Fluid statics

Fluid statics is the branch of fluid mechanics which studies fluids at rest.

Equations

Pressure variation in a static Fluid

Applying Newton's laws of motions to a fluid at rest, it can be determined that the sum of forces must equal zero throughout the fluid, which means any arbitrary element of the fluid is subjected to forces that sums to zero. Since the fluid is not deforming while it is at rest, the only forces acting on the fluid are those due to gravity and pressure.

For a given fluid element, the pressure is given by the differential equation

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho\mathbf{g} =\nabla P}

where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho} is the density of the fluid, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} is the gravitational acceleration (vector), and $P$ is the pressure at the fluid element.

For small changes in altitude, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} can be assumed to be constant (A more accurate description for large change in altitude can be found here). Simplifications can also be made for constant density, and reducing the analysis to only the y dimension:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P-P_0=\rho g\left( y-y_0\right)}

Examples

For system filled with an isothermal (constant temperature throughout) ideal gas, a relation between pressure altitude can be established using the following method: For ideal gases, the density is given by

$\rho ={\frac {PM}{RT}}$

Where M is the molar mass, R is the ideal gas constant and T is the temperature (Absolute temperature, in Kelvin or Rankine) of the gas. we can then set up the equation as follows:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dP}{dy}=-\rho g=-\frac{PMg}{RT}}

Note that we are using the scalar value of gravity (g), so the minus sign is included due to gravity is downwards, in the negative direction of the y-axis. using the method of separation of variables, we can rearrange the equation so

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dP}{P}=-\frac{Mg}{RT}dy}

Integrating both sides gives

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln\frac{P}{P_0}=-\frac{Mg\left(y-y_0\right)}{RT}}

Where P₀ is the reference pressure at point y₀ (often taken at the point which P₀ is the atmospheric pressure).

Rearranging gives

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P=P_0 \exp{\left(-\frac{Mg\left(y-y_0\right)}{RT}\right)}}