Hydrostatic Balance

Hydrostatic Balance#

The Navier-Stokes equations in three-dimensional Cartesian coordinates for a Newtonian fluid are expressed as follows:

General Form (Momentum Equation)#

\[ \rho \left( \frac{\partial \mathbf{V}}{\partial t} + \mathbf{V} \cdot \nabla \mathbf{V} \right) = -\nabla p + \mu \nabla^2 \mathbf{V} + \mathbf{f} \]

where:

  • \(\mathbf{V} = (u, v, w)\) is the velocity vector,

  • \(\rho\) is the fluid density,

  • \(p\) is the pressure,

  • \(\mu\) is the dynamic viscosity,

  • \(\mathbf{f}\) is the body force per unit volume, such as gravity.

Assumptions for Hydrostatic Balance#

To derive hydrostatic balance:

  1. Steady-state condition: \(\frac{\partial \mathbf{V}}{\partial t} = 0\) (no time-dependent changes).

  2. No motion: \(\mathbf{V} = 0\) (fluid is at rest, so no convective acceleration).

  3. Negligible viscous forces: \(\mu \nabla^2 \mathbf{V} = 0\) (dominance of pressure and body forces over viscous effects).

With these assumptions, the Navier-Stokes equations simplify to:

\[ -\nabla p + \mathbf{f} = 0 \]

Incorporating Gravitational Force#

The dominant body force in many scenarios is gravity, \(\mathbf{f} = -\rho g \mathbf{j}\), where \(\mathbf{j}\) is the unit vector in the vertical direction (\(z\)). Substituting:

\[ -\nabla p - \rho g \mathbf{j} = 0 \]

In component form, this becomes:

\[ \frac{\partial p}{\partial x} = 0, \quad \frac{\partial p}{\partial y} = 0, \quad \frac{\partial p}{\partial z} = -\rho g \]

The pressure only varies in the \(z\) direction, leading to:

\[ \frac{\partial p}{\partial z} = -\rho g \]

This is the equation of hydrostatic balance.

Integration of Hydrostatic Balance#

1. Incompressible Fluid (Liquids)#

For an incompressible fluid, \(\rho\) is constant. Integrating \(\frac{\partial p}{\partial z} = -\rho g\):

\[ p(z) = p_0 - \rho g z \]

where:

  • \(p_0\) is the pressure at a reference height \(z = 0\).

This is the well-known hydrostatic pressure formula for liquids.

2. Compressible Fluid (Atmosphere)#

For a compressible fluid, \(\rho\) varies with pressure and temperature. Using the ideal gas law:

\[ \rho = \frac{p}{R T} \]

where:

  • \(R\) is the specific gas constant,

  • \(T\) is the temperature.

Substitute \(\rho\) into \(\frac{\partial p}{\partial z} = -\rho g\):

\[ \frac{\partial p}{\partial z} = -\frac{p g}{R T} \]

Assuming isothermal conditions (\(T\) is constant), we can integrate:

\[ \frac{1}{p} \frac{\partial p}{\partial z} = -\frac{g}{R T} \]

Integrating:

\[ \ln p = -\frac{g z}{R T} + \ln p_0 \]

Exponentiate to solve for \(p(z)\):

\[ p(z) = p_0 e^{-\frac{g z}{R T}} \]

where \(p_0\) is the pressure at \(z = 0\). This is the barometric pressure formula for the atmosphere.

Summary#

  • For incompressible fluids, hydrostatic pressure increases linearly with depth.

  • For compressible fluids, pressure decreases exponentially with height under isothermal conditions. These simplifications and integrations illustrate the core principles of hydrostatic balance for different fluid states.

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