Consider a medium in which the finite difference formulation of a general interior node is given in its simplest form as $$ \frac{T_{m-1}-2 T_{m}+T_{m+1}}{\Delta x^{2}}+\frac{\dot{e}_{m}}{k}=0 $$ (a) Is heat transfer in this medium steady or transient? (b) Is heat transfer one-, two-, or three-dimensional? (c) Is there heat generation in the medium? (d) Is the nodal spacing constant or variable? (e) Is the thermal conductivity of the medium constant or variable?

Short Answer

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(b) Is the heat transfer one-, two-, or three-dimensional? (c) Is there heat generation in the medium? (d) Is the nodal spacing constant or variable? (e) Is the thermal conductivity of the medium constant or variable? Answer: (a) Heat transfer is steady. (b) Heat transfer is one-dimensional. (c) There is heat generation in the medium. (d) The nodal spacing is constant. (e) The thermal conductivity of the medium is constant.

Step by step solution

01

Break down the given equation

The finite difference formulation for a general interior node is provided as: $$ \frac{T_{m-1}-2 T_{m}+T_{m+1}}{\Delta x^{2}}+\frac{\dot{e}_{m}}{k}=0 $$ Here, \(T_{m-1}\), \(T_{m}\), and \(T_{m+1}\) are temperatures at adjacent nodes, \(\Delta x\) is the nodal spacing, \(\dot{e}_{m}\) is the heat generation rate in the medium, and \(k\) is the thermal conductivity of the medium.
02

Determine whether heat transfer is steady or transient

Since the given equation does not involve any time dependency (i.e., no derivatives with respect to time are present), we can conclude that heat transfer is steady. Answer (a): Heat transfer is steady.
03

Determine the dimensionality of heat transfer

The given equation only has one spatial variable (\(\Delta x\)), which indicates that heat transfer is occurring in only one direction. Therefore, heat transfer is one-dimensional. Answer (b): Heat transfer is one-dimensional.
04

Determine whether there is heat generation in the medium

The term \(\frac{\dot{e}_{m}}{k}\) in the given equation represents the heat generation rate per unit volume in the medium. Since this term is present in the equation, we can conclude that there is heat generation in the medium. Answer (c): There is heat generation in the medium.
05

Determine whether the nodal spacing is constant or variable

In the given equation, the nodal spacing \(\Delta x\) is constant, as the equation is written as \(\frac{T_{m-1}-2 T_{m}+T_{m+1}}{\Delta x^{2}}\). The constant \(\Delta x^{2}\) implies that the nodal spacing is uniform. Answer (d): The nodal spacing is constant.
06

Determine whether the thermal conductivity is constant or variable

The thermal conductivity, \(k\), appears as a denominator in the term \(\frac{\dot{e}_{m}}{k}\). Since the value of \(k\) does not change with respect to space (i.e., it does not have any subscripts related to node indices), we can conclude that the thermal conductivity of the medium is constant. Answer (e): The thermal conductivity of the medium is constant.

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Key Concepts

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

Steady Heat Transfer
When we discuss steady heat transfer, we're referring to a scenario where the temperature within a material does not change with time. In the provided exercise, the absence of any derivative with respect to time in the finite difference equation indicates that we're dealing with a steady-state problem. This means that, regardless of the time at which you measure the temperature at any point in the medium, you will get the same value.

In practical terms, if you're trying to keep your coffee at a constant warmth using a mug warmer, steady heat transfer would imply that, once your coffee reaches the desired temperature, it stays at that temperature, neither cooling down nor heating up further, as long as the mug warmer is functioning correctly.
One-Dimensional Heat Transfer
Concerning one-dimensional heat transfer, it simplifies the complex reality by assuming that temperature varies only in one direction and is uniform in the perpendicular directions. This assumption is valid when the geometry of the problem or the boundary conditions force the temperature to change significantly in one direction while reducing changes in other directions to a minimal.

In our textbook problem, the presence of a single spatial variable, \(\Delta x\), supports the idea that temperature changes occur along one dimension only. Imagine a long, thin rod being heated at one end; we would only be concerned about how the temperature varies along the length of the rod (one dimension), rather than across its width or depth.
Heat Generation in Medium
The term heat generation in medium refers to the production of heat within the material itself due to some internal processes, like chemical reactions, electrical resistance, or nuclear reactions. In the problem at hand, the inclusion of a term for heat generation rate, \(\dot{e}_m/k\), suggests that heat is being produced internally within the medium.

For instance, the heating element of an electric stove generates heat which then conducts to the pot on top of it. Our equation mathematically models such situations where a medium not only conducts heat but also internally generates it, contributing to the overall temperature profile.
Constant Nodal Spacing
The constant nodal spacing is a term we use to describe situations where the points at which we're measuring or calculating temperature—the nodes—are equally spaced. It's like having a ruler with marks at regular intervals; each mark represents a node. The equation provided has a uniform \(\Delta x^2\), indicating that the distance between nodes is consistent throughout the medium.

Consistent nodal spacing simplifies the mathematics of heat transfer problems and is often an assumption in one-dimensional analyses, which helps when creating computer models for predicting temperature distribution, like in the cooling of electronic components arranged in a regular grid pattern.
Constant Thermal Conductivity
Lastly, constant thermal conductivity within a medium implies that the material's ability to conduct heat does not change throughout the medium or with temperature. This is a common assumption in many heat transfer problems to simplify calculations. In the equation from our exercise, the thermal conductivity, \(k\), is a denominator applied uniformly across the medium.

This assumption would apply well to situations where temperature differences are not extreme, such as in walls of a house for insulation purposes. Here, the insulation has the same conductive properties in all areas, ensuring a predictable rate of heat loss through the walls.

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Most popular questions from this chapter

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).

Consider a long concrete dam \((k=0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), (es \(\alpha_{s}=0.7\) ) of triangular cross section whose exposed surface is subjected to solar heat flux of \(\dot{q}_{s}=\) \(800 \mathrm{~W} / \mathrm{m}^{2}\) and to convection and radiation to the environment at \(25^{\circ} \mathrm{C}\) with a combined heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The \(2-\mathrm{m}\)-high vertical section of the dam is subjected to convection by water at \(15^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and heat transfer through the 2-m-long base is considered to be negligible. Using the finite difference method with a mesh size of \(\Delta x=\Delta y=1 \mathrm{~m}\) and assuming steady two-dimensional heat transfer, determine the temperature of the top, middle, and bottom of the exposed surface of the dam.

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 \(54^{\circ} \mathrm{F}\) ).

Starting with an energy balance on a volume element, obtain the steady two- dimensional finite difference equation for a general interior node in rectangular coordinates for \(T(x, y)\) for the case of variable thermal conductivity and uniform heat generation.

Consider transient heat conduction in a plane wall whose left surface (node 0 ) is maintained at \(50^{\circ} \mathrm{C}\) while the right surface (node 6) is subjected to a solar heat flux of \(600 \mathrm{~W} / \mathrm{m}^{2}\). The wall is initially at a uniform temperature of \(50^{\circ} \mathrm{C}\). Express the explicit finite difference formulation of the boundary nodes 0 and 6 for the case of no heat generation. Also, obtain the finite difference formulation for the total amount of heat transfer at the left boundary during the first three time steps.

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