Ill 



(6) In increasing the kinetic energy of this mass because of the 

 increase (dv/dx)dx, in its velocity in the distance dx. The kinetic 

 energy of the moving mass is mi'^/2. The force required to increase 

 this energy is: 



b(m,v-/2) lbx=mvbvlc)x 



and the work done by this force over the distance dx is: 



7nv{c>vldx)dx. 



(c) In the acceleration of the mass of water in the section, with 

 respect to time. The work so done is: 



m{dv/dt)dx. 



(d) In overcoming frictional resistance in the section. The fric- 

 tional resistance in a channel is due to the turbulence which the flow 

 produces and is dependent upon the velocity of the current. The 

 turbulence created at any instant by the slowly varying velocity in a 

 tidal channel cannot differ sensibly from that wliich would be produced 

 by the same constant velocity. The work done in overcoming fric- 

 tional resistance in the section of length dx may then be taken as that 

 developed from the usually accepted formula for steady flow. This 

 work is mg{v^lC^r)dx. To become applicable to the reversing flow in 

 tidal channels, the algebraic sign of this expression must be considered, 

 since the work is positive when the flow is in the positive direction of 

 X, and negative when the flow is in the opposite direction. Since y- 

 does not change its sign in passing through zero, and the other quanti- 

 ties are not directional, this item of work will be written : 



±mg(v^/C''^r)dx. 



The positive sign is to be applied when v is positive, and the negative 

 sign when v is negative. 



220. Each of the items of work developed in the preceding para- 

 graph may be either positive or negative. The work done in raising 

 the mass of the discharge (item a) is positive if by/dx is positive, and 

 negative if negative. Item (6) is positive if the kinetic energy is 

 increasing in the positive direction of x, and negative if decreasing, 

 while item (c) changes its sign with br/dt, and item (d) with v. 



Since no external work is done by the flow in the channel, the sum 

 of all of the items must be zero, giving: 



mg{dyldx)dx-\-mv{dv/dx)dx-i-'m{dvldt)dx±mg(:v-IC^r)dx=0 



Dividing by mgdx, tliis equation reduces to : 



by/dx-\- (v/g) dvlbx+ (l/g) ()v/dt±vyC'r=0. (112) 



Thus is the basic equation of motion in a tidal channel. 



221. Discussion. — The first term, c>y/dx, in equation (112) is the 

 slope of the water surface in the channel at the given cross section and 



