# Finite Difference Solution for 2D Steady-State Plate ## Problem Overview A **steel plate** has dimensions: - Length \( L = 10 \) ft - Width \( W = 6 \) ft All flat (top/bottom) surfaces are insulated (2D heat conduction only). **Boundary Conditions:** - Left edge (\( x = \)): \( T = 100^\circ F \) - Right, top, bottom edges: \( T = 50^\circ F \) **Task:** a) Find interior temperatures using the finite difference method with grid spacing \( h = k = 4 \) inches. b) Plot the temperature profile and shade areas with \( T > 250^\circ F \). --- ## a) Finite Difference Discretization ### 1. Convert Units 1 ft = 12 in - Length: \( 10 \) ft \( = 120 \) in - Width: \( 6 \) ft \( = 72 \) in ### 2. Grid Layout - \( h = k = 4 \) in (spacing in both \( x \) and \( y \)) - Number of nodes: - \( x \): \( N_x = \frac{120}{4} + 1 = 31 \) - \( y \): \( N_y = \frac{72}{4} + 1 = 19 \) So, the grid is \( 31 \times 19 \) nodes. ### 3. Node Numbering Let \( (i, j) \) be the node at \( x_i = i \times h, \ y_j = j \times k \): - \( i = ,1,2,\ldots,30 \) (left to right) - \( j = ,1,2,\ldots,18 \) (bottom to top) ### 4. Discretized Equation For interior node \( T_{i,j} \): \[ T_{i,j} = \frac{1}{4} (T_{i+1,j} + T_{i-1,j} + T_{i,j+1} + T_{i,j-1}) \] ### 5. Apply Boundary Conditions - **Left edge** (\( i = \)): \( T_{,j} = 100 \) - **Right, Top, Bottom edges** (\( i = 30, \ j = ,18 \)): \( T = 50 \) ### 6. Assemble Linear System Form a system of equations for all interior nodes (\( 1 \leq i \leq 29, \ 1 \leq j \leq 17 \)), substituting the known boundary values where appropriate. **Total unknowns:** \( (N_x - 2) \times (N_y - 2) = 29 \times 17 = 493 \) unknowns. **For each interior node:** \[ T_{i,j} = \frac{1}{4} (T_{i+1,j} + T_{i-1,j} + T_{i,j+1} + T_{i,j-1}) \] or \[ 4T_{i,j} - T_{i+1,j} - T_{i-1,j} - T_{i,j+1} - T_{i,j-1} = \] - If a neighbor is on the boundary, use the boundary value (e.g., \( T_{,j} = 100 \) for the left neighbor at the left edge). ### 7. Solution Method Solve the linear system using: - Iterative method (Jacobi, Gauss-Seidel, or SOR) - Or direct solver (if memory allows) For practical purposes, use Python (NumPy/SciPy), MATLAB, or similar. --- ## b) Plotting and Shading ### 1. Temperature Profile Plot - Create a 2D plot or contour plot of \( T(x,y) \) over the domain. - Use a colorbar to show temperature values. ### 2. Highlight Area Where \( T > 250^\circ F \) - Overlay or shade regions where \( T > 250 \). - E.g., use a mask to shade these areas in a distinct color. --- ## Example Python Pseudocode ```python import numpy as np import matplotlib.pyplot as plt # Grid size Nx, Ny = 31, 19 T = np.ones((Ny, Nx)) * 50 # initialize to 50 F # Set left boundary T[:, ] = 100 # Iterate (Gauss-Seidel) for it in range(200): for j in range(1, Ny-1): for i in range(1, Nx-1): T[j,i] = .25 * (T[j,i+1] + T[j,i-1] + T[j+1,i] + T[j-1,i]) # Plot x = np.linspace(, 120, Nx) y = np.linspace(, 72, Ny) X, Y = np.meshgrid(x, y) plt.figure(figsize=(8,4)) cp = plt.contourf(X, Y, T, levels=20, cmap='hot') plt.colorbar(cp, label='Temperature (°F)') plt.xlabel('x (in)') plt.ylabel('y (in)') # Shade T > 250°F plt.contourf(X, Y, T, levels=[250, T.max()], colors='cyan', alpha=.4) plt.title('Temperature Profile in Plate\n(Shaded: $T > 250^\circ F$)') plt.show() ``` **[Alt text: Contour plot of temperature in a rectangular plate, with a shaded region showing where T > 250°F]** --- ## Summary Table | Parameter | Value | |-------------------|----------------------| | Plate length | 10 ft (120 in) | | Plate width | 6 ft (72 in) | | Grid spacing | 4 in | | Nodes (x, y) | 31, 19 | | Left boundary | 100°F | | Other boundaries | 50°F | --- ## References - Incropera & DeWitt, "Fundamentals of Heat and Mass Transfer" - Numerical Heat Transfer texts --- **Note:** The above approach outlines the method and provides sample code for solving and visualizing the temperature profile. For large systems, use optimized solvers. Adjust plotting as needed for your data.