Friction boilers

Converting the Kinetic and Potential Energies in Deep Ocean Waves to determine how much heat you can get from Friction Boilers
Calculating the True Power in deep oceanic waves.
Assuming a perfect sinusoidal oceanic wave, one metre high between crest and trough with a period of 10 seconds in deep water, its velocity is calculated using the formula
(gT/2π) = 15.613 m/s
The wavelength, λ vT (gT^2/2π) = 156.131 m
The potential energy passing per unit time, per unit length, is
Ppotential =m*gh/T
where m* is the mass per unit length and h is the change in height at the centreline of the sinusoidal mass of seawater.
The cross-section area of the wave’s amplitude is calculated using the following formula
Area (2 x base x height)/pi, (2 (λ/2) a)/ π because a h/2 this can be rewritten as (λh/2π).
Ppotential =m*g h/T (2 λ h g h ρ) / 4 π T (λ h^2 g ρ) / 2π T
and because λ/T is velocity, substituting we get,
Ppotential = h^2 g ρ v / 2π
(1 x 1 x 9.81 x 1025 x 15.613)/ 2π 24.986kW/m
The potential and kinetic energies in waves are exactly equal. I now will calculate the kinetic energy of deep water dispersive waves independently to see if the answers are the same.
P kinetic = ½ m v^2 (kW/m)
The simultaneous vertical descending and ascending velocity of all oceanic waves irrespective of their height is always √2 (m/s) which I will call vd and va respectively. Substituting we get
P kinetic = ½ [(vd h)(va h)(g v ρ)]/2π (kW/m)
= ½ [(√2 h) (√2 h) (g v ρ)]/2 π (kW/m)
= ½ 2h2 g v ρ)/2 π (kW/m)
Therefore, P kinetic = (h^2 g v ρ)/2π (kW/m)
= 24.986 (kW/m)
= P potential
P potential + P kinetic =24.986 + 24.986 = 49.972 kW/m
Therefore, P potential + P kinetic = (h^2 g v ρ)/ π (kW/m) so that for all deep pelagic ocean waves the following formula is true;
Ptotal = (h^2 g v ρ)/ π
Copyright Andrew H Mackay
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