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.