slidesWaveRegular.pdf
- 1. B. Waves 1. Linear wave theory 2. Wave transformation 3. Random waves and statistics 4. Wave loading on structures
- 2. Recommended Books ● Dean and Dalrymple, Water Wave Mechanics For Engineers and Scientists ● Kamphuis, Introduction to coastal engineering and management
- 3. 1. LINEAR WAVE THEORY 1.1 Main wave parameters 1.2 Dispersion relationship 1.3 Wave velocity and pressure 1.4 Wave energy 1.5 Group velocity 1.6 Energy transfer (wave power) 1.7 Particle motion 1.8 Shallow-water and deep-water behaviour 1.9 Waves on currents Linear Wave Theory
- 4. Linear Wave Theory Single-frequency (“monochromatic”, “regular”) progressive wave on still water: 𝜂 = 𝐴 cos 𝑘𝑥 − 𝜔𝑡 Linear wave theory: ● aka Airy wave theory ● assume amplitude small (compared with depth and wavelength) ‒ neglect powers and products of wave perturbations ‒ sum of any such wave fields also a solution L h A H (x,t) SWL (z=0) x z trough crest c bed (z= -h)
- 5. Amplitude and Height L h A H (x,t) SWL (z=0) x z trough crest c bed (z= -h) ● Amplitude 𝐴 is the maximum displacement from still-water level (SWL) ● Wave height 𝐻 is the vertical distance between neighbouring crest and trough ● For sinusoidal waves, 𝑯 = 𝟐𝑨 ● For regular waves, formulae more naturally expressed in terms of 𝐴 ● For irregular waves, 𝐻 is the more measurable quantity 𝜂 = 𝐴 cos 𝑘𝑥 − 𝜔𝑡
- 6. Wavenumber and Wavelength L h A H (x,t) SWL (z=0) x z trough crest c bed (z= -h) ● 𝑘 is the wavenumber ● Wavelength 𝐿 is the horizontal distance over which wave form repeats: 𝜂 = 𝐴 cos 𝑘𝑥 − 𝜔𝑡 𝐿 = 2π 𝑘 𝑘𝐿 = 2π
- 7. Frequency and Period L h A H (x,t) SWL (z=0) x z trough crest c bed (z= -h) ● 𝜔 is the wave angular frequency ● Period 𝑇 is the time over which the wave form repeats: 𝜂 = 𝐴 cos 𝑘𝑥 − 𝜔𝑡 𝑇 = 2π 𝜔 𝜔𝑇 = 2π ● The actual frequency 𝑓 (cycles per second, or Hertz) is 𝑓 = 1 𝑇 = 𝜔 2𝜋
- 8. Wave Speed L h A H (x,t) SWL (z=0) x z trough crest c bed (z= -h) ● 𝑐 is the phase speed or celerity ● 𝑐 is the speed at which the wave form translates 𝜂 = 𝐴 cos 𝑘𝑥 − 𝜔𝑡 = 𝐴 cos 𝑘 𝑥 − 𝜔 𝑘 𝑡 = 𝐴 cos 𝑘 𝑥 − c𝑡 𝑐 = 𝜔 𝑘 = 𝐿 𝑇 wavelength period = 𝑓𝐿 frequency × wavelength
- 9. Summary L h A H (x,t) SWL (z=0) x z trough crest c bed (z= -h) 𝜂 = 𝐴 cos 𝑘𝑥 − 𝜔𝑡 Surface elevation: Wavenumber 𝑘, wavelength 𝐿: 𝐿 = 2π 𝑘 Angular frequency 𝜔, period 𝑇: 𝑇 = 2π 𝜔 Phase speed (celerity) 𝑐: 𝑐 = 𝜔 𝑘 = 𝐿 𝑇
- 10. Hyperbolic Functions sinh 𝑥 ≡ e𝑥 − 𝑒−𝑥 2 cosh 𝑥 ≡ e𝑥 + 𝑒−𝑥 2 tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥
- 11. Hyperbolic Functions Trigonometric-like formulae: cosh2 𝑥 − sinh2 𝑥 = 1 cosh 2𝑥 = cosh2 𝑥 + sinh2 𝑥 = 2 cosh2 𝑥 − 1 sinh 2𝑥 = 2 sinh 𝑥 cosh 𝑥 Derivatives: d d𝑥 sinh 𝑥 = cosh 𝑥 d d𝑥 cosh 𝑥 = sinh 𝑥 d d𝑥 tanh 𝑥 = sech2 𝑥 Asymptotic behaviour: Small 𝑥: sinh 𝑥 ~ tanh 𝑥 ~ 𝑥, Large 𝑥: sinh 𝑥 ~ cosh 𝑥 ~ 1 2 e𝑥, cosh 𝑥 → 1 as 𝑥 → 0 tanh 𝑥 → 1 as 𝑥 → ∞
- 12. Fluid-Flow Equations Continuity: Irrotationality: Time-dependent Bernoulli equation: 𝜕𝑢 𝜕𝑥 + 𝜕𝑤 𝜕𝑧 = 0 𝜕𝑢 𝜕𝑧 − 𝜕𝑤 𝜕𝑥 = 0 or: 𝑢 = 𝜕𝜙 𝜕𝑥 , 𝑤 = 𝜕𝜙 𝜕𝑧 𝜙 is a velocity potential 𝜌 𝜕𝜙 𝜕𝑡 + 𝑝 + 1 2 𝜌𝑈2 + 𝜌𝑔𝑧 = 𝐶 𝑡 , along a streamline
- 13. Continuity 𝑢𝑒Δ𝑧 − 𝑢𝑤Δ𝑧 + 𝑤𝑛Δ𝑥 − 𝑤𝑠Δ𝑥 = 0 Net volume outflow: Divide by Δ𝑥Δ𝑧: 𝑢𝑒 − 𝑢𝑤 Δ𝑥 + 𝑤𝑛 − 𝑤𝑠 Δ𝑧 = 0 Δ𝑢 Δ𝑥 + Δ𝑤 Δ𝑧 = 0 Δ𝑥, Δ𝑧 → 0: 𝜕𝑢 𝜕𝑥 + 𝜕𝑤 𝜕𝑧 = 0 ue uw wn ws z x
- 14. Irrotationality Circulation: Divide by Δ𝑥Δ𝑧: Δ𝑥, Δ𝑧 → 0: we ww un us z x Pressure forces act normal to surfaces, so can cause no rotation. 𝑢𝑛Δ𝑥 − 𝑤𝑒Δ𝑧 − 𝑢𝑠Δ𝑥 + 𝑤𝑤Δ𝑧 = 0 𝑢𝑛 − 𝑢𝑠 Δ𝑧 − 𝑤𝑒 − 𝑤𝑤 Δ𝑥 = 0 Δ𝑢 Δ𝑧 − Δ𝑤 Δ𝑥 = 0 𝜕𝑢 𝜕𝑧 − 𝜕𝑤 𝜕𝑥 = 0
- 15. Velocity Potential The no-circulation condition makes the following well-defined: For any 2-d function: we ww un us z x d𝜙 = 𝑢 d𝑥 + 𝑤 d𝑧 d𝜙 = 𝜕𝜙 𝜕𝑥 d𝑥 + 𝜕𝜙 𝜕𝑧 d𝑧 The velocity components are the gradient of the velocity potential 𝜙: 𝑢 = 𝜕𝜙 𝜕𝑥 , 𝑤 = 𝜕𝜙 𝜕𝑧 Aim: solve a single scalar equation for 𝜙, then derive everything else from it.
- 16. Time-Dependent Bernoulli Equation U s 𝜌 𝜕𝑈 𝜕𝑡 + 𝑈 𝜕𝑈 𝜕𝑠 = − 𝜕𝑝 𝜕𝑠 − 𝜌𝑔 sin 𝜃 mass acceleration = force sin 𝜃 = Τ 𝜕𝑧 𝜕𝑠 𝑈 = 𝜕𝜙 𝜕𝑠 𝜌 𝜕2𝜙 𝜕𝑡 𝜕𝑠 + 𝜕 𝜕𝑠 (1 2𝑈2 ) = − 𝜕𝑝 𝜕𝑠 − 𝜌𝑔 𝜕𝑧 𝜕𝑠 𝜕 𝜕𝑠 𝜌 𝜕𝜙 𝜕𝑡 + 1 2 𝜌𝑈2 + 𝑝 + 𝜌𝑔𝑧 = 0 𝜌 𝜕𝜙 𝜕𝑡 + 𝑝 + 1 2 𝜌𝑈2 + 𝜌𝑔𝑧 = 𝐶(𝑡), along a streamline Special case: if steady-state then 𝑝 + 1 2 𝜌𝑈2 + 𝜌𝑔𝑧 = 𝐶, along a streamline
- 17. Recap of Fluid-Flow Equations Continuity Velocity potential Bernoulli equation 𝜕𝑢 𝜕𝑥 + 𝜕𝑤 𝜕𝑧 = 0 𝑢 = 𝜕𝜙 𝜕𝑥 , 𝑤 = 𝜕𝜙 𝜕𝑧 Laplace’s equation 𝜌 𝜕𝜙 𝜕𝑡 + 𝑝 + 1 2 𝜌𝑈2 + 𝜌𝑔𝑧 = 𝐶(𝑡) 𝜕2 𝜙 𝜕𝑥2 + 𝜕2 𝜙 𝜕𝑧2 = 0
- 18. Boundary Conditions ● Kinematic boundary condition: no net flow through boundary ● Dynamic boundary condition: stress continuous at interface 𝑧 = 𝑧surf 𝑥, 𝑡 D D𝑡 𝑧 − 𝑧surf = 0 on surface KBC D D𝑡 ≡ 𝜕 𝜕𝑡 + 𝑢 𝜕 𝜕𝑥 + 𝑤 𝜕 𝜕𝑧 𝑤 − 𝜕𝑧surf 𝜕𝑡 − 𝑢 𝜕𝑧surf 𝜕𝑥 = 0 on the surface 𝑤 = 𝜕𝑧surf 𝜕𝑡 + 𝑢 𝜕𝑧surf 𝜕𝑥 on 𝑧 = 𝑧surf 𝑥, 𝑡
- 19. Boundary Conditions KBBC – Kinematic Bed Boundary Condition 𝑤 = 0 on 𝑧 = −ℎ KFSBC – Kinematic Free-Surface Boundary Condition 𝑤 = 𝜕𝜂 𝜕𝑡 + 𝑢 𝜕𝜂 𝜕𝑥 on 𝑧 = 𝜂 𝑥, 𝑡 DFSBC – Dynamic Free-Surface Boundary Condition 𝑝 = 0 on 𝑧 = 𝜂 𝑥, 𝑡 L h A H (x,t) SWL (z=0) x z trough crest c bed (z= -h) 𝑤 = 𝜕𝑧surf 𝜕𝑡 + 𝑢 𝜕𝑧surf 𝜕𝑥 on 𝑧 = 𝑧surf
- 20. Linearised Equations 𝑦 = 𝑎 + 𝑏𝜀 + 𝑐𝜀2 + ⋯ ● If 𝜀 is small, ignore quadratic and higher powers: 𝑦 = 𝑎 + 𝑏𝜀 + ⋯ ● Boundary conditions on 𝑧 = 𝜂(𝑥, 𝑡) can be applied on 𝑧 = 0
- 21. Boundary Conditions KBBC – Kinematic Bed Boundary Condition 𝑤 = 0 on 𝑧 = −ℎ KBBC – Kinematic Free-Surface Boundary Condition 𝑤 = 𝜕𝜂 𝜕𝑡 + 𝑢 𝜕𝜂 𝜕𝑥 on 𝑧 = 𝜂 𝑥, 𝑡 DFSBC – Dynamic Free-Surface Boundary Condition 𝑝 = 0 on 𝑧 = 𝜂 𝑥, 𝑡 L h A H (x,t) SWL (z=0) x z trough crest c bed (z= -h) 𝑤 = 𝜕𝑧surf 𝜕𝑡 + 𝑢 𝜕𝑧surf 𝜕𝑥 on 𝑧 = 𝑧surf 𝜕𝜙 𝜕𝑧 = 0 on 𝑧 = −ℎ 𝜕𝜙 𝜕𝑧 = 𝜕𝜂 𝜕𝑡 on 𝑧 = 0 𝜕𝜙 𝜕𝑡 + 𝑔𝜂 = 𝐶(𝑡) on 𝑧 = 0 𝜌 𝜕𝜙 𝜕𝑡 + 𝑝 + 1 2 𝜌𝑈2 + 𝜌𝑔𝑧 = 𝐶(𝑡)
- 22. Summary of Equations and BCs Laplace’s equation 𝜕2𝜙 𝜕𝑥2 + 𝜕2𝜙 𝜕𝑧2 = 0 KBBC KFSBC DFSBC 𝜕𝜙 𝜕𝑧 = 0 on 𝑧 = −ℎ 𝜕𝜙 𝜕𝑧 = 𝜕𝜂 𝜕𝑡 on 𝑧 = 0 𝜕𝜙 𝜕𝑡 + 𝑔𝜂 = 𝐶(𝑡) on 𝑧 = 0
- 23. Solution For Velocity Potential, 𝝓 𝜂 = 𝐴 cos 𝑘𝑥 − 𝜔𝑡 Surface displacement: Look for solution by separation of variables: 𝜙 = 𝑋 𝑥, 𝑡 𝑍 𝑧 KFSBC: 𝜕𝜙 𝜕𝑧 = 𝜕𝜂 𝜕𝑡 on 𝑧 = 0 𝑋 ቤ d𝑍 d𝑧 𝑧=0 = 𝐴𝜔 sin 𝑘𝑥 − 𝜔𝑡 Hence: 𝑋 ∝ sin 𝑘𝑥 − 𝜔𝑡 WLOG: 𝑋 = sin 𝑘𝑥 − 𝜔𝑡 ቤ d𝑍 d𝑧 𝑧=0 = 𝐴𝜔 Laplace’s equation: 𝜕2𝜙 𝜕𝑥2 + 𝜕2𝜙 𝜕𝑧2 = 0 −𝑘2𝑋𝑍 + 𝑋 d2𝑍 d𝑧2 = 0 d2𝑍 d𝑧2 = 𝑘2𝑍 𝑍 = 𝛼e𝑘𝑧 + 𝛽e−𝑘𝑧 General solution:
- 24. Solution For Velocity Potential, 𝝓 𝑍 = 𝛼e𝑘𝑧 + 𝛽e−𝑘𝑧 So far: 𝜙 = 𝑍 𝑧 sin 𝑘𝑥 − 𝜔𝑡 KFSBC: d𝑍 d𝑧 = 𝐴𝜔 on 𝑧 = 0 KBBC: d𝑍 d𝑧 = 0 on 𝑧 = −ℎ 𝑍 = 𝐴𝜔 𝑘 cosh 𝑘 ℎ + 𝑧 sinh 𝑘ℎ Solution: 𝜙 = 𝐴𝜔 𝑘 cosh 𝑘 ℎ + 𝑧 sinh 𝑘ℎ sin 𝑘𝑥 − 𝜔𝑡
- 25. Dispersion Relationship 𝜙 = 𝐴𝜔 𝑘 cosh 𝑘 ℎ + 𝑧 sinh 𝑘ℎ sin 𝑘𝑥 − 𝜔𝑡 How is wavenumber (𝑘) related to wave angular frequency (𝜔)? 𝜕𝜙 𝜕𝑡 + 𝑔𝜂 = 𝐶(𝑡) on 𝑧 = 0 DFSBC: − 𝐴𝜔2 𝑘 cosh 𝑘ℎ sinh 𝑘ℎ cos 𝑘𝑥 − 𝜔𝑡 + 𝐴𝑔 cos 𝑘𝑥 − 𝜔𝑡 = 𝐶(𝑡) LHS has zero space average ... so 𝐶(𝑡) must be zero − 𝜔2 𝑘 cosh 𝑘ℎ sinh 𝑘ℎ + 𝑔 = 0 𝜔2 = 𝑔𝑘 tanh 𝑘ℎ 𝜔 𝑘 = 𝑔 𝜔 sinh 𝑘ℎ cosh 𝑘ℎ 𝜙 = 𝐴𝑔 𝜔 cosh 𝑘 ℎ + 𝑧 cosh 𝑘ℎ sin 𝑘𝑥 − 𝜔𝑡
- 26. Summary of Solution 𝜔2 = 𝑔𝑘 tanh 𝑘ℎ 𝜙 = 𝐴𝑔 𝜔 cosh 𝑘 ℎ + 𝑧 cosh 𝑘ℎ sin 𝑘𝑥 − 𝜔𝑡 Velocity potential: Dispersion relation: This is all we need!!! 𝑢 ≡ 𝜕𝜙 𝜕𝑥 𝑤 ≡ 𝜕𝜙 𝜕𝑧 𝑝 = −𝜌𝑔𝑧 − 𝜌 𝜕𝜙 𝜕𝑡 Velocity: Pressure: 𝜂 = 𝐴 cos 𝑘𝑥 − 𝜔𝑡 Surface displacement:
- 27. 1. LINEAR WAVE THEORY 1.1 Main wave parameters 1.2 Dispersion relationship 1.3 Wave velocity and pressure 1.4 Wave energy 1.5 Group velocity 1.6 Energy transfer (wave power) 1.7 Particle motion 1.8 Shallow-water and deep-water behaviour 1.9 Waves on currents Linear Wave Theory
- 28. Dispersion Relationship 𝜔2 = 𝑔𝑘 tanh 𝑘ℎ 𝜔2ℎ 𝑔 = 𝑘ℎ tanh 𝑘ℎ or 𝐿 = 2π 𝑘 𝑇 = 2π 𝜔 𝑐 ≡ 𝜔 𝑘 = 𝑔 𝑘 tanh 𝑘ℎ
- 29. Variation of Phase Speed With Depth 𝜔2 = 𝑔𝑘 tanh 𝑘ℎ When waves propagate into shallower water: Period 𝑇 - and hence 𝜔 - are unchanged Depth ℎ decreases … so wavenumber 𝑘 increases Wavelength 𝐿 decreases Speed 𝑐 decreases This is VERY important !
- 30. Solving the Dispersion Relationship 𝜔2 = 𝑔𝑘 tanh 𝑘ℎ 1. Know wavelength (𝑳) … find period (𝑻) 𝑘 = 2π 𝐿 𝑇 = 2π 𝜔 Substitute: gives 𝜔
- 31. Solving the Dispersion Relationship 𝜔2 = 𝑔𝑘 tanh 𝑘ℎ 2. Know period (𝑻) … find wavelength (𝑳) 𝐿 = 2π 𝑘 Rewrite as 𝜔 = 2π 𝑇 𝜔2 ℎ 𝑔 = 𝑘ℎ tanh 𝑘ℎ 𝑌 = 𝑋 tanh 𝑋 𝑋 = 𝑌 tanh 𝑋 Iterate or 𝑋 = 1 2 𝑋 + 𝑌 tanh 𝑋 Gives 𝑋 = 𝑘ℎ and hence 𝑘
- 32. Example Find, in still water of depth 15 m: (a) the period of a wave with wavelength 45 m; (b) the wavelength of a wave with period 8 s. In each case write down the phase speed (celerity).
- 33. Find, in still water of depth 15 m: (a) the period of a wave with wavelength 45 m; (b) the wavelength of a wave with period 8 s. In each case write down the phase speed (celerity). ℎ = 15 m 𝐿 = 45 m 𝑘 = 2π 𝐿 = 0.1396 m−1 𝜔 = 1.153 rad s−1 𝑇 = 2π 𝜔 𝑐 = 𝜔 𝑘 wavelength: wavenumber: angular frequency: period: phase speed (celerity): or 𝐿 𝑇 = 𝟖. 𝟐𝟓𝟗 𝐦 𝐬−𝟏 𝜔2 = 𝑔𝑘 tanh 𝑘ℎ = 𝟓. 𝟒𝟒𝟗 𝐬
- 34. Find, in still water of depth 15 m: (a) the period of a wave with wavelength 45 m; (b) the wavelength of a wave with period 8 s. In each case write down the phase speed (celerity). ℎ = 15 m = 𝟖𝟏. 𝟖𝟏 𝐦 𝐿 = 2π 𝑘 𝑘 = 0.0768 m−1 = 0.7854 rad s−1 𝜔 = 2π 𝑇 𝑐 = 𝜔 𝑘 wavelength: wavenumber: angular frequency: period: phase speed (celerity): = 𝟏𝟎. 𝟐𝟑 𝐦 𝐬−𝟏 𝜔2 = 𝑔𝑘 tanh 𝑘ℎ 𝑇 = 8 s 𝜔2ℎ 𝑔 = 𝑘ℎ tanh 𝑘ℎ 𝑘ℎ tanh 𝑘ℎ = 0.9432 𝑘ℎ = 0.9432 tanh 𝑘ℎ or 𝑘ℎ = 1 2 𝑘ℎ + 0.9432 tanh 𝑘ℎ 𝑘ℎ = 1.152
- 35. 1. LINEAR WAVE THEORY 1.1 Main wave parameters 1.2 Dispersion relationship 1.3 Wave velocity and pressure 1.4 Wave energy 1.5 Group velocity 1.6 Energy transfer (wave power) 1.7 Particle motion 1.8 Shallow-water and deep-water behaviour 1.9 Waves on currents Linear Wave Theory
- 36. Velocity 𝜂 = 𝐴 cos 𝑘𝑥 − 𝜔𝑡 Surface displacement: Velocity potential: 𝜙 = 𝐴𝑔 𝜔 cosh 𝑘 ℎ + 𝑧 cosh 𝑘ℎ sin 𝑘𝑥 − 𝜔𝑡 𝑢 ≡ 𝜕𝜙 𝜕𝑥 = 𝐴𝑔𝑘 𝜔 cosh 𝑘 ℎ + 𝑧 cosh 𝑘ℎ cos 𝑘𝑥 − 𝜔𝑡 𝑤 ≡ 𝜕𝜙 𝜕𝑧 = 𝐴𝑔𝑘 𝜔 sinh 𝑘 ℎ + 𝑧 cosh 𝑘ℎ sin 𝑘𝑥 − 𝜔𝑡
- 37. Pressure 𝜂 = 𝐴 cos 𝑘𝑥 − 𝜔𝑡 Surface displacement: Velocity potential: 𝜙 = 𝐴𝑔 𝜔 cosh 𝑘 ℎ + 𝑧 cosh 𝑘ℎ sin 𝑘𝑥 − 𝜔𝑡 Bernoulli equation: 𝑝 = −𝜌𝑔𝑧 − 𝜌 𝜕𝜙 𝜕𝑡 𝑝 = −𝜌𝑔𝑧 hydrostatic + 𝜌𝑔𝐴 cosh 𝑘 ℎ + 𝑧 cosh 𝑘ℎ cos 𝑘𝑥 − 𝜔𝑡 ) hydrodynamic (i.e. wave 𝜌 𝜕𝜙 𝜕𝑡 + 𝑝 + 𝜌𝑔𝑧 = 0 = −𝜌𝑔𝑧 + 𝜌𝑔𝜂 × cosh 𝑘 ℎ + 𝑧 cosh 𝑘ℎ
- 38. Example A pressure sensor is located 0.6 m above the sea bed in a water depth ℎ = 12 m. The pressure fluctuates with period 15 s. A maximum gauge pressure of 124 kPa is recorded. (a) What is the wave height? (b) What are the maximum horizontal and vertical velocities at the surface?
- 39. A pressure sensor is located 0.6 m above the sea bed in a water depth ℎ = 12 m. The pressure fluctuates with period 15 s. A maximum gauge pressure of 124 kPa is recorded. (a) What is the wave height? 𝑝 = −𝜌𝑔𝑧 + 𝜌𝑔𝐴 cosh 𝑘 ℎ + 𝑧 cosh 𝑘ℎ cos(𝑘𝑥 − 𝜔𝑡) ℎ = 12 m 𝑧 = −11.4 m 𝑇 = 15 s 𝑝max = 124000 Pa 124000 = 114630 + 10060𝐴 cosh(𝑘 × 0.6) cosh 𝑘ℎ 𝜔2 = 𝑔𝑘 tanh 𝑘ℎ 𝑘 = 0.04005 m−1 = 0.4189 rad s−1 𝜔 = 2π 𝑇 𝜔2ℎ 𝑔 = 𝑘ℎ tanh 𝑘ℎ 𝑘ℎ tanh 𝑘ℎ = 0.2147 𝑘ℎ = 0.2147 tanh 𝑘ℎ or 𝑘ℎ = 1 2 𝑘ℎ + 0.2147 tanh 𝑘ℎ 𝑘ℎ = 0.4806 124000 = 114630 + 10060𝐴 × 0.8949 𝐴 = 1.041 m 𝐻 = 2𝐴 = 𝟐. 𝟎𝟖𝟐 𝐦
- 40. (b) What are the maximum horizontal and vertical velocities at the surface? 𝑧 = 0 𝑘 = 0.04005 m−1 𝜔 = 0.4189 rad s−1 𝑘ℎ = 0.4806 𝑢 = 𝐴𝑔𝑘 𝜔 cosh 𝑘 ℎ + 𝑧 cosh 𝑘ℎ cos 𝑘𝑥 − 𝜔𝑡 𝑤 = 𝐴𝑔𝑘 𝜔 sinh 𝑘 ℎ + 𝑧 cosh 𝑘ℎ sin 𝑘𝑥 − 𝜔𝑡 𝐴 = 1.041 m 𝑢max = 𝐴𝑔𝑘 𝜔 𝑤max = 𝐴𝑔𝑘 𝜔 tanh 𝑘ℎ = 𝟎. 𝟗𝟕𝟔𝟒 𝐦 𝐬−𝟏 = 𝟎. 𝟒𝟑𝟔𝟐 𝐦 𝐬−𝟏 (surface)
- 41. 1. LINEAR WAVE THEORY 1.1 Main wave parameters 1.2 Dispersion relationship 1.3 Wave velocity and pressure 1.4 Wave energy 1.5 Group velocity 1.6 Energy transfer (wave power) 1.7 Particle motion 1.8 Shallow-water and deep-water behaviour 1.9 Waves on currents Linear Wave Theory
- 42. Wave Energy ● Wave energy density 𝐸 is average energy per unit horizontal area. ● Found by integrating over the water column, and averaging over a wave cycle. ● Kinetic energy: ● Potential energy: ● (Under linear theory) average wave-related KE and PE are the same. ● Total energy: 𝐸 = 1 2 𝜌𝑔𝐴2 = 1 8 𝜌𝑔𝐻2 KE = න 𝑧=−ℎ 𝜂 1 2 𝜌 𝑢2 + 𝑤2 d𝑧 = 1 4 𝜌𝑔𝐴2 PE = න 𝑧=−ℎ 𝜂 𝜌𝑔𝑧 d𝑧 = 1 4 𝜌𝑔𝐴2 + constant
- 43. Kinetic Energy (Appendix A4) KE = 1 2 𝜌 න 𝑧=−ℎ 𝜂 𝑢2 + 𝑤2 d𝑧 𝑢2 + 𝑤2 = 𝐴𝑔𝑘 𝜔 cosh 𝑘ℎ 2 cosh2 𝑘 ℎ + 𝑧 cos2 𝑘𝑥 − 𝜔𝑡 + sinh2 𝑘 ℎ + 𝑧 sin2 𝑘𝑥 − 𝜔𝑡 KE = 1 2 𝜌 න 𝑧=−ℎ 0 𝑢2 + 𝑤2 d𝑧 = 1 2 𝜌 𝐴𝑔𝑘 𝜔 cosh 𝑘ℎ 2 × 1 2 න −ℎ 0 cosh 2𝑘 ℎ + 𝑧 d𝑧 = 1 2 𝜌 𝐴𝑔𝑘 𝜔 cosh 𝑘ℎ 2 × 1 2 sinh 2𝑘 ℎ + 𝑧 2𝑘 −ℎ 0 = 1 2 𝜌 𝐴𝑔𝑘 𝜔 cosh 𝑘ℎ 2 × 1 2 × sinh 2𝑘ℎ 2𝑘 = 1 2 𝜌 𝐴𝑔𝑘 𝜔 cosh 𝑘ℎ 2 × 1 2 × 2 sinh 𝑘ℎ cosh 𝑘ℎ 2𝑘 = 1 4 𝜌𝐴2 𝑔2 𝑘 tanh 𝑘ℎ 𝜔2 𝜔2 = 𝑔𝑘 tanh 𝑘ℎ KE = 1 4 𝜌𝑔𝐴2 = 1 2 𝜌 𝐴𝑔𝑘 𝜔 cosh 𝑘ℎ 2 × 1 2 න −ℎ 0 cosh2 𝑘 ℎ + 𝑧 + sinh2 𝑘(ℎ + 𝑧) d𝑧
- 44. Potential Energy (Appendix A4) PE = න 𝑧=−ℎ 𝜂 𝜌𝑔𝑧 d𝑧 = 1 2 𝜌𝑔 𝑧2 −ℎ 𝜂 PE = 1 4 𝜌𝑔𝐴2 = 1 2 𝜌𝑔(𝜂2 − ℎ2 ) PE = 1 2 𝜌𝑔 × 1 2 𝐴2 Only the wave component is needed = 1 2 𝜌𝑔(𝐴2 cos2 𝑘𝑥 − 𝜔𝑡 + constant)
- 45. 1. LINEAR WAVE THEORY 1.1 Main wave parameters 1.2 Dispersion relationship 1.3 Wave velocity and pressure 1.4 Wave energy 1.5 Group velocity 1.6 Energy transfer (wave power) 1.7 Particle motion 1.8 Shallow-water and deep-water behaviour 1.9 Waves on currents Linear Wave Theory
- 46. Phase and Group Velocities Phase velocity 𝑐 ≡ 𝜔 𝑘 – velocity at which the waveform translates – really only meaningful for a regular wave, or single frequency component Group velocity 𝑐𝑔 ≡ d𝜔 d𝑘 – velocity at which energy propagates – more appropriate for a wave packet comprised of multiple frequency components
- 47. Combination of Frequency Components 𝜂 = 𝑎 cos (𝑘 + Δ𝑘)𝑥 − (𝜔 + Δ𝜔)𝑡 component 1 + 𝑎 cos (𝑘 − Δ𝑘)𝑥 − (𝜔 − Δ𝜔)𝑡 component 2 Amplitude modulation: 𝜂 = 2𝑎 cos 𝑘𝑥 − 𝜔𝑡 cos Δ𝑘. 𝑥 − Δ𝜔. 𝑡 Two components: frequencies ω ± Δω and wavenumbers 𝑘 ± Δ𝑘 𝐴 𝑡 = 2𝑎 cos Δ𝑘 𝑥 − Δ𝜔 𝑡 Speed of amplitude envelope: Δ𝜔 Δ𝑘 Group velocity 𝑐𝑔 ≡ d𝜔 d𝑘 cos 𝛼 + cos 𝛽 = 2 cos 𝛼 + 𝛽 2 cos 𝛼 − 𝛽 2
- 48. Group Velocity (Appendix A5) Group velocity 𝑐𝑔 ≡ d𝜔 d𝑘 Dispersion relation 𝜔2 = 𝑔𝑘 tanh 𝑘ℎ 2𝜔 d𝜔 d𝑘 = 𝑔 tanh 𝑘ℎ + 𝑔𝑘ℎ sech2 𝑘ℎ = 𝜔2 𝑘 + 𝜔2 tanh 𝑘ℎ ℎ cosh2 𝑘ℎ = 𝜔2 𝑘 1 + 𝑘ℎ sinh 𝑘ℎ cosh 𝑘ℎ d𝜔 d𝑘 = 1 2 1 + 2𝑘ℎ sinh 2𝑘ℎ 𝜔 𝑘 𝑐𝑔 = 𝑛𝑐 𝑐 ≡ 𝜔 𝑘 𝑛 = 1 2 1 + 2𝑘ℎ sinh 2𝑘ℎ 1 2 < 𝑛 < 1 group velocity < phase velocity
- 49. 1. LINEAR WAVE THEORY 1.1 Main wave parameters 1.2 Dispersion relationship 1.3 Wave velocity and pressure 1.4 Wave energy 1.5 Group velocity 1.6 Energy transfer (wave power) 1.7 Particle motion 1.8 Shallow-water and deep-water behaviour 1.9 Waves on currents Linear Wave Theory
- 50. Wave Power 𝑃 = 𝐸𝑐𝑔 𝐸 = 1 2 𝜌𝑔𝐴2 𝑐𝑔 = 𝑛𝑐 (energy density) Power (group velocity) Wave power 𝑃 is the (average) rate of energy transfer per unit length of wave crest. It can be calculated from the rate of working of pressure forces.
- 51. Wave Power (Appendix A6) Wave power = (time-averaged) rate of working of pressure forces (pressure area velocity) Per unit length of wave crest: power = න 𝑧=−ℎ 𝜂 𝑝𝑢 d𝑧 𝑝𝑢 = 𝜌𝑔𝐴 cosh 𝑘 ℎ + 𝑧 cosh 𝑘ℎ cos 𝑘𝑥 − 𝜔𝑡 × 𝐴𝑔𝑘 𝜔 cosh 𝑘 ℎ + 𝑧 cosh 𝑘ℎ cos 𝑘𝑥 − 𝜔𝑡 = 𝜌𝑔2 𝐴2 𝑘 𝜔 cosh2 𝑘 ℎ + 𝑧 cosh2 𝑘ℎ cos2 𝑘𝑥 − 𝜔𝑡 power = 𝜌𝑔2 𝐴2 𝑘 𝜔 cosh2 𝑘ℎ × න −ℎ 0 cosh2 𝑘 ℎ + 𝑧 d𝑧 × 1 2 1 2 න −ℎ 0 cosh 2𝑘 ℎ + 𝑧 + 1 d𝑧 = 1 2 sinh 2𝑘 ℎ + 𝑧 2𝑘 + 𝑧 −ℎ 0 = 1 2 sinh 2𝑘ℎ 2𝑘 + ℎ pressure (𝑝) area (1 × 𝑑𝑧) velocity (𝑢)
- 52. Wave Power power = 𝜌𝑔2 𝐴2 𝑘 𝜔 cosh2 𝑘ℎ × 1 2 sinh 2𝑘ℎ 2𝑘 + ℎ × 1 2 = 1 2 𝜌𝑔𝐴2 × 𝑔𝑘 𝜔 cosh2 𝑘ℎ × sinh 2𝑘ℎ 2𝑘 1 + 2𝑘ℎ sinh 2𝑘ℎ × 1 2 = 1 2 𝜌𝑔𝐴2 × 𝑔𝑘 𝜔 cosh2 𝑘ℎ × 2 sinh 𝑘ℎ cosh 𝑘ℎ 2𝑘 1 + 2𝑘ℎ sinh 2𝑘ℎ × 1 2 = 1 2 𝜌𝑔𝐴2 × 𝑔𝑘 tanh 𝑘ℎ 𝜔2 × 1 2 1 + 2𝑘ℎ sinh 2𝑘ℎ × 𝜔 𝑘 𝐸 1 𝑛 𝑐 𝑃 = 𝐸𝑐𝑔 𝐸 = 1 2 𝜌𝑔𝐴2 𝑐𝑔 = 𝑛𝑐 energy density power group velocity
- 53. Example A sea-bed pressure transducer in 9 m of water records a sinusoidal signal with amplitude 5.9 kPa and period 7.5 s. Find the wave height, energy density and wave power per metre of crest.
- 54. A sea-bed pressure transducer in 9 m of water records a sinusoidal signal with amplitude 5.9 kPa and period 7.5 s. Find the wave height, energy density and wave power per metre of crest. 𝑝wave = 𝜌𝑔𝐴 cosh 𝑘 ℎ + 𝑧 cosh 𝑘ℎ cos(𝑘𝑥 − 𝜔𝑡) ℎ = 9 m 𝑧 = −ℎ 𝑇 = 7.5 s Δ𝑝wave = 5900 Pa 𝜔2 = 𝑔𝑘 tanh 𝑘ℎ 𝑘 = 0.09993 m−1 = 0.8378 rad s−1 𝜔 = 2π 𝑇 𝜔2ℎ 𝑔 = 𝑘ℎ tanh 𝑘ℎ 𝑘ℎ tanh 𝑘ℎ = 0.6440 𝑘ℎ = 0.6440 tanh 𝑘ℎ or 𝑘ℎ = 1 2 𝑘ℎ + 0.6440 tanh 𝑘ℎ 𝑘ℎ = 0.8994 (sea bed) 𝐸 = 1 2 𝜌𝑔𝐴2 𝑃 = 𝐸𝑐𝑔 𝑐𝑔 = 𝑛𝑐 𝑛 = 1 2 1 + 2𝑘ℎ sinh 2𝑘ℎ (amplitude)
- 55. A sea-bed pressure transducer in 9 m of water records a sinusoidal signal with amplitude 5.9 kPa and period 7.5 s. Find the wave height, energy density and wave power per metre of crest. ℎ = 9 m 𝑧 = −ℎ 𝑇 = 7.5 s Δ𝑝wave = 5900 Pa 5900 = 10055𝐴 × 1 cosh 0.8994 𝑘 = 0.09993 m−1 𝜔 = 0.8378 rad s−1 𝑘ℎ = 0.8994 𝐻 = 2𝐴 = 𝟏. 𝟔𝟖𝟏 𝐦 (sea bed) 𝐸 = 1 2 𝜌𝑔𝐴2 𝑃 = 𝐸𝑐𝑔 𝑐𝑔 = 𝑛𝑐 𝑛 = 1 2 1 + 2𝑘ℎ sinh 2𝑘ℎ 𝑝wave = 𝜌𝑔𝐴 cosh 𝑘 ℎ + 𝑧 cosh 𝑘ℎ cos(𝑘𝑥 − 𝜔𝑡) 𝐴 = 0.8405 m 𝐸 = 1 2 𝜌𝑔𝐴2 = 𝟑𝟓𝟓𝟐 𝐉 𝐦−𝟐 𝑐 = 𝜔 𝑘 = 8.384 m s−1 𝑛 = 1 2 1 + 2𝑘ℎ sinh 2𝑘ℎ = 0.8061 𝑐𝑔 = 𝑛𝑐 = 6.758 m s−1 𝑃 = 𝐸𝑐𝑔 = 𝟐𝟒𝟎𝟎𝟎 𝐖 𝐦−𝟏
- 56. 1. LINEAR WAVE THEORY 1.1 Main wave parameters 1.2 Dispersion relationship 1.3 Wave velocity and pressure 1.4 Wave energy 1.5 Group velocity 1.6 Energy transfer (wave power) 1.7 Particle motion 1.8 Shallow-water and deep-water behaviour 1.9 Waves on currents Linear Wave Theory
- 57. Particle Motion = 𝜔𝐴 cosh 𝑘 ℎ + 𝑧 sinh 𝑘ℎ cos 𝑘𝑥 − ω𝑡 Velocity: 𝑤 = 𝐴 𝑔𝑘 𝜔 sinh 𝑘 ℎ + 𝑧 cosh 𝑘ℎ sin 𝑘𝑥 − ω𝑡 Dispersion relation: 𝜔2 = 𝑔𝑘 tanh 𝑘ℎ → 𝜔 sinh 𝑘ℎ = 𝑔𝑘 𝜔 cosh 𝑘ℎ d𝑋 d𝑡 = 𝑢 d𝑍 d𝑡 = 𝑤 𝑎 = 𝐴 cosh 𝑘 ℎ + 𝑍0 sinh 𝑘ℎ 𝑏 = 𝐴 sinh 𝑘 ℎ + 𝑍0 sinh 𝑘ℎ 𝑋 = 𝑋0 − 𝑎 sin 𝑘𝑋0 − 𝜔𝑡 𝑍 = 𝑍0 + 𝑏 cos 𝑘𝑋0 − 𝜔𝑡 sin2 𝜃 + cos2 𝜃 = 1 𝑋 − 𝑋0 𝑎 = − sin 𝑘𝑋0 − 𝜔𝑡 𝑍 − 𝑍0 𝑏 = cos 𝑘𝑋0 − 𝜔𝑡 = 𝑎𝜔 cos 𝑘𝑋0 − ω𝑡 = 𝑏𝜔 sin 𝑘𝑋0 − ω𝑡 𝑢 = 𝐴 𝑔𝑘 𝜔 cosh 𝑘 ℎ + 𝑧 cosh 𝑘ℎ cos 𝑘𝑥 − ω𝑡 = 𝜔𝐴 sinh 𝑘 ℎ + 𝑧 sinh 𝑘ℎ sin 𝑘𝑥 − ω𝑡
- 58. Particle Motion 𝑎 = 𝐴 cosh 𝑘 ℎ + 𝑍0 sinh 𝑘ℎ 𝑏 = 𝐴 sinh 𝑘 ℎ + 𝑍0 sinh 𝑘ℎ 𝑋 − 𝑋0 2 𝑎2 + 𝑍 − 𝑍0 2 𝑏2 = 1 Ellipse, centre (𝑋0, 𝑍0) and semi-axes 𝑎 and 𝑏 intermediate depth shallow water
- 59. 1. LINEAR WAVE THEORY 1.1 Main wave parameters 1.2 Dispersion relationship 1.3 Wave velocity and pressure 1.4 Wave energy 1.5 Group velocity 1.6 Energy transfer (wave power) 1.7 Particle motion 1.8 Shallow-water and deep-water behaviour 1.9 Waves on currents Linear Wave Theory
- 60. Shallow-Water / Deep-Water Limits Dispersion relationship: 𝜔2 = 𝑔𝑘 tanh 𝑘ℎ Asymptotic behaviour: ) tanh 𝑘ℎ ~𝑘ℎ (as 𝑘ℎ → 0 ) tanh 𝑘ℎ → 1 (as 𝑘ℎ → ∞ Shallow water (or long waves): 𝑘ℎ ≪ 1 𝜔2 ≈ 𝑘2𝑔ℎ 𝑐 = 𝑐𝑔 = 𝑔ℎ (non-dispersive) Deep water (or short waves): 𝑘ℎ ≫ 1 𝜔2 ≈ 𝑔𝑘 𝐿 = 𝑔𝑇2 2π 𝑐 = 𝐿 𝑇 = 𝑔𝑇 2π , 𝑛 = 1 2 , 𝑐𝑔 = 1 2 𝑐 (dispersive) 𝑛 = 1 𝑘ℎ = 2π ℎ 𝐿 𝜔 ≈ 𝑘 𝑔ℎ or
- 61. Shallow / Deep Limits Deep: 𝑘ℎ < 𝜋 10 ℎ < 1 20 𝐿 Shallow: 𝑘ℎ > π ℎ > 1 2 𝐿 𝜔2 = 𝑔𝑘 tanh 𝑘ℎ 𝜔2ℎ 𝑔 = 𝑘ℎ tanh 𝑘ℎ
- 62. Shallow / Deep Particle Motions 𝑎 = 𝐴 cosh 𝑘 ℎ + 𝑍0 sinh 𝑘ℎ , 𝑏 = 𝐴 sinh 𝑘 ℎ + 𝑍0 sinh 𝑘ℎ Ellipses: deep water intermediate depth shallow water Deep: Shallow: 𝑘ℎ ≫ 1 𝑎 = 𝑏 ≈ 𝐴e−𝑘 𝑍0 Circles diminishing in size over half a wavelength 𝑘ℎ ≪ 1 𝑎 ≈ 𝐴 𝑘ℎ , 𝑏 𝑎 ≪ 1 Highly-flattened ellipses; horizontal excursion almost independent of depth
- 63. Shallow / Deep Pressure Deep: Shallow: 𝑘ℎ ≫ 1 Perturbation decays over half a wavelength 𝑘ℎ ≪ 1 𝑝 = −𝜌𝑔𝑧 − 𝜌 𝜕𝜙 𝜕𝑡 = −𝜌𝑔𝑧 hydrostatic + 𝜌𝑔𝜂 cosh 𝑘 ℎ + 𝑧 cosh 𝑘ℎ hydrodynamic 𝑝 ≈ −𝜌𝑔𝑧 + 𝜌𝑔𝜂e−𝑘 𝑧 ) 𝑝 ≈ 𝜌𝑔(𝜂 − 𝑧 Hydrostatic
- 64. Example (a) Find the deep-water speed and wavelength of a wave of period 12 s. (b) Find the speed and wavelength of a wave of period 12 s in water of depth 3 m. Compare with the shallow-water approximation.
- 65. 𝜔2 = 𝑔𝑘 tanh 𝑘ℎ Deep water: 𝑘𝒉 → ∞ tanh 𝑘ℎ → 1 𝜔2 = 𝑔𝑘 2π 𝑇 2 = 𝑔 2π 𝐿 𝐿 = 𝑔𝑇2 2π 𝑐 = 𝑔𝑇 2π Shallow water: 𝑘𝒉 → 𝟎 tanh 𝑘ℎ ~𝑘ℎ 𝜔2 = 𝑔𝑘2 ℎ 𝜔 𝑘 2 = 𝑔ℎ 𝐿 = 𝑐𝑇 𝑐 = 𝑔ℎ Reminder of Deep and Shallow Limits
- 66. (a) Find the deep-water speed and wavelength of a wave of period 12 s. (b) Find the speed and wavelength of a wave of period 12 s in water of depth 3 m. Compare with the shallow-water approximation. Deep: 𝑇 = 12 s 𝑐 = 𝑔𝑇 2π = 𝟏𝟖. 𝟕𝟒 𝐦 𝐬−𝟏 𝐿 = 𝑔𝑇2 2π = 𝟐𝟐𝟒. 𝟖 𝐦 Exact, with ℎ = 3 m: 𝜔2 = 𝑔𝑘 tanh 𝑘ℎ 𝑐 = 𝜔 𝑘 = 0.5236 rad s−1 𝜔 = 2π 𝑇 𝜔2 ℎ 𝑔 = 𝑘ℎ tanh 𝑘ℎ 𝑘ℎ tanh 𝑘ℎ = 0.08384 𝑘ℎ = 0.08384 tanh 𝑘ℎ or 𝑘ℎ = 1 2 𝑘ℎ + 0.08384 tanh 𝑘ℎ 𝑘ℎ = 0.2937 𝑘 = 0.09790 m−1 = 𝟓. 𝟑𝟒𝟖 𝐦 𝐬−𝟏 𝐿 = 2π 𝑘 = 𝟔𝟒. 𝟏𝟖 𝐦 Shallow: 𝑐 = 𝑔ℎ = 𝟓. 𝟒𝟐𝟓 𝐦 𝐬−𝟏 𝐿 = 𝑐𝑇 = 𝟔𝟓. 𝟏𝟎 𝐦
- 67. 1. LINEAR WAVE THEORY 1.1 Main wave parameters 1.2 Dispersion relationship 1.3 Wave velocity and pressure 1.4 Wave energy 1.5 Group velocity 1.6 Energy transfer (wave power) 1.7 Particle motion 1.8 Shallow-water and deep-water behaviour 1.9 Waves on currents Linear Wave Theory
- 68. Waves on Currents ● Waves co-exist with background current 𝑈 ● Formulae hold in relative frame moving with the current: 𝑥𝑟 = 𝑥 − 𝑈𝑡 𝜂 = 𝐴 cos 𝑘𝑥𝑟 − 𝜔𝑟𝑡 = 𝐴 cos 𝑘𝑥 − 𝜔𝑟 + 𝑘𝑈 𝑡 = 𝑐𝑟 + 𝑈 𝜔𝑎 = 𝜔𝑟 + 𝑘𝑈 ● Dispersion relationship: 𝜔𝑎 − 𝑘𝑈 2 = 𝜔𝑟 2 = 𝑔𝑘 tanh 𝑘ℎ 𝑐𝑎 = 𝜔𝑎 𝑘 = 𝐴 cos 𝑘𝑥 − 𝜔𝑎𝑡
- 69. Example An acoustic depth sounder indicates regular surface waves with apparent period 8 s in water of depth 12 m. Find the wavelength and absolute phase speed of the waves when there is: (a) no mean current; (b) a current of 3 m s–1 in the same direction as the waves; (c) a current of 3 m s–1 in the opposite direction to the waves.
- 70. An acoustic depth sounder indicates regular surface waves with apparent period 8 s in water of depth 12 m. Find the wavelength and absolute phase speed of the waves when there is: (a) no mean current; (b) a current of 3 m s–1 in the same direction as the waves; (c) a current of 3 m s–1 in the opposite direction to the waves. ℎ = 12 m 𝑇𝑎 = 8 s (absolute) 𝜔𝑎 = 2π 𝑇𝑎 𝜔𝑎 − 𝑘𝑈 2 = 𝜔𝑟 2 = 𝑔𝑘 tanh 𝑘ℎ 𝑘 = 0.7854 − 𝑘𝑈 2 9.81 tanh 12𝑘 𝑘 = 1 2 𝑘 + 0.7854 − 𝑘𝑈 2 9.81 tanh 12𝑘 or 𝑈 = 0 0.08284 = 𝜔𝑎 𝑘 = 2π 𝑘 𝟕𝟓. 𝟖𝟓 𝟗. 𝟒𝟖𝟏 𝑈 = +3 m s−1 0.06024 𝟏𝟎𝟒. 𝟑 𝟏𝟑. 𝟎𝟒 𝑈 = −3 m s−1 0.1951 𝟑𝟐. 𝟐𝟎 𝟒. 𝟎𝟐𝟔 = 0.7854 rad s−1 𝑘 (m−1 ) 𝐿 (m) 𝑐𝑎 (m s−1 ) 𝑘 = 1 2 𝑘 + 0.78542 9.81 tanh 12𝑘 𝑘 = 1 2 𝑘 + 0.7854 − 3𝑘 2 9.81 tanh 12𝑘 𝑘 = 1 2 𝑘 + 0.7854 + 3𝑘 2 9.81 tanh 12𝑘 Iteration: