University:
Massachusetts Institute of Technology
Course:
2.24 | Ocean Wave Interaction with Ships and Offshore Energy Systems (13.022)
Academic year:

2019
Views:

310

Pages:

10
Author:

Rachel Pace

13.022 – FALL 99 QUIZ #1 - SOLUTIONS 1. [Diagram: A graph showing S(ω) (m²/s) on the y-axis and ω on the x-axis. The graph is a piecewise linear function with four segments. The first segment starts at S(ω) = 0 and rises to S(ω) = 20 at ω = 2Ω. The second segment is horizontal at S(ω) = 20 from ω = 2Ω to ω = 3Ω. The third segment decreases linearly from S(ω) = 20 at ω = 3Ω to S(ω) = 0 at ω = 4Ω. The fourth segment remains at S(ω) = 0 for ω > 4Ω.] a) σ² = ∫₀^∞ dω S(ω) = 2 ∫₀^2Ω dω S(ω) = 2 × 0.2 × 20 m² σ = 2.82 m b) [Diagram: A zoomed-in view of the graph from part (a), focusing on the region between ω = 2Ω and ω = 4Ω. The graph is divided into three intervals: ω1 to ω2, ω2 to ω3, and ω3 to ω4. The areas under each interval are labeled as S/2.] ω1 = 3Ω/2, ω2 = 5Ω/2, ω3 = 7Ω/2 ΔΩ = Ω S(ω1) = S/2, S(ω2) = S, S(ω3) = S/2 ξ(x,t) = ∑ᵢ³ Aᵢ cos(kᵢx - ωᵢt + φᵢ) • ½ Aᵢ² = S(ωᵢ) ΔΩ => Aᵢ = {2Ω S(ωᵢ)}¹ᐟ² • kᵢ = ωᵢ²/g • φᵢ: RANDOM PHASE ANGLE UNIFORMLY DISTRIBUTED IN (0, 2π) [Diagram: A horizontal line with an arrow pointing to the right. Below the line, "φ" is written, and above the line, "2π" is written. An arrow points upwards from the midpoint of the line, labeled "f(φ)".] HENCE: ξ(x,t) = ∑ᵢ³ {2Ω S(ωᵢ)}¹ᐟ² cos(kᵢx - ωᵢt + φᵢ) c) P = -ρ ∂φ/∂t (LINEAR TERM ONLY - PROP QUADRATIC + HYDROSTATIC) φ = ∑ᵢ³ Re{ i g Aᵢ / ωᵢ e^(kᵢz - ikᵢx + iωᵢt - iφᵢ) } CORRESPONDS TO WAVE ELEVATION DEFINED ABOVE. CHECK USING: J = - 1/g ∂φ/∂t |_(z=0) d) Φ = ∑ᵢ³ Re{ i g Aᵢ e^(kᵢz - ikᵢx + iωᵢt - iφᵢ) } ENERGY FLUX ACROSS PLANE AT X=0 dE/dt = ∫₋∞^∞ p ∂φ/∂t ∂φ/∂x dx dt , H.H. Eq. (71) ∂φ/∂t = ∑ᵢ³ Re{ -g Aᵢ ωᵢ e^(kᵢz + iωᵢt - iφᵢ) } |_x=0 ∂φ/∂x = ∑ᵢ³ Re{ -i g Aᵢ kᵢ e^(kᵢz + iωᵢt - iφᵢ) } |_x=0 ∫₋∞^∞ ∂φ/∂t ∂φ/∂x dx = ∫₋∞^∞ dz ∑ᵢ³ Re{ -g Aᵢ ωᵢ e^(kᵢz + iωᵢt - iφᵢ) } ∑ⱼ³ Re{ -i g Aⱼ kⱼ e^(kⱼz + iωⱼt - iφⱼ) } = ∑ᵢ³ ∑ⱼ³ Re{ -g² Aᵢ Aⱼ ωᵢ kⱼ e^(i(ωᵢ-ωⱼ)t - i(φᵢ-φⱼ)) ∫₋∞^∞ dz e^(i(kᵢ-kⱼ)z) } WHEN MEAN TIME VALUE IS TAKEN CROSS PRODUCTS WITH FREQUENCIES ωᵢ ≠ ωⱼ AVERAGE OUT SINCE e^(i(ωᵢ-ωⱼ)t) TERMS ωᵢ ≠ ωⱼ SURVIVE. HENCE ONLY THE DIAGONAL TERMS SURVIVE LEADING TO AN ENERGY FLUX dE/dt = ½ ρg²|A₁|² ω₁ + ½ ρg²|A₂|² ω₂ + ½ ρg²|A₃|² ω₃ V_g1 = ω₁/2k₁, V_g2 = ω₂/2k₂, V_g3 = ω₃/2k₃, ...