User:Andrew H Mackay

I do not believe that Wikipedia understands that some of the stuff is ALL OF MY OWN WORK so how can I CITE it to MYSELF? Here is the correct formula and proof for wave power - it is all my own work and is verified by me!

Calculating Wave Power

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)]/2 π (kW/m)

= ½ [(√2 h) (√2 h) (g v ρ)]/2 π (kW/m)

= ½ 2h^2 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

The formula given by Professor Mackay on page 309 of his book ‘Sustainable Energy – without the hot air’ is given as;

Ptotal = ¼ ρgh^2v

His formula gives answer of 39.248kW/m compared with my result 49.972 kW/m so where did the good professor go wrong? Well for a start, he assumed that his formula for Ppotential was correct and simply doubled it. Had he calculated Pkinetic separately he would have found out that the correct denominator is π and not 4 which gives him a result some 27% too low.

On page 308 of his book he writes that area of half of a sine wave “is roughly 1⁄2h(λ/2) (approximating ….. by the area of a triangle)” This triangle is shown as two triangles back-to-back in green on the figure above and is calculated to have an area of 19.516 m2

The area under the sine curve, as we have already calculated, is 24.849 m2 so that the cross-hatched area shown in red is 5.333 m2 has been excluded in Professor MacKay’s rough approximation of a sin x centroid where the centre of mass rises 393mm perpendicular to the centre point of the baseline instead of the full 500mm in this example. This accounts for an error of exactly 27.3% and as we have seen by calculating P kinetic independently it is equal to the P potential therefore proving the ‘centroid methodology’ adopted by Prof. David MacKay et al is flawed mathematics.

Therefore, the following formula is correct for a dispersant oceanic wave in deep water.

Ptotal = (h^2 g v ρ)/ π

References: Sustainable Energy – without the Hot Air. David JC Mackay http://www.inference.phy.cam.ac.uk/withouthotair/cF/page_307.shtml