Question

A large heated steel block \(\left(\rho=7832 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=\right.\) \(434 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=63.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), 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 minutes of cooling.

Short answer

The temperature at the edge of the steel block after 10 minutes of cooling is 450°C.

Step-by-step solution

Gather given information

We are given the following information in the problem: - Initial temperature: \(T_{i} = 450^{\circ} \mathrm{C}\) - Room temperature: \(T_{\infty} = 25^{\circ} \mathrm{C}\) - Thermal diffusivity: \(\alpha = 18.8 \times 10^{-6} \mathrm{m}^2/\mathrm{s}\) - Time: \(t = 10 \,\text{minutes}\) We have the distance from the edge \(x=0\) since we are looking for the temperature at the edge of the steel block.

Convert given time to seconds

The time in the problem is given in minutes, so we need to convert it to seconds: \(t = 10 \,\text{minutes} \times \frac{60\,\text{seconds}}{1\,\text{minute}} = 600\,\text{seconds}\)

Calculate the temperature at the edge of the steel block

Now, we can use the formula for the temperature at the edge of the steel block: \(T(s) = T_{i} + (T_{\infty} - T_{i}) erf(\frac{x}{2\sqrt(\alpha t)}) \) Substitute the values into the formula: \(T(s) = 450 + (25 - 450) erf(\frac{0}{2\sqrt(18.8 \times 10^{-6} \times 600)}) \) Since we have \(x=0\), the formula simplifies to: \(T(s) = 450 + (25 - 450) \cdot 0 = 450\) So, the temperature at the edge of the steel block after 10 minutes of cooling is \(450^{\circ} \mathrm{C}\).

Key concepts

Understanding Thermal Diffusivity

Thermal diffusivity is a property that measures how quickly heat spreads through a material. It is represented by the symbol \( \alpha \) and is defined by the equation \( \alpha = \frac{k}{\rho \cdot c_p} \) where \( k \) is the thermal conductivity, \( \rho \) is the density, and \( c_p \) is the specific heat capacity at constant pressure. An intuitive way to think of thermal diffusivity is to imagine pouring a cup of hot coffee into a lake. The rate at which the heat from the coffee spreads throughout the water is akin to a material's thermal diffusivity.

Materials with high thermal diffusivity, like our steel block in the exercise, can transfer heat rapidly throughout their volume. This is crucial when considering the cooling or heating processes, as it allows for a quick response to temperature changes in the environment.

Convection Heat Transfer Coefficient

The convection heat transfer coefficient, denoted by \( h \) is a measure of the convective heat transfer between a surface and a fluid moving over it. It is part of the Newton's Law of Cooling and is used in the equation \( Q = hA(T_{surface} - T_{fluid}) \) where \( Q \) is the heat transfer rate, \( A \) is the surface area, and \( T_{surface} \) and \( T_{fluid} \) are the temperatures of the surface and fluid, respectively.

In the context of our steel block problem, the convection heat transfer coefficient plays a critical role. The higher this coefficient, the more efficient the heat removal from the block's surface into the surrounding air, which influences the cooling rate. The given value of \( 25 \, \mathrm{W/m^2 \cdot K} \) indicates the amount of heat that can be transferred per unit area and per degree temperature difference between the block's surface and the ambient air.

Transient Heat Conduction

Transient heat conduction refers to the time-dependent process of heat transfer within a material as it moves towards thermal equilibrium with its surroundings. Unlike steady-state conduction, where temperatures remain constant over time, in transient heat conduction, temperatures can vary both in space and time.

When addressing problems involving transient heat conduction, such as the cooling of the steel block, it is essential to know the initial temperature distribution, properties of the material, and boundary conditions. The solution presented in the exercise assumes that the steel block can be treated as a quarter-infinite medium, which means it is semi-infinite in reality, and only a quarter is considered for simplification. Mathematical tools like the error function (erf) are often used to solve these kinds of heat transfer problems. The heat conduction in the block changes over time, leading us to determine the temperature at a specific moment, which in this case is 10 minutes after the cooling process begins.