34 



HYDRAULIC JUMP IN RECTANGULAR CHANNELS 



increased. Sloping the apron has the effect of giving satisfactory opera- 

 tion over an elongated area, in place of the narrow line AB, and greatly 

 facilitates fitting the natural tailwater rating curve in certain cases. 



Laboratory research of a fundamental nature is needed to test means 

 of adapting the hydraulic jump energy dissipator for satisfactory opera- 

 tion over a wider range of conditions than is now considered possible. 



The hydraulic jump in a channel with sloping bottom. In the pre- 

 ceding section, reference was made to the expedient of sloping the apron 

 on which the jump forms in order to gain satisfactory operation over a 

 wider range of tailwater conditions. As long as the slope is flat, this 

 procedure is satisfactory, but when the slope is too steep, the high 

 velocity water shoots out under the pool with little localized impact, 

 making dangerous erosion likely. Laboratory experiments seem to in- 

 dicate that the limit between the two types of action occurs at a slope of 

 5 or 6 horizontal to 1 vertical. 



840 

 830 



c 

 o 



1 820 



a> 



UJ 



810 



800 



10,000 20,000 



Discharge in c.f.s. 



30,000 



Fig. 305. Typical Rating Curve 

 for Depths Downstream from a 

 Hydraulic Jump. 



Fig. 306. Element of a 

 Stream on a Steep Slope. 



The mathematical analysis which gives results agreeing so precisely 

 with observed data, in the case of the jump on a horizontal or gently 

 sloping floor, leads to no such useful result for the jump on a steeply 

 sloping floor. The results of an analysis are instructive, however. 



Consider first the pressure against the slope, for a stream of vertical 

 depth D and thickness z perpendicular to the slope. The weight of an 

 elementary volume of unit width and length dl is wzdl. The depth is 

 assumed not to vary appreciably within the length dl, so that the forces 

 on the ends and sides of the elementary volume are equal in magnitude 

 and opposite in direction. The unit pressure against the plane is 

 therefore 



wz cos a = wD cos^ a 



The factor cos^ a is of course nearly equal to unity unless the slope is 

 great. The difference is less than 1 per cent until a is nearly 6 degrees, 

 a slope of about 1 in 10. 



