ON THE LOSS OF ENERGY BY FRICTION OF THE TIDES. 73 



Consider now a section of the basin 1 foot wide, and by means of this 

 formula compute the loss of energy in this strip in one second of time with 

 such a tide as has been supposed. 



SECTION "a 



>f 



Let us begin with section " a. " We will suppose the surface of the water 

 is level throughout the section. A tide of 4yt feet in 6 hours means a 

 constant flow throughout the section of 600 feet per hour or one-sixth 

 foot per second, i.e., v = J. The wetted perimeter is the bottom only, 

 that is p=l, and this also is constant throughout the section. We have, 

 therefore, 



»528000 



/ 528000 



^^^ 62.3X0^007409 , ,,^,+0,920,^)1,, 



where E^^ is the work done in section "a", the length of the strip being 

 528,000 feet. Evaluating the above expression there results 



E^ = 37.939 foot-pounds per second 



SECTION "b" 



The flow in section "a" shows that the surface sinks at the rate of 

 j^-g foot per second. We will suppose that the surface of section "b" 

 sinks at the same rate, remaining always level. If I be measured from the 



beginning of the section the velocity at the distance I is given by r =— , 



when q is the volume of water flowing by the point I and a is the area of 

 the cross-section. Since the surface sinks ^^^ foot per second 



g = 100 + ^47^ a = 600 • ^^ 



5280 ' 220 



Therefore 



1 ,, 528000 +Z 



6^ 528000 + 28Z 



The wetted perimeter is constant throughout the strip, so that p = l. 

 The length of the strip is 50 miles, or 264,000 feet. The dissipation of 

 energy in this strip is then 



/ 264000 

 ( ir 528000+? Y 0.1920 r 528000+Z 1- \ 

 \ 6^528000 + 28d 6^ 1528000 + 280/' 



•264000 

 J-, oJ.o XU.007409 / I -^ I u^(j\jyj\j-ri' i , yj.xv^yjt u^ouuu-rfr i' i „ 



By putting Z = 264000a; this expression becomes 





^^^ 62.3X0.007409X264000 , ^ ,^_^;^, +,.,52L_iX±_|- U. 



