264 HYDEAULICS AND ITS APPLICATIONS 



Also if all the pipes run full we have for continuity of flow 



A a v a = A b v b + A c v c + A d v d . (5) 



Since, in these five equations, the only unknowns are the four velocities 

 and the pressure at J (the losses A Hj, jII B , etc., being determinate in 

 terms of the velocities), the equations are perfectly determinate and a 

 solution will give the velocities in the various pipes in terms of known 

 quantities. The quantities discharged through each pipe may then be 

 determined. 



EXAMPLE. 



A reservoir A supplies three supplementary reservoirs B, C, D through 

 a single 24-inch pipe divided at J into two 12-inch and one 18-inch pipe 

 leading respectively to reservoirs B, C, and D. If the lengths of these 

 pipes are l a = 500 feet, l b = 1,000 feet, l c = 1,500 feet, l d = 3,000 feet ; 

 and if Z A = 50 feet, Z B = 80 feet, Z G = 10 feet, Z = 20 feet, Z 3 = 

 40 feet, determine the velocities of flow in each pipe. Take / = *005, 

 and neglect all except friction losses. 



Here A Hj in equation (1) above becomes 



005 X 500 X 4 vl _ 5 v a * 

 2 2 g ~~- 2 g ' 



Determining all such values, and substituting, equations (1) to (5) 

 above become 



W+ 62 = 10 (D 



4 v a = r b + r c + 2-25 v d . (5) 



Subtracting (2) from (1), (3) from (4), and (3) from (2) we eliminate 



f and get 



6r ft a +19r 6 a = 400 (6) 



29 t' c 2 - 39 vf = 20g (7) 



29 r c 2 - 19 v? = Wg (8) 



Determining v b , v v , and r d in terms of v a from equations (6), (7), and (8), 



and substituting these values in (5), this becomes, on writing g = 82'2, 

 and simplifying 



4 v a = V 67-75 - -316 v? + V 88'8 - '207 v a 2 + 



2-25 V 49-5 -154 v*. (9) 



