STEADY UNIFORM FLOW 



Since by the Manning formula 



Q^h^ ^^213^112 



it is seen that for a triangular channel 



The function in the brackets is easily evaluated. 



V 1 + z^\ 



[104] 



With this function known, the determination of the depth of flow 

 becomes direct. Assume, for example, that Q = 2 c.f.s., n = 0.014, 

 S = 0.02, and z = 1.25. Substitution in equation (104), using the 

 tabulated value of the function in the brackets corresponding to z = 1 .25, 

 gives 



2 X 0.014 



2)8/: 



from which 



0.142 X 0.992 

 D = 0.55 foot 



0.199 



An equation similar to equation (104) may be obtained for other 

 simple geometrical shapes, but the quantity in the brackets will usually 

 be a function of more than one variable. Tables 102 to 104 give this 

 function for trapezoidal, circular, and round -bottomed sections, permit- 

 ting a quick solution for any of the variables listed. The table headings 

 are self-explanatory. 



Illustrative Example 



Find the depth of flow for a discharge of 160 c.f.s. in a trapezoidal channel 

 with 8 ft. bottom width, side slopes of 1 on 1, and a uniform bottom slope of 

 2 feet per thousand, if the coefficient of roughness may be taken as 0.017. 



Summarizing the given information, Q = 160, b = S, z = 1, S = 0.002, and 

 n = 0.017. Since b is known, the use of Table 102A is indicated. 



160 X 0.017 



68/351/2 256 X 0.0447 



Qn 



= 0.238 



Opposite this value in the column z = 1 of Table 102A, we find that D/b = 0.326. 

 Hence D, the depth of flow, is 8 X 0.326 = 2.61 feet. 



