FLOW IN PRECIPITOUS CHANNELS 



73 



cipitous channels except when the distance is so short that the roll waves 

 do not have time to develop. 



Illustrative Example 



A river has a uniform slope of So = 0.0004 and is at a flood stage of y„ = 10 

 feet. Determine the backwater curve produced by a dam at which y = 25 feet. 

 Use n = 0.022. 



From inspection of the given data, it seems most likely that the curve will 

 be case Ml, To verify this, compute the value of yc. 



1.486 



V = 



0.022 



X 102/3 X 0.00041/2 = 6.3 ft./sec. 



Q per foot = 63 c.f.s. 



yc 



=%/!=^ 



63 X 63 

 32.2 



= 5.0 ft. 



Since jc < >'n < y, the curve is an Ml curve, and is given by equation (608) 



X = — z — y 

 5o 



*G. J 



The value of C corresponding to the depth of 10 feet is 



1.486 



C = 



0.022 



X 10i/« = 99 



by Manning's formula. Similarly, C is 115 at the depth of 25 feet. Since the 

 greater part of the curve lies in the range of the shallower depths, and since, 

 furthermore, the flow is expanding and thus is subject to greater loss than uni- 

 form flow at the same depth, let us use C = 105 in applying Bresse's formula. 

 Substituting the given values, with distances in miles, equation (608) becomes 



X = 4.74z - 4.09$[2] 

 The curve can be plotted from data in the following table: 



