A full journal bearing having clearance to radius ratio of 1/100, using a lubricant with absolute viscosity μ  = 28 × 10^-3 Pas, supports the shaft journal running at N= 2400 rpm. If the bearing pressure is 1.4 MPa, then the Somerfield number is:

Explanation:

Concept:

Somerfield number is the dimensionless number which is used in hydrodynamic lubrication analysis. It remains constant for a given bearing.

\(S=\frac{\mu{n}}{p}(\frac{r}{c})^2\)

Where, S = Somerfield number, \(\mu\) = absolute viscosity (Pas), n = speed (rps), p = bearing pressure (Pa), r = radius of journal bearing (m), c = radial clearance (m).

Calculation:

Given:

\(\frac{c}{r}=\frac{1}{100}\)

\(\mu\) = 28 \(\times\) 10^-3 Pas

N = 2400 rpm

p = 1.4 MPa

So, Somerfield number is:

\(S=\frac{0.028\times2400}{1.4\times10^6\times60}(\frac{100}{1})^2\) = 8\(\times\)10^-3.

Thus, option (2) is the correct answer.