Marine Hydrodynamics (Lecture 20)

2.20 - Marine Hydrodynamics, Spring 2005 Lecture 20 2.20 - Marine Hydrodynamics Lecture 20 Chapter 6 - Water Waves 6.1 Exact (Nonlinear) Governing Equations for Surface Gravity Waves, Assuming Potential Flow Free surface definition y = η ( x, z , t ) or F ( x, y, z , t ) = 0 y y x z B(x, y, z,t) = 0 Unknown variables Velocity field:  v (x, y, z, t) = ∇φ (x, y, z, t) Position of free surface: y = η (x, z, t) or F (x, y, z, t) = 0 Pressure field: p (x, y, z, t) Governing equations Continuity: ∇2 φ = 0 y < η or F < 0 Bernoulli for P-Flow: ∂φ ∂t Far way, no disturbance: ∂φ/∂t, ∇φ → 0 and p = + 12 |∇φ|2 + p−pa ρ + gy = 0; y < η or F < 0 pa − ρgy   atmospheric 1 hydrostatic Boundary Conditions 1. On an impervious boundary B (x, y, z, t) = 0, we have KBC: v · n ˆ = ∇φ · n ˆ= ∂φ  (x, t) · n =U ˆ (x, t) = Un on B = 0 ∂n Alternatively: a particle P on B remains on B, i.e., B is a material surface. For example if P is on B at t = t0 , P stays on B for all t. B(xP , t0 ) = 0, then B(xP (t), t) = 0 for all t, so that, following P B is always 0. ∴ DB ∂B = + (∇φ · ∇) B = 0 on B = 0 Dt ∂t For example, for a flat bottom at y = −h ⇒ B = y + h = 0 ⇒    DB ∂φ ∂ ∂φ = (y + h) = 0 ⇒ = 0 on B = y + h = 0 Dt ∂y ∂y ∂y    =1 2. On the free surface, y = η or F = y − η(x, z, t) = 0 we have KBC and DBC. KBC: free surface is a material surface, no normal velocity relative to the free surface. A particle on the free surface remains on the free surface for all times. DF D =0= (y − η) = Dt Dt ∂φ ∂η ∂φ ∂η ∂φ ∂η − − − on y = η  ∂z  ∂y ∂t ∂x  ∂x ∂z  still vertical slope slope velocity of f.s. of f.s. unknown DBC: p = pa on y = η or F = 0. Apply Bernoulli equation at y = η: ∂φ + ∂t 1 |∇φ|2 2   + g η  still unknown non-linear term 2 = pa on y = η 6.2 Linearized (Airy) Wave Theory Assume small wave amplitude compared to wavelength, i.e., small free surface slope A << 1 λ Wave height H crest H = A/2 Wave amplitude A SWL Water depth h trough wavelength λ Wave period T Consequently φ λ2 /T , η << 1 λ We keep only linear terms in φ, η. ∂ For example: ()|y=η = ()y=0 + η ()| + . . . Taylor series     ∂y  y=0 keep discard 3 6.2.1 BVP In this paragraph we state the Boundary Value Problem for linear (Airy) waves. ∂ 2φ ∂φ =0 +g ∂y ∂t 2 y=0 ∇ 2φ = 0 y = -h ∂φ =0 ∂y GE: BKBC: Finite depth h = const Infinite depth ∇2 φ = 0, −h < y < 0 ∇2 φ = 0, y < 0 ∂φ ∂y = 0, y = −h FSKBC: FSDBK: ∂φ ∂t ∇φ → 0, y → −∞ ⎫ ∂φ ∂η ⎬ 2 ∂y = ∂t , y = 0 → ∂∂t2φ + g ∂φ ∂y = 0 + gη = 0, y = 0 ⎭ Introducing the notation {} for infinite depth we can rewrite the BVP: Constant finite depth h ∇2 φ = 0, −h < y < 0 ∂φ = 0, y = −h ∂y ∂φ ∂2φ + g = 0, y = 0 ∂y ∂t2 {Infinite depth} ∇2 φ = 0, y < 0 (1) {∇φ → 0, y → −∞} (2)  ∂ 2φ ∂φ +g = 0, y = 0 ∂t2 ∂y Given φ calculate:  1 ∂φ  η (x, t) = − g ∂t y=0    1 ∂φ  η (x, t) = − g ∂t y=0 (3)  ∂φ p − pa = −ρ − ρgy   ∂t ∂φ p − pa = −ρ − ρgy   ∂t hydrostatic hydrostatic dynamic dynamic 4 (4)  (5) 6.2.2 Solution Solution of 2D periodic plane progressive waves, applying separation of variables. We seek solutions to Equation (1) of the form eiωt with respect to time. Using the KBC (2), after some algebra we find φ. Upon substitution in Equation (4) we can also obtain η. gA cosh k (y + h) φ= sin (kx − ωt) ω cosh kh η  = A cos (kx − ωt) using (4)  gA ky φ= sin (kx − ωt) e ω ⎧ ⎫ ⎨ ⎬ η  = A cos (kx − ωt) ⎩ ⎭ using(4) where A is the wave amplitude A = H/2. Exercise Verify that the obtained values for φ and η satisfy Equations (1), (2), and (4). 6.2.3 Review on plane progressive waves (a) At t = 0 (say), η = A cos kx → periodic in x with wavelength: λ = 2π/k Units of λ : [L] λ k = wavenumber = 2 /λ x K [L-1] (b) At x = 0 (say), η = A cos ωt → periodic in t with period: T = 2π/ω Units of T : [T ] T t ω = frequency = 2 /T    (c) η = A cos k x − ωk t [T-1], e.g. rad/sec ω Units of : k   L T ω  ω Following a point with velocity , i.e., xp = t + const, the phase of η does k k ω λ not change, i.e., = ≡ Vp ≡ phase velocity. k T 5 6.2.4 Dispersion Relation So far, any ω, k combination is allowed. However, recall that we still have not made use of the FSBC Equation (3). Upon substitution of φ in Equation (3) we find that the following relation between h, k, and ω must hold: ∂ 2φ ∂φ + g =0 ∂t2 ∂y −→ ↑ φ= gA sin(kx−ωt)f (z) ω −ω 2 cosh kh + gk sinh kh = 0 ⇒ ω 2 = gk tanh kh • This is the Dispersion Relation ω 2 = gk tanh kh ω 2 = gk (6) Given h, the Dispersion Relation (6) provides a unique relation between ω and k, i.e., ω = ω(k; h) or k = k(ω; h). • Proof C kh C ≡ ω2h = (kh) tanh (kh) g  from (6) 1 tanh kh C kh kh kh =f(c) = tanh kh → obtain unique solution for k • Comments - General As ω ↑ then k ↑, or equivalently as T ↑ then λ ↑.    λ ω g g - Phase speed Vp ≡ = = tanh kh Vp = T k k k Therefore as T ↑ or as λ ↑, then Vp ↑, i.e., longer waves are ‘faster’ in terms of phase speed. - Water depth effect For waves the same k (or λ), at different water depths, as h ↑ then Vp ↑, i.e., for fixed k Vp is fastest in deep water. - Frequency dispersion Observe that Vp = Vp (k) or Vp (ω). This means that waves of different frequencies, have different phase speeds, i.e., frequency dispersion. 6 6.2.5 Solutions to the Dispersion Relation : ω 2 = gk tanh kh Property of tanh kh: long waves shallow water sinh kh 1 − e−2kh ∼ tanh kh = = = cosh kh 1 + e−2kh kh for kh << 1. In practice 1 for kh >∼ 3. In practice    h < λ/20 h > λ2    short waves deep water Shallow water waves Intermediate depth Deep water waves or long waves or wavelength or short waves kh << 1 Need to solve ω 2 = gk tanh kh kh >> 1 ∼ h < λ/20 √ ω2 ∼ = gk · kh → ω = gh k √ λ = gh T given ω, h for k ∼ h > λ/2 (given k, h for ω - easy!) (a) Use tables or graphs (e.g.JNN fig.6.3) ω 2 = gk g 2 ω 2 = gk tanh kh = gk∞ λ= T 2π k∞ λ Vp   λ(in ft.) ≈ 5.12T 2 (in sec.) ⇒ = tanh kh = = k λ∞ Vp∞ (b) Use numerical approximation (hand calculator, about 4 decimals ) i. Calculate C = ω 2 h/g ii. If C > 2: ”deeper” ⇒ kh ≈ C(1 + 2e−2C − 12e−4C + . . .) If C < 2: ”shallower” ⇒ √ kh ≈ C(1 + 0.169C + 0.031C 2 + . . .) No frequency dispersion  Vp = gh Frequency dispersion  g tanh kh Vp = k 7 Frequency dispersion  g λ Vp = 2π 6.3 Characteristics of a Linear Plane Progressive Wave y λ Vp 2π k= λ 2π ω= T H = 2A η(x,t) = y A x MWL h Define U ≡ ωA Linear Solution: η = A cos (kx − ωt) ; φ= Ag cosh k (y + h) sin (kx − ωt) , where ω 2 = gk tanh kh ω cosh kh 6.3.1 Velocity field Velocity on free surface v (x, y = 0, t) u(x, 0, t) ≡ Uo = Aω 1 cos (kx − ωt) tanh kh v(x, 0, t) ≡ Vo = Aω sin (kx − ωt) = ∂η ∂t Velocity field v (x, y, t) Agk cosh k (y + h) ∂φ cos (kx − ωt) = ω cosh kh ∂x cosh k (y + h) cos (kx − ωt) ⇒ =  Aω sinh kh u= U u = Uo ⎧ ⎪ ⎨ ∼ eky cosh k (y + h) cosh kh ⎪ ⎩ ∼1 deep water shallow water • u is in phase with η ∂φ Agk sinh k (y + h) = sin (kx − ωt) ∂y ω cosh kh sinh k (y + h) =  Aω sin (kx − ωt) ⇒ sinh kh v= U v = Vo ⎧ ⎪ ky sinh k (y + h) ⎨ ∼ e sinh kh ⎪ ⎩ ∼1+ • v is out of phase with η 8 deep water y h shallow water Shallow water Velocity field v (x, y) Intermediate water 9 Deep water 6.3.2 Pressure field • Total pressure p = pd − ρgy. • Dynamic pressure pd = −ρ ∂φ . ∂t • Dynamic pressure on free surface pd (x, y = 0, t) ≡ pdo Pressure field Shallow water Intermediate water Deep water cosh k (y + h) cos (kx − ωt) cosh kh cosh k (y + h) = ρg η cosh kh pd same picture as Uuo p do pd = ρgη pd (−h) 1 = cosh kh pdo pd (−h) = 1 (no decay) p do pd (−h) = e−ky p do   p = ρg ηeky − y ρg(η − y)    p= pd = ρgeky η pd = ρgA “hydrostatic” approximation V p = gλ V p = gh y y y y y y x x ∇ ∇ kh >> 1 kh << 1 pd p ( − h) pd o pd o p (− h) p (− h) pd o p (− h) Pressure field in deep water Pressure field in shallow water 10 2π 6.3.3 Particle Orbits (‘Lagrangian’ concept) Let xp (t), yp (t) denote the position of particle P at time t. Let (x̄; ȳ) denote the mean position of particle P. The position P can be rewritten as xp (t) = x̄ + x (t), yp (t) = ȳ + y  (t), where (x (t), y  (t)) denotes the departure of P from the mean position. In the same manner let v ≡ v(¯ x, y, ¯ t) denote the velocity at the mean position and vp ≡ v (xp , yp , t) denote the velocity at P. (x' , y ' ) P (x , y ) P P vp = v (x̄ + x , ȳ + y  , t) =⇒ TSE vp = v (¯ x, y, ¯ t) + (x, y) ∂v ∂v (¯ x, y, ¯ t) x + (x̄, y, ¯ t) y  + . . . ⇒ ∂x ∂y    ignore - linear theory vp ∼ = v To estimate the position of P, we need to evaluate (x (t), y  (t)): x = dt u (¯ x, y, ¯ t) = = −A y = = A dt ωA cosh k (ȳ + h) cos (kx¯ − ωt) ⇒ sinh kh cosh k (ȳ + h) sin (kx̄ − ωt) sinh kh dt v (¯ x, y, ¯ t) = dt ωA sinh k (ȳ + h) sin (kx̄ − ωt) ⇒ sinh kh sinh k (ȳ + h) cos (kx̄ − ωt) sinh kh Check: On ȳ = 0, y  = A cos (kx̄ − ωt) = η, i.e., the vertical motion of a free surface particle (in linear theory) coincides with the vertical free surface motion. It can be shown that the particle motion satisfies x2 y 2 (xp − x̄)2 (yp − ȳ)2 + = 1 ⇔ + =1 b2 b2 a2 a2 cosh k (ȳ + h) sinh k (ȳ + h) where a = A and b = A , i.e., the particle orbits form sinh kh sinh kh closed ellipses with horizontal and vertical axes a and b. 11 crest Vp A ky (a) deep water kh >> 1: a = b = Ae ky circular orbits with radii Ae decreasing exponentially with depth A trough ky Ae A Vp = (b) shallow water kh << 1: a= y A = const. ; b = A(1+ ) kh h decreases linearly with depth A/kh Vp (c) Intermediate depth Vp P A Q R Q S R λ S R gh 6.3.4 Summary of Plane Progressive Wave Characteristics f (y) Deep water/ short waves Shallow water/ long waves kh > π (say) kh << 1 cosh k(y+h) cosh kh = f1 (y) ∼ eky 1 cosh k(y+h) sinh kh = f2 (y) ∼ eky 1 kh sinh k(y+h) sinh kh = f3 (y) ∼ eky 1+ e.g.pd e.g.u, a e.g. v, b 13 y h C (x) = cos (kx − ωt) S (x) = sin (kx − ωt) (in phase with η) (out of phase with η) η A = C (x) u Aω = C (x) f2 (y) pd ρgA = C (x) f1 (y) y� A v Aω x� A = C (x) f3 (y) a A = S (x) f3 (y) = −S (x) f2 (y) b A = f2 (y) b a 14 = f3 (y)