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Hello. As we all know, the energy, the electrical energy in particular, is among the primary sources we utilize when meeting our needs in daily life. Lighting, communication, air conditioning, numerous machines we use in our houses, elevators, transportation, etc. We know that the electrical energy we require in order to use such means is generated via diverse methods and conversions. Here we are aiming to conduct a study on the utilization of the wave energy, which is among such methods, but is not so popular First of all, this form of energy is renewable with an unlimited source. Moreover, it is entirely environment-friendly, making it possible to provide cycles that will not result in any contribution and waste with negative effects on the earth. In that case, if we can find the proper answers to the questions like the potential, the conversion efficiency and the cost of this energy, we can consider ourselves having discovered a treasure that would positively influence our life in every respect. As is known, 2/3 of the earth is covered by waters and this gigantic mass of water is in a state of continuous motion under the influence of various atmospheric phenomena and of the various magnetic interactions between the earth and the moon. This mass of water either has a dynamic motion as in the case of a current or a potential energy as in the case of the tidal effect or both potential and dynamic energy as in the case of wave motions. Only 1/100 of the potential power resulting from the motions of these gigantic masses of water in the oceans is more than 5 times the global energy demand. What we aim is to subject the energy of this gigantic mass of water in a state of continuous motion though in different forms, particularly the energy of the same in the form of waves, to the conversion in a way to enable its utilization in our life. According to the studies, the data on the wave power acting on the unit width are as follows: • North Atlantic 100 KW / mt • England – Scotland 70 KW / mt • South Africa 10-14 KW / mt • Portugal coasts 5-26 KW / mt • Canada 0.6-101 KW / mt • China 0.7-4.5 KW / mt • Average for the Mediterranean coasts outside Turkey 8.4 – 15.5 KW / mt • Average for the Mediterranean coasts 13 KW / mt • Average for Çeşme coasts 14.84 KW / mt • Kalkan 6.6 – 7.6 KW / mt (Wave height is 1.21 m – wave period is 6.09 s; these data are valid for Kalkan through most of the year.) • Average for the coasts outside the oceans 10 - 40 KW / mt Average global wave power 36 KW / m, exceptionally 700 KW / m Note: These values may exhibit partial deviations; it is necessary to conduct reliable measurements and observations for reliable results. The countries that utilize this energy : • South Korea 252 MW % 36 • France 240 MW % 34 • United Kingdom (England-Ireland-Scotland) 110 MW % 15 • India 30 MW % 4 • Canada 20 MW % 3 • USA 10 MW % 1.5 • Australia 5 MW % 0.5 As for Turkey, even if we assume that we may not utilize the wave energy of the coasts of the inland sea of Marmara and of the coastlines occupied by settlements and military zones, etc. and that we use only 1/5 of our coastlines for this purpose, we could still utilize the wave energy from 1650 km of coastline for the conversion (the overall coastline length of Turkey is 8210 km). Within such a coastline extent, it could be possible to obtain 18.5 billion KWH of energy per year from the waters with a wave power of 4-7 kw/m, by means of just a small-scale converter in series. This value corresponds to 13% of our total energy need. The power in the sea waves is calculated to be 10-15 times greater than the other renewable energy sources. Assuming that an average energy of 100 Watt/m² is obtained in connection with the solar energy; For 1 KW energy →→ 10 m² of solar energy field is required For 1 KW energy →→ 2 m² of wind energy field is required For 1 KW energy →→ 1 m² of wave power energy field is required With the system we envisage, we intend to absorb, by means of a flexible surface (energy carpet), the potential energy of the waves likely to impact from any direction and generate the electricity by way of mechanical rotation. It is our aim to absorb on the flexible surface almost all of the dynamic forces contained by the motion in the waves and the static forces caused by the wave height. Our energy carpet (flexible surface) is a three-dimensional material (having a width, thickness and length), which is able to stay on the water surface or even above the water surface by some extent owing to a plurality of floating buoys connected thereto and which has a certain weight. By its nature, this material also features the ability to bend along the axes x, y and z under the influence of the action of the wave. This being the case, the wave acting upon the floating carpet from any direction sets the energy carpet in motion along the axes x, y and z with an influence being greater at the point of initial impact and decreasing with the advance of the motion (as its energy will be reduced) and the wave thus releases its energy; the wave has even become devoid of energy as it leaves the energy carpet and the sea surface behind the energy carpet has attained a state free of chaps. All the energy carried by the wave is transferred to the energy carpet. The system is designed in such a way that even the smallest motion in the energy carpet can be converted into the cyclic motion, wherein said motions are converted into the electricity generation. With such an approach, we estimate that the partial leakages and the mechanical losses, even to a little extent, would occur and the system efficiency would be in the range of 0.8—0.9. (The efficiency is around 0.9 in the Pelton turbines.) Of course, the exact results may be obtained only from the measurements taken after the installation of the system. If we take the value of 7 KW/m as the average wave energy in Kalkan coastline, a 1 MW system could be installed on a 170 m coastal strip with an efficiency of 0.85. I believe this is a significant value worth considering. Note: The weight and the width of the energy carpet will be calculated in a way to allow for the complete absorption of the wave energy. 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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