Derivation of Bulk Modulus Formula

A substance's bulk modulus is a measurement of how resistant it is to compression. It's the ratio of the infinitesimal pressure rise to the volume's resultant relative decrease. To put it another way, the bulk modulus is a numerical constant that is used to measure and quantify a solid's or fluid's elastic properties when pressure is applied to all surfaces.

The bulk modulus of elasticity is one of the mechanical properties of solids that may be measured. Young's modulus and Shear modulus are two other elastic modules. In any event, a material's bulk elastic characteristics are used to calculate how much it will compress when subjected to a certain amount of external pressure. It's crucial to find and quantify the ratio of pressure change to fractional volume compression.

Bulk modulus is denoted by the symbol K or B. The SI unit of measurement of the bulk modulus is N/m².

Bulk modulus is nothing but the ratio of volumetric strain to the volumetric strain.

The bulk modulus formula is given as follows:

\[K = - V \frac{\mathrm{dp} }{\mathrm{d} v}\]

Where K is the bulk modulus.

V is the initial volume of the substance.

dp is the change in pressure applied per unit area on the material.

dv is the change in volume due to compression.

The bulk modulus formula in terms of density is as follows:

\[K = - \rho  \frac{\mathrm{dp} }{\mathrm{d} \rho }\]

Where K is the bulk modulus.

\[\rho \]is the initial density.

dp is the change in pressure applied per unit area on the material.

\[d\rho\] is the change in density due to compression.

Derivation of Bulk Modulus of Elasticity

In this section, we will see the step by step bulk modulus formula derivation.

Consider a block of the cube which is having the initial volume ‘V’. Now if the pressure applied on the block is dp which is changing per unit area on the material.

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The bulk modulus equation is given as follows:

Bulk modulus, K = - (Pressure applied) / (Fractional change in the volume)

A decrease in volume is associated with an increase in pressure, as indicated by the negative sign. meaning that changing the volume of water by a tiny quantity necessitates huge pressure. Water, like many other liquids, is viewed as incompressible for all practical purposes.

\[K = -\frac{dP}{\frac{dV}{V}}\]

\[K = -V \frac{dP}{dv}\]

Now if we consider a unit mass of a substance that is m = 1 kg. Then the volume in terms of density will be as follows:

\[V = \frac{m}{\rho }\]

Since mass is the unity we get

\[V = \frac{1}{\rho }\]

Now differentiating we get

\[dV=\frac{1}{d\rho }\]

This means that volume is inversely proportional to density. So the bulk modulus formula in terms of density will be the alternate form to express the bulk modulus.

\[K = \frac{dP}{\frac{d\rho }{\rho }}\]

\[K = \rho \frac{dP}{d\rho }\]

Till now we learned what is bulk modulus formula and step by step derivation of the bulk modulus formula. Let us discuss some solved examples to understand the concept clearly.

Exercise Problems on Bulk Modulus Formula

1. The pressure is applied to the copper tube is found to be 155 MPa. If the bulk modulus of copper is 1.38 x 10¹¹ Pa then calculate the change in volume of the piece of the copper when subjected to this pressure. 

Ans: Here the change in pressure is dP = 155 MPa = 155 x 10⁶ Pa.

The bulk modulus of the copper is given as K = 1.38 x 10¹¹ Pa.

Now substituting the values in the bulk modulus formula we get

\[K = \frac{dP}{\frac{dV}{V}}\]

\[1.38\times 10^{11}= \frac{155\times 10^{6}}{\frac{dV}{V}}\] 

\[\frac{dV}{V}= \frac{155\times 10^{6}}{1.38\times 10^{11}}\]

On solving we get

\[\frac{dV}{V} = 1.123\times 10^{-3}\]

Therefore the change in volume is \[1.123\times 10^{-3}\]

Conclusion

Bulk modulus is a numerical constant that describes a solid's or fluid's elastic properties when all surfaces are under pressure. When pressure is applied to a material, the volume is reduced; however, when the pressure is removed, the volume returns to its previous state. The compressibility of a fluid is measured by its bulk modulus. With many hydraulic systems currently operating at pressures of 5000 psi and greater, disregarding bulk modulus might cause a system's response time to be compromised. Rather than compressing the fluid, applied pressure should affect the system's activity directly.