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6 December, 22:45

The journals in a high speed oil engine are 80 mm in diameter. and 40 mm long. The radial clearance is 0.060mm. Each supports a load of 9 kN when the shaft is rotating at 3600 rpm. The bearing is lubricated with SAE 40 oil supplied at atmospheric pressure and average operating temperature is about 65oC. Using Raimondi - Boyd charts analyze the bearing under steady state operation.

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  1. 6 December, 22:56
    0
    Answer: S = 0.284

    Explanation:

    dа ta: d = 80 mm;

    l = 40 mm;

    c = 0.06 mm;

    F = 9kN;

    n = 3600rpm = 60 rps

    SAE 40 oil

    T = 65°C

    Therefore:

    p = F / ld

    = 9 x1000 / 40 x 80

    = 2.813 MPa

    μ = 30 cp at 65°C for SAE 40 oil

    S = r^2 x μ x n / c^2 x p

    S = (40) ^2 x 30*10^-3 x 60 / (0.06) ^2 x 2.813*10^6

    S = 2880 / 10,126.6

    S = 0.284

    l/d = ½,

    h o / c = 0.38

    ε = e / c = 0.62

    h o = 0.38 x C

    = 0.382 x 0.06

    =0.023mm

    = 23µm

    e = 0.62 x C

    = 0.62 x 0.06

    = 0.037 mm

    Viscosity temperature curves of SAE graded oils

    (r / c) f = 7.5,

    S = 0.284

    l / d = ½

    f = 7.5 x (c / r)

    = 7.5x (0.06/40)

    = 0.0113
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