Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel's energy can be released quickly to accomplish a task that demands high power. a) An industrial flywheel has a 2.0 diameter and a mass of 240 . Its maximum angular velocity is 1400 A motor spins up the flywheel with a constant torque of 55 . How long does it take the flywheel to reach top speed? b) How much energy is stored in the flywheel? C)The flywheel is disconnected from the motor and connected to a machine to which it will deliver energy. Half the energy stored in the flywheel is delivered in 2.5 . What is the average power delivered to the machine?D)How much torque does the flywheel exert on the machine?
Explanation / Answer
Mass of the wheel M = 250 Kg Radius of the wheel R = Diameter/2 = 1.5 m/2 = 0.75 m Moment of Inertia of the wheel I = (1/2)MR^2 = 0.5 * (250 Kg) * (0.75 m)^2 = 70.31 Kg m^2 Angular velocity = 2 (12000 rpm) = 2(12000/60 rps) = 1256.63 rad/s (a) Torque = 50 Nm Angular acceleration = /I = (50 Nm)/(70.31 Kg m^2) = 0.711 rad/s^2 time taken t = / = (1256.63 rad/s)/( 0.711 rad/s^2) = 1767.07 s or = 29.45 min (b) Energy stored in the wheel E = (1/2)I^2 = 0.5 * (70.31 Kg m^2) * (1256.63 rad/s)^2 = 55513926.92 J or = 55.51 MJ (c) The average power delivered to the machine P = (E/2)/time = (55513926.92 J /2)/2.0 s = 13878481.73 watt = 13.878 MW (d) E/2=1/2*I2 = Sqrt[E/I] = 888.57 rad/s We get angular acceleration =/t = 444.28 rad/s^2 The torque is =I = 31237.73 Nm = 13.878 MW (d) E/2=1/2*I2 = Sqrt[E/I] = 888.57 rad/s We get angular acceleration =/t = 444.28 rad/s^2 The torque is =I = 31237.73 Nm E/2=1/2*I2 = Sqrt[E/I] = 888.57 rad/s We get angular acceleration =/t = 444.28 rad/s^2 = Sqrt[E/I] = 888.57 rad/s We get angular acceleration =/t = 444.28 rad/s^2 The torque is =I = 31237.73 Nm