# Difference between revisions of "Fluid statics"

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For ideal gases, the density is given by | For ideal gases, the density is given by | ||

:<math>\rho =\frac{PM}{RT}</math> | :<math>\rho =\frac{PM}{RT}</math> | ||

− | 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. | + | Where <math>M</math> is the molar mass, <math>R</math> is the ideal gas constant and <math>T</math> is the [[temperature]] (Absolute temperature, in Kelvin or Rankine) of the gas. |

we can then set up the equation as follows: | we can then set up the equation as follows: | ||

+ | |||

:<math>\frac{dP}{dy}=-\rho g=-\frac{PMg}{RT}</math> | :<math>\frac{dP}{dy}=-\rho g=-\frac{PMg}{RT}</math> | ||

− | 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. | + | |

+ | Note that we are using the scalar value of gravity (<math>g</math>), 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 | using the method of [[separation of variables]], we can rearrange the equation so | ||

:<math>\frac{dP}{P}=-\frac{Mg}{RT}dy</math> | :<math>\frac{dP}{P}=-\frac{Mg}{RT}dy</math> | ||

+ | |||

Integrating both sides gives | Integrating both sides gives | ||

+ | |||

:<math>\ln\frac{P}{P_0}=-\frac{Mg\left(y-y_0\right)}{RT}</math> | :<math>\ln\frac{P}{P_0}=-\frac{Mg\left(y-y_0\right)}{RT}</math> | ||

− | Where | + | |

+ | Where <math>P_0</math> is the reference pressure at point <math>y_0</math> (often taken at the point which <math>P_0</math> is the atmospheric pressure). | ||

Rearranging gives | Rearranging gives | ||

:<math>P=P_0 \exp{\left(-\frac{Mg\left(y-y_0\right)}{RT}\right)}</math> | :<math>P=P_0 \exp{\left(-\frac{Mg\left(y-y_0\right)}{RT}\right)}</math> | ||

+ | |||

==References== | ==References== | ||

{{Reflist}} | {{Reflist}} |

## Latest revision as of 09:16, 19 December 2016

**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^{[1]}

where is the density of the fluid, is the gravitational acceleration (vector), and is the pressure at the fluid element.

For small changes in altitude, 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:

## 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

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

Note that we are using the scalar value of gravity (), 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

Integrating both sides gives

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

Rearranging gives

## References

- ↑ James R. Welty, Charles E. Wicks, Robert E. Wilson, and Gregory L. Rorrer,
*Fundamentals of Momentum, Heat, and Mass Transfer*, 4th Ed. Toronto: John Wiley & Sons Inc, 2001.