F. F. Cipher — N on- condensing Steam Engine. 287 



Multiplying (15) through by we have 



-8hp= — • • (IQ) 



33000 33000 ^ ' 



This is the equation of a parabola and corresponds to (10), 

 which is likewise a parabola when P' is constant. 



Differentiating (16), the condition of maximum output is 

 found to be 



»'-7*y • • • • (17 ) 



or the speed must be one-half that which the engine should 

 have at the same boiler pressure if w — o. 



The condition for maximum, according to this, would be 



r'=a + 2b' n . . . . (18) 



It will be observed that this is a right-line, tangent to the para- 

 bola (11) where n — o. According to (11), these values of n 

 represented by (17) would be somewhat too large. 



Solving (15) for n and multiplying by 



2nrk 2nr a /in , 



HP = to — w . . . (19) 



The condition for maximum obtained from this equation is 



to' = \ Jc . . . . (20) 



where w' is one-half the load which will bring the engine to 

 rest at that pressure. 



This value of w is somewhat too small to satisfy (11), although 

 as stated the error is probably alway too small to have any im- 

 portance. Substituting these two values of n and w' in (3) 

 and we undoubtedly have a very close approximation to the 

 maximum output at any pressure P', 



Jc 

 33000 Tc 1 



B hp =i _ _ hr 



k 

 where -— is the speed at that pressure when w = o as com- 

 k 



puted from (7) and hr = the turning moment which for that 

 pressure must be applied to the shaft in order to bring the 

 engine to rest. This can be computed from (9). It is wr 

 when n — o. 



In a similar manner we may represent indicated horse-power 

 as a function of boiler pressure. Solving (6) for P and substi- 

 tuting, as before, 



P -_ h+ (A + P„) (A + .F) ■ (21) 



h + P + F+(b-+e)n " V ; 



