which enters the estuary on the flooding tide is, except for the river 

 water contributed in a half tidal cycle, equal to the tidal prism volume 

 P. The average excursion E of the sea water near the seaward boundary 

 a of the estuary is, therefore, approximately equal to: 



E = ^ (24) 



where a = hd + ed (see Fig. 2). 



Now, for the James River estuary, a relation existed between the 

 maximum current velocity and the net salt transport, which will 

 be assumed to hold in general. This is to the effect that the net volume 

 transport velocity is approximately equal to one-fifth the maximum 

 current velocity. Multiplying (24) by'^' /2 gives the maximum excursion 



and multiplying (25) by one-fifth gives 



IT? 



Enet - loa • (26) 



The net volume of sea water entering the estuary is 



Uneta = -|u^3xa=|^Ua = ^« 2Qe, or (27) 



Therefore, 



/ 



and 



~ jlZ** 



, ^%_ ^ Qb (28) 



B 1/2 (A + B 



rA=R + QB- (29) 



*U|r = Umax^'" ^ .where Ovaries between o andTrduring the flooding tide 

 interval tp . The area under the velocity profile is /q"^ Umax sin ^ d ^= 

 _ Urnaxf- 1 - l] = 2Umax- I^ U is the average flood velocity, U7r=2Umax or 

 Umax = |U- Also U^a^tp = E^a^ , etc. 



** It is assumied that actual current velocities are not known. 



