Fluid Mechanics for MAP/Fluid Statics

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Definitions

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

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.

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

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: $\displaystyle F(x+\Delta x)=F+{\frac {\partial F}{\partial x}}+{\frac {1}{2}}{\frac {\partial ^{2}F}{\partial x^{2}}}\Delta x^{2}+\ldots$

P is the pressure in the center of the fluid element, therefore the pressure on the surface in direction of $x_{i}$ is $P+{\frac {\partial P}{\partial x_{i}}}{\frac {dx_{i}}{2}}$.

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

$\displaystyle =-{\frac {\partial P}{\partial x_{2}}}dx_{1}dx_{2}dx_{3}=-{\frac {\partial P}{\partial x_{2}}}dV$

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,$\displaystyle -{\frac {\partial P}{\partial x_{i}}}dV+\rho \ dV\ g_{i}=\rho \ a_{i}$

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

(a) pressure at location A and B (b)Depth is measured from the free surface (c) Pressure variation with depth of water column

$\displaystyle {\frac {\partial P}{\partial x_{3}}}=-\rho g\ ;\ x_{3}=z\rightarrow {\frac {\partial P}{\partial z}}=-\rho g$ 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.

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. Remember: $\displaystyle {\frac {\partial P}{\partial x_{i}}}+\rho \ g_{i}=\rho \ a_{i}$

similar depth in same fluid experience similar hydrostatic pressure

For incompressible flow, $\displaystyle p_{2}-p_{1}=-\rho \ g\ (z_{2}-z_{1})\rightarrow p_{1}=p_{atm}+\rho \ g\ h$ Consider 3 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

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

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.

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

The volume rate of flow, 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, $\displaystyle \rho _{1}$, the specific weight of the gage fluid, $\displaystyle \rho _{2}$, and the various heights indicated. (b) For $\displaystyle \rho _{1}=9.80{\frac {kN}{m^{3}}}$, $\displaystyle \rho _{2}=15.6{\frac {kN}{m^{3}}}$, $\displaystyle h_{1}=1.0m$ 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 $\displaystyle \rho _{1}h_{1}$ 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 $\displaystyle \rho _{2}h_{2}$. The pressures at levels (4) and (5) are equal, and as we move from (5) to B the pressure will increase by $\displaystyle \rho _{1}(h_{1}+h_{2})$. Thus, in equation form $\displaystyle p_{A}-\rho _{1}h_{1}-\rho _{2}h_{2}+\rho _{1}(h_{1}+h_{2})=p_{B}$ or $\displaystyle p_{A}-p_{B}=h_{2}(\rho _{2}-\rho _{1})$ 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 $\displaystyle \rho _{1}$ and $\displaystyle \rho _{2}$ 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 $\displaystyle p_{A}+\rho _{1}h_{1}-\rho _{2}l_{2}\sin \theta -\rho _{3}h_{3}=p_{B}$ or $\displaystyle p_{A}-p_{B}=\rho _{2}l_{2}\sin \theta +\rho _{3}h_{3}-\rho _{1}h_{1}$ where the pressure difference between points (1) and (2) is due to the vertical distance between the points, which can be expressed as $\displaystyle l_{2}\sin \theta$. 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 $\displaystyle p_{A}-p_{B}=\rho _{2}l_{2}\sin \theta$ or $\displaystyle l_{2}={\frac {p_{A}-p_{B}}{\rho _{2}\sin \theta }}$ 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

Other related topics

Buoyancy and calculation of forces on the submerged surfaces are topics related to fluid statics.