91Ó°ÊÓ

The energy balance equation for a rectangular volume element of dimensions \(\Delta x\), \(\Delta y\), and \(\Delta z\) with constant thermal conductivity \(k\) and no heat generation can be written as: \({(q_x - q_{x+\Delta x})\Delta y \Delta z + (q_y - q_{y+\Delta y})\Delta x \Delta z = \rho c_p \Delta x \Delta y \Delta z \frac{\partial T}{\partial t}}\) where \(\rho\) is the density, \(c_p\) is the specific heat capacity, and \(q_x\) and \(q_y\) are the heat fluxes in the x and y directions, respectively. We can write heat fluxes using Fourier's law: \(q_x = -k \frac{\partial T}{\partial x}\) and \(q_y = -k \frac{\partial T}{\partial y}\). Substitute these into the energy balance equation: \({(-k \frac{\partial T}{\partial x}+k \frac{\partial T}{\partial x + \Delta x})\Delta y \Delta z + (-k \frac{\partial T}{\partial y}+k \frac{\partial T}{\partial y + \Delta y})\Delta x \Delta z = \rho c_p \Delta x \Delta y \Delta z \frac{\partial T}{\partial t}}\) Divide both sides by \(\Delta x \Delta y \Delta z\): \({[\frac{-k}{\Delta x}(\frac{\partial T}{\partial x} - \frac{\partial T}{\partial x + \Delta x})] + [\frac{-k}{\Delta y}(\frac{\partial T}{\partial y} - \frac{\partial T}{\partial y + \Delta y})] = \rho c_p \frac{\partial T}{\partial t}}\) For simplicity, we can drop the +\(\Delta x\) and +\(\Delta y\) terms in the derivatives, assuming they represent small variations. Thus, we get the following form of the heat equation: \({\frac{k}{\Delta x^2} \frac{\partial^2 T}{\partial x^2} + \frac{k}{\Delta y^2} \frac{\partial^2 T}{\partial y^2} = \rho c_p \frac{\partial T}{\partial t}}\)

Now, we need to apply finite difference approximations for the second-order spatial derivatives. Using central difference, we have: \(\frac{\partial^2 T}{\partial x^2} \approx \frac{T(x+\Delta x, y) - 2T(x, y) + T(x-\Delta x, y)}{(\Delta x)^2}\) and \(\frac{\partial^2 T}{\partial y^2} \approx \frac{T(x, y+\Delta y) - 2T(x, y) + T(x, y-\Delta y)}{(\Delta y)^2}\) Substitute these approximations back into the heat equation: \({\frac{k}{\Delta x^2} (\frac{T(x+\Delta x, y) - 2T(x, y) + T(x-\Delta x, y)}{(\Delta x)^2}) + \frac{k}{\Delta y^2} (\frac{T(x, y+\Delta y) - 2T(x, y) + T(x, y-\Delta y)}{(\Delta y)^2}) = \rho c_p \frac{\partial T}{\partial t}}\) Simplify the equation: \({\left[\frac{k(\Delta y)^2}{(\Delta x)^2(\Delta y)^2}\right](T(x+\Delta x, y) - 2T(x, y) + T(x-\Delta x, y))+\left[\frac{k(\Delta x)^2}{(\Delta x)^2(\Delta y)^2}\right](T(x, y+\Delta y) - 2T(x, y) + T(x, y- \Delta y)) = \rho c_p \frac{\partial T}{\partial t}}\) Further simplify the equation: \({\left[\frac{k}{(\Delta x)^2}\right](T(x+\Delta x, y) - 2T(x, y) + T(x-\Delta x, y))+\left[\frac{k}{(\Delta y)^2}\right](T(x, y+\Delta y) - 2T(x, y) + T(x, y- \Delta y)) = \rho c_p \frac{\partial T}{\partial t}}\)

To obtain the explicit finite difference equation, we will approximate the time derivative using a forward difference scheme, which gives: \(\frac{\partial T}{\partial t} \approx \frac{T^{n+1}(x, y) - T^n(x, y)}{\Delta t}\) Substitute this back into the simplified equation: \({\left[\frac{k}{(\Delta x)^2}\right](T^n(x+\Delta x, y) - 2T^n(x, y) + T^n(x-\Delta x, y))+\left[\frac{k}{(\Delta y)^2}\right](T^n(x, y+\Delta y) - 2T^n(x, y) + T^n(x, y- \Delta y)) = \rho c_p \frac{T^{n+1}(x, y) - T^n(x, y)}{\Delta t}}\) Now isolate the term \(T^{n+1}(x, y)\) on the left-hand side: \(T^{n+1}(x, y) = \frac{\Delta t \left[\frac{k}{(\Delta x)^2}\right](T^n(x+\Delta x, y) - 2T^n(x, y) + T^n(x-\Delta x, y))+\frac{\Delta t \left[\frac{k}{(\Delta y)^2}\right](T^n(x, y+\Delta y) - 2T^n(x, y) + T^n(x, y- \Delta y))}{\rho c_p} + T^n(x, y)\) This is the final two-dimensional transient explicit finite difference equation for a general interior node in rectangular coordinates with constant thermal conductivity and no heat generation.