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    Flat Belt Engineering Question

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    **Please see the attached file for the diagram **

    You are to design a Flat pulley drive for a refrigeration system with the compressor pulley diameter being 450mm and the motor pulley diameter being 200mm. The two shafts will be 1.95m apart and connected to a closed belt. Find
    a. The angle of contact /angle of lap for the two pulleys

    b. Find the required length of belt

    c. Select a suitable belt and, ignoring the elongation at fitting but using the guide mass/unit length, calculate the power transmitted if the maximum permissible tension is 1kN and the coefficient of friction is 0.25 if the motor rotates at 1450rpm

    d. Also find the minimum initial tension require at commissioning stage

    Show calculations and working.

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    Solution Preview

    ** Please see the attached document for a Word formatted copy of the solution **

    d.a) Factor β = Sin-1 [(D_l - D_s )/ 2 C] where D_l is the diameter of the large pulley = 450mm = 0.45m, D_s is the diameter of the motor pulley = 200mm = 0.20 m and C the distance between the two pulley centers == 1.95 m

    = Sin^-1 [ (0.45 -0. 20) / 2*1.95]
    = Sin^-1 [0.25 / 3.9]
    = Sin^-1 [0.0641]
    = ...

    Solution Summary

    This solution explains how to find information about two pulley, based on their diameters and distance apart.