6 ME. G. I. TAYLOR ON TIDAL FRICTION IN THE IRISH SEA. 



Let ds be an element of length of the curve, s, 



v = velocity of current at any point on s, 

 6 = angle between element ds and direction of current, 

 D = depth of bottom below mean sea-level. 

 h = height of tide above mean sea-level, 

 p = density of sea-water, 

 g = acceleration due to gravity. 



First consider the rate at which energy is communicated to the portion of the sea 

 which, at time, t, was enclosed by the surface. 8. Let 8' be the moving surface 

 which encloses this water. 



Fig. 1. Diagram showing successive positions of a surface, S, formed by the vertical lines through 

 the curve, s as that surface moves with the current. 



The mean hydrostatic pressure on a vertical strip of 8. of height !) + /(' and 

 width ds is pff$(D+h), Its area is (D + h)ds. The work done by hydrostatic 

 pressure on the portion of the sea originally enclosed in the surface is therefore 



-}(v8iu6dt)(V+h)da (I I) 



This then is the amount of energy which has flowed during the time, dt, through 

 the surface, .8', which originally coincided with the fixed surface. 8, but which moves 

 with the fluid. 



To find the amount of energy which has crossed the original surface. 8, during the 

 time, dt, it is necessary to take account of the energy contained in the fluid which 

 has actually crossed the fixed surface, 8. The gravitational potential energy of a 

 vertical column of fluid, of height D + h and horizontal cross-section v sin 6 dt ds, is 



evidently pg (D + // 



A-D 



r sin 6 dt ds, mean sea-level being regarded as the surface 



of zero potential. The kinetic energy of the same column is 



%pv* (D + h) v sin Q dt ds. 



