Question

Consider one-dimensional transient heat conduction in a composite plane wall that consists of two layers \(A\) and \(B\) with perfect contact at the interface. The wall involves no heat generation and initially is at a specified temperature. The nodal network of the medium consists of nodes 0,1 (at the interface), and 2 with a uniform nodal spacing of \(\Delta x\). Using the energy balance approach, obtain the explicit finite difference formulation of this problem for the case of insulation at the left boundary (node 0 ) and radiation at the right boundary (node 2 ) with an emissivity of \(\varepsilon\) and surrounding temperature of \(T_{\text {surr }}\).

Short answer

Answer: The main steps in solving this problem include understanding the structure of the composite wall, applying the initial conditions, using the energy balance approach, and considering the boundary conditions at node 0 and 2. The explicit finite difference formulation is set up for nodes within the wall as well as nodes at the insulated and radiative boundaries.

Step-by-step solution

Understanding the problem

The composite wall consists of two layers A and B with perfect contact at the interface. There are nodes 0, 1, and 2 with a uniform nodal spacing \(\Delta x\). We need to calculate the temperature at each node using the energy balance approach considering the boundary conditions of insulation at node 0 and radiation at node 2.

Energy balance approach

The energy balance approach states that the rate of heat transfer to a node equals the rate of heat transfer out of that node. For the nodes (node 1) within the wall, we can use the following equation: \(\frac{\rho C_p \Delta x}{\Delta t} \left( T_{i,new}-T_{i,old} \right) = k_{A} \frac{T_{i-1,old}-T_{i,old}}{\Delta x} - k_{B} \frac{T_{i,old}-T_{i+1,old}}{\Delta x}\) Where \(T_{i,new}\) and \(T_{i,old}\) represent the new and old temperatures, respectively, \(\rho\) is the density, \(C_p\) is the specific heat, and \(k_A\) and \(k_B\) are the thermal conductivities of layers A and B, respectively.

Insulation boundary condition at node 0

Since the left boundary (node 0) has insulation, there is no heat flux through it. This condition implies: \(\frac{\partial T_0}{\partial x} = 0\)

Radiation boundary condition at node 2

The heat loss at the right boundary (node 2) happens by radiation. Therefore, we need to include radiation heat loss in the energy balance equation. The radiation heat loss can be calculated using the Stefan-Boltzmann law: \(q_{rad} = \varepsilon \sigma (T_2^4 - T_{surr}^4)\) So, the energy balance for node 2 becomes: \(\frac{\rho C_p \Delta x}{\Delta t} \left( T_{2,new}-T_{2,old} \right) = k_{B} \frac{T_{1,old}-T_{2,old}}{\Delta x} - \varepsilon \sigma (T_{2,old}^4 - T_{surr}^4)\)

Finite difference formulation

Using the energy balance approach and considering the boundary conditions at each node, we can set up the following explicit finite difference formulation: 1. For the node at the interface (node 1): \(T_{1,new}=\frac{k_{A} \Delta t}{\rho C_{p} \Delta x} \left(T_{0,old}-T_{1,old}\right)+\frac{k_{B} \Delta t}{\rho C_{p} \Delta x} \left(T_{2,old}-T_{1,old}\right)+ T_{1,old}\) 2. For the node at the right boundary (node 2): \(T_{2,new}=\frac{k_{B} \Delta t}{\rho C_{p} \Delta x}\left( T_{1,old}-T_{2,old}\right)+\frac{\varepsilon \sigma \Delta t}{\rho C_{p}}\left( T_{surr}^4-T_{2,old}^4\right)+T_{2,old}\) With these equations, we can calculate the new temperatures at each node for the given initial and boundary conditions.