(b) With Hq = 3.13 meters and z = -7.5 meters (-24.6 feet), evaluate the 

 exponent of e for use in equation (2-24), noting that L = L , 



2ttz _ 2it(-7.5) 

 L 

 thus, 



156 



-0.302 



Therefore, 



e-0.302 = 0.739 



o 2^2/ 3,13 

 A = B= — e L= — — (0.739) = 1.16 m (3.79 ft) 



The maximum displacement or diameter of the orbit circle would be 2(1.16) = 

 2.32 meters (7.61 feet). 



L -156 

 (c) z - - -2. - — Y- " ~ ^S'O ™ (255.9 ft) 



Therefore, 



^ . ^■<-^«' . -3.142 

 L 156 



e-3.m2 = 0.043 



and, 



A = B = -| e ^^^^ = ^j^ (0.043) = 0.067 m (0.221 ft) 



Thus, the maximum displacement of the particle is 0.067 meters which is 

 small when compared with the deepwater height, H = 3.13 meters (10.45 

 feet). 



*************************************** 



f. Subsurface Pressure . Subsurface pressure under a wave is the summa- 

 tion of two contributing components, dynamic and static pressures, and is 

 given by 



cosh[2Tr(2 + d)/L] H / 2Trx 2Trt \ 



P' = Pg .,. .,^. cos — — - pgz + p (2-26) 



cosh(2Trd/L) 2 \ L T / a 



where p' is the total or absolute pressure, p the atmospheric pressure 

 and p = w/g the mass density of water (for salt water, p = 1025 kilograms 

 per cubic meter (2.0 slugs per cubic foot); for fresh water, p = 1000 

 kilograms per cubic meter (1.94 slugs per cubic foot). The first term of 

 equation (2-26) represents a dynamic component due to acceleration, while the 

 second term is the static component of pressure. For convenience, the 

 pressure is usually taken as the gage pressure defined as 



cosh [2Tr(z + d)/L] H _.... _.. 



P = P - P = pg . ,- .,,. -r cos — ;;;- - pgZ (2-27) 



a cosh (2TTd/L) 2 ^ "^ ' 



2-21 



