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 \(\Delta x=\Delta y=10 \mathrm{~cm}\).

Step by step solution

01

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: \(\Delta x = \Delta y = 10 \mathrm{~cm}\)

02

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: \(T_{i,j}^{n+1} = T_{i,j}^{n} + \frac{\alpha \Delta t}{(\Delta x)^2}\left(T_{i+1,j}^{n} - 2T_{i,j}^{n} + T_{i-1,j}^{n}\right) + \frac{\alpha \Delta t}{(\Delta y)^2}\left(T_{i,j+1}^{n} - 2T_{i,j}^{n} + T_{i,j-1}^{n}\right)\)

03

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.

04

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.

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