Question

Steel rods, \(2 \mathrm{~m}\) in length and \(60 \mathrm{~mm}\) in diameter, are being drawn through an oven that maintains a temperature of $800^{\circ} \mathrm{C}\( and convection heat transfer coefficient of \)128 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. The steel rods \)\left(\rho=7832 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=434 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=63.9\right.\( \)\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\(, and \)\alpha=18.8 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}$ ) were initially in uniform temperature of \(30^{\circ} \mathrm{C}\). Using the analytical one-term approximation method, determine the amount of heat transferred to the steel rod after \(133 \mathrm{~s}\) of heating.

Short answer

Question: Determine the amount of heat transferred to a steel rod after being heated for 133 seconds in an oven. Answer: The amount of heat transferred to the steel rod after 133 seconds of heating is 20,874 W.

Step-by-step solution

Analyze the necessary formulas

The following formula will be used to estimate the heat transfer per length, obtained from the analytical one-term approximation method: \(q_{L}=\frac{k}{r} q_{1}\left[1-e^{-n_{1}^{2} \frac{\alpha t}{r^{2}}}\right]\). In this formula, \(q_{L}\) is the heat transfer per unit length. \(k\) is the thermal conductivity of the steel. \(r\) is the radius of the steel rod. \(q_{1}\) is the first Fourier sine series coefficient. \(n_{l}\) is the first non-dimensional eigenvalue. \(t\) is the time of heating. \(\alpha\) is the thermal diffusivity of the steel.

Calculate values for constants

We can find the values for the given parameters first. Given the diameter of the steel rod, we can find its radius as: \(r = \frac{60}{2}\times 10^{-3}\,\mathrm{m} = 0.03\,\mathrm{m}\). As per the one-term approximation method, we have \(n_{1} = \frac{2}{\pi}\) and \(q_{1} = \frac{2Nu_{\infty}\delta}{\pi}\), where \(Nu_{\infty}\) is the Nusselt number in a fully developed Biot number asymptotic limit and \(\delta\) is the difference in temperature between the oven and the initial rod temperature \(=800^{\circ}\mathrm{C} - 30^{\circ}\mathrm{C} = 770\,\mathrm{K}\). Using the given convection heat transfer coefficient, \(h = 128\,\mathrm{W}/\mathrm{m}^2\cdot\,\mathrm{K}\), we can find the Nusselt number as \(Nu_{\infty} = \frac{hr}{k} = \frac{128\times0.03}{63.9} = 0.06\). Finally, we have the values for all the constants in the formula.

Calculate the heat transfer per length after 133 seconds

We can now substitute all the values into the formula to find the heat transfer per length after \(133\,\mathrm{s}\) of heating: \(q_{L}=\frac{63.9}{0.03} \left(\frac{2\times0.06\times770}{\pi}\right)\left[1-\exp\left(-\left(\frac{2}{\pi}\right)^{2} \frac{18.8\times10^{-6}\times133}{(0.03)^2}\right)\right]\). Calculating this, we get \(q_{L}=10437\,\mathrm{W}/\mathrm{m}\).

Calculate the total heat transferred after 133 seconds

Now that we have found the heat transfer per length, we can multiply it by the total length of the rod to find the total heat transferred. Total heat transferred \((Q) = q_{L}\times\mathrm{Length} = 10437\,\mathrm{W}/\mathrm{m}\times2\,\mathrm{m} = 20874\,\mathrm{W}\). Thus, the amount of heat transferred to the steel rod after \(133\,\mathrm{s}\) of heating is \(20874\,\mathrm{W}\).