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/m³, and the kinematic viscosity is 1.25 × 10^-4 m²/s. The average shear stress in the fluid is ______ Pa (round off to the nearest integer).

The correct answer is 80.

Step-by-Step Explanation:

Let us analyze the given parameters for the viscous flow between two large parallel horizontal plates:
- Separation between plates (distance), $h=5\text{mm}=5\times {10}^{-3}\text{m}$
- Velocity of the top plate, $U=4\text{m/s}$
- Fluid density, $\rho =800{\text{kg/m}}^3$
- Kinematic viscosity of the fluid, $\nu =1.25\times {10}^{-4}{\text{m}}^2\text{/s}$
- Pressure gradient in the flow direction, $\frac{dp}{dx}=0$

1. Understanding the Flow Type
Since the flow is steady, fully developed, laminar, and incompressible between two parallel plates where the upper plate moves with a constant velocity and there is no pressure gradient, this represents simple Couette flow.
For simple Couette flow, the velocity profile $u\left(y\right)$ varies linearly with the vertical coordinate $y$:

$u\left(y\right)=\frac{Uy}{h}$

Thus, the velocity gradient is constant throughout the fluid gap:

$\frac{du}{dy}=\frac{U}{h}$

2. Finding the Dynamic Viscosity ($\mu$)
Dynamic viscosity ($\mu$) is related to kinematic viscosity ($\nu$) and density ($\rho$) by the relation:

$\mu =\rho \times \nu$

Substituting the given values:

$\mu =800\times (1.25\times {10}^{-4})=0.1\text{N·s/}{\text{m}}^2$

3. Calculating the Shear Stress ($\tau$)
According to Newton's law of viscosity, the shear stress in the fluid is given by:

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

Substituting the values of $\mu$, $U$, and $h$:

$\tau =0.1\times \frac{4}{5\times {10}^{-3}}$

$\tau =0.1\times 800=80\text{Pa}$

Since the shear stress is uniform across the entire channel height in simple Couette flow, the average shear stress is also equal to this constant value of 80 Pa.