Chemical application

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Stokes force

Equation

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The resistance is defined in terms of the fluid viscosity and the sphere's velocity as follows:

$ F_v = b v $



Stokes explicitly calculated the resistance experienced by the sphere and determined that viscosity is proportional to the sphere's radius and its velocity, leading to the following equation for resistance:

$ F_v =6 \pi \eta r v $

$\pi$
Pi
3.1415927
$rad$
5057
$r$
Radius of a sphere
$m$
10331
$v$
Speed
$m/s$
6029
$F_v$
Viscose force
$N$
4979
$\eta$
Viscosity
$Pa s$
5422

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Generación de gotitas

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Al pulverizar los líquidos se obtiene los droplets:

ID:(12892, 0)



Distribuidor de líquidos

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En caso de que se busca introducir el químico como liquido en el suelo se trabaja con un sistema que lleva un estanque y trabaja con un cuchillo de abre la tierra para depositar el liquido:

ID:(12893, 0)



Darcy's law and hydraulic resistance

Equation

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Darcy rewrites the Hagen Poiseuille equation so that the pressure difference ($\Delta p$) is equal to the hydraulic resistance ($R_h$) times the volume flow ($J_V$):

$ \Delta p = R_h J_V $

$R_h$
Hydraulic resistance
$kg/m^4s$
5424
$\Delta p$
Variación de la Presión
$Pa$
6673
$J_V$
Volume flow
$m^3/s$
5448

The volume flow ($J_V$) can be calculated from the hydraulic conductance ($G_h$) and the pressure difference ($\Delta p$) using the following equation:

$ J_V = G_h \Delta p $



Furthermore, using the relationship for the hydraulic resistance ($R_h$):

$ R_h = \displaystyle\frac{1}{ G_h }$



results in:

$ \Delta p = R_h J_V $

ID:(3179, 0)



Hydraulic resistance of a tube

Equation

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Since the hydraulic resistance ($R_h$) is equal to the inverse of the hydraulic conductance ($G_h$), it can be calculated from the expression of the latter. In this way, we can identify parameters related to geometry (the tube length ($\Delta L$) and the tube radius ($R$)) and the type of liquid (the viscosity ($\eta$)), which can be collectively referred to as a hydraulic resistance ($R_h$):

$ R_h =\displaystyle\frac{8 \eta | \Delta L | }{ \pi R ^4}$

$R_h$
Hydraulic resistance
$kg/m^4s$
5424
$\pi$
Pi
3.1415927
$rad$
5057
$\Delta L$
Tube length
$m$
5430
$R$
Tube radius
$m$
5417
$\eta$
Viscosity
$Pa s$
5422

Since the hydraulic resistance ($R_h$) is equal to the hydraulic conductance ($G_h$) as per the following equation:

$ R_h = \displaystyle\frac{1}{ G_h }$



and since the hydraulic conductance ($G_h$) is expressed in terms of the viscosity ($\eta$), the tube radius ($R$), and the tube length ($\Delta L$) as follows:

$ G_h =\displaystyle\frac{ \pi R ^4}{8 \eta | \Delta L | }$



we can conclude that:

$ R_h =\displaystyle\frac{8 \eta | \Delta L | }{ \pi R ^4}$

ID:(3629, 0)