Question

16-5 A flywheel in the form of a disc 25 ft in diameter and 10 ft thick runs normally at 3600 r/min. It is made of an anisotropic filament composite material of a uniform density 160 lbft'. Calculate (a) the energy of the flywheel, in kilowatt hours, (b) the change in rotational speed and corresponding maximum energy that may be extracted from the flywheel if the coefficient of speed Auctuation may not exceed 0.01.

Answer 1

page-① lovestiom 16-5 Solution | briven that A fly wheel of some thickness is gotating at game R.P.M speed diameter of flywheel (Dire)= d Ida 12 asft (72629 mil loft = (3.048) m -> thickness > ง meam 3600 Dev I min 70 W meam ar & Nomeam bo 2 sex 3600 bro (1208) radis Womeam (376.gg) sadis density of fly wheel = = 1601/2015

page-12 we have to calculate (a) The energy of the flywheel t = 4 flynteed front view side view mass of flywheel = SxVolume = 45xfxx02]at = box Ex(251* x 10 40 400 2625X r. m 785398.16] db "I do = 0.454 kg 705398.162 0.454 = 13565707) Kg|

Paox-3) Since, flywheel in rotating, so energy stored in flywheel is in form of rotatiomal kimetic energy. 2 2 and fm = energy of flywheel -I (Wam Ipływreel mg2: Hexe, aga 212= 762 =(3.811.com &=(3.811 m m = (35 65 70.77 kg E = 2 x moh x Woml? 24x356570.7) XB-01) 2 (396.9932 € =[(12940 03.98) 2(376.999%] Joule 729400 3.98 x (37.6.9999 2 2 KJ xhr = ) lvo X360o E 510 84.92 ) Rwany As

Page 6 the change im rotational speed and corresponding maximum energy, Exxt if the coefficient of speed fluctuation may mot exceed 0.01 ie,... Cg - coefficient of <o.u1 fluctuation of speed gu, fan maximum limit 0.01 >> Nmax N min - Onan Nomeam >> Norman Nimeon = (0.01) Nomera Nomeam= 3600 RPM & given od 01x3648 => a'i Woman Nmax Nomin Nomin = 36

page-6 saj change in Rotation and speed - Mornim) (36) P.Pan (Nomas La Nimeam Narmad C. + Wornion 2 =) Nomoa + Nomin = 22 Wonean * Nomoa + Tomin = 2x3600 Nmaa + Mornim 7200) R-P.M solve equation and 2 TI Naman + Nonym 7200 36 Nanan pomin = - 7236 2 Nomar Norman 7236 = (3618] RPM .2 Domax = 13618] R.P.M]

page at Woman ar Nomax = 21x 3618 bo Wmax = (378-87) sadis go, maximum emergy stored in flywheel maximum Rotational Kimetic energy. Emax £ I (W.max)? = 1 x mg ² x (Woman? x (356570-49%82811% 8272.7773 2 1294003.98 M(3 786 87) rear for for a 360 Ermax (51595.704 9. kwam As