Question

A power press makes steel pressings. The energy required for each pressing is 5000J with each pressing taking 1s. The time taken to remove the pressing and get the material ready for the next pressing is 4s.If the press is operated by an electric motor which runs continuously with a flywheel used to supply the extra torque required for the pressing. Determine the power of the motor that is required, the energy that has to be supplied by the flywheel and the moment of inertia of the flywheel if it’s speed is not to drop by more than 30 RPM from a speed of 600 RPM.

Answer 1

Q1

Given,

Energy required for each pressing = E₁ = 5000 J

Total Time of Pressing and Removing operation = 1 sec + 4 sec = 5 sec

Maximum Speed Of Flywheel = N₁ =600 RPM

Minimum Speed Of Flywheel = N₂ =570 RPM

Total Energy Required per Second = Energy Required / Total Time of Operation

                                                               = 5000 J / 5 sec = 1000 W

Power of motor = 1000W =1KW

Each pressing operation takes 1 sec, therefore energy supplied by the motor in 1 sec

                                                             E₂ = 1sec*1000 W

                                                                  = 1000 J

Energy Supplied By the flywheel during pressing

                                                    

                                                                  = 5000-1000 = 4000 J

Mean Speed Of Flywheel                   N = (N₁ + N₂)/2 = (600+570)/2 = 585 RPM

                                                                 = 61.2605 rad/sec

Now

C_s = Coefficient Of Fluctuation Of Speed

       =(N₁-N₂) / N

       =(600-570)/585

       = 0.05128

= ²C_s

4000 = * 61.2605² * 0.0512

= 20.78 Kg m²

The Moment of Inertia Of the Flywheel is 20.78 Kg m²