Question

Consider fully developed, steady state incompressible laminar flow of a viscous fluid between two large parallel horizontal plates. The bottom plate is fixed and the top plate moves with a constant velocity of U = 4 m/s. Separation between the plates is 5 mm. There is no pressure gradient in the direction of flow. The density of fluid is 800 kg/m$^3$, and the kinematic viscosity is \(1.25 \times 10^{-4}\) m$^2$/s. The average shear stress in the fluid is _________ Pa (round off to the nearest integer).

Hint:

In laminar flow between two parallel plates, the shear stress is calculated using the formula \(\tau = \frac{\mu U}{H}\), where \(\mu\) is dynamic viscosity.

Answer 1

Correct Answer: 79

For fully developed laminar flow between two parallel plates, the average shear stress \(\tau\) is given by the formula: \[ \tau = \frac{\mu U}{H}, \] where:
- \(\mu = \rho \nu\) is the dynamic viscosity,
- \(\rho = 800 \, \text{kg/m}^3\) is the density of the fluid,
- \(\nu = 1.25 \times 10^{-4} \, \text{m}^2/\text{s}\) is the kinematic viscosity,
- \(U = 4 \, \text{m/s}\) is the velocity of the top plate,
- \(H = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m}\) is the separation between the plates.
The dynamic viscosity is: \[ \mu = 800 \times 1.25 \times 10^{-4} = 0.1 \, \text{Pa.s}. \] Thus, the average shear stress is: \[ \tau = \frac{0.1 \times 4}{5 \times 10^{-3}} = \frac{0.4}{5 \times 10^{-3}} = 80 \, \text{Pa}. \] Therefore, the average shear stress in the fluid is: \[ \boxed{79 \, \text{to} \, 81 \, \text{Pa}}. \]