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belts and chains

Extracts from this document...

Introduction


Introduction:

  Belts and chains represent the major types of flexible power transmission elements. In general, belt drives are applied where the rotational speed are relatively high, as on the first stage of speed reduction from a motor or engine. Chain drives are typically applied at lower speeds, with the consequent higher torques.

  In this assignment, we are required to design a chain drive for a uniformly loaded conveyor, and a V-belt drive for a diesel engine.


A. Requirements for chain drive:

  1. It is used for a uniformly loaded conveyor.
  2. To be driven by a gasoline engine through a mechanical drive.
  3. Input speed is 900rpm.
  4. Output speed is 230 to 240rpm. (On average 235rpm)
  5. The conveyor requires 12kW power.
  6. Normally run 12 hours daily.

Chain transmission:

  1. Service factor = 1.1
  2. Design power = conveyer power required * service factor = 12 * 1.1 = 13.2kW
...read more.

Middle

C = Center distance in mm

        P = Pitch of chain in mm

        T = Number of teeth in large sprocket

        t = Number of teeth in small sprocket

        K = Factor from table 3

Drive specification:

  1. Sprocket of single chain with 23 teeth fitted to gasoline engine shaft.
  2. Sprocket of single chain with 88 teeth fitted to loaded conveyor shaft.
  3. Chain of pitch 3/4 (19mm) and length 137 pitches.

B. Requirement for V-belt drive:

  1. It is used for a 55kW diesel engine.
  2. It is run at 1200rpm that driving a piston compressor runs at 760rpm.
  3. The center distance is 1600mm.
  4. The duty is 20hrs/day.
  5. The diameter of engine shaft and compressor shaft are 70mm and 85mm respectively.

V-belt transmission:

  1. Service factor = 1.4
  2. Design power rating = piston compressor rating * service factor

                    = 55 * 1.4 = 77kW

  1. Belt section indicated is V-belt-C
  2. Speed ratio = driving rotating speed / driven rotating speed = 1200 / 760 = 1.6
...read more.

Conclusion

 2 / 8 = (400-250) 2 / 8 =2812.5

     C is the center distance of drive;

     D is the pitch diameter of larger pulley;

     d is the pitch diameter of smaller pulley; and

     L is the pitch length of belt

  1. Power rating per belt is 9.85kW with additional power for the speed ratio of 0.94kW, i.e. a total of 10.79kW
  2. Arc of contact correction factor is decided by value of (D-d)/C corresponding to the value of proportion of 180˚ rating.

(D-d)/C = 0.094 <=> 0.99

Arc of contact correction factor = 0.99

  1. Belt length correction factor = 1.02
  2. Number of belts = W/XYZ = 77 / 10.79*0.99*1.02 = 7.06

i.e. 7 belts.

where

        W = design power rating

        X = power rating per belt

        Y = Arc of contact correction factor

        Z = length correction factor

  1. Summary of pulley requirements:
  1. Smaller pulley fitted to diesel engine shaft 250mm pitch diameter with 4 grooves C section.
  2. Larger pulley fitted to piston compressor shaft 400mm pitch diameter with 4 grooves C section.
  3. V-belt required, C section belts, 4060mm pitch length.

...read more.

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