PROBLEMS 



41 



general, it is worth while to select the method of solution carefully, and 

 if a large number of solutions has to be made, spend some time in plan- 

 ning a method that will minimize the work. 



Table 401 will facilitate the solution of problems involving the critical 

 depth and alternate depths in channels of circular cross section. 



Illustrative Examples 



Find the critical depth for a discharge of 75 c.f.s. flowing in a trapezoidal 

 channel which has a bottom width of 10 feet and side slopes of 1 vertical to 

 1 horizontal. 



The problem will be solved by the cut-and-try procedure. The successive 

 trials are arranged in columns. 



The critical depth is 1.16 feet, to the nearest hundredth, for with flow at 

 that depth, the ratio of the velocity head to the average depth is most nearly 

 one-half. 



Find the critical depth for a discharge of 25 c.f.s. in a circular conduit 4 feet 

 in diameter. 



Table 401 lists values of Qc/D^'"^ corresponding to values of d/D, where d is 

 the depth of flow and D the diameter of the section. 



^ = - = 0-'81 



By interpolation in the table, d/D = 0.369, so that the desired value of the 

 critical depth is 0.369 X 4 = 1.47 feet. 



PROBLEMS 



401. (a) What is the critical depth for a discharge of 400 c.f.s. in a trapezoidal 

 channel with bottom width of 20 feet and side slopes of 1 vertical to 2 horizontal? 



(6) What is the critical depth in the same channel, when the total head is 15 feet? 

 (c) What is the depth alternate to a depth of \ foot, if the discharge is 400 c.f.s.? 



402. (a) What is the critical depth for a discharge of 20 c.f.s. in a circular pipe 

 culvert 3 feet in diameter? 



(6) What is the depth alternate to a depth of 1 foot for the same discharge in 

 the same channel? 



