Ocean waves, with a period of \(8 \mathrm{~s}\) and amplitude \(0.6 \mathrm{~m}\) in deep water, approach the shore in the normal direction. A device \(80 \mathrm{~m}\) long for extracting power from the waves is installed parallel to the shore in water \(5 \mathrm{~m}\) deep. If there is negligible dissipation of energy before the waves reach the device and its efficiency is \(50 \%\), what power is produced? What is the amplitude of the waves immediately before this position? (Density of sea-water \(\left.=1025 \mathrm{~kg} \cdot \mathrm{m}^{-3} .\right)\)

Understanding wave speed is essential for studying wave energy. Waves travel at different speeds depending on water depth. The formula for wave speed in deep water is \[\text{Wave speed} = \frac{\text{Wavelength}}{\text{Period}}\]. For this exercise, we used the period (T) of 8 seconds. In deep water, the wavelength approximates to 1.56 times the period squared, giving: \[\text{Wavelength} = 1.56 \times (8 \text{ s})^2 = 100 \text{ m}\]. Thus, the wave speed is: \[\text{Wave speed} = \frac{100 \text{ m}}{8 \text{ s}} = 12.5 \text{ m/s}\].
When the waves approach shallower water (depth d = 5 m), they slow down. Shallow water wave speed is derived using: \[\text{Wave speed in shallow water} = \sqrt{gd}\], where g is the gravitational constant (9.81 m/s²). Plugging in the values, the shallow water speed becomes: \[\text{Wave speed in shallow water} = \sqrt{9.81 \times 5} = 7.0 \text{ m/s}\].
Understanding how wave speed changes helps in designing efficient wave energy devices.

Calculating the power in ocean waves involves understanding several factors. The power per meter of wave front in deep water is given by: \[P = \frac{1}{16} \rho g H^2 c\], where \(\rho\) is sea-water density (1025 kg/m³), g is the gravitational constant (9.81 m/s²), H is wave height (twice the amplitude), and c is wave speed. With a wave height of 1.2 meters (2 \times 0.6 meters) and deep-water speed of 12.5 m/s, we find: \[P = \frac{1}{16} \times 1025 \times 9.81 \times 1.2^2 \times 12.5 \approx 943.2 \text{ W/m}\].
To find the total power for the device, multiply by its length (80 meters): \[\text{Total power} = 943.2 \times 80 \approx 75,456 \text{ W}\]. Given the device's efficiency is 50%, the power produced is: \[\text{Power produced} = 0.5 \times 75,456 = 37,728 \text{ W}\].
These calculations help estimate how much energy wave power devices can harness.

As waves move into shallower water, their amplitude changes to conserve energy. The relationship between amplitudes in deep and shallow water follows: \[A_{shallow} = A_{deep} \sqrt{\frac{c_{deep}}{c_{shallow}}}\]. For this exercise, with a deep water amplitude of 0.6 meters, deep water speed of 12.5 m/s and shallow water speed of 7.0 m/s, we get: \[A_{shallow} = 0.6 \times \sqrt{\frac{12.5}{7.0}} \approx 0.8 \text{ m}\].
This increase is because wave energy is conserved as speed decreases, resulting in a rise in wave height. This has implications for wave energy extraction devices positioned in variable water depths.

Efficiency is critical in wave energy devices. It represents the ratio of usable power to the total power available. Given the calculated total power of 75,456 W and an efficiency of 50%, the device yields: \[\text{Power produced} = 0.5 \times 75,456 \approx 37,728 \text{ W}\].
Maximizing efficiency involves optimizing device design to capture the most energy possible from waves while minimizing losses. Material choice, placement, and mechanics all play important roles in improving efficiency. Understanding and enhancing these factors can significantly impact the power output of wave energy devices.