CALCULATION OF THE POTENTIAL POWER IN A WAVE Note: The biggest wave ever measured in the world was observed in Alaska in the year of 1933. The data for this wave were as follows: h : 34 m ---- (wave height) L : 342 m ---- (wave length) T : 14.8 m ---- (wave period), in terms of velocity, this value corresponds to V : 23.1 m/s (velocity) or Vg: 11.5 m/s (group velocity) 1 ) -----Wave power = P = 5.5 h² T L d : depth of water ---------- m g : gravitational acceleration ------ 9.81 m/s² h : wave height ------------ m T : period ------------------- s V : velocity ( l/T ) -------- m/s l : wave length ------------ m Vg : group velocity ------- m/s a : wave crest ------------- m H : wave circle diameter – m Around the wave surface, H = h. • d ( l/2 ) corresponds to deep waters. Generally, the seas with a depth greater than 40 m • d ( l/20 ) corresponds to the shallow waters. 2 ) -----According to another approach; w = G.q. ( h / 2 )² ……………………J/m² g : 9.81 m/sn² P = Vg.w ……………………………watt/mt q : 1025 kg/m³ PERSONAL NOTES: w = m/sn² x kg/m³ x m² = Nt/m x m²/m² = Nm/m² = Joull / m² P = Vg.w →→→ P = m/sn x Joull/m² = Joull/ms = watt/m İş = w = F.s = 10 Nt. X 2 mt = 20 Nmt = 20 Joull ( 10 Nt luk yükün 2 mt kaldırılması ) F = m.g →→ Here, half the wave height (h/2) is taken for the calculation of the wave mass, m. For the average wave weight m = L.( h/2 ). ( h/2 ) . q w = F = m.g F = w = g ( h/2 )² . q. L ……………Joull Dividing both sides by “L” w = g.q. ( h/2 )² ………………Joull/m 3 ) -----According to another approach the power may be calculated using the constant and known values in the following power equation, according to the unit length of the three-dimensional sea waves; P = Q . g . H² . Cg / 4 = Q . g . H² . L / 8 T ………..( KW ) P = 1256.91 H² . L / T P = 1256.91 H² . V P = 2514 H² . Cg ……………………( w/m ) Cg = group velocity m/s ……………. Cg = V/2 = L / 2T H = average wave height … m Q = 1025 kg / m³ g = 9.81 m / s² Note: The observed values for the wave measurement data for the coastlines of Turkey are rather inadequate. The data used in order to calculate the wave energy densities on the coastlines of Turkey could be obtained from the studies conducted for NATO and could be reflected on the map of Turkey with the aid of the software Surfer. EXAMPLE SOLUTION: For the biggest wave observed in 1933; h = 34 mt V = 23.1 m/sn L = 342 mt Vg = 11.5 m/sn T = 14.8 sn According to the 1st approach P = 5.5 h² T L P =5.5 x 34² x 14.8 x 342 = 32.2 x 106 w/mt According to the 2nd approach w = g . q . ( h/2 )² →→ P = Vg . w w = 9.81 x 1025 x 34² / 4 = 2.906 x 1000000 P = 2.906 x 106 x 11.5 = 33.41 x 106 w/mt According to the 3th approach P = Q . g . H² . Cg / 4 P = Q . g . H² .L / 8T = 1256.91 H² . L /T P = 1256.91 x 34² x 342 / 14.8 = 33.5 x 106 w/mt P = 1256.91 H² . V P = 1256.91 x 34² x 23.1 = 33.5 x 106 w/mt P = 2514 . H² . Cg P = 2514 x 34² x 11.5 = 33.4 x 106 w/mt In cases where it is not possible to perform reliable observations and measurements for the wave motions, the wave energy may be calculated based on the current wind velocity data for such region, by using such data in semi-empirical relations; W = Wave energy spectrum ( m²s ) W = 8.1 x g² x 10-3 x e-0.74[ g / ɯ .Vr ] / ɯ5 g = gravitational acceleration = 9.81 m/sn2 Vr = wind velocity m/sn ɯ = Wave cyclic frequency rad / sn Note: In practice, the wave cyclic frequencies in the range of 0.2—3 rad/s contribute to the wave energy acting upon the unit area. Wave Power = P = E . b . Cg ………….. ( watt ) B = Wave width Cg = Wave velocity For deep water, Cg = C/2 c = wave velocity For shallow water, Cg = √ g.d d = water depth OBSERVATION CHART
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