This satisfies equation (2) and the solution may proceed. The average 

 depth, d = 16.5 feet, and the average friction factor, ff = 0.108. For 

 d = 16.5 feet and H^ = 3.17 feet (from Fig. 1). 



f e = 5,400 feet ; 



for d = 16.5 feet, H^ = 3.17 feet, if = 0.108, Ax = 3,000 feet, and T 

 = 3.31 seconds {from Fig. 14), 



££* 0.294 

 gT 2 



Kf.01 = °- 988 £or f f = °- 01 and f f H i Ax/d2 = °' 349 

 Y.f a •= 0.88 for f f = 0.108 and ff H^ Ax/d 2 = 3.77 . 



Using equation (4), a = 0.1 and 



F a = a Ax = 0.1 (3,000) = 300 feet 



F = F e + F a = 5,400 + 300 = 5,700 feet . 



For d = 16.5 feet (from Figs. 1 and 2), 



Iff = 3.27 ^feet and T = 3.41 seconds . 



The remaining 2,000 feet of the fetch can then be treated as a third 

 segment. The average depth, d = 14 feet, and the average friction 

 factor is f f = 0.13. 



For d = 14 feet and H^ = 3.27 feet (from Fig. 1), 

 F e = 7,200 feet ; 



for d = 14 feet, H-£ = 3.27 feet, £» = 0.13 (from Fig. 13) 



Ax = 2,000 feet, T = 3.41 seconds, and 2iTd/(gT 2 ) = 0.235 



K f.01 = °- 98 for f f = °' 01 and £ f H i Ax / d2 = 0-334 

 Kf a = 0.80 for ff = 0.13 and ff W^ Ax/d 2 = 4.34 . 

 Using equation (4), a = 0.1 and 



F a = a Ax = 0.1 (2,000) = 100 feet 



F = F g + F a = 7,200 + 200 = 7,400 feet . 



For d = 14 feet, U = 70 miles per hour, and F = 7,400 feet (from Figs. 

 1 and 2) 



31 



