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The speed of a journal in a journal bearing is 1500 r.p.m. The absolute viscosity of the lubricant is 1.2 kg-m/s and the bearing pressure on the projected bearing area is 1N/m2. Determine the bearing characteristics number
1. 30
2. 25
3. 15
4. 20

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Correct Answer - Option 1 : 30

Concept:

The bearing characteristic number is calculated by,

\(\frac{{μ N_s}}{P}=\rm Bearing~characteristic~number\)

Where, μ = absolute viscosity of lubricant (kg/m.s), N = speed of the journal (rps), P = bearing pressure (N/mm2)

Calculation:

Given:

μ = 1.2 kg-m/s, N = 1500 r.p.m, P = 1 N/m2

\(\rm Bearing~characteristic~number=\frac{{μ N_s}}{P}\;\)

\(= \frac{1.2\times1500}{60\times1} = 30\;\)

The bearing characteristics number is 30.

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