Question

Driven by a pressure gradient of 100 kPa/m, a fluid of dynamic viscosity 0.1 Pa·s flows between two fixed infinitely large parallel plates under steady, incompressible, and fully developed laminar conditions. The average velocity of the flow is 2 m/s. The gap between the parallel plates in mm (rounded off to 2 decimal places) is ............

Hint:

In plane Poiseuille flow, average velocity is proportional to pressure gradient and square of plate spacing: \(V_{avg} \propto \tfrac{h^2}{\mu}\). Always verify whether $h$ is full gap or half gap.

Answer 1

Correct Answer: 4.8

Step 1: Recall formula for velocity profile.
For plane Poiseuille flow between two stationary parallel plates separated by gap $h$: \[ u(y) = \frac{1}{2\mu}\left(-\frac{dp}{dx}\right)\left(\frac{h^2}{4} - y^2\right). \] The average velocity is: \[ V_{avg} = \frac{h^2}{12\mu}\left(-\frac{dp}{dx}\right). \]

Step 2: Insert known values.
Given: \[ -\frac{dp}{dx} = 100 \,\text{kPa/m} = 100 \times 10^3 \,\text{Pa/m}, \mu = 0.1 \,\text{Pa·s}, V_{avg}=2 \,\text{m/s}. \]

Step 3: Solve for $h$.
\[ 2 = \frac{h^2}{12(0.1)} (100 \times 10^3). \] \[ 2 = \frac{h^2}{1.2}(100000). \] \[ 2 = \frac{100000}{1.2} h^2. \] \[ h^2 = \frac{2 \times 1.2}{100000} = \frac{2.4}{100000} = 2.4 \times 10^{-5}. \] \[ h = \sqrt{2.4 \times 10^{-5}} = 4.899 \times 10^{-3} \, \text{m}. \] So the gap between plates = $h = 4.90 \,\text{mm}$. But note: $h$ is the full gap, not half gap. Re-check formula: The correct formula is: \[ V_{avg} = \frac{h^2}{12\mu}\left(-\frac{dp}{dx}\right), \] where $h$ = full gap between plates. So result is correct: $h = 4.90 \,\text{mm}$. Final Answer:
\[ \boxed{4.90 \,\text{mm}} \]