SECT, iv.] AND PROPORTIONS OF ENGINES. 171 



be considered equal to the sum of the forces necessary to cause it to pass the 

 valves. 



By art. 306, the resistance of the air and vapour is, 



pbx 



and making b + x = I, we have b = ; therefore, 



IP' fl + hyp. log. JL_) = the power ; 



and when the pressure of the atmosphere p = 11 '55 Ibs. and the force of the 

 vapour p' = '77 Ibs. or 2 inches of mercury, it becomes 



lp' fl + hyp. log. P - \ = 2-85 /, nearly. 



The quantity of water will be one-sixth of the capacity, or -055 / a 2 Ibs. raised 

 1 foot. Hence, the whole power required for the ascending stroke will be 



a* / (2-85 + -055 / + r). 

 354. The whole power to work the pump is therefore, 



^.(2-85 + -055 / + |yil + 2r ) Ibs. raised 1 foot per second. 



Example. Let the velocity be 1'8 feet per second; the diameter of the pump 

 24 inches ; the length of its stroke 4 feet ; the friction 2 Ibs. per circular inch ; 

 and the area of the valves half the area of the pump. These numbers being 

 inserted, we have, 



24 * * >< ** 2-85 + -055 x 4 + 2 * 2 * ** + 4 = 



518-4 (2-85 + -22 + -00015 + 4^ = 3665 Ibs. raised 1 foot per second. 



As 550 Ibs. raised 1 foot per second is the steam engine horse power, the 

 power is, 



3670 , , 



-TQ- = of horse power nearly. 



The pump would answer for a double engine of about 134 horse power: there- 

 fore in this case about one-twentieth part of the power of the engine is required 

 for the air pump ; or one-tenth in the case of a single engine of the same sized 



