External flow: the boundary layer decides the drag
引入| English | 释义 | 拼音 |
|---|---|---|
| boundary layer | 边界层 | biān jiè céng |
| skin friction | 摩擦阻力 | mó cā zǔ lì |
| pressure drag | 压差阻力 | yā chà zǔ lì |
| flow separation | 流动分离 | liú dòng fēn lí |
| Strouhal number | 斯特劳哈尔数 | sī tè láo hā ěr shù |
The wind on the vent pipe
- Wind on an exposed pipe and drag on a vehicle are external flows: a body sits in a stream. At the wall the fluid does not slip, so a thin boundary layer 边界层 forms, where the speed rises from zero to the free stream.
- On a flat plate the laminar layer grows as $\delta = 5x/\sqrt{Re_x}$ and turns turbulent near $Re_x \approx 5 \times 10^5$.
The fluid slides along a solid wall at the free-stream speed.
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No slip: the speed at the wall is zero.
Two kinds of drag
- Skin friction 摩擦阻力 is shear on the surface, and it uses the wetted area. Pressure drag 压差阻力 comes from flow separation 流动分离 behind blunt bodies, and it uses the frontal area.
- Both are summed into $F_D = C_D\cdot\tfrac12\rho U^2A$; lift is $F_L = C_L\cdot\tfrac12\rho U^2A$.
Skin friction or pressure drag?
Decide which kind of drag dominates each body.
Skin-friction drag is calculated with:
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Shear acts over every wetted surface.
The plate in water
- A plate 0.5 m long and 0.3 m wide in water at 0.4 m/s: $Re_L = \mathbf{1.99 \times 10^5}$, laminar throughout.
- Edge thickness $\delta = 5 \times 0.5/\sqrt{1.99 \times 10^5} = \mathbf{5.6}$ mm.
- Blasius $C_f = 1.328/\sqrt{Re_L} = 0.00298$, so one side carries $\mathbf{0.036}$ N. A turbulent layer would more than double it.
Plate 0.5 m long in water at 0.4 m/s, nu = 1.004e-6 (in units of 10^5, two decimals)?
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UL over nu.
Its laminar boundary-layer thickness at the trailing edge (one decimal, mm)?
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5x over root Re.
Blasius C_f at that Re (five decimals)?
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1.328 over root Re_L.
The vent pipe in wind
- A 0.1 m pipe, 3 m tall, in a 15 m/s wind: $Re = \mathbf{9.89 \times 10^4}$, where a cylinder has $C_D \approx 1.2$ because separation dominates.
- $F = 1.2 \times 0.5 \times 1.204 \times 225 \times 0.3 = \mathbf{48.8}$ N, and the base bending moment is $\mathbf{73.1}$ N·m: a load for the bending check.
Drag on the 0.1 m by 3 m vent pipe in 15 m/s wind, C_D = 1.2 (one decimal, N)?
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C_D times half rho U squared times the frontal area 0.3 m2.
Base bending moment if the load acts at mid-height (one decimal, N m)?
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48.8 N times 1.5 m.
The drag crisis
- Near $Re \approx 3 \times 10^5$ a cylinder's boundary layer turns turbulent before it separates. It clings further round the back, the wake narrows, and $C_D$ falls from about 1.2 to about 0.3.
- For the 0.1 m vent pipe that needs $U = 3 \times 10^5 \times 1.516 \times 10^{-5}/0.1 = \mathbf{45.5}$ m/s, a storm. A golf ball's dimples trigger the same transition at a lower speed.
Wind speed at which the 0.1 m pipe reaches $Re = 3 \times 10^5$ in air (one decimal, m/s)?
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$U = Re\,\nu/D = 3 \times 10^5 \times 1.516 \times 10^{-5}/0.1 = 45.5$ m/s.
Past the drag crisis a smooth cylinder's drag coefficient rises.
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It falls, from about 1.2 to about 0.3: the turbulent layer separates later and the wake narrows.
Worked example: the pipe that sings
- Behind the cylinder, vortices shed alternately from each side at $f = St\,U/D$, with Strouhal number 斯特劳哈尔数 $St \approx 0.2$.
- At 15 m/s: $f = 0.2 \times 15/0.1 = \mathbf{30}$ Hz of side-to-side forcing. If the pipe's own natural frequency is near 30 Hz, it resonates. Helical strakes on chimneys break up that shedding.
Vortex shedding from the 0.1 m pipe in 15 m/s wind, $St = 0.2$ (Hz)?
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$f = 0.2 \times 15/0.1 = 30$ Hz.
Three drag slips
- Frontal area for skin friction. Skin friction uses the wetted area.
- A constant $C_D$ across the drag crisis. It drops sharply near $Re = 3 \times 10^5$.
- A flow picture without its $Re$. The same body looks different at another $Re$.
Recap
- No slip makes the boundary layer; laminar $\delta = 5x/\sqrt{Re_x}$, 5.6 mm at the plate's edge.
- $C_f = 1.328/\sqrt{Re_L}$; blunt bodies are dominated by separation.
- $F = C\cdot\tfrac12\rho U^2A$: 48.8 N and 73.1 N·m on the vent pipe.