Storyboard

It is considered a column with water with a hole in its lower part. The emptying is monitored obtaining an output speed depending on the height of the column.

If the data is modeled with Bernoulli but the exit through the hole is modeled with Hagen Poiseville, which corrects the problem of the case in which it was assumed without viscosity.

>Model

ID:(1428, 0)

Concept

>Top

ID:(15492, 0)

Description

>Top

If a small sphere with radius $a$ is dropped into a medium with viscosity $\eta$, it will accelerate until the gravitational force,

$mg=\displaystyle\frac{4\pi}{3}a^3\rho_sg$

where $\rho_s$ is the density of the sphere material, is equal to the viscous force,

$6\pi \eta a v$

where $v$ is the velocity.

Hence, it is possible to estimate the viscosity by measuring the velocity, since

$\displaystyle\frac{4\pi}{3}a^3\rho_sg =6\pi \eta a v$

where the radius can be directly measured.

Here you can observe the behavior of the sphere:

ID:(9871, 0)

Description

>Top

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)

Description

>Top

If we analyze the equation

which describes the application of Hagen-Poiseuille, we observe that the curve only matches the experimental data under the following conditions:

The velocity is low (when the column is nearly empty).

The radius of the evacuation channel must be reduced from 1.5 mm to 0.6 mm.

This shows that the flow is primarily turbulent and that only at low velocity levels is the velocity low enough for the Reynolds number to be low and the flow to be laminar.

ID:(11065, 0)

Image

>Top

If we observe the path that the little ball takes over time, we can see that it mostly moves at a constant speed of about 0.31 meters in 25 seconds, which is equivalent to 0.0124 m/s.

By rearranging the equality between the gravitational force and the Stokes' resistance force:

$\displaystyle\frac{4\pi}{3}a^3\rho_sg =6\pi \eta a v$

we obtain:

$\eta = \displaystyle\frac{2 a^2\rho_sg}{9 v}$

Considering that the ball has a radius of 2 mm and weighs 8 mg, we can determine its density to be approximately $\rho_s\sim 2.38 g/cm^3$. Therefore, the viscosity is estimated to be around $\eta\sim 1.67, Pa s$.

ID:(9881, 0)

Concept

>Top

Selected parameter

Calculations

ID:(15494, 0)

Equation

>Top, >Model

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

we can condense the constants into a characteristic time unit:

$ \tau_{hp} = \displaystyle\frac{S \Delta L}{\pi R^4}\displaystyle\frac{8 \eta}{\rho g}$

Conditions

Variables

$\tau_{hp}$

Characteristic time column with Hagen Pouseuille

$s$

$R$

Cylinder radio

$m$

$\rho$

Density

$kg/m^3$

$g$

Gravitational Acceleration

9.8

$m/s^2$

$\pi$

Pi

3.1415927

$rad$

$S$

Section Tube

$m^2$

$\Delta L$

Tube length

$m$

$\eta$

Viscosity

$Pa s$

Relation

This value becomes a characteristic time for the draining of a column with cross-sectional area $S$ of a viscous liquid with viscosity $\eta," as it flows through a tube of radius $R$.

ID:(14521, 0)

Equation

>Top, >Model

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}$

ID:(9882, 0)

Description

>Top

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:

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

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.

ID:(11063, 0)

Equation

>Top, >Model

Given the model for the flow of a viscous liquid through a tube and considering that the height of the column determines the pressure, we can estimate the velocity as a function of the column height:

$ v = \displaystyle\frac{ \rho g R ^2}{8 \eta \Delta L } h $

Conditions

Variables

$h$

Column height

$m$

$R$

Cylinder radio

$m$

$v_s$

Flow speed

$m/s$

$g$

Gravitational Acceleration

9.8

$m/s^2$

$\rho_w$

Liquid density

$kg/m^3$

$\Delta L$

Tube length

$m$

$\eta$

Viscosity

$Pa s$

Relation

Development

If we consider that the drainage channel presents hydraulic resistance, we can model it using the Hagen-Poiseuille equation:

$ J_V =-\displaystyle\frac{ \pi R ^4}{8 \eta }\displaystyle\frac{ \Delta p }{ \Delta L }$

where the pressure difference is determined by the water column:

$ p = \rho_w g h $

and the velocity is obtained through the flow:

$ j_s = \displaystyle\frac{ J_V }{ S }$

In this way, we obtain the relationship for calculating the velocity as a function of height:

$ v = \displaystyle\frac{ \rho g R ^2}{8 \eta \Delta L } h $

ID:(11064, 0)

Equation

>Top, >Model

The laminar flow of a fluid with viscosity $\eta$ through a tube of radius $R$ is described by Hagen-Poiseuille's law:

The pressure difference is determined by the height of the column $\Delta h$:

which decreases as the liquid flows out. By applying the continuity equation, we can demonstrate that the height decreases over time as follows:

$ S \displaystyle\frac{dh}{dt} = - \displaystyle\frac{\pi R^4}{\Delta L}\displaystyle\frac{\rho g}{8 \eta} h $

Conditions

Variables

$h$

Column height

$m$

$R$

Cylinder radio

$m$

$\rho$

Density

$kg/m^3$

$g$

Gravitational Acceleration

9.8

$m/s^2$

$\pi$

Pi

3.1415927

$rad$

$S$

Section Tube

$m^2$

$t$

Time

$s$

$\Delta L$

Tube length

$m$

$\eta$

Viscosity

$Pa s$

Relation

Development

If the flow through the tube is described by the equation:

$ J_V =-\displaystyle\frac{ \pi R ^4}{8 \eta }\displaystyle\frac{ \Delta p }{ \Delta L }$

and the pressure difference $\Delta p$ is proportional to the height of the column $\Delta h = h:

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

we can apply the conservation of flow $J_{V1}=J_V$ between the tube and the column $J_{V2}$:

$ J_{V1} = J_{V2} $

,

where the flow in column $J_{V2}$ with cross-sectional area $S$ is given by:

$ j_s = \displaystyle\frac{ J_V }{ S }$

Here, the flux density $j_s$ corresponds to the average velocity, which is equal to the rate of change of height over time:

$j_s = \displaystyle\frac{dh}{dt}$

In this way, we obtain the equation for the height of the column as a function of time:

$ S \displaystyle\frac{dh}{dt} = - \displaystyle\frac{\pi R^4}{\Delta L}\displaystyle\frac{\rho g}{8 \eta} h $

ID:(14520, 0)

Equation

>Top, >Model

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

We can rewrite this equation using the characteristic time:

Upon integration, this yields:

$h = h_0 e^{-t/\tau_{hp}}$

Conditions

Variables

$\tau_{hp}$

Characteristic time column with Hagen Pouseuille

$s$

$h$

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} = - \displaystyle\frac{\pi R^4}{\Delta L}\displaystyle\frac{\rho g}{8 \eta} h $

the constants are replaced by

$ \tau_{hp} = \displaystyle\frac{S \Delta L}{\pi R^4}\displaystyle\frac{8 \eta}{\rho g}$

we obtain the first-order linear differential equation

$\displaystyle\frac{dh}{dt}=\displaystyle\frac{1}{\tau_{hp}} h$

whose solution is

$h = h_0 e^{-t/\tau_{hp}}$

where $h_0$ represents the initial height.

ID:(14522, 0)