*************** EXAMPLE PROBLEM 2*************** 



GIVEN : A mathematical model is to be used to simulate storm wave effects for 

 water depths shoreward of 9 meters at Nags Head, North Carolina, with the 

 threshold for storm waves taken to be the wave height exceeded 10 percent of 

 the time. The wave climate at this site has been defined by several rela- 

 tively complete years of data from a pier-mounted gage located in mean water 

 depth of 5.2 meters (Thompson, 1977): wave height exceeded in 10 percent of 

 these measurements is about 1.7 meters and the typical wave period for this 

 wave height is about 8.5 seconds. The shore-normal distance between 5.2- and 

 9-raeter water depth is 600 meters; representative sand size for the inter- 

 vening bottom is D = 0.20 millimeter. 



FIND : The wave height at dj = 9 meters corresponding to Hj = 1.7 meters and 

 Ti = 8.5 seconds at di = 5.2 meters. 



SOLUTION : For Ti = 8.5 seconds and d^ = 5.2 meters, 



d, 2Trdi (2Tr)(5.2) 



-*- = 5- = — -r = 0.04610 



Lo gTf (9.81) (8.5)2 



Table C-1 in SPM gives n^ = 0.9074, (Hj/H,^) = 1.038 = Kg^ and d^/Li = 

 0.09002, so that Li = 57.76 meters. With Hj = 1.7 meters, equation (2) is 



?! = I p g h2 c^ n^ = I (1026) (9. 81) (1.7) 2X57.^ (0.9074) 



= 2.24 • 10^ kilogram-meter per second cubed 



Since d = /d,d. = /(5.2)(9) =6.84 meters 

 m 1 J 



•^m _ (2Tr)(6.84) . _ ,. ,, 



L^ - (9.81)(8.5)2 = ^-^^^^"^ 



and Table C-1 gives (dm/Lm) = 0.1049, sinh (2TTdm/Lm) = 0.7082, (Hm/Hj^) = 

 0.9916 = Ksm, and n^, = 0.8799. Thus, R^ = (H-^ Kg^/Kg^) = (1.7) (0.9916)/ 

 1.038 = 1.624 meters and according to equation (9), 



^ Hm ^ 1.624 ^1-15 meters 



^ /2Trdm\ 2(0.7082) 



With Dm = 0.20 millimeter, equation (8) is 



fem = exp {-5.882 + 14.57 (Dm/Cm) ° " ^^'*> 



= exp {-5.882 + 14.57 [2 • 10-'*/1.15] • l^**} = 0.0422 



