114 OF FLUID PRESSURE USED AS A MECHANICAL POWER. 



It is required to find the length of the tube, such, that 

 when it and the vessel are filled with an incompressible fluid, 

 the pressure on the bottom of the vessel may be equal to any 

 number of times the fluid's weight. 



In resolving this problem, it will be sufficient to refer to the pre- 

 ceding diagram, because a separate construction would exhibit no 

 variety ; for this purpose then, 



Put D zz DC, the diameter of the base of the cylindrical vessel, 

 A zz the area of its base, 



h zz A D, the altitude, or perpendicular depth of the vessel, 

 C the capacity, or solid content, 

 P the pressure on the bottom, 



d ed, the diameter of the tube inserted in the top of the vessel, 

 a zz the area of its horizontal section, 

 c zz the capacity, or solid content of the tube, 

 w ~~ the weight of the fluid in the vessel, and w' the weight of 



that in the tube ; 



s zz the specific gravity of the fluid, 



n zz the number of times the pressure exceeds the weight, 

 and x zz the required length of the tube. 



Then, according to the principles of mensuration, the area of the 

 bottom of the vessel becomes 



AZZ.7854D 2 , 

 and that of a horizontal section of a tube, is 



azz.7854d 2 ; 

 and again, by the geometry of solids, the capacity of the vessel is 



Czz.7854D 2 A, 

 and for the capacity of the tube, we have 



the respective weights being 



wzz.7854D 2 As, and w 1 zz .7854d 2 a-s; 



consequently, the whole weight of the fluid in the vessel and tube, is 



w + w' = .7854s (D 2 A, -f d*x). 



Now, we have shown above, that the pressure on the bottom of the 

 vessel is equal to the weight of a fluid cylinder, whose diameter is 

 DC and altitude <ZD; consequently, the pressure on the bottom is 

 expressed by 



