OF THE HYDROSTATIC PRESS. 125 



Here we have given d 1 inch; Ji=^ of an inch, and p 750 

 Ibs. ; consequently, by substitution, the equation (94) becomes 



P X w = .25* X 750 ; that is, w = 46.875 Ibs,, the 

 very same value as we derived from the fifth example. 



If both sides of the equation marked (94) be divided by d*, the 

 general expression for the value of w becomes 



_3> 



~d 2 * (99). 



And the practical rule supplied by this equation, may be expressed 

 in words at length in the following manner. 



RULE. Multiply the force with which the plunger descends 

 by the square of the diameter of the safety valve, and divide 

 the product by the square of the diameter of the plunger ; 

 then the quotient will express the load upon the safety valve. 



134. EXAMPLE 10. The diameter of the safety valve is J of an 

 inch, that of the forcing pump is one inch, and the load upon the 

 safety valve is 46.875 Ibs. ; what is the power applied, or the force 

 with which the plunger in the forcing pump descends ? 



Here we have given 3=J of an inch, d=:l inch, and w=r46.875 

 Ibs. ; consequently, by substitution, equation (94) becomes 



.25 2 j9=46.875 X I 2 , 

 and from this, by division, we obtain 



' ' 



The general expression for the value of p, as obtained from the 

 equation marked (94), becomes 



_d*w 



: ~F* (100). 



from which we derive the following rule. 



RULE. Multiply the load on the safety valve by the square 

 of the diameter of the forcing pump ; then, divide the product 

 by the square of the diameter of the safety valve, and the 

 quotient will give the force with which the piston descends. 



135. EXAMPLE 11. The diameter of the plunger or the piston of 

 the forcing pump is one inch, the force with which it descends is 

 equivalent to 750 Ibs., and the load on the safety valve is 46.875 Ibs.; 

 what is its diameter ? 



