Torque and power loss in a shaft bearing

Given data

$D = 0.1\ \text{m}$, $L = 0.15\ \text{m}$, $c = 0.2\ \text{mm} = 2\times10^{-4}\ \text{m}$, $N = 300\ \text{rpm}$, $\mu = 0.08\ \text{Pa s}$.

Since the clearance is very small compared with the diameter, the oil film can be treated as flat with a linear velocity profile, so $du/dy = U/c$.

Surface speed of the shaft

$$U = \frac{\pi D N}{60} = \frac{\pi \times 0.1 \times 300}{60} = 1.571\ \text{m/s}$$

Shear stress

$$\tau = \mu \frac{U}{c} = 0.08 \times \frac{1.571}{2\times10^{-4}} = 628.3\ \text{Pa}$$

Resisting torque

Shear area $A = \pi D L = \pi \times 0.1 \times 0.15 = 0.04712\ \text{m}^2$

$$\begin{aligned} F &= \tau A = 628.3 \times 0.04712 = 29.61\ \text{N} \\ T &= F \times \frac{D}{2} = 29.61 \times 0.05 = 1.480\ \text{N m} \end{aligned}$$

Power lost

$$\begin{aligned} \omega &= \frac{2\pi N}{60} = 31.42\ \text{rad/s} \\ P &= T\omega = 1.480 \times 31.42 = 46.5\ \text{W} \end{aligned}$$

Answer: $\tau = 628\ \text{Pa}$, $T = 1.48\ \text{N m}$, $P = 46.5\ \text{W}$