# Wave Propagation over Variable Bottom Topography Given Data: - Deep water wavelength, \(\lambda_ = 300\,m\) - Deep water wave height,H_ = 2\,m\) - Depth at site, \(d = 30\,m\) --- ## Part (a): Wave Properties at \(d=30\,m\) ### Step 1: Determine the deep water wave number \(k_\) \[ k_ = \frac{2\pi}{\lambda_} = \frac{2\pi}{300} \approx .02094\, \text{rad/m} \] ### Step 2: Find the wave number \(k\) at \(d=30\,m\) - **Dispersion relation**: \[ \omega^2 = gk \tanh(kd) \] - **Deep water condition**: \[ \omega^2 = gk_ \] - **Calculate \(\omega\)**: \[ \omega = \sqrt{gk_} = \sqrt{9.81 \times .02094} \approx \sqrt{.2058} \approx .453\, \text{rad/sec} \] - **Solve for \(k\) at \(d=30\,m\)**: \[ \omega^2 = gk \tanh(kd) \] \[ .453^2 = 9.81 \times k \times \tanh(30k) \] \[ .2058 = 9.81 \times k \times \tanh(30k) \] This transcendental equation can be solved iteratively or approximately. ### Step 3: Approximate \(k\) - For intermediate depths, assume \(k d\) is moderate. - Guess \(k \approx .05\, \text{rad/m}\): \[ 30k = 1.5 \] \[ \tanh(1.5) \approx .905 \] \[ LHS: .2058 \] \[ RHS: 9.81 \times .05 \times .905 \approx .444 \] Too high; need smaller \(k\). - Try \(k=.02\): \[ 30 \times .02 = .6 \] \[ \tanh(.6) \approx .537 \] \[ 9.81 \times .02 \times .537 \approx .105 \] Too low; increase \(k\). - Interpolating, approximate \(k \approx .035\, \text{rad/m}\): \[ 30 \times .035 = 1.05 \] \[ \tanh(1.05) \approx .785 \] \[ 9.81 \times .035 \times .785 \approx .27 \] Close to .2058; slightly high, so refine to \(k \approx .03\, \text{rad/m}\). ### **Final estimate**: \[ k \approx .03\, \text{rad/m} \] ### Step 4: Compute wavelength at \(d=30\,m\) \[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{.03} \approx 209\,m \] ### Step 5: Wave height at \(d=30\,m\) - **Wave shoaling relation**: \[ \frac{H}{H_} = \sqrt{\frac{C_g}{C_{g}}} \] - **Group velocity** (deep water): \[ C_{g} = \frac{1}{2} \sqrt{\frac{g}{k_}} = \frac{1}{2} \times \frac{\omega}{k_} = \frac{1}{2} \times \frac{.453}{.02094} \approx 10.8\, \text{m/sec} \] - **At finite depth**: \[ C_g = \frac{1}{2} \left( \frac{\omega}{k} \right) \left[ \tanh(kd) + kd \ \text{sech}^2(kd) \right] \] Approximate \(\tanh(kd) \approx .785\) and \(\text{sech}^2(kd) \approx 1 - \tanh^2(kd) \approx 1 - .785^2 \approx .383\). \[ C_g \approx \frac{1}{2} \times \frac{.453}{.03} \times (.785 + 1.05 \times .383) \] \[ = .5 \times 15.1 \times (.785 + .402) = 7.55 \times 1.187 \approx 8.96\, \text{m/sec} \] ### Step 6: Shoaling factor: \[ \frac{H}{H_} = \sqrt{\frac{C_g}{C_{g}}} = \sqrt{\frac{8.96}{10.8}} \approx \sqrt{.83} \approx .91 \] **Wave height at \(d=30\,m\)**: \[ H \approx 2\,m \times .91 \approx 1.82\,m \] --- ## **Summary for Part (a):** | Property | Value | |---|---| | Wavelength \(\lambda\) | **209 m** | | Wave height \(H\) | **1.82 m** | | Group velocity \(C_g\) | **8.96 m/sec** | --- ## Part (b): Energy per Unit Surface Area ### Step 1: Wave energy density \[ E = \frac{1}{8} \rho g H^2 \] where: - \(\rho = 1025\,kg/m^3\) (sea water density) - \(g = 9.81\,m/sec^2\) - \(H = 1.82\,m\) ### Step 2: Calculate \[ E = \frac{1}{8} \times 1025 \times 9.81 \times (1.82)^2 \] \[ E \approx .125 \times 1025 \times 9.81 \times 3.312 \] \[ E \approx .125 \times 1025 \times 32.48 \approx .125 \times 33328 \approx 4166\, \text{J/m}^2 \] **Average energy per unit surface area**: \[ \boxed{\approx 4166\, \text{J/m}^2} \] --- ## Part (c): Effect of 60° Crest Orientation ### Explanation: - **Wave crests oriented at 60° to bottom contours** influence the shoaling process. - When wave crests are oblique (\( \theta = 60^\circ \)), the component of the wave propagating perpendicular to the shoreline is reduced: \[ C_{g,\perp} = C_g \cos \theta \] - The effective energy flux normal to the shoreline decreases, affecting shoaling and wave height. ### Final adjustment: \[ H_{oblique} \approx H_{initial} \times \frac{1}{\cos 60^\circ} = H_{initial} \times 2 \] - Because the wave energy flux component normal to the shoreline is halved. ### **Adjusted wave height:** \[ H_{60^\circ} \approx 1.82\,m \times 2 = 3.64\,m \] ### **Adjusted energy per unit surface area:** \[ E_{60^\circ} \propto H^2 \Rightarrow (3.64)^2 \approx 13.25\, \text{m}^2 \] - New energy density: \[ E_{60^\circ} \approx 4166 \times \left(\frac{3.64}{1.82}\right)^2 \approx 4166 \times 4 \approx 16664\, \text{J/m}^2 \] --- # **Final Summary** | Part | Result | |---|---| | (a) Wavelength at 30 m depth | **209 m** | | (a) Wave height at 30 m depth | **1.82 m** | | (a) Group velocity at 30 m depth | **8.96 m/sec** | | (b) Energy per unit surface area | **4166 J/m²** | | (c) Wave height with 60° crest orientation | **3.64 m** | | (c) Energy per unit surface area with 60° orientation | **16664 J/m²** | --- **Note:** All calculations are approximate, based on typical wave theory and iterative estimations.