ILLUSTRATIVE EXAMPLE 77 



By using empirical equations to represent the variable factors making 

 upfiy), the integration may be accomplished. Mononobe gives tables 

 for complete evaluation of the integral.^ Unfortunately his method is 

 marred by inadmissible approximations. Bakhmeteff gives tables 

 which completely evaluate the integral only when the effect of velocity 

 head changes is neglected.^ Though his method and tables are 

 accurate, the correction which must be made to take the velocity head 

 changes into account destroys the convenience of the method except 

 when the effect of the velocity head changes is insignificant. 



Illustrative Example 



A trapezoidal canal with bottom width of 100 feet, side slopes of 1 on 1, and 

 a gradient So of 0.0004 delivers a discharge of 6220 c.f.s. to a large reservoir. 

 Determine the depth of flow at the entrance of the canal, 7.0 miles upstream 

 from the reservoir, if the depth of flow at its lower end is 25 feet. Use n = 0.022 

 throughout. 



The data are chosen to provide a comparison with the results of the illus- 

 trative example solved by Bresse's method. In that example the channel was 

 considered to have an infinitely wide rectangular cross section. The normal 

 depth of flow, 10 feet, is the same in each case, and the curve is an Ml curve, as 

 before. The computations for the graph oi f{y) are made in Table 701, and 

 the results are plotted in Fig. 701(a). The values of depth chosen for the com- 

 putations of Table 701 are selected so that the points for plotting the graph will 

 be more closely spaced where the curvature is small. Had points closer to the 

 normal depth been needed, smaller increments of depth would have been neces- 

 sary for the interval between 10 and 11 feet of depth. 



The depth at the upper end of the canal is found by determining a value of 

 y such that the area under the curve between y = y and ^^ = 25 is 5,280 X 7 = 

 36,960 feet. This was done by first planimetering the area from y = 25 to 

 y = 13, which was found to be 34,900 feet. Then the position of the line to 

 enclose the remaining area on the diagram was estimated, and after checking, 

 was found to be 3' = 12.55, which is, then, the required depth of flow 7 miles 

 upstream from the lower end of the channel. 



The depth at the same point, as determined by interpolating in the results 

 of the example solved by Bresse's method, is 12.87, a difi"erence of less than 

 3 per cent. The close correspondence is due to the fact that the cross section 

 is comparatively broad and shallow. 



When more than one depth is required it is advantageous to plot a mass 

 curve to give the area under /(y). The mass curve for the curve of Fig. 701(a) 

 is shown in Fig. 701(&), which shows directly that the depth is 12.55 feet 7 miles 

 upstream from where it is 25 feet. The answer to any other such problem 

 within the range of depths covered, and for the same discharge, may be read 



2 Ibid. 



' Hydraulics of Open Channels, Boris A. Bakhmeteff, McGraw-Hill, 1932. 



