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 of static pressure and gravitational force)

The boundary conditions are :

(i) at r = 0, $r \frac{dw}{dr}= 0\hspace{0.9cm}$ (ii) at r =$R_0$, w = 0

Using symmetry, two linear elements and Rayleigh - Ritz method over general element determine :

(i) Element Matrix Equation

(ii) Global matrix Equation

(iii) Velocity distribution

Compare the velocity distribution with exact solution, at least at the nodes.

Write all steps clearly, reflecting preprocessing, processing and post processing therein.

Solution : To develop element matrix equation.

(i) Governing differential equation is

$\hspace{4cm} - \frac{1}{r} \frac{d}{dr} \bigg(r \mu \frac{dw}{dr}\bigg)= f_0$

$\hspace{3cm} i.e. \frac{1}{r} \frac{d}{dr} \bigg(r \mu \frac{dw}{dr}\bigg) + f_0=0$

$\hspace{3.2cm}r \frac{dw}{dr}$ $\begin{vmatrix} \ =0 & w \\ \ r=0 \\ \end{vmatrix}_{r = R_0} $= 0

(ii) General element and governing differential equation in local coordinates

$\hspace{4cm} r=\bar{r} + r_A$

$\hspace{3.8cm} \therefore dr = d \bar{r}$

$\therefore$ Local boundary conditions

$\hspace{3.5cm} \therefore Q_1^e = -2 \pi \bar{r} \mu \frac{dw}{dr}\bigg|_{\bar{r}= 0} $

$\hspace{4cm}Q_2^e = -2 \pi \bar{r} \mu \frac{dw}{dr}\bigg|_{\bar{r}= h_e}$

$\therefore$ The governing differential equation in local coordinates is written as

$\hspace{2cm}\frac{1}{\bar{r} = r_A}\frac{d}{d\bar{r}}\bigg((\bar{r} = r_A)\mu \frac{dw}{d\bar{r}}\bigg) + f_0 = 0$

$\hspace{1.6cm} i.e.\frac{d}{d\bar{r}}\bigg((\bar{r} = r_A)\mu \frac{dw}{d\bar{r}}\bigg) + f_0 (\bar{r} = r_A)= 0$

with local boundary conditions as

$\hspace{3.8cm} (i) (\bar{r} + r_A) \mu \frac{dw}{d\bar{r}}\bigg|_{\bar{r} = 0 }=-\frac{Q_1}{2 \pi}$

$\hspace{3.8cm} (ii) (\bar{r} + r_A) \mu \frac{dw}{d\bar{r}}\bigg|_{\bar{r} = h_e }=\frac{Q_2}{2 \pi}$

(iii) Residue R

$\hspace{0.3cm} R = \frac{d}{d \bar{r}}\bigg((\bar{r} + r_A) \mu \frac{dw}{d\bar{r}}\bigg) + f_0 (\bar{r} + r_A)\neq 0 \hspace{2cm}(\therefore \text{solution is approximate})$

(iv) Weighted integral from

$\hspace{3cm} 0 = \int\limits_0^{h_e} w_i R d\bar{r}$

$\hspace{3cm} 0 =\int\limits_0^{h_e} w_i\bigg[\frac{d}{d\bar{r}}\bigg((\bar{r} + r_A)\mu \frac{dw}{d\bar{r}}\bigg) + f_0(\bar{r} + r_A)\bigg]d\bar{r}$

$\hspace{3cm} 0 =\int\limits_0^{h_e} w_i\frac{d}{d\bar{r}}\bigg((\bar{r} + r_A)\mu \frac{dw}{d\bar{r}}\bigg)d\bar{r} + f_0 \int\limits_0^{h_e}w_i(\bar{r} + r_A)d\bar{r}$

(v) Weak formulation

Solving $1^{st}$ integral by parts, get

$\hspace{1cm} 0 = \bigg[w_i(\bar{r} + r_A)\mu \frac{dw}{d\bar{r}}\bigg]_0^{h_e}- \int\limits_0^{h_e}\frac{dw_i}{d\bar{r}}(\bar{r} + r_A)\mu \frac{dw}{d\bar{r}}d\bar{r}+ f_0 \int\limits_0^{h_e} w_i (\bar{r}+ r_A)d\bar{r}$

$\hspace{1cm} 0 = \big[w_i\big]_{h_e}\bigg[(\bar{r} + r_A)\mu \frac{dw}{d\bar{r}}\bigg]_{h_e}- \big[w_i\big]_0\bigg[(\bar{r}+r_A)\mu\frac{dw}{d\bar{r}}\bigg]_0$

$\hspace{4cm}-\int\limits_0^{h_e}\frac{dw_i}{d\bar{r}}(\bar{r} + r_A)\mu \frac{dw}{d\bar{r}}d\bar{r}+ f_0 \int\limits_0^{h_e} w_i (\bar{r}+ r_A)d\bar{r}\hspace{2cm}----(4.48)$

(vi) Approximate solution and lagrange's shape functions.

Take linear element

$\hspace{5cm} w = \sum\limits_{j=1}^{j=n}w_j \phi_j$

Since elemen , is linear N = 2

$\hspace{5cm}\therefore w = w_1 \phi_1 +w_2 \phi_2$

$\hspace{4.2cm} Where\hspace{0.2cm}\phi_1= 1-\frac{\bar{r}}{h_e} and\hspace{0.2cm} \phi_2 = \frac{\bar{r}}{h_e}$

$\hspace{4.9cm} \therefore \frac{d \phi_1}{d \bar{r}}=- \frac{1}{h_e}and \frac{d \phi_2}{d \bar{r}}=\frac{1}{h_e}$

(vii) For Rayleigh-Ritz method,$w_i= \phi_ i$

$\therefore$ Rewriting equation (4.48), we get

$\hspace{0.9cm}0 = \big[\phi _i\big]_{h_e}\bigg[(\bar{r} + r_A)\mu \frac{dw}{d \bar{r}}\bigg]_{h_e}- \big[\phi _i\big]_0\bigg[(\bar{r} + r_A)\mu \frac{dw}{d \bar{r}}\bigg]_o$

$\hspace{3.3cm}-\mu \int\limits_0^{h_e}(\bar{r}+r_A)\frac{d \phi_i}{d\bar{r}}\frac{dw_i}{d\bar{r}}d\bar{r}+f_0 \int\limits_0^{h_e}\phi _i (\bar{r}=r_A)d\bar{r}\hspace{2cm}...(4.49)$

for i = 1

$\hspace{0.7cm} \phi_i = \phi_1 = 1 -\frac{\bar{r}}{h_e}$

$\hspace{0.7cm} 0 = 0 -\bigg[-\frac{Q_1}{2 \pi}\bigg]- \mu\int\limits_0^{h_e}\bigg(-\frac{1}{h_e}\bigg)(\bar{r}+r_A)\frac{d}{d\bar{r}}\bigg[\sum \limits_{j=1}^2w_j\phi_j\bigg]d\bar{r}$

$\hspace{3.7cm}+f_0\int\limits_0^{h_e}\bigg(1-\frac{\bar{r}}{h_e}\bigg)(\bar{r} + r_A)d\bar{r}$

Rearranging the terms

Consider integral $I_1$

$\hspace{2.5cm}I_1 = - \frac{\mu}{h_e}\int\limits_0^{h_e}(\bar{r} +r_A) \sum\limits_{j =1}^2w_j\frac{d\phi_1}{d\bar{r}}d\bar{r}$

$\hspace{2.9cm} = - \frac{\mu}{h_e}\int\limits_0^{h_e}(\bar{r} +r_A) \bigg[w_1\frac{d \phi _1}{d \bar{r}}+w_2\frac{d \phi _2}{d \bar{r}}\bigg]d\bar{r}$

$\hspace{2.9cm} = - \frac{\mu}{h_e}\int\limits_0^{h_e}(\bar{r} +r_A) \bigg[-\frac{w_1}{h_e}+\frac{w_2}{h_e}\bigg]d\bar{r}$

$\hspace{2.9cm} = - \frac{\mu}{h_e} \bigg[\frac{w_1}{h_e}-\frac{w_2}{h_e}\bigg]\int\limits_0^{h_e}(\bar{r} +r_A) d\bar{r}$

$\hspace{2.9cm} = \frac{\mu}{h_e^2}(w_1-w_2)\bigg[\frac{(\bar{r}+r_A)^2}{2}\bigg]\bigg|_0^{h_e}$

$\hspace{2.9cm} = \frac{\mu}{2h_e^2}(w_1-w_2)\bigg[(h_e+r_A)_2-r_A^2\bigg]= \frac{\mu}{2h_e^2}(w_1-w_2)(h_e^2+2h_er_A)$

$\hspace{2.9cm} =\frac{\mu(2r_A+h_e)}{2h_e}.(w_1-w_2)$

Now consider integral $I_2$

$\hspace{2.5cm}I_2=f_0\int\limits_0^{h_e}\bigg(1-\frac{\bar{r}}{h_e}\bigg)(\bar{r}+r_A)d \bar{r}$

$\hspace{3cm} = f_0 \bigg[\frac{\bar{r}^2}{2}+r_A\bar{r}-\frac{\bar{r}^3}{3h_e}-\frac{r_A\bar{r}^2}{2h_e}\bigg]_0^{h_e}$

$\hspace{3cm} = f_0\bigg[\frac{h_e^2}{2}+r_Ah_e -\frac{h_0^2}{3}- \frac{r_Ah_e}{2}\bigg]=f_0\bigg[\frac{h_0^2}{6}+ \frac{r_A h_e}{2}\bigg]$

$\hspace{3cm} = \frac{f_0h_e}{6}(h_e + 3r_A)$

$\hspace{3cm} = \frac{Q_1}{2\pi}= \frac{\mu(2r_A+h_e)}{2h_e}.(w_1-w_2)-\frac{f_0h_e}{6}(h_e +3r_A)\hspace{2cm}...(4.50)$

For i = 2

$\hspace{0.8cm}\phi_i = \phi_2 = \frac{\bar{r}}{h_e}$

$\therefore 0= \frac{Q_2}{2\pi}-\mu \int\limits_0^{h_e}\frac{1}{h_e}(\bar{r}+r_A)\frac{d}{d\bar{r}}\bigg[\sum\limits_{j=1}^2 w_j \phi_j\bigg]d\bar{r}+f_0\int\limits_0^{h_e}\frac{\bar{r}}{h_e}(\bar{r}+r_A)d\bar{r}$

Rearranging the terms

Consider integral $I_1$

$\hspace{0.7cm} I_1=\frac{\mu}{h_e}\int\limits_0^{h_e}(\bar{r}+ r_A)\sum\limits_{j=1}^2 w_j \frac{d \phi_j}{d\bar{r}}d\bar{r}$

$\hspace{1.1cm} =\frac{\mu}{h_e}\int\limits_0^{h_e}(\bar{r}+ r_A)\bigg[w_1\frac{d \phi_i}{d\bar{r}}+w_2\frac{d \phi_2}{d\bar{r}}\bigg]d\bar{r}$

$\hspace{1.1cm}=\frac{\mu}{h_e}\int\limits_0^{h_e}(\bar{r}+r_A)\bigg[-\frac{w_1}{h_e}+\frac{w_2}{h_e}\bigg]d\bar{r}= \frac{\mu}{h_e^2}\int\limits_0^{h_e}(\bar{r}+r_A)(w_1+w_2)d\bar{r}$

$\hspace{1.1cm}= \frac{\mu}{h_e^2}(-w_1+w_2)\bigg[\frac{(\bar{r}+r_A)^2}{2}\bigg]\bigg|_0^{h_e}=\frac{\mu}{2h_e^2}(-w_1+w_2)\bigg[(h_e+r_A)^2-r_A^2\bigg]$

$\hspace{1.1cm}= \frac{\mu}{2h_e^2}(-w_1+w_2)(h_e^2+2h_er_A)$

$\hspace{1.1cm}= \frac{\mu}{2h_e}(h_e+2r_A)(-w_1+w_2)$

Now, consider integral $I_2$

$\hspace{0.8cm}I_2 =\frac{f_0}{h_e}\int\limits_0^{h_e}\bar{r}(\bar{r}+r_A)d\bar{r} =\frac{f_0}{h_e}\bigg[\frac{\bar{r}^3}{3}+\frac{r_A\bar{r}^2}{2}\bigg]_0^{h_e}$

$\hspace{1.2cm}=\frac{f_0}{h_e}\bigg[\frac{h_e^3}{3}+\frac{r_Ah_e^2}{2}\bigg] = \frac{f_0h_e}{6}(2h_e+3r_A)$

$\hspace{0.6cm}\therefore\frac{Q_2}{2\pi} =\frac{\mu(2r_A+h_e)}{2h_e}(-w_1+w_2)-\frac{f_0h_e}{6}(2h_e+3r_A)\hspace{2cm}...(4.51)$

(viii) To, get element matrix equation, put equation (4.50) and (4.51) in matrix from as

$\hspace{0.7cm}\frac{\mu(2r_A+h_e)}{2h_e}$ $\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix}$ $\begin{Bmatrix} \ w_1 \\ \ w_2 \\ \end{Bmatrix}$ $=\frac{f_0h_e}{6}$ $\begin{bmatrix} \ 3r_A & + & h_e \\ \ 3r_A & + & 2h_e \\ \end{bmatrix}$ $+\frac{1}{2\pi}$ $\begin{Bmatrix} \ Q_1 \\ \ Q_2 \\ \end{Bmatrix}$ $\hspace{2cm}...(4.52)$

(ix) Element matrix equations for each element.

Taking two linear elements of equal size.

For element number 1 : $r_A = 0 ; h_e = \frac{R_0}{2}$

$\hspace{2.8cm}\therefore \frac{2r_A+h_e}{2h_e} =\frac{\frac{R_0}{2}}{2\frac{R_0}{2}}=\frac{1}{2}$

$\hspace{3.3cm}3r_A+h_e = \frac{R_0}{2}\\ \hspace{3.3cm}3r_A+2h_e = R_0$

$\therefore$ Element matrix equation for element number 1 is

$\hspace{0.6cm}\frac{\mu}{2}$ $\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix}$ $\begin{Bmatrix} \ w_1^1 \\ \ w_2^2 \\ \end{Bmatrix}$ $\frac{f_0R_0}{12}$ $\begin{bmatrix} \ R_0/2 \\ \ R_0 \\ \end{bmatrix}$ $+\frac{1}{2\pi}$ $\begin{Bmatrix} \ Q_1 \\ \ Q_2 \\ \end{Bmatrix}$ $\hspace{2cm}...(4.53)$

For element number 2 : $r_A= h_e = \frac{R_0}{2}$

$\hspace{3cm}\therefore\frac{2r_A+h_e}{2h_e}=\frac{R_0\frac{R_0}{2}}{2\frac{R_0}{2}}=\frac{3}{2}\\ \hspace{3.5cm}3r_A+h_e = \frac{3R_0}{2}+\frac{R_0}{2}=2R_0\\ \hspace{3.5cm}3r_A+2h_e = \frac{3R_0}{2}+R_0=\frac{5R_0}{2}$

$\therefore$ Element matrix equation for element number 1 is

$\hspace{1.5cm}\frac{3\mu}{2}$ $\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix}$ $\begin{Bmatrix} \ w_1^2 \\ \ w^2_2 \\ \end{Bmatrix}$ $= \frac{f_0R_0}{12}$ $\begin{bmatrix} \ 2R_0 \\ \ \frac{5R_0}{2} \\ \end{bmatrix}$ $+\frac{1}{2\pi}$ $\begin{Bmatrix} \ Q_1^2 \\ \ Q^2_2 \\ \end{Bmatrix}$ $\hspace{2cm}...(4.54)$

(x) Global matrix equation : We can assemble the two elementary matrix equations giving the global matrix equation as

$\hspace{1.5cm}\frac{\mu}{2}$ $\begin{bmatrix} \ 1 & -1 & 0 \\ \ -1 & 4 & -3 \\ \ 0 & -3 & 3 \\ \end{bmatrix}$ $\begin{Bmatrix} \ w_1 \\ \ w_2 \\ \ w_3 \\ \end{Bmatrix}$ $= \frac{f_0R_0}{12}$ $\begin{bmatrix} \ \frac{R_0}{2} \\ \ 3R_0 \\ \ \frac{5R_0}{2} \\ \end{bmatrix}$ $+\frac{1}{2\pi}$ $\begin{Bmatrix} \ Q_1 \\ \ Q_2 \\ \ Q_3 \\ \end{Bmatrix}$

(xi) Impose global boundary conditions

$\hspace{0.6cm} r = 0 ; r\frac{dw}{dr}=0 \hspace{0.3cm} i. e. Q_1 = 0\\ \hspace{0.6cm}r=R_0 ; w = 0 \hspace{0.3cm} i.e. w_3 = 0 $

Applying condition for balance of secondary variables at intermediate nodes,

$\hspace{0.6cm}Q_2 = 0$

$\therefore$ The global matrix equation becomes

$\hspace{1.5cm}\frac{\mu}{2}$ $\begin{bmatrix} \ 1 & -1 & 0 \\ \ -1 & 4 & -3 \\ \ 0 & -3 & 3 \\ \end{bmatrix}$ $\begin{Bmatrix} \ w_1 \\ \ w_2 \\ \ 0 \\ \end{Bmatrix}$ $= \frac{f_0R_0^2}{24}$ $\begin{bmatrix} \ 1 \\ \ 6 \\ \ 5 \\ \end{bmatrix}$ $+\frac{1}{2\pi}$ $\begin{Bmatrix} \ 0 \\ \ 0 \\ \ Q_3 \\ \end{Bmatrix}$

(xi) Impose global boundary conditions

$\hspace{0.6cm} r = 0 ; r\frac{dw}{dr}=0 \hspace{0.3cm} i. e. Q_1 = 0$

$\hspace{0.6cm}r=R_0 ; w = 0 \hspace{0.3cm} i.e. w_3 = 0$

Applying condition for balance of secondary variables at intermediate nodes,

$\hspace{0.6cm}Q_2 = 0$

$\therefore$ The global matrix equation becomes

$\hspace{1.5cm}\frac{\mu}{2}$ $\begin{bmatrix} \ 1 & -1 & 0 \\ \ -1 & 4 & -3 \\ \ 0 & -3 & 3 \\ \end{bmatrix}$ $\begin{Bmatrix} \ w_1 \\ \ w_2 \\ \ 0 \\ \end{Bmatrix}$ $= \frac{f_0R_0^2}{24}$ $\begin{bmatrix} \ 1 \\ \ 6 \\ \ 5 \\ \end{bmatrix}$ $+\frac{1}{2\pi}$ $\begin{Bmatrix} \ 0 \\ \ 0 \\ \ Q_3 \\ \end{Bmatrix}$

(xii) Framing the equations, we get

$\hspace{5cm}w_1-w_2 = \frac{f_0R_0^2}{12\mu}\\ \hspace{5cm}-w_1+4w_2 = \frac{f_0R_0^2}{2\mu}\\ \hspace{5.8cm}\therefore w_2 = \frac{7f_0R_0^2}{36\mu} =0.194 \frac{f_0R_0^2}{\mu}\\ \hspace{6.2cm} w_2 = \frac{5f_0R_0^2}{18\mu} = 0.273 \frac{f_0R_0^2}{\mu}$

Preprocessing - Steps number (i) and (xi)

Processing$\hspace{0.6cm}$ -Steps number (x) and (xii)

Postprocessing - Tallying of answers with expected answers.

Governing differential equation is

$\hspace{4cm}-\frac{1}{r} \frac{d}{dr}\bigg(r \mu\frac{dw}{dr}\bigg) =f_0$

$\hspace{4cm} i.e. \frac{d}{dr}\bigg(r \mu\frac{dw}{dr}\bigg) = f_0r$

$\therefore$ Integrating, we get $\hspace{2cm}\mu r\frac{dw}{dr} = - f_0 \frac{r^2}{2}+c$

$\hspace{1.6cm}At r = 0, r\frac{dw}{dr} = 0\hspace{1.4cm} \therefore c = 0$

$\hspace{5cm}\therefore \mu r\frac{dw}{dr} = - \frac{f_0r^2}{2}$

$\hspace{5cm}\therefore \mu\hspace{0.2cm} dw = - \frac{f_0r}{2}dr $

Integrating again $\hspace{2.6cm} \mu\hspace{0.2cm}w= \frac{f_0r^2}{4}+c$

$\hspace{1.5cm}At\hspace{0.2cm}r = R_0, w = 0 \hspace{0.9cm}\therefore c = \frac{f_0}{4}R_0^2$

$\hspace{5.5cm}\therefore \mu\hspace{0.2cm}w = \frac{f_0}{4}(R_0^2-r^2)$

$\hspace{1.3cm}$ At node 1, r = 0,

$\hspace{5.5cm}\therefore w_1 = 0.25\frac{f_0}{\mu}R_0^2$

$\hspace{1.3cm}$ At node 2,$r = \frac{R_0}{2}$

$\hspace{5.5cm}\therefore w_2 = \frac{f_0}{4\mu}\bigg(R_0^2 -\frac{R^2_0}{4}\bigg)$

$\hspace{5.9cm}w_2 = \frac{3f_0R_0^2}{16\mu}=0.1875 \frac{f_0}{\mu}R_0^2$