27.5: Compressibility of a Fluid

When the pressure is uniform on all sides of an object in a fluid, the pressure will squeeze the object resulting in a smaller volume. When we increase the pressure by $ΔP$ on a material of volume $v_{o}$, then the volume of the material will change by $ΔV < 0$ and consequently the density of the material will also change. Define the bulk stress by the increase in pressure change

$$\sigma_{B} \equiv \Delta P \nonumber$$

Define the bulk strain by the ratio

$$\varepsilon_{B} \equiv \frac{\Delta V}{V_{0}} \nonumber$$

For many materials, for small pressure changes, the bulk stress is linearly proportional to the bulk strain,

$$\Delta P=-B \frac{\Delta V}{V_{0}} \label{27.5.3}$$

where the constant of proportionality $B$ is called the bulk modulus. The SI unit for bulk modulus is the pascal. If the bulk modulus of a material is very large, a large pressure change will result in only a small volume change. In that case the material is called incompressible. In Table 27.2, the bulk modulus is tabulated for various materials. $$\text {Table 27.2 Bulk Modulus for Various Materials} \nonumber$$

$$\begin{array}{|l|l|} \hline \text { Material } & \text { Bulk Modulus, } Y \text { , (Pa) } \\ \hline \text { Diamond } & 4.4 \times 10^{11} \\ \hline \text { Iron } & 1.6 \times 10^{11} \\ \hline \text { Nickel } & 1.7 \times 10^{11} \\ \hline \text { Steel } & 1.6 \times 10^{11} \\ \hline \text { Copper } & 1.4 \times 10^{11} \\ \hline \text { Brass } & 6.0 \times 10^{10} \\ \hline \text { Aluminum } & 7.5 \times 10^{10} \\ \hline \text { Crown Glass } & 5.0 \times 10^{10} \\ \hline \text { Lead } & 4.1 \times 10^{10} \\ \hline \begin{array}{l} \text { Water (value increases } \\ \text { at higher pressure) } \end{array} & 2.2 \times 10^{9} \\ \hline \begin{array}{l} \text { Air (adiabatic bulk } \\ \text { modulus) } \end{array} & 1.42 \times 10^{5} \\ \hline \begin{array}{l} \text { Air (isothermal bulk } \\ \text { modulus) } \end{array} & 1.01 \times 10^{5} \\ \hline \end{array} \nonumber$$

Example 27.3: Compressibility of Water

Determine the percentage decrease in a fixed volume of water at a depth of 4 km where the pressure difference is 40 Mpa, with respect to sea level.

Solution

The bulk modulus of water is $2.2 \times 10^{9} \text Pa$. From Equation $\ref{27.5.3}$,

$$\frac{\Delta V}{V_{0}}=-\frac{\Delta P}{B}=-\frac{40 \times 10^{6} \mathrm{Pa}}{2.2 \times 10^{9} \mathrm{Pa}}=-0.018. \nonumber$$

There is only a 1.8% decrease in volume. Water is essentially incompressible even at great depths in ocean, justifying our assumption that the density of water is uniform in the ocean in Example 27.1.