Consider the steady laminar flow of a viscous fluid through a long circular cylindrical tube. The governing equation is

$\hspace{1.6cm} -\frac{1}{r} \frac{d}{dr}\bigg(r \mu \frac{dw}{dr}\bigg) = \frac{P_0-P_L}{L} = f_0$

Where w is the axial (i.c.; z) component of velocity, $\mu$ is the viscosity, and $f_0$ is the gradient of pressure (which includes the combined effect of static pressure an gravitational force). The boundary conditions are

$\hspace{1.7cm}\bigg(r \frac{dw}{dr}\bigg)\bigg|_{r =o}=0,\hspace{0.7cm}w(R_a) = 0$

Using the symmetry and two linear elements, determine the velocity field and compare with the exact solution at the nodes:

$\hspace{1.6cm}W(r) = \frac {f_0R_0^2}{4 \mu}\bigg[1-\bigg(\frac{r}{R_0}\bigg)^2\bigg]$

Mumbai University > Mechanical Engineering > Sem 6 > Finite Element Analysis

Marks: 15M

Year: Day 2016

The governing differential equation, representing steady laminar flow a viscous fluid through a long circular cylindrical tube is given as

$-\frac{1}{r} \frac{d}{dr}\bigg(r \mu \frac{dw}{dr}\bigg)=f_0$

$Where \hspace{0.8cm} \omega $ is axial component of velocity

$\hspace{1.9cm}\mu$ is the constant viscosity

$\hspace{1.9cm} f_0$ is the constant pressure gradient

(which includes the combined effect …