In liquid column

Storyboard

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

ID:(1427, 0)



In liquid column

Storyboard

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.

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$\tau_b$
tau_b
Characteristic time column with Bernoulli
s
$h$
h
Column height
m
$\rho$
rho
Density
kg/m^3
$r$
r
Disc radius
m
$v_s$
v_s
Flow speed
m/s
$j_s$
j_s
Flux density
m/s
$j_{s1}$
j_s1
Flux density 1
m/s
$j_{s2}$
j_s2
Flux density 2
m/s
$\Delta h$
Dh
Height of liquid column
m
$h_0$
h_0
Initial height of liquid column
m
$\rho_w$
rho_w
Liquid density
kg/m^3
$\Delta p_s$
Dp_s
Pressure difference
Pa
$S_1$
S_1
Section in point 1
m^2
$S_2$
S_2
Section in point 2
m^2
$S$
S
Section Tube
m^2
$S$
S
Surface of a disk
m^2
$t$
t
Time
s
$\Delta t$
Dt
Time elapsed
s
$\Delta s$
Ds
Tube element
m
$R$
R
Tube radius
m
$\Delta p$
Dp
Variación de la Presión
Pa
$\Delta V$
DV
Volume element
m^3
$J_V$
J_V
Volume flow
m^3/s

Calculations


First, select the equation:   to ,  then, select the variable:   to 

Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

 Variable   Given   Calculate   Target :   Equation   To be used



Equations

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

equation=4252

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

equation=4250

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:

equation=4251

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

equation

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

equation=4345

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

equation=4835

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:

equation=4350

Therefore, the equation that describes this situation is:

equation

If in the equation

equation=9882

the constants are replaced by

equation=14523

we obtain the first-order linear differential equation

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



whose solution is

equation

In this case, it can be assumed that the mean Speed of Fluid in Point 2 ($v_2$) represents zero velocity and the mean Speed of Fluid in Point 1 ($v_1$) corresponds to the flow speed ($v_s$). Therefore, for the speed difference between surfaces ($\Delta v$) the following is established:

$\Delta v = v_2 - v_1 = 0 - v_s = - v_s$



and for the average speed ($\bar{v}$) it is calculated:

$\bar{v} = \displaystyle\frac{v_1 + v_2}{2} = \frac{v_s}{2}$



Consequently, with the variación de la Presión ($\Delta p$), which equals the pressure difference ($\Delta p_s$), we obtain:

equation=4835

resulting in:

$\Delta p_s = \displaystyle\frac{1}{2} \rho v_s^2$



leading to:

equation

The volume ($V$) for a tube with constant the section Tube ($S$) and a position ($s$) is

equation=4876

If the section Tube ($S$) is constant, the temporal derivative will be

$\displaystyle\frac{dV}{dt} = S\displaystyle\frac{ds}{dt}$



thus, with the volume flow ($J_V$) defined by

equation=12713

and with the flux density ($j_s$) associated with the position ($s$) via

equation=12714

it is concluded that

equation


Examples


mechanisms

If there is a column height ($h$) of liquid with the liquid density ($\rho_w$) under the effect of gravity, using the gravitational Acceleration ($g$), the variación de la Presión ($\Delta p$) is generated according to:

equation=4345

This the variación de la Presión ($\Delta p$) produces a flow through the outlet tube with the tube length ($\Delta L$), the tube radius ($R$), and the viscosity ($\eta$) of a volume flow 1 ($J_{V1}$) according to the Hagen-Poiseuille law:

equation=3178

Since this equation includes the section in point 2 ($S_2$), the flux density 2 ($j_{s2}$) can be calculated using:

equation=4349

With this, we obtain:

equation=11064

which corresponds to an average velocity.

video

To model the system, the key parameters are:

• Interior diameter of the container: 93 mm

• Interior diameter of the evacuation channel: 3.2 mm

• Length of the evacuation channel: 18 mm

The initial liquid height is 25 cm.

Let's consider the system of a cylindrical bucket with a drainage hole. When the plug is removed, the water starts to flow due to the existing pressure. According to Bernoulli's principle, inside the bucket ($v\sim 0$), the velocity is zero, and we have:

$\displaystyle\frac{1}{2}\rho v^2 + \rho g h\sim \rho g h$



while outside the bucket ($h=0$), only the kinetic component exists:

$\displaystyle\frac{1}{2}\rho v^2 + \rho g h\sim \displaystyle\frac{1}{2}\rho v^2$



Since both expressions are equal, we have:

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



which gives the velocity as:

$v=\sqrt{2 g h}$



To compare with the experiment, we can use this expression to estimate, with:

equation=9882

the range that the stream should have. If we plot it graphically, we observe:

image

where:

• the red dots correspond to the experimental measurements,

• the blue dots correspond to the calculated range using a factor of 0.11,

• the transparent dots correspond to the calculated range using a factor of 0.09.

Therefore, we can conclude that Bernoulli's model overestimates the velocity at which the bucket empties. This is because in the vicinity of the drainage hole, the effects of viscosity are not negligible, and therefore, the velocity is lower.

If the Tracker program is used, the height of the meniscus of the column and the range of the jet can be measured. The relationship between the two is shown in the following graph:

image

The recorded data, which can be downloaded as an Excel table from the following link excel table, are as follows:

Time [s]Height [m]Reach [m]
02.23E-011.89E-01
42.14E-011.86E-01
82.04E-011.82E-01
121.94E-011.77E-01
161.86E-011.72E-01
201.79E-011.68E-01
241.71E-011.66E-01
281.63E-011.62E-01
321.54E-011.58E-01
361.46E-011.52E-01
401.39E-011.48E-01
441.32E-011.44E-01
481.24E-011.39E-01
521.18E-011.35E-01
561.11E-011.31E-01
601.06E-011.27E-01
649.88E-021.23E-01
689.29E-021.18E-01
728.70E-021.15E-01
768.11E-021.12E-01
807.52E-021.06E-01
847.12E-021.02E-01
886.51E-029.69E-02
926.00E-029.42E-02
965.58E-028.94E-02
1005.09E-028.52E-02
1044.70E-028.13E-02
1084.34E-027.63E-02
1123.97E-027.22E-02
1163.49E-026.79E-02
1203.15E-026.28E-02
1242.91E-025.96E-02
1282.58E-025.33E-02
1322.23E-024.92E-02
1361.98E-024.31E-02
1401.71E-023.85E-02
1441.54E-023.38E-02
1481.28E-022.85E-02
1521.11E-022.23E-02
1569.17E-031.54E-02
1607.15E-037.95E-03

Note: E indicates scientific notation (ej. 1.2E+3 = 1.2x10^3 = 1200, y 1.2E-3 = 1.2x10^-3 = 0.0012)


model

If we have a tube with a the section Tube ($S$) moving a distance the tube element ($\Delta s$) along its axis, having displaced the volume element ($\Delta V$), then it is equal to:

kyon

The volume flow ($J_V$) corresponds to the volume flowing ($\Delta V$) flowing through the channel at the time elapsed ($\Delta t$). Therefore, we have:

kyon

The flux density ($j_s$) is related to the distance traveled in a time ($\Delta s$), which is the distance that the fluid travels in the time elapsed ($\Delta t$), as follows:

kyon

A flux density ($j_s$) can be expressed in terms of the volume flow ($J_V$) using the section or Area ($S$) through the following formula:

kyon

The surface of a disk ($S$) of ERROR:5275.1 is calculated as follows:

kyon

When applying the Bernoulli equation relative to a stationary point in the flow, it is established that the flow speed ($v_s$) is associated with the pressure difference ($\Delta p_s$) concerning the pressure at this point. With the density ($\rho$), the following is observed:

kyon

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:

kyon

The principle of continuity dictates that the flow at the first point, which is equal to the flux density 1 ($j_{s1}$) times the section in point 1 ($S_1$), must be equal to the flow at the second point, given by the flux density 2 ($j_{s2}$) times the section in point 2 ($S_2$). From this, it follows that:

kyon

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

equation=4345

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

equation=4835

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:

kyon

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

equation=9882

we can condense the constants into a characteristic time unit:

kyon

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

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

equation=9882

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

equation=14523

Upon integrating, we obtain:

kyon

Where $h_0$ represents the initial height.


>Model

ID:(1427, 0)



Mechanisms

Definition


ID:(15487, 0)



Static and dynamic pressure

Image

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.

ID:(11092, 0)



Column emptying experiment

Note

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.

ID:(9870, 0)



Model

Quote


ID:(15490, 0)