# 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.
We can apply Newton's second law of motion to evaluate the reaction of the particle to the applied forces.
Force balance in $\displaystyle i^{th}$ direction:
$\displaystyle F_{i}^{net}=m\cdot a_{i}$
We can also say,$\displaystyle F_{i}^{body}+F_{i}^{surf}=m\cdot a_{i}$
Force created by pressure is :
$\displaystyle F^{surf}=F^{pressure}=-p\cdot {\vec {A}}$
$\displaystyle {\vec {A}}$ is the vector having the surface area as magnitude and surface normal as direction.
Thus,
$\displaystyle F^{pressure}=-p\cdot {\vec {A}}=-p\ A\ {\vec {n}}$
Force caused by the pressures opposite to the surface normal.
For a differential fluid element: $\displaystyle dF_{i}^{body}+dF_{i}^{surf}=dm\cdot a_{i}$
Remember Taylor Series expansion of a function : $\displaystyle f(x+\Delta x)=f+{\frac {\partial f}{\partial x}}\Delta x+{\frac {1}{2}}{\frac {\partial ^{2}f}{\partial x^{2}}}\Delta x^{2}+\ldots$
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 $P+{\frac {\partial P}{\partial x_{i}}}{\frac {dx_{i}}{2}}$. The force created by the pressure along $x_{2}$ direction can be written as:

### $\displaystyle dF_{2}^{pressure}=\left[P-{\frac {\partial P}{\partial x_{2}}}{\frac {dx_{2}}{2}}\right]dx_{1}\ dx_{3}-\left[P+{\frac {\partial P}{\partial x_{2}}}{\frac {dx_{2}}{2}}\right]dx_{1}\ dx_{3}$

    $\displaystyle =-{\frac {\partial P}{\partial x_{2}}}dx_{1}dx_{2}dx_{3}=-{\frac {\partial P}{\partial x_{2}}}dV$
Thus,
$\displaystyle dF_{i}^{pressure}=-{\frac {\partial P}{\partial x_{i}}}dV$
$\displaystyle dF_{i}^{body}=dm\ g_{i}$
Thus,
$\displaystyle -{\frac {\partial P}{\partial x_{i}}}dV+dm\ g_{i}=dm\ a_{i}$
or,
$\displaystyle -{\frac {\partial P}{\partial x_{i}}}dV+\rho \ dV\ g_{i}=\rho \ dV\ a_{i}$
or,
$\displaystyle -{\frac {\partial P}{\partial x_{i}}}+\rho \ g_{i}=\rho a_{i}$
or,$\displaystyle -{\frac {\partial P}{\partial x_{1}}}=\rho a_{1}\ ;\ -{\frac {\partial P}{\partial x_{2}}}=\rho a_{2}\ ;\ -{\frac {\partial P}{\partial x_{3}}}-\rho g=\rho a_{3}$
for $\displaystyle a_{i}=0$
$\displaystyle {\frac {\partial P}{\partial x_{1}}}=0\ ;\ {\frac {\partial P}{\partial x_{2}}}=0\ ;\ {\frac {\partial P}{\partial x_{3}}}=-\rho g$
Pressure changes only in $\displaystyle x_{3}$ direction.

## Pressure variation in an incompressible and static fluid

$\displaystyle {\frac {\partial P}{\partial x_{3}}}=-\rho g\ ;\ x_{3}=z\rightarrow {\frac {\partial P}{\partial z}}=-\rho g$ is constant since $\displaystyle \rho$ and $\displaystyle g$ are constants.
$\displaystyle \int _{p_{1}}^{p_{2}}{dP}=-\int _{z_{1}}^{z_{2}}{\rho gdz}$
$\displaystyle p_{2}-p_{1}=-\rho g\ (z_{2}-z_{1})$
If we take $\displaystyle p_{2}$ at the surface, then:
$\displaystyle p_{atm}-p_{1}=-\rho g\ (z_{2}-z_{1})$
$\displaystyle p_{1}=p_{atm}+\rho gh\ ;\ \ h=z_{2}-z_{1}$
h is measured from the surface. For incompressible flow,
$\displaystyle p_{2}-p_{1}=-\rho \ g\ (z_{2}-z_{1})\rightarrow p_{1}=p_{atm}+\rho \ g\ h$
| 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. $\displaystyle P_{C}=?$ $\displaystyle P_{A}=P_{0}\ +\ \rho _{A}\ g\ h_{A}$ $\displaystyle P_{B}=P_{A}\ +\ \rho _{B}\ g\ (h_{B}-h_{A})$ $\displaystyle \ P_{C}=P_{B}\ +\ \rho _{C}\ g\ (h_{C}-h_{B})$ $\displaystyle \Rightarrow P_{C}=\rho _{C}\ g\ (h_{C}-h_{B})+\rho _{B}\ g\ (h_{B}-h_{A})+\rho _{A}\ g\ h_{A}+P_{0}$ $\displaystyle P_{C}=\rho _{C}\ g\ H_{C}+\rho _{B}\ g\ H_{B}+\rho _{A}\ g\ H_{A}+P_{0}$ | 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: $\displaystyle P_{1}=P_{2}\Rightarrow {\frac {F_{1}}{A_{1}}}={\frac {F_{2}}{A_{2}}}\Rightarrow F_{2}={\frac {A_{2}}{A_{1}}}F_{1}$ 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 $\displaystyle P_{01}+\rho _{1}\ g\ (H_{1}-h_{1})=P_{02}+\rho _{2}\ g\ (H_{2}-h_{2})$ (of course,when we calculate the small 'h', it should be measured at the height of connecting valve for both column distinctively.) for $\displaystyle \rho _{1}=\rho _{2}=\rho$ $\displaystyle P_{01}+\rho \ g\ (H_{1}-h_{1})=P_{02}+\rho \ g\ (H_{2}-h_{2})$ So from the picture above, we can understand that the pressure in the right column is higher than the left column. For open Containers, $\displaystyle \Rightarrow P_{01}=P_{02}=P_{atm}$ If both fluid columns are at the same level $\displaystyle \Rightarrow H_{1}-h_{1}=H_{2}-h_{2}$ 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), $\displaystyle P_{0}=\rho _{f}\ g\ h$ for $\displaystyle P_{0}=1$atm = $\displaystyle 101,3$ kPa water height will be: $\displaystyle 101,3\cdot 10^{3}=1000\cdot 9.79\cdot h_{water}$ $\displaystyle h_{water}\approx 10m$ where as height of mercury will be around $\displaystyle {\frac {10m}{13,6}}\approx 0,7m$ 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 $\displaystyle Q$, 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, $\displaystyle p_{A}-p_{B}$, along the pipe which is related to the flow through the equation $\displaystyle Q=K{\sqrt {p_{A}-p_{B}}}$ 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 $\displaystyle p_{A}-p_{B}$ in terms of the specific weight of the flowing fluid, $\displaystyle \gamma _{1}=g\rho _{1}$, the specific weight of the gage fluid, $\displaystyle \gamma _{2}=g\rho _{2}$, and the various heights indicated. (b) For $\displaystyle \gamma _{1}=9.80{\frac {kN}{m^{3}}}$, $\displaystyle \gamma _{2}=15.6{\frac {kN}{m^{3}}}$, $\displaystyle h_{1}=1.0m$ and $\displaystyle h_{2}=0.5m$, what is the value of the pressure drop, $\displaystyle p_{A}-p_{B}$? 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 $\displaystyle \gamma _{1}h_{1}$ 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 $\displaystyle \gamma _{2}h_{2}$. The pressures at levels (4) and (5) are equal, and as we move from (5) to B the pressure will increase by $\displaystyle \gamma _{1}(h_{1}+h_{2})$. Thus, in equation form $\displaystyle p_{A}-\gamma _{1}h_{1}-\gamma _{2}h_{2}+\gamma _{1}(h_{1}+h_{2})=p_{B}$ or $\displaystyle p_{A}-p_{B}=h_{2}(\gamma _{2}-\gamma _{1})=h_{2}g(\rho _{2}-\rho _{1})$ It is to be noted that the only column height of importance is the differential reading, $\displaystyle h_{2}$. The diferential manometer could be placed 0.5 m or 5.0 m above the pipe $\displaystyle (h_{1}=0.5m$ or $\displaystyle h_{1}=5.0m)$ and the value of $\displaystyle h_{2}$ would remain the same. Relatively large values for the differential reading $\displaystyle h_{2}$ can be obtained for small pressure differences, $\displaystyle p{A}-p_{B}$,if the difference between $\displaystyle \gamma _{1}$ and $\displaystyle \gamma _{2}$ is small. (b) The specific value of the pressure drop for the data given is $\displaystyle p_{A}-p_{B}=(0.5m)(15.6\ {\frac {kN}{m^{3}}}-9.8\ {\frac {kN}{m^{3}}})=2.90\ kPa$ | 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 $\displaystyle \theta$, and the differential reading $\displaystyle l_{2}$ is measured along the inclined tube. The difference in pressure $\displaystyle p_{A}-p_{B}$ can be expressed as $\displaystyle p_{A}+\rho _{1}gh_{1}-\rho _{2}gl_{2}\sin \theta -\rho _{3}gh_{3}=p_{B}$ or $\displaystyle p_{A}-p_{B}=\rho _{2}gl_{2}\sin \theta +\rho _{3}gh_{3}-\rho _{1}gh_{1}$ where the pressure difference between points (1) and (2) is due to the vertical distance between the points, which can be expressed as $\displaystyle l_{2}\sin \theta$. 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 $\displaystyle p_{A}-p_{B}=\rho _{2}gl_{2}\sin \theta$ or $\displaystyle l_{2}={\frac {p_{A}-p_{B}}{\rho _{2}g\sin \theta }}$ where the contributions of the gas columns $\displaystyle h_{1}$ and $\displaystyle h_{3}$ have been neglected. The equation above shows that the differential reading $\displaystyle l_{2}$ (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 $\displaystyle {\frac {1}{\sin \theta }}$. Recall that $\displaystyle \sin \theta \rightarrow 0$ as $\displaystyle \theta \rightarrow 0$. | 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.
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
$dF_{3}=(p_{0}+\rho _{f}gh_{2})dA-(p_{0}+\rho _{f}gh_{1})dA$
$dF_{3}=\rho _{f}g\underbrace {(h_{2}-h1)dA} _{dV}$
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:
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
$F_{R}=p_{0}A+\rho gsin\theta y_{c}A=(p_{0}+\rho gh_{c})A=p_{c}A$
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 $(x',y')$ 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:
$y'F_{R}=\int _{A}ypdA=\int _{A}y(p_{0}+\rho gh)dA$
and the first moment about $y-axis$ is:
$x'F_{R}=\int _{A}xpdA=\int _{A}x(p_{0}+\rho gh)dA$
Now, one can determine $y'$. Using $h=ysin\theta$
$y'F_{R}=\int _{A}y(p_{0}+\rho gysin\theta )dA=\int _{A}(p_{0}y+\rho gy^{2}sin\theta )dA$
$y'F_{R}=p_{0}\int _{A}ydA+\rho gsin\theta \int _{A}y^{2}dA$
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:
$I_{xx}=I_{{\hat {x}}{\hat {x}}}+Ay_{c}^{2}$
Inserting this into the above equation yields:
$y'F_{R}=p_{0}y_{c}A+\rho gsin\theta (I_{{\hat {x}}{\hat {x}}}+Ay_{c}^{2})=y_{c}(p_{0}+\rho gy_{c}sin\theta )A+\rho gsin\theta I_{{\hat {x}}{\hat {x}}}$
and
$y'F_{R}=y_{c}(p_{0}+\rho gh_{c})A+\rho gsin\theta I_{{\hat {x}}{\hat {x}}}=y_{c}F_{R}+\rho gsin\theta I_{{\hat {x}}{\hat {x}}}$
This equation yields $y'$ coordinate of the line of action of the resultant force.
$y'=y_{c}+{\frac {\rho gsin\theta I_{{\hat {x}}{\hat {x}}}}{F_{R}}}$
Similarly, $x'$ can be found by equating the moment of $F_{R}$ about $y-axis$ to its integral equivalent:
$x'F_{R}=\int _{A}xpdA=\int _{A}x(p_{0}+\rho gh)dA$
$x'F_{R}=\int _{A}x(p_{0}+\rho gysin\theta )dA=\int _{A}(p_{0}x+\rho gxysin\theta )dA$
$x'F_{R}=p_{0}\int _{A}xdA+\rho gsin\theta \int _{A}xydA=p_{0}x_{c}A+\rho gsin\theta I_{xy}$
Since according to the parallel axis theorem, $I_{xy}=I_{{\hat {x}}{\hat {y}}}+Ax_{c}y_{c}$, this equation can be written as.
$x'F_{R}=x_{c}(p_{0}+\rho gy_{c}sin\theta )A+\rho gsin\theta I_{{\hat {x}}{\hat {y}}}=x_{c}F_{R}+\rho gsin\theta I_{{\hat {x}}{\hat {y}}}$
Hence,
$x'=x_{c}+{\frac {\rho gsin\theta I_{{\hat {x}}{\hat {y}}}}{F_{R}}}$

### 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.
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 ($A_{x}$) .