Fluid Mechanics for Mechanical Engineers/Fluid Statics

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Definitions

Fluid statics is the study of fluids which are either at rest or in rigid body motion with respect to a fixed frame of reference. Rigid body motion means that there is no relative velocity between the fluid particles.

In a fluid at rest, there is no shear stress, i. e. fluid does not deform, but fluid sustains normal stresses.

Fluid at rest(left) and in rigid body motion (right)

We can apply Newton's second law of motion to evaluate the reaction of the particle to the applied forces.

Forces created by pressure on the surfaces of a differential fluid volume in a static fluid

Force balance in $i^{th}$ direction:

$F_i^{net}=m\cdot a_i$

We can also say,$F_i^{body}+F_i^{surf}=m\cdot a_i$

Force created by pressure is :

$F^{surf}=F^{pressure}=-p\cdot \overrightarrow{A}$

$\overrightarrow{A}$ is the vector having the surface area as magnitude and surface normal as direction.

Thus,

$F^{pressure}=-p\cdot \overrightarrow{A}=-pA\overrightarrow{n}$

Force caused by the pressures opposite to the surface normal.

For a differential fluid element: $d{F}_i^{body}+d{F}_i^{surf}=dm\cdot a_i$

Remember Taylor Series expansion of a function : $f(x+\mathrm{\Delta }x)=f+\frac{\partial f}{\partial x}\mathrm{\Delta }x+\frac{1}{2}\frac{{\partial }^{2}f}{\partial x^2}\mathrm{\Delta }{x}^2+\dots$

Taylor series expansion can be use to approximate the pressure on each face. Hence, let $P$ the pressure in the center of the fluid element, therefore the pressure on the surface in direction of $x_i$ is $P+\frac{\partial P}{\partial x_i}\frac{d{x}_i}{2}$. The force created by the pressure along $x_2$ direction can be written as:

$d{F}_2^{pressure}=[P-\frac{\partial P}{\partial x_2}\frac{d{x}_2}{2}]d{x}_{1}d{x}_3-[P+\frac{\partial P}{\partial x_2}\frac{d{x}_2}{2}]d{x}_{1}d{x}_3$

$=-\frac{\partial P}{\partial x_2}d{x}_{1}d{x}_{2}d{x}_3=-\frac{\partial P}{\partial x_2}dV$

Thus,

$d{F}_i^{pressure}=-\frac{\partial P}{\partial x_i}dV$

$d{F}_i^{body}=dm{g}_i$

Thus,

$-\frac{\partial P}{\partial x_i}dV+dm{g}_i=dm{a}_i$

or,

$-\frac{\partial P}{\partial x_i}dV+\rho dV{g}_i=\rho dV{a}_i$

or,

$-\frac{\partial P}{\partial x_i}+\rho g_i=\rho a_i$

or,$-\frac{\partial P}{\partial x_1}=\rho a_1;-\frac{\partial P}{\partial x_2}=\rho a_2;-\frac{\partial P}{\partial x_3}-\rho g=\rho a_3$

for $a_i=0$

$\frac{\partial P}{\partial x_1}=0;\frac{\partial P}{\partial x_2}=0;\frac{\partial P}{\partial x_3}=-\rho g$

Pressure changes only in $x_3$ direction.

Pressure variation in an incompressible and static fluid

$\frac{\partial P}{\partial x_3}=-\rho g;x_3=z\to \frac{\partial P}{\partial z}=-\rho g$ is constant since $\rho$ and $g$ are constants.

Pressure variation with depth of water column

${\int }_{p_1}^{p_2}dP=-{\int }_{z_1}^{z_2}\rho gdz$

$p_2-p_1=-\rho g(z_2-z_1)$

If we take $p_2$ at the surface, then:

$p_{atm}-p_1=-\rho g(z_2-z_1)$

$p_1=p_{atm}+\rho gh;h=z_2-z_1$

h is measured from the surface. For incompressible flow,

$p_2-p_1=-\rho g(z_2-z_1)\to p_1=p_{atm}+\rho gh$

Any two points at the same elevation in a continuous length of the same liquid are at the same pressure. Pressure increases as one goes down in a liquid column.

Equal depths in the same fluid experience equal hydrostatic pressure

Consider three immiscible fluids in a container and find out a relation for the pressure at the bottom of the fluid shown in the schematics besides. $P_C=?$ $P_A=P_0+{\rho }_{A}g{h}_A$ $P_B=P_A+{\rho }_{B}g(h_B-h_A)$ $P_C=P_B+{\rho }_{C}g(h_C-h_B)$ $\Rightarrow P_C={\rho }_{C}g(h_C-h_B)+{\rho }_{B}g(h_B-h_A)+{\rho }_{A}g{h}_A+P_0$ $P_C={\rho }_{C}g{H}_C+{\rho }_{B}g{H}_B+{\rho }_{A}g{H}_A+P_0$

Hydrostatic pressure profile for fluids with different densities and height

Transmission of Pressure

Concept of transmission of pressure is very important for hydraulic and pneumatics system. Neglecting elevation changes the following relation can be written: $P_1=P_2\Rightarrow \frac{F_1}{A_1}=\frac{F_2}{A_2}\Rightarrow F_2=\frac{A_2}{A_1}{F}_1$ which could be stated in the famous Pascal's law like below: A change in pressure at any point in an enclosed fluid at rest is transmitted undiminished to all points in the fluid

Pascal's law

Communicating containers

Lets consider two closed containers(which means the free surface pressure could be different than atmospheric pressure) both contain same fluid are connected via a connector valve. When the valve is open, the heights of the fluid columns can give an indication about the pressure in both chamber. For closed container $P_{01}+{\rho }_{1}g(H_1-h_1)=P_{02}+{\rho }_{2}g(H_2-h_2)$ (of course,when we calculate the small 'h', it should be measured at the height of connecting valve for both column distinctively.) for ${\rho }_1={\rho }_2=\rho$ $P_{01}+\rho g(H_1-h_1)=P_{02}+\rho g(H_2-h_2)$ So from the picture above, we can understand that the pressure in the right column is higher than the left column. For open Containers, $\Rightarrow P_{01}=P_{02}=P_{atm}$ If both fluid columns are at the same level $\Rightarrow H_1-h_1=H_2-h_2$ so, the depth of the fluid from free surface in both column will be the same. This nice principle was used for Water-based Barometer a.k.a 'Storm Barometer' or 'Goethe Barometer'. Try to see if you understand the device.

variation of height of fluid column due to difference in pressure from free surface Goethe Barometer

Pressure Measurement Equipments

Barometer

From the equations which we derived before , it is also possible to measure the pressure exerted by almost 100 km thick earth atmosphere which is above us. Since the constituents and the density varies over the height of the atmosphere , we will consider a fluid column which have free surface with no atmospheric pressure but connected with a fluid which experience atmospheric pressure like communicating container. Let consider first (from previous section), $P_0={\rho }_{f}gh$ for $P_0=1$atm = $101,3$ kPa water height will be: $101,3\cdot 10^3=1000\cdot 9.79\cdot h_{water}$ $h_{water}\approx 10m$ where as height of mercury will be around $\frac{10m}{13,6}\approx 0,7m$ So now taking this barometer to desired location and observing the mercury column height, the atmospheric pressure could be measured.

Barometer

U-tube manometer

(see Pressure measurement)

The volume flow rate $Q$, through a pipe can be determined by means of a flow nozzle located in the pipe as illustrated in the figure. The nozzle creates a pressure drop, $p_A-p_B$, along the pipe which is related to the flow through the equation $Q=K\sqrt{p_A-p_B}$ where K is a constant depending on the pipe and nozzle size. The pressure drop is frequently measured with a differential U-tube manometer of the type illustrated. (a) Determine an equation for $p_A-p_B$ in terms of the specific weight of the flowing fluid, ${\gamma }_1=g{\rho }_1$, the specific weight of the gage fluid, ${\gamma }_2=g{\rho }_2$, and the various heights indicated. (b) For ${\gamma }_1=9.80\frac{kN}{m^3}$, ${\gamma }_2=15.6\frac{kN}{m^3}$, $h_1=1.0m$ and $h_2=0.5m$, what is the value of the pressure drop, $p_A-p_B$? Solution: (a) Although the fluid in the pipe is moving, the fluids in the columns of the manometer are at rest so that the pressure variation in the manometer tubes is hydrostatic. If we start at point A and move vertically upward to level (1), the pressure will decrease by ${\gamma }_1{h}_1$ and will be equal to the pressure at (2) and (3). We can move from (3) to (4) where the pressure has been further reduced by ${\gamma }_2{h}_2$. The pressures at levels (4) and (5) are equal, and as we move from (5) to B the pressure will increase by ${\gamma }_1(h_1+h_2)$. Thus, in equation form $p_A-{\gamma }_1{h}_1-{\gamma }_2{h}_2+{\gamma }_1(h_1+h_2)=p_B$ or $p_A-p_B=h_2({\gamma }_2-{\gamma }_1)=h_{2}g({\rho }_2-{\rho }_1)$ It is to be noted that the only column height of importance is the differential reading, $h_2$. The diferential manometer could be placed 0.5 m or 5.0 m above the pipe $(h_1=0.5m$ or $h_1=5.0m)$ and the value of $h_2$ would remain the same. Relatively large values for the differential reading $h_2$ can be obtained for small pressure differences, $pA-p_B$,if the difference between ${\gamma }_1$ and ${\gamma }_2$ is small. (b) The specific value of the pressure drop for the data given is $p_A-p_B=(0.5m)(15.6\frac{kN}{m^3}-9.8\frac{kN}{m^3})=2.90kPa$

Application of U tube manometer to measure pressure difference

Inclined-Tube Manometer

To measure small pressure changes, a monometer of the type shown in the figure, is frequently used. One leg of the manometer is inclined at an angle $\theta$, and the differential reading $l_2$ is measured along the inclined tube. The difference in pressure $p_A-p_B$ can be expressed as $p_A+{\rho }_{1}g{h}_1-{\rho }_{2}g{l}_2\sin \theta -{\rho }_{3}g{h}_3=p_B$ or $p_A-p_B={\rho }_{2}g{l}_2\sin \theta +{\rho }_{3}g{h}_3-{\rho }_{1}g{h}_1$ where the pressure difference between points (1) and (2) is due to the vertical distance between the points, which can be expressed as $l_2\sin \theta$. Thus, for relatively small angles the differential reading along the inclined tube can be made large even for small pressure differences. The inclined-tube manometer is often used to measure small differences in gas pressures so that if pipes A and B contain a gas then $p_A-p_B={\rho }_{2}g{l}_2\sin \theta$ or $l_2=\frac{p_A-p_B}{{\rho }_{2}g\sin \theta }$ where the contributions of the gas columns $h_1$ and $h_3$ have been neglected. The equation above shows that the differential reading $l_2$ (for a given pressure difference) of the inclined-tube manometer can be increased over that obtained with a conventional U-tube manometer by the factor $\frac{1}{\sin \theta }$. Recall that $\sin \theta \to 0$ as $\theta \to 0$.

inlclined-Tube manometer to measure small difference

Buoyancy

Static pressure rises in fluids as the location of interest gets closer to the earth due to gravitational force. Owing to this change of pressure at different elevations, a lifting force, namely buoyancy, occurs.

Derivation of buoyancy on a body immersed in a fluid

Buoyancy can be best explained when a body is immersed in a liquid. Consider a differential column volume $dV$ of the immersed body. At the bottom surface, a higher hydrostatic force is applied than that on the top surface because of higher hydrostatic pressure on the bottom surface. Hence, the net differential force in $x_3$direction is

$d{F}_3=(p_0+{\rho }_{f}g{h}_2)dA-(p_0+{\rho }_{f}g{h}_1)dA$

$d{F}_3={\rho }_{f}g\underset{dV}{\underbrace{(h_2-h1)dA}}$

Hence, the total force becomes: $F_3=F_{buoyancy}={\int }_V{\rho }_{f}gdV={\rho }_{f}gV=\text{weight of displaced fluid}$

This is the buoyancy force and it exists on all bodies immersed in a fluid which is under the effect of gravitation.

Hydrostatic Forces on Submerged Surfaces

When an object is submerged into liquid, forces due to hydrostatic pressure act on the surface of the body. These forces are distributed on the surface of the object and their magnitude and direction change with the local depth and the surface normal, respectively. When designing technical applications, it is essential to know:

• The magnitude of the resultant force on the surface (integrated force)
• The direction of the resultant force
• Line of action of the resultant force

so that the structure can be designed to sustain the hydrostatic surface forces. Examples technical applications are: under water tunnels, buildings, gates, submarines, etc..

Magnitude of Resultant Force

Consider the submerged flat surface. The magnitude of the resultant hydrostatic force on the liquid side can be calculated by integrating distributed hydrostatic force over the surface:

Resultant hydrodynmaic force, its direction and line of action (center of pressure) on a submerged plane surface.

Hence, the resultant force is the integral of distributed force on the surface created by the pressure:

$F_R={\int }_{A}pdA={\int }_A(p_0+\rho gh)dA$

since $h=ysin\theta$

$F_R={\int }_A(p_0+\rho gysin\theta )dA={\int }_A{p}_{0}dA+\rho gsin\theta {\int }_{A}ydA$ where ${\int }_{A}ydA=y_{c}A$ is the first moment of the area about $x-axis$. Hence

$F_R=p_{0}A+\rho gsin\theta y_{c}A=(p_0+\rho g{h}_c)A=p_{c}A$

where $h_c$is the depth of the centroid of the submerged surface and $p_c$is the pressure at the centroid of the surface. Note that the resultant force is independent of the angle ($\theta$) at which it is slanted and the shape of the surface. Though the resultant force proportional to the pressure at the centroid, the line of action does not pass through the centroid.

Line of action of the resultant force

The location where the resultant force acts $(x^{\prime },y^{\prime })$ can be found by using the first moments of the forces about $x$ and $y-axis$. In other words, the resultant force should act at such a location that the moment created by the resultant force should be same as the moment created by the distributed force due to fluid pressure.

The first moment about $x-axis$ is:

$y^{\prime }{F}_R={\int }_{A}ypdA={\int }_{A}y(p_0+\rho gh)dA$

and the first moment about $y-axis$ is:

$x^{\prime }{F}_R={\int }_{A}xpdA={\int }_{A}x(p_0+\rho gh)dA$

Now, one can determine $y^{\prime }$. Using $h=ysin\theta$

$y^{\prime }{F}_R={\int }_{A}y(p_0+\rho gysin\theta )dA={\int }_A(p_{0}y+\rho g{y}^{2}sin\theta )dA$

$y^{\prime }{F}_R=p_0{\int }_{A}ydA+\rho gsin\theta {\int }_A{y}^{2}dA$

where ${\int }_A{y}^{2}dA$ is the second moment of area about the $x-axis$, $I_{xx}$.

Let the $\hat{x}$ and $\hat{y}$ are the coordinates about the orthogonal axis passing through the centroid of $A$. According to the parallel axis theorem:

$I_{xx}=I_{\hat{x}\hat{x}}+A{y}_c^2$

Inserting this into the above equation yields:

$y^{\prime }{F}_R=p_0{y}_{c}A+\rho gsin\theta (I_{\hat{x}\hat{x}}+A{y}_c^2)=y_c(p_0+\rho g{y}_{c}sin\theta )A+\rho gsin\theta I_{\hat{x}\hat{x}}$

and

$y^{\prime }{F}_R=y_c(p_0+\rho g{h}_c)A+\rho gsin\theta I_{\hat{x}\hat{x}}=y_c{F}_R+\rho gsin\theta I_{\hat{x}\hat{x}}$

This equation yields $y^{\prime }$ coordinate of the line of action of the resultant force.

$y^{\prime }=y_c+\frac{\rho gsin\theta I_{\hat{x}\hat{x}}}{F_R}$

Similarly, $x^{\prime }$ can be found by equating the moment of $F_R$ about $y-axis$ to its integral equivalent:

$x^{\prime }{F}_R={\int }_{A}xpdA={\int }_{A}x(p_0+\rho gh)dA$

$x^{\prime }{F}_R={\int }_{A}x(p_0+\rho gysin\theta )dA={\int }_A(p_{0}x+\rho gxysin\theta )dA$

$x^{\prime }{F}_R=p_0{\int }_{A}xdA+\rho gsin\theta {\int }_{A}xydA=p_0{x}_{c}A+\rho gsin\theta I_{xy}$

Since according to the parallel axis theorem, $I_{xy}=I_{\hat{x}\hat{y}}+A{x}_c{y}_c$, this equation can be written as.

$x^{\prime }{F}_R=x_c(p_0+\rho g{y}_{c}sin\theta )A+\rho gsin\theta I_{\hat{x}\hat{y}}=x_c{F}_R+\rho gsin\theta I_{\hat{x}\hat{y}}$

Hence,

$x^{\prime }=x_c+\frac{\rho gsin\theta I_{\hat{x}\hat{y}}}{F_R}$

Direction of the Resultant Force

The direction of the resultant force is opposite to the surface normal. If a surface is composed of several subsurfaces having different surface normal, either those surfaces should be treated individually or the vector sum of the surface forces on each surface will give the direction of the resultant force.

Hydrostatic Forces on Curved Submerged Surfaces

The hydrostatic force on curved surfaces can be calculated after decomposing it into forces on projected surfaces onto orthogonal planes along cartesian coordinates.

Resultant vertical and horizontal forces and line of actions on a submerged curved surface

Consider a submerged surface curved in $xy$ plane: Vertical resultant force is equal to the weight of the liquid above the curved surface plus the force created by the ambient air pressure above the liquid surface. The line of action of this force passes through the center of gravity of the fluid above the submerged surface. The horizontal resultant force and its line of action are equal to those of the projected plane surface ($A_x$) .