304 HYDKAULICS AND ITS APPLICATIONS 



Considering the section of the complete channel to one side of the 

 is 00', we have 



A a 4- / x d y 

 y- = J _ . = constant = m 

 p + s 



.'. a + / x d y = m (p -\- s). 



d s 

 Differentiating, this gives x = m -= 



d x \ a 



1/1 / d x \ 

 .-. x 2 = m? ] I +(-, ) 

 V d y / 



J 



d y _ d x 

 m " V x* m* 

 Integrating this we have 



= m cosh~ l --- \- D 



= m log e { x + x* - m* } + D'. 



But x = b when y = 



/. D =^ racosh- 1 



' = m log e { 



?/ = m ( cosh" x -- cosh" 1 

 \ 



(1) 



b + 



From equation (2) the curve of the side may be plotted by calculating 

 values of ?/, corresponding to a series of values of x. 



Since v = C V m i this velocity with any given gradient may be 

 adjusted to any given value by designing the small channel so as to give 

 the required value of m. The only restriction is that m cannot exceed 

 b -f- 2, this being its value when the lower channel is semicircular, or 

 rectangular with a breadth equal to twice its depth. 



EXAMPLE. 



To design a channel to give an uniform velocity of flow of 4 feet per 

 second, the half breadth b being 2'5 feet, and C having a value 90. 



Here, assuming a rectangular section for the lower channel, of depth 

 2'5 feet, we have m = T25 feet. 



.-. 4 = 90 V 1-25 i 

 .-. i = -00158. 



