Bulk modulus: Information from Answers.com

"Incompressibility" redirects here. For the topic in fluid dynamics, see Compressibility.

Illustration of uniform compression

The bulk modulus (K) of a substance measures the substance's resistance to uniform compression. It is defined as the ratio of the infinitesimal pressure increase to the resulting relative decrease of the volume. Its base unit is the pascal.^[1]

Contents

1 Definition
2 Thermodynamic relation
3 Measurement
4 Selected values
5 References

Definition

The bulk modulus K>0 can be formally defined by the equation:

$K=-V\frac{\mathrm d P}{\mathrm d V}$

where P is pressure, V is volume, and dP/dV denotes the derivative of pressure with respect to volume. Equivalently

$K=\rho \frac{\partial P}{\partial \rho}$

where ρ is density and dp/dρ denotes the derivative of pressure with respect to density. The inverse of the bulk modulus gives a substance's compressibility.

Other moduli describe the material's response (strain) to other kinds of stress: the shear modulus describes the response to shear, and Young's modulus describes the response to linear stress. For a fluid, only the bulk modulus is meaningful. For an anisotropic solid such as wood or paper, these three moduli do not contain enough information to describe its behaviour, and one must use the full generalized Hooke's law.

Thermodynamic relation

Strictly speaking, the bulk modulus is a thermodynamic quantity, and it is necessary to specify how the temperature varies in order to specify a bulk modulus: constant-temperature (isothermal $K_T$ ), constant-entropy (adiabatic $K_S$ ), and other variations are possible. In practice, such distinctions are usually only relevant for gases.

For an ideal gas, the adiabatic bulk modulus $K_S$ is given by

$K_S=\gamma\, P$

and the isothermal bulk modulus $K_T$ is given by

$K_T=P\,$

where

γ is the adiabatic index, sometimes called κ.

P is the pressure.

When the gas is not ideal, these equations give only an approximation of the bulk modulus. In a fluid, the bulk modulus K and the density ρ determine the speed of sound c (pressure waves), according to the Newton-Laplace formula

$c=\sqrt{\frac{K}{\rho}}.$

Solids can also sustain transverse waves: for these materials one additional elastic modulus, for example the shear modulus, is needed to determine wave speeds.

Measurement

It is possible to measure the bulk modulus using powder diffraction under applied pressure.

Selected values

Approximate bulk modulus (K) for common materials
Material	Bulk modulus in Pa	Bulk modulus in ksi
Glass (see also diagram below table)	3.5×10¹⁰ to 5.5×10¹⁰	5.8×10³
Steel	1.6×10¹¹	23×10³
Diamond^[2]	4.42×10¹¹	64×10³

Influences of selected glass component additions on the bulk modulus of a specific base glass.^[3]

Material with bulk modulus value of 35GPa needs external pressure of 0.35 GPa (~3500Bar) to reduce the volume by one percent.

Approximate bulk modulus (K) for other substances
Water	2.2×10⁹ Pa (value increases at higher pressures)
Air	1.42×10⁵ Pa (adiabatic bulk modulus)
Air	1.01×10⁵ Pa (constant temperature bulk modulus)
Solid helium	5×10⁷ Pa (approximate)

References

^ "Bulk Elastic Properties". hyperphysics. Georgia State University. http://hyperphysics.phy-astr.gsu.edu/hbase/permot3.html.
^ Cohen, Marvin (1985). "Calculation of bulk moduli of diamond and zinc-blende solids". Phys. Rev. B 32: 7988–7991. Bibcode 1985PhRvB..32.7988C. doi:10.1103/PhysRevB.32.7988.
^ Fluegel, Alexander. "Bulk modulus calculation of glasses". glassproperties.com. http://www.glassproperties.com/bulk_modulus/.

v t e Elastic moduli for homogeneous isotropic materials

Bulk modulus ( $K$ ) · Young's modulus ( $E$ ) · Lamé's first parameter ( $\lambda$ ) · Shear modulus ( $G$ ) · Poisson's ratio ( $\nu$ ) · P-wave modulus ( $M$ )

Conversion formulas
Homogeneous isotropic linear elastic materials have their elastic properties uniquely determined by any two moduli among these, thus given any two, any other of the elastic moduli can be calculated according to these formulas.
	$(K,\,E)$	$(K,\,\lambda)$	$(K,\,G)$	$(K,\, \nu)$	$(E,\,G)$	$(E,\,\nu)$	$(\lambda,\,G)$	$(\lambda,\,\nu)$	$(G,\,\nu)$	$(G,\,M)$
$K=\,$					$\tfrac{EG}{3(3G-E)}$	$\tfrac{E}{3(1-2\nu)}$	$\lambda+ \tfrac{2G}{3}$	$\tfrac{\lambda(1+\nu)}{3\nu}$	$\tfrac{2G(1+\nu)}{3(1-2\nu)}$	$M - \tfrac{4G}{3}$
$E=\,$		$\tfrac{9K(K-\lambda)}{3K-\lambda}$	$\tfrac{9KG}{3K+G}$	$3K(1-2\nu)\,$			$\tfrac{G(3\lambda + 2G)}{\lambda + G}$	$\tfrac{\lambda(1+\nu)(1-2\nu)}{\nu}$	$2G(1+\nu)\,$	$\tfrac{G(3M-4G)}{M-G}$
$\lambda=\,$	$\tfrac{3K(3K-E)}{9K-E}$		$K-\tfrac{2G}{3}$	$\tfrac{3K\nu}{1+\nu}$	$\tfrac{G(E-2G)}{3G-E}$	$\tfrac{E\nu}{(1+\nu)(1-2\nu)}$			$\tfrac{2 G \nu}{1-2\nu}$	$M - 2G\,$
$G=\,$	$\tfrac{3KE}{9K-E}$	$\tfrac{3(K-\lambda)}{2}$		$\tfrac{3K(1-2\nu)}{2(1+\nu)}$		$\tfrac{E}{2(1+\nu)}$		$\tfrac{\lambda(1-2\nu)}{2\nu}$
$\nu=\,$	$\tfrac{3K-E}{6K}$	$\tfrac{\lambda}{3K-\lambda}$	$\tfrac{3K-2G}{2(3K+G)}$		$\tfrac{E}{2G}-1$		$\tfrac{\lambda}{2(\lambda + G)}$			$\tfrac{M - 2G}{2M - 2G}$
$M=\,$	$\tfrac{3K(3K+E)}{9K-E}$	$3K-2\lambda\,$	$K+\tfrac{4G}{3}$	$\tfrac{3K(1-\nu)}{1+\nu}$	$\tfrac{G(4G-E)}{3G-E}$	$\tfrac{E(1-\nu)}{(1+\nu)(1-2\nu)}$	$\lambda+2G\,$	$\tfrac{\lambda(1-\nu)}{\nu}$	$\tfrac{2G(1-\nu)}{1-2\nu}$

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