Question

Consider a nuclear fuel element \((k=57 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that can be modeled as a plane wall with thickness of \(4 \mathrm{~cm}\). The fuel element generates \(3 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\) of heat uniformly. Both side surfaces of the fuel element are cooled by liquid with temperature of \(80^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(8000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Using a uniform nodal spacing of \(8 \mathrm{~mm},(a)\) obtain the finite difference equations, \((b)\) determine the nodal temperatures by solving those equations, and (c) compare the surface temperatures of both sides of the fuel element with analytical solution.

Short answer

#Answer: (a) The finite difference equations for the given system are: At node 1: $\frac{k}{\Delta x^2}(T_2 - T_1) + q'' - \frac{h \Delta x}{2}(T_1 - T_\infty) = 0$ At nodes 2 - 5: $\frac{k}{\Delta x^2}(T_{i+1} - 2T_i + T_{i-1}) + q'' = 0$ At node 6: $\frac{k}{\Delta x^2}(T_6 - T_5) + q'' + \frac{h \Delta x}{2}(T_6 - T_\infty) = 0$ (b) The nodal temperatures, $T_1, T_2, T_3, T_4, T_5,$ and $T_6$, can be obtained by solving the system of linear equations detailed in the solution. (c) To compare the surface temperatures with the analytical solution, we need to find the temperatures $T_0$ and $T_L$ from the conduction-convection equation for a plane wall with volumetric heat generation. Then, compare these values with the obtained nodal temperatures, $T_1$ and $T_6$. The difference between these values shows the accuracy of our numerical solution.

Step-by-step solution

Problem Setup

First, we need to assign some variables for the given input parameters and constants and also divide the plane wall into equal segments of the specified nodal spacing. Let the nodal spacing be \(\Delta x\), where \(\Delta x = 8 \times 10^{-3} \, \text{m}\). Let the thickness of the wall be \(L = 4 \times 10^{-2} \, \text{m}\). Based on the thickness and the nodal spacing, we can find the number of nodes, n: $$n = \frac{L}{\Delta x} + 1 = \frac{4 \times 10^{-2}}{8 \times 10^{-3}} + 1 = 6$$ Now, let's assign some variables for the given data: $$k = 57 \, \mathrm{W/m \cdot K}$$ $$q'' = 3 \times 10^7 \, \mathrm{W/m^3}$$ $$T_\infty=80^{\circ} \mathrm{C}$$ $$h = 8000 \, \mathrm{W/m^2 \cdot K}$$

a) Obtain Finite Difference Equations

Using the energy balance in a one-dimensional cartesian coordinate system, we can write the discretized form of the heat conduction equation for each node i as: $$\frac{k}{\Delta x^2}(T_{i+1} - 2T_i + T_{i-1}) + q'' = 0$$ We can rewrite the equation for all nodes, including the boundary conditions at nodes 1 and 6. At node 1, we have to consider the convection from the liquid: $$\frac{k}{\Delta x^2}(T_2 - 2T_1 + T_1) + q'' - \frac{h \Delta x}{2}(T_1 - T_\infty) = 0$$ At node 6, we have to consider the convection from the liquid: $$\frac{k}{\Delta x^2}(T_6 - 2T_5 + T_4) + q'' + \frac{h \Delta x}{2}(T_6 - T_\infty) = 0$$

b) Determine Nodal Temperatures by Solving Equations

We can now set up a system of linear equations for the nodal temperatures \(T_1, T_2, T_3, T_4, T_5,\) and \(T_6\). We have six equations: 1. \(\frac{k}{\Delta x^2}(T_2 - T_1) + q'' - \frac{h \Delta x}{2}(T_1 - T_\infty) = 0\) 2. \(\frac{k}{\Delta x^2}(T_3 - 2T_2 + T_1) + q'' = 0\) 3. \(\frac{k}{\Delta x^2}(T_4 - 2T_3 + T_2) + q'' = 0\) 4. \(\frac{k}{\Delta x^2}(T_5 - 2T_4 + T_3) + q'' = 0\) 5. \(\frac{k}{\Delta x^2}(T_6 - 2T_5 + T_4) + q'' = 0\) 6. \(\frac{k}{\Delta x^2}(T_6 - T_5) + q'' + \frac{h \Delta x}{2}(T_6 - T_\infty) = 0\) By solving these equations (e.g, using matrix inversion, Gaussian elimination, or any suitable numerical solver), we can obtain the nodal temperatures \(T_1, T_2, T_3, T_4, T_5,\) and \(T_6\).

c) Compare Surface Temperatures with Analytical Solution

To find the analytical solution for this problem, we can use the conduction-convection equation for a plane wall with volumetric heat generation: $$k \frac{dT}{dx} = h \cdot (T - T_\infty) + \frac{q'' x}{2}$$ Solving for the temperature distribution in both sides, we can find the surface temperatures at \(x = 0\) and \(x = L\), let's call them \(T_0\) and \(T_L\). Then, we can compare these values with the obtained nodal temperatures at \(T_1\) and \(T_6\), which are the numerical surface temperatures. By comparing these values, we can determine the accuracy of our numerical solution.

Key concepts

Finite Difference Method

The Finite Difference Method is a numerical technique used to approximate solutions to differential equations, which are often encountered in engineering problems like heat transfer. This method involves breaking down a domain, such as a plane wall, into smaller discrete points called nodes. In this exercise, the wall is divided into segments with a nodal spacing of \(8\text{ mm}\).

By applying the finite difference method, we can write a discrete version of the continuous heat conduction equation. For any node \(i\), the equation becomes:

• \(\frac{k}{\Delta x^2}(T_{i+1} - 2T_i + T_{i-1}) + q'' = 0\)

This equation allows us to determine temperatures at each node using known values from neighboring nodes. The boundary nodes, however, require special handling to incorporate boundary conditions like convection, making them slightly different, as shown in the original solution. This approximation provides a practical way to find temperature distributions without a complete analytical solution.

Convection Heat Transfer

Convection heat transfer occurs when a fluid, such as a liquid or gas, carries heat away from a surface. In this problem, the nuclear fuel element's sides are cooled by a liquid at \(80^{\circ} \text{C}\) with a convection coefficient of \(8000 \text{ W/m}^2 \cdot \text{K}\). This coefficient reflects the efficiency of heat removal by convection.

To incorporate convection into our finite difference equations, the terms at the boundary nodes are adjusted to include the heat exchange with the surrounding liquid. This is crucial for accurately capturing how the solid material loses heat to the convection process. At the edges, the convection effect is added or subtracted from the nodal energy balance, ensuring that the model represents reality effectively.

Nodal Analysis

Nodal analysis is a technique used to simplify complex systems into a manageable set of equations. In the context of this heat transfer problem, it involves calculating the temperature at several discrete points, known as nodes, across the plane wall. Each node represents a specific position in the material where temperatures are computed.

The analysis typically sets up a system of linear equations correlating the nodes. Here, nodes are labeled \(T_1, T_2, T_3, ..., T_6\) across the wall. By solving these equations, either manually or with computational tools, you determine the temperature distribution within the material. This method is valuable for exploring how heat moves through complex geometries.

Thermal Conduction

Thermal conduction is the process by which heat energy is transferred within a material due to temperature differences. In this exercise, the nuclear fuel element wall has a thermal conductivity \(k = 57 \text{ W/m} \cdot \text{K}\), highlighting how easily heat moves through it. The heat conduction equation, \(\frac{k}{\Delta x^2}(T_{i+1} - 2T_i + T_{i-1})\), allows us to model this transfer numerically.

The material's ability to conduct heat directly affects how quickly thermal energy spreads throughout the object, and thus influences the nodal temperatures calculated using the finite difference approach. Incorporating conduction is essential for understanding the thermal behavior of nuclear materials, where precise temperature management is critical.

Volumetric Heat Generation

Volumetric heat generation refers to the production of heat within a material, as opposed to heat entering from its surfaces. In the case of the nuclear fuel element, this internal heat generation is uniform and is quantified as \(3 \times 10^7 \text{ W/m}^3\). This value represents the energy contributed by nuclear reactions within the fuel.

In our finite difference model, this volumetric heat source term \(q''\) is included in the equations for each node, ensuring that the internal heating of the material is considered when calculating temperature distribution. Understanding how this generated heat impacts the equilibrium within the material is crucial for designing effective cooling systems and ensuring safety in nuclear applications.