PIPE FLOW 



265 



Writing this as 4 v a </> (v n ) = F (v a ) (where </> (v a ) is the right-hand 

 side of this equation) and giving v a the value . 



6-0 this makes (F v a ) = 7'32. 



Putting v a = 6-5 F (v n ) = - 5-18. 



*. = 7'5 F(O= + '26. 



Plotting these values of v a and F (v a ) we find that v a = 7*48 makes 

 F (v a ) = 0, and therefore satisfies the above equation (9). 



Substituting this value of v a in equations (6), (7) and (8) gives v b = 

 7-08, v c = 8'79, v d = 6'39, while from equation (1) we have 



2 



j. 10 6 = 4-78 feet of water. 





AET. 77. MULTIPLE SUPPLY. 



Where more than one reservoir or source of supply feeds into one 

 common pipe, the surfaces of the 

 water in the supply reservoirs 

 being at different levels, it be- 

 comes necessary to determine 

 what share of the total flow each 

 of these sources of supply con- 

 tributes. E.y., in the case 

 illustrated in Fig. 115 the two 

 reservoirs A and B, with surface 

 levels Z A and Z B above datum, 

 feed through pipes of areas A a 

 and A b , into a common pipe of 

 area A e , at J, the joint flow 

 passing into a reservoir at C, 

 with its surface at a height Z 6 above datum level. 



Here we have 



' Hj representing the 

 loss from A to J. 



FIG. 115. 





+**'* 



(1) 

 (2) 



Zc ~h jH-c Z j -f- JT^ ~h 2"" (3) 



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 . (4) 



Then since A Hj, B Hj, jH c are respectively proportional to v a 2 , v b *, v?, 



and may be determined in terms of these velocities when the construe- 



