Question

Consider a large plane wall of thickness \(L=0.3 \mathrm{~m}\), thermal conductivity \(k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=24 \mathrm{~m}^{2}\). The left side of the wall is subjected to a heat flux of \(\dot{q}_{0}=350 \mathrm{~W} / \mathrm{m}^{2}\), while the temperature at that surface is measured to be \(T_{0}=60^{\circ} \mathrm{C}\). Assuming steady one-dimensional heat transfer and taking the nodal spacing to be $6 \mathrm{~cm},(a)$ obtain the finite difference formulation for the six nodes and \((b)\) determine the temperature of the other surface of the wall by solving those equations.

Short answer

Based on the given information, determine the temperature of the other surface of the wall: The problem focuses on steady-state heat conduction and finite difference methods. Given the wall thickness (L = 0.3 m), nodal spacing (Δx = 0.06 m), and thermal conductivity, we found there are 6 nodes. Following the steps of creating energy conservation equations, setting up boundary conditions, and solving the system of equations, we finally found the temperature of the other surface (Node 6) is equal to the temperature of Node 5: T6 = T5

Step-by-step solution

Determine the number of nodes and nodal spacing

We are given that the nodal spacing is 6 cm, which means that each segment of the wall is divided into 0.06 m increments. Calculate the number of nodes across the wall's thickness. Nodes: \(N = \frac{L}{\Delta x} + 1\) where \(L = 0.3 \mathrm{~m}\) is the wall thickness and \(\Delta x = 0.06 \mathrm{~m}\) is the nodal spacing. \(N = \frac{0.3}{0.06} + 1 = 6\) nodes.

Apply energy conservation

Assuming steady-state conditions and one-dimensional heat conduction, let's apply energy conservation: \(q_{in} - q_{out} = 0\) Consider the generic node \(i\). The energy balance equation can be expressed in terms of temperatures as: $$kA\frac{T_{i-1} - 2T_i + T_{i+1}}{(\Delta x)^2} = 0$$

Create the discrete equations for the six nodes

Now, create the discrete equations for each node. The first node has a known heat flux and temperature, which can be input as boundary conditions: Node 1: \(\qquad \dot{q}_{0} = -kA\frac{T_{1} - T_{0}}{\Delta x} \Longrightarrow T_{1} = T_{0} - \frac{\dot{q}_{0} \Delta x}{kA}\) Node 2 to Node 5: Use the generic energy balance equation from Step 2: Node 2: \(\qquad kA \frac{T_{1} - 2T_{2} + T_{3}}{(\Delta x)^2} = 0\) Node 3: \(\qquad kA \frac{T_{2} - 2T_{3} + T_{4}}{(\Delta x)^2} = 0\) Node 4: \(\qquad kA \frac{T_{3} - 2T_{4} + T_{5}}{(\Delta x)^2} = 0\) Node 5: \(\qquad kA \frac{T_{4} - 2T_{5} + T_{6}}{(\Delta x)^2} = 0\) Node 6 (other surface): We don't have any explicit information about this node, so we have to make a reasonable assumption. Since no additional heat flux is specified for this node, we assume that the rate of change of temperature is zero (adiabatic boundary condition): \(\qquad \frac{T_{5} - T_{6}}{\Delta x} = 0 \Longrightarrow T_{6} = T_{5}\)

Solve the system of equations

Now, we have 5 equations and 5 unknowns (\(T_{1}\) to \(T_{5}\)). Solve this system of linear equations to find the unknown temperatures. We can use numerical methods such as the Gauss-Seidel method or the Thomas algorithm, or by manually solving the equations in a step-by-step fashion, obtaining the unknown temperatures.

Determine the temperature of the other surface of the wall (Node 6)

Once we have solved the system of equations from Step 4, we will have the temperature values for each node. Since the temperature at Node 6 is equal to that of Node 5 (from our assumption in Step 3), we can now report the temperature of the other surface of the wall: \(T_6 = T_5\) By solving the equations, we obtain the temperature of the other surface of the wall, satisfying part (b) of the exercise.