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

Marks: 5M

Year: may 2016

$\theta(L) = 470^0 \hspace{3cm}\theta(L) =170^0c$

$\hspace{0.8cm}=(T_L-T_a)\hspace{3.cm} = (T_L-T_a)$

Governing d .e is,

$KA\frac{d^2\theta}{dx^2}-hp\theta = 0 \hspace{0.9cm} where , \theta = T_x-T_a$

$ \hspace{0.6cm}\frac{d^2\theta}{dx^2}-m^2\theta = 0 \hspace{0.9cm} where ,m^2 = \frac{hp}{KA}$

$Q_1^e\hspace{1.6cm}Q_2^e$

$\hspace{0.6cm}Q_1^e =-KA\frac{d\theta}{dx}$

$\hspace{0.6cm}Q_2^e =KA\frac{d\theta}{dx}$

Residue ,$ R = \frac{d^2\theta}{dx^2}-m^2\theta$

weighted integral form ,

$\int\limits_0^{he} w_i \hspace{0.1cm}R \hspace{0.1cm}dx = 0$

$\int\limits_0^{he}w_i \bigg(\frac{d^2\theta}{dx^2}-m^2 \theta\bigg)dx = 0 $

$\int\limits_0^{he}w_i \frac{d^2\theta}{dx^2}dx - \int\limits_0^{he} w_i M^2\theta dx = 0$

$\bigg[w_i\frac{d\theta}{dx}\bigg]_0^{he}- \int\limits_0^{he}\frac{dw_i}{dx}\frac{d \theta}{dx}dx-\int\limits_0^{he} w_i m^2 \theta dx$

$w_i \frac{d\theta}{dx}\bigg|_{x=he} - w_i \frac{d \theta}{dx}\bigg|_{x=0}- \int\limits_0^{he}\frac{dw_i}{dx}\frac{d\theta}{dx}dx -\int\limits_0^{he}w_i m^2 \theta dx \hspace{2cm}---(1)$

$\frac{d \theta}{dx}\bigg|_{x=0}= - \frac{Q_1}{KA}\hspace{0.6cm} $ & $\hspace{0.6cm} \frac{d \theta}{dx}\bigg|_{x=he}= \frac{Q_2}{KA}$

$\hspace{0.9cm} \theta= \theta_1 \phi_1+ \theta_2 \phi_2 \hspace{0.9cm} (\text{As element is linear})$

$\hspace{0.9cm} \therefore = w_1 = \phi_1 =1 -\frac{x}{he}\hspace{0.6cm} $ & $\hspace{0.6cm} w_2 = \phi_2 = \frac{x}{he}$

$\hspace{0.9cm} \frac{dw_1}{dx}= \frac{d \theta_1}{dx}=-\frac{1}{he}\hspace{0.6cm} $ & $\hspace{0.6cm} \frac{dw_2}{dx}= \frac{d \theta_2}{dx}=\frac{1}{he}$

$\frac{d\theta}{dx}=\theta ,(-\frac{1}{he})+\theta_2(\frac{1}{he})$

for i = 1 Eq' (1) becomes,

$0- \bigg[\frac{-Q_1}{KA}\bigg]- \int\limits_0^{he}\bigg(-\frac{1}{he}\bigg)\bigg(-\frac{1}{he}\theta_1 \frac{1}{he}\theta_2\bigg)dx-m^2 \int\limits_0^{he}\bigg(1-\frac{x}{he}\bigg)\bigg(\bigg(1-\frac{x}{he}\bigg)\theta_1+\frac{x}{he}\theta_2\bigg)dx$

$\frac{Q_1}{KA}=-\frac{1}{he}\bigg[-\frac{\theta_1}{he}x+\frac{\theta_2}{he}x\bigg]_0^{he}+m^2\int\limits_0^{he}\bigg(1-\frac{x}{he}\bigg)^2\theta_1+\bigg(\frac{x}{he}-\frac{x^2}{h^2e}\bigg)\theta_2dx$

$\hspace{0.8cm}=-\frac{1}{he}[-\theta_1+\theta_2]+m^2\bigg[\frac{-\theta_1}{3}\bigg(1-\frac{x}{he})^3he+\theta_2\bigg(\frac{x^2}{2he}-\frac{x^3}{3he^2}\bigg)\bigg]_0^{he}$

$\hspace{0.8cm}=\frac{1}{he}[-\theta_1-\theta_2]+m^2\bigg(\theta_1\frac{he}{3}+\theta_2\frac{he}{6}\bigg)$

$\hspace{0.8cm}\frac{KA}{he}[\theta_1-\theta_2]+\frac{KA.m^2he}{6}[2\theta_1+\theta_2]+Q_1\hspace{2cm}---(2)$

for i = 2 : Eq' (1) becomes ,

$\frac{Q_2}{KA}- \int\limits_0^{he}-\frac{1}{he}\bigg(-\frac{1}{he}\theta_1 \frac{1}{he}\theta_2\bigg)dx-m^2 \int\limits_0^{he}\frac{x}{he}\bigg(1-\frac{x}{he}\bigg)\bigg(\bigg(1-\frac{x}{he}\bigg)\theta_1+\frac{x}{he}\theta_2\bigg)dx$

$\frac{Q_2}{KA}=-\frac{1}{he}\bigg[-\frac{\theta_1}{he}x+\frac{\theta_2}{he}x\bigg]_0^{he}+m^2\int\limits_0^{he}\bigg(\frac{x}{he}-\frac{x^2}{he^2}\bigg)\theta_1+\frac{x^2}{he^2}\theta_2 dx$

$\hspace{0.8cm}=-\frac{1}{he}[-\theta_1+\theta_2]+m^2\bigg[ \theta_1\bigg(1-\frac{x^2}{2he}\frac{x^3}{3he^2}\bigg)+\frac{-\theta_1}{3}\bigg(1-\frac{x}{he})\bigg)^3he\bigg ]_0^{he}$

$\hspace{0.8cm}=\frac{1}{he}[-\theta_1+\theta_2]+m^2\bigg(\theta_1\frac{he}{6}+\theta_2\frac{he}{3}\bigg)$

$\hspace{0.3cm}\frac{KA}{he}[-\theta_1+\theta_2]+\frac{KA.m^2he}{6}[\theta_1+2\theta_2]+Q_2\hspace{2cm}---(3)$

Put Eq' (2) & (3) in matrix from ,

$\bigg\{\frac{KA}{he}$ $\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix}+\frac{KAm^2he}{6}$ $\begin{bmatrix} \ 2 & 1 \\ \ 1 & 2 \\ \end{bmatrix}\bigg\}$ $\begin{Bmatrix} \ Q_1 \\ \ Q_2 \\ \end{Bmatrix}$= $\begin{Bmatrix} \ Q_1 \\ \ Q_2 \\ \end{Bmatrix}$

Taking 2 linear elements of equal length,

$h_e= \frac{L}{2} = \frac{0.06}{2} = 0.03m$

$A= \frac{\pi}{4}.d^2 = \frac{\pi}{4}\times0.02^2=0.000314 m^2$

$K=100 w/m^0c$

$h = 25 w/m^2 \hspace{0.2cm}{}^0C$ $m^2 = \frac{hp}{KA}= \frac{h \times \pi \times D \times 4}{K \times \pi \times D^2}= \frac{4h}{KD}$

$\hspace{0.6cm}= \frac{4 \times 25}{100 \times 0.02 } = 50$

$\therefore \frac{KA}{he} = \frac{100\times0.000314}{0.03} = 1.046$

$\frac{KAm^2he}{6}= \frac{100 \times 0.000314 \times 50 \times 0.03}{6}$

$\hspace{1.4cm}=0.00785$

Element matrix Eq' is,

$\bigg\{1.46$ $\begin{bmatrix} \ 1 & -1 \\ \ -1 & 1 \\ \end{bmatrix}+0.00785$ $\begin{bmatrix} \ 2 & 1 \\ \ 1 & 2 \\ \end{bmatrix}\bigg\}$ $\begin{Bmatrix} \ Q_1 \\ \ Q_2 \\ \end{Bmatrix}$= $\begin{Bmatrix} \ Q_1 \\ \ Q_2 \\ \end{Bmatrix}$

$\begin{bmatrix} \ 1.0617 & -1.03815 \\ \ -1.03815 & 1.0617 \\ \end{bmatrix}$ $\begin{Bmatrix} \ Q_1 \\ \ Q_2 \\ \end{Bmatrix}$= $\begin{Bmatrix} \ Q_1 \\ \ Q_2 \\ \end{Bmatrix}$

$\begin{bmatrix} \ 1.0617 & -1.03815 & 0 \\ \ -1.03815 & 2.1234 & -1.03815 \\ \ 0 & -1.03815 & 1.0617 \\ \end{bmatrix}$ $\begin{Bmatrix} \ Q_1 \\ \ Q_2 \\ \ Q_3 \\ \end{Bmatrix}$= $\begin{Bmatrix} \ Q_1 \\ \ Q_2 \\ \ Q_3 \\ \end{Bmatrix}$

B.cs

$\theta_1= \theta(0) = 470^0C$

$\theta_4 = \theta(L) = 170 ^0C \hspace{2cm}$& $Q_2 =0$

$\begin{bmatrix} \ 1.0617 & -1.03815 & 0 \\ \ -1.03815 & 2.1234 & -1.03815 \\ \ 0 & -1.03815 & 1.0617 \\ \end{bmatrix}$ $\begin{Bmatrix} \ 470 \\ \ Q_2 \\ \ 170 \\ \end{Bmatrix}$= $\begin{Bmatrix} \ Q_1 \\ \ 0 \\ \ Q_3 \\ \end{Bmatrix}$

$\hspace{4cm}\therefore \theta _2 = 312.91^0C \hspace{0.6cm} at x=3 cms$

$\hspace{2cm}T_2 =312.91 +91+30= 340.90^0C $

Rate of haet input $Q_1=174.15 watts$

Exact sol' :

$KA\frac{d^2\theta}{dx^2}- hp \theta = 0$

$KA = 100 \times \frac{\pi}{4}\times 0.02^2 = 0.0314$

$hp = 25 \times \pi \times D = 25 \times \pi \times 0.02 = 1.57$

$\therefore$ diff eq' becomes ,

$\hspace{0.6cm} 0.0314 \frac{d^2\theta}{dx^2}-1.57\theta = 0 $

$\hspace{0.2cm} i. e \frac{d^2\theta}{dx^2} - 50.02\theta = 0 $

$(D^2-50.02)\theta = 0 $

A. E : $D^2-50.02 = 0 $

$\hspace{1.5cm}D = I7.07$

$\therefore cf = c_1 e^{7.07x}+c_2 e^{-7.07x} $

$PI=0$ $\therefore \theta=c_1 e^{7.07x}+c_2 e ^{-7.07x}$

At $x = 0' \theta = 470 {}^0C$

$\hspace{0.6cm}470 = c_1 +c_2\hspace{2cm} ---(4) $

At $x=0.06, \theta = 170 ^0C$

$\hspace{0.6cm}170 = 1.528C_1 +0.6543C_2\hspace{2cm} ---(5) $

from (4) & (5) =$\big\gt C_1 = -157.4$ and $C_2 = 627.4$

$\hspace{1.6cm} \therefore \theta = -157.4\hspace{0.2cm} e^{7.07x}+627.4\hspace{0.2cm} e^{-7.07x}$

At$x=0.03 \hspace{0.2cm}\theta = 312.90{}^0C\hspace{0.6cm}\therefore T_2=312.90+30=342.90{}^0C$