Question

Steel rods $\left(\rho=7832 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=434 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\(, and \)k=63.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( ) are heated in a furnace to \)850^{\circ} \mathrm{C}\( and then quenched in a water bath at \)50^{\circ} \mathrm{C}$ for a period of \(40 \mathrm{~s}\) as part of a hardening process. The convection heat transfer coefficient is \(650 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\). If the steel rods have diameter of \(40 \mathrm{~mm}\) and length of $2 \mathrm{~m}$, determine their average temperature when they are taken out of the water bath.

Short answer

Answer: The average temperature of the steel rods after being quenched in the water bath for 40 seconds is approximately 210.1°C.

Step-by-step solution

\(d = 40\,\mathrm{mm} \times \frac{1\,\mathrm{m}}{1000\,\mathrm{mm}} = 0.04\,\mathrm{m}\) #Step 2: Calculate the external surface area and the volume of the rod# Use the diameter and length of the rod to calculate its external surface area and volume.

\(A = \pi d L = \pi(0.04\,\mathrm{m})(2\,\mathrm{m}) = 0.2513\,\mathrm{m^2}\) \(V = \frac{\pi d^2}{4} L = \frac{\pi(0.04\,\mathrm{m})^2}{4}(2\,\mathrm{m}) = 0.002513\,\mathrm{m^3}\) #Step 3: Calculate the temperature after quenching time# Use the lumped capacitance equation as described above and given values to calculate the temperature after 40 seconds

\(T(t)=T_{\infty} + (T_{i}-T_{\infty})(\mathrm{e}^{-\frac{hAt}{\rho Vc_p}})\) Plug in the given values, and calculate the temperature after \(40\,\mathrm{s}\).

\(T(40)=50+(850-50)\mathrm{e}^{-\frac{650\cdot0.2513\cdot40}{7832\cdot0.002513\cdot434}}\) \(T(40) \approx 210.1^{\circ}\mathrm{C}\) The average temperature of the steel rods after being quenched in the water bath for \(40\,\mathrm{s}\) is approximately \(210.1^{\circ}\mathrm{C}\).