Question

A stainless steel plate is connected to an insulation plate by square ASTM A479 904L stainless steel bars. Each square bar has a thickness of $1 \mathrm{~cm}\( and a length of \)5 \mathrm{~cm}$. The bars are exposed to convection heat transfer with a hot gas. The temperature of the hot gas is \(300^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of $25 \mathrm{~W} / \mathrm{m}^{2}$. . The thermal conductivity for ASTM A479 904L stainless steel is known to be \(12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The stainless steel plate maintains a uniform temperature of $100^{\circ} \mathrm{C}$. According to the ASME Code for Process Piping (ASME B31.3-2014, Table A-1M), the maximum use temperature for ASTM A479 904L stainless steel bar is \(260^{\circ} \mathrm{C}\). Using the finite difference method with a uniform nodal spacing of \(\Delta x=5 \mathrm{~mm}\) along the bar, determine the temperature at each node. Compare the numerical results with the analytical solution. Plot the temperature distribution along the bar. Would any part of the ASTM A479 904L bars be above the maximum use temperature of \(260^{\circ} \mathrm{C}\) ?

Short answer

Answer: The finite difference method is used to solve the heat conduction problem for the stainless steel bar. In the end, we will evaluate if any part of the steel bars exceed the maximum use temperature of \(260^{\circ}\mathrm{C}\) according to the ASME Code.

Step-by-step solution

Set up the differential equation

First, we need to set up the differential equation for heat conduction through the steel bars. We will use Fourier's law of heat conduction, which states that the heat flux \(q_x\) through a material is proportional to the temperature gradient: \(q_x = -k\frac{dT}{dx}\) where \(k\) is the thermal conductivity. Now let's consider the energy balance for a small volume element \(\Delta x\) along the bar: \(-q_x A - q_c A + q_{x+\Delta x} A = 0\) where \(q_c\) is the convective heat flux through the surface in contact with the hot gas, and \(A\) is the cross-sectional area of the bar. For the convection term, we can use Newton's law of cooling: \(q_c = h(T_{gas} - T)\) where \(h\) is the convection heat transfer coefficient and \(T_{gas}\) is the hot gas temperature. Combining the above equations and dividing by \(A\Delta x\), we get the differential equation for the temperature \(T\) as a function of position \(x\): \(\frac{d^2T}{dx^2} = \frac{h}{k}(T_{gas} - T)\) with boundary conditions \(T(0)=100^{\circ}\mathrm{C}\) and \(T(5)=T_{insulation}\), where \(T_{insulation}\) is the temperature at the insulation plate.

Apply the finite difference method

To solve this differential equation numerically, we will use the finite difference method with a uniform nodal spacing \(\Delta x=5\mathrm{~mm}\). We first need to discretize the equation and represent the derivatives using finite differences: \(\frac{T_{i+1} - 2T_{i} + T_{i-1}}{(\Delta x)^{2}} = \frac{h}{k}(T_{gas} - T_{i})\) where \(T_i\) is the temperature at node \(i\). We can write this equation for each node in the bar, solving for the unknown temperatures.

Solve the system of equations and find temperatures

Next, we need to solve the system of equations obtained by applying the finite difference method to find the temperatures at each node. We could use any method to solve this system of linear equations, such as Gaussian elimination.

Analytical solution

In order to compare our numerical results with the analytical solution, we need to derive the analytical solution for the temperature distribution. The differential equation we derived in Step 1 is a linear second-order ordinary differential equation with constant coefficients, which can be solved by the method of separation of variables. The analytical solution for the temperature distribution can be derived as: \(T(x) = T_{0} + \frac{T_{gas} - T_{0}}{L}x + \frac{h}{2k}(L - x)x\) where \(T_{0}\) is the initial temperature of the steel plate (\(100^{\circ}\mathrm{C}\)) and \(L\) is the length of the bar (\(5\mathrm{~cm}\)).

Plot temperature distribution and compare to the analytical solution

Now, we can plot the numerical temperature distribution we obtained from the finite difference method along the bar and compare it to the analytical solution derived in Step 4. This will help us visualize the temperature distribution and assess the accuracy of our numerical results.

Evaluate the maximum use temperature

Lastly, we need to check whether any part of the steel bars would be above the maximum use temperature of \(260^{\circ}\mathrm{C}\) according to the ASME Code. We can do this by finding the maximum temperature in the numerical temperature distribution and compare it to the maximum use temperature: \(\max(T_{i}) \leq 260^{\circ}\mathrm{C}\) If this condition is met, none of the steel bars would be above the maximum use temperature.