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In the case of a column of liquid, Bernoulli's law can be applied along with the hydrostatic pressure term. However, it's important to note that when viscosity of the liquid is not considered, the reduction in level occurs uniformly. In this regard, it can be modeled using the continuity equation to determine the downward velocity of the cylinder.

For a column of liquid with an outlet at the bottom, the behavior is similar to what is estimated with Bernoulli's principle. Differences arise due to the formation of small vortices at the outlet, effectively reducing the outlet area and obstructing the flow. However, the flow of a low-viscosity liquid can be modeled in the zone without vortices using Bernoulli's principle.

>Model

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Concept

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Description

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When you have four columns with different cross-sections interconnected, the liquid will assume the same level in all of them. If you open the interconnecting channel, the liquid will start to flow towards the opening where the pressure is equal to the ambient pressure. In the first cylinder, the pressure is equal to the pressure of the water column plus the atmospheric pressure, so the difference with respect to the pressure at the exit is the pressure of the first column. The liquid begins to gain velocity while the dynamic pressure starts to decrease, which is evident in the increasingly smaller columns.

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Description

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This means that as the column empties and the height $h$ decreases, the velocity $v$ also decreases proportionally.

The key parameters are:

• Inner diameter of the vessel: 93 mm

• Inner diameter of the evacuation channel: 3 mm

• Length of the evacuation channel: 18 mm

These parameters are important to understand and analyze the process of column emptying and how the exit velocity varies with height.

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Concept

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Selected parameter

Calculations

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Equation

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Since a fluid or gas is a continuum, the concept of energy can no longer be associated with a specific mass. However, it is possible to consider the energy contained in a volume of the continuum, and by dividing it by the volume itself, we obtain the energy density ($e$). Therefore, with the density ($\rho$), the speed on a cylinder radio ($v$), the column height ($h$), the gravitational Acceleration ($g$), and the water column pressure ($p_t$), we have:

$ e_1 =\displaystyle\frac{1}{2} \rho v_1 ^2+ \rho g h_1 + p_1 $

$ e =\displaystyle\frac{1}{2} \rho v ^2+ \rho g h + p $

Conditions

Variables

$h$

$h_1$

Height or depth 1

$m$

$\rho$

Density

$kg/m^3$

$e$

$e_1$

Energy density in 1

$J/m^3$

$g$

Gravitational Acceleration

9.8

$m/s^2$

$v$

$v_1$

Mean Speed of Fluid in Point 1

$m/s$

$p$

$p_1$

Pressure in column 1

$Pa$

Relation

Development

Another useful equation is the one corresponding to the conservation of energy, which is applicable in cases where viscosity, a process that leads to energy loss, can be neglected. If we consider the classic energy equation $E$, which takes into account kinetic energy, gravitational potential energy, and an external force displacing the liquid over a distance $\Delta z$, it can be expressed as:

$E=\displaystyle\frac{m}{2}v^2+mgh+F\Delta x$

If we consider the energy within a volume $\Delta x\Delta y\Delta z$, we can replace the mass with:

$m=\rho \Delta x\Delta y\Delta z$

And since pressure is given by:

$F=p \Delta S =p \Delta y\Delta z$

We obtain the equation for energy density:

$ e =\displaystyle\frac{1}{2} \rho v ^2+ \rho g h + p $

which corresponds to the Bernoulli equation.

In the absence of viscosity, the conservation of energy implies that the energy density ($e$) is constant at any point in the fluid. Therefore, knowing the velocity and/or pressure at any location in the fluid is sufficient to establish a relationship between velocity and pressure at any point in the fluid.

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Equation

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If energy is conserved within the flowing volumes, then the energy density in 1 ($e_1$) and the energy density in 2 ($e_2$) must be equal:

$ e_1 = e_2 $

Conditions

Variables

$e_1$

Energy density in 1

$J/m^3$

$e_2$

Energy density in 2

$J/m^3$

Relation

This is only possible if viscosity is negligible, as it is associated with energy diffusion, and there are no vortices present, which themselves exhibit energy differences due to varying tangential velocities along the vortex radius.

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Equation

>Top, >Model

Since a fluid or gas is a continuum, the concept of energy can no longer be associated with a specific mass. However, it is possible to consider the energy contained in a volume of the continuum, and by dividing it by the volume itself, we obtain the energy density ($e$). Therefore, with the density ($\rho$), the speed on a cylinder radio ($v$), the column height ($h$), the gravitational Acceleration ($g$), and the water column pressure ($p_t$), we have:

$ e_2 =\displaystyle\frac{1}{2} \rho v_2 ^2+ \rho g h_2 + p_2 $

$ e =\displaystyle\frac{1}{2} \rho v ^2+ \rho g h + p $

Conditions

Variables

$h$

$h_2$

Height or depth 2

$m$

$\rho$

Density

$kg/m^3$

$e$

$e_2$

Energy density in 2

$J/m^3$

$g$

Gravitational Acceleration

9.8

$m/s^2$

$v$

$v_2$

Mean Speed of Fluid in Point 2

$m/s$

$p$

$p_2$

Pressure in column 2

$Pa$

Relation

Development

Another useful equation is the one corresponding to the conservation of energy, which is applicable in cases where viscosity, a process that leads to energy loss, can be neglected. If we consider the classic energy equation $E$, which takes into account kinetic energy, gravitational potential energy, and an external force displacing the liquid over a distance $\Delta z$, it can be expressed as:

$E=\displaystyle\frac{m}{2}v^2+mgh+F\Delta x$

If we consider the energy within a volume $\Delta x\Delta y\Delta z$, we can replace the mass with:

$m=\rho \Delta x\Delta y\Delta z$

And since pressure is given by:

$F=p \Delta S =p \Delta y\Delta z$

We obtain the equation for energy density:

$ e =\displaystyle\frac{1}{2} \rho v ^2+ \rho g h + p $

which corresponds to the Bernoulli equation.

In the absence of viscosity, the conservation of energy implies that the energy density ($e$) is constant at any point in the fluid. Therefore, knowing the velocity and/or pressure at any location in the fluid is sufficient to establish a relationship between velocity and pressure at any point in the fluid.

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Equation

>Top, >Model

Since a fluid or gas is a continuum, the concept of energy can no longer be associated with a specific mass. However, it is possible to consider the energy contained in a volume of the continuum, and by dividing it by the volume itself, we obtain the energy density ($e$). Therefore, with the density ($\rho$), the speed on a cylinder radio ($v$), the column height ($h$), the gravitational Acceleration ($g$), and the water column pressure ($p_t$), we have:

$ e =\displaystyle\frac{1}{2} \rho v ^2+ \rho g h + p $

Conditions

Variables

$h$

Column height

$m$

$\rho$

Density

$kg/m^3$

$e$

Energy density

$J/m^3$

$g$

Gravitational Acceleration

9.8

$m/s^2$

$v$

Speed on a cylinder radio

$m/s$

$p$

Water column pressure

$Pa$

Relation

Development

Another useful equation is the one corresponding to the conservation of energy, which is applicable in cases where viscosity, a process that leads to energy loss, can be neglected. If we consider the classic energy equation $E$, which takes into account kinetic energy, gravitational potential energy, and an external force displacing the liquid over a distance $\Delta z$, it can be expressed as:

$E=\displaystyle\frac{m}{2}v^2+mgh+F\Delta x$

If we consider the energy within a volume $\Delta x\Delta y\Delta z$, we can replace the mass with:

$m=\rho \Delta x\Delta y\Delta z$

And since pressure is given by:

$F=p \Delta S =p \Delta y\Delta z$

We obtain the equation for energy density:

$ e =\displaystyle\frac{1}{2} \rho v ^2+ \rho g h + p $

which corresponds to the Bernoulli equation.

In the absence of viscosity, the conservation of energy implies that the energy density ($e$) is constant at any point in the fluid. Therefore, knowing the velocity and/or pressure at any location in the fluid is sufficient to establish a relationship between velocity and pressure at any point in the fluid.

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Equation

>Top, >Model

If energy is conserved and the medium flows without deformation, the density between two points must be equal, resulting in the well-known Bernoulli's equation:

$\displaystyle\frac{1}{2} \rho v_1 ^2+ \rho g h_1 + p_1 =\displaystyle\frac{1}{2} \rho v_2 ^2+ \rho g h_2 + p_2 $

Conditions

Variables

$g$

Gravitational Acceleration

9.8

$m/s^2$

$h_1$

Height or depth 1

$m$

$h_2$

Height or depth 2

$m$

$\rho$

Liquid density

$kg/m^3$

$v_1$

Mean Speed of Fluid in Point 1

$m/s$

$v_2$

Mean Speed of Fluid in Point 2

$m/s$

$p_1$

Pressure in column 1

$Pa$

$p_2$

Pressure in column 2

$Pa$

Relation

Development

Assuming that the energy density is conserved, for a cell where the average velocity is v, the density is \rho, the pressure is p, the height is h, and the gravitational acceleration is g, we have:

$ e =\displaystyle\frac{1}{2} \rho v ^2+ \rho g h + p $

At point 1, this equation will be equal to the same equation at point 2:

$e(v_1,p_1,h_1)=e(v_2,p_2,h_2)$

where v_1, h_1, and p_1 represent the velocity, height, and pressure at point 1, respectively, and v_2, h_2, and p_2 represent the velocity, height, and pressure at point 2, respectively. Thus, we have:

$\displaystyle\frac{1}{2} \rho v_1 ^2+ \rho g h_1 + p_1 =\displaystyle\frac{1}{2} \rho v_2 ^2+ \rho g h_2 + p_2 $

It is important to bear in mind the following assumptions:

Energy is conserved, particularly assuming the absence of viscosity.

There is no deformation in the medium, hence the density remains constant.

There is no vorticity, meaning no swirling motion leading to circulation in the medium. The fluid must exhibit laminar behavior.

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Equation

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When two liquid columns are connected with the height of liquid column 1 ($h_1$) and the height of liquid column 2 ($h_2$), a the height difference ($\Delta h$) is formed, which is calculated as follows:

$ \Delta h = h_2 - h_1 $

Conditions

Variables

$h_1$

Height or depth 1

$m$

$h_2$

Height or depth 2

$m$

$\Delta h$

Height or depth difference

$m$

Relation

the height difference ($\Delta h$) will generate the pressure difference that will cause the liquid to flow from the higher column to the lower one.

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Equation

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When two liquid columns are connected with the pressure in column 1 ($p_1$) and the pressure in column 2 ($p_2$), a the pressure difference ($\Delta p$) is formed, which is calculated according to the following formula:

$ \Delta p = p_2 - p_1 $

Conditions

Variables

$p_1$

Pressure in column 1

$Pa$

$p_2$

Pressure in column 2

$Pa$

$\Delta p$

Variación de la Presión

$Pa$

Relation

the pressure difference ($\Delta p$) represents the pressure difference that will cause the liquid to flow from the taller column to the shorter one.

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Equation

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The area the section ($S$) of a disk with a diameter of ($$) is calculated as follows:

$ S_1 = \pi r_1 ^2$

$ S = \pi r ^2$

Conditions

Variables

$\pi$

Pi

3.1415927

$rad$

$r$

$r_1$

Radius of Section 1

$m$

$S$

$S_1$

Section in point 1

$m^2$

Relation

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Equation

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The area the section ($S$) of a disk with a diameter of ($$) is calculated as follows:

$ v_{max} = \pi r_2 ^2$

$ S = \pi r ^2$

Conditions

Variables

$\pi$

Pi

3.1415927

$rad$

$r$

$r_2$

Radius of Section 2

$m$

$S$

$v_{max}$

Maximum flow velocity through a cylinder

$m^2$

Relation

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Equation

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The height difference, denoted by the height difference ($\Delta h$), implies that the pressure in both columns is distinct. In particular, the pressure difference ($\Delta p$) is a function of the liquid density ($\rho_w$), the gravitational Acceleration ($g$), and the height difference ($\Delta h$), as follows:

$ \Delta p = \rho_w g \Delta h $

Conditions

Variables

$g$

Gravitational Acceleration

9.8

$m/s^2$

$\Delta h$

Height or depth difference

$m$

$\rho_w$

Liquid density

$kg/m^3$

$\Delta p$

Variación de la Presión

$Pa$

Relation

Development

If there is the pressure difference ($\Delta p$) between two points, as determined by the equation:

$ \Delta p = p_2 - p_1 $

we can utilize the water column pressure ($p_t$), which is defined as:

$ p = p_0 + \rho_w g h $

This results in:

$\Delta p=p_2-p_1=p_0+\rho_wh_2g-p_0-\rho_wh_1g=\rho_w(h_2-h_1)g$

As the height difference ($\Delta h$) is:

$ \Delta h = h_2 - h_1 $

the pressure difference ($\Delta p$) can be expressed as:

$ \Delta p = \rho_w g \Delta h $

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Equation

>Top, >Model

Continuity

leads to that the average speed per section multiplied by the section

is constant. Therefore, if the product is compared in points 1 and 2, it is obtained that

$ S_1 j_{s1} = S_2 j_{s2} $

Conditions

Variables

$j_{s1}$

Flux density 1

$m^3/s$

$j_{s2}$

Flux density 2

$m^3/s$

$S_1$

Section in point 1

$m^2$

$S_2$

Section in point 2

$m^2$

Relation

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Equation

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For the case of a non-viscous liquid flowing in a laminar fashion, the pressure difference generated by the column is:

This results in a velocity flow $v$ through a tube according to Bernoulli's principle:

Given the velocity and the tube's radius, we can calculate the flow, which is related to the flow within the column through the law of continuity. In turn, this is connected to the variation in height $h," as described in:

$ S\displaystyle\frac{dh}{dt} = \pi R^2 \sqrt{2gh}$

Conditions

Variables

$h$

Column height

$m$

$R$

Cylinder radio

$m$

$g$

Gravitational Acceleration

9.8

$m/s^2$

$\pi$

Pi

3.1415927

$rad$

$S$

Section Tube

$m^2$

$t$

Time

$s$

Relation

Development

Using Bernoulli's equation, we can analyze the case of a column of water that generates a pressure difference:

$ \Delta p = \rho_w g \Delta h $

and induces a velocity flow $v$ through a tube, in accordance with:

$ \Delta p = - \rho \bar{v} \Delta v $

Thus, we can estimate the velocity as:

$v = \sqrt{2 g h}$

This velocity, through a tube section of radius $R$, results in a flow:

$J = \pi R^2 v$

If the column has a cross-sectional area $S$, and its height decreases with respect to the variation in height $h$ over time $t$, we can apply the law of continuity, which states:

$ S_1 j_{s1} = S_2 j_{s2} $

Therefore, the equation that describes this situation is:

$ S\displaystyle\frac{dh}{dt} = \pi R^2 \sqrt{2gh}$

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Equation

>Top, >Model

If we examine the equation for the draining of a non-viscous liquid column:

we can condense the constants into a characteristic time unit:

$ \tau_b = \displaystyle\frac{S}{\pi R^2}\sqrt{\displaystyle\frac{h_0}{g}}$

Conditions

Variables

$\tau_b$

Characteristic time column with Bernoulli

$s$

$R$

Cylinder radio

$m$

$g$

Gravitational Acceleration

9.8

$m/s^2$

$h_0$

Initial height of liquid column

$m$

$\pi$

Pi

3.1415927

$rad$

$S$

Section Tube

$m^2$

Relation

This value becomes the time it takes for the column to completely empty, and it depends on the initial height.

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Equation

>Top, >Model

The equation describing the evolution of the draining viscous liquid column is as follows:

We can rewrite this equation in terms of the characteristic time:

Upon integrating, we obtain:

$ h = h_0\left(1-\displaystyle\frac{t}{\tau_b}\right)^2$

Conditions

Variables

$\tau_b$

Characteristic time column with Bernoulli

$s$

$h$

Column height

$m$

$h_0$

Initial height of liquid column

$m$

$t$

Time

$s$

Relation

Development

If in the equation

$ S\displaystyle\frac{dh}{dt} = \pi R^2 \sqrt{2gh}$

the constants are replaced by

$ \tau_b = \displaystyle\frac{S}{\pi R^2}\sqrt{\displaystyle\frac{h_0}{g}}$

we obtain the first-order linear differential equation

$\displaystyle\frac{dh}{dt}=\displaystyle\frac{1}{\tau_b} \sqrt{h_0 h}$

whose solution is

$ h = h_0\left(1-\displaystyle\frac{t}{\tau_b}\right)^2$

Where $h_0$ represents the initial height.

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