SOLUTION : 



a. Assume k^^ = 0.1, k^^ = 1.0, and f = 0.03. With B = 600 feet 

 and d = 12 feet. 



Ag = BL = 600(12) = 7,200 square feet (669 square meters) 



^n 7 2nn 



^ = ^ - />^uu ^ 11.54 feet (3.51 meters) 



(B + 2d) (600 + 2(12)) 



F = k^^ . k,, + fk = 1.0 . 0.1 . 0-0^^3'600) ^ 

 en ex 4^ 4(11.54) 



^ ^sA&F ^ (4.4/2)(2)(108) 3.43 _ 

 1 2LA^ 2(3,600) (7,200) ~ '^^ ' ^ 



and 



K = ll /-^ = 2^ / 3, 600(2) iP" 

 2 T \ gA^ 12.4(60)(60) ^32.2(7,200) '^•^^• 



From Figures 2, 3, and 4 with the above values of K^ and K2 



V^ = 0.66 



Hi 



— = 0.78 

 and 



E = 53° 

 Therefore, from equation (1) 



m A^T 



0.66(2)(3.14)(2.2)(2) 10^ , ,„ ^ ,, .,, 

 V^ = 7 2QQf 12 41 r3 6001 ~ 5.68 feet (1.73 meters) per second. 



Since a^/a^ = 0.78, a^ = 0.78(2.2) = 1.72 feet (0.52 meter) and the 

 bay tidal range is 1,72(2) or 3.44 feet (1.05 meters). The tidal 

 prism- is 2a2,A2, = 2 (1 .72) (2) (10^) = 6.86 x lO^ cubic feet (6.37 x 10^ 

 cubic meters) and the bay tidal phase lag is 53° or (53/360) (12.4) ^ 

 1.83 hours. 



18 



