When the wave crests are not parallel to the bottom contours, some parts of 

 the wave will be traveling at different speeds and the wave will be refracted; 

 equation (2-43) does not apply (see Sec. III). 



The following problem illustrates some basic principles of wave energy and 

 energy flux : 



*************** EXAMPLE PROBLEM 6*************** 



GIVEN ; A deepwater oscillatory wave with a wavelength L = 156 meters (512 

 feet), a height H^ = 2 meters (6.56 feet), and a celerity Cq = 15.6 meters 

 per second, moving shoreward with its crest parallel to the depth contours. 

 Any effects due to reflection from the beach are negligible. 



FIND: 



(a) Derive a relationship between the wave height in any depth of water and 

 the wave height in deep water, assuming that wave energy flux per unit 

 crest width is conserved as a wave moves from deep water into shoaling 

 water. 



(b) Calculate the wave height for the given wave when the depth is 3 meters 

 (9.84 feet). 



(c) Determine the rate at which energy per unit crest width is transported 

 toward the shoreline and the total energy per unit width delivered to 

 the shore in 1 hour by the given waves. 



SOLUTION: 



(a) Since the wave crests are parallel to the bottom contours, refraction 

 does not occur, therefore H^ = H^ (see Sec. III). 



From equation (2-43), 



- E C = nEC 

 2 o o 



The expressions for E and E are 



and 



E = 



o 8 



i = fl^ 



where H' represents the wave height in deep water if the wave is not 

 refracted. 



Substituting into the above equation gives 



1 ^ PgH;2 



I ^o "8 "^ 8 



2-27 



