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to rotate by a differential angle , the differential work done is

      (2-26)equation

Schematic illustration of the definition of torque, work, and power.

      If this differential rotation takes place in a differential time dt, the power can be expressed as

      This stored kinetic energy can be recovered by making the power p(t) reverse direction, that is, by making p(t) negative.

      EXAMPLE 2‐5

       Solution

equation

      (2-30)equation

      where B is the coefficient of viscous friction or viscous damping.

       Solution

      Traveling at 50 km/h, compared to 100 km/h, requires 1/8th of the power, but it takes twice as long to reach the destination. Therefore, the energy required at 50 km/h would be 1/4th that at 100 km/h.

Schematic illustration of actual and linearized friction characteristics.

Vehicle u = 50 km/h u = 100 km/h
Cw = 0.3 fL = 62.06 N P = 0.86 kW fL = 248.2 N P = 6.9 kW
Cw = 0.5 fL = 103.4 N P = 1.44 kW fL = 413.7 N P = 11.5 kW

      This torque at the load‐end overcomes the load torque

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