Flywheel Tutorial Problem & Solution (Dynamics of Machines, ME22004)

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The resisting torque on the crank of a riveting machine is shown below. The duration of cycle is 2 sec. The motor driving the machine has a speed of 1500 rpm and it delivers constant torque. The crank-shaft of the machine is geared to the motor shaft. Neglecting frictional losses, determine: 

a) the power of the motor 

b) the moment of inertia of the flywheel mounted on the motor shaft to keep the speed fluctuation within 2% of the average speed. 

c) the maximum angular acceleration of the flywheel Solution: In this problem, torque supplied is constant and demand torque is fluctuating. 

The demand torq  ue is shown in figure. a) The energy required per cycle 3 1 200 1600 (1600 200) 200 2 4 2 4 2 4 2904.5 Nm The duration of cycle is 2 seconds 2904.5 Energy required per second 1452.25 2 Power of motor required, p 1.452 kW 1452.25 W ) 2904.5 The average torque 462.5 Nm 2 2 462.5 Nm The shaded portion is fluctuation of energy. Therefore, 3 (1600 462.5) (1600 462.5) (1600 462.5) 4 4 2 2 (1600 200) 1618.45 Nm 2 2 1500 157.08 rad/s 60 60 2 ( 2) Coefficient of speed fluctuation, 0.04 100 But we have, 1256.33 0.04 (157.08) 1.64 kgm Moment of inertia of the flywheel is 1.273 kgm c) Maximum s s N k E Ik I I max acceleration shall occur where torque is maximum and minimum acceleration will occur where torque is minimum Maximum excess torque ( ) 1600 462.5 1137.5 Nm Let angular acceleration be ' ' excess 2 max 2 max 2 1137.5 893.6 rad/s 1.273 893.6 rad/s Maximum angular acceleration of the flywheel is 893.6 rad/s excess excess 

The resisting torque on the crank of a riveting machine is 200 Nm for first 90o, from 90o to 135o is 1600 Nm then it drops linearly to 200 Nm upto 180o and remains the same upto 360o. The duration of cycle is 2 sec. The motor driving the machine, however, has a speed of 1450 rpm and it delivers constant torque. The crank shaft of the machine is geared to the motor shaft. The speed fluctuation is limited to  2% of mean speed. Determine : 133 (a) power of the motor, and Flywheel (b) moment of inertia of the flywheel mounted on the motor shaft. Solution In this problem, torque supplied is constant and demand torque is fluctuating. The demand torque is shown in Figure 4.14. (a) The energy required per cycle 3 1 200 1600 (1600 200)  3 200 1600 1400 200 2 4 8 4 700 300 400 50 (750 175) 925 4= 2904.5 Nm The duration of the cycle is 2 seconds  Energy required per second 2904.5 1452.25 2  Power of motor required, p = 1.452 kW. Figure 4.14 : Figure for Example 4.5 (b) The average torque 2904.5 2 2 av E M or, Mav = 462.5 Nm The shaded portion is fluctuation of energy. Therefore, (1600 462.5) 3 (1600 462.5) 4 (1600 462.5) 4 2 2 (1600 200) E 2 1137.5 1137.5 4 1400 4  (1137.5 924.22) 1618.45 Nm 4 1600 462.5 200 Mav M 900 1350 134 Theory of Machines  2 2 1450 151.76 r/s 60 60  Coefficient of speed fluctuation 2 2 0.04 100  2 1618.45 151.76) or, 1618.45 2 1.76 kg m 0.04 23032.1

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