In a journal bearing, the diameter of the journal is 0.15 m, its speed is 900 r.p.m. and the load on the bearing is 40 kN. Considering μ = 0.0072, the heat generated will be nearly

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ESE Mechanical 2018 Official Paper

Answer 1

Concept:

The heat generated in journal bearing is given by,

H_g= μ × W × V

Where, μ = coefficient of friction, W= load on the bearing,

V = linear speed of bearing = \(\frac{{\pi \times D \times N}}{{60}}\) where D is the diameter of the journal and N is the speed in r.p.m.

Calculation:

Given, D = 0.15m, N = 900 r.p.m. W = 40 kN, μ = 0.0072

Now, \(V = \frac{{\pi \times 0.15 \times 900}}{{60}} = 7.07\;m/s\)

H_g= μ × W × V = 0.0072 × 40 × 7.07 = 2.036 kW