Starting with an energy balance on the volume element, obtain the three- dimensional transient explicit finite difference equation for a general interior node in rectangular coordinates for \(T(x, y, z, t)\) for the case of constant thermal conductivity and no heat generation.

Short Answer

Expert verified
Based on the solution above, the three-dimensional transient explicit finite difference equation for a general interior node in rectangular coordinates for the temperature function T(x, y, z, t) in the case of constant thermal conductivity and no heat generation is: $$T_{i,j,k}^{n+1} = T_{i,j,k}^n + \frac{k \Delta t}{\rho c_p V} \left(\frac{A}{\Delta x^2}(T_{i+1,j,k}^{n} - 2T_{i,j,k}^n + T_{i-1,j,k}^n) + \frac{A}{\Delta y^2}(T_{i,j+1,k}^{n}-2T_{i,j,k}^n+T_{i,j-1,k}^n) + \frac{A}{\Delta z^2}(T_{i,j,k+1}^n-2T_{i,j,k}^n+T_{i,j,k-1}^n)\right)$$

Step by step solution

01

Energy Balance on the Volume Element

Since there is no heat generation, the energy balance on a volume element can be expressed as: $$\frac{\partial(\rho c_p T)}{\partial t} V = \dot{Q}_{in} - \dot{Q}_{out}$$ Where: - \(\rho\) is density - \(c_p\) is specific heat capacity at constant pressure - T is the temperature - V is the volume of the element - \(\dot{Q}_{in}\) and \(\dot{Q}_{out}\) are the heat fluxes entering and leaving the volume, respectively.
02

Heat Fluxes

The heat fluxes can be expressed as: $$\dot{Q}_{in} = kA\left(\frac{\partial T}{\partial x}\Bigg|_{x-\Delta x/2} + \frac{\partial T}{\partial y}\Bigg|_{y-\Delta y/2} + \frac{\partial T}{\partial z}\Bigg|_{z-\Delta z/2}\right)$$ $$\dot{Q}_{out} = kA\left(\frac{\partial T}{\partial x}\Bigg|_{x+\Delta x/2} + \frac{\partial T}{\partial y}\Bigg|_{y+\Delta y/2} + \frac{\partial T}{\partial z}\Bigg|_{z+\Delta z/2}\right)$$ Where: - k is the thermal conductivity (constant in this case) - A is the cross-sectional area - \(\Delta x\), \(\Delta y\), and \(\Delta z\) are the dimensions of the volume element in x, y, and z, respectively.
03

Applying Finite Difference Approximations

We can approximate the partial derivatives with finite difference approximations: $$\frac{\partial T}{\partial x}\Bigg|_{x\pm\Delta x/2} \approx \frac{T_{i\pm1,j,k} - T_{i,j,k}}{\Delta x}$$ $$\frac{\partial T}{\partial y}\Bigg|_{y\pm\Delta y/2} \approx \frac{T_{i,j\pm1,k} - T_{i,j,k}}{\Delta y}$$ $$\frac{\partial T}{\partial z}\Bigg|_{z\pm\Delta z/2} \approx \frac{T_{i,j,k\pm1} - T_{i,j,k}}{\Delta z}$$
04

Substituting Finite Difference Approximations into Energy Balance

Substituting the finite difference approximations into the energy balance equation and rearranging terms, we obtain: $$\frac{\rho c_p V}{\Delta t}(T_{i,j,k}^{n+1}-T_{i,j,k}^n) = kA\left(\frac{T_{i+1,j,k}^{n}-T_{i,j,k}^n}{\Delta x} - \frac{T_{i-1,j,k}^n-T_{i,j,k}^n}{\Delta x} + \frac{T_{i,j+1,k}^n-T_{i,j,k}^n}{\Delta y} - \frac{T_{i, j-1, k}^n-T_{i, j, k}^n}{\Delta y}+\frac{T_{i,j,k+1}^n-T_{i,j,k}^n}{\Delta z} - \frac{T_{i,j,k-1}^n-T_{i,j,k}^n}{\Delta z}\right)$$
05

Simplifying Equation

Now we can simplify the equation: $$T_{i,j,k}^{n+1} = T_{i,j,k}^n + \frac{k \Delta t}{\rho c_p V} \left(\frac{A}{\Delta x^2}(T_{i+1,j,k}^{n} - 2T_{i,j,k}^n + T_{i-1,j,k}^n) + \frac{A}{\Delta y^2}(T_{i,j+1,k}^{n}-2T_{i,j,k}^n+T_{i,j-1,k}^n) + \frac{A}{\Delta z^2}(T_{i,j,k+1}^n-2T_{i,j,k}^n+T_{i,j,k-1}^n)\right)$$ This is the three-dimensional transient explicit finite difference equation for a general interior node in rectangular coordinates for \(T(x, y, z, t)\) in the case of constant thermal conductivity and no heat generation.

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Key Concepts

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

Energy Balance
Understanding the concept of energy balance is fundamental in many fields of engineering and physics, especially concerning heat transfer problems. Energy balance is a principle stating that energy cannot be created or destroyed, only transferred or converted from one form to another. In the context of heat transfer, and more specifically within the finite difference method, an energy balance involves accounting for the energy entering and leaving a control volume.

For a volume element within a material, the balance would equate the change in internal energy of the volume with the net heat flux: energy entering the volume minus energy leaving it. When we say that energy is 'balanced,' we are essentially saying that all the heat entering a volume element minus all the heat leaving it, must equal the change in energy stored within that volume due to temperature changes over time.

The mathematical representation of this concept, as shown in the exercise, involves the density \(\rho\), specific heat \(c_p\), and temperature \(T\), which together define the energy content of the volume element. The thermal energy change in the volume over time \(\partial(\rho c_p T)/\partial t\ times V\) corresponds to the net heat flux difference \(\dot{Q}_{in} - \dot{Q}_{out}\). This equation forms the basis for deriving the finite difference equations used in modeling heat transfer in materials.
Thermal Conductivity
Thermal conductivity is a material property that quantifies the ability of a material to conduct heat. It is denoted by the symbol \(k\) and is often considered constant in many simple heat transfer problems — although in reality it can depend on factors such as temperature and the material phase.

In the context of the finite difference method for heat transfer, thermal conductivity plays a crucial role in defining how heat is exchanged between adjacent nodes in a material. For a given temperature gradient (the rate of temperature change with respect to distance), \(k\) determines the amount of heat flux through a material. The higher the thermal conductivity, the more efficient the material is in transferring heat.

Mathematically, in the equation used for heat flux, \(k\) multiplies the area \(A\) and the temperature gradient, thus scaling the heat transfer rate. In our exercise, the assumption of constant thermal conductivity simplifies the equation by allowing us to treat \(k\) as a constant rather than a variable, streamlining the calculation process for transient heat conduction analysis in a material.
Transient Heat Conduction
The term transient heat conduction refers to the situation where the temperature within a material changes with time. Unlike steady-state heat conduction where temperatures are constant over time, in transient heat conduction scenarios, we must consider how temperatures evolve as a result of thermal energy storage and transfer.

Finite difference methods are particularly well-suited for solving transient heat conduction problems. They convert partial differential equations into algebraic equations that can be solved iteratively for each point in a discretized domain, advancing in time steps. The explicit finite difference scheme used in our exercise is a straightforward approach where the temperature at the next time step is calculated based on the current temperatures and the energy balance at that node.

The core of this method is an approximation that replaces continuous spatial derivatives with ratios of temperature differences at discrete points, allowing us to express the change in temperature at a point as a function of the surrounding temperatures, the material properties, and the time step. This is essential for predicting how heat will flow through a material over time and for understanding how quickly a material will respond to changes in temperature or heat input.

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Most popular questions from this chapter

Consider a house whose windows are made of \(0.375\)-in-thick glass \(\left(k=0.48 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\) and \(\alpha=\) \(4.2 \times 10^{-6} \mathrm{ft}^{2} / \mathrm{s}\) ). Initially, the entire house, including the walls and the windows, is at the outdoor temperature of \(T_{o}=35^{\circ} \mathrm{F}\). It is observed that the windows are fogged because the indoor temperature is below the dew-point temperature of \(54^{\circ} \mathrm{F}\). Now the heater is turned on and the air temperature in the house is raised to \(T_{i}=72^{\circ} \mathrm{F}\) at a rate of \(2^{\circ} \mathrm{F}\) rise per minute. The heat transfer coefficients at the inner and outer surfaces of the wall can be taken to be \(h_{i}=1.2\) and \(h_{o}=2.6 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\), respectively, and the outdoor temperature can be assumed to remain constant. Using the explicit finite difference method with a mesh size of \(\Delta x=0.125\) in, determine how long it will take for the fog on the windows to clear up (i.e., for the inner surface temperature of the window glass to reach \(54^{\circ} \mathrm{F}\) ).

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 one-dimensional heat conduction in a pin fin of constant diameter \(D\) with constant thermal conductivity. The fin is losing heat by convection to the ambient air at \(T_{\infty}\) with a heat transfer coefficient of \(h\) and by radiation to the surrounding surfaces at an average temperature of \(T_{\text {surr }}\). The nodal network of the fin consists of nodes 0 (at the base), 1 (in the middle), and 2 (at the fin tip) with a uniform nodal spacing of \(\Delta x\). Using the energy balance approach, obtain the explicit finite difference formulation of this problem for the case of a specified temperature at the fin base and negligible heat transfer at the fin tip.

What is a practical way of checking if the discretization error has been significant in calculations?

Design a fire-resistant safety box whose outer dimensions are \(0.5 \mathrm{~m} \times 0.5 \mathrm{~m} \times 0.5 \mathrm{~m}\) that will protect its combustible contents from fire which may last up to \(2 \mathrm{~h}\). Assume the box will be exposed to an environment at an average temperature of \(700^{\circ} \mathrm{C}\) with a combined heat transfer coefficient of \(70 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the temperature inside the box must be below \(150^{\circ} \mathrm{C}\) at the end of \(2 \mathrm{~h}\). The cavity of the box must be as large as possible while meeting the design constraints, and the insulation material selected must withstand the high temperatures to which it will be exposed. Cost, durability, and strength are also important considerations in the selection of insulation materials.

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