Page 416 - Marks Calculation for Machine Design
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P1: Naresh
                          January 4, 2005
                                      15:28
        Brown.cls
                 Brown˙C09
                                           APPLICATION TO MACHINES
                  398
                    By analogy with the discussion for internal combustion engines where the torque varies
                  with the angle of rotation, the change in the inertial energy levels of the punch press system
                  is equal to the work done on or by the system. Therefore, Eq. (9.57) is applicable here where
                  the torque varies with angular velocity. Expanding the difference in angular velocities in
                  Eq. (9.57) as it was done in Eq. (9.58), the work done becomes
                                 1      2    2     1       2  2
                          Work =  I sys ω max  − ω min  =  I sys ω − ω 1
                                                          2
                          1→2    2                 2
                                 1                           ω 2 + ω 1
                              =   I sys (ω 2 + ω 1 )(ω 2 − ω 1 ) = I sys  (ω 2 − ω 1 )  (9.73)
                                 2                             2

                              = I sys ω m (ω 2 − ω 1 ) = I sys ω 1 C f ω 1
                              = I sys C f ω 2
                                       1
                    Solving for the mass moment of inertia of the system (I sys ) in Eq. (9.73) gives
                                                   Work
                                                    1→2
                                              I sys =  2                       (9.74)
                                                   C f ω
                                                      1
                  which is very similar to Eq. (9.62) for internal combustion engines. However, the mass
                  moment of inertia of a punch press system (I sys ) is actually determined using Eq. (9.69),
                  once the torque (T 2 ) has been found by trial and error using Eq. (9.67).
                            U.S. Customary                       SI/Metric
                  Example 4. A punch press requires a certain  Example 4. A punch press requires a certain
                  punching torque during only 4 percent of the  punching torque during only 4 percent of the
                  punching cycle, which is (1 s). Determine the  punching cycle, which is (1 s). Determine the
                  required mass moment of inertia of the system  required mass moment of inertia of the system
                  (I sys ) and the coefficient of fluctuation (C f ),  (I sys ) and the coefficient of fluctuation (C f ),
                  where                              where
                    T punch = 150 ft · lb              T punch = 225 N · m
                       t 2 = 1s                          t 2 = 1s
                    P rated = 5hp                      P rated = 4kW
                    ω rated = 1,725 rpm                ω rated = 1,725 rpm
                     ω syn = 1,800 rpm                  ω syn = 1,800 rpm
                  solution                           solution
                  Step 1. Convert the rated power (P rated ) from  Step 1. Convert the rated power (P rated ) from
                  horsepower (HP) to (ft · lb/s).    kilowatts (kW) to (N · m/s).
                                       ft · lb                            N · m
                                    550                               1,000
                                        s                                  s
                        P rated = 5HP ×                   P rated = 4kW ×
                                      HP                                 kW
                                  ft · lb                           N · m
                            = 2,750                           = 4,000
                                    s                                 s
                  Step 2. Convert the rated angular velocity  Step 2. Convert the rated angular velocity
                  (ω rated ) from (rpm) to (rad/s).  (ω rated ) from (rpm) to (rad/s).
                              rev  2 π rad  1 min                rev  2 π rad  1 min
                    ω rated = 1,725  ×  ×              ω rated = 1,725  ×  ×
                              min   rev    60 s                  min   rev    60 s
                        = 180.6 rad/s                      = 180.6 rad/s
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