Large steel plates \(1.0\)-cm in thickness are quenched from \(600^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) by submerging them in an oil reservoir held at \(30^{\circ} \mathrm{C}\). The average heat transfer coefficient for both faces of steel plates is \(400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Average steel properties are \(k=45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7800 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=470 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Calculate the quench time for steel plates.

First, we need to find the difference between the initial plate temperature (\(T_1\)) and the oil reservoir temperature (\(T_\infty\)): $$\Delta T = T_1 - T_\infty = 600^{\circ}\mathrm{C} - 30^{\circ}\mathrm{C} = 570^{\circ}\mathrm{C}$$

The Biot number (Bi) is a dimensionless number used to determine if we can use the lumped capacitance method. $$Bi = \frac{hL_{c}}{k}$$ Where: \(h\) is the average heat transfer coefficient, \(L_{c}\) is the characteristic length (in this case, half the plate's thickness), and \(k\) is the thermal conductivity of the steel. Calculating the Biot number: $$L_{c} = \frac{1}{2} \times 0.01\, \mathrm{m} = 0.005\,\mathrm{m}$$ $$Bi = \frac{400 \,\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K} \times 0.005\, \mathrm{m}}{45\, \mathrm{W} / \mathrm{m} \cdot \mathrm{K}} = 0.0444$$ Since the Biot number is less than 0.1, we can use the lumped capacitance method.

We can determine the heat transfer resistance (\(R_{h}\)) as follows: $$R_{h} = \frac{1}{hA}$$ Where: \(A\) is the surface area of the steel plate. We have not been given the dimensions of the steel plate, so we cannot determine the surface area. However, we can still find the heat transfer resistance in terms of \(A\): $$R_{h} = \frac{1}{400 \,\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K} \times A}$$

The volume (\(V\)) and mass (\(m\)) of the steel plate can also be determined in terms of the unknown dimensions: $$V = A \times 0.01\, \mathrm{m}$$ $$m = \rho V = 7800\, \mathrm{kg} / \mathrm{m}^{3} \times (A \times 0.01\, \mathrm{m}) = 78A\, \mathrm{kg}$$

Using the lumped capacitance method, the quench time (\(t\)) is given by: $$t = \frac{(T_1 - T_\infty)R_{h}mc_{p}}{(T_1 - T_2)}$$ Substituting the known values: $$t = \frac{(600^{\circ}\mathrm{C} - 30^{\circ}\mathrm{C})\left(\frac{1}{400 \,\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K} \times A}\right)(78A\,\mathrm{kg})(470 \,\mathrm{J} / \mathrm{kg} \cdot \mathrm{K})}{(600^{\circ}\mathrm{C} - 100^{\circ}\mathrm{C})}$$ The surface area (\(A\)) cancels out, simplifying the expression for quench time: $$t = \frac{570\, \mathrm{K} \times 0.0025 \, \mathrm{m} \times 7800 \, \mathrm{kg} / \mathrm{m}^{3} \times 470 \, \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}}{500\, \mathrm{K}} = 1000\, \mathrm{s}$$ The quench time for the steel plates is 1000 seconds.