DRIVES AND CONTROLS 209 



formly the body from rest to a speed A'^ r.p.m. in t sec, by using an average speed N/2 



TxN/2 

 " = 5,250 (^9^) 



WR'N' WR'N' 



10,500 X 308i 3,234 X 10' X t 



In mechanical systems with a number of rotating parts, the energy E^ stored in 

 the moving system is the sum of the energies stored in each part, or 



p W:R\N\ + W,R\N\ + W,RhNh + - • ■ + W,.R\N\ ,,„. 



^' = 5;873 (^^) 



In power-drive and motor-application problems, it is advantageous to express 

 the energy Es in the system in terms of an " equivalent WR"^," which will be designated 

 here as W^R^s, at the drive or motor shaft having a speed of Nd, such that 



W R^ N-^ 

 E. = 1!^' (41) 



By combining Eqs. (40) and (41), it will be seen that 



WM\ = W^R\ (^y + W,RS (^^y + W^R\ {^^y + ■ • ■ + Wr^RJ (^)' (42) 



The torque T^ necessary to accelerate uniformly a system at rest to a required 

 speed in t sec. can be obtained by substituting WsR'\ for WR^, and Nd for A^" in Eq. (37), 

 which then becomes 



^ WsR'sNd 



dOSt 



(43) 



The horsepower H^ required to accelerate the system from the drive shaft at rest 

 to a speed of Nd r.p.m. in t sec. can be determined by substituting WsR^sN^d for WR'N^ 

 in Eq. (39), which then becomes 



W R^ AT^ 

 ^ 3,234 X 10^ X i ^ ^ 



or from Eq. (39a) by substituting A''^ for N, and for T the value of T, as given by Eq. 

 (43) which then becomes 



WsR'sNd Nd ^ WM'N'd 



308t 5,250 X 2 3,234 X 10^ X ^ 



jj _ VV s-t^ 5-tVd y iVd ^ *y s-tl- -tV d /j^rN 



Sometimes complex systems are encountered involving both linear and rotating 

 motion. The equivalent WR- of the linearly moving parts can also be reduced to 

 the motor-shaft speed by the equation 



"Equivalent WR'" = Tf (p^^Y (46) 



\ZTrI\ d' 



where W = weight of the body Y = velocity, in feet per min 



'N d = r.p.m. of the drive or motor shaft 



