User:
Efecto de la Viscosidad
Storyboard 
Variables
Calculations
to 
,
then, select the variable: 
to 
Calculations
Variable
Given
Calculate
Target :
Equation
To be used
Equations
If we consider the profile of ERROR:5449,0 for a fluid in a cylindrical channel, where the speed on a cylinder radio ($v$) varies with respect to ERROR:10120,0 according to the following expression:
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:
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 ERROR:10120,0 from $0$ to ERROR:5417,0. 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:
As the viscous force is
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
where $\eta$ represents the viscosity and $dv/dr$ is the velocity gradient between the wall and the flow.
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:
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 ($r$):
Where:
is the maximum flow rate ($v_{max}$) at the center of the flow.
Examples
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:
In the case of a cylinder, the surface is defined by ERROR:5430,0, 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:
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:
$\eta_n=\eta_p\left(1+\displaystyle\frac{2}{5}Ht\right)$
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 ($R$):
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:
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.
$\eta_{fl}=\displaystyle\frac{\eta_n}{\left(1-\displaystyle\frac{d}{R}\right)^4}$
$\eta_d=\displaystyle\frac{\eta_n}{1+C_{\sigma}\displaystyle\frac{d\sigma}{dt}}$
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:
ID:(328, 0)