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Hydrostatics and pressure

Pressure in a fluid at rest, forces on dams, Archimedes' principle and the exponential atmosphere.

If the fluid is at rest, \(\mathbf{u} = 0\), Cauchy's equation with \(\sigma = -pI\) and gravity \(\mathbf{g} = -g\,\mathbf{e}_z\) reduces to \[ \nabla p = \rho\mathbf{g}, \qquad \frac{\partial p}{\partial z} = -\rho g, \quad \frac{\partial p}{\partial x} = \frac{\partial p}{\partial y} = 0. \] Pressure depends only on height and grows with depth. For a liquid of constant density, a depth \(d\) below a free surface at atmospheric pressure \(p_0\) has \(p = p_0 + \rho g d\). Ten metres of water add \(1000\times9.81\times10 = 98\,100\) Pa, almost exactly one more atmosphere.

Forces on submerged surfaces are integrals of pressure. A vertical dam of width \(W\) holding water of depth \(H\) feels a gauge pressure \(\rho g(H-z)\) at height \(z\), so the total horizontal force is \[ F = \int_0^H \rho g (H - z)\,W\,dz = \tfrac12\rho g W H^2. \] It grows with the square of the depth, which is why dams are thick at the bottom. The force acts at one third of the depth from the bottom, not at the middle.

Archimedes' principle comes from the divergence theorem. The net pressure force on a submerged body occupying \(V\) is \[ -\oint_{\partial V} p\,\mathbf{n}\,dS = -\int_V \nabla p\,dV = -\int_V \rho_{\text{fluid}}\,\mathbf{g}\,dV = \rho_{\text{fluid}}\,g\,|V|\,\mathbf{e}_z, \] an upward force equal to the weight of the displaced fluid. A floating body sinks until the displaced weight matches its own, so wood of density 800 floats with four fifths of its volume under water.

For a gas the density depends on pressure. An ideal gas at constant temperature has \(\rho = pM/(RT)\), so \(dp/dz = -p/H\) with scale height \(H = RT/(Mg)\), and \[ p(z) = p_0\,e^{-z/H}. \] At \(288\) K, \(H \approx 8.4\) km: the pressure halves roughly every \(5.8\) km of altitude. Finally, taking the curl of \(\nabla p = \rho\mathbf{g}\) gives \(\nabla\rho\times\mathbf{g} = 0\): equilibrium is only possible when surfaces of constant density are horizontal. A tilted density layer cannot sit still; it must flow.

Picture it: a column of fluid above each point, whose weight per unit area is the pressure. Think it: hydrostatics is the statement that \(\rho\mathbf{g}\) is a gradient; when it is not (tilted density), no pressure can balance it and motion is forced. This is the seed of every buoyancy-driven flow.

Ohatra · 101325 + 1000*9.81*10

Evaluate 101325 + 1000*9.81*10

199425.0

Dingana amin'ny dingana

  1. 199425.0 = 98100 + 101325

    Multiply: 1000·9.81·10 = 98100.

  2. 98100 + 101325 = 199425

    Add: 98100 + 101325 = 199425.

Asehoy ny valinteny
199425.0

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Ahoana ny fomba: Hydrostatics and pressure

  1. Choose z upward and write dp/dz = -rho g.
  2. For a liquid of constant density integrate once: p = p0 + rho g (depth).
  3. For a force on a wall, integrate pressure times width over the wetted height.
  4. For a gas, use the equation of state to write rho in terms of p and solve the resulting ODE.

Fanontaniana napetrak'ireo olona

Why does the shape of the container not matter?

Because pressure depends only on depth: the hydrostatic paradox. A narrow tube and a wide tank filled to the same height have the same pressure at the bottom.

Is the buoyant force the same at every depth?

For an incompressible liquid and a rigid body, yes, because only the difference of pressure between top and bottom matters, and that depends only on the height of the body.

What do I need before starting fluid dynamics?

Multivariable calculus (divergence, curl, the divergence and Stokes theorems), linear algebra (symmetric matrices and eigenvalues) and partial differential equations (the heat and Laplace equations). The last lessons also use Sobolev spaces, which are introduced where they are needed.

What are the Navier-Stokes equations in one sentence?

Newton's second law for each particle of a viscous incompressible fluid: acceleration equals the pressure force plus viscous diffusion of momentum, with the constraint that the velocity field is divergence-free.

Why is two-dimensional flow easier than three-dimensional flow?

In 2D the vorticity is a scalar that is only carried and diffused, so its maximum never grows; in 3D vortex lines can be stretched, which amplifies vorticity, and no known bound rules out unlimited growth.

Is this course physics or mathematics?

Both, in order. The first half derives the equations from physical principles and solves classical flows; the second half treats the equations as mathematical objects and studies which of their properties can be proved.

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