SM EATON'S AIR-PUMP. 



279 



quired to find the density of the air in the receiver at the 

 end of the loth stroke, we have from (1) 



Pi 6 =P(i) 15 = 0.03552 p. 



If the air originally had an elastic force equal to the 

 pressure of 30 in. of mercury, this would give the elastic 

 force of the air remaining in the receiver as equal to a 

 pressure of 1.05G in. of mercury. In this case, it is custom- 

 ary to say that the vacuum pressure is one of 1.056 in. of 

 mercury. 



Sen. It is evident from (1) that p n can never become 

 zero as long as n is finite, and therefore, even if the machine 

 were mechanically perfect, we could not by any number of 

 strokes completely remove the air; for, after every stroke 

 there would be a certain fraction left of that which occupied 

 it before. 



In working the instrument, the force required is that 

 which will overcome the friction, together with the differ- 

 ence of the pressures on the under surfaces of the pistons, 

 the pressures on their upper surfaces being the same. 



151. Smeaton's Air-Pump. This instrument con- 

 sists of a cylinder AB in which a piston is worked by a rod 

 passing through an air-tight collar at the 

 top ; a pipe BD passes from B to the 

 glass receiver C, and three valves, open- 

 ing upwards, are placed at B, A, and in 

 the piston. 



Suppose the receiver and cylinder to 

 be filled with atmospheric air, and the 

 piston at B. Eaising the piston, the 

 valve A is opened by the compressed air 

 in AM which flows out through it, while at the same time 

 a portion of the air in C flows through the pipe DB to fill 

 the partial vacuum formed in MB, so that when the piston 

 arrives at A, the air which at first occupied C now fills both 



ti"f- i 



&= 



Fig. 80 



