For section B (clear = 6 inches) 

 z = 4.5 feet 

 z 4.5 



= 0.0106 



424 



From tab 



2lTZ 



les, cosh — j— = 1 



0022 



T\ H ''°^"'- L -" TT(IO) (1.0022) ^ 1, . , , 



"If . ^,2ud, = (10) (1.481) =2-^^ ^^^^ P^^ ^^^""'^ 



smh (— j— ) ^ J ^ J 



component of u^^^j, perpendicular to the pipeline axis is 

 Umax Ccos 30°) = (2.13) (0.866) =1.84 feet per second 

 (a) Use Figure 57 to determine a value for k 



clear 





Dia 



u T 

 max 



u T 



max 



(0-5) (8) 



(1.84) (10) (1.84) (10) 



= 0.0118 



so from Figure 57, k = 0.67 



and from Figure 58, ()> = 45°. 



Alternatively, either (J) or k could be determined from Fig. 39, 

 once the other is known. 



From Figure 46, for k = 0.67, 



so Cl 



CL(l-k) = 2.75 

 2.75 



(1 - 0.67) 



= 8.3. 



(b) Maximum positive lift (per unit length) 

 F^d-k) = I Cl P A u^^/ (1-k) 



= \ (8.3)(2)(8)(1.84)2 (1 - 0.67) 



= 74.2 pounds per foot 



116 



