Starting with an energy balance on a disk volume element, derive the one- dimensional transient implicit finite difference equation for a general interior node for \(T(z, t)\) in a cylinder whose side surface is subjected to convection with a convection coefficient of \(h\) and an ambient temperature of \(T_{\infty}\) for the case of constant thermal conductivity with uniform heat generation.

Short Answer

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#Summary# In this exercise, we derived a one-dimensional transient implicit finite difference equation for a general interior node to find the temperature, \(T(z, t)\), within a cylinder. We began by defining the cylinder's geometry and establishing the energy balance equation for a disk volume element. We then applied the transient heat conduction governing equation and discretized it using the finite difference method. Finally, we applied the convection boundary condition at the nodes adjacent to the surface. The resulting equation represents the temperature within the cylinder, taking into account the thermal conductivity, heat generation, and convection heat transfer from the cylinder's side surface.

Step by step solution

01

Define cylinder's geometry and establish the energy balance equation

We start by considering a cylindrical coordinate system for our problem. Let \(z\) represent the spatial coordinate in the axial direction and \(t\) denote time. A disk volume element has a thickness of \(\Delta z\) in the \(z\) direction. The energy balance equation for the disk volume element can be given as: $$\frac{\partial}{\partial z}\left( k\frac{\partial T}{\partial z} \right) - q_g + hP(T_{\infty} - T) = \rho c_p \frac{\partial T}{\partial t}$$ where: - \(k\) is the constant thermal conductivity, - \(T\) is the temperature, - \(z\) is the axial coordinate, - \(q_g\) is the uniform heat generation, - \(h\) is the convection coefficient, - \(T_{\infty}\) is the ambient temperature, - \(P\) is the perimeter of the disk, - \(\rho\) is the density of the material, - \(c_p\) is the specific heat capacity, and - \(t\) is time.
02

Apply the transient heat conduction governing equation

The equation derived in Step 1 represents the transient heat conduction governing equation with heat generation and convection heat transfer on the side surface of the cylinder. It can be further simplified to: $$\frac{\partial^2 T}{\partial z^2} = \frac{1}{\alpha}\frac{\partial T}{\partial t} + \frac{q_g - hP(T_{\infty} - T)}{k}$$ where \(\alpha = \frac{k}{\rho c_p}\) is the thermal diffusivity.
03

Discretize the governing equation

Now we need to discretize the governing equation using the finite difference method. We can replace the second derivative in space with a central difference approximation: $$\frac{\partial^2 T}{\partial z^2} \approx \frac{T_{i+1}-2T_i+T_{i-1}}{(\Delta z)^2}$$ and the time derivative with a forward difference approximation: $$\frac{\partial T}{\partial t} \approx \frac{T_i^{n+1}-T_i^n}{\Delta t}$$ where \(i\) is the index for spatial nodes and \(n\) is the index for time steps. Substitute the approximated derivatives back into the governing equation and rearrange: $$T_{i+1}^{n+1} - 2T_i^{n+1} + T_{i-1}^{n+1} = \frac{\alpha \Delta t}{(\Delta z)^2}(T_i^{n+1} - T_i^n) + \frac{\Delta t}{k}(q_g - hP(T_{\infty} - T_i^{n+1}))$$
04

Apply the convection boundary condition

Since the side surface of the cylinder is subjected to convection, we need to apply a convection boundary condition at the nodes adjacent to the surface. This can be done by using the Robin boundary condition: $$k\frac{\partial T}{\partial z} + h(T-T_{\infty}) = 0$$ The discretized equation derived in Step 3, along with the convection boundary condition, forms the one-dimensional transient implicit finite difference equation for a general interior node for \(T(z, t)\) in the cylinder.

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Most popular questions from this chapter

Explain why the local discretization error of the finite difference method is proportional to the square of the step size. Also explain why the global discretization error is proportional to the step size itself.

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