Question

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A cast iron flywheel used for a four stroke I.C. engine is developing 187.5 kW at 250 r.p.m. The hoop stress developed in the flywheel is 5.2 MPa. The total fluctuation of speed is to be limited to 3% of the mean speed. If the work done during the power stroke is 1/3 times more than the average workdone during the whole cycle, find: 1. mean diameter of the flywheel, 2. mass of the flywheel and 3. cross-sectional dimensions of the rim when the width is twice the thickness. The density of cast iron may be taken as 7220 kg/m3 [Ans 2.05m; 4561 kg; 440 mm, 220 mm)

Answer 1

Qui Guvenir P= 187.5kw N=250 rpm " = 5.2 MPa N-N2 = 0.03N : 8 = 7220 kg/m² solun- P = 27N Tmean 60 Tmean 187.5x10²x60 2 X x 250 7161.97 Nom Work clone percycle Timeony O= 7161.97X 4 X = 90,000 N-m 90.000 + $x 90,000 Work done during the power stroke = & 120,000 Nm Tman Torque B & Timean D G E A F с 27 4T 30 Work done during pouus stroke - xnx Tman - 120,000 Tman 120,000 x 2 76,394.37 Nm # From geometric relation area of AB DE area of A ABC (BG) 2 (BF) Man fluctuation DE of energy of ABDE area (DABC) x (BG j² BP AF = 120,000 76,394.37 -7 161.97 76,394.37 DE = 98,554 7 Nom

(1) Mean dio of flywhul! - a = Pv² ☆ v= F V 5.2x106 26.83 mis V 7220 VE TDN 60 box v TN 60 x 26.83 2.05019 TX 250 D=2.0502 m (2) Mars of the flywheel! - 2 x 250 60 W = 20N 60 = 26. 18 radla Ni-N2 = = 0.03 DE = m (Rw? (s) 98554.7 = mx ** (20592)x(2616)*X0.03 m= 4561.25 kg (3) Dimensions of rimi - m= AX (AP) & 8 4561.25 il TDS TX 2.0502 x 7220 bit = 0.09808 A = b=27 2+? 0.09808 t = 0.22 1 m * *PLS RATE ! - t=221mm b=2x+= 2x221 = 442 mm