18_ch 16 Mechanical Design budynas_SM_ch16

18_ch 16 Mechanical - I = md 2 4 W = 4 gI d 2 = 4(386(239 6 48 2 = 161 lbf Ans 16-25 Use Ex 16-6 and Table 16-6 data for one cylinder of a

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 16 413 16-24 (a) The useful work performed in one revolution of the crank shaft is U = 35(2000)(8)(0 . 15) = 84(10 3 )in · lbf Accounting for friction, the total work done in one revolution is U = 84(10 3 ) / (1 0 . 16) = 100(10 3 )in · lbf Since 15% of the crank shaft stroke is 7.5% of a crank shaft revolution, the energy fluctuation is E 2 E 1 = 84(10 3 ) 100(10 3 )(0 . 075) = 76 . 5(10 3 )in · lbf Ans. (b) For the flywheel n = 6(90) = 540 rev/min ω = 2 π n 60 = 2 π (540) 60 = 56 . 5 rad/s Since C s = 0 . 10 I = E 2 E 1 C s ω 2 = 76 . 5(10 3 ) 0 . 10(56 . 5) 2 = 239 . 6 lbf · in · s 2 Assuming all the mass is concentrated at the effective diameter, d ,
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: I = md 2 4 W = 4 gI d 2 = 4(386)(239 . 6) 48 2 = 161 lbf Ans . 16-25 Use Ex. 16-6 and Table 16-6 data for one cylinder of a 3-cylinder engine. C s = . 30 n = 2400 rev/min or 251 rad/s T m = 3(3368) 4 π = 804 in · lbf Ans. E 2 − E 1 = 3(3531) = 10 590 in · lbf I = E 2 − E 1 C s ω 2 = 10 590 . 30(251 2 ) = . 560 in · lbf · s 2 Ans . 16-26 (a) (1) ( T 2 ) 1 = − F 21 r P = − T 2 r G r P = T 2 − n Ans . r P r G T 1 F 12 F 21 T 2...
View Full Document

This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

Ask a homework question - tutors are online