/*! 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 13 A wall is made from an inhomogen... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 13

A wall is made from an inhomogeneous (nonuniform) material for which the thermal conductivity varies through the thickness according to \(k=a x+b\), where \(a\) and \(b\) are constants. The heat flux is known to be constant. Determine expressions for the temperature gradient and the temperature distribution when the surface at \(x=0\) is at temperature \(T_{1}\).

Short Answer

Expert verified
The temperature gradient and temperature distribution expressions for an inhomogeneous material with a constant heat flux and thermal conductivity varying according to \(k = ax + b\) are: Temperature gradient: \(\frac{dT}{dx} = - \frac{q}{ax + b}\) Temperature distribution: \(T = - \frac{q}{a} \ln{(ax + b)} + T_1 + \frac{q}{a} \ln{(b)}\)

Step by step solution

01

Identify the governing equation

We'll start with the heat conduction equation, which is given by the Fourier's law: \(q = -k \frac{dT}{dx}\) where q is heat flux, k is the thermal conductivity, and \(dT/dx\) is the temperature gradient. Since the thermal conductivity is given as a function of x, we can write: \(k = ax + b\) The heat flux is constant, so we can rewrite the heat conduction equation as: \(q = - (ax + b) \frac{dT}{dx}\)
02

Solve for the temperature gradient

Now, we'll solve for the temperature gradient, \(dT/dx\). Rearrange the equation to isolate the temperature gradient: \(\frac{dT}{dx} = - \frac{q}{ax + b}\)
03

Integrate to find the temperature distribution

To find the temperature distribution, we'll now integrate both sides of the equation with respect to x: \(\int dT = -q \int \frac{dx}{ax + b}\) First, we can solve the right-hand side integral using substitution. Let \(u = ax + b\) \(du = adx\) \(dx = \frac{du}{a}\) The integral becomes: \(-q \int \frac{du}{au}\) Now, let's integrate: \(\int dT = - \frac{q}{a} \int \frac{du}{u}\) Integrate both sides to find T as a function of x: \(T = - \frac{q}{a} \ln{(|u|)} + C\) where C is the constant of integration. Now we can substitute u back and get the temperature distribution formula: \(T = - \frac{q}{a} \ln{(ax + b)} + C\)
04

Apply the boundary condition

We're given the boundary condition that when x = 0, the temperature T = \(T_1\). We can use this information to find C: \(T_1 = - \frac{q}{a} \ln{(a(0) + b)} + C\) \(T_1 = - \frac{q}{a} \ln{(b)} + C\) Rearrange to find the value of the constant C: \(C = T_1 + \frac{q}{a} \ln{(b)}\)
05

Write the final expressions

Now, we have the expressions for the temperature gradient and the temperature distribution: Temperature gradient: \(\frac{dT}{dx} = - \frac{q}{ax + b}\) Temperature distribution: \(T = - \frac{q}{a} \ln{(ax + b)} + T_1 + \frac{q}{a} \ln{(b)}\) And these are the expressions for the temperature gradient and temperature distribution for the given inhomogeneous material with a constant heat flux.

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!

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

A freezer compartment is covered with a 2 -mm-thick layer of frost at the time it malfunctions. If the compartment is in ambient air at \(20^{\circ} \mathrm{C}\) and a coefficient of \(h=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) characterizes heat transfer by natural convection from the exposed surface of the layer, estimate the time required to completely melt the frost. The frost may be assumed to have a mass density of \(700 \mathrm{~kg} / \mathrm{m}^{3}\) and a latent heat of fusion of \(334 \mathrm{~kJ} / \mathrm{kg}\).

The thermal conductivity of a sheet of rigid, extruded insulation is reported to be \(k=0.029 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The measured temperature difference across a 20 -mm-thick sheet of the material is \(T_{1}-T_{2}=10^{\circ} \mathrm{C}\). (a) What is the heat flux through a \(2 \mathrm{~m} \times 2 \mathrm{~m}\) sheet of the insulation? (b) What is the rate of heat transfer through the sheet of insulation?

The free convection heat transfer coefficient on a thin hot vertical plate suspended in still air can be determined from observations of the change in plate temperature with time as it cools. Assuming the plate is isothermal and radiation exchange with its surroundings is negligible, evaluate the convection coefficient at the instant of time when the plate temperature is \(225^{\circ} \mathrm{C}\) and the change in plate temperature with time \((d T / d t)\) is \(-0.022 \mathrm{~K} / \mathrm{s}\). The ambient air temperature is \(25^{\circ} \mathrm{C}\) and the plate measures \(0.3 \times 0.3 \mathrm{~m}\) with a mass of \(3.75 \mathrm{~kg}\) and a specific heat of \(2770 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

The temperature controller for a clothes dryer consists of a bimetallic switch mounted on an electrical heater attached to a wall-mounted insulation pad. The switch is set to open at \(70^{\circ} \mathrm{C}\), the maximum dryer air temperature. To operate the dryer at a lower air temperature, sufficient power is supplied to the heater such that the switch reaches \(70^{\circ} \mathrm{C}\left(T_{\text {set }}\right)\) when the air temperature \(T\) is less than \(T_{\text {set. }}\). If the convection heat transfer coefficient between the air and the exposed switch surface of \(30 \mathrm{~mm}^{2}\) is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how much heater power \(P_{e}\) is required when the desired dryer air temperature is \(T_{\infty}=50^{\circ} \mathrm{C}\) ?

A square silicon chip \((k=150 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is of width \(w=5 \mathrm{~mm}\) on a side and of thickness \(t=1 \mathrm{~mm}\). The chip is mounted in a substrate such that its side and back surfaces are insulated, while the front surface is exposed to a coolant. If \(4 \mathrm{~W}\) are being dissipated in circuits mounted to the back surface of the chip, what is the steady-state temperature difference between back and front surfaces?
See all solutions

Recommended explanations on Physics Textbooks

Translational Dynamics

Read Explanation

Electricity and Magnetism

Read Explanation

Quantum Physics

Read Explanation

Kinematics Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Dynamics

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.