Definition and Analysis method

(a) Fluid, Newtonian and non-Newtonian fluids

A fluid is a substance that deforms continuously under the action of a shear stress, however small. Liquids and gases are fluids.

A Newtonian fluid obeys Newton's law of viscosity: shear stress is directly proportional to the rate of shear strain (velocity gradient).

$$\tau = \mu \frac{du}{dy}$$

Effect of temperature:

• Liquids: viscosity decreases as temperature rises, because cohesive forces between molecules weaken.
• Gases: viscosity increases as temperature rises, because molecular momentum exchange between layers increases.

(b) Lagrangian and Eulerian methods

Fluid mechanics mostly uses the Eulerian method because it is easier to handle mathematically.

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