GRAPHICAL SOLUTION OF HYDRAULIC JUMP EQUATIONS 55 



or as it may be written by the aid of equation (502) 



k(A' - ^i') = ^1(^1 - ^o)Fi - D2{V2 - Fo)F2 = 

 D,{V, - Fo)(Fi - V2) = D2iV2 - Fo)(Fi - V2) [503] 



The pair of simultaneous equations (502) and (503) represent the 

 two laws, namely, the law of continuity and the law of change of mo- 

 mentum and contain the whole theory for the moving hydraulic jump 

 for the case being considered. These equations contain five variables, 

 any three of which may be considered the independent variables. If 

 values for three variables are given or known, the values of the other 

 two will be fixed. 



From equations (502) and (503) any one of the five variables may be 

 eliminated, giving five different equations each containing four variables. 

 Each of these equations may be solved for any one of the variables, 

 giving a total of twenty different forms in which these equations might 

 conceivably be written. To write all these forms requires in some 

 cases the solution of a quadratic, and in others the solution of a cubic 

 equation. Furthermore, various other variables may be introduced 

 into these equations, such as the velocity heads Vi^/2g, Y^ llg, the 

 total heads, Dx + (Fi^/2g), £>2 + (F2^/2g), the height of the jump 

 Di — Di, the difference between the total heads, or the loss of energy 

 in the jump, as well as ratios of these variables. Evidently a great 

 number of different equations can be thus formed, if desired or needed 

 for any purpose, by ordinary algebraic transformations. 



Graphical solution of hydraulic jump equations. It is a difficult 

 matter, in general, to construct a diagram from which functions of 

 three independent variables may be conveniently read. In this case by 

 taking advantage of the symmetrical nature of equations (502) and 

 (503) it is possible to construct such a diagram in a form convenient 

 for general use, as shown in Fig. 502. The process is as follows: Elimi- 

 nating D2 and solving for V2 — Vq gives 



Taking first when Vq = 0, we have 



V2 = 



\ + ^hD.+{^y [505] 



Similarly, eliminating Di and solving for V2 — Vq gives 



feg^d^. - ^0) [506] 



