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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Finite Difference Method
The Finite Difference Method (FDM) is a numerical approach commonly used in engineering and physics to solve differential equations. In the context of heat conduction, it provides a way to approximate the temperature distribution in a material over time without requiring an explicit analytical solution.
The main idea behind FDM is to discretize the continuous domain into a grid or mesh of nodes. Instead of dealing with derivatives directly, you approximate them using differences between the nodes. For example, the second derivative of temperature with respect to space can be approximated using a central difference formula like \[\frac{T_{i+1} - 2T_i + T_{i-1}}{(\Delta z)^2}\]where \(T_{i+1}\), \(T_i\), and \(T_{i-1}\) are the temperatures at adjacent nodes. This transformation turns the differential equations into algebraic equations that are easier to solve numerically.
  • Central Difference is used for spatial derivatives, providing higher accuracy.
  • Forward Difference may be used for time derivatives, often involving a time stepping scheme.
By applying these approximations and integrating boundary conditions, FDM helps simulate how heat spreads in different materials under various conditions, aiding in engineering designs and analyses.
Convection Heat Transfer
Convection Heat Transfer occurs when heat is transferred between a solid surface and a fluid (such as air or water) that is in motion across the surface. This movement enhances the heat transfer process compared to conduction alone. In many practical scenarios, including the one described in the original exercise, convection plays a significant role in defining temperature changes.
The effectiveness of convection is often characterized by a convection coefficient \(h\), which indicates how easily the material exchanges heat with the surrounding fluid. The convection equation is \[q = hA(T_s - T_{\infty})\]where \(q\) is the heat transfer rate, \(A\) is the surface area, \(T_s\) is the surface temperature, and \(T_{\infty}\) is the fluid temperature. This equation is essential when imposing boundary conditions in heat conduction problems, ensuring that surface temperature changes are considered.
  • Convection enhances heat dissipation from surfaces.
  • The convection coefficient \(h\) is a crucial parameter in calculations, reflecting fluid properties and flow characteristics.
By understanding and modeling convection, engineers can better predict and manage thermal systems, ensuring efficiency and safety.
Thermal Diffusivity
Thermal Diffusivity \(\alpha\) is a key property that measures how quickly heat spreads through a material. It combines the effects of thermal conductivity, density, and specific heat capacity into a single number, making it easier to compare different materials' responses to heat.
The formula for thermal diffusivity is given by\[\alpha = \frac{k}{\rho c_p}\]where \(k\) is the thermal conductivity, \(\rho\) is the density, and \(c_p\) is the specific heat capacity. Materials with high thermal diffusivity can adjust quickly to temperature changes, whereas low diffusivity materials require more time to adapt.
In transient heat conduction problems, \(\alpha\) plays a critical role in defining how temperature changes propagate through the material over time.
  • High \(\alpha\) means rapid heat spreading, while low \(\alpha\) indicates slow spreading.
  • It helps predict time-dependent temperature fields in response to boundary or initial changes.
Understanding thermal diffusivity is vital for selecting appropriate materials in thermal management applications and simulating their behavior accurately.

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Most popular questions from this chapter

Consider transient one-dimensional heat conduction in a plane wall that is to be solved by the explicit method. If both sides of the wall are subjected to specified heat flux, express the stability criterion for this problem in its simplest form.

Consider transient two-dimensional heat conduction in a rectangular region that is to be solved by the explicit method. If all boundaries of the region are either insulated or at specified temperatures, express the stability criterion for this problem in its simplest form.

The wall of a heat exchanger separates hot water at \(T_{A}=90^{\circ} \mathrm{C}\) from cold water at \(T_{B}=10^{\circ} \mathrm{C}\). To extend the heat transfer area, two-dimensional ridges are machined on the cold side of the wall, as shown in Fig. P5-76. This geometry causes non-uniform thermal stresses, which may become critical for crack initiation along the lines between two ridges. To predict thermal stresses, the temperature field inside the wall must be determined. Convection coefficients are high enough so that the surface temperature is equal to that of the water on each side of the wall. (a) Identify the smallest section of the wall that can be analyzed in order to find the temperature field in the whole wall. (b) For the domain found in part \((a)\), construct a twodimensional grid with \(\Delta x=\Delta y=5 \mathrm{~mm}\) and write the matrix equation \(A T=C\) (elements of matrices \(A\) and \(C\) must be numbers). Do not solve for \(T\). (c) A thermocouple mounted at point \(M\) reads \(46.9^{\circ} \mathrm{C}\). Determine the other unknown temperatures in the grid defined in part (b).

A circular fin \((k=240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of uniform cross section, with diameter of \(10 \mathrm{~mm}\) and length of \(50 \mathrm{~mm}\), is attached to a wall with surface temperature of \(350^{\circ} \mathrm{C}\). The fin tip has a temperature of \(200^{\circ} \mathrm{C}\), and it is exposed to ambient air condition of \(25^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assume steady one-dimensional heat transfer along the fin and the nodal spacing to be uniformly \(10 \mathrm{~mm}\), (a) using the energy balance approach, obtain the finite difference equations to determine the nodal temperatures, and (b) determine the nodal temperatures along the fin by solving those equations and compare the results with the analytical solution.

A hot surface at \(100^{\circ} \mathrm{C}\) is to be cooled by attaching 3 -cm- long, \(0.25\)-cm-diameter aluminum pin fins \((k=\) \(237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) with a center-to-center distance of \(0.6 \mathrm{~cm}\). The temperature of the surrounding medium is \(30^{\circ} \mathrm{C}\), and the combined heat transfer coefficient on the surfaces is \(35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming steady one-dimensional heat transfer along the fin and taking the nodal spacing to be \(0.5 \mathrm{~cm}\), determine \((a)\) the finite difference formulation of this problem, \((b)\) the nodal temperatures along the fin by solving these equations, \((c)\) the rate of heat transfer from a single fin, and \((d)\) the rate of heat transfer from a \(1-\mathrm{m} \times 1-\mathrm{m}\) section of the plate.

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