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:

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

$$ \displaystyle dF_{i}^{pressure} = -\frac{\partial P}{\partial x_{i}}dV$$

$$ \displaystyle dF_{i}^{body} = dm\ g_{i}$$

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.

$$ \displaystyle \int^{p_{2}}_{p_{1}}{dP} = -\int^{z_{2}}_{z_{1}}{\rho g dz}$$

$$ \displaystyle p_{2} - p_{1} = -\rho g\ (z_{2}-z_{1})$$

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.

U-tube manometer
(see Pressure measurement)

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: $$\ F_3=F_{buoyancy}=\int_{V}\rho_f g dV=\rho_f g V=\text{weight of displaced fluid} $$

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: so that the structure can be designed to sustain the hydrostatic surface forces. Examples technical applications are: under water tunnels, buildings, gates, submarines, etc..
 * The magnitude of the resultant force on the surface (integrated force)
 * The direction of the resultant force
 * Line of action of the resultant force

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:

$$F_R=\int_A p dA=\int_A (p_0 +\rho gh)dA$$

since $$h=ysin\theta$$

$$F_R=\int_A (p_0 +\rho gysin\theta)dA=\int_A p_0 dA+\rho g sin\theta\int_A ydA$$ where $$\int_A y dA=y_c A$$ is the first moment of the area about $$x-axis$$. Hence

$$F_R=p_0 A+\rho g sin\theta y_c A=(p_0+\rho g h_c)A=p_cA$$

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:

$$y'F_R=\int_A ypdA=\int_A y(p_0+\rho g h)dA$$

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

$$x'F_R=\int_A xpdA=\int_A x(p_0+\rho g h)dA$$

Using $$h=ysin\theta$$

$$y'F_R=\int_A y(p_0+\rho g y sin\theta)dA=\int_A (p_0y+\rho g y^2 sin\theta)dA$$

$$y'F_R=p_0\int_A ydA+\rho g sin\theta\int_A y^2 dA$$

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:

$$y'F_R=p_0 y_c A+\rho g sin\theta(I_{\hat{x}\hat{x}}+A y_c^2)=y_c(p_0+\rho g y_c sin\theta)A+\rho g sin\theta I_{\hat{x}\hat{x}}$$

and

$$y'F_R=y_c(p_0+\rho g h_c)A+\rho g sin\theta I_{\hat{x}\hat{x}}=y_c F_R+\rho g sin\theta I_{\hat{x}\hat{x}}$$

This equation yields $$y'$$ coordinate of the line of action of the resultant force.

$$y'=y_c +\frac{\rho g sin\theta I_{\hat{x}\hat{x}}}{F_R}$$

Similarly, $$x'$$ can be found by equating the moment of $$F_R $$ about $$y-axis$$ to its integral equivalent:

$$x'F_R=\int_A xpdA=\int_A x(p_0+\rho g h)dA$$

$$x'F_R=\int_A x(p_0+\rho g y sin\theta)dA=\int_A (p_0x+\rho g xy sin\theta)dA$$

$$x'F_R=p_0\int_A xdA+\rho g sin\theta\int_A xy dA=p_0x_cA+\rho g sin\theta I_{xy}$$

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

$$x'F_R=x_c(p_0+\rho g y_c sin\theta)A+\rho g sin\theta I_{\hat{x}\hat{y}}=x_c F_R+\rho g sin\theta I_{\hat{x}\hat{y}}$$

Hence,

$$x'=x_c + \frac{\rho g sin\theta I_{\hat{x}\hat{y}}}{F_R}$$

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 ($$A_x$$).