Numerical - Rise in temperature of lubricating oil in hydrodynamic bearing

Q.  Lubricating oil used in hydrodynamic bearing has total flow rate of 0.340 l/min and side leakage of 0.1520 l/min. If mass density of oil is 600 kg/m3  and specific heat is 1.05 kJ/kg oC, what is the rise in temperature if power lost in friction is 0.05 kW?
- Published on 19 Oct 15

a. 11.23o C
b. 15.11o C
c. 18.03o C
d. 22.23o C

ANSWER: 18.03o C

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    Discussion

  • Sravanthi   -Posted on 05 Oct 15
    Given: Flow rate (Q) = 0.340 l/min, side leakage of lubricant (Qs) = 0.1520 l/min, mass density of oil (ρ) = 600 kg/m3 , specific heat of oil = 1.05 kJ/kgoC, power lost in friction = 0.05 kW

    If effect of leakage is considered the following formula is used to calculate rate of heat dissipated or power lost due to friction:

    Pf = {ρ Q Cp ΔT [1 – 0.5 (Qs / Q)]} / 109
    where, Qs = side leakage of lubricant & Q = total flow rate of lubricant

    Convert flow rate of lubricant from l/min into mm3 / s

    Flow rate (Q) = 0.340 l/min = (0.340 x 10002) / 60 = 5666.66 mm3 / s
    (Qs) = 0.1520 l/min = (0.1520 x 10002) / 60 = 2533.33 mm3 / s
    Substituting the given values we get,

    0.05 = {600 x 5666.66 x 1.05 x ΔT [ 1 – 0.5 (0.447)]} / 109

    Substituting the given values we get,

    ΔT = 18.03o C

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