Introduction to - and definition of - Bulk Modulus Elasticity commonly used to characterize 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 / V0)
= - (p1 - p0) / ((V1 - V0) / V0) (1)
K = bulk modulus elasticity (Pa, N/m2)
dp = differential change in pressure on the object (Pa, N/m2)
dV = differential change in volume of the object (m3)
V0 = initial volume of the object (m3)
p0 = initial pressure (Pa, N/m2)
p1 = final pressure (Pa, N/m2)V1 = final volume (m3)
The Bulk Modulus Elasticity can alternatively be expressed as
K = dp / (dρ / ρ0)
= (p1 - p0) / ((ρ1 - ρ0) / ρ0) (2)
dρ = differential change in density of the object (kg/m3)
ρ0 = initial density of the object (kg/m3)
ρ1 = final density of the object (kg/m3)
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/m2 (Pa)
- The imperial (BG) unit is lbf/in2 (psi)
- 1 lbf/in2 (psi) = 6.894 103 N/m2 (Pa)
A large Bulk Modulus indicates a relative incompressible fluid.
Bulk Modulus for some common fluids:
|Imperial Units - BG|
|ISO 32 mineral oil||2.6||1.8|
|Petrol||1.55 - 2.16||1.07 - 1.49|
|SAE 30 Oil||2.2||1.5|
|Water - glycol||5||3.4|
|Water in oil emulsion||3.3||2.3|
- 1 GPa = 109 Pa (N/m2)
Stainless steel with Bulk Modulus 163 109 Pa is aprox. 80 times harder to compress than water with Bulk Modulus 2.15 109 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
p1 = (1022 kg/m3) (9.81 m/s2) (10994 m)
= 110 106 Pa (110 MPa)
The initial pressure at sea-level is 105 Pa and the density of seawater at sea level is 1022 kg/m3.
The density of seawater in the deep can be calculated by modifying (2) to
ρ1 = ((p1 - p0) ρ0 + K ρ0) / K
= (((110 106 Pa) - (1 105 Pa) (1022 kg/m3)) + (2.34 109 Pa) (1022 kg/m3)) / ((2.34 109 Pa))
= 1070 kg/m3
Note! - since the density of the seawater varies with dept the pressure calculation could be done more accurate by calculating in dept intervals.