HYDRAULIC JUMP 25 



condition of the water reduces its specific gravity and accordingly the 

 surface at the top of the rise is above the normal level, but as soon as 

 all the air bubbles reach the surface, so that the water is again trans- 

 parent, the surface becomes comparatively smooth and level. 



The moving mass of water loses much momentum in passing from the 

 position ahfe to the position cdhg. According to Newton's second law 

 of motion, the rate of loss of momentum must be equal to the unbalanced 

 force acting on the moving mass to retard its motion. 



Against the face ah is the static pressure of the water acting towards 

 the right. Opposed to this force are the static pressure acting against 

 the face ef and the surface friction along the bottom hf. The latter is 

 small and may be neglected. 



Let Di = depth of stream entering hydraulic jump, 

 Vi = velocity of same stream, 

 D2 = depth of stream leaving hydraulic jump, 

 V2 = velocity of same stream, 

 Dc = depth of the stream at critical flow, 

 Vc = velocity at critical flow, Vj^ = gDc, 

 Q = quantity in second-feet of flow per unit width of stream, 



Q = v,D, = V2D2 = VcDc = VgD? 



wQ 

 Mass of water flowing per second = — , 



g 



Change of velocity = Vi — V2, 



wQ 

 Change of momentum per second = — (Fi — V2), 



wQ 





2 



Dl 

 2 



Static pressure acting on face ah = 



Static pressure acting on face ef = 

 Therefore, 



gD2 2 



Dividing both members by w{D2 — Di) 



QVx Di-^D2 



[301] 

 gD2 2 



