# Bulk Modulus and Fluid Elasticity

## Introduction to - and definition of - Bulk Modulus Elasticity commonly used to characterize the compressibility of fluids.

The Bulk Modulus Elasticity - or Volume Modulus - is a material property characterizing the compressibility of a fluid - how easy a unit volume of a fluid can be changed when changing the pressure working upon it.

The Bulk Modulus Elasticity can be calculated as

K = - dp / (dV /)V_{0}

= - (p) / ((V_{1}- p_{0}) / V_{1}- V_{0}_{0}) (1)

where

K = Bulk Modulus of Elasticity (Pa, N/m^{2})

dp = differential change in pressure on the object (Pa, N/m^{2})

dV = differential change in volume of the object (m^{3})

V_{0}= initial volume of the object (m^{3})

p_{0}= initial pressure (Pa, N/m)^{2}

p_{1}= final pressure (Pa, N/m)^{2}V_{1}= final volume (m)^{3}

The Bulk Modulus Elasticity can alternatively be expressed as

K = dp / (dρ /)ρ_{0}

=(((p) /_{1}- p_{0}ρ-_{1}ρ_{0}) /ρ_{0}) (2)

where

dρ = differential change in density of the object (kg/m^{3})

= initial density of the objectρ_{0}(kg/m^{3})

= final density of the object (ρ_{1}kg/m^{3})

An increase in the pressure will decrease the volume *(1).* A decrease in the volume will increase the density *(2)*.

- The SI unit of the bulk modulus elasticity is
*N/m*^{2}(Pa) - The imperial (BG) unit is
*lb*_{f}/in^{2}(psi) *1 lb*_{f}/in^{2}(psi) = 6.894 10^{3}N/m^{2}(Pa)

A large Bulk Modulus indicates a relative incompressible fluid.

Bulk Modulus for some common fluids:

Fluid | Bulk Modulus - K - | |
---|---|---|

Imperial Units - BG(10 psi, lb^{5}_{f}/in^{2}) | SI Units(10 Pa, N/m^{9}^{2}) | |

Acetone | 1.34 | 0.92 |

Benzene | 1.5 | 1.05 |

Carbon Tetrachloride | 1.91 | 1.32 |

Ethyl Alcohol | 1.54 | 1.06 |

Gasoline | 1.9 | 1.3 |

Glycerin | 6.31 | 4.35 |

ISO 32 mineral oil | 2.6 | 1.8 |

Kerosene | 1.9 | 1.3 |

Mercury | 41.4 | 28.5 |

Paraffin Oil | 2.41 | 1.66 |

Petrol | 1.55 - 2.16 | 1.07 - 1.49 |

Phosphate ester | 4.4 | 3 |

SAE 30 Oil | 2.2 | 1.5 |

Seawater | 3.39 | 2.34 |

Sulfuric Acid | 4.3 | 3.0 |

Water | 3.12 | 2.15 |

Water - glycol | 5 | 3.4 |

Water in oil emulsion | 3.3 | 2.3 |

*1 GPa = 10*^{9}Pa (N/m^{2})

Stainless steel with Bulk Modulus *163 10 ^{9} Pa* is aprox.

*80 times*harder to compress than water with Bulk Modulus

*2.15 10*.

^{9}Pa### Example - Density of Seawater in the Mariana Trench

- the deepest known point in the Earth's oceans - *10994 m*.

The hydrostatic pressure in the Mariana Trench can be calculated as

*p _{1} = (1022 kg/m^{3}) (9.81 m/s^{2}) (10994 m)*

* = 110 10 ^{6} Pa (110 MPa) *

The initial pressure at sea-level is *10 ^{5} Pa* and the density of seawater at sea level is

*1022 kg/m*.

^{3}The density of seawater in the deep can be calculated by modifying *(2)* to

*ρ _{1}* = (

*(**p*)_{1}- p_{0}*+ K**ρ*_{0}*) / K**ρ*_{0}* = (((110 10 ^{6} Pa) - (1 10^{5} Pa)) (1022 kg/m^{3}) + (2.34 10^{9} Pa) (1022 kg/m^{3})) / (2.34 10^{9} Pa) *

* = 1070 kg/m ^{3 }*

Note! - since the density of the seawater varies with dept the pressure calculation could be done more accurate by calculating in dept intervals.