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5 November, 22:46

The Power Input at Gear P is 1.7802 kW, rotating at 222.222 RPM. Assume the the Module m is the same for all gears. a. What is the Input Torque at gear P? b. Assuming 100% efficiency at all gear pairs, what is the Output Power, Torque and RPM at gear Q? c. If the efficiencies of the gear pairs are: ηP-B = 90%, ηC-Q = 95%, what is the Output Power and Torque at gear Q? d. At 100% efficiency, what are the gear tooth tangential and Normal forces at the P-B and C-Q contacts: FT-B, FN-B, FT-C & FN-C?

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  1. 5 November, 22:54
    0
    From the example the answer are (a) 76.499 N-M (b) 179.43N-M (c) 159.88 N-M (d) 867.93N

    Explanation:

    Solution

    From the example given we solve the following problems.

    Recall

    The power input at given P (Pp) = 1.7802 kW

    The speed of the gear P (Np) = 222.222 RPM

    Let the module m be the same for all the gears

    (a) Torque input gear at P/Pp =

    Tp * 2πNp/60 =

    1.7802 * 10^3 = Tp * 2π * 222.222/60

    Tp = 76.499 N-M

    (b) Let assume that the there is 100 % efficiency at all pair of gears,

    Pp = Pb = Pc = Pq

    = 1.7802 * 10 ^3 = Tb * 2πNb/60 = Tc * 2πNc/60 = Tq * 2πNc/60

    now,

    dp = 18cm,

    db = 40cm

    dc = 20cm

    dq = 44cm

    where d = mƵ, Ƶ = Number of teeth

    Then

    Ƶp / Ƶq = Nq/Np

    dp/m/dq/m = Nq/222.222

    Nq = 90. 908RPM

    The output power which is Pq = 1.7802kW

    The output torque Tq = 179.43N-M

    (c) = Pq = Tq * 2πNq/60 = 1.522071 * 10^3

    = Tq * 2πNq / 60 = 159.88 N-M

    Tq = 159.88 N-M

    (d) Now, at efficiency of 100%

    (Ftb) * (db/2) = Tb

    Ft-b = Ft-p

    Ft-p * dp/2 = Tp

    Ftp = 849.989 N

    Ft-b = 849.989 N

    Ft-b = Ft-n cos 20%

    Ft-n = 904.54 N

    so,

    Ft-c = Ft-q

    (Ft - q) * dq/2 = Tq

    Ft-q = 815.59 N

    where,

    Ft-c = 815.59 N

    then,

    Ft-c = Ft-n cos 20

    Ft-n = 867. 93N
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