Consider a medium in which the finite difference formulation of a general interior node is given in its simplest form as $$ T_{\text {left }}+T_{\text {top }}+T_{\text {right }}+T_{\text {bottom }}-4 T_{\text {node }}+\frac{\dot{e}_{\text {node }} l^{2}}{k}=0 $$ (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

Expert verified
Question: Analyze the given finite difference formulation for a general interior node in a medium and determine the properties of heat transfer in this medium. Answer: Based on the analysis, the properties of heat transfer in the medium are as follows: (a) The heat transfer is steady. (b) Heat transfer is two-dimensional. (c) There is heat generation in the medium. (d) Nodal spacing is constant. (e) Thermal conductivity is constant.

Step by step solution

01

(a) Determining whether heat transfer is steady or transient

To determine if the heat transfer is steady or transient, we need to see if there's a time derivative term in the given equation. A time derivative term would indicate that the properties of the medium change with time, making the heat transfer transient. In this case, the given equation has no time derivative term, so the heat transfer is steady.
02

(b) Determining the dimensionality of heat transfer

To determine the dimensionality of heat transfer, we need to look at the terms in the equation. The terms T_left, T_top, T_right, and T_bottom indicate that heat is conducted in multiple directions: left and right as well as top and bottom. Since heat transfer occurs in both horizontal and vertical directions, we can conclude that it is two-dimensional. If heat transfer occurred in only one direction, it would be one-dimensional, and if it happened in all three directions, it would be three-dimensional.
03

(c) Determining if there's heat generation in the medium

To determine if there's heat generation in the medium, we have to look for a term in the equation that represents heat generation. In this case, the term \(\frac{\dot{e}_{\text {node }} l^{2}}{k}\) represents heat generation. Since this term is present in the given equation, we can conclude that there is heat generation in the medium.
04

(d) Determining if nodal spacing is constant or variable

The nodal spacing is the distance between nodes in the medium. If the distance was changing, the terms in the equation should be in the form of ratios or expressions representing different distances. However, in the given equation, there are no such expressions or ratios. It is a simple linear equation indicating that nodal spacing is constant throughout the medium.
05

(e) Determining if thermal conductivity is constant or variable

To determine if the thermal conductivity is constant or variable, we must look at the given equation. In this equation, the thermal conductivity, denoted by 'k', is present and has no variables associated with it. Since it doesn't appear to be changing with respect to any parameters, we can conclude that the thermal conductivity is constant in the medium.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Understanding Steady-State Heat Transfer
When we discuss steady-state heat transfer, we are referring to a condition where the temperature at any given point in the medium does not change with time. Unlike transient heat transfer, where temperatures vary with time due to changes in the environment or operating conditions, steady-state implies a consistent temperature field.

In our exercise, the lack of a time derivative in the finite difference equation indicates that the temperatures at the nodes are constant over time. This is crucial for simplifying the calculations required for determining temperatures within the system because once the steady state is reached, the system's behavior becomes predictable. This is often desired in engineering applications such as heat sinks or insulation materials where maintaining a constant temperature is essential.
Deciphering Two-Dimensional Heat Transfer
Heat transfer can occur in one, two, or three dimensions. Two-dimensional heat transfer involves the flow of heat across two perpendicular axes—in this case, the left-right and the top-bottom axes.

In our exercise, each term (T_left, T_top, T_right, T_bottom) represents the temperature at each of the four sides of a node, suggesting that the heat is being conducted in a plane along both axes. Knowledge of two-dimensional heat transfer is useful in applications like printed circuit boards or floor heating systems, where heat spreads over a surface rather than just in a line or throughout a volume.
Heat Generation within the Medium
When we speak of heat generation in the context of a medium, we are referring to the internal production of heat within the material, which could be due to chemical reactions, electrical energy, or radioactive decay, to name a few.

The term \(\frac{\dot{e}_{\text {node }} l^{2}}{k}\) in our exercise signifies the heat being generated at the node. This concept is pivotal in understanding and managing thermal behavior in mediums such as nuclear fuel rods where heat is generated from fission reactions or in electronic devices where electrical resistance turns energy into heat.
Constant Nodal Spacing
Constant nodal spacing means that the distance between the nodes in a grid used for the finite difference model is the same throughout the medium.

For the problem we are analyzing, the simplicity of the finite difference equation implies that the distance between each node, and consequently between each temperature measurement point, is uniform. This assumption is particularly important for simplifying the calculations involved in modeling the heat transfer process, as variable spacing would require a more complex treatment of the heat transfer equation.
Constant Thermal Conductivity
The assumption of constant thermal conductivity simplifies the analysis of heat transfer problems. Thermal conductivity, denoted by the symbol 'k', is a measure of a material's ability to conduct heat.

In the equation from our exercise, the thermal conductivity is considered to be a constant value across the medium. This implies that the heat transfer properties of the material do not change with temperature or position within the material. This assumption is valid for a wide range of engineering materials under certain conditions, and it allows for a more straightforward application of the finite difference method in calculating temperatures within the medium.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is the basis of the energy balance method? How does it differ from the formal finite difference method? For a specified nodal network, will these two methods result in the same or a different set of equations?

Consider steady heat conduction in a plane wall whose left surface (node 0 ) is maintained at \(30^{\circ} \mathrm{C}\) while the right surface (node 8 ) is subjected to a heat flux of \(1200 \mathrm{~W} / \mathrm{m}^{2}\). Express the finite difference formulation of the boundary nodes 0 and 8 for the case of no heat generation. Also obtain the finite difference formulation for the rate of heat transfer at the left boundary.

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).

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.

What are the limitations of the analytical solution methods?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free