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ID:(328, 0)

Equation

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The viscose force ($F_v$) can be calculated from the parallel surfaces ($S$), the viscosity ($\eta$), the speed difference between surfaces ($\Delta v$), and the distance between surfaces ($\Delta z$) using the following method:

$F_v =- S \eta \displaystyle\frac{ \Delta v }{ \Delta z }$

Conditions

Variables

$\Delta z$

Distance between surfaces

$m$

5436

$S$

Section

$m^2$

5205

$\Delta v$

Speed difference between surfaces

$m/s$

5556

$F_v$

Viscose force

$N$

4979

$\eta$

Viscosity

$Pa s$

5422

Relation

ID:(3622, 0)

Equation

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In the case of a cylinder, the surface is defined by tube length ($\Delta L$), and by the perimeter of each of the internal cylinders, which is calculated by multiplying $2\pi$ by the radius of position in a tube ($r$). With this, the cylinder resistance force ($F_v$) is calculated using the viscosity ($\eta$) and the variation of speed between two radii ($dv$) for the width of the cylinder the radius variation in a tube ($dr$), resulting in:

$F_v =-2 \pi r \Delta L \eta \displaystyle\frac{ dv }{ dr }$

Conditions

Variables

$r$

Cylinder radial position

$m$

5420

$\pi$

Pi

3.1415927

$rad$

5057

$v$

Speed on a cylinder radio

$m/s$

5449

$\Delta L$

Tube length

$m$

5430

$R$

Tube radius

$m$

5417

$F_v$

Viscose force

$N$

4979

$\eta$

Viscosity

$Pa s$

5422

Relation

Development

As the viscous force is

$F_v =- S \eta \displaystyle\frac{ \Delta v }{ \Delta z }$

and the surface area of the cylinder is

$S=2\pi R L$

where $R$ is the radius and $L$ is the length of the channel, the viscous force can be expressed as

$F_v =-2 \pi r \Delta L \eta \displaystyle\frac{ dv }{ dr }$

where $\eta$ represents the viscosity and $dv/dr$ is the velocity gradient between the wall and the flow.

ID:(3623, 0)

Equation

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To describe the flow, a coordinate system is defined in which the liquid flows from the position at the beginning of the tube ($L_i$) to the position at the end of the tube ($L_e$), indicating that the pressure at the pressure in the initial position ($p_i$) is greater than at the pressure in end position (e) ($p_e$). This movement depends on the tube length ($\Delta L$), which is calculated as follows:

$\Delta L = L_e - L_i$

Conditions

Variables

$L_i$

Position at the beginning of the tube

$m$

6274

$L_e$

Position at the end of the tube

$m$

6275

$\Delta L$

Tube length

$m$

5430

Relation

ID:(3802, 0)

Image

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ID:(1694, 0)

Equation

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$\eta_n=\eta_p\left(1+\displaystyle\frac{2}{5}Ht\right)$

ID:(3636, 0)

Equation

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When solving the flow equation with the boundary condition, we obtain the speed on a cylinder radio ($v$) as a function of the curvature radio ($r$), represented by a parabola centered at the maximum flow rate ($v_{max}$) and equal to zero at the tube radius ($R$):

$v = v_{max} \left(1-\displaystyle\frac{ r ^2}{ R ^2}\right)$

Conditions

Variables

$r$

Cylinder radial position

$m$

5420

$v_{max}$

Maximum flow rate

$m/s$

5421

$v$

Speed on a cylinder radio

$m/s$

5449

$R$

Tube radius

$m$

5417

Relation

Development

When a the pressure difference ($\Delta p_s$) acts on a section with an area of $\pi R^2$, with the tube radius ($R$) as the curvature radio ($r$), it generates a force represented by:

$\pi r^2 \Delta p$

This force drives the liquid against viscous resistance, given by:

$F_v =-2 \pi r \Delta L \eta \displaystyle\frac{ dv }{ dr }$

By equating these two forces, we obtain:

$\pi r^2 \Delta p = \eta 2\pi r \Delta L \displaystyle\frac{dv}{dr}$

Which leads to the equation:

$\displaystyle\frac{dv}{dr} = \displaystyle\frac{1}{2\eta}\displaystyle\frac{\Delta p}{\Delta L} r$

If we integrate this equation from a position defined by the curvature radio ($r$) to the edge where the tube radius ($R$) (taking into account that the velocity at the edge is zero), we can obtain the speed on a cylinder radio ($v$) as a function of the curvature radio ($r$):

$v = v_{max} \left(1-\displaystyle\frac{ r ^2}{ R ^2}\right)$

Where:

$v_{max} =-\displaystyle\frac{ R ^2}{4 \eta }\displaystyle\frac{ \Delta p }{ \Delta L }$

is the maximum flow rate ($v_{max}$) at the center of the flow.

.

ID:(3627, 0)

Equation

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The value of the maximum flow rate ($v_{max}$) at the center of a cylinder depends on the viscosity ($\eta$), the tube radius ($R$), and the gradient created by the pressure difference ($\Delta p_s$) and the tube length ($\Delta L$), as represented by:

$v_{max} =-\displaystyle\frac{ R ^2}{4 \eta }\displaystyle\frac{ \Delta p }{ \Delta L }$

Conditions

Variables

$v_{max}$

Maximum flow rate

$m/s$

5421

$\Delta L$

Tube length

$m$

5430

$R$

Tube radius

$m$

5417

$\Delta p$

Variación de la Presión

$Pa$

6673

$\eta$

Viscosity

$Pa s$

5422

Relation

The negative sign indicates that the flow always occurs in the direction opposite to the gradient, meaning from the area of higher pressure to the area of lower pressure.

ID:(3628, 0)

Image

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ID:(1896, 0)

Equation

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$\eta_{fl}=\displaystyle\frac{\eta_n}{\left(1-\displaystyle\frac{d}{R}\right)^4}$

ID:(3638, 0)

Image

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ID:(1895, 0)

Equation

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$\eta_d=\displaystyle\frac{\eta_n}{1+C_{\sigma}\displaystyle\frac{d\sigma}{dt}}$

ID:(3637, 0)

Image

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ID:(1695, 0)

Equation

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The volume flow ($J_V$) can be calculated with the Hagen-Poiseuille law that with the parameters the viscosity ($\eta$), the pressure difference ($\Delta p$), the tube radius ($R$) and the tube length ($\Delta L$) is:

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

Conditions

Variables

$\pi$

Pi

3.1415927

$rad$

5057

$\Delta L$

Tube length

$m$

5430

$R$

Tube radius

$m$

5417

$\Delta p$

Variación de la Presión

$Pa$

6673

$\eta$

Viscosity

$Pa s$

5422

$J_V$

Volume flow

$m^3/s$

5448

Relation

Development

If we consider the profile of speed on a cylinder radio ($v$) for a fluid in a cylindrical channel, where the speed on a cylinder radio ($v$) varies with respect to radius of position in a tube ($r$) according to the following expression:

$v = v_{max} \left(1-\displaystyle\frac{ r ^2}{ R ^2}\right)$

involving the tube radius ($R$) and the maximum flow rate ($v_{max}$). We can calculate the maximum flow rate ($v_{max}$) using the viscosity ($\eta$), the pressure difference ($\Delta p$), and the tube length ($\Delta L$) as follows:

$v_{max} =-\displaystyle\frac{ R ^2}{4 \eta }\displaystyle\frac{ \Delta p }{ \Delta L }$

If we integrate the velocity across the cross-section of the channel, we obtain the volume flow ($J_V$), defined as the integral of $\pi r v(r)$ with respect to radius of position in a tube ($r$) from $0$ to tube radius ($R$). This integral can be simplified as follows:

$J_V=-\displaystyle\int_0^Rdr \pi r v(r)=-\displaystyle\frac{R^2}{4\eta}\displaystyle\frac{\Delta p}{\Delta L}\displaystyle\int_0^Rdr \pi r \left(1-\displaystyle\frac{r^2}{R^2}\right)$

The integration yields the resulting Hagen-Poiseuille law:

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

ID:(3178, 0)