Fluid statics: the gate, the gauge and the oil film
Diperkenalkan| English | Makna |
|---|---|
| centre of pressure | The point where the resultant pressure force on a submerged surface acts, always below the centroid on a vertical wall. |
| viscosity/vɪˈskɒsɪti/ | A measure of a fluid's resistance to flow or to relative motion between adjacent layers. |
| kinematic viscosity | Dynamic viscosity divided by density, measured in square metres per second, the form used in the Reynolds number. |
| Reynolds number/ˈrenəldz ˈnʌmbə/ | The dimensionless ratio of inertial to viscous forces, ρVD/μ, which decides whether a flow is laminar or turbulent. |
| buoyancy/ˈbɔɪənsi/ | The upward force a fluid exerts on an immersed body, equal to the weight of the fluid displaced. |
Why the latch sits low
- A tank gate does not feel its load at its middle. Pressure in a fluid at rest grows with depth, $p = p_0 + \rho g h$, so the lower half of the gate is pushed harder than the upper half.
- The resultant is $F = \rho g h_c A$, the pressure at the centroid depth times the area, but it acts at the centre of pressure 压力中心, which is always below the centroid. That is where the latch must hold.
Gauge pressure 2.5 m below a water surface (one decimal, kPa)?
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rho g h = 998 x 9.81 x 2.5.
On a submerged vertical gate, the centre of pressure is:
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Pressure grows with depth, so the resultant sits lower than the centroid.
The tank gate
- A vertical gate $1.2\ \mathrm{m}$ wide and $2\ \mathrm{m}$ high has its top edge $1.5\ \mathrm{m}$ below the surface, so $h_c = 2.5\ \mathrm{m}$ and $A = 2.4\ \mathrm{m^2}$.
- $F = 998 \times 9.81 \times 2.5 \times 2.4 = \mathbf{58.7}$ kN.
- With $I_c = bh^3/12 = 0.8\ \mathrm{m^4}$: $y_{cp} = y_c + I_c/(y_c A) = \mathbf{2.633}$ m, which is 133 mm below the centroid.
Force on the 1.2 m by 2 m gate with its centroid 2.5 m deep (one decimal, kN)?
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rho g h_c A.
Its centre-of-pressure depth (three decimals, m)?
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y_c + I_c/(y_c A) = 2.5 + 0.8/6.
A height becomes a pressure
- A mercury U-tube (SG 13.6) across a fitting reads $\Delta h = 120$ mm. The water above the mercury cancels on both sides except over $\Delta h$, so $\Delta p = (\rho_m - \rho) g \Delta h = \mathbf{14.8}$ kPa.
- Using $\rho_m$ alone forgets that water column and overstates the reading by about 8%.
Which statics rule answers it?
Route each question to the rule that answers it.
Mercury manometer, dh = 120 mm, water above it: pressure difference (one decimal, kPa)?
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(13.6 - 1) x 998 x 9.81 x 0.12.
When the fluid moves: viscosity 黏度
- Statics ignores viscosity, but motion does not: $\tau = \mu\,du/dy$. A plate sliding at $2\ \mathrm{m/s}$ on a $0.5\ \mathrm{mm}$ oil film with $\mu = 0.29\ \mathrm{Pa\,s}$ feels $\tau = \mathbf{1160}$ Pa.
- The kinematic viscosity 运动黏度 $\nu = \mu/\rho$ is the one that appears in Reynolds numbers 雷诺数: $1.004 \times 10^{-6}\ \mathrm{m^2/s}$ for water.
Plate at 2 m/s on a 0.5 mm oil film, mu = 0.29 Pa s: shear stress (Pa)?
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mu times the velocity gradient 2/0.0005.
The same law floats things
- Pressure that grows with depth pushes harder on a body's underside than its top. The net is the buoyancy 浮力 force, $\rho g V$: the weight of the displaced fluid (Archimedes).
- A 0.2 m³ float of 50 kg, held fully under water, feels $998 \times 9.81 \times 0.2 = 1958$ N up against 491 N of weight: its tether must hold $\mathbf{1468}$ N.
A 0.2 m³, 50 kg float held under water: tether tension (N, nearest whole)?
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Buoyancy $998 \times 9.81 \times 0.2 = 1958$ N; minus weight $50 \times 9.81 = 491$ N gives 1468 N.
Worked example: hinge on top, latch below
- Hinge the 2 m gate along its top edge, 1.5 m deep, and latch it at the bottom. The 58.7 kN acts $2.633 - 1.5 = 1.133$ m below the hinge.
- Moments about the hinge: $R_{latch} \times 2 = 58.7 \times 1.133$, so $R_{latch} = \mathbf{33.3}$ kN, and the hinge takes the other 25.5 kN. The centre of pressure is what loads the latch more.
The gate hinged at its top edge (1.5 m deep), latched at the bottom: latch force (one decimal, kN)?
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Moments about the hinge: $58.7 \times (2.633 - 1.5)/2 = 33.3$ kN.
The hinge and the latch each carry half of the 58.7 kN.
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The force acts below the centroid, at the centre of pressure, so the bottom latch carries more: 33.3 kN against 25.5 kN.
Three statics slips
- The force at the centroid. It acts lower, at the centre of pressure.
- Gauge and absolute in one equation. Choose one reference and keep it.
- The missing water column. A manometer reads $\rho_m - \rho$, not $\rho_m$.
A fluid at rest can carry a shear stress.
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With no motion there is no velocity gradient, so pressure acts equally in all directions.
Recap
- $p = p_0 + \rho g h$; $F = \rho g h_c A = 58.7$ kN on the gate, acting at $y_{cp} = 2.633$ m.
- $\Delta p = (\rho_m - \rho) g \Delta h = 14.8$ kPa for a 120 mm mercury reading.
- $\tau = \mu\,du/dy$; $\nu = \mu/\rho$.