Design of Machine Elements - 2 - Mechanical Engineering (MCQ) questions and answers for Q. 11777

Q.  Lubricating oil of mass density 800 kg/m3  used in 360o hydrodynamic bearing has a flow rate of 6000 mm3. Neglecting side leakage if temperature rises to 10 oC and specific heat is 1.55 kJ/kg oC, what is the rate of heat dissipated in the bearing?
- Published on 19 Oct 15

a. 7.4 W
b. 236 W
c. 0.236 kW
d. 0.0744 kW

ANSWER: 0.0744 kW
 

    Discussion

  • Sravanthi   -Posted on 05 Oct 15
    Given: Mass density of lubricating oil (ρ) = 800 kg / m3 , Flow rate (Q) = 6000 mm3 / s, ΔT =10oC, specific heat of lubricant (Cp) = 1.55 kJ/kgoC

    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

    When leakage of lubricant is neglected, rate of heat dissipated is calculated neglecting Qs and the formula is modified as shown below,

    Pf = {ρ Q Cp ΔT [1 – 0]} / 109
    i.e Pf = (ρ Q Cp ΔT ) / 109 ------- (1)

    Therefore, substituting the given values in equation (1) we get,

    Pf = (ρ Q Cp ΔT ) / 109
    Pf = ( 800 x 6000 x 1.55 x 10) / 109
    Pf = 0.0744 kW

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