130 BENDS, TRANSITIONS, AND OBSTRUCTIONS 



Yarnell computed Nagler, D'Aubuisson, and Rehbock coefficients for 

 his tests, which covered all three classes of flow. The Nagler coefficient 

 showed the least dispersion in Classes 1 and 2, and the D'Aubuisson 

 coefficient the least dispersion in Class 3. Some variation with velocity 

 was observed, and since the highest velocities reached were only about 

 6 feet per second, the need for further tests, especially field measure- 

 ments, is indicated. It should be noted that Yarnell 's tests did not 

 include percentages of contraction of less than 11.7 per cent. For 

 smaller percentages the D'Aubuisson formula should be used. 



In computing the backwater due to a bridge or trestle, the formulas 

 must be applied to the average properties of the whole cross section, and 

 not to individual sections, no matter how much they differ, for the height 

 of the backwater is the same at all parts of the cross section. 



Illustrative Examples 



1. A stream discharging 50,000 cubic feet per second, at flood stage, flows 

 under a bridge having 5 rectangular piers 12 by 50 feet which reduce the normal 

 width of waterway from 400 feet to 340 feet. If the average normal depth of 

 flow in the uncontracted section is 20 feet, what will be the height of the back- 

 water caused by the bridge? 



First summarize the known factors, and compute those needed to classify 

 the flow. 



Q = 50,000, Wi = 400, W2 = 340, a = ^ = 0.15, Dz = 20, 



VI 



^ J^^ ^ VI ^ ^^^ , = 2L = ^^ = 0.0302. 



20 X 400 2g Ds 20 



From Fig. 1004 the flow is seen to be in Class 1, so that the Nagler formula 

 should be used. 



Q = KW2VYglD,-^-^ yJHz + ^ [10111 



The value of K, from Fig 1007, is taken as 0.88, and /3, from Fig. 1006, as 1.5. 

 Using 6 = 0.3, substituting the known factors, and solving, 



Vi^ 

 Hz + 1.5 -i- = 1.02 

 2g 



The value of A is not known, so that Vi is not known. However, Vx^/2g cannot 

 be much less than Vz^/2g, so let Vx^/2g = 0.60 for a first approximation. Then 



iTs = 0.12 feet 



