Flywheels are large, massive wheels used to store energy. They can be spun up sl

Question

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. An industrial flywheel has a 1.5 m diameter and a mass of 250 Kg. Its maximum angular velocity is 12000 rpm.

a) A motor spins up the flywheel with a constant torque of 50 Nm. 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 of the energy stored in the flywheel is delivered in 2.0 s. 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                                                                      

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---