of a quartz sand bed when viscous effects are negligible: 



da = HT (g/50(X) D)0-5 (11) 



If d^ computed from H^ , Tj , and Djjj is iiuch larger than the maximum water 

 depth of interest, the requirement for strong bed agitation may be considered 

 satisfied. \ 



To assess whether flow is likely to be rough turbulent, another simple 

 computation can be performed . Incorporating the same approximation for C in 

 intermediate water depth (2Trd/L near unity) as is utilized in equation (11), 

 fundamental results reviewed in Hallermeier (in preparation, 1983) support 



HT >_ d {metric units} (12) 



as an approximate criterion for rough turbulent flow at a strongly agitated bed 

 of quartz sand. If equation (12) is true according to Hj , T^, and the 

 maximum water depth of interest, the requirement for rough turbulent near-bed 

 flow may be considered satisfied. 



The following example problems demonstrate the use of the present procedure 

 in calculating nearshore wave shoaling with energy dissipation due to a strongly 

 agitated sand bed. 



*************** EXAMPLE PROBLEM 1*************** 



GIVEN ; At the CERC Field Research Facility, significant wave height exceeding 

 3.5 meters was recorded during three 1981 storms by a Waverider buoy located 

 in an 18-meter mean water depth. Wave periods associated with these ex- 

 tremely high waves ranged from 9.3 to 14.0 seconds. Nearshore bathymetry at 

 this site is regularly surveyed to the 9-meter water depth contour, and the 

 wave characteristics at the seaward boundary to the survey region are of 

 interest. The shore-normal distance between the water depths is 1800 meters, 

 and the representative sand size for the intervening bottom is D = 0.12 

 millimeter. 



FIND : The wave height at dj = 9 meters corresponding to Hj = 3.5 meters at 

 d] = 18 meters for: (a) T^ = 9.3 seconds and (b) Tj = 14.0 seconds. 



SOLUTION: 



(a) For 



Tl 



= 9.3 



seconds and dj = 18 meters 



dl 



2Trdi 



= (2Tf)(18) = 0.1333 



Lo 



ST? 



9.81 (9.3)2 



Table C-1 in the SPM gives n^ = 0.7570, (H^/H^) = 0.9160 = Kgj and 

 di/Li = 0.1694, so that Li = 106.3 meters. With Hi = 3.5 meters and 

 p = 1026 kilograms per cubic meter for saltwater, equation (2) becomes 



10 



