Question

Consider steady one-dimensional heat conduction in a plane wall with variable heat generation and variable thermal conductivity. The nodal network of the medium consists of nodes 0,1 , and 2 with a uniform nodal spacing of $\Delta x$. Using the energy balance approach, obtain the finite difference formulation of this problem for the case of specified heat flux \(\dot{q}_{0}\) to the wall and convection at the left boundary (node 0 ) with a convection coefficient of \(h\) and ambient temperature of \(T_{\infty^{+}}\) and radiation at the right boundary (node 2 ) with an emissivity of \(\varepsilon\) and surrounding surface temperature of \(T_{\text {surr }}\)

Short answer

i. For node 1: $$k_{1} \frac{T_{2} - T_{1}}{\Delta x} - k_{0} \frac{T_{1} - T_{0}}{\Delta x} - q_{1} \Delta x = 0$$ ii. For node 0: $$-k_{0} \frac{T_{1} - T_{0}}{\Delta x} = \dot{q}_{0} - h \left(T_{0}-T_{\infty^{+}} \right)$$ iii. For node 2: $$k_{1} \frac{T_{2} - T_{1}}{\Delta x} = \varepsilon \sigma \left(T_{2}^{4} - T_{\text{surr}}^{4} \right)$$

Step-by-step solution

General expression for energy conservation per unit volume

Considering the variable thermal conductivity \(k(x)\) and variable heat generation rate \(q(x)\), we can write the energy conservation equation on a per-unit volume basis as: $$\frac{\mathrm{d}}{\mathrm{dx}} \left( k(x) \frac{\mathrm{dT}}{\mathrm{dx}} \right) - q(x) = 0$$

Discretized form of energy balance equation

To obtain the finite difference formulation, we will discretize the conservation equation using the central difference approach. The discrete form of the energy balance around node 1 can be written as: $$k_{1} \frac{T_{2} - T_{1}}{\Delta x} - k_{0} \frac{T_{1} - T_{0}}{\Delta x} - q_{1} \Delta x = 0$$, where \(k_{1} = k\left(\frac{3}{2} \Delta x\right)\) and \(k_{0} = k\left(\frac{1}{2}\Delta x\right)\) are the thermal conductivities at the midpoint between nodes 1 and 2 and nodes 0 and 1, respectively, and \(q_{1}=q(\Delta x)\) is the heat generation rate at node 1.

Set up boundary conditions

We have three boundary conditions given in the problem: specified heat flux \(q_0\) at node 0, convection at node 0, and radiation at node 2. For convection (node 0), we can write: $$\dot{q}_{0} = h \left(T_{0} - T_{\infty^{+}}\right)$$ For radiation (node 2), we can write: $$q_{\text{rad,2}}=\varepsilon \sigma \left( T_{2}^{4} - T_{\text{surr}}^{4} \right)$$

Finite difference formulation for each node

For node 1, we can express the finite difference formulation from step 2 as: $$k_{1} \frac{T_{2} - T_{1}}{\Delta x} - k_{0} \frac{T_{1} - T_{0}}{\Delta x} - q_{1} \Delta x = 0$$ For node 0, we can express the finite difference formulation by replacing the convection boundary condition: $$-k_{0} \frac{T_{1} - T_{0}}{\Delta x} = \dot{q}_{0} - h \left( T_{0}-T_{\infty^{+}} \right)$$ For node 2, we can express the finite difference formulation by replacing the radiation boundary condition: $$k_{1} \frac{T_{2} - T_{1}}{\Delta x} = \varepsilon \sigma \left( T_{2}^{4} - T_{\text{surr}}^{4} \right)$$

Assemble the final set of equations to be solved

Finally, we have a system of three nonlinear equations for the three unknowns (\(T_0, T_1,\) and \(T_2\)): i. For node 1: $$k_{1} \frac{T_{2} - T_{1}}{\Delta x} - k_{0} \frac{T_{1} - T_{0}}{\Delta x} - q_{1} \Delta x = 0$$ ii. For node 0: $$-k_{0} \frac{T_{1} - T_{0}}{\Delta x} = \dot{q}_{0} - h \left(T_{0}-T_{\infty^{+}} \right)$$ iii. For node 2: $$k_{1} \frac{T_{2} - T_{1}}{\Delta x} = \varepsilon \sigma \left(T_{2}^{4} - T_{\text{surr}}^{4} \right)$$ These nonlinear equations can be further manipulated and solved numerically using the Newton-Raphson iterative method or another suitable method to obtain the temperatures at each node.