/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 162 The conduction equation boundary... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 162

The conduction equation boundary condition for an adiabatic surface with direction \(n\) being normal to the surface is (a) \(T=0\) (b) \(d T / d n=0\) (c) \(d^{2} T / d n^{2}=0\) (d) \(d^{3} T / d n^{3}=0\) (e) \(-k d T / d n=1\)

Short Answer

Expert verified
Answer: \(\frac{dT}{dn}=0\)

Step by step solution

01

Understand the concept of adiabatic surface

An adiabatic surface is defined as a surface with no heat transfer across it. In other words, the rate of heat transfer through the surface is zero. To find the correct boundary condition for an adiabatic surface, we need to analyze each option in terms of heat transfer.
02

Analyze each option

Using the concept of no heat transfer across an adiabatic surface, we can determine which option corresponds to this condition. (a) \(T=0\): This states that the temperature of the surface is zero. This does not necessarily imply that there is no heat transfer across the surface, as the temperature could be non-zero inside the material. (b) \(\frac{dT}{dn}=0\): This states that the temperature gradient in the direction normal to the surface (n) is zero. As Fourier's Law states that the rate of heat transfer through a material is proportional to the temperature gradient, a zero gradient means no heat transfer across the adiabatic surface. This is the correct choice. (c) \(\frac{d^2T}{dn^2}=0\): This refers to the curvature of the temperature profile in the direction normal to the surface. This has no direct relation to the heat transfer across the surface. (d) \(\frac{d^3T}{dn^3}=0\): This refers to the rate of change of curvature of the temperature profile in the direction normal to the surface. This has no direct relation to the heat transfer across the surface. (e) \(-k\frac{dT}{dn}=1\): This states that the heat flux (rate of heat transfer per unit area) across the surface is equal to 1. This contradicts the definition of an adiabatic surface, which has no heat transfer.
03

Identify the correct boundary condition

Based on the analysis in Step 2, we can conclude that the correct conduction equation boundary condition for an adiabatic surface with direction \(n\) being normal to the surface is: (b) \(\frac{dT}{dn}=0\)

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 91影视!

Key Concepts

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

Heat Transfer
Heat transfer is a process in which thermal energy is exchanged between physical systems. There are three primary modes of heat transfer: conduction, convection, and radiation. In this discussion, we'll focus on conduction.

Conduction occurs when heat is transferred through a material without the movement of the material itself. It relies on the interaction between particles in a substance, such as atoms or molecules. These particles collide, transferring energy from the hotter region to the cooler one.

In an adiabatic surface, which is a surface that does not permit any heat to pass through, the heat transfer rate is effectively zero. This means the system is perfectly insulated, preventing any loss or gain of heat. Understanding this concept is crucial in fields such as chemistry and engineering, as it allows for the design of efficient insulation systems.
Temperature Gradient
The temperature gradient is a physical quantity that describes the direction and rate of temperature change in a particular region. It is defined mathematically as the difference in temperature per unit distance. In simpler terms, it tells us how the temperature varies from one point to another.

A high temperature gradient indicates a sharp change in temperature, which could cause faster heat transfer, while a low temperature gradient shows a more gradual change. Typically, in the presence of heat transfer through a material, you'll find that the temperature gradient guides the flow of heat, moving from hot to cold regions. This gradient is crucial because it represents the driving force behind heat conduction between materials.
  • In an adiabatic surface, the temperature gradient normal to the surface is zero, ensuring no heat transfer.
  • Understanding gradients helps in analyzing thermal systems, where controlling temperature is critical for efficiency.
Fourier's Law
Fourier's Law is a fundamental principle that describes the conduction of heat within solid materials. Formulated by Jean-Baptiste Joseph Fourier, this law relates the rate of heat transfer through a material to its temperature gradient and its thermal conductivity, denoted as \( k \).

The mathematical expression of Fourier's Law is given by:\[ q = -k \frac{dT}{dx} \]where:
  • \( q \) is the heat flux, the heat transfer per unit area over time.
  • \( k \) represents the thermal conductivity of the material, indicating its ability to conduct heat.
  • \( \frac{dT}{dx} \) denotes the temperature gradient in the direction of heat flow.
In the context of adiabatic surfaces, applying Fourier's Law helps confirm the absence of heat transfer since the temperature gradient, \( \frac{dT}{dn} \), is zero, leading to zero flux. This theory is pivotal in engineering and physics, as it helps predict how heat diffuses through different materials under thermal gradients.

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 difference between an algebraic equation and a differential equation?

What is heat generation in a solid? Give examples.

When a long section of a compressed air line passes through the outdoors, it is observed that the moisture in the compressed air freezes in cold weather, disrupting and even completely blocking the air flow in the pipe. To avoid this problem, the outer surface of the pipe is wrapped with electric strip heaters and then insulated. Consider a compressed air pipe of length \(L=6 \mathrm{~m}\), inner radius \(r_{1}=3.7 \mathrm{~cm}\), outer radius \(r_{2}=4.0 \mathrm{~cm}\), and thermal conductivity \(k=14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) equipped with a 300 -W strip heater. Air is flowing through the pipe at an average temperature of \(-10^{\circ} \mathrm{C}\), and the average convection heat transfer coefficient on the inner surface is \(h=30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming 15 percent of the heat generated in the strip heater is lost through the insulation, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the pipe, \((b)\) obtain a relation for the variation of temperature in the pipe material by solving the differential equation, and \((c)\) evaluate the inner and outer surface temperatures of the pipe.

Starting with an energy balance on a ring-shaped volume element, derive the two-dimensional steady heat conduction equation in cylindrical coordinates for \(T(r, z)\) for the case of constant thermal conductivity and no heat generation.

In a nuclear reactor, heat is generated uniformly in the 5 -cm-diameter cylindrical uranium rods at a rate of \(2 \times 10^{8} \mathrm{~W} / \mathrm{m}^{3}\). If the length of the rods is \(1 \mathrm{~m}\), determine the rate of heat generation in each rod. Answer: \(393 \mathrm{~kW}\)
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Work Energy and Power

Read Explanation

Mechanics and Materials

Read Explanation

Magnetism

Read Explanation

Solid State Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation
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.