The bulk modulus measures a material’s resistance to compression; it quantifies how much a substance will decrease in volume under increased pressure. For instance, squeezing a sponge decreases its volume, and the bulk modulus characterises how much the sponge resists compression.
Daily Life Examples of Bulk Modulus
- Car Tires: The bulk modulus of air in car tires (typically around 1.4 x 105 Pa) affects how much the tire compresses under varying pressures.
- Balloon Inflation: When inflating a balloon, the rubber’s bulk modulus (around 6 x 108 Pa) determines how much the balloon’s volume changes with added air.
- Hydraulic Systems: In hydraulic brakes, the bulk modulus of hydraulic fluid (about 2.2 x 109 Pa) is crucial for transmitting pressure efficiently.
- Deep-Sea Diving: The bulk modulus of water (2.2 x 109 Pa) is significant for understanding pressure changes at different depths during deep-sea diving.
- Ear Protection: The bulk modulus of ear protection materials (e.g., foam, rubber) influences their ability to absorb sound waves, providing a comfortable environment.
Mathematical Relation
The bulk modulus (K) is mathematically related to the pressure (P) and the fractional change in volume (ΔV/V) of a material under pressure. The formula for bulk modulus is given by:
K=−V(ΔP/ΔV)
Where:
- K is the bulk modulus,
- ΔP is the change in pressure,
- ΔV is the change in volume,
- V is the original volume of the material.
In some cases, this formula is also expressed using the bulk modulus of elasticity (Y) and the material’s original volume (V):
K=Y(1/3−2ν)
Where:
- Y is the Young’s modulus of the material,
- ν is Poisson’s ratio (a dimensionless ratio describing the lateral contraction to longitudinal extension)
This second formula is often used when the Young’s modulus and Poisson’s ratio are known.
Simple Case Study: Bulk modulus of the material used in the diving suit.
Given:
- Original volume of the diving suit material, V=0.05 m3
- Initial pressure at sea level, P1=1 atm≈101.325 kPa
- Final pressure at depth, P2=20 atm≈2026.5 kPa
- Change in volume, ΔV=−0.01 m3
Using the bulk modulus formula:
K=−V(ΔP/ΔV)
Bulk Modulus (K)=0.05 m3{(2026.5 kPa−101.325 kPa)/0.01 m3) = −96258.75 N/m2
