*************** EXAMPLE PROBLEM 14*************** 



GIVEN : A semi-infinite breakwater is sited in 8 meters (26.2 feet) of 



water. The incident wave spectrum has a significant height of 2 meters 



(6.56 feet) and a period of maximum energy density of 8 seconds. The waves 

 approach generally perpendicular to the breakwater. 



FIND : The significant wave height and period of maximum energy density at a 

 point 500 meters (1640 feet) behind and 500 meters in the lee of the 

 breakwater for wave conditions characteristic of wide directional spreading 

 and for narrow directional spreading. 



SOLUTION : Calculate the deepwater wavelength, L , associated with the 

 period of maximum energy density, T 



L = 1.56 T^ = 1.56(64) 

 o p 



L = 99.84 m (327.6 ft) 

 o 



Therefore, d/L^ = 8/(99.84) = 0.0801 . Entering Table C-1 with d/L^ = 

 0.0801 yields d/L = 0.1233 . The wavelength at the breakwater tip is, 

 therefore, 



L = d/(0.1233) 



L = 8/(0.1233) = 64.9 m (212.9 ft) 



The 500-meter (1640-foot) distance, therefore, translates to 500/64.9 = 

 7.7 wavelengths. From Figure 7-61a, for S^^^ = 10 (wide directional 

 spreading) for a point 7.7 wavelengths behind the tip and 7.7 wavelengths 

 behind the breakwater, read the diffraction coefficient K' equals 0.32 and 

 the period ratio equals 0.86 . The significant wave height is, therefore. 



H = 0.32(2) = 0.6 m (2.1 ft) 

 and the transformed period of maximum energy density is 



T = 0.86(8) = 6.9 s 

 P 



From Figure 7-61, for ^xaax ~ ^^ (narrow spreading), read K' = 0.15 and 

 the period ratio = 0.75 . Therefore, 



and 



H = 0.15(2) = 0.3 m (1.0 ft) 



T = 0.75(8) = 6.0 s 

 P 



7-93 



