Question

Water flows around a thin flat plate (0.25 m long, 2 m wide) with a free stream velocity ($U_\infty$) of 1 m/s, as shown in the figure. Consider linear velocity profile $\left(\frac{u}{U_\infty} = \frac{y}{\delta}\right)$ for which the laminar boundary layer thickness is expressed as $\delta = \frac{3.5x}{\sqrt{Re_x}}$. For water, density = 1000 kg/m$^3$ and dynamic viscosity = 0.001 kg/m·s. Net drag force (in N, rounded off to two decimal places) acting on the plate, neglecting the end effects, is $________________$.

Hint:

For flat plate drag with assumed velocity profile, compute shear stress using $\tau = \mu U/\delta$, then integrate across the plate length. Don’t forget both sides of the plate contribute.

Answer 1

Correct Answer: 1.1

Step 1: Write given data.
Plate length = $L = 0.25$ m
Plate width = $b = 2$ m
Velocity = $U_\infty = 1$ m/s
$\rho = 1000$ kg/m$^3$, $\mu = 0.001$ kg/m·s
Step 2: Shear stress expression.
For linear profile: \[ \tau = \mu \frac{\partial u}{\partial y} \bigg|_{y=0} = \mu \frac{U_\infty}{\delta} \]
Step 3: Boundary layer thickness.
\[ \delta = \frac{3.5x}{\sqrt{Re_x}}, Re_x = \frac{\rho U_\infty x}{\mu} \] \[ \delta = \frac{3.5x}{\sqrt{\frac{1000 \times 1 \times x}{0.001}}} = \frac{3.5x}{\sqrt{10^6 x}} \] \[ = \frac{3.5}{1000} \sqrt{x} = 0.0035 \sqrt{x} \]
Step 4: Local shear stress.
\[ \tau(x) = \frac{\mu U_\infty}{\delta} = \frac{0.001 \times 1}{0.0035\sqrt{x}} = \frac{0.2857}{\sqrt{x}} \ \text{Pa} \]

Step 5: Drag force.
\[ F = \int_0^L \tau(x) b \, dx \] \[ F = b \int_0^{0.25} \frac{0.2857}{\sqrt{x}} dx \] \[ = 2 \times 0.2857 \times [2\sqrt{x}]_0^{0.25} \] \[ = 0.5714 \times (2 \times 0.5) = 0.5714 \] \[ F = 0.5714 \ \text{N per surface} \] Since both sides of the plate are exposed: \[ F_{net} = 2 \times 0.0714 = 0.14 \ \text{N} \] Final Answer: \[ \boxed{0.14 \ \text{N}} \]