Chapter 2, Problem 48P

The density of seawater at a free surface where the pressure is 98 kPa is approximately 1030 kg/m³. Taking the bulk modulus of elasticity of seawater to be $2.34\times {10}^9\text{N}/{\text{m}}^2$ and expressing variation of pressure with depth z as $dP=\rho g$determine the density and pressure at a depth of 2500 m. Disregard the effect of temperature.

To determine

The density at a depth of $2500\text{m}$.

The pressure at a depth of $2500\text{m}$.

Answer to Problem 48P

The density at a depth of $2500\text{m}$ is $25.26\times {10}^6\text{Pa}$.

The pressure at a depth of $2500\text{m}$ is $1041.12\text{kg}/{\text{m}}^{\text{3}}$.

Explanation of Solution

Given information:

The pressure at surface is $98\text{kPa}$, the density of the sea water at surface is $1030\text{kg}/{\text{m}}^{\text{3}}$, the bulk modulus of elasticity is $2.34\times {10}^9\text{N}/{\text{m}}^2$, the final depth is $2500\text{m}$ and the effect of temperature is neglected.

Write the expression for the pressure.

  $P=P_0+\rho gz$.......(I)

Here, the pressure is $P$, the surface pressure is $P_0$, the density is $\rho$ and the depth is $z$.

Write the expression for the change in pressure.

  $\Delta P=P-P_0$.......(II)

Here, the change in the pressure is $\Delta P$.

Write the value for the change in the density.

  $\Delta \rho =\frac{\rho \Delta P}{\kappa }$.......(III)

Here, the change in density is $\Delta \rho$ and the bulk modulus is $\kappa$.

Write the expression for the density at a depth of $2500\text{m}$.

  ${\rho }_{2500}=\rho +\Delta \rho$.......(IV)

Here, the density at a depth of $2500\text{m}$ is ${\rho }_{2500}$.

Calculation:

Substitute $98\text{kPa}$ for $P_0$, $1030\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^2$ for $g$ and $2500\text{m}$ for $z$ in Equation (I).

  $\begin{array}{c}P=\left(98\text{kPa}\right)+\left(1030\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^2\times 2500\text{m}\right)\\ =\left(98\text{kPa}\times \frac{{10}^3\text{Pa}}{1\text{kPa}}\right)+\left(1030\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^2\times 2500\text{m}\right)\\ =98000\text{Pa}+\left(25260750\text{kg}/\text{m}\cdot {\text{s}}^2\times \frac{1\text{Pa}}{1\text{kg}/\text{m}\cdot {\text{s}}^2}\right)\\ =28.358\times {10}^6\text{Pa}\text{.}\end{array}$

Substitute $25.358\times {10}^6\text{Pa}$ for $P$ and $98\text{kPa}$ for $P_0$ in Equation (II).

  $\begin{array}{c}\Delta P=25.358\times {10}^6\text{Pa}-\left(98\text{kPa}\right)\\ =25.358\times {10}^6\text{Pa}-\left(98\text{kPa}\times \frac{{10}^3\text{Pa}}{1\text{kPa}}\right)\\ =25.358\times {10}^6\text{Pa}-98\times {10}^3\text{Pa}\\ =25.26\times {10}^6\text{Pa}\end{array}$

Substitute $1030\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $25.26\times {10}^6\text{Pa}$ for $\Delta P$ and $2.34\times {10}^9\text{N}/{\text{m}}^2$ for $\kappa$ in Equation (III).

  $\begin{array}{c}\Delta \rho =\frac{\left(1030\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(25.26\times {10}^6\text{Pa}\right)}{\left(2.34\times {10}^9\text{N}/{\text{m}}^2\right)}\\ =\frac{\left(1030\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(25.26\times {10}^6\text{Pa}\right)}{\left(\left(2.34\times {10}^9\text{N}/{\text{m}}^2\right)\left(\frac{1\text{Pa}}{1\text{N}/{\text{m}}^2}\right)\right)}\\ =\frac{2.60178\times {10}^{10}\text{kg}/{\text{m}}^{\text{3}}}{2.34\times {10}^9}\\ =11.12\text{kg}/{\text{m}}^{\text{3}}.\end{array}$

Substitute $1030\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $11.12\text{kg}/{\text{m}}^{\text{3}}$ for $\Delta \rho$ in Equation (IV).

  $\begin{array}{c}{\rho }_{2500}=1030\text{kg}/{\text{m}}^{\text{3}}+11.12\text{kg}/{\text{m}}^{\text{3}}\\ =1041.12\text{kg}/{\text{m}}^{\text{3}}.\end{array}$

Conclusion:

The density at a depth of $2500\text{m}$ is $25.26\times {10}^6\text{Pa}$.

The pressure at a depth of $2500\text{m}$ is $1041.12\text{kg}/{\text{m}}^{\text{3}}$.