A plate of 500 mm × 200 mm has been placed longitudinally in a stream of an oil which flows with a velocity of 1.0 m/s. The kinematic viscosity of the oil was 0.5 stoke. The thickness of the boundary layer at the trailing edge is _______.
25 mm
The question asks us to determine the thickness of the boundary layer at the trailing edge of a flat plate placed in a stream of oil. The boundary layer is the thin layer of fluid near the surface of the plate where viscous effects are significant.
To solve this problem, we need to use the concept of the Reynolds number to determine the flow regime (laminar or turbulent) and then apply the appropriate formula for boundary layer thickness.
The kinematic viscosity is given in stokes. We need to convert it to standard SI units (m²/s).
We know that:
Therefore,
\(\nu = 0.5 \, \text{stoke} = 0.5 \times 10^{-4} \, \text{m}^2/\text{s}\)
The Reynolds number (\(Re\)) is a dimensionless quantity that helps predict flow patterns. For flow over a flat plate, it is calculated as:
\(Re_x = \frac{U \times x}{\nu}\)
where x is the distance from the leading edge.
We need the boundary layer thickness at the trailing edge, which is at \(x = L = 0.5 \, \text{m}\). So, we calculate the Reynolds number at the trailing edge:
\(Re_L = \frac{U \times L}{\nu}\)
\(Re_L = \frac{1.0 \, \text{m/s} \times 0.5 \, \text{m}}{0.5 \times 10^{-4} \, \text{m}^2/\text{s}}\)
\(Re_L = \frac{0.5}{0.5 \times 10^{-4}}\)
\(Re_L = 10^4\)
For flow over a flat plate, the transition from laminar to turbulent flow typically occurs around a critical Reynolds number of \(Re_{crit} \approx 5 \times 10^5\). Since our calculated \(Re_L = 10^4\), which is much less than \(5 \times 10^5\), the flow over the entire plate is laminar.
For a laminar boundary layer over a flat plate, the thickness (\(\delta\)) at a distance x from the leading edge is given by the Blasius solution (often approximated as):
\(\delta(x) = \frac{5x}{\sqrt{Re_x}}\)
We want the thickness at the trailing edge, so we use \(x = L\):
\(\delta_L = \frac{5L}{\sqrt{Re_L}}\)
Substituting the values:
\(\delta_L = \frac{5 \times 0.5 \, \text{m}}{\sqrt{10^4}}\)
\(\delta_L = \frac{2.5}{\sqrt{10000}}\)
\(\delta_L = \frac{2.5}{100}\)
\(\delta_L = 0.025 \, \text{m}\)
The options are given in millimeters, so we convert the thickness from meters to millimeters:
\(\delta_L = 0.025 \, \text{m} \times \frac{1000 \, \text{mm}}{1 \, \text{m}}\)
\(\delta_L = 25 \, \text{mm}\)
| Step | Description | Calculation | Result |
|---|---|---|---|
| 1 | Convert kinematic viscosity | \(0.5 \, \text{stoke} = 0.5 \times 10^{-4} \, \text{m}^2/\text{s}\) | \(0.5 \times 10^{-4} \, \text{m}^2/\text{s}\) |
| 2 | Calculate Reynolds number at trailing edge | \(Re_L = \frac{1.0 \times 0.5}{0.5 \times 10^{-4}}\) | \(10^4\) |
| 3 | Confirm flow regime | \(Re_L < 5 \times 10^5\) | Laminar |
| 4 | Calculate boundary layer thickness (m) | \(\delta_L = \frac{5 \times 0.5}{\sqrt{10^4}}\) | \(0.025 \, \text{m}\) |
| 5 | Convert thickness to mm | \(0.025 \, \text{m} \times 1000 \, \text{mm/m}\) | \(25 \, \text{mm}\) |
Based on the calculations, the thickness of the boundary layer at the trailing edge of the plate is 25 mm.
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