# 4.3.3.2: Liquid Phase Under Hydrostatic Pressure


The bulk modulus was defined in Introduction Chapter. The simplest approach is to assume that the bulk modulus is constant (or has some representative average). For these cases, there are two differential equations that needed to be solved. Fortunately, here, only one hydrostatic equation depends on density equation. So, the differential equation for density should be solved first. The governing differential density equation is

$\rho = B_T \dfrac{\partial \rho}{\partial P} \label{static:eq:rhoSolutionG}$

The variables for equation (58) should be separated and then the integration can be carried out as

$\int_{P_0}^{P}d\,P = \int_{\rho_0}^{\rho} B_T \dfrac{d \rho}{\rho} \label{static:eq:rhoSolutionIF}$ The integration of equation (59) yields

$P- {P_0} = B_T \ln \dfrac{\rho}{\rho_0} \label{static:eq:rhoSolutionI}$ Equation (60) can be represented in a more convenient form as

Density variation

$\label{static:eq:rhoSolutionF} {\rho} = {\rho_0} {e}^{\dfrac{P- {P_0}}{ B_T}}$

Equation (61) is the counterpart for the Utilizing equation (61) in equation (11) transformed into

$\dfrac{\partial P } {\partial z} = - g\, {\rho_0} {e}^{\dfrac{P- {P_0}}{ B_T}} \label{static:eq:dzRhoLiquidG}$

Equation (62) can be integrated to yield

$\dfrac{B_T }{g\, \rho_0} \text{e}^{\dfrac{P- {P_0}}{ B_T}} = z + Constant \label{static:eq:dzRhoLiquidI}$ It can be noted that $$B_T$$ has units of pressure and therefore the ratio in front of the exponent in equation (63) has units of length. The integration constant, with units of length, can be evaluated at any specific point. If at $$z=0$$ the pressure is $$P_0$$ and the density is $$\rho_0$$ then the constant is

$Constant = \dfrac{B_T }{g\, \rho_0} \label{static:eq:dzRhoLiquidc}$

Fig. 4.10 Hydrostatic pressure when there is compressibility in the liquid phase.

This constant, $$B_T/g\,\rho_0$$, is a typical length of the problem. Additional discussion will be presented in the dimensionless issues chapter (currently under construction). The solution becomes

$\dfrac{B_T }{g\, \rho_0} \left( \text{e}^{\dfrac{P- {P_0}}{ B_T}} - 1\right) = z \label{static:eq:dzRhoLiquidF}$

Or in a dimensionless form

Density in Liquids

$\label {static:eq:dzRhoLiquidFF} \left( \text{e}^{\dfrac{P- {P_0}}{ B_T}} - 1\right) = \dfrac{z \,g\, \rho_0}{B_T}$

The solution is presented in equation (65) and is plotted in Figure 4.10. The solution is a reverse function (that is not $$P = f(z)$$ but z = f (P)) it is a monotonous function which is easy to solve for any numerical value (that is only one $$z$$ corresponds to any Pressure). Sometimes, the solution is presented as

$\dfrac{P}{P_0} = \dfrac{B_T}{P_0}\,\ln\left(\dfrac{g\,\rho_0 z}{B_T} + 1 \right) + 1 \label{static:eq:BTreversed}$

An approximation of equation (66) is presented for historical reasons and in order to compare the constant density assumption. The exponent can be expanded as It can be noticed that equation (??) is reduced to the standard equation when the normalized pressure ratio, $$P/B_T$$ is small ($$<< 1$$). Additionally, it can be observed that the correction is on the left hand side and not as the "traditional'' correction on the piezometric pressure side. After the above approach was developed, new approached was developed to answer questions raised by hydraulic engineers. In the new approach is summarized by the following example.

Example 4.6

The hydrostatic pressure was neglected in example . In some places the ocean depth is many kilometers (the deepest places is more than 10 kilometers). For this example, calculate the density change in the bottom of 10 kilometers using two methods. In one method assume that the density is remain constant until the bottom. In the second method assume that the density is a function of the pressure.

Solution 4.6

For the the first method the density is

$\label{hydrostatic:gov} B_T \cong \dfrac{\Delta P }{ \Delta V /V} \Longrightarrow \Delta V = V \dfrac{\Delta P }{B_T}$

The density at the surface is $$\rho = m/V$$ and the density at point $$x$$ from the surface the density is

$\label{hydrostatic:rhoX} \rho(x) = \dfrac{m}{ V - \Delta V} \Longrightarrow \rho (x) = \dfrac{m}{ V - V \dfrac{\Delta P }{B_T}}$
In this Chapter it was shown (integration of equation (8)) that the change pressure for constant gravity is

$\label{hydrostatic:DeltaP} \Delta P = g\, \int_0^z \rho (z) dz$ Combining equation (69) with equation (70) yields

$\label{hydrostatic:eqGov} \rho (z) = \dfrac{m}{ V - \displaystyle \dfrac{V\, g}{B_T} \,\int_0^z \rho (z) dz }$ Equation can be rearranged to be

$\label{hydrostatic:eqGovr} \rho (z) = \dfrac{m}{ V \left( 1 - \dfrac{g} {B_T} {\displaystyle\int_0^z \rho (z) dz } \right) } \Longrightarrow \rho (z) = \dfrac{\rho_0}{ \left( 1 - \dfrac{g} {B_T} {\displaystyle\int_0^z \rho (z) dz } \right) }$ Equation (72) is an integral equation which is discussed in the appendix. It is convenient to rearrange further equation (72) to

$\label{hydrostatic:eqGovrf} 1 - \dfrac{g} {B_T} {\displaystyle\int_0^z \rho (z) dz } = \dfrac{\rho_0}{ \rho (z) }$ The integral equation (73) can be converted to a differential equation form when the two sides are differentiated as

$\label{hydrostatic:govDif} \dfrac{g} {B_T} \rho (z) + \dfrac{\rho_0} {\rho(z)^2} \dfrac{d\,\rho(z)}{dz} = 0$ equation (74) is first order non-linear differential equation which can be transformed into

$\label{hydrostatic:govDif1} \dfrac{g\,\rho (z)^3} {B_T\,\rho_0} + \dfrac{d\,\rho(z)}{dz} = 0$ The solution of equation (75) is

$\label{hydrostatic:gSol} \dfrac{\rho_0\,B_T}{2\,g\,{\rho}^{2}} = z + c$ or rearranged as

$\label{hydrostatic:gSolr} \rho = \sqrt{\dfrac{\rho_0\,B_T}{2\,g\,\left( z + c\right) } }$ The integration constant can be found by the fact that at $$z=0$$ the density is $$\rho_0$$ and hence

$\label{hydrostatic:ic0} \rho_0 = \sqrt{\dfrac{\rho_0\,B_T}{2\,g\,\left( c\right) } } \Longrightarrow c = \dfrac{B_T} {2\,g\,{\rho_0}}$ Substituting the integration constant and opening the parentheses, the solution is

$\label{hydrostatic:iniConPlus} \rho = \sqrt{\dfrac{\rho_0\,B_T} {2\,g\,z + \dfrac{\cancel{2\,g}\,B_T} {\cancel{2\,g}\,{\rho_0}}} }$ Or

$\label{hydrostatic:fSol} \rho = \sqrt{\dfrac{\cancel{\dfrac{1}{\rho_0}}\,{\rho_0}^2\,B_T} { \cancel{\dfrac{1}{\rho_0}} \,\left( 2\,g\,\rho_0\, z + B_T \right) } } \Longrightarrow \dfrac{\rho}{\rho_0} = \sqrt{\dfrac{B_T} { \left( 2\,g\,\rho_0\, z + B_T \right) } }$ Equation (80) further be rearranged to a final form as

$\label{hydrostatic:rhoC} \dfrac{\rho}{\rho_0} = \sqrt{ \dfrac{\cancelto{1}{B_T} } { \cancel {B_T}\, \left( \dfrac{2\,g\,\rho_0\, z}{ B_T} + 1 \right) } } \Longrightarrow \dfrac{\rho}{\rho_0} = \sqrt{ \dfrac{1 } { \left( \dfrac{2\,g\,\rho_0\, z}{ B_T} + 1 \right) } }$ The parameter $$\dfrac{2\,g\,\rho_0\, z}{ B_T}$$ represents the dimensional length controlling the problem. For small length the expression in (81) is similar to

$\label{hydrostatic:taylor} f(x) = \sqrt{\dfrac{1}{x+1}} = 1-\dfrac{x}{2}+\dfrac{3\,{x}^{2}}{8}-\dfrac{5\,{x}^{3}}{16} +\, \dots$ hence it can be expressed as

$\label{hydrostatic:smallRhoC} \dfrac{\rho}{\rho_0} = 1 - \dfrac{2\,g\,\rho_0\, z} {2\, B_T} + \dfrac{3\,g^2\,{\rho_0}^2\, z^2} {8\, {B_T}^2} - \dfrac{5\,g^3\,{\rho_0}^3\, z^3} {16\, {B_T}^3} \, +\, \dots$

Example 4.7

Water in deep sea undergoes compression due to hydrostatic pressure. That is the density is a function of the depth. For constant bulk modulus, it was shown in "Fundamentals of Compressible Flow'' by this author that the speed of sound is given by

$\label{liquidBT:sound} c = \sqrt{\dfrac{B_T}{\rho}}$

Calculate the time it take for a sound wave to propagate perpendicularly to the surface to a depth D (perpendicular to the straight surface). Assume that no variation of the temperature exist. For the purpose of this exercise, the salinity can be completely ignored.

Solution 4.7

The equation for the sound speed is taken here as correct for very local point. However, the density is different for every point since the density varies and the density is a function of the depth. The speed of sound at any depth point, x, is to be continue ??????????????

$\label{liquidBT:cx} c = \sqrt{\dfrac{B_T}{\dfrac{\rho_0\,B_T}{B_T-g\,\rho_0\,z}} } = \sqrt{\dfrac{B_T-g\,\rho_0\,z} {\rho_0}}$

The time the sound travel a small interval distance, $$dz$$ is

$\label{liquidBT:dt} d\tau = \dfrac{dz}{\sqrt{\dfrac{B_T-g\,\rho_0\,z} {\rho_0}}}$ The time takes for the sound the travel the whole distance is the integration of infinitesimal time

$\label{liquidBT:ds} t = \int_0^D \dfrac{dz}{\sqrt{\dfrac{B_T-g\,\rho_0\,z} {\rho_0}}}$ The solution of equation (87) is

$\label{liquidBT:sol} t = \sqrt{\rho_0}\,\left( 2\,\sqrt{B_T}-2\,\sqrt{B_T-D}\right)$ The time to travel according to the standard procedure is

$\label{liquidBT:oldTime} t = \dfrac{D}{\sqrt{\dfrac{B_T}{\rho_0}}} = \dfrac{D\,\sqrt{\rho_0}}{\sqrt{B_T}}$ The ratio between the corrected estimated to the standard calculation is

$\label{liquidBT:ratioTimes} \text{correction ratio } = \dfrac{\sqrt{\rho_0}\,\left( 2\,\sqrt{B_T}-2\,\sqrt{B_T-D}\right)} {\dfrac{D\,\sqrt{\rho_0}}{\sqrt{B_T}}}$

In Example 4.6 ratio of the density was expressed by equations qref{hydrostatic:fSol} while here the ratio is expressed by different equations. The difference between the two equations is the fact that Example 4.6 use the integral equation without using any "equation of state.'' The method described in the Example 4.6 is more general which provided a simple solution. The equation of state suggests that $$\partial P = g\,\rho_0\,f(P)\, dz$$ while the integral equation is $$\Delta P = g\, \int \rho\, dz$$ where no assumption is made on the relationship between the pressure and density. However, the integral equation uses the fact that the pressure is function of location. The comparison between the two methods will be presented.

Example 4.8