From equation (3-49) 



F = F + F = 3800 + 153 = 3953 m (13,000 ft) 

 e a 



For d = 6.25 meters, U^ = 31.31 meters per second , and F = 3953 

 meters (from Figs. 3-21 and 3-22) 



H^ = 0.92 m and T = 2.84 sec 



AH = 0.92 - 0.9 = 0.02 meter (0.7 ft) 



thus 



AH < 0.5 H. (3-44) 



V 



This satisfies the third basic requirement (eq. 3-44), and the solution 

 may proceed to the next segment which is the remaining 1525 meters of the 

 area, with the water depth varying from 5.5 to 4 meters. 



0.25 d . = 0.25 (5.5) = 1.38 m 



1* 



Since Ad= 5.5-4= 1.5 meters > 0.25 d . , which does not satisfy equa- 

 tion (3-42) a shorter segment is required. For a 1000-meter segment, 

 assuming a uniform depth variation, the depth will vary from 5.5 to 4.5 

 meters. This satisfies equation (3-42), and the solution can then 

 proceed as above for a 1000-meter segment and then for a 525-meter 

 segment . 



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

 *************** EXAMPLE PROBLEM 7*************** 



GIV EN ; A coastal area is flooded by a storm surge so that the water 

 dep^th over the area is 3 meters. The actual distance across the area in 

 the direction of wave travel is 1000 meters. The area is covered with 

 thick stands of tall grass and a small to moderate amount of brush or 

 low, bushy trees in an even distribution. The windstress factor is 40 

 meters per second, and the initial wave height at the seaward edge of the 

 area is 2 meters; the wave period is 4.7 seconds. 



FIND ; The wave height and period at the landward end of the area. 



SOLUTION : Because of the constant depth and uniform friction effects, the 

 first two fetch segment conditions are met. The third condition is 

 tested after the wave height is determined. From the long dashline in 

 Figure 3-21, for the windspeed of 40 meters per second and the water 

 depth of 3 meters 



gd ^ 1,8x3^ 0.0184 

 U^ (40)^ 



giving (at the intersection of the above line with the long dashline) 



^ = 0.0075 



"a 



3-74 



