SOLUTION: Calculate, 



H 10 



= 0.0031, 



gT2 (32.2) (10) 

 d 15 



gT^ (32.2) (10)^ 



-= 0.0047, 



and the average Keulegan-Carpenter number ^rnax ^/^ using the maximum 

 horizontal velocity at the SWL and at the bottom to obtain Umax • 

 Therefore, from Equation 7-16 with z = -d , 



H gT 1 



'^^} bottom 2 L^ coshp'fd^ 1 ' 



10 (32.2) (1 0) 

 bottom 2 215 



i^^max] ... = '^ '~^ (0-89) = 6.66ft./sec.. 



where L^ is found from Figure 7-40 by entering with d/gT^ and read- 

 ing L^/L^ = 2TTL^/gT2 = 0.42. Also, 1/cosh [2Trd/L] is the K value 

 on Figure 7-40. Then, from Equation 7-16 with z = , 



H gT 



'SWL 2 L 



\ "^ISWL 2 215 



The average velocity is therefore. 



A 



10 (32.2) (10) ^ ^ , 



= 7.49ft./sec. 



l"""^) bottom "^ i'''^'^)sWL 

 2 



6.66 + 7.49 14.15 



= 7.08 ft./sec. 



max 2 2 



and the average Keulegan-Carpenter number is 



^max^ 7.08(10) 



= ^ — = 70.8. 



D 1 



The computed value of u^ax ^/'-* ^^ well beyond the range of Figure 

 7-56, and the lift coefficient should be taken to be equal to the 

 drag coefficient (.for a rigid struoture) . Therefore, 



C, = Cr> = 0.7. 



7-98 



