OF FLUID PRESSURE USED AS A MECHANICAL POWER. 115 



and this, according to the conditions of the problem, is equal to n 

 times the entire weight ; hence we have 



.7854 D 2 s (h + x) = .7854 ns (tfh + d*x), 

 therefore, by casting out the common factors, we get 



or by separating the terms and transposing, we get 



( D 2 n (T ) x = D 2 h ( n 1 ) , 

 from which, by division, we obtain 

 _D 8 A(n--l) 

 1 D'^-nd- ' (87). 



119. It may be perhaps proper to illustrate this case by an example; 

 but in the first place, it becomes necessary to give the rule by which 

 the operation is to be performed. 



RULE. From the number of times which the pressure on the 

 bottom of the vessel, is proposed to exceed the weight of the 

 fluid, subtract unity ; multiply the remainder by the square 

 of the vessel 's diameter, drawn into its depth or perpendicular 

 altitude, and the result will be the dividend. 



Then, from the square of the vessel's diameter, subtract 

 n times the square of the diameter of the tube, and divide the 

 above dividend by the remainder for the length of the tube 

 required. 



120. EXAMPLE. If the perpendicular height of a cylindric vessel be 

 18 inches, its diameter 5 inches; the diameter of a tube fixed to the 

 top of the vessel one inch ; and if the vessel and tube be filled with an 

 incompressible and non- elastic fluid, till the pressure on the bottom 

 of the vessel is equal to twelve times the entire weight of the fluid; 

 what is the length of the tube into which the fluid is poured ? 



Here, by proceeding according to the rule, it is 



n 1=12 1~11, 



DA=z5X5x 18=450; 



therefore, by multiplication, we obtain 



2 /* (n 1) = 450X11 == 4950 dividend. 



Again, to determine the divisor, we have 



D 2 Bd=z5X5 12=13; 



consequently, by division, we obtain 



^ = 49504- 13 = 380|f inches. 



i 2 



