72 THE TIDAL PROBLEM. 



take g = 32 feet per second per second. The coefficient $ itself is dependent 

 upon the velocity. As a result of many experiments the following value is 

 assigned : 



f = 0.007409(1+^^) 



where v is the mean velocity. From this formula it will be observed that 

 ^ increases as the velocity decreases. 



This formula is applicable to a canal in which the cross-section is uni- 

 form throughout its length. In order to adapt it to a canal of variable 

 cross-section and velocity an integration is necessary. Consider an element 

 of the canal between parallel cross-sections at distances I and l + dl from the 

 upper end, and put 



r = velocity a = area of cross-section p = wetted perimeter 



We have then, from the above formula, 



dh _^ v^p 

 dl 2ga 

 or, since 2g = 64 and 



this may be written 



f = . 007409(1 + ^) 



dh .007409, , , ,„„,,? 



This expression represents the slope at the point I necessary to main- 

 tain the flow. The rate of fall of the water in this element of the canal is 

 obtained by multiplying the slope by the velocity, that is 



v—ir- = T8ite of fall of the water 

 dl 



The distance through which the water falls multiplied by its weight gives 

 the amount of work done expressed in foot-pounds. Consequently 



E = I iw 



Jl^ 



dl 



where E is the amount of work done per unit time and w is the weight of 

 water in the element considered. The volume of water is equal to adl and 

 the weight of a cubic foot of water is approximately 62.3 pounds. Therefore 



w;=62.3 adl 



These values substituted in the expression for E give 





„ 62.3 X. 007409 , , , , 1non.2^ j/ 



E= ^ / {v^ -{■ .l^2Qm^) 'pdl 



