Question

Consider a cubic block whose sides are \(5 \mathrm{~cm}\) long and a cylindrical block whose height and diameter are also \(5 \mathrm{~cm}\). Both blocks are initially at \(20^{\circ} \mathrm{C}\) and are made of granite $\left(k=2.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\( and \)\left.\alpha=1.15 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)$. Now both blocks are exposed to hot gases at \(500^{\circ} \mathrm{C}\) in a furnace on all of their surfaces with a heat transfer coefficient of $40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Determine the center temperature of each geometry after 10 , 20 , and \(60 \mathrm{~min}\). Solve this problem using the analytical oneterm approximation method.

Short answer

Based on the provided information and methodology, calculate the center temperature values of both the cubic and cylindrical granite blocks after exposure to 10, 20, and 60-minute intervals in a furnace. Provide the values for each time interval and shape.

Step-by-step solution

Identify thermal properties and constants

First, we collect the provided values as constants that we'll use to solve the problem: - Thermal conductivity (k): \(2.5 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}}\) - Thermal diffusivity (α): \(1.15 \times 10^{-6} \frac{\mathrm{m^2}}{\mathrm{s}}\) - Initial temperature: \(20^{\circ}\mathrm{C}\) - Furnace temperature: \(500^{\circ}\mathrm{C}\) - Heat transfer coefficient (h): \(40 \frac{\mathrm{W}}{\mathrm{m^2} \cdot \mathrm{K}}\) - Times to evaluate: 10, 20, and 60 minutes. Remember to convert time in minutes to seconds when solving for the temperature.

Use the one-term approximation method for the cubic block

Using the one-term approximation method for a cubic block, the formula for the center temperature (\(T(0,t)\)) is: \(T(0,t) = T_\infty + (T_i - T_\infty) \sum_{n=1}^\infty exp\left(\frac{-n^2 \pi^2 \alpha t}{L^2}\right)\) Where: - \(T_\infty\) is the furnace temperature - \(T_i\) is the initial temperature - \(L\) is the half-thickness of the block (in this case, \(L = \frac{5}{2} \mathrm{~cm}\), convert this to meters) Since we're using the one-term approximation, we'll only consider the first term in the sum when \(n=1\). Calculate the center temperature for t=10, 20, and 60 minutes (convert to seconds for calculation).

Use the one-term approximation method for the cylindrical block

For the cylindrical block, the formula for the center temperature (\(T(0,t)\)) is: \(T(0,t) = T_\infty + (T_i - T_\infty) \sum_{n=1}^\infty \frac{J_0\left(\alpha_n'\right)}{\alpha_n' J_1\left(\alpha_n'\right)} exp\left(\frac{-n^2 \pi^2 \alpha t}{R^2}\right)\) Where: - \(J_0\) and \(J_1\) are the zeroth and first-order Bessel functions - \(\alpha_n'\) is the nth root of the derivative of the first-order Bessel function - \(R\) is the cylinder radius, half the diameter (in this case, \(R = \frac{5}{2} \mathrm{~cm}\), convert this to meters) Again, using the one-term approximation, we will only use the first term in the sum when \(n=1\). Calculate the center temperature for t=10, 20, and 60 minutes (convert to seconds for calculation).

Calculate the center temperature values

Calculate the center temperature values for both shapes at each given time using the formulas above. You will obtain three temperature values for the cubic block and three for the cylindrical block. These temperatures represent the center temperature of each geometry at 10, 20, and 60 minutes.