K is (H t ./H i ) 2 where (Hj./^) can be calculated from equation (12). Kj. 

 is more difficult to evaluate. Seelig (1979) provides a computer program to 

 calculate the combined transmission coefficient. 



Wave heights in the deposition basin vary with the tidal stage as the weir 

 crest submerges and emerges from the water. Maximum wave transmission usually 

 occurs at high tide. Ebb and flood tidal currents flowing across the weir 

 also influence the level of wave action in the deposition area. During flood- 

 flows, the waves are generally lower and longer; during ebb flows, the waves 

 steepen, becoming higher and shorter for the same incident wave conditions. 



*************** EXAMPLE PROBLEM 3 * 



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



GIVEN : A sinusoidally varying tide with an amplitude of 5.0 feet at a 

 vertical sheet-pile weir. The water depth below MTL at the weir is 7.5 

 feet. The weir crest is 6 feet above the bottom (1.5 feet below MTL). 

 The wave height and period are H b = 6.0 feet and T = 8.0 seconds. 



FIND: The wave height variation in the deposition basin over a tidal cycle 

 assuming waves approach the weir perpendicularly. 



SOLUTION : The time history of water level at the weir is shown in Figure 

 Hh For an impermeable sheet-pile weir, equation (11) with a = 1.8 and B 

 =0.1 can be used. The transmission coefficient is given by 



-£• = 0.5 



-£■- 0.5 



H_. 



1 - sin 



1 - sin 



2a \ Hj_ / 



tt /6.0 - d_ V 



[ §■+ 0.1) 



3.6 V % J 



since h = 6.0 feet and 2a = 3.6. The solution is given in Table 5 and is 

 presented graphically in Figure 20. Maximum wave transmission occurs at 

 high tide (d = 10.0 feet) with H t = 4.42 feet. Table 5 is calculated by 

 first determining the incident wave height. Since the water depth at the 

 weir is only 5 feet, the 6-foot-high incident wave will break seaward of 

 the weir. The maximum wave height that can occur at the weir is given 

 approximately by the condition that 1^ < 0.78 d b> Therefore, R ± = 0.78 

 (5.0) = 3.90 feet. Substituting into equation (11) 



-^ = 0.5 

 H, 



1 - sin 



it /6.0 - 5.0 \" 



( + 0.1) 



3.6 V 3.90 ) 



-^ = 0.5 [1 - sin(0.3110)] = 0.347 

 H, 



Thus, Hj. = 0.347 ^ = 0.347(3.90) = 1.35 feet. Note that equation (11) 

 is valid even though the water level is 1 foot below the weir crest. The 



equation is equally valid for conditions when 

 (See Seelig, 1976.) 



is greater than h. 



37 



