Page 462 - Mechatronics with Experiments
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                              and no leakage in the transmission line. Hence, the input pressure and flow rate to the
                              motor are
                                                      p = p − 200 psi = 1800 psi                (7.116)
                                                            p
                                                       m
                                                                       3
                                                      Q = Q = 10 800 in ∕min                    (7.117)
                                                            p
                                                       m
                                   The volumetric efficiency of the motor indicates that less than 100% of the input flow
                              is converted to displacement,
                                                     Q a
                                                   =                                            (7.118)
                                                 v
                                                     Q m
                                                                               3
                                                Q =    ⋅ Q = 0.9 ⋅ 10 800 = 9720 in ∕min        (7.119)
                                                 a   v   m
                              and we can determine the output speed of the motor,
                                                        Q =    ⋅ Q = D ⋅ w m                    (7.120)
                                                         a
                                                                       m
                                                                 m
                                                              v
                                                             Q a
                                                        w =     = 243 rpm                       (7.121)
                                                         m
                                                             D
                                                              m
                              The output power of the motor is determined from the ratio of its input power and output
                              power, given the overall efficiency of the motor,
                                                            T ⋅ w m
                                                             m
                                                   =    ⋅    =                                  (7.122)
                                                     v
                                                        m
                                                 o
                                                            p ⋅ Q m
                                                             m
                                                       ⋅ p ⋅ Q m
                                                        m
                                                     o
                                               T =                                              (7.123)
                                                m
                                                        w m
                                                                      2
                                                                               3
                                                    0.9 ⋅ 0.85 ⋅ 1800 lb∕in ⋅ 10 800 in ∕min
                                                  =                                             (7.124)
                                                          243 rev∕min ⋅ 2   rad∕rev
                                                  = 61 200∕(2  )lb ⋅ in = 9740 lb ⋅ in          (7.125)
                              Hence, the mechanical power delivered at the output shaft of the motor is
                                                               (61 200∕(2  )) ⋅ 243 ⋅ 2  
                                       MotorPower out  = T ⋅ w =                   = 37.55 HP   (7.126)
                                                       m
                                                           m
                                                                       6600
                              7.3.3 Pump Control
                              The pump control element is the actuation mechanism that controls the angular position
                              of the swash plate. The mechanism which controls the swash plate angle is called the
                              compensator or the pump controller.
                                   A pump control system has the following components (Figure 7.33),
                                1. One or a pair of small control cylinders (also called displacement pistons)usedto
                                   move the swash plate.
                                2. A proportional valve which meters flow to the control cylinders, and is controlled by
                                   electrical or hydro-mechanical means for a certain control objective (i.e., constant
                                   pressure output from the pump, load sensing pump).
                                3. Sensor or sensors (either hydraulic pressure sensing lines or electrical sensors) used
                                   to measure the variables of interest, such as pressure, flow rate, position.
                                4. A controller (i.e., a digital electronic control module (ECM) or a hydro-mechanical
                                   control logic configured to control the proportional valve) used to implement the
                                   control logic (i.e., if pump is controlled to provide a constant output pressure).
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