﻿Problem of the Weir. 115 



By the continuity of flow, 



udz = u'dz f =zd\jr. 

 Also, since the vo'rticity is constant along a stream-line, 



du/dz = dv! jdz' . 

 Hence, du 2 /dyfr = du' 2 ldy\r, 



QT» 1#2 It 2 



Taking the surface stream-line we find 



u' 2 =<+U' 2 -U 2 , (3) 



where u and v! are the velocities above and below the weir 

 on the same stream-line. 



Substituting this value of v! in udz — u'dz' we have 



dz' = udz/(ic* + U h2 -XJ 2 )i (4) 



Put 



W 2 = U 2 <K0, wnere ?= ( z ~ d )IK and let t,' = z'lh f , . (5) 

 From (4) 



When ^ = d; «'=0, i. «., when f=0, ?' = 0. 

 When z = d + A, jBr , = 7t / ; 2. e. } when f = 1, f = l- 

 Hence, Pf <M?) "]*,«. 



Since, by (2), »' 2 — 1 = k(S— tc), we have, finally, 



Since zt and id, for physical reasons, have the same sign, it 

 follows that the sign of v / (w 2 + U' 2 — U 2 ) will be the same as 

 that of u. Hence, the positive sign must be attached to the 

 integrand in (6). 



When u is given as a function of z, <£(f) is determined 

 from (5). Equation (6) will then determine w as a function 

 of a and 3, and thence y is given by (2). U / and h' are 

 found from (1) and u is determined from (3), viz., 



«"-=u 2 {<Kr) + v 5 -i}. 



The circumstances of the motion along the lower bed are 

 thus completely found. 



