Efecto de la Viscosidad
Model 
Variables
Calculations
Calculations
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:
| $ 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 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:
| $ J_V =-\displaystyle\frac{ \pi R ^4}{8 \eta }\displaystyle\frac{ \Delta p }{ \Delta L }$ |
(ID 3178)
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)
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)
Examples
(ID 1896)
(ID 1895)
(ID 1695)
ID:(328, 0)
