ω↑ A2 r A1 ω1 r ω1 A1 < A2 ω1 > ω 2 L ∀ ∴ as L ↑ Example 2: r Γ L ∀ U ∝Γ ω2 A2 A1 Γ A √ ∀/L → r∝ Γ1 L1 < L 2 A1 > A2 Γ1 = Γ2 ω1 < ω 2 U1 < U 2 L1 L2 Γ2 r ω2 Given From continuity only From Kelvin’s theorem From Kelvin’s theorem + continuity From Kelvin’s theorem + continuity 8 ∂ 3.4 Bernoulli Equation for Steady ( ∂t = 0), Ideal(ν = 0), Rotational flow   p=f v p = f (|v |) Viscous flow: Navier-Stokes’ Equations (Vector Equations) Ideal flow: Bernoulli Equation (Scalar equation) Steady, inviscid Euler equation (momentum equation):   p + gy v · ∇v = −∇ ρ (1) From Vector Calculus we have ∇ (u · v ) = (u · ∇) v + (v · ∇) u + u × (∇ × v ) + v × (∇ × u) ⇒ ∇( 12 |v |2 ) = v · ∇v + v × (∇ × v ) ⇒  2 v v · ∇v = ∇ − v × (∇ × v ) where v 2 ≡ v · v = |v |2 2 From the previous identity and Equation (1) we obtain  2   p v − v · v × (∇ × v ) = −v · ∇ + gy v · (1) → v · ∇    2 ρ  ⊥v  v · momentum (1) → energy Therefore, v · ∇ i.e., In general, v2 2 + p ρ  v2 2 v2 2 + p ρ 0  + gy = 0 = D Dt  v2 2 + p ρ  + gy streamline pathline p + ρ + gy = constant on a streamline + gy = F (Ψ) where Ψ is a tag for a particular streamline. Assumptions: Ideal fluid, Steady flow, Rotational in general. 9 3.4.1 Example: Contraction in Water or Wind Tunnel Contraction Ratio: γ = R1 /R2 >> 1 ( γ = O(10) for wind tunnel ; γ = O(5) for water tunnel) Let U¯1 and U¯2 denote the average velocities at sections 1 and 2 respectively.  2   U¯2 R1 2 2 = γ 2 >> 1 1. From continuity: Ū1 πR1 = Ū2 πR2 → ¯ = R2 U1 2. Since ∂u =0,ω  = 0 → vortex ring. ∂r 10 ⎧ ω1 ω2 ω2 R2 1 ⎪ ⎪ = → = ∼ << 1 ⎪ ⎪ ⎨ 2πR1 2πR2 ω1 R1 γ ω/L = constant ⇒     ⎪ ⎪ ∂u ∂u ∂u ⎪ ⎪ since ω ∼ → << ⎩ ∂r ∂r 2 ∂r 1 i.e., Section 2 Section 1 3. Near the center, let U1 = U¯1 (1 + ε1 ) and U2 = U¯2 (1 + ε2 ) where ε1 and ε2 measure the relative velocity fluctuations. Apply the Bernoulli equation along a reference average streamline P1 + 12 ρŪ12 = P2 + 12 ρŪ22 Apply Bernoulli Equation to a particular streamline  2  2 P1 + 12 ρ U¯1 (1 + ε1 ) = P2 + 12 ρ Ū2 (1 + ε2 ) From (2) and (3) we obtain ε1 Ū12 = ε2 Ū22 + O(ε2 ) → 11 ε2 Ū 2 1 ∼ 12 ∼ 4 << 1 ε1 γ Ū2 (2)