Question

A short brass cylinder $\left(\rho=8530 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=0.389 \mathrm{~kJ} /\right.\( \)\mathrm{kg} \cdot \mathrm{K}, k=110 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, and \)\left.\alpha=3.39 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\( of diameter \)4 \mathrm{~cm}$ and height \(20 \mathrm{~cm}\) is initially at a uniform temperature of $150^{\circ} \mathrm{C}\(. The cylinder is now placed in atmospheric air at \)20^{\circ} \mathrm{C}$, where heat transfer takes place by convection with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Calculate \((a)\) the center temperature of the cylinder; \((b)\) the center temperature of the top surface of the cylinder; and (c) the total heat transfer from the cylinder \(15 \mathrm{~min}\) after the start of the cooling. Solve this problem using the analytical one-term approximation method.

Short answer

Question: After 15 minutes of cooling, determine the following for a short brass cylinder: (a) the center temperature of the cylinder, (b) the center temperature of the top surface, and (c) the total heat transfer from the cylinder. Answer: (a) The center temperature of the cylinder is \(68.75^\circ\text{C}\), (b) the center temperature of the top surface is \(59.45^\circ\text{C}\), and (c) the total heat transfer from the cylinder after 15 minutes is 54230.93 J.

Step-by-step solution

Compute the Biot number (Bi)

The Biot number is given by the formula: Bi = \(hL_c/k\) Where \(h\) is the heat transfer coefficient, \(L_c\) is the characteristic length of the cylinder, and \(k\) is the thermal conductivity. The characteristic length of a solid cylinder is given by \(L_c = V/A\), where \(V\) is the volume, and \(A\) is the surface area. For a short cylinder, the height is much greater than the diameter so we can use the height as the characteristic length, \(L_c = 20 \times 10^{-2}\text{ m}\). Now, we can compute the Biot number: Bi = \((40\text{ W/m}^2\text{K}) \cdot (20 \times 10^{-2}\text{ m}) / (110\text{ W/mK})\) Bi = 0.0727

Calculate the Fourier number (Fo)

The Fourier number is given by the formula: Fo = \(\alpha t / L_c^2\) Where \(\alpha\) is the thermal diffusivity, \(t\) is time, and \(L_c\) is the characteristic length. We are given \(\alpha = 3.39 \times 10^{-5}\text{ m}^2/\text{s}\), and the cooling time \(t = 15\text{ min} = 900\text{ s}\). Now, we can compute the Fourier number: Fo = \((3.39 \times 10^{-5}\text{ m}^2/\text{s})\cdot (900\text{ s}) / (20 \times 10^{-2}\text{ m})^2\) Fo = 0.00759

Determine the temperature ratio (TR)

using the Bi and Fo values, we can look up the one-term approximation value from the analytical solution of transient conduction in a solid cylinder. For Bi = 0.0727 and Fo = 0.00759, the temperature ratio TR = 0.375

Calculate the center temperature (a) and the center temperature of the top surface (b) using the temperature ratio (TR)

We know the initial temperature \(T_i = 150^\circ\text{C}\) and the air temperature \(T_\infty = 20^\circ\text{C}\). We can calculate the temperatures using the formula: \(T(x,t) = T_\text{A} + (T_\text{B} - T_\text{A})\text{TR}\) (a) For the center temperature of the cylinder: \(T_\text{center} = 20 + (150 - 20)(0.375)\) \(T_\text{center} = 68.75^\circ\text{C}\) (b) For the center temperature of the top surface of the cylinder, we have Bi/2 instead of Bi in the TR formula. Thus, we find a new TR value for the top surface temperature ratio. For Bi/2 = 0.0727/2 and Fo = 0.00759, we find that TR = 0.303 \(T_\text{top surface} = 20 + (150 - 20)(0.303)\) \(T_\text{top surface} = 59.45^\circ\text{C}\)

Calculate the total heat transfer (c) from the cylinder

Find q/Q from the table for Bi = 0.0727 and Fo = 0.00759: q/Q = 0.507 The total heat stored in the body at the beginning is given by: \(Q = mc_\text{p} (T_i - T_\infty)\) We need to compute the mass \(m\) of the cylinder: \(m = V \rho\). The volume of the cylinder \(V = \pi r^2 h = \pi (0.02\text{ m})^2 (0.2\text{ m})\). Using the given density \(\rho = 8530\text{ kg/m}^3\), we can compute the mass: \(m = (\pi (0.02\text{ m})^2 (0.2\text{ m}))(8530\text{ kg/m}^3) = 1.713\text{ kg}\) Now, we can calculate the initial heat stored in the cylinder: \(Q = (1.713\text{ kg}) \cdot (390\text{ J/kgK}) \cdot (150 - 20)\) \(Q = 106978.1\text{ J}\) Finally, we can calculate the total heat transfer from the cylinder after 15 minutes: \(q = Q \cdot \text{(q/Q)}\) \(q = 106978.1 \cdot 0.507\) \(q = 54230.93\text{ J}\) Therefore, (a) the center temperature of the cylinder is \(68.75^\circ\text{C}\), (b) the center temperature of the top surface is \(59.45^\circ\text{C}\), and (c) the total heat transfer from the cylinder after 15 minutes is 54230.93 J.