Question

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The turning moment diagram for the engine is drawn to the following scales: Turning moment, 1 mm = 1000 N-m and crank angle, 1 mm = 6º. The areas above and below the mean turning moment line taken in order are : 530, 330, 380, 470, 180, 360, 350 and 280 mm2. The mean speed of the engine is 150 r.p.m. and the total fluctuation of speed must not exceed 3.5% of mean speed. Determine the diameter and mass of the flywheel rim, assuming that the total energy of the flywheel to be 15/14 that of rim. The peripheral velocity of the flywheel is 15 m/s. Find also the suitable cross-scetional area of the rim of the flywheel. Take density of the material of the rim as 7200 kg/m3. [Ans. 1.91 m; 8063 kg; 0.1866 m²]

Answer 1

Turning moment diagram 530mm² 380mm² Isomm? •350mm² A a I В 2 330mm 470mm 36omm? 280mm start from Energy at a EAFE EB =E+530 Ec= E +530-330=E+200 Epa E +200 +380 = E +580 EE E +580 – 470 = E +110 Ef = Etllo +180 = E +290 EG= E +290-360 = E-70 EH E-70 +35o= E +280 E +280 - 299=E EI= Maximum fluctuation of energy ДЕ Е Emax Emin ДЕ Ер – En - - (€+589)-(6-70) 2 SE = 650 mm

Amma - Hence looon-m & Imm 6' So, after unit conversion ДЕ 6 50 x 1000X EX (18x) 68067.84 ivm Total energy of flywheel AEZ 2 = 1507 rade GO Wmean = 27N zax 150 Go 3.5% on 0.035 Csa # To determine the diameter of flywheel v=TON 60 15 = TXDX 150 60 Dz of Flywheel = 15 of rim) Total energy legogm Diameter of ř logim Am flywheel By using the given the given relationship (total energy Energy fluctuation of rim 14X 68067.84 15 63529.98 N-m # Po determine mass of flywheel I w²es where I (mass moment of inertia) for rim type Flywheel Where R=% DE Rima DE Rim me? [ where -

2 (DE) Rim 63529.98 = mx m (2) xw²x( (1,243 x 15.7**0.035 8074.29 kg ñ 8063 ky(approx) An m = # Po determine the cross-sectional area of the fly when m As TDP A= 8063 1.91 x 7200 2 A = 0:1866 mi Any
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