Question

A horizontal belt of negligible weight moves with velocity \(V = 2.5\,\text{m/s}\) over an oil film of depth \(h = 3\,\text{cm} = 0.03\,\text{m}\). The belt has length \(L = 2\,\text{m}\) and width \( b = 0.6\,\text{m} \). Find the viscosity of the oil (Pa·s) if minimum power required is 100 W. Neglect end effects. (Round off to two decimal places)

Hint:

For a moving plate over a viscous film, use Couette flow: \( \tau = \mu V/h \). Power equals shear force times velocity.

Answer 1

Correct Answer: 0.39

The shear stress on the moving belt is:
\[ \tau = \mu \frac{V}{h} \] Shear force on belt:
\[ F = \tau A = \mu \frac{V}{h}(Lb) \] Power required to move the belt:
\[ P = FV = \mu \frac{V}{h}(Lb)V \] Thus viscosity:
\[ \mu = \frac{Ph}{LbV^2} \] Substitute values:
\[ \mu = \frac{100 \times 0.03}{(2)(0.6)(2.5^2)} \] \[ \mu = \frac{3}{(1.2)(6.25)} = \frac{3}{7.50} = 0.40\,\text{Pa·s} \] Rounded to two decimals:
\[ \mu = 0.40\,\text{Pa·s} \]