Obviously, if n is large, t will be too great to warrant consideration 

 of the flushing problem, when the region near the head of the bay 

 is considered alone. We will, therefore, consider the average situation 

 in the bay as a whole. 



Let S be the sunn of all the exchange ratios. Then 



nrav=S = r + r2+ +r". 



Multiplying by r gives, 



rS=r2 + r3+ +r<" + ^> 



from which S(l-r) =r — r(n + i)or S = r/(l-r), since r^" + ^^ ,is very small. 

 Since n segments were involved in the sum S the average exchange 

 ratio for the bay is 



r - J_ - _P 

 where P is the tidal prism volume and V, the high tide volume. 



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