Theory exercise 3
Feb 17, 2019
Problem 1: Heat transfer in conical cooling rib
Figure 1: Conical cooling rib.
Fig. 1 shows a conical cooling rib with radius \( R \) for \( X=L \). The stationary, one-dimensional heat transfer equation for the rib, written in terms of dimensionless variables, is given by: $$ \begin{align} \frac{d}{dx}\left(x^2\frac{d\theta}{dx}\right) -x\beta^2\theta= 0\,, \label{eq:coolingRibDimLess} \end{align} $$
or $$ \begin{align} x\frac{d^2\theta}{dx^2} + 2\frac{d\theta}{dx}-\beta^2\theta = 0\,, \label{eq:coolingRibDimLess2} \end{align} $$
where \( x=\frac{X}{L} \) is the dimensionless coordinate, based on the coordinate, \( X \) and length, \( L \) both of which have dimensions (see. Fig. 1 ). Furthermore, \( \theta = \frac{T - T_{\infty}}{T_L - T_{\infty}} \) is a dimensionless temperature with \( T_{\infty} \) the surrounding temperature, and \( T_L \) the temperature at \( X=L \). \( T_{\infty} \) and \( T_L \) are assumed to be constant. Finally, \( \beta^2=\frac{2\bar{h} \cdot L}{k \cdot R} \), where \( \bar{h} \) is the heat transfer coefficient and \( k \) is the heat conduction coefficient.
a) Discretize Eq. \eqref{eq:coolingRibDimLess2} with central differences. Set \( x_i \) = \( i \cdot h \), \( i = 0,1, \dots N + 1 \), where \( h=\frac{1}{N + 1} \). Write down the resulting equation system. Note that \( \theta_0 \) has to be found separately (see next question).
By introducing \( x_i \) = \( i \cdot h \) and discretizing Eq. \eqref{eq:coolingRibDimLess2} using central differences we obtain: $$ \begin{align} i\,h\frac{\theta_{i + 1} - 2\,\theta_i + \theta_{i - 1}}{h^2} + 2 \frac{\theta{i+1}-\theta{i-1}}{2h}- \beta\, \theta^2\ = 0,, \label{eq:coolingRibDimLess2_discretized} \end{align} $$ and by collecting terms we obtain: $$ \begin{align} \left(i-1\right)\theta_{i-1} - \left(2\,i+\,\beta^2\,h\right)\theta_i + \left(i+1\right)\theta_{i+1} = 0 \,. \label{eq:coolingRibDimLess2_discretized_collected} \end{align} $$
For \( i=1 \) and \( i = N \) we get: $$ \begin{align} - \left(2+\,\beta^2\,h\right)\theta_1 + 2\,\theta_{2} = 0 \,, \label{_auto1}\\ \left(N-1\right)\theta_{N-1} - \left(2\,N+\,\beta^2\,h\right)\theta_N = -\left(N+1\right)\theta_{N+1} = -\left(N+1\right) \,. \label{eq:coolingRibDimLess2_discretized_collected_1_N} \end{align} $$
b) \( \theta_0 \) may be found using Taylor expansions around \( x=0 \): $$ \begin{align} \theta \left(x\right) = \theta_0 + x \cdot \theta_0^\prime + \frac{x^2}{2} \theta_0^{\prime\prime}\,. \label{_auto2} \end{align} $$
Find \( \theta_0^\prime \) and \( \theta_0^{\prime\prime} \) by using Eq. \eqref{eq:coolingRibDimLess2} and show that: $$ \begin{align} \theta_0 = \frac{\theta_1}{1 + \frac{\beta^2}{2}h + \frac{\beta^4}{12}h^2} \label{eq:theta0} \end{align} $$
Evaluating the taylor expansion at \( x=h \), we get: $$ \begin{align} \theta \left(h\right) = \theta_1 = \theta_0 + h \cdot \theta_0^\prime + \frac{h^2}{2} \theta_0^{\prime\prime}\,. \label{eq:taylor} \end{align} $$
For simplicity rewrite Eq. \eqref{eq:coolingRibDimLess2} as: $$ \begin{align*} x\,\theta^{\prime\prime} + 2\,\theta^{\prime}-\beta^2\theta = 0\,, \end{align*} $$
Evaluating the equation at \( x=0 \) we may express \( \theta^{\prime}_0 \) in terms of \( \theta_0 \): $$ \begin{align*} 0\cdot\theta^{\prime\prime}_0 + 2\,\theta^{\prime}_0-\beta^2\theta_0 = 0\,, \quad \rightarrow \quad\theta^{\prime}_0 = \frac{\beta^2}{2}\theta_0\,. \end{align*} $$
Further, by differentiation of \eqref{eq:coolingRibDimLess2} we get: $$ \begin{align*} \left(x\,\theta^{\prime\prime}\right)^{\prime} + 2\,\theta^{\prime\prime}-\beta^2\theta^{\prime} = \left(x^{\prime}\,\theta^{\prime\prime} + x\,\theta^{\prime\prime\prime}\right) + 2\,\theta^{\prime\prime}-\beta^2\theta^{\prime} = x\,\theta^{\prime\prime\prime} + 3\,\theta^{\prime\prime}-\beta^2\theta^{\prime}=0\,, \end{align*} $$ Evaluating the equation at \( x=0 \) we may express \( \theta^{\prime\prime}_0 \) in terms of \( \theta_0 \): $$ \begin{align*} 0\cdot\theta^{\prime\prime\prime}_0 + 3\,\theta^{\prime\prime}_0-\beta^2\theta^{\prime}_0 = 0 \,, \quad \rightarrow \quad \theta^{\prime\prime}_0 = \frac{\beta^2}{3}\theta^{\prime}_0=\frac{\beta^4}{6}\theta_0\,. \end{align*} $$ Finally, by substituting \( \theta^{\prime}_0 \) and \( \theta^{\prime\prime}_0 \) into Eq. \eqref{eq:taylor} we get $$ \begin{align*} \theta_0 = \frac{\theta_1}{1 + \frac{\beta^2}{2}h + \frac{\beta^4}{12}h^2} \end{align*} $$
c) Set \( \beta^2 = 4 \) and \( h=0.25 \) and solve the system of algebraic equations in a). Find \( \theta_0 \) by using the formula found in b).
\( \theta_0 = 0.2077 \), \( \theta_1=0.3288 \), \( \theta_2=0.4932 \), \( \theta_3=0.7123 \).
With \( h = 0.25 \) we have that \( N=3 \) and the equation system may be written as: $$ \begin{equation} \left[\begin{matrix}-3 & 2 & 0 \\ 1 & -5 & 3 \\ 0 & 2 & -7\end{matrix}\right] \cdot \left[\begin{matrix} \theta_1 \\ \theta_2 \\ \theta_3 \end{matrix}\right] = \left[\begin{matrix}0\\ 0\\ -4\end{matrix}\right] \label{_auto3} \end{equation} $$ which gives \( \theta_1=0.3288 \), \( \theta_2=0.4932 \), \( \theta_3=0.7123 \). By substituting the solution for \( \theta_1 \) into \eqref{eq:theta0} we get \( \theta_0=0.3288/\left(1 + 1/2 + 1/12\right)=0.2077 \).
Problem 2: Linearization of non-linear ODE
In this problem we will look at the following ordinary differential equation (ODE): $$ \begin{align} \frac{d^2y}{dx^2} = -2\,\left(y\,\frac{dy}{dx}\right), \label{eq:ODE} \end{align} $$
with \( x \in [x_{min},x_{max}] \), and the following boundaries: $$ \begin{align} y\left(x_{\min}\right) = y\left(1\right) = 1 \qquad \text{and} \qquad y\left(x_{\max}\right) = y\left(2\right) = 0.5. \label{eq:BCs} \end{align} $$
a) Discretize Eq. \eqref{eq:ODE} with central differences for \( y^{\prime\prime} \) and \( y^\prime \), and linearize the non-linear term in the discretized equation using Newton linearization.
Show that the equation system may be written in the following tri-diagonal form: $$ \begin{align} \left( 1 - \Delta x \, \alpha_i \right) y_{i-1}^{m+1} + \left( \Delta x \, \beta_i -2 \right) y_i^{m+1} + \left( 1 + \Delta x \, \alpha_i\right) y_{i+1}^{m+1} = \Delta x \, \alpha_i \cdot \beta_i \,, \label{eq:tridiagonal} \end{align} $$
and write down the expressions for \( \alpha_i \) and \( \beta_i \).
Figure 2: Schematic of the discretized domain with boundaries.
Newton linearization is given by: $$ \begin{align} F\left(y_{i-1}, y_{i}, y_{i+1}\right)_{m+1} = F_m &+ \frac{\partial F_m}{\partial y_{i-1}^m}\;\delta y_{i-1} + \frac{\partial F_m}{\partial y_{i}^m}\;\delta y_{i} + \frac{\partial F_m}{\partial y_{i+1}^m}\;\delta y_{i+1}\,, \label{eq:newton} \end{align} $$ where \( F_m = F\left(y_{i-1}^m, y_{i}^m, y_{i+1}^m\right) \) is the non-linear term, and $$ \begin{align*} \delta y_{i-1} = y_{i-1}^{m+1} - y_{i-1}^{m}, \qquad \delta y_{i} = y_{i}^{m+1} - y_{i}^{m}, \qquad \delta y_{i+1} = y_{i+1}^{m+1} - y_{i+1}^{m}\,, \end{align*} $$ where \( m \) is an iteration index.
We use central differences such that \( y^{\prime\prime} \approx \frac{y_{i-1} - 2 \, y_i + y_{i+1}}{{\Delta x}^2} \) and \( y^{\prime} \approx \frac{y_{i+1}-y_{i-1}}{2 \, \Delta x} \), which by subtsitution into \eqref{eq:ODE} gives: $$ \begin{align} \frac{y_{i-1} - 2 \, y_i + y_{i+1}}{{\Delta x}^2} + 2 \, y_i \frac{y_{i+1}-y_{i-1}}{2 \, \Delta x} = 0 \label{_auto4}\\ \rightarrow y_{i-1} - 2 \, y_i + y_{i+1} + \Delta x \, y_i \left( y_{i+1}-y_{i-1} \right) = 0\,. \label{eq:discretized} \end{align} $$
The nonlinear term know becomes \( F_{m+1} = y_i^{m+1} \left( y_{i+1}^{m+1}-y_{i-1}^{m+1} \right) \), and by use of \eqref{eq:newton} we get: $$ \begin{align} F_{m+1} \approx y_i^{m} \left( y_{i+1}^{m}-y_{i-1}^{m} \right) - \left( y_{i-1}^{m+1}-y_{i-1}^m \right) y_{i}^m + \left( y_{i}^{m+1}-y_{i}^m\right) \left( y_{i+1}^{m}-y_{i-1}^{m}\right) + \left( y_{i+1}^{m+1}-y_{i+1}^m\right) y_{i}^m \nonumber \\ = - y_i^m \, y_{i-1}^{m+1} + \left( y_{i+1}^m - y_{i-1}^m \right) \, y_i^{m+1} + y_i^m \, y_{i+1}^{m+1} - y_i^m \left( y_{i+1}^m - y_{i-1}^m\right) \label{_auto5}\\ = -\alpha_i \, y_{i-1}^{m+1} + \beta_i \, y_i^{m+1} + \alpha_i \, y_{i+1}^{m+1} - \alpha_i \cdot \beta_i \nonumber \,, \end{align} $$ where we have used \( \alpha_i = y_i^m \) and \( \beta_i = y_{i+1}^m - y_{i-1}^m \), which in addition gives \( \alpha_i \cdot \beta_i = F_m \). By further substitution into \eqref{eq:discretized} we obtain: $$ \begin{align} \left( 1 - \Delta x \, \alpha_i \right) y_{i-1}^{m+1} + \left(\Delta x \, \beta_i -2 \right) y_i^{m+1} + \left( 1 + \Delta x \, \alpha_i\right) y_{i+1}^{m+1} = \Delta x \, \alpha_i \cdot \beta_i \label{_auto6} \end{align} $$
b) Write down Eq. \eqref{eq:tridiagonal} for the first and last unknown.
For \( i=1 \) we get: $$ \begin{align} \left(\Delta x \, \beta_1 -2 \right) y_1^{m+1} + \left( 1 + \Delta x \, \alpha_1\right) y_{2}^{m+1} = \Delta x \, \alpha_1 \cdot \beta_1 - \left( 1 - \Delta x \, \alpha_1 \right) y(x_{\min})\,, \label{_auto7} \end{align} $$
and for \( i=N-1 \): $$ \begin{align} \left( 1 - \Delta x \, \alpha_{N-1} \right) y_{N-2}^{m+1} + \left(\Delta x \, \beta_{N-1} -2 \right) y_{N-1}^{m+1} = \Delta x \, \alpha_{N-1} \cdot \beta_{N-1} - \left( 1 + \Delta x \, \alpha_{N-1}\right) y(x_{\max}) \label{_auto8} \end{align} $$
c) Describe the process of how you would solve Eqs. \eqref{eq:ODE} - \eqref{eq:BCs} using the suggested linearization.
Solution of Eqs. \eqref{eq:ODE} - \eqref{eq:BCs} using linearization may be obtained by performing the following steps
- Discretize ODE
- Linearize discrete version of ODE
- Provide an initial guess for the solution of \( y(x) \)
- Solve (the resulting algebraic system of equations) iteratively until convergence