Journal bearing design with film thickness check

Data: $d = 60$ mm, $l = 60$ mm, $r = 30$ mm, $c = 0.06$ mm, so $r/c = 500$; $N = 1500/60 = 25$ rev/s; $W = 4500$ N; $\mu = 0.02$ Pa.s.

Bearing pressure

$$p = \frac{W}{ld} = \frac{4500}{0.06 \times 0.06} = 1.25\ \text{MPa}$$

Sommerfeld number

$$S = \left(\frac{r}{c}\right)^2\frac{\mu N}{p} = 500^2 \times \frac{0.02 \times 25}{1.25\times10^6} = 0.10$$

Minimum film thickness

$$h_0 = 0.35\,c = 0.35 \times 0.06 = 0.021\ \text{mm}$$

Coefficient of friction

$$f = \frac{2.9}{r/c} = \frac{2.9}{500} = 0.0058$$

Friction torque and power loss

$$\begin{aligned} T &= f W r = 0.0058 \times 4500 \times 0.03 = 0.783\ \text{N.m} \\ P_{loss} &= 2\pi N T = 2\pi \times 25 \times 0.783 = 123\ \text{W} \end{aligned}$$

Oil flow

$$\begin{aligned} Q &= 3.6\,(r\,c\,N\,l) = 3.6 \times 0.03 \times 0.06\times10^{-3} \times 25 \times 0.06 \\ &= 9.72\times10^{-6}\ \text{m}^3/\text{s} = 9.72\ \text{cm}^3/\text{s} \end{aligned}$$

Trumpler's criterion

$$h_{0,min} = 0.005 + 0.00004 \times 60 = 0.0074\ \text{mm}$$

The actual $h_0 = 0.021$ mm is nearly three times this, so the film is thick enough to cover the surface roughness and small particles: the design is safe. (The oil temperature rise, from the heat generated, should still be checked.)

Answer: p = 1.25 MPa; S = 0.10; h0 = 0.021 mm (greater than 0.0074 mm); f = 0.0058; power loss = 123 W; oil flow = 9.7 cm$^3$/s.