Question

A hot brass plate is having its upper surface cooled by an impinging jet of air at a temperature of \(15^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(220 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The 10 -cm-thick brass plate $\left(\rho=8530 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=380 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=110 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\(, and \)\alpha=33.9 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}$ ) had a uniform initial temperature of \(650^{\circ} \mathrm{C}\), and the lower surface of the plate is insulated. Using a uniform nodal spacing of \(\Delta x=2.5 \mathrm{~cm}\) and time step of \(\Delta t=10 \mathrm{~s}\), determine \((a)\) the implicit finite difference equations and \((b)\) the nodal temperatures of the brass plate after 10 seconds of cooling.

Short answer

Question: Determine the nodal temperatures of the brass plate after 10 seconds of cooling using implicit finite difference equations for the given scenario. Answer: To find the nodal temperatures after 10 seconds, we need to follow these steps: 1. List down the given information, including material properties, geometry, and time step. 2. Discretize the heat conduction equation implicitly using the finite difference method. 3. Apply initial and boundary conditions to the discretized equation. 4. Create a system of equations for nodal temperatures. 5. Solve the system of equations for nodal temperatures after 10 seconds. By solving the system of equations, we will obtain the nodal temperatures of the brass plate after 10 seconds of cooling, i.e., temperatures at Node 1, Node 2, Node 3, and Node 4.

Step-by-step solution

List down given information

We are given the following information: - Air temperature: \(T_\infty = 15^\circ C\) - Convection heat transfer coefficient: \(h = 220\, W/(m^2 K)\) - Brass material properties: \(\rho = 8530\, kg/m^3\), \(c_p = 380\, J/(kg\cdot K)\), \(k = 110\, W/(m\cdot K)\) - Brass plate thickness and geometry: \(d = 10 \, cm\), \(\Delta x = 2.5\, cm\) - Time step: \(\Delta t = 10\, s\) - Initial temperature: \(T_0 = 650\,^{\circ}C\)

Discretize the heat conduction equation implicitly

The heat conduction equation is: \(\frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial x^2}\) Using implicit finite difference method, we approximate the equation as: \(\frac{T^{n+1}_i - T^n_i}{\Delta t} = \alpha \left(\frac{T^{n+1}_{i+1} - 2T^{n+1}_i + T^{n+1}_{i-1}}{(\Delta x)^2}\right) \) Rearranging the equation: \(T^{n+1}_{i} = T^{n}_{i} + \frac{\alpha\Delta t}{(\Delta x)^2}(T^{n+1}_{i+1} - 2T^{n+1}_i + T^{n+1}_{i-1}) \)

Apply initial and boundary conditions

Initial condition: The initial temperature of the brass plate is uniform and constant, \(T_i^n = T_0 = 650^{\circ} C\). Boundary conditions: 1. Upper surface (cooling by air impinging): Convection heat transfer, \(q(x=0, t) = h(T_0 - T_\infty)\) where \(T_0 = T^{n+1}_1\) 2. Lower surface (insulated): No heat loss, \(q(x=10cm, t) = 0\) which is equivalent to \(T^{n+1}_{i+1} = T^{n+1}_{i-1}\)

Create the system of equations for nodal temperatures

We consider 4 nodes with uniform spacing of \(\Delta_x = 2.5 cm\). Equation for node 1: \(T^{n+1}_{1} = T^{n}_{1} + \frac{\alpha\Delta t}{(\Delta x)^2}(T^{n+1}_{2}- 2T^{n+1}_1 + T^{n+1}_{0}) \) Using the boundary condition \(q(x=0, t) = h(T^{n+1}_1 - T_\infty)\) at node 1, we obtain \(T^{n+1}_{0} = 2\frac{T_\infty - T^{n+1}_1}{(2h\Delta x / k)} + T^{n+1}_1\) Equation for node 2: \(T^{n+1}_{2} = T^{n}_{2} + \frac{\alpha\Delta t}{(\Delta x)^2}(T^{n+1}_{3} - 2T^{n+1}_2 + T^{n+1}_{1})\) Equation for node 3: \(T^{n+1}_{3} = T^{n}_{3} + \frac{\alpha\Delta t}{(\Delta x)^2}(T^{n+1}_{4} - 2T^{n+1}_3 + T^{n+1}_{2})\) Using the boundary condition \(q(x=10\, cm, t) = 0\) at node 4, we obtain \(T^{n+1}_{4} = T^{n+1}_{2}\).

Solve the system for nodal temperatures after 10 seconds

We can now use the initial temperature and the equations from the previous step to solve for the nodal temperatures after 10 seconds, i.e. when \(n+1 = 1\). Temperature at all nodes initially \(T^n_i = 650 ^\circ C\). Substitute the initial temperatures and material properties into each equation and solve using any numerical method, like Gauss-Seidel Iterations or Tridiagonal matrix method to find: - \(T^{1}_1\) (Temperature at Node 1) - \(T^{1}_2\) (Temperature at Node 2) - \(T^{1}_3\) (Temperature at Node 3) - \(T^{1}_4\) (Temperature at Node 4) After solving the system of equations, we will obtain the nodal temperatures of the brass plate after 10 seconds of cooling.