Chapter 2, Problem 118P

Although liquids, in general, are hard to compress, the compressibility effect (variation in the density) may become unavoidable at the great depths in the oceans due to enormous pressure increase. At a certain depth the pressure is reported to be $100\text{MPa}$ and the average coefficient of compressibility is about $\text{2350}\text{MPa}\text{.}$
(a) Taking the liquid density at the free surface to be ${\rho }_0=1030{\text{kg/m}}^{\text{3}}$ obtain an analytical relation between density and pressure, and determine the density at the specified pressure. Answer: $1074{\text{kg/m}}^{\text{3}}$
(b) Use Eq. 2-13 to estimate the density for the specified pressure and compare your result with that of part (a).

To determine

(a)

The density of water at the specified depth.

Answer to Problem 118P

The density of water at the specified depth is $1078.744\text{kg}/{\text{m}}^3$.

Explanation of Solution

Given information:

The pressure at a certain depth is $100\text{MPa}$, the average coefficient of compressibility is $2350\text{MPa}$ and the liquid density at the free surface is ${\rho }_{\circ }=1030\text{kg}/{\text{m}}^3$.

Write the expression for density at a depth.

  $\alpha =\frac{1}{\rho }{\left(\frac{\partial \rho }{\partial P}\right)}_T$...... (I)

Here, the partial derivative of density is $\partial \rho$, the partial derivative of density is $\partial P$ and isothermal compressibility of water is $\alpha$.

Write the expression for ratio of partial derivative of density with respect to pressure.

  $\frac{\partial \rho }{\partial P}={\left(\frac{{\rho }_1-{\rho }_{\circ }}{P_1-P_{\circ }}\right)}_T$

Here, the pressure at the specified depth is $P_1$, the initial pressure of water at the free surface is $P_{\circ }$, the density at the free surface is ${\rho }_{\circ }$ and the density of the liquid at the specified depth is ${\rho }_1$.

Substitute $\frac{{\rho }_1-{\rho }_{\circ }}{P_1-P_{\circ }}$ for $\frac{\partial \rho }{\partial P}$ in Equation (I).

  $\begin{array}{l}\alpha =\frac{1}{\rho }{\left(\frac{{\rho }_1-{\rho }_{\circ }}{P_1-P_{\circ }}\right)}_T\\ \left({\rho }_1-{\rho }_{\circ }\right)=\alpha {\rho }_{\circ }\left(P_1-P_{\circ }\right)\\ {\rho }_1=\left(\alpha {\rho }_{\circ }\left(P_1-P_{\circ }\right)\right)+{\rho }_{\circ }\end{array}$...... (II)

Substitute, $4.80\times {10}^{-5}{\text{atm}}^{-1}$ for $\alpha$, $986.9232\text{atm}$ for $P_1$, $0.986923\text{atm}$ for $P_{\circ }$ and $1030\text{kg}/{\text{m}}^3$ for ${\rho }_{\circ }$ in Equation (II).

  $\begin{array}{c}{\rho }_1=\left[\begin{array}{l}\left(\left(4.80\times {10}^{-5}{\text{atm}}^{-1}\right)\times 1030\text{kg}/{\text{m}}^3\left(986.9232\text{atm}-\text{0}\text{.986923}\text{atm}\right)\right)\\ +1030\text{kg}/{\text{m}}^3\end{array}\right]\\ =\left(4.80\times {10}^{-5}{\text{atm}}^{-1}\times 1030\text{kg}/{\text{m}}^3\left(985.936277\text{atm}\right)\right)+1030\text{kg}/{\text{m}}^3\\ =48.7446\text{kg}/{\text{m}}^3+1030\text{kg}/{\text{m}}^3\\ =1078.744\text{kg}/{\text{m}}^3\end{array}$

Conclusion:

The density of water at the specified depth is $1078.744\text{kg}/{\text{m}}^3$.

To determine

(b)

The density of water at the specified depth and its comparison with part (a).

Answer to Problem 118P

The density of water at the specified depth is $1073.7859\text{kg}/{\text{m}}^3$.

Explanation of Solution

Write the expression for the density of water using Equation 2-13.

  $k=\frac{\Delta P}{\Delta \rho /{\rho }_o}$...... (III)

Write the expression for pressure change.

  $\Delta P=P_1-P_o$

Write the expression for change in density.

  $\Delta \rho ={\rho }_1-{\rho }_o$

Substitute $P_1-P_o$ for $\Delta P$ and ${\rho }_1-{\rho }_o$ for $\Delta \rho$ in Equation (III).

  $\begin{array}{l}k=\frac{\left(P_1-P_{\circ }\right)}{\left(\frac{{\rho }_1-{\rho }_{\circ }}{{\rho }_{\circ }}\right)}\\ \frac{{\rho }_1}{{\rho }_{\circ }}-1=\frac{\left(P_1-P_{\circ }\right)}{k}\\ {\rho }_1=\left(\frac{P_1-P_{\circ }}{k}+1\right)\times {\rho }_{\circ }\end{array}$...... (IV)

Here, the coefficient of compressibility is $k$, the pressure at the specified depth is $P_1$, the initial pressure of water at the free surface is $P_{\circ }$, the density at the free surface is ${\rho }_{\circ }$ and the density of the liquid at the specified depth is ${\rho }_1$.

Substitute, $2350\text{MPa}$ for $k$, $100\times {10}^6\text{Pa}$ for $P_1$, $100\times {10}^3\text{Pa}$ for $P_{\circ }$ and $1030\text{kg}/{\text{m}}^3$ for ${\rho }_{\circ }$ in Equation (IV).

  $\begin{array}{c}{\rho }_1=\left(\frac{100\text{MPa}-100\text{kPa}}{2350\text{MPa}}+1\right)\times 1030\text{kg}/{\text{m}}^3\\ =\left(\frac{100\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)-100\text{kPa}\left(\frac{{10}^3\text{Pa}}{1\text{kPa}}\right)}{2350\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)}+1\right)\times 1030\text{kg}/{\text{m}}^3\\ =\left(0.4251+1\right)\times 1030\text{kg}/{\text{m}}^3\\ =1073.7859\text{kg}/{\text{m}}^3\end{array}$

The value of density obtained in part (a) that is $1078.744\text{kg}/{\text{m}}^3$ is greater than the value obtained in part (b) that is $1073.7859\text{kg}/{\text{m}}^3$.Thus, the density calculated in part (b) is more exact as the isothermal compressibility of water in part (a) is an approximate value.

Conclusion:

The density of water at the specified depth is $1073.7859\text{kg}/{\text{m}}^3$.

The density calculated in part (b) is more exact as the isothermal compressibility of water in part (a) is an approximate value.