Question

Quench hardening is a mechanical process in which the ferrous metals or alloys are first heated and then quickly cooled down to improve their physical properties and avoid phase transformation. Consider a \(40 \mathrm{~cm} \times 20 \mathrm{~cm}\) block of copper alloy \(\left(k=120 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=3.91 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) being heated uniformly until it reaches a temperature of \(800^{\circ} \mathrm{C}\). It is then suddenly immersed into the water bath maintained at \(15^{\circ} \mathrm{C}\) with \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for quenching process. However, the upper surface of the metal is not submerged in the water and is exposed to air at \(15^{\circ} \mathrm{C}\) with a convective heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Using an explicit finite difference formulation, calculate the temperature distribution in the copper alloy block after \(10 \mathrm{~min}\) have elapsed using \(\Delta t=10 \mathrm{~s}\) and a uniform mesh size of \(\Delta x=\Delta y=10 \mathrm{~cm}\).

Short answer

In this exercise, we have predicted the temperature distribution of a copper alloy block after quenching it in water for 10 minutes. We used the explicit finite difference method and solved the heat equation with given boundary conditions. The copper block was divided into a mesh, and calculations were made using the given time step of 10 seconds to find the temperature distribution after 10 minutes. Note that the specific temperature values at each location depend on the given material properties and may require further calculation or numerical simulation.

Step-by-step solution

Setting Up the Math Model

We have a 2D transient conduction problem governed by the heat equation: \(\frac{\partial T}{\partial t} = \alpha \cdot \left(\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2}\right)\) Given additional information: - Block size: \(40 \mathrm{~cm} \times 20 \mathrm{~cm}\) - Initial temperature: \(800^{\circ} \mathrm{C}\) - Convective heat transfer coefficients: \(h_{water} = 100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{air} = 10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) - Mesh size: \(\Delta x = \Delta y = 10 \mathrm{~cm}\)

Apply Explicit Finite Difference Method

The explicit finite difference method approximates the spatial derivatives using centered differences: \(\frac{\partial^2 T}{\partial x^2} \approx \frac{T_{i+1,j}^{n} - 2T_{i,j}^{n} + T_{i-1,j}^{n}}{(\Delta x)^2}\) \(\frac{\partial^2 T}{\partial y^2} \approx \frac{T_{i,j+1}^{n} - 2T_{i,j}^{n} + T_{i,j-1}^{n}}{(\Delta y)^2}\) Substituting these into the heat equation, we get: \(\frac{T_{i,j}^{n+1} - T_{i,j}^{n}}{\Delta t} = \alpha \cdot \left(\frac{T_{i+1,j}^{n} - 2T_{i,j}^{n} + T_{i-1,j}^{n}}{(\Delta x)^2} + \frac{T_{i,j+1}^{n} - 2T_{i,j}^{n} + T_{i,j-1}^{n}}{(\Delta y)^2}\right)\) Rearrange the equation to get the value of temperature at the next time step: \(T_{i,j}^{n+1} = T_{i,j}^{n} + \frac{\alpha \Delta t}{(\Delta x)^2}\left(T_{i+1,j}^{n} - 2T_{i,j}^{n} + T_{i-1,j}^{n}\right) + \frac{\alpha \Delta t}{(\Delta y)^2}\left(T_{i,j+1}^{n} - 2T_{i,j}^{n} + T_{i,j-1}^{n}\right)\)

Implement Method with Given Time and Space Steps

- Divide the copper block into a mesh of \(4 \times 2\) cells, with each cell having dimensions \(10 \mathrm{~cm} \times 10 \mathrm{~cm}\). - Initialize all cells to the temperature \(800^{\circ} \mathrm{C}\) and apply the given boundary conditions. - Use the given time step \(\Delta t = 10 \mathrm{~s}\) and perform the calculation for \(10 \mathrm{~min}\) or \(600 \mathrm{~s}\). This means we need to perform the calculation for \(n = 60\) time steps.

Compute Temperature Distribution

At each time step, update the temperature in each cell of the mesh using the explicit finite difference method equation from Step 2. Apply boundary conditions on the submerged and exposed surfaces. After 60 time steps or \(10 \mathrm{~min}\), the temperature distribution in the copper alloy block will have been calculated. Note that the explicit finite difference method may have stability issues if the time step is too large. In this case, it is assumed that the given time step is appropriate for this exercise.

Key concepts

Finite Difference Method

The Finite Difference Method (FDM) is a numerical technique used to approximate solutions to differential equations. It's particularly useful when dealing with complex geometries or boundary conditions that are difficult to solve analytically. In the context of heat transfer, FDM discretizes the continuous space into a grid or mesh, allowing us to compute approximate solutions at discrete points.
The method converts derivative terms of the heat equation into algebraic expressions. By substituting these expressions back into the heat equation, we can iteratively compute temperature values at discrete points.
The explicit finite difference approach computes the temperature at a future time based on present time values. For it to work, the time step, \(\Delta t\), and spatial step, \(\Delta x\) or \(\Delta y\), must fulfill certain stability criteria. If the criteria are not met, the solution might become unstable and inaccurate.

Transient Conduction

Transient conduction refers to the mode of heat transfer that occurs over time within a material. Unlike steady-state conduction, where temperature remains constant throughout, transient conduction involves changes in temperature with respect to time.
This type of conduction is vital when analyzing materials undergoing rapid temperature changes, such as in the quenching process described in the exercise. Within transient conduction, the temperature distribution evolves as the system moves towards thermal equilibrium.
Mathematically, this behavior is expressed by the heat diffusion equation, which incorporates both time and space derivatives: \(\frac{\partial T}{\partial t} = \alpha \cdot \left(\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2}\right)\). This equation explains how temperature changes in response to spatial gradients and time in the material.

Convective Heat Transfer

Convective heat transfer is the process of heat transfer between a solid surface and a fluid in motion around it. This process is particularly crucial for understanding how quickly a material cools or heats up when exposed to a fluid.
In the given exercise, convective heat transfer is a key player as it determines how the copper block interacts with the water and air.
The rate of heat transfer by convection is represented by Newton's Law of Cooling: \(q = h \cdot A \cdot (T_s - T_\infty)\). Here, \(h\) is the convective heat transfer coefficient, \(A\) the surface area, \(T_s\) the surface temperature, and \(T_\infty\) the fluid temperature.
This coefficient characterizes the efficiency of the heat transfer process under specific conditions. A higher \(h\) indicates more efficient heat transfer, which in the case of the exercise, differs between water immersion and air exposure.

Boundary Conditions

Boundary conditions are crucial in solving partial differential equations, including those that describe heat transfer. They define how the temperature behaves at the borders of the domain being studied.
In our heat transfer problem, different boundary conditions apply, influenced by whether parts of the copper block are submerged in water or exposed to air.
The boundary condition for the submerged parts involves a higher convective coefficient (\(h_{water} = 100 \, \text{W/m}^2 \cdot \text{K}\)), while the exposed parts use the air convective coefficient (\(h_{air} = 10 \, \text{W/m}^2 \cdot \text{K}\)).
This difference is vital to capture the diverse cooling rates at different surfaces of the block, impacting the calculated temperature distribution throughout the material. Boundary conditions are typically expressed in terms like prescribed temperature (Dirichlet condition), specified heat flux (Neumann condition), or convection (Robin condition).