224 HYDRAULICS AND ITS APPLICATIONS 



As an example, consider the uniformly retarded flow through a pipe of 

 uniform cross-sectional area of and length 1. Let -y- = a, and let the 



suffixes v and i refer to the pipe immediately behind the valve and to 

 the inlet at the top respectively. Let x be measured from the inlet 

 (Fig. 102), and let v a be the velocity in the pipe line. 

 Then at the inlet, where x = 0, we have p = pi, z z\, 



(x 2 



Q iP.dx = 0, so that c = ^ + |^ + Zl from (3) 



When x = I, i.e., behind the valve, we have p = p v , z = z vt 



( l 2 A 27 P dv . ( l , 



v*dx = v a *l, I - d x = a \ d x = al ; 

 Jo JQ at Jo 



But ^ -j- ^ -f 2" -- ^ is the head equivalent of the statical pressure p s 

 at the valve with no flow through the pipe, so that we get 



water, (5) 



w g w 



or p v = j) s + \ a I l -~r ( I + ) [ Ibs. per square foot, 



f/ I Z \ m / ) 



the result previously obtained from general considerations. Obviously 

 this expression has its maximum value when v a = 0, i.e., at the instant 

 the valve reaches its seat. 



In order to get uniform retardation of a column by closing a valve at its 

 lower end, the rate of closure of this valve would, however, need to be some- 

 what complicated. If a is the pipe area, and a the effective valve area at any 

 instant (the effective valve area is the actual area multiplied by the coefficient 

 of discharge), and if v is the corresponding velocity of efflux, v a being 



the corresponding velocity of pipe flow, we have v a = - , so that 



If i is the effective valve area when the valve begins to close, and if vi 

 is the corresponding velocity of pipe flow, the value of a Q after t seconds 

 is given by 



+ * 



= a 



See " Water Hammer in Hydraulic Pipe Lines," p. 8. Gibson, Constable <Sr Co., 1908. 



