126 THEORY OF CONSTRUCTION AND SCIENTIFIC DESCRIPTION 



Here we have given dnz 1 inch, p =. 750 Ibs., and w zz 46.875 Ibs. ; 

 consequently, by substitution, we have 



750 3"= 1 s X 46.875, 

 and from this, by division, we obtain 



and finally, by evolution, we have 



a = V.0625 = .25 of an inch. , 



Let both sides of the equation marked (94) be divided by p, the 

 power or force with which the piston of the forcing pump descends, 

 and we shall have 



and by extracting the square root, we get 



(101). 

 Hence, the following practical rule. 



RULE. Multiply the weight or load upon the safety valve, 

 by the square of the diameter of the forcing pump, and divide 

 the product by the force with which the plunger or piston 

 of the forcing pump descends ; then, the square root of the 

 quotient will be the diameter of the safety valve. 



136. EXAMPLE 12. The diameter of the safety valve is one fourth 

 of an inch, the weight upon it is 46.875 Ibs., and the power applied, 

 or the force with which the plunger descends, is 750 Ibs ; what is the 

 diameter of the forcing pump ? 



Here we have given J of an inch, w> 46.875 Ibs., and /?=z750 

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



46.875d 2 =.25*x750; 



therefore, by division, we obtain 



.25^X750 



: 46.875 Z 



and finally, by extracting the square root, we get 



<frz 1 inch. 



The general expression for the value of the diameter of the forcing 

 pump, as derived from the equation (94), is 



d =^ (102). 



