%- 



7,000 



= 0.73 



(0.5)(1025.2)(9.8)(0.3)(3) (0.71) 



If Airy theory had been used (H = 0), Figure 7-72 with d/gT^ = 0.0046 

 would give Kp^ = 0.23 , and therefore 



Dm 



■^D^ Airy' 



^ D^H = 0.87 H^ 



H = 0.87 



% 



Dm 



0.71 



= 0.73 = 2.25 



0.235 



'Airy (H « 0) 



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



*************** EXAMPLE PROBLEM 25************** 



GIVEN : Same conditions as the preceding example, but with a wave height H = 

 15.0 m (49.2 ft) , a depth d = 30.0 m (98.4 ft) , and F = F/^ = 130,000 N 

 (29,225 lb) . 



FIND ; The appropriate value of Cp . 



SOLUTION ; From Figure 7-75 1^ = 20.6 m (68 ft) ; then H/H^ = 15.0/20.6 = 

 0.73 . Entering Figure 7-72 with d/gl^ = 0.031 , K^p;^ = 0.38 is found. 

 Therefore, from equation (7-33), 



CZ) = 



'Dm 



1/2 pgDH^Kj 



■Dm 



C = 



D 



130,000 



= 1.01 



0.5( 1025. 2)(9.8)(0.3)( 15.0) (0.38) 



If Airy theory had been used, Kn_, = 0.17 and 



(C ) = (C ) 

 O'Airy Z? H = 0.73 H 



H = 0.73 H 



K 



Dm] 



(0.38) 



(1.01) = 2.26 



(0.17) 



'Airy (H « 0) 



Some of the difference between the two values of Cp exists because the SWL 

 (instead of the wave crest) was the upper limit of the integration performed 

 to obtain Kp^ for Airy theory. The remaining difference occurs because 

 Airy theory is unable to describe accurately the water-particle velocities 

 of finite-amplitude waves. 



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



7-142 



