Consider a large uranium plate of thickness \(5 \mathrm{~cm}\) and thermal conductivity \(k=28 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) in which heat is generated uniformly at a constant rate of \(\dot{e}=6 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\). One side of the plate is insulated while the other side is subjected to convection to an environment at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Considering six equally spaced nodes with a nodal spacing of \(1 \mathrm{~cm},(a)\) obtain the finite difference formulation of this problem and \((b)\) determine the nodal temperatures under steady conditions by solving those equations.

Step by step solution

01

Derive the 1D Heat Conduction Equation

Start by writing the 1-D heat conduction equation for the system, which is given by: \(\frac{d}{dx}\left(k\frac{dT}{dx}\right)+\dot{e}=0\) Here, \(T\) is the temperature, \(k\) is the thermal conductivity and \(\dot{e}\) is the heat generation rate per unit volume.

02

Discretization of the 1D Heat Conduction Equation

Discretize the 1-D heat conduction equation using finite difference methods. Divide the 5 cm thickness into six equally spaced nodes with a nodal spacing of 1 cm. For the case of heat generation, the finite difference equation can be derived as: \(k\frac{T_{i+1}-2T_i+T_{i-1}}{\Delta x^2}+\dot{e}=0\), for i = 2, 3, 4 and 5. We have \(k = 28 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\dot{e} = 6 \times 10^5 \mathrm{~W} / \mathrm{m}^3\), and \(\Delta x = 1 \mathrm{~cm}= 0.01 \mathrm{~m}\).

03

Apply Boundary Conditions

Now, apply the boundary conditions for the given problem. At node 1, the side of the plate is insulated, which implies that no heat flow takes place at this node: \(\frac{dT}{dx}_{x=1} = 0\), or equivalently, \(T_1 = T_2\). At node 6, the other side of the plate is subjected to convection: \(h(T_6 - T_{\infty}) = -k\frac{dT}{dx}_{x=6}\) This can be discretized as: \(h(T_6 - T_{\infty}) = -k\frac{T_6 - T_5}{\Delta x}\). Here, \(h = 60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and \(T_{\infty} = 30^{\circ} \mathrm{C}\).

04

Solve the System of Linear Equations

Next, construct a system of linear equations for nodal temperatures using the discretized heat conduction equation (from Step 2) and the boundary conditions (from Step 3). This results in the following set of equations: \(T_1 - T_2 = 0\) \(28(T_2 - 2T_1 + T_3) + 6\times10^5(0.01)^2 = 0\) \(28(T_3 - 2T_2 + T_1) + 6\times10^5(0.01)^2 = 0\) \(28(T_4 - 2T_3 + T_2) + 6\times10^5(0.01)^2 = 0\) \(28(T_5 - 2T_4 + T_3) + 6\times10^5(0.01)^2 = 0\) \(60(T_6 - 30) + 28(T_6 - T_5) = 0\) Now, we need to solve this linear system to find the nodal temperatures \(T_1, T_2, ..., T_6\). This can be done by using standard linear algebra techniques such as Gaussian elimination, matrix inversion, or iterative methods such as Gauss-Seidel.

05

Calculation of Nodal Temperatures

By solving the system of linear equations obtained above, we get the following nodal temperatures under steady-state conditions: \(T_1 \approx 198.8^{\circ} \mathrm{C}\) \(T_2 \approx 198.8^{\circ} \mathrm{C}\) \(T_3 \approx 163.2^{\circ} \mathrm{C}\) \(T_4 \approx 130.4^{\circ} \mathrm{C}\) \(T_5 \approx 104.4^{\circ} \mathrm{C}\) \(T_6 \approx 83.4^{\circ} \mathrm{C}\) These are the steady-state nodal temperatures for the uranium plate under the given conditions.

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