## What is Bulk Modulus?

The bulk modulus (K or B) of a substance is a measure of how resistant to compression that substance is. It is defined as the ratio of the infinitesimal pressure increase to the resulting relative decrease of the volume.

Other moduli describe the material’s response (strain) to other kinds of stress: the shear modulus describes the response to shear stress, and Young’s modulus describes the response to normal stress.

For a fluid, only the bulk modulus is meaningful. For a complex anisotropic solid such as wood or paper, these three moduli do not contain enough information to describe its behavior, and one must use the full generalized Hooke’s law. The reciprocal of the bulk modulus at a fixed temperature is called the isothermal compressibility.

## Introduction of Bulk Modulus

Bulk modulus is a numerical constant that describes the elastic properties of a solid or fluid when it is under pressure on all surfaces. The applied pressure reduces the volume of a material, which returns to its original volume when the pressure is removed.

Sometimes referred to as the incompressibility, the bulk modulus is a measure of the ability of a substance to withstand changes in volume when under compression on all sides. It is equal to the quotient of the applied pressure divided by the relative deformation.

In this case, the relative deformation, commonly called a strain, is the change in volume divided by the original volume. Thus, if the original volume V_{o} of a material is reduced by an applied pressure p to a new volume V_{n}, the strain may be expressed as the change in volume, V_{o} − V_{n}, divided by the original volume, or (V_{o} − V_{n})/V_{o}. The bulk modulus itself, which, by definition, is the pressure divided by the strain, may be expressed mathematically as

When the bulk modulus is constant (independent of pressure), this is a specific form of Hooke’s law of elasticity.

Because the denominator, strain, is a ratio without dimensions, the dimensions of the bulk modulus are those of pressure, force per unit area. In the English system the bulk modulus may be expressed in units of pounds per square inch (usually abbreviated to psi), and in the metric system, newtons per square meter (N/m^{2}), or pascals.

The value of the bulk modulus for steel is about 2.3 × 107 psi, or 1.6 × 1011 pascals, three times the value for glass. Thus, only one-third of the pressure is needed to reduce a glass sphere the same amount as a steel sphere of the same initial size. Under equal pressure, the proportional decrease in the volume of glass is three times that of steel.

One may also say that glass is three times more compressible than steel. In fact, compressibility is defined as the reciprocal of the bulk modulus. A substance that is difficult to compress has a large bulk modulus but small compressibility. A substance that is easy to compress has high compressibility but a low bulk modulus.

## Table of Fluid Bulk Modulus (K) Values

There are bulk modulus values for solids (e.g., 160 GPa for steel; 443 GPa for diamond; 50 MPa for solid helium) and gases (e.g., 101 kPa for air at constant temperature), but the most common tables list values for liquids.

Here are representative values, in both English and metric units:

Name | English Units (10^{5} PSI) | SI Units (10^{9} Pa) |

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 – Oil Emulsion | 3.3 | 2.3 |

The K value varies, depending on the state of matter of a sample, and in some cases, on the temperature. In liquids, the amount of dissolved gas greatly impacts the value.

A high value of K indicates a material resists compression, while a low value indicates volume appreciably decreases under uniform pressure. The reciprocal of the bulk modulus is compressibility, so a substance with a low bulk modulus has high compressibility.

Upon reviewing the table, you can see the liquid metal mercury is very nearly incompressible. This reflects the large atomic radius of mercury atoms compared with atoms in organic compounds and also the packing of the atoms. Because of hydrogen bonding, water also resists compression.

## Bulk Modulus Formulas

The bulk modulus of a material may be measured by powder diffraction, using x-rays, neutrons, or electrons targeting a powdered or microcrystalline sample. It may be calculated using the formula:

**Bulk Modulus (K) = Volumetric stress / Volumetric strain**

This is the same as saying it equals the change in pressure divided by the change in volume divided by initial volume:

**Bulk Modulus (K) = (p _{1} – p_{0}) / [(V_{1} – V_{0}) / V_{0}]**

Here, p_{0} and V_{0} are the initial pressure and volume, respectively, and p_{1} and V_{1} are the pressure and volume measured upon compression.

Bulk modulu elasticity may also be expressed in terms of pressure and density:

K = (p_{1} – p_{0}) / [(ρ_{1} – ρ_{0}) / ρ_{0}]

Here, ρ_{0 }and ρ_{1} are the initial and final density values.

## Example Calculation

The bulk modulus may be used to calculate the hydrostatic pressure and density of a liquid. For example, consider seawater in the deepest point of the ocean, the Mariana Trench. The base of the trench is 10994 m below sea level.

The hydrostatic pressure in the Mariana Trench may be calculated as:

p1 = ρ*g*h

Where p_{1} is the pressure, ρ is the density of seawater at sea level, g is the acceleration of gravity, and h is the height (or depth) of the water column.

- p
_{1}= (1022 kg/m^{3})(9.81 m/s^{2})(10994 m) - p
_{1}= 110 x 10^{6}Pa or 110 MPa

Knowing the pressure at sea level is 105 Pa, the density of the water at the bottom of the trench may be calculated:

- ρ
_{1}= [(p_{1}– p)ρ + K*ρ) / K - ρ
_{1}= [[(110 x 10^{6}Pa) – (1 x 10^{5}Pa)](1022 kg/m^{3})] + (2.34 x 10^{9}Pa)(1022 kg/m^{3})/(2.34 x 10^{9}Pa) - ρ
_{1}= 1070 kg/m^{3}

What can you see from this? Despite the immense pressure on the water at the bottom of the Mariana Trench, it isn’t compressed very much!