Question

Air (density = 1.2 kg/m³, kinematic viscosity = 1.5×10⁻⁵ m²/s) flows over a flat plate with a free-stream velocity of 2 m/s. The wall shear stress at a location 15 mm from the leading edge is \(\tau_w\). What is the wall shear stress at a location 30 mm from the leading edge?

• A. \(\tau_w / 2\)
• B. \(\sqrt{2} \tau_w\)
• C. \(2 \tau_w\)
• D. \(\tau_w / \sqrt{2}\)

Hint:

Memorize the key dependencies for flat plate boundary layers. For laminar flow: boundary layer thickness \(\delta \propto \sqrt{x}\) and wall shear stress \(\tau_w \propto 1/\sqrt{x}\). For turbulent flow: \(\delta \propto x^{4/5}\) and \(\tau_w \propto 1/x^{1/5}\). Knowing these proportionalities allows for quick ratio calculations.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem involves boundary layer flow over a flat plate. We first need to determine the nature of the flow (laminar or turbulent) by calculating the Reynolds number. Then, we can use the appropriate relationship between wall shear stress (\(\tau_w\)) and the distance from the leading edge (\(x\)).
Step 2: Key Formula or Approach:
1. Calculate the Reynolds number (\(Re_x\)) to determine the flow regime: \[ Re_x = \frac{\rho u_\infty x}{\mu} = \frac{u_\infty x}{\nu} \] where \(u_\infty\) is the free-stream velocity, \(x\) is the distance from the leading edge, and \(\nu\) is the kinematic viscosity. Flow is considered laminar if \(Re_x<5 \times 10^5\).
2. For laminar flow over a flat plate (Blasius solution), the wall shear stress is given by: \[ \tau_w(x) = \frac{0.332 \rho u_\infty^2}{\sqrt{Re_x}} = 0.332 \rho u_\infty^2 \sqrt{\frac{\nu}{u_\infty x}} \] From this, we can see the key relationship: \[ \tau_w \propto \frac{1}{\sqrt{x}} \] Step 3: Detailed Calculation:
1. Check the flow regime: Let's calculate the Reynolds number at the farthest point, \(x = 30\) mm = 0.03 m. \[ Re_{x=0.03m} = \frac{(2 \text{ m/s})(0.03 \text{ m})}{1.5 \times 10^{-5} \text{ m}^2/\text{s}} = \frac{0.06}{1.5 \times 10^{-5}} = 4000 \] Since \(4000<5 \times 10^5\), the flow is laminar over the entire region of interest.
2. Apply the shear stress relationship:
We have the relation \(\tau_w \propto x^{-1/2}\). Let \(\tau_{w1}\) be the shear stress at \(x_1\) and \(\tau_{w2}\) be the shear stress at \(x_2\). \[ \frac{\tau_{w2}}{\tau_{w1}} = \left(\frac{x_2}{x_1}\right)^{-1/2} = \sqrt{\frac{x_1}{x_2}} \] Given:
- \(x_1 = 15\) mm
- \(x_2 = 30\) mm
- \(\tau_{w1} = \tau_w\)
We need to find \(\tau_{w2}\). \[ \frac{\tau_{w2}}{\tau_w} = \sqrt{\frac{15 \text{ mm}}{30 \text{ mm}}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \] Therefore, \[ \tau_{w2} = \frac{\tau_w}{\sqrt{2}} \] Step 4: Final Answer:
The wall shear stress at a location 30 mm from the leading edge is \(\tau_w / \sqrt{2}\).
Step 5: Why This is Correct:
The calculation confirms the flow is laminar. For laminar boundary layer flow, the wall shear stress is inversely proportional to the square root of the distance from the leading edge. Doubling the distance from 15 mm to 30 mm therefore reduces the shear stress by a factor of \(\sqrt{2}\).