Basics of Fluid Mechanics
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Chapter Static (continue)

4.6 Buoyancy and Stability

Immersed Cylinder

Fig. 4.34 Schematic of Immersed Cylinder.

One of the oldest known scientific research on fluid mechanics relates to buoyancy due to question of money was carried by Archimedes. Archimedes principle is related to question of density and volume. While Archimedes did not know much about integrals, he was able to capture the essence. Here, because this material is presented in a different era, more advance mathematics will be used. While the question of the stability was not scientifically examined in the past, the floating vessels structure (more than 150 years ago) show some understanding. The total forces the liquid exacts on a body are considered as a buoyancy issue. To understand this issue, consider a cubical and a cylindrical body that is immersed in liquid and center in a depth of, $h_0$ as shown in Figure 4.34. The force to hold the cylinder at the place must be made of integration of the pressure around the surface of the square and cylinder bodies. The forces on square geometry body are made only of vertical forces because the two sides cancel each other. However, on the vertical direction, the pressure on the two surfaces are different. On the upper surface the pressure is $\rho \, g \, (h_0-a/2)$. On the lower surface the pressure is $\rho \, g \, (h_0+a/2)$. The force due to the liquid pressure per unit depth (into the page) is \begin{align} F = \rho \, g \, \left( (h_0-a/2) - (h_0+a/2) \right)\,ll\, b = - \rho\, g\, a\, b\, ll = -\rho \, g \, V \label{static:eq:qPforce} \end{align} In this case the $ll$ represents a depth (into the page). Rearranging equation \eqref{static:eq:qPforce} to be \begin{align} \dfrac{F} {V} = \rho \, g \label{static:eq:qPforcePerell} \end{align} The force on the immersed body is equal to the weight of the displaced liquid. This analysis can be generalized by noticing two things. All the horizontal forces are canceled. Any body that has a projected area that has two sides, those will cancel each other. Another way to look at this point is by approximation. For any two rectangle bodies, the horizontal forces are canceling each other. Thus even these bodies are in contact with each other, the imaginary pressure make it so that they cancel each other. On the other hand, any shape is made of many small rectangles. The force on every rectangular shape is made of its weight of the volume. Thus, the total force is made of the sum of all the small rectangles which is the weight of the sum of all volume.

Immersed Cylinder Force

Fig. 4.35 The floating forces on Immersed Cylinder.

In illustration of this concept, consider the cylindrical shape in Figure 4.34. The force per area (see Figure 4.35) is \begin{align} dF = \overbrace{\rho\,g\, \left(h_0 - r\, \sin \theta\right)}^{P} \overbrace{\sin \theta\, r\, d\theta }^{dA_{vertical}} \label{static:eq:cylinderElement} \end{align} The total force will be the integral of the equation qref {static:eq:cylinderElement} \begin{align} F = \int_{0}^{2\pi} {\rho\,g\,\left(h_0 - r\, \sin \theta\right)} {\,r\, d\theta\, \sin \theta} \label{static:eq:cylinderElementIa} \end{align} Rearranging equation \eqref{static:eq:cylinderElement} transforms it to \begin{align} F = r\,g\,\rho\int_{0}^{2\pi} { \left(h_0 - r\, \sin \theta\right)} {\,\sin \theta\, d\theta } \label{static:eq:cylinderElementI} \end{align} The solution of equation \eqref{static:eq:cylinderElementI} is \begin{align} F = -\pi \,{r}^{2}\,\rho\,g \label{static:eq:cylinderElementISolution} \end{align} The negative sign indicate that the force acting upwards. While the horizontal force is \begin{align} F_v = \int_{0}^{2\,\pi} { \left(h_0 - r\, \sin \theta\right)} \cos\theta\,d\theta = 0 \label{static:eq:cylinderhorizontalForce} \end{align}

Example 4.19

To what depth will a long log with radius, $r$, a length, $ll$ and density, $\rho_w$ in liquid with density, $\rho_l$. Assume that $\rho_l>\rho_w$. You can provide that the angle or the depth.

Floating Thin Body

Fig. 4.36 Schematic of a thin wall floating body.

Typical examples to explain the buoyancy are of the vessel with thin walls put upside down into liquid. The second example of the speed of the floating bodies. Since there are no better examples, these examples are a must.

Example 4.20

A cylindrical body, shown in Figure 4.36 ,is floating in liquid with density, $\rho_{l}$. The body was inserted into liquid in a such a way that the air had remained in it. Express the maximum wall thickness, $t$, as a function of the density of the wall, $\rho_s$ liquid density, $\rho_{l}$ and the surroundings air temperature, $T_1$ for the body to float. In the case where thickness is half the maximum, calculate the pressure inside the container. The container diameter is $w$. Assume that the wall thickness is small compared with the other dimensions ($t << w$ and $t<

Solution

The air mass in the container is \[ m_{air} = \overbrace{\pi\,w^2\,h}^{V} \overbrace{\dfrac{P_{atmos}}{R\,T}} ^{\rho_{air}} \] The mass of the container is \[ m_{container} = \left(\overbrace{\pi\,w^2 + 2\,\pi\,w\,h}^{A}\right) \,t \,\rho_{s} \] The liquid amount enters into the cavity is such that the air pressure in the cavity equals to the pressure at the interface (in the cavity). Note that for the maximum thickness, the height, $h_1$ has to be zero. Thus, the pressure at the interface can be written as \[ P_{in} = \rho_{l}\,g\,h_{in} \] On the other hand, the pressure at the interface from the air point of view (ideal gas model) should be \[ P_{in} = \dfrac{m_{air}\,R\,T_1} {\underbrace{ h_{in}\,\pi\,w^2}_{V}} \] Since the air mass didn't change and it is known, it can be inserted into the above equation. \[ \rho_{l}\,g\,h_{in}+ P_{atmos} = P_{in} = \dfrac{{\left(\pi\,w^2\,h\right) \overbrace{\dfrac{P_{atmos}}{R\,T_1}}^{\rho} }\,R\,T_1} { h_{in}\,\pi\,w^2} \] The last equation can be simplified into \[ \rho_{l}\,g\,h_{in} + P_{atmos} = \dfrac{h \, {P_{atmos}} } {h_{in}} \] And the solution for $h_{in}$ is \[ h_{in}= - \dfrac{P_{atmos} +\sqrt{4\,g\,h\,P_{atmos}\,\rho_l+{P_{atmos}}^{2}} }{2\,g\,\rho_l} \] and \[ h_{in} = \dfrac {\sqrt{4\,g\,h\,P_{atmos}\,\rho_l+{P_{atmos}}^{2}}-P_{atmos}} {2\,g\,\rho_l} \] The solution must be positive, so that the last solution is the only physical solution.


Advance Material

Example 4.21

Calculate the minimum density an infinitely long equilateral triangle (three equal sides) has to be so that the sharp end is in the water.

Solution

The solution demonstrates that when $h \longrightarrow 0$ then $h_{in} \longrightarrow 0$. When the gravity approaches zero (macro gravity) then \begin{align*} h_{in}= \dfrac{P_{atmos}}{\rho_l\,g}+h -\dfrac{{h}^{2}\,\rho_l\,g}{P_{atmos}} +\dfrac{2\,{h}^{3}\,{\rho_l}^{2}\,{g}^{2}}{{P_{atmos}}^{2}} -\dfrac{5\,{h}^{4}\,{\rho_l}^{3}\,{g}^{3}}{{P_{atmos}}^{3}} +\cdots \end{align*} This ``strange'' result shows that bodies don't float in the normal sense. When the floating is under vacuum condition, the following height can be expanded into \begin{align*} h_{in}=\sqrt{\dfrac{h\,P_{atmos}} {{g\,\rho_l}}} +\dfrac{P_{atmos}}{2\,g\,\rho_l} + \cdots \end{align*} which shows that the large quantity of liquid enters into the container as it is expected. Archimedes theorem states that the force balance is at displaced weight liquid (of the same volume) should be the same as the container, the air. Thus, \begin{align*} \overbrace{\pi\, w^2\, (h-h_{in}) \,g}^{\text{net displayed
water}}= \overbrace{\left(\pi\,w^2 + 2\,\pi\,w\,h\right)\,t\,\rho_{s}\,g} ^{\text{container}} + \overbrace{{\pi\,w^2\,h}\, \left(\dfrac{P_{atmos}}{R\,T_1} \right) \,g} ^{\text{air}} \end{align*} If air mass is neglected the maximum thickness is \begin{align*} t_{max} = \dfrac{ 2\,g\,h\,w\,\rho_l+P_{atmos}\,w -w\,\sqrt{4\,gh\,P_{atmos}\,\rho_l+{P_{atmos}}^{2}} } {\left( 2\,g\,w+4\,g\,h\right) \,\rho_l\,\rho_s} \end{align*} The condition to have physical value for the maximum thickness is \begin{align*} 2\,g\,h\,\rho_l+P_{atmos} \ge \sqrt{4\,gh\,P_{atmos}\,\rho_l+{P_{atmos}}^{2}} \end{align*} The full solution is \begin{align*} \begin{array}{cc} t_{max} = & -\dfrac{\left( w\,R\,\sqrt{4\,gh\,P_{atmos}\,\rho_l+{P_{atmos}}^{2}} -2\,g\,h\,w\,R\,\rho_l-P_{atmos}\,w\,R\right) \,T_1+2\,g\,h\,P_{atmos}\,w\,\rho_l} {\left( 2\,g\,w+4\,g\,h\right) \,R\,\rho_l\,\rho_s\,T_1} \end{array} \end{align*} In this analysis the air temperature in the container immediately after insertion in the liquid has different value from the final temperature. It is reasonable as the first approximation to assume that the process is adiabatic and isentropic. Thus, the temperature in the cavity immediately after the insertion is \begin{align*} \dfrac{T_i}{T_f} = \left( \dfrac{P_i}{P_f} \right) \end{align*} The final temperature and pressure were calculated previously. The equation of state is \begin{align*} P_i = \dfrac{m_{air}\,R\,T_i}{V_i} \end{align*} The new unknown must provide additional equation which is \begin{align*} V_i = \pi\,w^2\,h_{i} \end{align*}

Thickness Below The Maximum

For the half thickness $t= \dfrac{t_{max}}{2}$ the general solution for any given thickness below maximum is presented. The thickness is known, but the liquid displacement is still unknown. The pressure at the interface (after long time) is \begin{align*} \rho_l \,g \, h_{in} +P_{atmos} = \dfrac{\pi\,w^2\,h \dfrac{P_{atmos}}{R\,T_1} R\, T_1} {\left(h_{in}+h_1\right)\,\pi\,w^2} \end{align*} which can be simplified to \begin{align*} \rho_l \, g\,h_{in} + P_{atmos} = \dfrac{h\,P_{atmos}}{h_{in}+h_1} \end{align*} The second equation is Archimedes' equation, which is \begin{align*} \pi\,w^2\left(h-h_{in} -h_1\right) = \left( \pi\,w^2 +2\,\pi\,w\,h)\,t\,\rho_s\,g \right) +\pi\,w^2\,h\,\left( \dfrac{P_{atmos}}{R\,T_1}\right)\,g \end{align*}


End Advance Material

Example 4.22

A body is pushed into the liquid to a distance, $h_0$ and left at rest. Calculate acceleration and time for a body to reach the surface. The body's density is $\alpha\, \rho_{l}$ , where $\alpha$ is ratio between the body density to the liquid density and ($0 < \alpha < 1$). Is the body volume important?

Solution

The net force is \begin{align*} F = \overbrace{V\,g\,\rho_l}^{\text{liquid weight}} - \overbrace{V\,g\,\alpha\,\rho_l}^{\text{body weight}} = V\,g\, \rho_l \,( 1 -\alpha) \end{align*} But on the other side the internal force is \begin{align*} F = m\,a = \overbrace{V\,\alpha \rho_l}^{m}\, a \end{align*} Thus, the acceleration is \begin{align*} a = g \left( \dfrac{1-\alpha}{\alpha}\right) \end{align*} If the object is left at rest (no movement) thus time will be ($h=1/2\,a\,t^2$) \begin{align*} t = \sqrt{\dfrac{2\,h \alpha}{g(1-\alpha)}} \end{align*} If the object is very light ($\alpha \longrightarrow 0$) then \begin{align*} t_{min} = \sqrt{\dfrac{2\,h\,\alpha}{g}} +\dfrac{\sqrt{2\,g\,h}\;{\alpha}^{\dfrac{3}{2}}} {2\,g} +\dfrac{3\,\sqrt{2\,g\,h}\,{\alpha}^{\dfrac{5}{2}}}{8\,g} +\dfrac{5\,\sqrt{2\,g\,h}\,{\alpha}^{\dfrac{7}{2}}}{16\,g} +\cdots \end{align*} From the above equation, it can be observed that only the density ratio is important. This idea can lead to experiment in ``large gravity'' because the acceleration can be magnified and it is much more than the reverse of free falling.

Example 4.23

In some situations, it is desired to find equivalent of force of a certain shape to be replaced by another force of a ``standard'' shape. Consider the force that acts on a half sphere. Find equivalent cylinder that has the same diameter that has the same force.

Solution

The force act on the half sphere can be found by integrating the forces around the sphere. The element force is \[ dF = (\rho_L - \rho_S) \, g\, \overbrace{r\, \cos\phi\, \cos\theta}^{h} \overbrace{\cos\theta\,\cos\phi\, \overbrace{r^2\,d\theta\,d\phi}^{dA} }^{dA_x} \] The total force is then \[ F_x = \int_0^{\pi} \int_0^{\pi} (\rho_L - \rho_S) \, g\, {\cos^2\phi \cos^2\theta} \, {r^3\,d\theta\,d\phi} \] The result of the integration the force on sphere is \[ F_s = \dfrac{{\pi}^{2}\, (\rho_L - \rho_S)\, r^3 }{4} \] The force on equivalent cylinder is \[ F_c = \pi\,r^2 \, (\rho_L - \rho_S)\,h \] These forces have to be equivalent and thus \[ \dfrac{{\pi}^{\cancel{2}}\, \cancel{(\rho_L-\rho_S)}\,r^{\cancelto{1}{3}}}{4} = \cancel{\pi}\,\cancel{r^2} \, \cancel{(\rho_L - \rho_S)}\,h \] Thus, the height is \[ \dfrac{h}{r} = \dfrac{\pi}{4} \]

Example 4.24

In the introduction to this section, it was assumed that above liquid is a gas with inconsequential density. Suppose that the above layer is another liquid which has a bit lighter density. Body with density between the two liquids, $\rho_l < \rho_s < rho_h$ is floating between the two liquids. Develop the relationship between the densities of liquids and solid and the location of the solid cubical. There are situations where density is a function of the depth. What will be the location of solid body if the

Solution

In the discussion to this section, it was shown that net force is the body volume times the the density of the liquid. In the same vein, the body can be separated into two: one in first liquid and one in the second liquid. In this case there are two different liquid densities. The net force down is the weight of the body $\rho_c\, h\, A$. Where $h$ is the height of the body and $A$ is its cross section. This force is balance according to above explanation by the two liquid as \begin{align*} \rho_c\, \cancel{h\, A} = \cancel{A \,h}\,\left( \alpha\,\rho_l + (1-\alpha) \rho_h \right) \end{align*} Where $\alpha$ is the fraction that is in low liquid. After rearrangement it became \begin{align*} \alpha = \dfrac{ \rho_c - \rho_h}{\rho_l - \rho_h} \end{align*} the second part deals with the case where the density varied parabolically. The density as a function of $x$ coordinate along $h$ starting at point $\rho_h$ is \begin{align*} \rho (x) = \rho_h - \left( \dfrac{x}{h} \right)^2 \left( \rho_h - \rho_l \right) \end{align*} Thus the equilibration will be achieved, $A$ is canceled on both sides, when \begin{align*} \rho_c\, h = \int_{x_1}^{x_1+h} \left[ \rho_h - \left( \dfrac{x}{h} \right)^2 \left(\rho_h-\rho_l\right) \right]dx \end{align*} After the integration the equation transferred into \begin{align*} \rho_c\, h = \dfrac{\left( 3\,\rho_l-3\,\rho_h\right) \,{x1}^{2}+ \left( 3\,h\,\rho_l-3\,h\,\rho_h\right) \,x1+{h}^{2}\,\rho_l+2\,{h}^{2}\,\rho_h} {3\,h} \end{align*} And the location where the lower point of the body (the physical), $x_1$, will be at \begin{align*} X_1 = \dfrac{\sqrt{3}\,\sqrt{3\,h^2\,{\rho_l}^{2}+\left( 4\,\rho_c-6\,{h}^{2}\,\rho_h\right) \,\rho_l+3\,{h}^{2}\,{\rho_h}^{2}-12\,\rho_c\,\rho_h}+3\,h\,\rho_l-3\,h\,\rho_h} {6\,\rho_h-2\,\rho_l} \end{align*} For linear relationship the following results can be obtained. \begin{align*} x_1=\dfrac{h\,\rho_l+h\,\rho_h-6\,\rho_c}{2\,\rho_l-2\,\rho_h} \end{align*} In many cases in reality the variations occur in small zone compare to the size of the body. Thus, the calculations can be carried out under the assumption of sharp change. However, if the body is smaller compare to the zone of variation, they have to accounted for.

Example 4.25

A hollow sphere is made of steel ($\rho_s/\rho_w \cong 7.8$) with a $t$ wall thickness. What is the thickness if the sphere is neutrally buoyant? Assume that the radius of the sphere is $R$. For the thickness below this critical value, develop an equation for the depth of the sphere.

Solution

The weight of displaced water has to be equal to the weight of the sphere \begin{align} \label{sphere:gov} \rho_s\,\cancel{g} \, \dfrac{4\,\pi\, R^3}{3} = \rho_w \,\cancel{g} \, \left( \dfrac{4\,\pi\, R^3}{3} - \dfrac{4\,\pi\, \left(R-t\right)^3}{3} \right) \end{align} after simplification equation \eqref{sphere:gov} becomes \begin{align} \label{sphere:govR} \dfrac{\rho_s\,R^3 }{\rho_w} = 3\,t\,{R}^{2}-3\,{t}^{2}\,R+{t}^{3} \end{align} Equation \eqref{sphere:govR} is third order polynomial equation which it's solution (see the appendix) is \begin{align*} \label{sphere:completSol} t_1&=&\left( -\dfrac{\sqrt{3}\,i}{2}-\dfrac{1}{2}\right) \,{\left( {\dfrac{\rho_s}{\rho_w}R}^{3}- {R}^{3}\right) }^{\dfrac{1}{3}}+R \\ t_2&=&\left( \dfrac{\sqrt{3}\,i}{2}-\dfrac{1}{2}\right) \,{\left( {\dfrac{\rho_s}{\rho_w} R}^{3}- {R}^{3}\right) }^{\dfrac{1}{3}}+R\\ t_3&=& R\,\left( \sqrt[3]{ \dfrac{\rho_s}{\rho_w} - 1 } + 1 \right) \end{align*} The first two solutions are imaginary thus not valid for the physical world. The last solution is the solution that was needed. The depth that sphere will be located depends on the ratio of $t/R$ which similar analysis to the above. For a given ratio of $t/R$, the weight displaced by the sphere has to be same as the sphere weight. The volume of a sphere cap (segment) is given by \begin{align} \label{sphere:capV} V_{cap} = \dfrac{\pi\,h^2\,(3R-h)}{3} \end{align} Where $h$ is the sphere height above the water. The volume in the water is \begin{align} \label{sphere:waterV} V_{water} = \dfrac{4\,\pi\, R^3}{3} - \dfrac{\pi\,h^2\,(3R-h)}{3} = \dfrac{4\,\pi\,\left( R^3 -3\,R\,h^2 + h^3 \right) }{3} \end{align} When $ V_{water}$ denotes the volume of the sphere in the water. Thus the Archimedes law is \begin{align} \label{sphere:archimedes1} \dfrac{\rho_w\,4\,\pi\,\left( R^3 -3\,R\,h^2 + h^3 \right) }{3} = \dfrac{\rho_s\,4\,\pi\,\left( 3\,t\,{R}^{2}-3\,{t}^{2}\,R+{t}^{3} \right)}{3} \end{align} or \begin{align} \label{sphere:archimedes} \left( R^3 -3\,R\,h^2 + h^3 \right) = \dfrac{\rho_w}{\rho_s} \left( 3\,t\,{R}^{2}-3\,{t}^{2}\,R+{t}^{3} \right) \end{align} The solution of \eqref{sphere:archimedes} is \begin{multline} \label{sphere:solArc} h = \left( \dfrac{\sqrt{-fR\,\left( 4\,{R}^{3}-fR\right) }}{2}-\dfrac{fR-2\,{R}^{3}}{2}\right)^{\dfrac{1}{3}} \\ + \dfrac{{R}^{2}} {{\left( \dfrac{\sqrt{-fR\,\left( 4\,{R}^{3}-fR\right) }}{2}-\dfrac{fR-2\,{R}^{3}}{2}\right)}^{\dfrac{1}{3}}} \end{multline} Where $-fR = R^3- \dfrac{\rho_w}{\rho_s}\,(3\,t\,R^2-3\,t^2\,R+t^3)$ There are two more solutions which contains the imaginary component. These solutions are rejected.

Example 4.26

One of the common questions in buoyancy is the weight with variable cross section and fix load. For example, a wood wedge of wood with a fix weight/load. The general question is at what the depth of the object (i.e. wedge) will be located. For simplicity, assume that the body is of a solid material.

Solution

It is assumed that the volume can be written as a function of the depth. As it was shown in the previous example, the relationship between the depth and the displaced liquid volume of the sphere. Here it is assumed that this relationship can be written as \begin{align} \label{FixVariableW:d-V} V_w = f(d,\mbox{other geometrical parameters}) \end{align} The Archimedes balance on the body is \begin{align} \label{FixVariableW:archimedes1} \rho_{ll} V_{a}= \rho_{w} V_{w} \end{align} \begin{align} \label{FixVariableW:archimedes} d = f^{-1} \dfrac{\rho_{ll} V_{a}}{ \rho_{w}} \end{align}

Example 4.27

In example 4.26 a general solution was provided. Find the reverse function, $f^{-1}$ for cone with $30^{\circ}$ when the tip is in the bottom.

Solution

First the function has to built for $d$ (depth). \begin{align} \label{woodenCone:gov} V_{w} = \dfrac{\pi\,d\,\left(\dfrac{d}{\sqrt{3}} \right)^2}{3} = \dfrac{\pi\,d^3}{9} \end{align} Thus, the depth is \begin{align} \label{woodenCone:d} d = \sqrt[3]{\dfrac{9\,\pi\, \rho_w}{\rho_{ll}\,V_a} } \end{align}

Stability of Floating Bodies

Fig. 4.37 Schematic of floating bodies.

4.6.1 Stability

Figure 4.37 shows a body made of hollow balloon and a heavy sphere connected by a thin and light rod. This arrangement has mass centroid close to the middle of the sphere. The buoyant center is below the middle of the balloon. If this arrangement is inserted into liquid and will be floating, the balloon will be on the top and sphere on the bottom. Tilting the body with a small angle from its resting position creates a shift in the forces direction (examine Figure 4.37b). These forces create a moment which wants to return the body to the resting (original) position. When the body is at the position shown in Figure 4.37c ,the body is unstable and any tilt from the original position creates moment that will further continue to move the body from its original position. This analysis doesn't violate the second law of thermodynamics. Moving bodies from an unstable position is in essence like a potential.

Stability of Cubic Body

Fig. 4.38 Schematic of floating cubic.

A wooden cubic (made of pine, for example) is inserted into water. Part of the block floats above water line. The cubic mass (gravity) centroid is in the middle of the cubic. However the buoyant center is the middle of the volume under the water (see Figure 4.38). This situation is similar to Figure 4.37c. However, any experiment of this cubic wood shows that it is stable locally. Small amount of tilting of the cubic results in returning to the original position. When tilting a larger amount than $\pi/4$ , it results in a flipping into the next stable position. The cubic is stable in six positions (every cubic has six faces). In fact, in any of these six positions, the body is in situation like in 4.37c. The reason for this local stability of the cubic is that other positions are less stable. If one draws the stability (later about this criterion) as a function of the rotation angle will show a sinusoidal function with four picks in a whole rotation.

Stability of Floating Body

Fig. 4.39 Stability analysis of floating body.

So, the body stability must be based on the difference between the body's local positions rather than the ``absolute'' stability. That is, the body is ``stable'' in some points more than others in their vicinity. These points are raised from the buoyant force analysis. When the body is tilted at a small angle, $\beta$, the immersed part of the body center changes to a new location, B' as shown in Figure . The center of the mass (gravity) is still in the old location since the body did not change. The stability of the body is divided into three categories. If the new immerse volume creates a new center in such way that couple forces (gravity and buoyancy) try to return the body, the original state is referred as the stable body and vice versa. The third state is when the couple forces do have zero moment, it is referred to as the neutral stable. The body, shown in Figure 4.39, when given a tilted position, move to a new buoyant center, B'. This deviation of the buoyant center from the old buoyant center location, B , should to be calculated. This analysis is based on the difference of the displaced liquid. The right green area (volume) in Figure 4.39 is displaced by the same area (really the volume) on left since the weight of the body didn't change so the total immersed section is constant. For small angle, $\beta$, the moment is calculated as the integration of the small force shown in the Figure 4.39 as $\Delta F$. The displacement of the buoyant center can be calculated by examining the moment these forces creates. The body weight creates opposite moment to balance the moment of the displaced liquid volume. \begin{align} \overline{BB'}\, W = \mathbf{M} \label{static:eq:momentBouyant} \end{align} Where $\mathbf{M}$ is the moment created by the displaced areas (volumes), $\overline{BB'}$ is the distance between points B and point B' , and, $W$ referred to the weight of the body. It can be noticed that the distance $\overline{BB'}$ is an approximation for small angles (neglecting the vertical component.). So the perpendicular distance, $\overline{BB'}$, should be \begin{align} \overline{BB'} = \dfrac{ \mathbf{M}}{W} \label{static:eq:momentBouyantD} \end{align} The moment $\mathbf{M}$ can be calculated as \begin{align} \mathbf{M} = \int_{A} \overbrace{g\,\rho_l\,\underbrace{x\,\beta\,dA}_{dV}}^{\delta F}\,x = g \,\rho_l\, \beta \int_{A} x^2 dA \label{static:eq:staticMoment} \end{align} The integral in the right side of equation qref{static:eq:staticMoment} is referred to as the area moment of inertia and was discussed in Chapter 3. The distance, $\overline{BB'}$ can be written from equation qref{static:eq:staticMoment} as \begin{align} \overline{BB'} = \dfrac{g\,\rho_l\, I_{xx} } {\rho_{s} V_{body} } \label{static:eq:tiltdeX} \end{align} The point where the gravity force direction is intersecting with the center line of the cross section is referred as metacentric point, M. The location of the metacentric point can be obtained from the geometry as \begin{align} \overline{BM} = \dfrac{\overline{BB'}} {\sin \beta} \label{static:eq:metacentricP} \end{align} And combining equations \eqref{static:eq:tiltdeX} with qref{static:eq:metacentricP} yields \begin{align} \overline{BM} = \dfrac{\cancel{g}\, \rho_l \beta I_{xx}} {\cancel{g}\,\rho_{s}\,\sin\beta\,V_{body}} = \dfrac{ \rho_l\,I_{xx}}{\rho_{s}\,V_{body}} \label{static:eq:GMIntermidiateA} \end{align} For small angle ($\beta \sim 0$) \begin{align} \lim_{\beta \rightarrow 0} \dfrac{\sin \beta}{ \beta} \sim 1 \label{static:eq:lhopitalRule} \end{align} It is remarkable that the results is independent of the angle. Looking at Figure 4.39, the geometrical quantities can be related as \begin{align} \overline{GM} = \overbrace{\dfrac{\rho_l\,I_{xx}}{\rho_{s} V_{body}}} ^{\overline{BM}} - \overline{BG} \label{static:eq:GMIntermidiate} \end{align}

Example 4.28

A solid cone floats in a heavier liquid (that is $\rho_l/\rho_c> 1$). The ratio of the cone density to liquid density is $\alpha$. For a very light cone $\rho_{c}/\rho_{l} \sim 0$, the cone has zero depth. At this condition, the cone is unstable. For middle range, $1 > \rho_{c}/\rho_{l} > 0$ there could be a range where the cone is stable. The angle of the cone is $\theta$. Analyze this situation.

Solution

The floating cone volume is $\dfrac{\pi\,d\,r^2}{3}$ and the center of gravity is D/4. The distance $\overline{BG}$ depend on $d$ as \begin{align} \label{coneStability:BG} \overline{BG} = D/4 - d/4 \end{align} Where $D$ is the total height and $d$ is the height of the submerged cone. The moment of inertia of the cone is circle shown in Table . The relationship between the radius the depth is \begin{align} \label{coneStability:d-r} r = d\,\tan\theta \end{align} \begin{align} \label{coneStability:GMini} \overline{GM} = \dfrac{\rho_l\,\overbrace{\dfrac{\pi\,\left( d\,\tan\theta\right)^4}{64}} ^{I_{xx}}}{\rho_{s} \underbrace{\dfrac{\pi\,d\,\left( d\,\tan\theta\right)^2}{3} }_{V_{body}} } - \overbrace{\left(\dfrac{D}{4} - \dfrac{d}{4} \right)}^{\overline{BG}} \end{align} Equation \eqref{coneStability:GMini} can be simplified as \begin{align} \label{coneStability:GM} \overline{GM} = \dfrac{\rho_l\,d\, \tan^2\theta }{\rho_{s}\,192} - \left(\dfrac{D}{4} - \dfrac{d}{4} \right) \end{align} The relationship between $D$ and $d$ is determined by the density ratio ( as displaced volume \begin{align} \label{coneStability:d-D} \rho_l\,d^3 = \rho_c\, D^3 \Longrightarrow D = d \sqrt[3]{\dfrac{\rho_l}{\rho_c}} \end{align} Substituting equation \eqref{coneStability:d-D} into qref{coneStability:GM} yield the solution when $\overline{GM} = 0 $ \begin{align} \label{coneStability:sol} 0 = \dfrac{\rho_l\,d\, \tan^2\theta }{\rho_{s}\,192} - \left(\dfrac{d \sqrt[3]{\dfrac{\rho_l}{\rho_c}}}{4} - \dfrac{d}{4} \right) \Longrightarrow \dfrac{\rho_l\, \tan^2\theta} {\rho_{s}\,48} = \sqrt[3]{\dfrac{\rho_l}{\rho_c} - 1 } \end{align} Since $\rho_l > \rho_c$ this never happened.

Cubic Body

Fig. 4.40 Cubic body dimensions for stability analysis.

To understand these principles consider the following examples.

Example 4.29

A solid block of wood of uniform density, $\rho_s = \alpha\,\rho_{l}$ where ( $0\le\alpha\le1$ ) is floating in a liquid. Construct a graph that shows the relationship of the $\overline{GM}$ as a function of ratio height to width. Show that the block's length, $L$, is insignificant for this analysis.

Solution

Equation \eqref{static:eq:GMIntermidiate} requires that several quantities should be expressed. The moment of inertia for a block is given in Table and is $I_{xx}= \dfrac{La^3}{12}$. Where $L$ is the length into the page. The distance $\overline{BG}$ is obtained from Archimedes' theorem and can be expressed as \begin{align*} W = \rho_s \,\overbrace{a\,h\,L}^{V} = \rho_l \,\overbrace{a\,h_1\,L}^{\text{immersed
volume} } \Longrightarrow h_1 = \dfrac{\rho_s}{\rho_l} h \end{align*}

Stability of Cubic Body

Fig. 4.41 Stability of cubic body infinity long.

Thus, the distance $\overline{BG}$ is (see Figure 4.38) \begin{align*} \overline{BG} = \dfrac{h}{2} - \overbrace{\dfrac{\rho_s}{\rho_l}\, h}^{h_1}\,\dfrac{1}{2} = \dfrac{h}{2} \left(1 - \dfrac{\rho_s}{\rho_l} \right) \label{static:eq:BGbar} \end{align*} \begin{align*} GM = \dfrac{\cancel{g}\,\rho_l\, \overbrace{\dfrac{\cancel{L}\,a^3}{12}}^{I_{xx}} } {\cancel{g}\,\rho_s\,\underbrace{a\,h\,\cancel{L}}_V} - \dfrac{h}{2} \left(1 - \dfrac{\rho_s}{\rho_l} \right) \end{align*} Simplifying the above equation provides \begin{align*} \dfrac{\overline{GM}}{h} = \dfrac{1}{12\,\alpha} \left(\dfrac{a}{h}\right)^2 - \dfrac{1}{2} \left( 1 - \alpha \right) \end{align*} Notice that $\overline{GM}/{h}$ isn't a function of the depth, $L$. This equation leads to the condition where the maximum height above which the body is not stable anymore as \begin{align*} \dfrac{a}{h} \ge \sqrt {{6\,(1-\alpha)\alpha}} \label{static:eq:stabilityCritieriaCubic} \end{align*}

Maximum Value As A Function Density Ratio

Fig. 4.42 The maximum height reverse as a function of density ratio.

One of the interesting point for the above analysis is that there is a point above where the ratio of the height to the body width is not stable anymore. In cylindrical shape equivalent to equation qref{static:eq:stabilityCritieriaCubic} can be expressed. For cylinder (circle) the moment of inertia is $I_{xx} = \pi\,b^4/64$. The distance $\overline{BG}$ is the same as for the square shape (cubic) (see above \eqref{static:eq:BGbar}). Thus, the equation is \begin{align*} \dfrac{\overline{GM}}{h} = \dfrac{g}{64\,\alpha} \left(\dfrac{b}{h}\right)^2 - \dfrac{1}{2} \left( 1 - \alpha \right) \end{align*} And the condition for maximum height for stability is \begin{align*} \dfrac{b}{h} \ge \sqrt{{32\,(1-\alpha)\,\alpha}} \end{align*} This kind of analysis can be carried for different shapes and the results are shown for these two shapes in Figure 4.42. It can be noticed that the square body is more stable than the circular body shape.

Principle Main Axises

Any body has infinite number of different axises around which moment of inertia can be calculated. For each of these axises, there is a different moment of inertia. With the exception of the circular shape, every geometrical shape has an axis in which the moment of inertia is without the product of inertia. This axis is where the main rotation of the body will occur. Some analysis of floating bodies are done by breaking the rotation of arbitrary axis to rotate around the two main axises. For stability analysis, it is enough to find if the body is stable around the smallest moment of inertia. For example, a square shape body has larger moment of inertia around diagonal. The difference between the previous calculation and the moment of inertia around the diagonal is \[ \Delta I_{xx} = \overbrace{\dfrac{\sqrt{2}\,a\left( \dfrac{\sqrt{3}\,a}{2}\right)^3 }{6}}^{I\;diagonal\;axis} \;- \overbrace{\dfrac{a^4}{12}}^{``normal'' axis} \sim 0.07\,{a}^{4} \] Which show that if the body is stable at main axises, it must be stable at the ``diagonal'' axis. Thus, this problem is reduced to find the stability for principle axis.

Unstable Bodies

What happen when one increases the height ratio above the maximum height ratio? The body will flip into the side and turn to the next stable point (angle). This is not a hypothetical question, but rather practical. This happens when a ship is overloaded with containers above the maximum height. In commercial ships, the fuel is stored at the bottom of the ship and thus the mass center (point $G$) is changing during the voyage. So, the ship that was stable (positive $\overline{GM}$) leaving the initial port might became unstable (negative $\overline{GM}$) before reaching the destination port.

Stability of Two Triangles

Fig. 4.43 Stability of two triangles put tougher.

Example 4.30

One way to make a ship to be a hydrodynamic is by making the body as narrow as possible. Suppose that two opposite sides triangle (prism) is attached to each other to create a long ``ship'' see Figure 4.43. Supposed that $\mathbf{a/h}\longrightarrow \tilde 0$ the body will be unstable. On the other side if the $\mathbf{a/h}\longrightarrow \tilde \infty$ the body is very stable. What is the minimum ratio of $\mathbf{a/h}$ that keep the body stable at half of the volume in liquid (water). Assume that density ratio is $\rho_l / \rho_s = \bar{\rho}$.

Solution

The answer to the question is that the limiting case where $\overline{GM} = 0$. To find this ratio equation terms in \eqref{static:eq:GMIntermidiate} have to be found. The Volume of the body is \begin{align*} V = 2\;\left( \dfrac {a^2 \, h} {2} \right) = a^2 \, h \end{align*} The moment of inertia is triangle (see explanation in example \eqref{mech:ex:triangleIxx} is \begin{align*} I_{xx} = \dfrac{a\,h^3}{2} \end{align*} And the volume is \begin{align*} V_{body} = a^2 \; \sqrt{h^2 - \dfrac{a^2}{4} } = a^2\,h \; \sqrt{1 - \dfrac{1}{4}\, \dfrac{a^2}{h^2} } \end{align*} The point $\mathbf{B}$ is a function of the density ratio of the solid and liquid. Denote the liquid density as $\rho_l$ and solid density as $\rho_s$. The point $\mathbf{B}$ can be expressed as \begin{align*} B = \dfrac {a\, \rho_s} {2\, \rho_l} \end{align*} And thus the distance $\overline{BG}$ is \begin{align*} \overline{BG} = \dfrac{a}{2} \left( 1 - \dfrac{\rho_s}{\rho_l} \right) \end{align*} The limiting condition requires that $ \overline{GM} = 0$ so that \begin{align*} \dfrac{\rho_l\,I_{xx}}{\rho_{s} V_{body}} = \overline{BG} \end{align*} Or explicitly \begin{align*} \dfrac{\rho_l \,\dfrac{a\,h^3}{2}} { \rho_s \,a^2\,h \; \sqrt{1 - \dfrac{1}{4}\, \dfrac{a^2}{h^2} } } = \dfrac{a}{2} \left( 1 - \dfrac{\rho_s}{\rho_l} \right) \end{align*} After rearrangement and using the definitions of $\xi= h/a$ $\bar{\rho} \rho_l/\rho_s$ results in \begin{align*} \dfrac{\bar{\rho} \,\xi^2 }{ \sqrt{1 - \dfrac{\xi^2}{4} } } = \left( 1 - \dfrac{1}{\bar{\rho}} \right) \end{align*} The solution of the above solution is obtained by squaring both sides and defining a new variable such as $x=\xi^2$. After the above manipulation and selecting the positive value and to keep stability as \begin{align*} x < \dfrac{\sqrt{\dfrac{ \sqrt{64\,{\bar{\rho}}^{4}-64\,{\bar{\rho}}^{3}+{\bar{\rho}}^{2}-2\,\bar{\rho}+1} }{\bar{\rho}}+\dfrac{1}{\bar{\rho}}-1}}{2\,\sqrt{2}\,\bar{\rho}} \end{align*}

4.6.1.1 Stability of Body with Shifting Mass Centroid

GM Changed Due To Liquid

Fig. 4.44 The effects of liquid movement on the $\overline{GM}$.

Ships and other floating bodies carry liquid or have a load which changes the mass location during tilting of the floating body. For example, a ship that carries wheat grains where the cargo is not properly secured to the ship. The movement of the load (grains, furniture, and/or liquid) does not occur in the same speed as the body itself or the displaced outside liquid. Sometimes, the slow reaction of the load, for stability analysis, is enough to be ignored. Exact analysis requires taking into account these shifting mass speeds. However, here, the extreme case where the load reacts in the same speed as the tilting of the ship/floating body is examined. For practical purposes, it is used as a limit for the stability analysis. There are situations where the real case approaches to this extreme. These situations involve liquid with a low viscosity (like water, alcohol) and ship with low natural frequency (later on the frequency of the ships). Moreover, in this analysis, the dynamics are ignored and only the statics is examined (see Figure 4.44). A body is loaded with liquid ``B'' and is floating in a liquid ``A'' as shown in Figure 4.44. When the body is given a tilting position the body displaces the liquid on the outside. At the same time, the liquid inside is changing its mass centroid. The moment created by the inside displaced liquid is \begin{align} M_{in} = g\, {\rho_l}_B \beta {I_{xx}}_B \label{static:eq:momentL} \end{align} Note that ${I_{xx}}_B$ isn't the same as the moment of inertia of the outside liquid interface. The change in the mass centroid of the liquid ``A'' then is \begin{align} \overline{G_{1}G_{1}'} = \dfrac{\cancel{g}\, \cancel{{\rho_l}_B} \beta {I_{xx}}_B} {\underbrace{\cancel{g}\,V_B\,\cancel{{\rho_l}_B}} _{\text{Inside liquid weight }}} = \dfrac{{I_{xx}}_B}{V_B} \label{static:eq:GG'} \end{align} Equation \eqref{static:eq:GG'} shows that $\overline{GG^{'}}$ is only a function of the geometry. This quantity, $\overline{G_1G_1'}$, is similar for all liquid tanks on the floating body. The total change of the vessel is then calculated similarly to center area calculations. \begin{align} \cancel{g}\,m_{total}\, \overline{GG'} = \cancelto{0}{{g}\,m_{body}} + \cancel{g}\, m_f \overline{G_1G_1'} \label{static:eq:GGtotal} \end{align} For more than one tank, it can be written as \begin{align} \overline{GG'} = \dfrac{g}{W_{total}} \sum_{i=1}^n \overline{G_iG_i} {\rho_l}_i V_i = \dfrac{g}{W_{total}} \sum_{i=1}^n \dfrac{{{I_{xx}}_b}_i} {{V_b}_i} \label{static:eq:totalGG} \end{align} A new point can be defined as $G_c$. This point is the intersection of the center line with the vertical line from $G'$. \begin{align} \overline{G\,G_c} = \dfrac{\overline{GG'}} {\sin\beta} \label{static:eq:GGc} \end{align} The distance that was used before $\overline{GM}$ is replaced by the criterion for stability by $\overline{G_c\,M}$ and is expressed as \begin{align} \overline{G_c\,M} = {\dfrac{g\, \rho_A\,{I_{xx}}_A}{\rho_{s} V_{body}}} -\overline{BG} - \dfrac{1}{m_{total}}\, \dfrac{{I_{xx}}_b}{V_b} \label{static:eq:GcM} \end{align} If there are more than one tank partially filled with liquid, the general formula is \begin{align} \overline{G_c\,M} = {\dfrac{g\, \rho_A\,{I_{xx}}_A}{\rho_{s} V_{body}}} -\overline{BG} - \dfrac{1}{m_{total}} \sum_{i=1}^{n}\dfrac{{{I_{xx}}_b}_i}{{V_b}_i} \label{static:eq:GcMg} \end{align}

GM Measurement

Fig. 4.45 Measurement of GM of floating body.

One way to reduce the effect of the moving mass center due to liquid is done by substituting a single tank with several tanks. The moment of inertial of the combine two tanks is smaller than the moment of inertial of a single tank. Increasing the number of tanks reduces the moment of inertia. The engineer could design the tanks in such a way that the moment of inertia is operationally changed. This control of the stability, $\overline{GM}$, can be achieved by having some tanks spanning across the entire body with tanks spanning on parts of the body. Movement of the liquid (mostly the fuel and water) provides way to control the stability, $GM$, of the ship.

Metacentric Height, $\overline{GM}$, Measurement

The metacentric height can be measured by finding the change in the angle when a weight is moved on the floating body. Moving the weight, $T$ a distance, $d$ then the moment created is \begin{align} M_{weight} = T\,d \label{static:eq:Td} \end{align} This moment is balanced by \begin{align} M_{righting} = W_{total} \overline{GM}_{new} \,\theta \label{static:eq:TdR} \end{align} Where, $W_{total}$, is the total weight of the floating body including measuring weight. The angle, $\theta$, is measured as the difference in the orientation of the floating body. The metacentric height is \begin{align} \overline{GM}_{new} = \dfrac{T\,d}{W_{total} \,\theta} \label{static:eq:GMmessured} \end{align} If the change in the $\overline{GM}$ can be neglected, equation \eqref{static:eq:GMmessured} provides the solution. The calculation of $\overline{GM}$ can be improved by taking into account the effect of the measuring weight. The change in height of $G$ is \begin{align} \cancel{g}\, m_{total}\, G_{new} = \cancel{g}\, m_{ship}\, G_{actual} + \cancel{g}\,T\,h \label{static:eq:deltaGMR} \end{align} Combining equation \eqref{static:eq:deltaGMR} with equation \eqref{static:eq:GMmessured} results in \begin{align} \overline{GM}_actual = \overline{GM}_{new}\, \dfrac{m_{total}}{m_{ship}} - h \, \dfrac{T}{m_{ship}} \label{static:eq:GMactual} \end{align} The weight of the ship is obtained from looking at the ship depth.

4.6.1.3 Stability of Submerged Bodies

The analysis of submerged bodied is different from the stability when the body lays between two fluid layers with different density. When the body is submerged in a single fluid layer, then none of the changes of buoyant centroid occurs. Thus, the mass centroid must be below than buoyant centroid in order to have stable condition. However, all fluids have density varied in some degree. In cases where the density changes significantly, it must be taken into account. For an example of such a case is an object floating in a solar pond where the upper layer is made of water with lower salinity than the bottom layer(change up to 20% of the density). When the floating object is immersed into two layers, the stability analysis must take into account the changes of the displaced liquids of the two liquid layers. The calculations for such cases are a bit more complicated but based on the similar principles. Generally, this density change helps to increase the stability of the floating bodies. This analysis is out of the scope of this book (for now).

4.6.1.4 Stability of None Systematical or ``Strange'' Bodies

Fluid Element in accelerate system

Fig. 4.46 Calculations of $\overline{GM}$ for abrupt shape body.

While most floating bodies are symmetrical or semi–symmetrical, there are situations where the body has a ``strange'' and/or un-symmetrical body. Consider the first strange body that has an abrupt step change as shown in Figure 4.46. The body weight doesn't change during the rotation that the green area on the left and the green area on right are the same (see Figure 4.46). There are two situations that can occur. After the tilting, the upper part of the body is above the liquid or part of the body is submerged under the water. The mathematical condition for the border is when $b=3\,a$. For the case of $b< 3\,a$ the calculation of moment of inertia are similar to the previous case. The moment created by change in the displaced liquid (area) act in the same fashion as the before. The center of the moment is needed to be found. This point is the intersection of the liquid line with the brown middle line. The moment of inertia should be calculated around this axis. For the case where $b < 3\,a$ $x$ some part is under the liquid. The amount of area under the liquid section depends on the tilting angle. These calculations are done as if none of the body under the liquid. This point is intersection point liquid with lower body and it is needed to be calculated. The moment of inertia is calculated around this point (note the body is ``ended'' at end of the upper body). However, the moment to return the body is larger than actually was calculated and the bodies tend to be more stable (also for other reasons).

4.6.1.5 Neutral frequency of Floating Bodies

This case is similar to pendulum (or mass attached to spring). The governing equation for the pendulum is \begin{align} ll \ddot{\beta} - g\,\beta = 0 \label{static:eq:govPendulum} \end{align} Where here $ll$ is length of the rode (or the line/wire) connecting the mass with the rotation point. Thus, the frequency of pendulum is $\dfrac{1}{2\,\pi}\sqrt{\dfrac{g}{ll}}$ which measured in $Hz$. The period of the cycle is $2\,\pi\,\sqrt{ll/g}$. Similar situation exists in the case of floating bodies. The basic differential equation is used to balance and is \begin{align} \overbrace{I\ddot{\beta}}^{rotation} - \overbrace{V\,\rho_s\,\overline{GM}\,\beta}^{rotating\;moment}=0 \label{static:eq:govFloat} \end{align} In the same fashion the frequency of the floating body is \begin{align} \dfrac{1}{2\,\pi} \sqrt{\dfrac {V\,\rho_s\,\overline{GM}}{I_{body}}} \label{static:eq:floatFreq} \end{align} and the period time is \begin{align} 2\,\pi \sqrt{\dfrac{I_{body}} {V\,\rho_s\,\overline{GM}}} \label{static:eq:periodFreq} \end{align} In general, the larger $\overline{GM}$ the more stable the floating body is. Increase in $\overline{GM}$ increases the frequency of the floating body. If the floating body is used to transport humans and/or other creatures or sensitive cargo it requires to reduce the $\overline{GM}$ so that the traveling will be smoother.

4.6.2 Surface Tension

The surface tension is one of the mathematically complex topic and related to many phenomena like boiling, coating, etc. In this section, only simplified topics like constant value will be discussed. In one of the early studies of the surface tension/pressure was done by Torricelli. In this study he suggest construction of the early barometer. In barometer is made from a tube sealed on one side. The tube is filled with a liquid and turned upside down into the liquid container. The main effect is the pressure difference between the two surfaces (in the tube and out side the tune). However, the surface tension affects the high. This effect is large for very small diameters.

Example 4.31

In interaction of the molecules shown in Figure ? describe the existence of surface tension. Explain why this description is erroneous?

Solution

The upper layer of the molecules have unbalanced force towards the liquid phase. Newton's law states when there is unbalanced force, the body should be accelerate. However, in this case, the liquid is not in motion. Thus, the common explanation is wrong.

Needle Floating on Liquid

Fig. 4.47 A heavy needle is floating on a liquid.

Example 4.32

Needle is made of steel and is heavier than water and many other liquids. However, the surface tension between the needle and the liquid hold the needle above the liquid. After certain diameter, the needle cannot be held by the liquid. Calculate the maximum diameter needle that can be inserted into liquid without drowning.

Solution

Under Construction

4.7 Rayleigh–Taylor Instability

Rayleigh–Taylor instability (or RT instability) is named after Lord Rayleigh and G. I. Taylor. There are situations where a heavy liquid layer is placed over a lighter fluid layer. This situation has engineering implications in several industries. For example in die casting, liquid metal is injected in a cavity filled with air. In poor designs or other situations, some air is not evacuated and stay in small cavity on the edges of the shape to be casted. Thus, it can create a situation where the liquid metal is above the air but cannot penetrate into the cavity because of instability. This instability deals with a dense, heavy fluid that is being placed above a lighter fluid in a gravity field perpendicular to interface. Example for such systems are dense water over oil (liquid–liquid), or water over air(gas–liquid). The original Rayleigh's paper deals with the dynamics and density variations. For example, density variations according to the bulk modulus (see section 4.3.3.2) are always stable but unstable of the density is in the reversed order. Supposed that a liquid density is arbitrary function of the height. This distortion can be as a result of heavy fluid above the lighter liquid. This analysis asks the question of what happen when a small amount of liquid from the above layer enter into the lower layer? Whether this liquid continue and will grow or will it return to its original conditions? The surface tension is the opposite mechanism that will returns the liquid to its original place. This analysis is referred to the case of infinite or very large surface. The simplified case is the two different uniform densities. For example a heavy fluid density, $\rho_L$, above lower fluid with lower density, $\rho_G$. For perfectly straight interface, the heavy fluid will stay above the lighter fluid. If the surface will be disturbed, some of heavy liquid moves down. This disturbance can grow or returned to its original situation. This condition is determined by competing forces, the surface density, and the buoyancy forces. The fluid above the depression is in equilibrium with the sounding pressure since the material is extending to infinity. Thus, the force that acting to get the above fluid down is the buoyancy force of the fluid in the depression.

Depression in Taylor Instability

Fig. 4.48 Description of depression to explain the Rayleigh–Taylor instability.

The depression is returned to its original position if the surface forces are large enough. In that case, this situation is considered to be stable. On the other hand, if the surface forces (surface tension) are not sufficient, the situation is unstable and the heavy liquid enters into the liquid fluid zone and vice versa. As usual there is the neutral stable when the forces are equal. Any continues function can be expanded in series of cosines. Thus, example of a cosine function will be examined. The conditions that required from this function will be required from all the other functions. The disturbance is of the following \begin{align} h = -h_{max} \cos \dfrac{2\,\pi\,x} {L} \label{static:eq:disturbance} \end{align} where $h_{max}$ is the maximum depression and $L$ is the characteristic length of the depression. The depression has different radius as a function of distance from the center of the depression, $x$. The weakest point is at $x=0$ because symmetrical reasons the surface tension does not act against the gravity as shown in Figure qref{static:fig:cos}. Thus, if the center point of the depression can ``hold'' the intrusive fluid then the whole system is stable. The radius of any equation is expressed by equation \eqref{intro:eq:radius}. The first derivative of $\cos$ around zero is $\sin$ which is approaching zero or equal to zero. Thus, equation \eqref{intro:eq:radius} can be approximated as \begin{align} \dfrac{1}{R} = \dfrac{d^2h}{dx^2} \label{static:eq:radiusAppx} \end{align} For equation \eqref{static:eq:disturbance} the radius is \begin{align} \dfrac{1}{R} =-\dfrac{4\,\pi^2\,h_{max}}{L^2} \label{static:eq:sinR} \end{align} According to equation \eqref{intro:eq:STbcylinder} the pressure difference or the pressure jump is due to the surface tension at this point must be \begin{align} P_H - P_L = \dfrac{4\,h_{max}\,\sigma\,\pi^2}{L^2} \label{static:eq:deltaP} \end{align} The pressure difference due to the gravity at the edge of the disturbance is then \begin{align} P_H - P_L = g\,\left( \rho_H-\rho_L \right) h_{max} \label{static:eq:detlaPe} \end{align} Comparing equations \eqref{static:eq:deltaP} and qref{static:eq:detlaPe} show that if the relationship is \begin{align} \dfrac{4\,\sigma\,\pi^2}{L^2} > g \,\left( \rho_H-\rho_L \right) \label{static:eq:noEq} \end{align} It should be noted that $h_{max}$ is irrelevant for this analysis as it is canceled. The point where the situation is neutral stable \begin{align} L_c = \sqrt {\dfrac{4\,\pi^2 \sigma}{g\left( \rho_H-\rho_L \right)} } \label{phase:eq:notEq} \end{align} An alternative approach to analyze this instability is suggested here. Consider the situation described in Figure 4.49. If all the heavy liquid ``attempts'' to move straight down, the lighter liquid will ``prevent'' it. The lighter liquid needs to move up at the same time but in a different place. The heavier liquid needs to move in one side and the lighter liquid in another location. In this process the heavier liquid ``enter'' the lighter liquid in one point and creates a depression as shown in Figure 4.49.

Instability Due To Depression

Fig. 4.49 Description of depression to explain the instability.

To analyze it, considered two control volumes bounded by the blue lines in Figure 4.49. The first control volume is made of a cylinder with a radius $r$ and the second is the depression below it. The ``extra'' lines of the depression should be ignored, they are not part of the control volume. The horizontal forces around the control volume are canceling each other. At the top, the force is atmospheric pressure times the area. At the cylinder bottom, the force is $\rho\,g\,h\times A$. This acts against the gravity force which make the cylinder to be in equilibrium with its surroundings if the pressure at bottom is indeed $\rho\,g\,h$. For the depression, the force at the top is the same force at the bottom of the cylinder. At the bottom, the force is the integral around the depression. It can be approximated as a flat cylinder that has depth of $r\,\pi/4$ (read the explanation in the example 4.23) This value is exact if the shape is a perfect half sphere. In reality, the error is not significant. Additionally when the depression occurs, the liquid level is reduced a bit and the lighter liquid is filling the missing portion. Thus, the force at the bottom is \begin{align} F_{bottom} \sim \pi\,r^2 \left[ \left( \dfrac{\pi\,r}{4} + h \right) \,\left( \rho_L - \rho_G\right) \,g + P_{atmos} \right] \label{static:eq:bottomDepression1} \end{align} The net force is then \begin{align} F_{bottom} \sim \pi\,r^2 \left(\dfrac{\pi\,r}{4} \right) \,\left( \rho_L - \rho_G\right) \,g \label{static:eq:bottomDepression} \end{align} The force that hold this column is the surface tension. As shown in Figure 4.49, the total force is then \begin{align} F_{\sigma} = 2\,\pi \, r \,\sigma\, \cos\theta \label{static:eq:sigmaF} \end{align} The forces balance on the depression is then \begin{align} 2\,\pi \, r \,\sigma \cos\theta \sim \pi\,r^2 \left(\dfrac{\pi\,r}{4} \right) \,\left( \rho_L - \rho_G\right) \,g \label{static:eq:balance} \end{align} The radius is obtained by \begin{align} r \sim \sqrt{ \dfrac { 2\,\pi\,\sigma\cos\theta}{ \,\left( \rho_L - \rho_G\right) \,g }} \label{static:eq:radiusMinTubeExI} \end{align} The maximum surface tension is when the angle, $\theta=\pi/2$. At that case, the radius is \begin{align} r \sim \sqrt{ \dfrac { 2\,\pi\,\sigma}{ \,\left( \rho_L - \rho_G\right) \,g }} \label{static:eq:radiusMinTubeEx} \end{align}

Max Radios of Lifting Liquid

Fig. 4.50 The cross section of the interface. The purple color represents the maximum heavy liquid raising area. The yellow color represents the maximum lighter liquid that ``goes down.''

The maximum possible radius of the depression depends on the geometry of the container. For the cylindrical geometry, the maximum depression radius is about half for the container radius (see Figure 4.50). This radius is limited because the lighter liquid has to enter at the same time into the heavier liquid zone. Since the ``exchange'' volumes of these two process are the same, the specific radius is limited. Thus, it can be written that the minimum radius is \begin{align} {r_{min}}_{tube} = 2\,\sqrt{\dfrac{2\,\pi\,\sigma}{ g\,{\left(\rho_L-\rho_G\right)} }} \label{static:eq:solutionD2} \end{align} The actual radius will be much larger. The heavier liquid can stay on top of the lighter liquid without being turned upside down when the radius is smaller than the equation ef{static:eq:solutionD2}. This analysis introduces a new dimensional number that will be discussed in a greater length in the Dimensionless chapter. In equation \eqref{static:eq:solutionD2} the angle was assumed to be 90 degrees. However, this angle is never can be obtained. The actual value of this angle is about $\pi/4$ to $\pi/3$ and in only extreme cases the angle exceed this value (considering dynamics). In Figure 4.50, it was shown that the depression and the raised area are the same. The actual area of the depression is only a fraction of the interfacial cross section and is a function. For example,the depression is larger for square area. These two scenarios should be inserting into equation 4.168 by introducing experimental coefficient.

Example 4.33

Estimate the minimum radius to insert liquid aluminum into represent tube at temperature of 600$[K]$. Assume that the surface tension is $400[mN/m]$. The density of the aluminum is $2400 kg/m^3$.

Solution

The depression radius is assume to be significantly smaller and thus equation \eqref{static:eq:radiusMinTubeEx} can be used. The density of air is negligible as can be seen from the temperature compare to the aluminum density. \[ r \sim \sqrt{\dfrac{8\,\pi\,\overbrace{0.4}^{\sigma}}{ 2400\times 9.81}} \] The minimum radius is $r \sim 0.02 [m]$ which demonstrates the assumption of $h>>r$ was appropriate.

Cylinder with Three Liquids Layers

Fig. 4.51 Three liquids layers under rotation with various critical situations.

Open Question by April 15, 2010

The best solution of the following question will win 18 U.S. dollars and your name will be associated with the solution in this book.

Example 4.34

A canister shown in Figure 4.51 has three layers of different fluids with different densities. Assume that the fluids do not mix. The canister is rotate with circular velocity, $\omega$. Describe the interface of the fluids consider all the limiting cases. Is there any difference if the fluids are compressible? Where is the maximum pressure points? For the case that the fluids are compressible, the canister top center is connected to another tank with equal pressure to the canister before the rotation (the connection point). What happen after the canister start to be rotated? Calculated the volume that will enter or leave, for known geometries of the fluids. Use the ideal gas model. You can assume that the process is isothermal. Is there any difference if the process is isentropic? If so, what is the difference?

4.8 Qualitative questions

These qualitative questions are for advance students and for those who would like to prepare themselves preliminary examination (Ph. D. examinations).

Additional Question

  1. 1.

    The atmosphere has different thickness in different locations. Where will be atmosphere thickness larger in the equator or the north pole? Explain your reasoning for the difference. How would you estimate the difference between the two locations.

  2. 2.

    The author's daughter (8 years old) stated that fluid mechanics make no sense. For example, she points out that warm air raise and therefor the warm spot in a house is the top floor (that is correct in a 4 story home). So why when there is snow on high mountains? It must be that the temperature is below freezing point on the top of the mountain (see for example Mount Kilimanjaro, Kenya). How would you explain this situation? Hint, you should explain this phenomenon using only concepts that where developed in this chapter.

  3. 3.

    The surface of the ocean has spherical shape. The stability analysis that was discussed in this chapter was based on the assumption that surface is straight. How in your opinion the effect of the surface curvature affects the stability analysis.

  4. 4.

    If the gravity was changing due to the surface curvature what is the effect on the stability.

  5. 5.

    A car is accelerated (increase of velocity) in an include surface upwards. Draw the constant pressure line. What will constant pressure lines if the car will be driven downwards.

  6. 6.

    A symmetrical cylinder filled with liquid is rotating around its center. What are the directions of the forces that acting on cylinder. What are the direction of the force if the cylinder is not symmetrical?

  7. 7.

    A body with a constant area is floating in the liquid. The body is pushed down of the equilibrium state into the liquid by a distance $ll$. Assume that the body is not totally immersed in the liquid. What are simple harmonic frequency of the body. Assume the body mass is $m$ its volume is, $V$. Additionally assume that the only body motion is purely vertical and neglect the add mass and liquid resistance.

Mass Integral Static1 Index TOC
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