Basics of Fluid Mechanics
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Chapter 4 Fluids Statics

4.1 Introduction

The simplest situation that can occur in the study of fluid is when the fluid is at rest or quasi rest. This topic was introduced to most students in previous study of rigid body. However, here this topic will be more vigorously examined. Furthermore, the student will be exposed to stability analysis probably for the first time. Later, the methods discussed here will be expanded to more complicated dynamics situations.

4.2 The Hydrostatic Equation

Fluid Element in accelerate system

Fig. 4.1 Description of a fluid element in accelerated system under body forces.

A fluid element with dimensions of $DC$, $dy$, and $dz$ is motionless in the accelerated system, with acceleration, $a$ as shown in Figure 4.1. The combination of an acceleration and the body force results in effective body force which is \begin{align} \mathbf{g}_{G} - a = g_{eff} \label{static:eq:geff} \end{align} Equation \eqref{static:eq:geff} can be reduced and simplified for the case of zero acceleration, $\mathbf{a} = 0$. In these derivations, several assumptions must be made. The first assumption is that the change in the pressure is a continuous function. There is no requirement that the pressure has to be a monotonous function e.g. that pressure can increase and later decrease. The changes of the second derivative pressure are not significant compared to the first derivative ($\partial P / \partial n \times d\ell >> \partial^2 P/ \partial n^2$). where $n$ is the steepest direction of the pressure derivative and $d\ell$ is the infinitesimal length. This mathematical statement simply requires that the pressure can deviate in such a way that the average on infinitesimal area can be found and expressed as only one direction. The net pressure force on the faces in the $x$ direction results in \begin{align} d\mathbf{F} = - \left( \dfrac{\partial P} {\partial x}\right)dydx \;\hat{i} \label{static:eq:ForceX} \end{align} In the same fashion, the calculations of the three directions result in the total net pressure force as \begin{align} \sum_{surface} F = - \left( \dfrac{\partial P}{\partial x} \;\hat{i} + \dfrac{\partial P}{\partial y} \;\hat{j} + \dfrac{\partial P}{\partial y} \;\hat{k} \right) \label{static:eq:ForceT} \end{align} The term in the parentheses in equation \eqref{static:eq:ForceT} referred to in the literature as the pressure gradient (see for more explanation in the Mathematics Appendix). This mathematical operation has a geometrical interpretation. If the pressure, $P$, was a two–dimensional height (that is only a function of $x$ and $y$) then the gradient is the steepest ascent of the height (to the valley). The second point is that the gradient is a vector (that is, it has a direction). Even though, the pressure is treated, now, as a scalar function (there no reference to the shear stress in part of the pressure) the gradient is a vector. For example, the dot product of the following is \begin{align} \widehat{i} \cdot \mathbf {grad} P = \widehat{i} \cdot \nabla P = \dfrac{\partial P} {\partial x} \label{static:eq:dotProcut} \end{align} In general, if the coordinates were to ``rotate/transform'' to a new system which has a different orientation, the dot product results in \begin{align} \overline{i_n} \cdot \mathbf {grad} P = \overline{i_n} \cdot \nabla P = \dfrac{\partial P} {\partial n} \label{static:eq:dotProdcutN} \end{align} where $i_n$ is the unit vector in the $n$ direction and $\partial / \partial n$ is a derivative in that direction. As before, the effective gravity force is utilized in case where the gravity is the only body force and in an accelerated system. The body (element) is in rest and therefore the net force is zero \begin{align} \sum_{total} \mathbf{F} = \sum_{surface} \mathbf{F} + \sum _{body} \mathbf{F} \label{static:eq:staticNetF} \end{align} Hence, the utilizing the above derivations one can obtain \begin{align} - \mathbf{grad} P dx\, dy\, dz + \rho\, g_{\text{eff}} dx\, dy\, dz = 0 \label{static:eq:exStaticNetF} \end{align} or

Pressure Gradient

\begin{align} \label{static:eq:staticEq} \mathbf{grad} P = \boldsymbol{\nabla} P = \rho\, g_{\text{eff}} \ \end{align}
Some refer to equation \eqref{static:eq:staticEq} as the Fluid Static Equation. This equation can be integrated and therefore solved. However, there are several physical implications to this equation which should be discussed and are presented here. First, a discussion on a simple condition and will continue in more challenging situations.

4.3 Pressure and Density in a Gravitational Field

In this section, a discussion on the pressure and the density in various conditions is presented.

Constant Density in Gravitational Field

The simplest case is when the density, $\rho$, pressure, $P$, and temperature, $T$ (in a way no function of the location) are constant. Traditionally, the $z$ coordinate is used as the (negative) direction of the gravity. The effective body force is \begin{align} g_{\text{eff}} = - \mathbf{g}\,\hat{k} \label{static:eq:gravity} \end{align}

Pressure lines Static Fluid

Fig. 4.2 Pressure lines in a static fluid with a constant density.

Utilizing equation \eqref{static:eq:gravity} and substituting it into equation \eqref{static:eq:staticEq} results into three simple partial differential equations. These equations are \begin{align} \dfrac {\partial P}{\partial x} = \dfrac {\partial P}{\partial y} = 0 \label{static:eq:dxy} \end{align} and

Pressure Change

\begin{align} \label{static:eq:dz} \dfrac{\partial P}{\partial z} = -\rho\, \mathbf{g} \end{align}
Equations \eqref{static:eq:dxy} can be integrated to yield \begin{align} P(x,y) = constant \label{static:eq:dxyS} \end{align} and constant in equation \eqref{static:eq:dxyS} can be absorbed by the integration of equation \eqref{static:eq:dz} and therefore \begin{align} P(x,y,z) = -\rho\, g\, z + constant \label{static:eq:dzS} \end{align} The integration constant is determined from the initial conditions or another point. For example, if at point $z_0$ the pressure is $P_0$ then the equation qref{static:eq:dzS} becomes \begin{align} P(z) -P_0 = -\rho \, g\, ( z - z_0) \label{static:eq:dzSExplisit} \end{align}
Parametric Pressure

Fig. 4.3 A schematic to explain the measure of the atmospheric pressure.

It is evident from equation \eqref{static:eq:dzS} that the pressure depends only on $z$ and/or the constant pressure lines are in the plane of $x$ and $y$. Figure 4.2 describes the constant pressure lines in the container under the gravity body force. The pressure lines are continuous even in area where there is a discontinuous fluid. The reason that a solid boundary doesn't break the continuity of the pressure lines is because there is always a path to some of the planes. It is convenient to reverse the direction of $z$ to get rid of the negative sign and to define $h$ as the dependent of the fluid that is $h quiv - ( z - z_0 ) $ so equation qref{static:eq:dzSExplisit} becomes

Pressure relationship

\begin{align} \label{static:eq:hStatic} P(h) - P_0 = \rho\, g\, h \end{align}
In the literature, the right hand side of the equation qref{static:eq:hStatic} is defined as piezometric pressure.

Example 4.1

Two chambers tank depicted in Figure 4.4 are in equilibration. If the air mass at

\hbox to \textwidth {chamber A is 1 Kg while the mass at chamber B is unknown. The difference in the}

Effective gravity of Accelerated Box

Fig. 4.4 The effective gravity is for accelerated cart.

liquid heights between the two chambers is $2[m]$. The liquid in the two chambers is water. The area of each chamber is $1[m^2]$. Calculate the air mass in chamber B. You can assume ideal gas for the air and the water is incompressible substance with density of $1000 [kg/m^2]$. The total height of the tank is $4[m]$.

Assume that the chamber are at the same temperature of $27^{\circ}C$.

Solution

The equation of state for the chamber A is \begin{align} \label{twoChambers:gasA} m_A = \dfrac{R\,T}{P_A\,V_A} \end{align} The equation of state for the second chamber is \begin{align} \label{twoChambers:gasB} m_B = \dfrac{R\,T}{P_B\,V_B} \end{align} The water volume is \begin{align} \label{twoChambers:totalV} V_{total} = h_1\, A + (h_1+h_2)A = (2\,h_1+ h_2)\, A \end{align} The pressure difference between the liquid interface is estimated negligible the air density as \begin{align} \label{twoChambers:DeltaP} P_A - P_B = \Delta P = h_2 \, \rho\,g \end{align} combining equations \eqref{twoChambers:gasA}, \eqref{twoChambers:gasB} results in \begin{align} \label{twoChambers:EeltaPe} \dfrac{R\,T}{m_A\,V_A} - \dfrac{R\,T}{m_B\,V_B} = h_2 \, \rho\,g \Longrightarrow \left( 1 - \dfrac{1}{\dfrac{m_B}{m_A} \,\dfrac{V_B}{V_A}} \right) = \dfrac{h_2 \, \rho\,g\,m_A\,V_A}{R\,T} \end{align} In equation the only unknown is the ratio of $m_B/m_A$ since everything else is known. Denoting $X= m_B/m_A $ results in \begin{align} \label{twoChambers:almostSol} \dfrac{1}{X} = 1 - \dfrac{h_2 \, \rho\,g\,m_A\,V_A}{R\,T} \Longrightarrow X = \dfrac{1} {1 - \dfrac{h_2 \, \rho\,g\,m_A\,V_A}{R\,T} } \end{align}

The following question is a very nice qualitative question of understanding this concept.

Example 4.2

Tank with Two Liquids Piezometric Tubes

Fig. 4.5 Tank and the effects different liquids.

A tank with opening at the top to the atmosphere contains two immiscible liquids one heavy and one light as depicted in Figure 4.5 (the light liquid is on the top of the heavy liquid). Which piezometric tube will be higher? why? and how much higher? What is the pressure at the bottom of the tank?

Solution

The common instinct is to find that the lower tube will contain the higher liquids. For the case, the lighter liquid is on the top the heavier liquid the the top tube is the same as the surface. However, the lower tube will raise only to (notice that $g$ is canceled) \begin{align} \label{whoIsHigh:second} h_L = \dfrac{\rho_1\,h_1 + \rho_2\,h_2}{\rho_2} \end{align} Since $\rho_1 > \rho_1$ the mathematics dictate that the height of the second is lower. The difference is \begin{align} \label{whoIsHigh:ratio} \dfrac{h_H - h_L}{h_2} = \dfrac{h_H}{h_2} - \left( \dfrac{\rho_1\,h_1 + \rho_2\,h_2}{h_r21\,\rho_2} \right) \end{align} It can be noticed that $h_H=h_1+h-2$ hence, \begin{align} \label{whoIsHigh:ratioC} \dfrac{h_H - h_L}{h_2} = \dfrac{h_1+h_2}{h_2} - \left( \dfrac{\rho_1\,h_1 + \rho_2\,h_2}{h_2\,\rho_2} \right) = \dfrac{h_1}{h_2} \left(1 - \dfrac{\rho_1}{\rho_2} \right) \end{align} or \begin{align} \label{whoIsHigh:ratioF} h_H - h_L = {h_1}\, \left(1 - \dfrac{\rho_1}{\rho_2} \right) \end{align} The only way the $h_L$ to be higher of $h_H$ is if the heavy liquid is on the top if the stability allow it. The pressure at the bottom is \begin{align} \label{whoIsHigh:pressure} P = P_{atmos} + g\,\left( \rho_1\,h_1 + \rho_2\,h_2 \right) \end{align}

Example 4.3

The effect of the water in the car tank is more than the possibility that water freeze in fuel lines. The water also can change measurement of fuel gage. The way the interpretation of an automobile fuel gage is proportional to the pressure at the bottom of the fuel tank. Part of the tank height is filled with the water at the bottom (due to the larger density). Calculate the error for a give ratio between the fuel density to the water.

Solution

The ratio of the fuel density to water density is $ arsigma= \rho_f/\rho_w$ and the ratio of the total height to the water height is $x= h_w/h_{total}$ Thus the pressure at the bottom when the tank is full with only fuel \begin{align} \label{waterCar:fullP} P_{full} = \rho_f\,h_{total}\,g \end{align} But when water is present the pressure will be the same at \begin{align} \label{waterCar:waterP} P_{full} = \left(\rho_w \,x\,+ \phi\,\rho_f\right) g\,h_{total} \end{align} and if the two are equal at \begin{align} \label{waterCar:Pequal} \rho_f\,\cancel{h_{total}}\,\cancel{g} = \left(\rho_w \,x\,+ \phi\,\rho_f\right) \cancel{g}\, \cancel {h_{total}} \end{align} where $\phi$ in this case the ratio of the full height (on the fake) to the total height. Hence, \begin{align} \label{waterCar:sol} \phi = \dfrac{\rho_f - x\,\rho_w}{\rho_f} \end{align}

4.3.2 Pressure Measurement

4.3.2.1 Measuring the Atmospheric Pressure

One of the application of this concept is the idea of measuring the atmospheric pressure. Consider a situation described in Figure 4.3. The liquid is filling the tube and is brought into a steady state. The pressure above the liquid on the right side is the vapor pressure. Using liquid with a very low vapor pressure like mercury, will result in a device that can measure the pressure without additional information (the temperature).

Example 4.4

{Calculate the atmospheric pressure at $20^{\circ}C$. The high of the Mercury is 0.76 [m] and the gravity acceleration is 9.82[$m/sec$]. Assume that the mercury vapor pressure is 0.000179264[kPa]. The description of the height is given in Figure 4.3. The mercury density is 13545.85[$kg/m^3$].

Solution

The pressure is uniform or constant plane perpendicular to the gravity. Hence, knowing any point on this plane provides the pressure anywhere on the plane. The atmospheric pressure at point a is the same as the pressure on the right hand side of the tube. Equation \eqref{static:eq:hStatic} can be utilized and it can be noticed that pressure at point a is \begin{align} P_{a} = \rho \, g \, h + P_{vapor} \label{static:eq:vaporP} \end{align} The density of the mercury is given along with the gravity and therefore, \[ P_a = 13545.85 \times 9.82 \times 0.76 \sim 101095.39 [Pa] \sim 1.01[Bar] \] The vapor pressure is about $1 \times 10^{-4}$ percent of the total results.

Gas Pressure Measurement

Fig. 4.6 Schematic of gas measurement utilizing the ``U'' tube.

The main reason the mercury is used because of its large density and the fact that it is in a liquid phase in most of the measurement range. The third reason is the low vapor (partial) pressure of the mercury. The partial pressure of mercury is in the range of the 0.000001793[Bar] which is insignificant compared to the total measurement as can be observed from the above example.

Example 4.5

A liquid a in amount $H_a$ and a liquid b in amount $H_b$ in to an U tube. The ratio of the liquid densities is $\alpha = \rho_1/\rho_2$. The width of the U tube is $L$. Locate the liquids surfaces.

Solution

The question is to find the equilibrium point where two liquids balance each other. If the width of the U tube is equal or larger than total length of the two liquids then the whole liquid will be in bottom part. For smaller width, $L$, the ratio between two sides will be as \[ \rho_1\,h_1 = \rho_2\,h_2 \rightarrow h_2 = \alpha \, h_1 \] The mass conservation results in \[ H_a + H_b = L + h_1 + h_2 \] Thus two equations and two unknowns provide the solution which is \[ h_1 = \dfrac{H_a + H_b - L}{1+\alpha} \] When $H_a > L$ and $\rho_a\left( H_a -L\right) \geq \rho _b$ (or the opposite) the liquid $\mathbf{a}$ will be on the two sides of the U tube. Thus, the balance is \[ h_1 \, \rho_b + h_2\, \rho_a = h_3\, \rho_a \] where $h_1$ is the height of liquid $\mathbf{b}$ where $h_2$ is the height of ``extra'' liquid $\mathbf{a}$ and same side as liquid $\mathbf{b}$ and where $h_3$ is the height of liquid $\mathbf{b}$ on the other side. When in this case $h_1$ is equal to $H_b$. The additional equation is the mass conservation as \[ H_a = h_2+L+h_3 \] The solution is \[ h_2 = \dfrac{(H_a-L)\,\rho_a - H_b\rho_b}{2\,\rho_a} \]

4.3.2.2 Pressure Measurement

Magnified Pressure Measurement

Fig. 4.7 Schematic of sensitive measurement device.

The idea describes the atmospheric measurement that can be extended to measure the pressure of the gas chambers. Consider a chamber filled with gas needed to be measured (see Figure 4.6). One technique is to attached ``U'' tube to the chamber and measure the pressure. This way, the gas is prevented from escaping and its pressure can be measured with a minimal interference to the gas (some gas enters to the tube). The gas density is significantly lower than the liquid density and therefore can be neglected. The pressure at point ``1'' is \begin{align} P_1 = P_{atmos} + \rho\, g \, h \label{static:eq:gageP} \end{align} Since the atmospheric pressure was measured previously (the technique was shown in the previous section) the pressure of the chamber can be measured.

4.3.2.3 Magnified Pressure Measurement

For situations where the pressure difference is very small, engineers invented more sensitive measuring device. This device is build around the fact that the height is a function of the densities difference. In the previous technique, the density of one side was neglected (the gas side) compared to other side (liquid). This technique utilizes the opposite range. The densities of the two sides are very close to each other, thus the height become large. Figure 4.7 shows a typical and simple schematic of such an instrument. If the pressure differences between $P_1$ and $P_2$ is small this instrument can ``magnified'' height, $h_1$ and provide ``better'' accuracy reading. This device is based on the following mathematical explanation. In steady state, the pressure balance (only differences) is \begin{align} P_1 + g\, \rho_1 (h_1 + h_2) = P_2 + g\, h_2\, \rho_2 \label{static:eq:pBalance} \end{align} It can be noticed that the ``missing height'' is canceled between the two sides. It can be noticed that $h_1$ can be positive or negative or zero and it depends on the ratio that two containers filled with the light density liquid. Additionally, it can be observed that $h_1$ is relatively small because $A_1 >> A_2$. The densities of the liquid are chosen so that they are close to each other but not equal. The densities of the liquids are chosen to be much heavier than the measured gas density. Thus, in writing equation \eqref{static:eq:pBalance} the gas density was neglected. The pressure difference can be expressed as \begin{align} P_1 - P_2 = g \left[ \rho_2\,h_2 - \rho_1 (h_1 + h_2) \right] \label{static:eq:pDiffB} \end{align} If the light liquid volume in the two containers is known, it provides the relationship between $h_1$ and $h_2$. For example, if the volumes in two containers are equal then \begin{align} - h_1\, A_1 = h_2 \,A_2 \longrightarrow h_1 = - \dfrac{h_2\,A_2}{A_1} \label{static:eq:twoContainers} \end{align} Liquid volumes do not necessarily have to be equal. Additional parameter, the volume ratio, will be introduced when the volumes ratio isn't equal. The calculations as results of this additional parameter does not cause a significant complications. Here, this ratio equals to one and it simplify the equation \eqref{static:eq:twoContainers}. But this ratio can be inserted easily into the derivations. With the equation for height \eqref{static:eq:twoContainers} equation \eqref{static:eq:pBalance} becomes \begin{align} P_1 - P_2 = g \,h_2 \left( \rho_2 - \rho_1 \left(1 - \dfrac{A_2}{A_1} \right) \right) \label{static:eq:balanceContA} \end{align} or the height is \begin{align} h_2 = \dfrac{P_1 - P_2} {g \left[ (\rho_2 - \rho_1) + \rho_1 \dfrac{A_2}{A_1} \right] } \label{static:eq:balanceCont} \end{align} For the small value of the area ratio, $A_2/A_1 \ll 1$, then equation qref{static:eq:balanceCont} becomes \begin{align} h_2 = \dfrac{P_1 - P_2} {g \left( \rho_2 - \rho_1 \right) } \label{static:eq:balanceContShort} \end{align} Some refer to the density difference shown in equation \eqref{static:eq:balanceContShort} as ``magnification factor'' since it replace the regular density,

Inclined Manometer

Inclined Manometer

Fig. 4.8 Inclined manometer.

One of the old methods of pressure measurement is the inclined manometer. In this method, the tube leg is inclined relatively to gravity (depicted in Figure 4.8). This method is an attempt to increase the accuracy by ``extending'' length visible of the tube. The equation \eqref{static:eq:gageP} is then \begin{align} \label{static:eq:inclindManometer} P_1 - P_{outside} = \rho\,g\,d\ell \end{align} If there is a insignificant change in volume (the area ratio between tube and inclined leg is significant), a location can be calibrated on the inclined leg as zero.

Inverted U-tube manometer

Inverted Manometer

Fig. 4.9 Schematic of inverted manometer.

The difference in the pressure of two different liquids is measured by this manometer. This idea is similar to ``magnified'' manometer but in reversed. The pressure line are the same for both legs on line $ZZ$. Thus, it can be written as the pressure on left is equal to pressure on the right legs (see Figure 4.9). \begin{align} \label{static:eq:govInvMano} \overbrace{P_2 - \rho_2\,(b+h) }^{\text{right leg}}\,g = \overbrace{P_1 - \rho_1\,a - \rho\,h) }^{\text{left leg}}\,g \end{align} Rearranging equation \eqref{static:eq:govInvMano} leads to \begin{align} \label{static:eq:reGovInvMano} P_2- P_1 = \rho_2\,(b+h)\,g - \rho_1\,a \,g - \rho\,h \,g \end{align} For the similar density of $\rho_1=\rho_2$ and for $a=b$ equation \eqref{static:eq:reGovInvMano} becomes \begin{align} \label{static:eq:finalInvMano} P_2- P_1 = (\rho_1 - \rho )\,g \,h \end{align} As in the previous ``magnified'' manometer if the density difference is very small the height become very sensitive to the change of pressure.

4.3.3 Varying Density in a Gravity Field

There are several cases that will be discussed here which are categorized as gases, liquids and other. In the gas phase, the equation of state is simply the ideal gas model or the ideal gas with the compressibility factor (sometime referred to as real gas). The equation of state for liquid can be approximated or replaced by utilizing the bulk modulus. These relationships will be used to find the functionality between pressure, density and location.

4.3.3.1 Gas Phase under Hydrostatic Pressure

Ideal Gas under Hydrostatic Pressure

The gas density vary gradually with the pressure. As first approximation, the ideal gas model can be employed to describe the density. Thus equation \eqref{static:eq:dz} becomes \begin{align} \dfrac{\partial P } {\partial z} = - \dfrac{g\, P}{ R\, T} \label{static:eq:dzRhoGas} \end{align} Separating the variables and changing the partial derivatives to full derivative (just a notation for this case) results in \begin{align} \dfrac{dP} {P} = - \dfrac{g\, dz}{ R\, T} \label{static:eq:sepDzRhoGas} \end{align} Equation \eqref{static:eq:sepDzRhoGas} can be integrated from point ``0'' to any point to yield \begin{align} \ln \dfrac{P} {P_0} = - \dfrac{g }{R\,T} \left(z - z_0\right) \label{static:eq:dzRhoGasSolutionA} \end{align} It is convenient to rearrange equation \eqref{static:eq:dzRhoGasSolutionA} to the following \begin{align} \dfrac{P} {P_0} = {e}^ {- \left( \dfrac{g (z -z_o) }{R\,T} \right)} \label{static:eq:dzRhosolutionT} \end{align} Here the pressure ratio is related to the height exponentially. Equation \eqref{static:eq:dzRhosolutionT} can be expanded to show the difference to standard assumption of constant pressure as \begin{align} \dfrac{P} {P_0} = 1 - \overbrace{\dfrac{\left(z - z_0\right) g }{R\,T}} ^{-\dfrac{h\,\rho_0\, g}{P_0}} + \dfrac{\left(z - z_0\right)^2 g }{6\,R\,T} + \cdots \label{static:eq:dzRhoGasSolutionExpantion} \end{align} Or in a simplified form where the transformation of $h = (z - z_0)$ to be \begin{align} \dfrac{P} {P_0} = 1 + \dfrac{\rho_0\, g}{P_0} \left( h - \overbrace{\dfrac{h^2}{6} + \cdots}^{\text{correction factor}} \right) \label{static:eq:dzRhoGasSolutionExpantionSimpleC} \end{align} Equation \eqref{static:eq:dzRhoGasSolutionExpantionSimpleC} is useful in mathematical derivations but should be ignored for practical use.

Real Gas under Hydrostatic Pressure

The mathematical derivations for ideal gas can be reused as a foundation for the real gas model ($P = Z \rho R T$). For a large range of $P/P_c$ and $T/T_c$, the value of the compressibility factor, $Z$, can be assumed constant and therefore can be swallowed into equations \eqref{static:eq:dzRhosolutionT} and qref{static:eq:dzRhoGasSolutionExpantion}. The compressibility is defined in Thermodyanimcs Chapter. The modified equation is \begin{align} \dfrac{P} {P_0} = {e}^{- \left( \dfrac{g\, (z -z_o) }{Z\,R\,T} \right)} \label{static:eq:dzRhosolutionTz} \end{align} Or in a series form which is \begin{align} \dfrac{P} {P_0} = 1 - \dfrac{\left(z - z_0\right) g }{Z\, R\,T} + \dfrac{\left(z - z_0\right)^2 g }{6\,Z\,R\,T} + \cdots \label{static:eq:dzRhoGasSolutionExpantionA} \end{align} Without going through the mathematics, the first approximation should be noticed that the compressibility factor, $Z$ enter the equation as $h/Z$ and not just $h$. Another point that is worth discussing is the relationship of Z to other gas properties. In general, the relationship is very complicated and in some ranges $Z$ cannot be assumed constant. In these cases, a numerical integration must be carried out.

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 \begin{align} \rho = B_T \dfrac{\partial \rho}{\partial P} \label{static:eq:rhoSolutionG} \end{align} The variables for equation \eqref{static:eq:rhoSolutionG} should be separated and then the integration can be carried out as \begin{align} \int_{P_0}^{P}d\,P = \int_{\rho_0}^{\rho} B_T \dfrac{d \rho}{\rho} \label{static:eq:rhoSolutionIF} \end{align} The integration of equation \eqref{static:eq:rhoSolutionIF} yields \begin{align} P- {P_0} = B_T \ln \dfrac{\rho}{\rho_0} \label{static:eq:rhoSolutionI} \end{align} Equation \eqref{static:eq:rhoSolutionI} can be represented in a more convenient form as

Density variation

\begin{align} \label{static:eq:rhoSolutionF} {\rho} = {\rho_0} {e}^{\dfrac{P- {P_0}}{ B_T}} \end{align}
Equation \eqref{static:eq:rhoSolutionF} is the counterpart for the Utilizing equation \eqref{static:eq:rhoSolutionF} in equation qref{static:eq:dz} transformed into \begin{align} \dfrac{\partial P } {\partial z} = - g\, {\rho_0} {e}^{\dfrac{P- {P_0}}{ B_T}} \label{static:eq:dzRhoLiquidG} \end{align} Equation \eqref{static:eq:dzRhoLiquidG} can be integrated to yield \begin{align} \dfrac{B_T }{g\, \rho_0} \text{e}^{\dfrac{P- {P_0}}{ B_T}} = z + Constant \label{static:eq:dzRhoLiquidI} \end{align} It can be noted that $B_T$ has units of pressure and therefore the ratio in front of the exponent in equation qref{static:eq:dzRhoLiquidI} 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 \begin{align} Constant = \dfrac{B_T }{g\, \rho_0} \label{static:eq:dzRhoLiquidc} \end{align}
BT effect on static pressure

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 \begin{align} \dfrac{B_T }{g\, \rho_0} \left( \text{e}^{\dfrac{P- {P_0}}{ B_T}} - 1\right) = z \label{static:eq:dzRhoLiquidF} \end{align} Or in a dimensionless form

Density in Liquids

\begin{align} \label {static:eq:dzRhoLiquidFF} \left( \text{e}^{\dfrac{P- {P_0}}{ B_T}} - 1\right) = \dfrac{z \,g\, \rho_0}{B_T} \end{align}
The solution is presented in equation \eqref{static:eq:dzRhoLiquidF} and is plotted in Figure ef{static:fig:BTP0}. 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 \begin{align} \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} \end{align} An approximation of equation \eqref{static:eq:dzRhoLiquidFF} 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 \eqref{static1:eq:dzRhoLiquidAprox} 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

For the the first method the density is \begin{align} \label{hydrostatic:gov} B_T \cong \dfrac{\Delta P }{ \Delta V /V} \Longrightarrow \Delta V = V \dfrac{\Delta P }{B_T} \end{align} The density at the surface is $\rho = m/V$ and the density at point $x$ from the surface the density is \begin{align} \label{hydrostatic:rhoX} \rho(x) = \dfrac{m}{ V - \Delta V} \Longrightarrow \rho (x) = \dfrac{m}{ V - V \dfrac{\Delta P }{B_T}} \end{align} In this Chapter it was shown (integration of equation \eqref{static:eq:staticEq}) that the change pressure for constant gravity is \begin{align} \label{hydrostatic:DeltaP} \Delta P = g\, \int_0^z \rho (z) dz \end{align} Combining equation \eqref{hydrostatic:rhoX} with equation qref{hydrostatic:DeltaP} yields \begin{align} \label{hydrostatic:eqGov} \rho (z) = \dfrac{m}{ V - \displaystyle \dfrac{V\, g}{B_T} \,\int_0^z \rho (z) dz } \end{align} Equation can be rearranged to be \begin{align} \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) } \end{align} Equation \eqref{hydrostatic:eqGovr} is an integral equation which is discussed in the appendix. It is convenient to rearrange further equation \eqref{hydrostatic:eqGovr} to \begin{align} \label{hydrostatic:eqGovrf} 1 - \dfrac{g} {B_T} {\displaystyle\int_0^z \rho (z) dz } = \dfrac{\rho_0}{ \rho (z) } \end{align} The integral equation \eqref{hydrostatic:eqGovrf} can be converted to a differential equation form when the two sides are differentiated as \begin{align} \label{hydrostatic:govDif} \dfrac{g} {B_T} \rho (z) + \dfrac{\rho_0} {\rho(z)^2} \dfrac{d\,\rho(z)}{dz} = 0 \end{align} equation \eqref{hydrostatic:govDif} is first order non–linear differential equation which can be transformed into \begin{align} \label{hydrostatic:govDif1} \dfrac{g\,\rho (z)^3} {B_T\,\rho_0} + \dfrac{d\,\rho(z)}{dz} = 0 \end{align} The solution of equation \eqref{hydrostatic:govDif1} is \begin{align} \label{hydrostatic:gSol} \dfrac{\rho_0\,B_T}{2\,g\,{\rho}^{2}} = z + c \end{align} or rearranged as \begin{align} \label{hydrostatic:gSolr} \rho = \sqrt{\dfrac{\rho_0\,B_T}{2\,g\,\left( z + c\right) } } \end{align} The integration constant can be found by the fact that at $z=0$ the density is $\rho_0$ and hence \begin{align} \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}} \end{align} Substituting the integration constant and opening the parentheses, the solution is \begin{align} \label{hydrostatic:iniConPlus} \rho = \sqrt{\dfrac{\rho_0\,B_T} {2\,g\,z + \dfrac{\cancel{2\,g}\,B_T} {\cancel{2\,g}\,{\rho_0}}} } \end{align} Or \begin{align} \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) } } \end{align} Equation \eqref{hydrostatic:fSol} further be rearranged to a final form as \begin{align} \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) } } \end{align} The parameter $\dfrac{2\,g\,\rho_0\, z}{ B_T}$ represents the dimensional length controlling the problem. For small length the expression in \eqref{hydrostatic:rhoC} is similar to \begin{align} \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 \end{align} hence it can be expressed as \begin{align} \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 \end{align}


Advance Material

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 \begin{align} \label{liquidBT:sound} c = \sqrt{\dfrac{B_T}{\rho}} \end{align} 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

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 ?????????????? \begin{align} \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}} \end{align} The time the sound travel a small interval distance, $dz$ is \begin{align} \label{liquidBT:dt} d\tau = \dfrac{dz}{\sqrt{\dfrac{B_T-g\,\rho_0\,z} {\rho_0}}} \end{align} The time takes for the sound the travel the whole distance is the integration of infinitesimal time \begin{align} \label{liquidBT:ds} t = \int_0^D \dfrac{dz}{\sqrt{\dfrac{B_T-g\,\rho_0\,z} {\rho_0}}} \end{align} The solution of equation \eqref{liquidBT:ds} is \begin{align} \label{liquidBT:sol} t = \sqrt{\rho_0}\,\left( 2\,\sqrt{B_T}-2\,\sqrt{B_T-D}\right) \end{align} The time to travel according to the standard procedure is \begin{align} \label{liquidBT:oldTime} t = \dfrac{D}{\sqrt{\dfrac{B_T}{\rho_0}}} = \dfrac{D\,\sqrt{\rho_0}}{\sqrt{B_T}} \end{align} The ratio between the corrected estimated to the standard calculation is \begin{align} \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}}} \end{align}

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.

4.3.4 The Pressure Effects Due To Temperature Variations

4.3.4.1 The Basic Analysis

There are situations when the main change of the density results from other effects. For example, when the temperature field is not uniform, the density is affected and thus the pressure is a location function (for example, the temperature in the atmosphere is assumed to be a linear with the height under certain conditions.). A bit more complicate case is when the gas is a function of the pressure and another parameter. Air can be a function of the temperature field and the pressure. For the atmosphere, it is commonly assumed that the temperature is a linear function of the height. Here, a simple case is examined for which the temperature is a linear function of the height as \begin{align} \dfrac{dT}{dh} = - C_x \label{static:eq:Tdz} \end{align} where $h$ here referred to height or distance. Hence, the temperature–distance function can be written as \begin{align} T = Constant - C_x\, h \label{static:eq:TxxIassss} \end{align} where the $Constant$ is the integration constant which can be obtained by utilizing the initial condition. For $\;h = 0$, the temperature is $T_0$ and using it leads to

Temp variations

\begin{align} \label{static:eq:TxxI} T = T_0 - C_x\, h \end{align}
Combining equation \eqref{static:eq:TxxI} with qref{static:eq:dz} results in \begin{align} \dfrac {\partial P} {\partial h} = - \dfrac{g \,P }{R \left( T_0 -C_x\, h\right) } \label{static:eq:TxxIaa} \end{align} Separating the variables in equation \eqref{static:eq:TxxIaa} and changing the formal $\partial$ to the informal $d$ to obtain \begin{align} \dfrac {d\,P} {P} = - \dfrac{g \,dh }{R \left( T_0 -C_x\, h\right) } \label{static:eq:TxxG} \end{align} Defining a new variable as $\xi = (T_0 -C_x\, h)$ for which $\xi_0 = T_0 -C_x\, h_0$ and $d/d\xi = - C_x\,d/dh$. Using these definitions results in \begin{align} \dfrac {d\,P} {P} = \dfrac{g }{R C_x} \dfrac{d\xi}{ \xi } \label{static:eq:ThhG} \end{align} After the integration of equation \eqref{static:eq:TxxG} and reusing (the reverse definitions) the variables transformed the result into \begin{align} \ln \dfrac {P} {P_0} = \dfrac{g }{R \,C_x} \ln \dfrac{ T_0 -C_x\, h}{T_0} \label{static:eq:TxxIbb} \end{align} Or in a more convenient form as

Pressure in Atmosphere

\begin{align} \label{static:eq:TxxSi} \dfrac {P} {P_0} = \left(\dfrac{ T_0 -C_x\, h}{T_0} \right) ^ {\left( \dfrac{g }{R \,C_x} \right) } \end{align}
It can be noticed that equation \eqref{static:eq:TxxSi} is a monotonous function which decreases with height because the term in the brackets is less than one. This situation is roughly representing the pressure in the atmosphere and results in a temperature decrease. It can be observed that $C_x$ has a ``double role'' which can change the pressure ratio. Equation \eqref{static:eq:TxxSi} can be approximated by two approaches/ideas. The first approximation for a small distance, $h$, and the second approximation for a small temperature gradient. It can be recalled that the following expansions are \begin{multline} \dfrac{P}{P_0} = \lim_{h -> 0} {\left(1 - \dfrac{C_x}{T_0}\,h\right) }^{\dfrac{g}{R\,C_x}} = \\ { 1-\overbrace{\dfrac{g\,h}{T_0\,R}}^{\dfrac{g\,h\,\rho_0}{P_0}}- \overbrace{\dfrac{\left( R\,g\,C_x-{g}^{2}\right) \,{h}^{2}}{2\,{T_0}^{2}\,{R}^{2}}}^{\text{correction factor}} } - \cdots \quad \label{static:eq:aproxTonPh} \end{multline} Equation \eqref{static:eq:aproxTonPh} shows that the first two terms are the standard terms (negative sign is as expected i.e. negative direction). The correction factor occurs only at the third term which is important for larger heights. It is worth to point out that the above statement has a qualitative meaning when additional parameter is added. However, this kind of analysis will be presented in the dimensional analysis chapter. The second approximation for small $C_x$ is \begin{multline} \dfrac{P}{P_0} = \lim_{C_x -> 0} {\left(1 - \dfrac{C_x}{T_0}\,h\right) }^{\dfrac{g}{R\,C_x}} = \\ \text{ e}^{-\dfrac{g\,h}{R\,T_0}} - \dfrac{g\,{h}^{2}\,C_x}{2 \,{T_0}^{2}\,R} \text{e}^{-\dfrac{g\,h}{R\,T_0}} - \cdots \quad \label{static:eq:aproxTonPCx} \end{multline} Equation \eqref{static:eq:aproxTonPCx} shows that the correction factor (lapse coefficient), $C_x$, influences at only large values of height. It has to be noted that these equations qref{static:eq:aproxTonPh} and qref{static:eq:aproxTonPCx} are not properly represented without the characteristic height. It has to be inserted to make the physical significance clearer. Equation \eqref{static:eq:TxxSi} represents only the pressure ratio. For engineering purposes, it is sometimes important to obtain the This relationship can be obtained from combining equations qref{static:eq:TxxSi} and \eqref{static:eq:TxxI}. The simplest assumption to combine these equations is by assuming the ideal gas model, equation \eqref{thermo:eq:idealGas}, to yield \begin{align} \dfrac{\rho}{\rho_0} = \dfrac{P\,T_0}{P_0\,T} = \overbrace{\left( 1 - \dfrac{C_x\, h}{T_0} \right) ^ {\left( \dfrac{g }{R \,C_x} \right) } } ^{\dfrac{P}{P_0}} \overbrace{\left( 1 + \dfrac{C_x\, h}{T} \right)} ^{\dfrac{T_0}{T}} \label{static:eq:TxxRho} \end{align}
Advance Material

4.3.4.2 The Stability Analysis

\label{static:stability}

Two adjoin layers

Fig. 4.11 Two adjoin layers for stability analysis.

It is interesting to study whether this solution qref{static:eq:TxxSi} is stable and if so under what conditions. Suppose that for some reason, a small slab of material moves from a layer at height, $h$, to layer at height $h + dh$ (see Figure 4.11) What could happen? There are two main possibilities one: the slab could return to the original layer or two: stay at the new layer (or even move further, higher heights). The first case is referred to as the stable condition and the second case referred to as the unstable condition. The whole system falls apart and does not stay if the analysis predicts unstable conditions. A weak wind or other disturbances can make the unstable system to move to a new condition. This question is determined by the net forces acting on the slab. Whether these forces are toward the original layer or not. The two forces that act on the slab are the gravity force and the surroundings pressure (buoyant forces). Clearly, the slab is in equilibrium with its surroundings before the movement (not necessarily stable). Under equilibrium, the body forces that acting on the slab are equal to zero. That is, the surroundings ``pressure'' forces (buoyancy forces) are equal to gravity forces. The buoyancy forces are proportional to the ratio of the density of the slab to surrounding layer density. Thus, the stability question is whether the slab density from layer $h$, $\rho^{'}(h)$ undergoing a free expansion is higher or lower than the density of the layer $h+dh$. If $\rho^{'}(h) > \rho(h+dh)$ then the situation is stable. The term $\rho^{'}(h)$ is slab from layer $h$ that had undergone the free expansion. The reason that the free expansion is chosen to explain the process that the slab undergoes when it moves from layer $h$ to layer $h+dh$ is because it is the simplest. In reality, the free expansion is not far way from the actual process. The two processes that occurred here are thermal and the change of pressure (at the speed of sound). The thermal process is in the range of [cm/sec] while the speed of sound is about 300 [m/sec]. That is, the pressure process is about thousands times faster than the thermal process. The second issue that occurs during the ``expansion'' is the shock (in the reverse case $[h+dh]\rightarrow h$). However, this shock is insignificant (check book on Fundamentals of Compressible Flow Mechanics by this author on the French problem). The slab density at layer $h+dh$ can be obtained using equation qref{static:eq:TxxRho} as following \begin{align} \dfrac{\rho(h +dh) }{\rho(h)} = \dfrac{P\,T_0}{P_0\,T} = {\left( 1 - \dfrac{C_x\, dh}{T_0} \right) ^ {\left( \dfrac{g }{R \,C_x} \right) } } {\left( 1 + \dfrac{C_x\, dh}{T} \right)} \label{static:eq:TxxRhodh} \end{align} The pressure and temperature change when the slab moves from layer at $h$ to layer $h+dh$. The process, under the above discussion and simplifications, can be assumed to be adiabatic (that is, no significant heat transfer occurs in the short period of time). The little slab undergoes isentropic expansion as following for which (see equation \eqref{thermo:eq:idealGas}) \begin{align} \dfrac{\rho'(h+dh)}{\rho(h)} = \left( \dfrac{P'(h+dh)}{P(h)} \right) ^{1/k} \label{static:eq:dhInsentropic} \end{align} When the symbol $'$ denotes the slab that moves from layer $h$ to layer $h+dh$. The pressure ratio is given by equation \eqref{static:eq:TxxSi} but can be approximated by equation \eqref{static:eq:aproxTonPh} and thus \begin{align} \dfrac{\rho'(h+dh)}{\rho(h)} = \left( 1 - \dfrac{g\, dh}{T(h) \;R } \right) ^{1/k} \label{static:eq:dhInsentropicPtmp} \end{align} Again using the ideal gas model for equation qref{static:eq:dhInsentropicP} transformed into \begin{align} \dfrac{\rho'(h+dh)}{\rho(h)} = \left( 1 - \dfrac{\rho\, g dh}{P} \right) ^{1/k} \label{static:eq:dhInsentropicP} \end{align} Expanding equation \eqref{static:eq:dhInsentropicP} in Taylor series results in \begin{align} \left( 1 - \dfrac{\rho\, g dh}{P} \right) ^{1/k} = 1-\dfrac{g\,\rho\,dh}{P\,k} -\dfrac{\left(g^2\,\rho^2\,k-g^2\,\rho^2\right)\,dh^2} {2\,P^2\,k^2} -... \label{static:eq:dhInsentropicPExpansion} \end{align} The density at layer $h+dh$ can be obtained from qref{static:eq:TxxRhodh} and then it is expanded in taylor series as \begin{multline} \dfrac{\rho(h +dh) }{\rho(h)} = {\left( 1 - \dfrac{C_x\, dh}{T_0} \right) ^ {\left( \dfrac{g }{R \,C_x} \right) } } {\left( 1 + \dfrac{C_x\, dh}{T} \right)} \\ \sim 1 - \left( \dfrac{g\,\rho }{P} - \dfrac{C_x}{T} \right) dh + \cdots \quad \label{static:eq:dhTxxRho} \end{multline} The comparison of the right hand terms of equations qref{static:eq:dhTxxRho} and qref{static:eq:dhInsentropicPExpansion} provides the conditions From a mathematical point of view, to keep the inequality for a small $dh$ only the first term need to be compared as \begin{align} \dfrac{g\,\rho}{P\,k} > \dfrac{g\,\rho }{P} - \dfrac{C_x}{T} \label{static:eq:termoStability} \end{align} After rearrangement of the inequality \eqref{static:eq:termoStability} and using the ideal gas identity, it transformed to \begin{align} \nonumber \dfrac{C_x}{T} > \dfrac{(k-1)\,g\,\rho}{k\, P}\ C_x < \dfrac{k - 1 }{ k} \dfrac{g}{R} \label{static:eq:thSbl} \end{align} The analysis shows that the maximum amount depends on the gravity and gas properties. It should be noted that this value should be changed a bit since the $k$ should be replaced by polytropic expansion $n$. When lapse rate $C_x$ is equal to the right hand side of the inequality, it is said that situation is neutral. However, one has to bear in mind that this analysis only provides a range and isn't exact. Thus, around this value additional analysis is needed . One of the common question this author has been asked is about the forces of continuation. What is the source of the force(s) that make this situation when unstable continue to be unstable? Supposed that the situation became unstable and the layers have been exchanged, would the situation become stable now? One has to remember that temperature gradient forces continuous heat transfer which the source temperature change after the movement to the new layer. Thus, the unstable situation is continuously unstable.

4.3.5 Gravity Variations Effects on Pressure and Density

Varying Gravity Effects

Fig. 4.12 The varying gravity effects on density and pressure.

Until now the study focus on the change of density and pressure of the fluid. Equation \eqref{static:eq:dz} has two terms on the right hand side, the density, $\rho$ and the body force, $g$. The body force was assumed until now to be constant. This assumption must be deviated when the distance from the body source is significantly change. At first glance, the body force is independent of the fluid. The source of the gravity force in gas is another body, while the gravity force source in liquid can be the liquid itself. Thus, the discussion is separated into two different issues. The issues of magnetohydrodynamics are too advance for undergraduate student and therefore,will not be introduced here.

4.3.5.1 Ideal Gas in Varying Gravity

In physics, it was explained that the gravity is a function of the distance from the center of the plant/body. Assuming that the pressure is affected by this gravity/body force. The gravity force is reversely proportional to $r^2$. The gravity force can be assumed that for infinity, $r \rightarrow \infty$ the pressure is about zero. Again, equation \eqref{static:eq:dz} can be used (semi one directional situation) when $r$ is used as direction and thus \begin{align} \dfrac{\partial P }{\partial r} = - \rho \dfrac {G}{r^2} \label{static:eq:gravityGchange} \end{align} where $G$ denotes the general gravity constant. The regular method of separation is employed to obtain \begin{align} \int_{P_{b}}^{P} \dfrac{dP}{P} = - \dfrac {G}{RT} \int_{r_{b}} ^r \dfrac {dr}{r^2} \label{static:eq:gravityGchangeB} \end{align} where the subscript $b$ denotes the conditions at the body surface. The integration of equation \eqref{static:eq:gravityGchangeB} results in \begin{align} \ln \dfrac{P}{P_{b}} = - \dfrac {G}{RT} \left( \dfrac{1}{r_{b}}-\dfrac{1}{r}\right) \label{static:eq:gravityGchangeA} \end{align} Or in a simplified form as \begin{align} \dfrac{\rho}{\rho_{b}} = \dfrac{P}{P_{b}} = \text{ e} ^{ -\dfrac{G}{RT} \dfrac{r-r_{b}}{r\,r_{b}} } \label{static:eq:gravityGchangeF} \end{align} Equation \eqref{static:eq:gravityGchangeF} demonstrates that the pressure is reduced with the distance. It can be noticed that for $r \rightarrow r_{b}$ the pressure is approaching $P \rightarrow P_{b}$. This equation confirms that the density in outer space is zero $\rho(\infty) = 0$. As before, equation \eqref{static:eq:gravityGchangeF} can be expanded in Taylor series as \begin{align} \dfrac{\rho}{\rho_{b}} = \dfrac{P}{P_{b}} = \overbrace{1 - \dfrac{G\,\left( r-r_b\right) }{R\,T} }^{\text{standard}} - \overbrace{\dfrac{\left( 2\,G\,{R\,T}+ {G}^{2}\,r_b\right) \,{\left( r-r_b\right) }^{2}}{2\,r_b\,{(R\,T)}^{2}} +\cdots}^{\text{correction factor}} \label{static:eq:gravityGchangeFexpended} \end{align} Notice that $G$ isn't our beloved and familiar $g$ and also that $G\,r_b/RT$ is a dimensionless number (later in the Chapter 9 a discution about the definition of the dimensionless number and its meaning).

4.3.5.2 Real Gas in Varying Gravity

The regular assumption of constant compressibility, $Z$, is employed. It has to remember when this assumption isn't accurate enough, numerical integration is a possible solution. Thus, equation \eqref{static:eq:gravityGchangeB} is transformed into \begin{align} \int_{P_{b}}^{P} \dfrac{dP}{P} = - \dfrac {G}{Z\,R\,T} \int_{r_{b}} ^r \dfrac {dr}{r^2} \label{static:eq:gravityGchangeBZ} \end{align} With the same process as before for ideal gas case, one can obtain \begin{align} \dfrac{\rho}{\rho_{b}} = \dfrac{P}{P_{b}} = \text{ e} ^{ -\dfrac{G}{Z\,R\,T} \dfrac{r-r_{b}}{r\,r_{b}} } \label{static:eq:gravityGchangeFZ} \end{align} Equation \eqref{static:eq:gravityGchangeF} demonstrates that the pressure is reduced with the distance. It can be observed that for $r \rightarrow r_{b}$ the pressure is approaching $P \rightarrow P_{b}$. This equation confirms that the density in outer space is zero $\rho(\infty) = 0$. As before Taylor series for equation \eqref{static:eq:gravityGchangeF} is \begin{multline} \dfrac{\rho}{\rho_{b}} = \dfrac{P}{P_{b}} = \overbrace{1 - \dfrac{G\,\left( r-r_b\right) }{Z\,R\,T} }^{\text{standard}} - \overbrace{\dfrac{\left( 2\,G\,{Z\,R\,T}+ \\ {G}^{2}\,r_b\right) \,{\left( r-r_b\right) }^{2}}{2\,r_b\,{(Z\,R\,T)}^{2}} +\cdots}^{\text{correction factor}} \label{static:eq:gravityGchangeFZexpended} \end{multline} It can be noted that compressibility factor can act as increase or decrease of the ideal gas model depending on whether it is above one or below one. This issue is related to Pushka equation that will be discussed later.

4.3.5.3 Liquid Under Varying Gravity

For comparison reason consider the deepest location in the ocean which is about 11,000 [m]. If the liquid ``equation of state'' \eqref{static:eq:rhoSolutionF} is used with the hydrostatic fluid equation results in \begin{align} \dfrac{\partial P}{\partial r} = - {\rho_0} \text{ e}^{\dfrac{P- {P_0}}{ B_T}} \dfrac{G}{r^2} \label{static:eq:liquidGhydro} \end{align} which the solution of equation \eqref{static:eq:liquidGhydro} is \begin{align} \text{e}^{\dfrac{P_0-P}{B_T}} =Constant -\dfrac{B_T\,g\;\rho_0}{r} \label{static:eq:liquidGhroS} \end{align} Since this author is not aware to which practical situation this solution should be applied, it is left for the reader to apply according to problem, if applicable.

4.3.6 Liquid Phase

While for most practical purposes, the Cartesian coordinates provides sufficient treatment to the problem, there are situations where the spherical coordinates must be considered and used. Derivations of the fluid static in spherical coordinates are

Pressure Spherical Coordinates

\begin{align} \label{static:eq:spherical} \dfrac{1} {r^2} \dfrac{d}{dr} \left( \dfrac{r^2}{\rho} \dfrac{dP}{dr}\right) + 4\,\pi\,G\rho = 0 \end{align}
Or in a vector form as \begin{align} \nabla \bullet \left( \dfrac{1}{\rho}\nabla P \right) + 4\,\pi\,G\rho = 0 \label{static:eq:vSpherical} \end{align}

4.4 Fluid in a Accelerated System

Up to this stage, body forces were considered as one-dimensional. In general, the linear acceleration have three components as opposed to the previous case of only one. However, the previous derivations can be easily extended. Equation \eqref{static:eq:staticEq} can be transformed into a different coordinate system where the main coordinate is in the direction of the effective gravity. Thus, the previous method can be used and there is no need to solve new three (or two) different equations. As before, the constant pressure plane is perpendicular to the direction of the effective gravity. Generally the acceleration is divided into two categories: linear and angular and they will be discussed in this order.

4.4.1 Fluid in a Linearly Accelerated System

For example, in a two dimensional system, for the effective gravity \begin{align} g_{eff} = a\, \hat{i} + g\, \hat{k} \label{static:eq:geff2d} \end{align} where the magnitude of the effective gravity is \begin{align} | g_{eff} | = \sqrt{g^2 + a^2} \label{static:eq:g_eff_mag} \end{align} and the angle/direction can be obtained from \begin{align} tan \beta = \dfrac {a} {g} \label{static:eq:g_effAngle} \end{align} Perhaps the best way to explain the linear acceleration is by examples. Consider the following example to illustrate the situation.

Example 4.9

Effective gravity of Accelerated Box

Fig. 4.13 The effective gravity is for accelerated cart.

A tank filled with liquid is accelerated at a constant acceleration. When the acceleration is changing from the right to the left, what happened to the liquid surface? What is the relative angle of the liquid surface for a container in an accelerated system of $a = 5[m/sec]$?

Solution

This question is one of the traditional question of the fluid static and is straight forward. The solution is obtained by finding the effective angle body force. The effective angle is obtained by adding vectors. The change of the acceleration from the right to left is like subtracting vector (addition negative vector). This angle/direction can be found using the following \[ \tan^{-1} \beta = \tan^{-1} \dfrac {a} {g} = \dfrac{5}{9.81} \sim 27.01^{\circ} \] The magnitude of the effective acceleration is \[ \left|g_{eff}\right| = \sqrt{5^2 + 9.81^2} = 11.015 [m/sec^2] \]

Cart sliding inclined plane

Fig. 4.14 A cart slide on inclined plane.

Example 4.10

A cart partially filled with liquid and is sliding on an inclined plane as shown in Figure 4.14. Calculate the shape of the surface. If there is a resistance, what will be the angle? What happen when the slope angle is straight (the cart is dropping straight down)?

Solution

(a)

The angle can be found when the acceleration of the cart is found. If there is no resistance, the acceleration in the cart direction is determined from \begin{align} a = \mathbf{g} \sin \beta \label{static:eq:cartAcc} \end{align} The effective body force is acting perpendicular to the slope. Thus, the liquid surface is parallel to the surface of the inclination surface.

(b)

In case of resistance force (either of friction due to the air or resistance in the wheels) reduces the acceleration of the cart. In that case the effective body moves closer to the gravity forces. The net body force depends on the mass of the liquid and the net acceleration is \begin{align} a = \mathbf{g} - \dfrac{F_{net} }{m} \label{static:eq:aCartFnet} \end{align} The angle of the surface, $\alpha < \beta$, is now \begin{align} \tan \alpha = \dfrac {\mathbf{g} - \dfrac{F_{net} }{m}} {\mathbf{g}\, cos{\beta}} \label{static:eq:FnetAngle} \end{align}

Forces Sliding Cart

Fig. 4.15 Forces diagram of cart sliding on inclined plane.

(c)

In the case when the angle of the inclination turned to be straight (direct falling) the effective body force is zero. The pressure is uniform in the tank and no pressure difference can be found. So, the pressure at any point in the liquid is the same and equal to the atmospheric pressure.

4.4.2 Angular Acceleration Systems: Constant Density

For simplification reasons, the first case deals with a rotation in a perpendicular to the gravity. That effective body force can be written as \begin{align} \mathbf{g}_{eff} = - g\, \hat{k}+ \omega^2 r\; \hat{r} \label{static:eq:GeffAngular} \end{align}

Angular Angle

Fig. 4.16 Schematic to explain the angular angle.

The lines of constant pressure are not straight lines but lines of parabolic shape. The angle of the line depends on the radius as \begin{align} \dfrac{dz}{dr} = - \dfrac{g}{\omega^2\, r} \label{static:eq:angularAngle} \end{align} Equation \eqref{static:eq:angularAngle} can be integrated as \begin{align} z - z_0 = \dfrac{\omega^2\, r^2}{2\, g} \label{static:eq:angularAngleI} \end{align} Notice that the integration constant was substituted by $z_0$. The constant pressure will be along

Angular Acceleration System

\begin{align} \label{static:eq:angularPressure} P - P_0 = \rho\, g \, \left[ (z_0 - z) + \dfrac{\omega^2\,r^2}{ 2\,g} \right] \end{align}
To illustrate this point, example 4.11 is provided.
Angular Rotating System

Fig. 4.17 Schematic angular angle to explain example .

Example 4.11

A ``U'' tube with a length of $(1+x)L$ is rotating at angular velocity of $\omega$. The center of rotation is a distance, $L$ from the ``left'' hand side. Because the asymmetrical nature of the problem there is difference in the heights in the U tube arms of $S$ as shown in Figure 4.17. Expresses the relationship between the different parameters of the problem.

Solution

It is first assumed that the height is uniform at the tube (see for the open question on this assumption). The pressure at the interface at the two sides of the tube is same. Thus, equation \eqref{static:eq:angularAngleI} represents the pressure line. Taking the ``left'' wing of U tube \[ \overbrace{z_l - z_0}^{\mbox{change in z direction}} = \overbrace{\dfrac{\omega^2\,L^2}{2\,g}}^{\mbox{change in r direction}} \] The same can be said for the other side \[ z_r - z_0 = \dfrac{\omega^2\,x^2\,L^2}{2\,g} \] Thus subtracting the two equations above from each each other results in \[ z_r - z_l = \dfrac{L\,\omega^2\,\left(1 - x^2\right)}{2\,g} \] It can be noticed that this kind equipment can be used to find the gravity.

Example 4.12

Assume that the diameter of the U tube is $R_t$. What will be the correction factor if the curvature in the liquid in the tube is taken in to account. How would you suggest to define the height in the tube?

Solution

In Figure 4.17 shows the infinitesimal area used in these calculations. The distance of the infinitesimal area from the rotation center is ?. The height of the infinitesimal area is ?. Notice that the curvature in the two sides are different from each other. The volume above the lower point is ? which is only a function of the geometry.

Example 4.13

In the U tube in example 4.11 is rotating with upper part height of $ll$. At what rotating velocity liquid start to exit the U tube? If the rotation of U tube is exactly at the center, what happen the rotation approach very large value?


Advance Material

4.4.3 Fluid Statics in Geological System

Acknoldgemnt

This author would like to express his gratitude to Ralph Menikoff for suggesting this topic.

In geological systems such as the Earth provide cases to be used for fluid static for estimating pressure. It is common in geology to assume that the Earth is made of several layers. If this assumption is accepted, these layers assumption will be used to do some estimates. The assumption states that the Earth is made from the following layers: solid inner core, outer core, and two layers in the liquid phase with a thin crust. For the purpose of this book, the interest is the calculate the pressure at bottom of the liquid phase.
Earth Layers

Fig. 4.18 Earth layers not to scale.

This explanation is provided to understand how to use the bulk modulus and the effect of rotation. In reality, there might be an additional effects which affecting the situation but these effects are not the concern of this discussion. Two different extremes can recognized in fluids between the outer core to the crust. In one extreme, the equator rotation plays the most significant role. In the other extreme, at the north–south poles, the rotation effect is demished since the radius of rotation is relatively very small (see Figure 4.19). In that case, the pressure at the bottom of the liquid layer can be estimated using the equation \eqref{static:eq:dzRhoLiquidFF} or in approximation of equation \eqref{hydrostatic:gSolr}. In this case it also can be noticed that $g$ is a function of $r$. Different Radii If the bulk modulus is assumed constant (for simplicity), the governing equation can be constructed starting with equation \eqref{intro:eq:bulkModulus_r}. The approximate definition of the bulk modulus is \begin{align} \label{static:eq:iniGov} B_T = \dfrac{\rho \, \Delta P }{ \Delta \rho} \Longrightarrow \Delta \rho = \dfrac{\rho \, \Delta P }{ B_T} \end{align} Using equation to express the pressure difference (see Example 4.6 for details explanation) as \begin{align} \label{static:eq:govBT1} \rho(r) = \dfrac{\rho_0}{1 - \displaystyle \int_{R_0}^r \dfrac{g(r)\rho(r)}{B_T(r)} dr } \end{align} In equation \eqref{static:eq:govBT1} it is assumed that $B_T$ is a function of pressure and the pressure is a function of the location. Thus, the bulk modulus can be written as a function of the location radius, $r$. Again, for simplicity the bulk modulus is assumed to be constant.


End Advance Material

4.5 Fluid Forces on Surfaces

The forces that fluids (at static conditions) extracts on surfaces are very important for engineering purposes. This section deals with these calculations. These calculations are divided into two categories, straight surfaces and curved surfaces.

4.5.1 Fluid Forces on Straight Surfaces

A motivation is needed before going through the routine of derivations. Initially, a simple case will be examined. Later, how the calculations can be simplified will be shown.

Example 4.14

Consider a rectangular shape gate as shown in Figure 4.20. Calculate the minimum forces, $F_1$ and $F_2$ to maintain the gate in position. Assuming that the atmospheric pressure can be ignored.

Solution

Rectangular Shape Under Pressure

Fig. 4.20 Rectangular area under pressure.

The forces can be calculated by looking at the moment around point ``O.'' The element of moment is $a\,d\xi$ for the width of the gate and is \begin{align*} dM = \overbrace{P\, \underbrace{a\, d\xi}_{dA}}^{dF} (ll + \xi) \end{align*} The pressure, $P$ can be expressed as a function $\xi$ as the following \begin{align*} P = g\, \rho \, (ll +\xi) sin \beta \end{align*} The liquid total moment on the gate is \begin{align*} M = \int_0^b g\, \rho \, (ll +\xi) \sin \beta\, a\, d\xi (ll + \xi) \end{align*} The integral can be simplified as \begin{align*} M = g\, a\,\rho\,\sin\beta\,\int_0^b (ll +\xi)^2 d\xi \end{align*} The solution of the above integral is \begin{align*} M = g\, \rho\,a\,\sin\beta\,\left( \dfrac{3\,b\,{l}^{2}+3\,{b}^{2}\,l+{b}^{3}}{3} \right) \end{align*} This value provides the moment that $F_1$ and $F_2$ should extract. Additional equation is needed. It is the total force, which is \begin{align*} F_{total} = \int_0^b g\, \rho \, (ll +\xi) \sin \beta\, a\, d\xi \end{align*} The total force integration provides \begin{align*} F_{total} = g\,\rho\,a\,\sin \beta\, \int_0^b (ll +\xi) d\xi = g\,\rho\,a\,\sin \beta\,\left(\dfrac{2\,b\,ll+{b}^{2}}{2}\right) \end{align*} The forces on the gate have to provide \begin{align*} F_1 + F_2 = g\,\rho\,a\,\sin \beta\,\left(\dfrac{2\,b\,ll+{b}^{2}}{2}\right) \end{align*} Additionally, the moment of forces around point ``O'' is \begin{align*} F_1\,ll + F_2 (ll + b) = g\, \rho\,a\,\sin\beta\,\left( \dfrac{3\,b\,{l}^{2}+3\,{b}^{2}\,l+{b}^{3}}{3} \right) \end{align*} The solution of these equations is \begin{align*} F_1 = \dfrac{\left( 3\,ll+b\right) \,a\,b\,g\,\rho\,\sin\beta}{6} \end{align*} \begin{align*} F_2 = \dfrac{\left( 3\,ll+2\,b\right) \,a\,b\,g\,\rho\,\sin\beta}{6} \end{align*}

Submerged area under Pressure

Fig. 4.21 Schematic of submerged area to explain the center forces and moments.

The above calculations are time consuming and engineers always try to make life simpler. Looking at the above calculations, it can be observed that there is a moment of area in equation \eqref{static:eq:momentEx} and also a center of area. These concepts have been introduced in Chapter 3. Several represented areas for which moment of inertia and center of area have been tabulated in Chapter 3. These tabulated values can be used to solve this kind of problems.

Symmetrical Shapes

Consider the two–dimensional symmetrical area that are under pressure as shown in Figure 4.21. The symmetry is around any axes parallel to axis $x$. The total force and moment that the liquid extracting on the area need to be calculated. First, the force is \begin{multline} F = \int_A P dA = \int (P_{atmos} + \rho \, g \, h ) dA \\ = A\,P_{atmos} + \rho \, g \, \int_{ll_0}^{ll_1} \overbrace{ (\xi+ll_0) \sin \beta}^{h(\xi)} dA \qquad \label{static:eq:F_g} \end{multline} In this case, the atmospheric pressure can include any additional liquid layer above layer ``touching'' area. The ``atmospheric'' pressure can be set to zero. The boundaries of the integral of equation qref{static:eq:F_g} refer to starting point and ending points not to the start area and end area. The integral in equation \eqref{static:eq:F_g} can be further developed as \begin{align} F_{total} = A\,P_{atmos} + \rho\, g \, \sin\beta\,\left( ll_0\, A + \overbrace{\int_{ll_0}^{ll_1} \xi dA}^{x_c\,A} \right) \label{static:eq:F_ga} \end{align} In a final form as

Total Force in Inclined Surface

\begin{align} \label{static:eq:F_gaF} F_{total} = A\,\left[ P_{atmos} + \rho\,g\,\sin\beta\,\left(ll_0 +x_c \right) \right] \end{align}
Submerged forces

Fig. 4.22 The general forces acting on submerged area.

The moment of the liquid on the area around point ``O'' is \begin{align} M_y = \int _{\xi_0}^{\xi_1} P(\xi) \xi dA \label{static:eq:monentsO} \end{align} \begin{align} M_y = \int _{\xi_0}^{\xi_1} (P_{atmos}+ g\,\rho\, \overbrace{h(\xi)}^{\xi\,\sin\beta}\,) \xi dA \label{static:eq:momentsInt} \end{align} Or separating the parts as \begin{align} M_y = P_{atmos}\,\overbrace{\int _{\xi_0}^{\xi_1} \xi dA}^{x_c\,A}+ g\,\rho\,\sin\beta\overbrace{\int _{\xi_0}^{\xi_1}\xi^2 dA} ^{I_{x^{'}x^{'}}} \label{static:eq:momentsIntA} \end{align} The moment of inertia, $I_{x^{'}x^{'}}$, is about the axis through point ``O'' into the page. Equation \eqref{static:eq:momentsIntA} can be written in more compact form as

Total Moment in Inclined Surface

\begin{align} \label{static:eq:momentsIntAF} M_y = P_{atmos}\,{x_c\,A}+ g\,\rho\,\sin\beta {I_{x^{'}x^{'}}} \end{align}
Example 4.14 can be generalized to solve any two forces needed to balance the area/gate. Consider the general symmetrical body shown in figure 4.22 which has two forces that balance the body. Equations \eqref{static:eq:F_gaF} and qref{static:eq:momentsIntAF} can be combined the moment and force acting on the general area. If the ``atmospheric pressure'' can be zero or include additional layer of liquid. The forces balance reads \begin{align} F_1 + F_2 = A\,\left[ P_{atmos} + \rho\,g\,\sin\beta\,\left(ll_0 +x_c \right) \right] \label{static:eq:SFbalance} \end{align} and moments balance reads \begin{align} F_1\,a + F_2\, b = P_{atmos}\,{x_c\,A}+ g\,\rho\,\sin\beta {I_{x^{'}x^{'}}} \label{static:eq:SMbalance} \end{align} The solution of these equations is \begin{multline} F_1=\dfrac{\left[ \left(\rho\,\sin\beta-\dfrac{P_{atmos}}{g\,b}\right) \,x_c+ll_0\,\rho\,\sin\beta+\dfrac{P_{atmos}}{g}\right] \,b\,A } {g\left(b-a\right)} \\ - \dfrac{I_{x^{'}x^{'}}\,\rho\,\sin\beta} {g\left(b-a\right)} \qquad \label{static:eq:f1G} \end{multline} and \begin{multline} F_2=\dfrac{ I_{x^{'}x^{'}}\,\rho\,\sin\beta } {g\left(b-a\right)} - \\ \dfrac{\left[ \left(\rho\,\sin\beta-\dfrac{P_{atmos}}{g\,a}\right) \,x_c+ll_0\,\rho\,\sin\beta+\dfrac{P_{atmos}}{g}\right]\,a \,A} {g\left(b-a\right)} \qquad \label{static:eq:f2G} \end{multline} In the solution, the forces can be negative or positive, and the distance $a$ or $b$ can be positive or negative. Additionally, the atmospheric pressure can contain either an additional liquid layer above the ``touching'' area or even atmospheric pressure simply can be set up to zero. In symmetrical area only two forces are required since the moment is one dimensional. However, in non–symmetrical area there are two different moments and therefor three forces are required. Thus, additional equation is required. This equation is for the additional moment around the $x$ axis (see for explanation in Figure 4.23). The moment around the $y$ axis is given by equation \eqref{static:eq:momentsIntAF} and the total force is given by \eqref{static:eq:F_gaF}. The moment around the $x$ axis (which was arbitrary chosen) should be \begin{align} M_x = \int_A y\, P dA \label{static:eq:moment_x} \end{align} Substituting the components for the pressure transforms equation qref{static:eq:moment_x} into \begin{align} M_x = \int_A y\, \left(P_{atmos} + \rho\,g\,\xi\,\sin\beta\right)dA \label{static:eq:moment_x1} \end{align} The integral in equation \eqref{static:eq:moment_x} can be written as \begin{align} M_x = P_{atmos}\,\overbrace{\int_A y\, dA}^{A\,y_c} + \rho\,g\,\sin\beta\,\overbrace{\int_A\xi\,y\,dA}^{I_{x^{'}y^{'}}} \label{static:eq:momShort} \end{align} The compact form can be written as

Moment in Inclined Surface

\begin{align} \label{static:eq:momShortF} M_x = P_{atmos}\,{A\,y_c} + \rho\,g\,\sin\beta\,{I_{x^{'}y^{'}}} \end{align}
Non Symmetrical Area Moments

Fig. 4.23 The general forces acting on non symmetrical straight area.

The product of inertia was presented in Chapter 3. These equations \eqref{static:eq:F_gaF}, qref{static:eq:momentsIntAF} and \eqref{static:eq:momShortF} provide the base for solving any problem for straight area under pressure with uniform density. There are many combinations of problems (e.g. two forces and moment) but no general solution is provided. Example to illustrate the use of these equations is provided.

Example 4.15

Calculate the forces which required to balance the triangular shape shown in the Figure 4.24.

Solution

The three equations that needs to be solved are \begin{align} F_1+F_2+F_3 = F_{total} \label{static:eq:triangleF1} \end{align} The moment around $x$ axis is \begin{align} F_1\,b = M_y \label{static:eq:triangleMy} \end{align} The moment around $y$ axis is \begin{align} F_1\,ll_1 +F_2\,(a +ll_0 ) +F_3\,ll_0 = M_x \label{static:eq:triangleMx} \end{align} The right hand side of these equations are given before in equations \eqref{static:eq:F_gaF}, qref{static:eq:momentsIntAF} and \eqref{static:eq:momShortF}. The moment of inertia of the triangle around $x$ is made of two triangles (as shown in the Figure qref{static:fig:triangleForeces} for triangle 1 and 2). Triangle 1 can be calculated as the moment of inertia around its center which is $ll_0+ 2*(ll_1-ll_0)/3$. The height of triangle 1 is $(ll_1 - ll_0)$ and its width $b$ and thus, moment of inertia about its center is $I_{xx} = b (ll_1 - ll_0)^3/36$. The moment of inertia for triangle 1 about $y$ is \[ {I_{xx}}_1 = \dfrac{b (ll_1 - ll_0)^3}{36} + \overbrace{\dfrac{b (ll_1 - ll_0)}{3}}^{A_1} \, \overbrace{\left(ll_0+ \dfrac{2(ll_1-ll_0)}{3}\right)^2} ^{{\Delta x_1}^2} \] The height of the triangle 2 is $a - (ll_1 - ll_0)$ and its width $b$ and thus, the moment of inertia about its center is \[ {I_{xx}}_2 = \dfrac{b [a - (ll_1 - ll_0)]^3}{36} + \overbrace{\dfrac{b [a - (ll_1 - ll_0)]}{3}}^{A_2} \, \overbrace{\left(ll_1 + \dfrac{[a - (ll_1 - ll_0)]}{3} \right)^2}^{{\Delta x_2}^2} \]

Non symmetrical Area Moments

Fig. 4.24 The general forces acting on a non symmetrical straight area.

and the total moment of inertia \begin{align*} I_{xx} = {I_{xx}}_1 + {I_{xx}}_2 \end{align*} The product of inertia of the triangle can be obtain by integration. It can be noticed that upper line of the triangle is $y = \dfrac{\left(ll_1 - ll_0\right)x}{b} +ll_0$. The lower line of the triangle is $y = \dfrac{\left(ll_1 - ll_0 - a \right)x}{b} +ll_0 + a$. \begin{align*} I_{xy} = \int_0^b \left[\int_{\dfrac{\left(ll_1 - ll_0\right)x}{b} +ll_0}^{\dfrac{\left(ll_1 - ll_0 - a \right)x}{b} +ll_0 + a} x\,y\, dx \right] dy =\dfrac {2\,a\,b^2\,ll_1+2\,a\,b^2\,ll_0+a^2\,b^2} {24} \end{align*} The solution of this set equations is \begin{align*} F_1=\overbrace{\left[\dfrac{a\,b}{3}\right]}^{A} \dfrac{\left( g\,\left( 6\,ll_1+3\,a\right) +6\,g\,ll_0\right) \,\rho\,\sin\beta+8\,P_{atmos}}{24}, \end{align*} \begin{multline*} \dfrac{F_2} {\left[\dfrac{a\,b}{3}\right]} =- \begin{array}{c} \dfrac{\left( \left( 3\,ll_1-14\,a\right) - ll_0\,\left( \dfrac{12\,ll_1}{a}-27\right) + \dfrac{12\,{ll_0}^{2}}{a}\right)\,g \,\rho\,\sin\beta} {72} - \\ \dfrac{\left( \left( \dfrac{24\,ll_1}{a}-24\right) +\dfrac{48\,ll_0}{a}\right) \,P_{atmos}} {72}, \end{array} \end{multline*} \begin{multline*} \dfrac{F_3} {\left[\dfrac{a\,b}{3}\right]} = \begin{array}{c} \dfrac{\left( \left( a-\dfrac{15\,ll_1}{a}\right) +ll_0\,\left( 27- \dfrac{12\,ll_1}{a}\right) +\dfrac{12\,{ll_0}^{2}}{a}\right)\,g\,\rho\,\sin\beta } {72}\\ + \dfrac{\left( \left(\dfrac{24\,ll_1}{a}+24\right) +\dfrac{48\,ll_0}{a}\right) \,P_{atmos} }{72} \end{array} \end{multline*}

4.5.1.1 Pressure Center

In the literature, pressure centers are commonly defined. These definitions are mathematical in nature and has physical meaning of equivalent force that will act through this center. The definition is derived or obtained from equation qref{static:eq:momentsIntAF} and equation qref{static:eq:momShortF}. The pressure center is the distance that will create the moment with the hydrostatic force on point ``O.'' Thus, the pressure center in the $x$ direction is \begin{align} x_p = \dfrac{1}{F} \int_A x\,P \,dA \label{static:eq:xp} \end{align} In the same way, the pressure center in the $y$ direction is defined as \begin{align} y_p = \dfrac{1}{F} \int_A y\,P \,dA \label{static:eq:yp} \end{align} To show relationship between the pressure center and the other properties, it can be found by setting the atmospheric pressure and $ll_0$ to zero as following \begin{align} x_p = \dfrac{g\,\rho\sin\beta\,I_{x^{'}x^{'}}} {A\,\rho\,g\,\sin\beta\,x_c} \label{static:eq:MyF} \end{align} Expanding $I_{x^{'}x^{'}}$ according to Chapter on Mechamics results in \begin{align} x_p = \dfrac{I_{xx}}{x_c \,A} + x_c \label{static:eq:xpPx} \end{align} and in the same fashion in $y$ direction \begin{align} y_p = \dfrac{I_{xy}}{y_c \,A} + y_c \label{static:eq:ypPy} \end{align} It has to emphasis that these definitions are useful only for case where the atmospheric pressure can be neglected or canceled and where $ll_0$ is zero. Thus, these limitations diminish the usefulness of pressure center definitions. In fact, the reader can find that direct calculations can sometimes simplify the problem.

4.5.1.2 Multiply Layers

In the previous sections, the density was assumed to be constant. For non constant density the derivations aren't ``clean'' but are similar. Consider straight/flat body that is under liquid with a varying density. If density can be represented by average density, the force that is acting on the body is \begin{align} F_{total} = \int_A g\,\rho\, h \, dA \sim \bar{\rho} \int_A g\, h \, dA \label{static:eq:vRhoStraight} \end{align} In cases where average density cannot be represented reasonably, the integral has be carried out. In cases where density is non–continuous, but constant in segments, the following can be said \begin{multline} F_{total} = \int_A g\,\rho\, h \, dA = \int_{A_1} g\,\rho_1\, h \, dA + \\ \int_{A_2} g\,\rho_2\, h \, dA + \cdots + \int_{A_n} g\,\rho_n\, h \, dA \qquad \label{static:eq:disConRhoIni} \end{multline} As before for single density, the following can be written \begin{align} F_{total} = g\, \sin\beta\, \left[ \rho_1\, \overbrace{\int_{A_1} \xi\, dA}^{{x_c}_1\,A_1} + \rho_2\, \overbrace{\int_{A_2} \xi\, dA}^{{x_c}_2\,A_2} + \cdots + \rho_n\, \overbrace{\int_{A_n} \xi\, dA}^{{x_c}_n\,A_n} \right] \label{static:eq:disConRhoMainAAA} \end{align} Or in a compact form and in addition considering the ``atmospheric'' pressure can be written as

Total Static Force

\begin{align} \label{static:eq:disConRhoMainA} F_{total} = P_{atmos}\,A_{total} + g\, \sin\beta\, \sum_{i=1}^{n} \rho_i\, {{x_c}_i\,A_i} \end{align}
where the density, $\rho_i$ is the density of the layer $i$ and $A_i$ and ${x_c}_i$ are geometrical properties of the area which is in contact with that layer. The atmospheric pressure can be entered into the calculation in the same way as before. Moreover, the atmospheric pressure can include all the layer(s) that do(es) not with the ``contact'' area. The moment around axis $y$, $M_y$ under the same considerations as before is \begin{align} M_y = \int_A g\,\rho \,\xi^2\, \,\sin\beta\, dA \label{static:eq:disConRhoMyIniA} \end{align} After similar separation of the total integral, one can find that \begin{align} M_y = g\,\sin\beta\, \sum_{i=1}^{n} \rho_i\, {{I_{x^{'}x^{'}}}_i} \label{static:eq:disConRhoMyIni} \end{align} If the atmospheric pressure enters into the calculations one can find that

Total Static Moment

\begin{align} \label{static:eq:disConRhoMyIniF} M_y = P_{atmos} \,x_c\, A_{total} + g\,\sin\beta\, \sum_{i=1}^{n} \rho_i\, {{I_{x^{'}x^{'}}}_i} \end{align}
In the same fashion one can obtain the moment for $x$ axis as

Total Static Moment

\begin{align} \label{static:eq:disConRhoMxIniF} M_x = P_{atmos} \,y_c\, A_{total} + g\,\sin\beta\, \sum_{i=1}^{n} \rho_i\, {{I_{x^{'}y^{'}}}_i} \end{align}
To illustrate how to work with these equations the following example is provided.

Example 4.16

Consider the hypothetical Figure 4.25. The last layer is made of water with density of $1000 [kg/m^3]$. The densities are $\rho_1 = 500[kg/m^3]$, $\rho_2 = 800[kg/m^3]$, $\rho_3 = 850[kg/m^3]$, and $\rho_4 = 1000[kg/m^3]$. Calculate the forces at points $a_1$ and $b_1$. Assume that the layers are stables without any movement between the liquids. Also neglect all mass transfer phenomena that may occur. The heights are: $h_1 = 1[m]$, $h_2 = 2[m]$, $h_3 = 3[m]$,and $h_4 = 4[m]$. The forces distances are $a_1=1.5[m]$, $a_2=1.75[m]$, and $b_1=4.5[m]$. The angle of inclination is is $\beta= 45^\circ$.

Multi Layers Density

Fig. 4.25 The effects of multi layers density on static forces.

Solution

Since there are only two unknowns, only two equations are needed, which are \eqref{static:eq:disConRhoMyIniF} and qref{static:eq:disConRhoMainA}. The solution method of this example is applied for cases with less layers (for example by setting the specific height difference to be zero). Equation \eqref{static:eq:disConRhoMyIniF} can be used by modifying it, as it can be noticed that instead of using the regular atmospheric pressure the new ``atmospheric'' pressure can be used as \[ {P_{atmos}}^{'} = P_{atmos} + \rho_1\,g\,h_1 \] The distance for the center for each area is at the middle of each of the ``small'' rectangular. The geometries of each areas are \begin{array}{lcr} {x_c}_1 = \dfrac{a_2 + \dfrac{h_2}{\sin\beta}}{2} & A_1 = ll \left( \dfrac{h_2}{\sin\beta} -a_2 \right) & {I_{x^{'}x^{'}}}_1 = \dfrac{ll\left(\dfrac{h_2}{\sin\beta}-a_2\right)^{3}}{36} + \left({x_c}_1\right)^2\, A_1 \\ {x_c}_2 = \dfrac{h_2 + h_3}{2\,\sin\beta} & A_2 = \dfrac{ll}{\sin\beta} \left(h_3 - h_2\right) & {I_{x^{'}x^{'}}}_2 = \dfrac{ll\left({h_3}-h_2\right)^{3}}{36\,\sin\beta} + \left({x_c}_2\right)^2\, A_2 \\ {x_c}_3 = \dfrac{h_3 + h_4}{2\,\sin\beta} & A_3 = \dfrac{ll}{\sin\beta} \left(h_4 - h_3\right) & {I_{x^{'}x^{'}}}_3 = \dfrac{ll\left({h_4}-h_3\right)^{3}}{36\,\sin\beta} + \left({x_c}_3\right)^2\, A_3 \end{array} After inserting the values, the following equations are obtained Thus, the first equation is \[ F_1 + F_2 = {P_{atmos}}^{'} \overbrace{ll (b_2-a_2)}^{A_{total}} + g\,\sin\beta\,\sum_{i=1}^{3}\rho_{i+1}\, {x_c}_i\, A_i \] The second equation is \eqref{static:eq:disConRhoMyIniF} to be written for the moment around the point ``O'' as \[ F_1\,a_1 + F_2\,b_1 = {P_{atmos}}^{'}\, \overbrace{\dfrac{(b_2+a_2)}{2}{ll (b_2-a_2)}}^{{x_c}{A_{total}}} + g\,\sin\beta\sum_{i=1}^{3}\rho_{i+1}\,{I_{x^{'}x^{'}}}_i \] The solution for the above equations is \begin{multline*} F1= \begin{array}{c} \dfrac{ 2\,b_1\,g\,\sin\beta\,\sum_{i=1}^{3}\rho_{i+1}\, {x_c}_i\, A_i -2\,g\,\sin\beta\,\sum_{i=1}^{3}\rho_{i+1}\,{I_{x^{'}x^{'}}}_i } {2\,b_1-2\,a_1} - \\ \qquad\dfrac{\left({b_2}^{2}-2\,b_1\,b_2+2\,a_2\,b_1-{a_2}^{2}\right) ll\,P_{atmos}} {2\,b_1-2\,a_1} \end{array} \end{multline*} \begin{multline*} F2= \begin{array}{c} \dfrac{ 2\,g\,\sin\beta\,\sum_{i=1}^{3}\rho_{i+1}\,{I_{x^{'}x^{'}}}_i -2\,a_1\,g\,\sin\beta\,\sum_{i=1}^{3}\rho_{i+1}\, {x_c}_i\, A_i} {2\,b_1-2\,a_1} + \\ \dfrac{ \left( {b_2}^{2}+2\,a_1\,b_2+{a_2}^{2}-2\,a_1\,a_2\right) ll\,P_{atmos}} {2\,b_1-2\,a_1} \end{array} \end{multline*} The solution provided isn't in the complete long form since it will makes things messy. It is simpler to compute the terms separately. A mini source code for the calculations is provided in the the text source. The intermediate results in SI units ([m], [$m^2$], [$m^4$]) are: \begin{align*} \begin{array}{lcr} x_{c1}=2.2892& x_{c2}=3.5355& x_{c3}=4.9497\\ A_1=2.696& A_2=3.535& A_3=3.535\\ {I_{x'x'}}_1=14.215& {I_{x'x'}}_2=44.292& {I_{x'x'}}_3=86.718 \end{array} \end{align*} The final answer is \[ F_1=304809.79[N] \] and \[ F_2=958923.92[N] \]

4.5.2 Forces on Curved Surfaces

Curved Area

Fig. 4.26 The forces on curved area.

The pressure is acting on surfaces perpendicular to the direction of the surface (no shear forces assumption). The element force is \begin{align} d\,\mathbf{F} = - P\, \hat{n}\, \mathbf{dA} \label{static:eq:dForc} \end{align} Here, the conventional notation is used which is to denote the area, $dA$, outward as positive. The total force on the area will be the integral of the unit force \begin{align} \mathbf{F} = - \int_{A} P\, \hat{n}\, \mathbf{dA} \label{static:eq:Force} \end{align} The result of the integral is a vector. So, if the $y$ component of the force is needed, only a dot product is needed as \begin{align} dF_y = d\mathbf{F} \bullet \hat{j} \label{static:eq:Fy} \end{align} From this analysis (equation \eqref{static:eq:Fy}) it can be observed that the force in the direction of $y$, for example, is simply the integral of the area perpendicular to $y$ as \begin{align} F_y = \int_A P\, dA_y \label{static:eq:FyA} \end{align} The same can be said for the $x$ direction. The force in the $z$ direction is \begin{align} F_z = \int_A h\,g\,\rho dA_z \label{static:eq:dzForce} \end{align}

Net Forces on Floating Body

Fig. 4.27 Schematic of Net Force on floating body.

The force which acting on the $z$ direction is the weight of the liquid above the projected area plus the atmospheric pressure. This force component can be combined with the other components in the other directions to be \begin{align} F_{total} = \sqrt{ {F_z}^2 + {F_x} ^2 + {F_y}^2} \label{static:eq:FcurvedTotal} \end{align} And the angle in ``$x\,z$'' plane is \begin{align} \tan \theta_{xz} = \dfrac{F_z}{F_x} \label{static:eq:FcurvedFanglezx} \end{align} and the angle in the other plane, ``$y\,z$'' is \begin{align} \tan \theta_{zy} = \dfrac{F_z}{F_y} \label{static:eq:FcurvedFanglezy} \end{align} The moment due to the curved surface require integration to obtain the value. There are no readily made expressions for these 3–dimensional geometries. However, for some geometries there are readily calculated center of mass and when combined with two other components provide the moment (force with direction line).

Cut–Out Shapes Effects

There are bodies with a shape that the vertical direction ($z$ direction) is ``cut–out'' aren't continuous. Equation \eqref{static:eq:dzForce} implicitly means that the net force on the body is $z$ direction is only the actual liquid above it. For example, Figure 4.27 shows a floating body with cut–out slot into it. The atmospheric pressure acts on the area with continuous lines. Inside the slot, the atmospheric pressure with it piezometric pressure is canceled by the upper part of the slot. Thus, only the net force is the actual liquid in the slot which is acting on the body. Additional point that is worth mentioning is that the depth where the cut–out occur is insignificant (neglecting the change in the density).

Example 4.17

Effective gravity of Accelerated Box

Fig. 4.28 Calculations of forces on a circular shape dam.

Calculate the force and the moment around point ``O'' that is acting on the dam (see Figure \eqref{static:fig:quaterCirc}). The dam is made of an arc with the angle of $\theta_0=45^{\circ}$ and radius of $r=2[m]$. You can assume that the liquid density is constant and equal to 1000 $[kg/m^3]$. The gravity is 9.8$[m/sec^2]$ and width of the dam is $b=4[m]$. Compare the different methods of computations, direct and indirect.

Solution

The force in the $x$ direction is \begin{align*} F_x = \int_{A} P \overbrace{r\,\cos\theta\,d\theta}^{dA_x} \end{align*} Note that the direction of the area is taken into account (sign). The differential area that will be used is, $b\,r\,d\theta$ where $b$ is the width of the dam (into the page). The pressure is only a function of $\theta$ and it is \begin{align*} P = P_{atmos} + \rho \, g \, r \sin\theta \end{align*} The force that is acting on the $x$ direction of the dam is $A_x \times P$. When the area $A_x$ is $b\,r\,d\theta \,\cos\theta$. The atmospheric pressure does cancel itself (at least if the atmospheric pressure on both sides of the dam is the same.). The net force will be \begin{align*} F_x = \int_0^{\theta_0} \overbrace{\rho \,g\,r\,\sin\theta}^{P} \overbrace{\,b\,r\,\cos \theta\,d\theta}^{dA_x} \end{align*} The integration results in

Area Arc Subtract the Triangle

Fig. 4.29 Area above the dam arc subtract triangle.

\begin{align*} F_x = \dfrac{\rho\,g\,b\,r^2}{2}\, \left(1-{cos^{2}\left( \theta_ 0\right) } \right) \end{align*} Alternative way to do this calculation is by calculating the pressure at mid point and then multiply it by the projected area, $A_x$ (see Figure 4.29) as \begin{align*} F_x = \rho\,g\,\overbrace{b\,r\sin\theta_0}^{A_x} \overbrace{\dfrac{r\sin\theta_0}{2}}^{x_c} = \dfrac{\rho\,g\,b\,r}{2}\, \sin^2\theta \end{align*} Notice that $dA_x$($\cos\theta$) and $A_x$ ($\sin\theta$) are different, why? The values to evaluate the last equation are provided in the question and simplify subsidize into it as \begin{align*} F_x = \dfrac{1000\times 9.8\times 4 \times 2}{2} \sin(45^{\circ}) = 19600.0 [N] \end{align*} Since the last two equations are identical (use the sinuous theorem to prove it $ \sin^2\theta + \cos^2 = 1$), clearly the discussion earlier was right (not a good proof The force in the $y$ direction is the area times width. \begin{align*} F_y = - \overbrace{\left(\overbrace{\dfrac{\theta_0\,r^2}{2}- \dfrac{r^2\sin\theta_0\cos\theta_0}{2}}^{A}\right) \,b}^{V} \,g\, \rho \sim 22375.216[N] \end{align*} The center area ( purple area in Figure 4.29) should be calculated as \begin{align*} y_c = \dfrac{{y_c\,A}_{arc} - {y_c\,A}_{triangle}}{A} \end{align*} The center area above the dam requires to know the center area of the arc and triangle shapes. Some mathematics are required because the shift in the arc orientation. The arc center (see Figure 4.30) is at \begin{align*} {y_c}_{arc} = \dfrac{4\,r \sin^2\left(\dfrac{\theta}{2} \right)} {3\,\theta} \end{align*}

Arc Center in the y direction

Fig. 4.30 Area above the dam arc calculation for the center.

All the other geometrical values are obtained from Tables and ef{mech:tab:momentInertia2}. and substituting the proper values results in \begin{align*} {y_c}_r = \dfrac{\overbrace{\dfrac{\theta\,r^2}{2}}^{A_{arc}} \overbrace{ \dfrac{4\,r \sin\left(\dfrac{\theta}{2} \right) \cos\left(\dfrac{\theta}{2}\right) } {3\,\theta} }^{y_c} - \overbrace{\dfrac{2\,r\,\cos\theta}{3}}^{y_c} \overbrace{\dfrac{\sin\theta\,r^2}{2}}^{A_{triangle}} } {\underbrace{\dfrac{\theta\,r^2}{2}}_{A_{arc}} - \underbrace {\dfrac{r^2\,\sin\theta\,\cos\theta}{2}}_{A_{triangle}}} \end{align*} This value is the reverse value and it is \begin{align*} {y_c}_r = 1.65174[m] \end{align*} The result of the arc center from point ``O'' (above calculation area) is \begin{align*} y_c = r - {y_c}_{r} = 2 - 1.65174 \sim 0.348[m] \end{align*} The moment is \begin{align*} M_v = y_c \,F_y \sim 0.348\times 22375.2 \sim 7792.31759 [N\times m] \end{align*} The center pressure for $x$ area is \begin{align*} x_p = x_c + \dfrac{I_{xx} }{x_c\,A} = \dfrac{r\,cos\theta_0}{2} + \dfrac{\overbrace{\dfrac{\cancel{b}\,\left(r\cos\theta_0\right)^3} {36}}^{I_{xx}}} {\underbrace{\dfrac{r\,cos\theta_0}{2}}_{x_c} \cancel{b}\left(r\,\cos\theta_0\right)} = \dfrac{5\,r\cos\theta_0}{9} \end{align*} The moment due to hydrostatic pressure is \begin{align*} M_h = x_p \, F_x = \dfrac{5\,r\,cos\theta_0}{9} \,F_x \sim 15399.21[N\times m] \end{align*} The total moment is the combination of the two and it is \begin{align*} M_{total} = 23191.5 [N\times m] \end{align*}

Moment on Arc Element

Fig. 4.31 Moment on arc element around Point ``O.''

For direct integration of the moment it is done as following \begin{align*} dF = P\,dA = \int_0^{\theta_0} \rho \, g\, \sin\theta \,b\,r\,d\theta \end{align*} and element moment is \begin{align*} dM = dF \times ll = dF \overbrace{2\,r\,\sin\left(\dfrac{\theta}{2} \right)}^{ll} \cos\left(\dfrac{\theta}{2} \right) \end{align*} and the total moment is \begin{align*} M = \int_0^{\theta_0} dM \end{align*} or \begin{align*} M = \int_0^{\theta_0} \rho \, g\, \sin\theta \,b\,r\, {2\,r\,\sin\left(\dfrac{\theta}{2} \right)} \cos\left(\dfrac{\theta}{2} \right) d\theta \end{align*} The solution of the last equation is \begin{align*} M = \dfrac{g\,r\,\rho\, \left( 2\,\theta_0 - \sin\left( 2\,\theta_0\right) \right) }{4} \end{align*} The vertical force can be obtained by \begin{align*} F_v = \int_0^{\theta_0}P\,dA_v \end{align*} or \begin{align*} F_v = \int_0^{\theta_0} \overbrace{\rho\,g\, r\,\sin\theta}^{P} \overbrace{r\,d\theta\,\cos\theta}^{dA_v} \end{align*} \begin{align*} F_v = \dfrac{g\,{r}^{2}\,\rho}{2} \,\left( 1 -{{cos\left( \theta_0\right) }^{2}}\right) \end{align*} Here, the traditional approach was presented first, and the direct approach second. It is much simpler now to use the second method. In fact, there are many programs or hand held devices that can carry numerical integration by inserting the function and the boundaries.

To demonstrate this point further, consider a more general case of a polynomial function. The reason that a polynomial function was chosen is that almost all the continuous functions can be represented by a Taylor series, and thus, this example provides for practical purposes of the general solution for curved surfaces.

Example 4.18

General Area  of Polynomial Shape

Fig. 4.32 Polynomial shape dam description for the moment around point ``O'' and force calculations.

For the liquid shown in Figure ,calculate the moment around point ``O'' and the force created by the liquid per unit depth. The function of the dam shape is $y = \sum_{i=1}^n a_i\,x^i$ and it is a monotonous function (this restriction can be relaxed somewhat). Also calculate the horizontal and vertical forces.

Solution

The calculations are done per unit depth (into the page) and do not require the actual depth of the dam. The element force (see Figure 4.32) in this case is \[ dF = \overbrace{\overbrace{(b-y)}^{h}\,g\,\rho}^{P} \overbrace{\sqrt{dx^2 + dy^2}}^{dA} \] The size of the differential area is the square root of the $dx^2$ and $dy^2$ (see Figure 4.32). It can be noticed that the differential area that is used here should be multiplied by the depth. From mathematics, it can be shown that \[ \sqrt{dx^2 + dy^2} = dx\,\sqrt{1+ \left(\dfrac{dy}{dx}\right)^2} \]

Difference Between Angles

Fig. 4.33 The difference between the slop and the direction angle.

The right side can be evaluated for any given function. For example, in this case describing the dam function is \begin{align*} \sqrt{1+ \left(\dfrac{dy}{dx}\right)^2} = \sqrt{ 1 + {\left( \sum_{i=1}^{n}i\,a\left( i\right) \,{x\left( i\right) } ^{i-1}\, \right) }^{2} } \end{align*} The value of $x_b$ is where $y=b$ and can be obtained by finding the first and positive root of the equation of \begin{align*} 0 = \sum_{i=1}^n a_i\,x^i - b \end{align*} To evaluate the moment, expression of the distance and angle to point ``O'' are needed (see Figure ef{static:fig:6Poly}). The distance between the point on the dam at $x$ to the point ``O'' is \begin{align*} ll(x) = \sqrt{(b-y)^2+ (x_b-x)^2 } \end{align*} The angle between the force and the distance to point ``O'' is \begin{align*} \theta(x) = \tan^{-1}\left(\dfrac{dy}{dx}\right) - \tan^{-1}\left(\dfrac{b-y}{x_b-x}\right) \end{align*} The element moment in this case is \begin{align*} dM = ll(x)\, \overbrace{(b-y)\,g\,\rho\,\sqrt{1+ \left(\dfrac{dy}{dx}\right)^2}}^{dF} \,\cos\theta(x)\, dx \end{align*} To make this example less abstract, consider the specific case of $y = 2\,x^6$. In this case, only one term is provided and $x_b$ can be calculated as following \begin{align*} x_b = \sqrt[6]{\dfrac{b}{2}} \end{align*} Notice that $ \sqrt[6]{\dfrac{b}{2}}$ is measured in meters. The number ``2'' is a dimensional number with units of [1/$m^5$]. The derivative at $x$ is \begin{align*} \dfrac{dy}{dx} = 12\,x^5 \end{align*} and the derivative is dimensionless (a dimensionless number). The distance is \begin{align*} ll = \sqrt{\left(b -2\,x^6\right)^2 + \left( \sqrt[6]{\dfrac{b}{2}} - x \right)^2} \end{align*} The angle can be expressed as \begin{align*} \theta = \tan^{-1}\left(12\,x^5\right) - \tan^{-1}\left(\dfrac{b- 2\,x^6}{\sqrt[6]{\dfrac{b}{2}} -x}\right) \end{align*} The total moment is \begin{align*} M = \int_0^{\sqrt[6]{b}} ll(x) \cos\theta(x) \left( b-2\,x^6 \right)\,g\,\rho \sqrt{1+12\,x^5}\;dx \end{align*} This integral doesn't have a analytical solution. However, for a given value $b$ this integral can be evaluate. The horizontal force is \begin{align*} F_h = b\,\rho\,g \,\dfrac{b}{2} = \dfrac{\rho\,g\,b^2}{2} \end{align*} The vertical force per unit depth is the volume above the dam as \begin{align*} F_v = \int_0^{\sqrt[6]{b}} \left(b - 2\,x^6\right) \rho \,g\,dx = \rho \,g\,\dfrac{5\,{b}^{\dfrac{7}{6}}}{7} \end{align*} In going over these calculations, the calculations of the center of the area were not carried out. This omission saves considerable time. In fact, trying to find the center of the area will double the work. This author find this method to be simpler for complicated geometries while the indirect method has advantage for very simple geometries.

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