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).

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: $$ \begin{align} \theta_0 = \frac{\theta_1}{1 + \frac{\beta^2}{2}h + \frac{\beta^4}{12}h^2} \label{eq:theta0} \end{align} $$

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).

Hint.

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 \( \beta_i \).

b) Write down Eq. \eqref{eq:tridiagonal} for the first and last unknown.

c) Describe the process of how you would solve Eqs. \eqref{eq:ODE} - \eqref{eq:BCs} using the suggested linearization.