Hydrostatic Equilibrium All Info ChE

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The principle of hydrostatic equilibrium is useful to determine the pressure exerted by the fluid on the wall of a column or container.

Consider a column filled with a stationary mass of static fluid. Take any cross-section of the column parallel to the earth’s surface. The pressure is the same on this cross-section, but if we take another cross-section of the column, the pressure is different than the first cross-section. So, the pressure is changing from height to height.

Hydrostatic Equilibrium Equation Derivation

Assume the following dimension of the column:

The cross-sectional area of the column is $S.$
Pressure is $p$ at height $z$ from the bottom of the column.
The density of the fluid is $\rho$

Now, we will analyze a small portion of the column with height $dz$ and area $S.$

The three forces acting on this small portion of the column are as follows:

Pressure $p$ is acting in the upward direction = $pS$
The force due to pressure $p+dp$ acts in a downward direction = $(p+dp)S$
The force due to gravity acting downward = $\rho \;S \; dz \; g$

Force due to gravity is equal to mass times acceleration = $mg$

Mass $(m)$ = Volume × Density = $V \rho$

Volume = Area × Height = $S \; dz$

Therefore, $m = S \; dz \; g$

And force due to gravity = $\rho \;S \; dz \; g$

The resultant force on a small portion must be zero, and forces acting upward are taken as positive, and forces acting downward are taken as negative.

$\begin{align*} +pS-(p+dp)S-\rho\; S\; dz\; g=0\\ pS-pS-Sdp-\rho \;S \;dz\; g=0 \end{align*}$

(1) $\begin{equation*} -Sdp-\rho \;S \;dz \;g=0 \end{equation*}$

Divide equation (1) by $S.$

(2) $\begin{equation*} dp+\rho \; dz \; g=0 \end{equation*}$

Equation (2) is the desired basic equation that can be used for obtaining the pressure
at any height.

Incompressible Fluid

For incompressible fluids, density is independent of pressure.

Integrating Equation (2), we get

$dp+\rho \; dz \; g=0$

$\int dp + \rho \; g \int dz = 0$

(3) $\begin{equation*} p + z \rho g = constant \end{equation*}$

Equation no. 3 shows that the pressure is at its maximum at the base of the column
or container of the fluid, decreasing as we move up the column.

Consider that the pressure at the bottom of the column is $p_1$ where $z = 0$ , and the pressure at any height $z$ above the base is $p_2$ such that $p_1 > p_2$ , then

(4) $\begin{equation*} \int_{p_1}^{p_2} dp = \rho \; g \int_{0}^{z} dz \end{equation*}$

Integrating the above equation (4), we get

(5) $\begin{equation*} (p_1 - p_2)dp = \rho \; g \; z \end{equation*}$

Where $p_1$ and $p_2$ are expressed in $\frac{N}{m^2}$ , $\rho$ in $\frac{kg}{m^3}$ , $z$ in $m$ , and $g$ in $\frac{m}{s^2}$ in SI Unit.

With the help of Equation (5), the pressure difference in a fluid between any two points can be obtained by measuring the height of the vertical column of the fluid.

Compressible Fluids

For compressible fluids, density changes with pressure.

For an ideal gas, the density is given by the following relation:

(6) $\begin{equation*} \[\rho = \frac{pM}{RT} \end{equation*}$

Here,

$p$ = absolute pressure
$M$ = molecular weight of the gas
$R$ = universal gas constant
$T$ = absolute temperature.

Putting the value of $\rho$ from Equation (6) into Equation (2),

(7) $\begin{equation*} dp + g\; \frac {pM}{RT} \; dz= 0 \end{equation*}$

Rearranging Equation (7),

(8) $\begin{equation*} \frac {dp}{p} + g \frac {M}{RT} \; dz= 0 \end{equation*}$

Integrating Equation (8), we get

(9) $\begin{equation*} \ln {p} + g \frac {M}{RT} \; dz= constant \end{equation*}$

Integrating the above equation between two heights $z_1$ and $z_2$ where the pressures acting are $p_1$ and $p_2$ , we get

$\ln {\frac{p_2}{p_1}}= - g \frac {M(z_2-z_1)}{RT}$

(10) $\begin{equation*} \frac{p_2}{p_1}=exp\; [-g \frac{M}{RT}(z_2-z_1)] \end{equation*}$

Equation (10) is known as the Barometric Equation, giving us the idea of pressure distribution within an ideal gas for isothermal conditions.

Hydrostatic Equilibrium in a Centrifugal Field

In a rotating centrifuge, a layer of liquid is thrown outward at the axis of rotation and held against the bowl’s wall by centrifugal force.

The surface of the liquid is shaped like the paraboloid of the revolution. Still, in industrial centrifuges, the rotational speed is so high and the centrifugal force is so much greater than the force of gravity that the liquid surface is virtually cylindrical and coaxial with the axis of rotation.

This situation is shown in Figure 2: Hydrostatic Equilibrium in a Centrifugal Field.

Here,
$r_1$ = Radial distance from the axis of rotation to the free liquid surface.
$r_2$ = The radius of the centrifugal bowl.

The entire mass of the liquid indicated in the figure is rotating like a rigid body, with no sliding of one layer of liquid over another.

Under these conditions, the pressure distribution in the liquid may be determined by the principle of fluid statics.

The pressure drop over any ring of rotating liquid is calculated as follows:

considering the ring of liquid shown in the figure above and the volume element of thickness $rho$ at radius $r.$

$dF = \omega^2 \; r \; dm$

Where,
$dF$ = Centrifugal Force.
$dm$ = Mass of liquid in an element.
$\omega$ = angular velocity $(rad/s)$

If $\rho$ is the density of liquid and “b” is the breadth of the ring,

$dF = 2\;\pi \; \rho \; r \; b \; dr$

Eliminating $dm$ gives,

$dF = 2 \;\pi \;\rho \; b \;\omega^2 \; r^2 \; dr$

The change in pressure over the element is the force exerted by the element of the liquid divided by the area of the ring

$dp = \frac{dF}{2\; \pi \; r \; b}= \omega^2 \; \rho \; r\; dr$

The pressure drop over the entire ring is

$p_2 - p_1 = \int_{r_1}{r-2} \omega^2 \; \rho \; r \; dr$

Assuming the density is constant and integrating gives

$p_2 - p_1 = \frac{\omega^2 \; \rho \; (r_2^2-r_1^2)}{2}$

Summary: Hydrostatic Equilibrium and Pressure Variation in Fluids

The concept of hydrostatic equilibrium delves into the behavior of fluids within columns or containers and how pressure changes with height. In a stationary fluid column, pressure remains constant along horizontal cross-sections parallel to the earth’s surface but varies from one cross-section to another as you ascend.

For incompressible fluids, the pressure decreases with increasing height, expressed as
p + zρg = constant, where 'z' is pressure, 'z' is height, ‘ρ‘ is density, and 'g’ is the acceleration due to gravity.

In contrast, compressible fluids like ideal gases exhibit a pressure-height relationship defined by the Barometric Equation: ln(p) + g(M/RT)dz = constant, where M represents molecular weight, R is the universal gas constant, and T is the absolute temperature.

The concept extends to hydrostatic equilibrium in rotating centrifuges, where a layer of liquid is pushed outward by centrifugal force. The pressure distribution follows fluid statics principles, and the pressure drop over a rotating liquid ring can be calculated using dp = ω²ρrdr.

These insights into hydrostatic equilibrium, pressure variations in fluids, and their application in rotating systems provide a comprehensive understanding of fluid behavior in various contexts.

FAQ’s

Give hydrostatic equilibrium definition.

The hydrostatic equilibrium involves understanding the pressure exerted by a fluid in a column or container, particularly how it varies with height.

How does pressure change within a stationary fluid column?

In a column filled with stationary fluid, pressure remains constant on any horizontal cross-section parallel to the Earth’s surface. However, as you move to a different cross-section, the pressure changes with height.

What is the basic equation for hydrostatic equilibrium?

The basic equation for hydrostatic equilibrium is derived as dp + ρdzg = 0, where p is pressure, ρ is density, dz is the height change, and g is the acceleration due to gravity.

How is pressure related to height in hydrostatic equilibrium for incompressible fluids?

For incompressible fluids, pressure decreases with increasing height. This relationship is described by the equation p + zρg = constant.

How is pressure related to height in hydrostatic equilibrium for compressible fluids, such as ideal gases?

In compressible fluids like ideal gases, the pressure-height relationship is given by ln(p) + g(M/RT)dz = constant, which is known as the Barometric Equation.

What happens in hydrostatic equilibrium within a rotating centrifuge?

In a rotating centrifuge, a layer of liquid is pushed outward by centrifugal force, creating a unique shape. The pressure distribution in this situation follows the principles of fluid statics.

How is the pressure drop calculated in a rotating liquid ring within a centrifuge?

The pressure drop over a rotating liquid ring is calculated using the equation dp = ω²ρrdr, where ω is the angular velocity, ρ is the density, r is the radial distance, and dr is the change in radius

What is the pressure difference across a rotating liquid ring within a centrifuge?

The pressure difference across a rotating liquid ring is given by p₂ - p₁ = (ω²ρ(r₂² - r₁²))/2, where p₁ and p₂ are pressures at different radial distances, and r₁ and r₂ are the corresponding radius.

What is hydrostatic equilibrium in the atmosphere?

Hydrostatic equilibrium in the atmosphere refers to the balanced state of forces and pressures that exist vertically in the Earth’s atmosphere. In simple terms, it’s the condition where the upward force due to air pressure at any given height is equal to the downward force due to the weight of the air above that height.

What is the principle of hydrostatic equilibrium?

The principle of hydrostatic equilibrium, often simply called hydrostatic equilibrium. It pertains to the equilibrium or balance of forces within a fluid, such as a liquid or gas, that is at rest or in a state of constant motion (i.e., not accelerating).

Read Also:

Fluid Flow Operations

Fluid Statics and Its Application

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