Question

A large heated steel block $\left(\rho=7832 \mathrm{~kg} / \mathrm{m}^{3}\right.\(, \)c_{p}=434 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=63.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, and \)\left.\alpha=18.8 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)$ is allowed to cool in a room at \(25^{\circ} \mathrm{C}\). The steel block has an initial temperature of \(450^{\circ} \mathrm{C}\), and the convection heat transfer coefficient is $25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Assuming that the steel block can be treated as a quarter-infinite medium, determine the temperature at the edge of the steel block after \(10 \mathrm{~min}\) of cooling.

Short answer

Answer: The temperature at the edge of the steel block after 10 minutes of cooling is 25°C.

Step-by-step solution

List the given data

We are given the following information for the large heated steel block: 1. Density (\(\rho\)): 7832 kg/m³ 2. Specific heat capacity (\(c_p\)): 434 J/kg⋅K 3. Thermal conductivity (k): 63.9 W/m⋅K 4. Thermal diffusivity (\(\alpha\)): 18.8 x 10⁻⁶ m²/s 5. Initial temperature (\(T_i\)): 450°C 6. Room temperature (\(T_\infty\)): 25°C 7. Convection heat transfer coefficient (h): 25 W/m²⋅K 8. Cooling time (t): 10 min

Convert the cooling time into seconds

Since the given cooling time is in minutes, we need to convert it into seconds: Time in seconds (t) = 10 min × 60 s/min = 600 s

Calculate the Biot number (Bi)

The Biot number (Bi) is a dimensionless quantity that is used to express convective heat transfer through a solid. It is defined as the ratio of the convection heat transfer coefficient (h) and the product of the thermal conductivity (k) and the thermal diffusivity (\(\alpha\)): Bi = h / (k * \(\alpha\)) Plugging in the given values, we get: Bi = 25 / (63.9 * 18.8 × 10⁻⁶) = 0.021 Since Bi << 1, we can treat the steel block as a quarter-infinite medium.

Use the one-dimensional transient heat conduction formula in a semi-infinite solid

For a semi-infinite solid with convective heat transfer from its surface, we can use the formula for transient heat conduction in one dimension: \(T(x, t) = T_\infty + (T_i - T_\infty) \operatorname{erf}(\frac{x}{2\sqrt{\alpha t}})\) Where T(x, t) is the temperature at the edge of the steel block after time t, erf is the error function, and x represents the distance from the edge of the steel block (x = 0 for the edge).

Calculate the temperature at the edge of the steel block after 10 minutes

To determine the temperature at the edge of the steel block after 10 minutes, we apply the formula from Step 4 using the given values: T(0, 600) = \(25 + (450 - 25)\operatorname{erf}(\frac{0}{2\sqrt{18.8 × 10^{-6} (600)}}) = 25 + 425 \operatorname{erf}(0)\) As the error function erf(0) = 0, we find that: T(0, 600) = 25°C Hence, the temperature at the edge of the steel block after 10 minutes of cooling is 25°C.