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Chapter 1: Problem 14

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 the temperature distribution for the inhomogeneous wall can be found using the following expressions: Temperature gradient: \(\frac{dT}{dx} = -\frac{q''}{ax+b}\) Temperature distribution: \(T(x) = T_1 - q''a\ln\left(\frac{ax+b}{b}\right)\)

Step by step solution

01

Write down Fourier's Law of Heat Conduction

Fourier's Law of heat conduction states that the heat flux, denoted as \(q''\), is proportional to the temperature gradient in the material. The mathematical expression for this law is: \(q'' = -k\frac{dT}{dx}\) Where \(q''\) is the heat flux, \(k\) is the thermal conductivity, and \(\frac{dT}{dx}\) is the temperature gradient with respect to \(x\).
02

Substitute the given thermal conductivity

Thermal conductivity varies according to \(k=ax+b\), where \(a\) and \(b\) are constants. Substitute this into Fourier's law: \(q'' = -(ax+b)\frac{dT}{dx}\)
03

Rearrange the equation and separate variables

Now, we need to separate the temperature and distance variables. Rearrange the equation: \(\frac{dT}{dx} = -\frac{q''}{ax+b}\) Now, separate the variables by moving the terms containing \(T\) to the left side and the terms containing \(x\) to the right side: \(\frac{dT}{q''} = -\frac{dx}{ax+b}\)
04

Integrate both sides

Integrate both sides of the equation to find the expression for temperature: \(\int_{T_1}^{T(x)} \frac{dT}{q''} = -\int_0^{x} \frac{dx'}{a x'+b}\) Let the integral on the left side be represented by \(I\) and the integral on the right side be represented by \(J\).
05

Solve the integrals

First, solve the integral \(I\): \(I = \int_{T_1}^{T(x)} \frac{dT}{q''} = \frac{1}{q''} (T(x) - T_1)\) Now, solve the integral \(J\): \(J = -\int_0^x \frac{dx'}{ax'+b} = -\frac{1}{a} \ln(|ax'+b|)\Big|_0^x = -\frac{1}{a}\ln\left(\frac{ax+b}{b}\right)\)
06

Combine the integrals

Now, equate the two integrals: \(\frac{1}{q''} (T(x) - T_1) = -\frac{1}{a}\ln\left(\frac{ax+b}{b}\right)\)
07

Solve for the temperature gradient and temperature distribution

First, solve for the temperature gradient: \(\frac{dT}{dx} = -\frac{q''}{ax+b}\) Next, rearrange the equation to solve for the temperature distribution: \(T(x) = T_1 - q''a\ln\left(\frac{ax+b}{b}\right)\) These are the expressions for the temperature gradient and temperature distribution through the inhomogeneous wall, given the provided boundary condition at \(x=0\) and varying thermal conductivity.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a measure of a material's ability to conduct heat. Imagine placing a hot iron rod against a stick of butter and a block of chocolate. Over time, you would notice the butter melt faster than the chocolate because butter has higher thermal conductivity.

Mathematically, we denote thermal conductivity as 'k'. It varies between different materials and can even change within a material under different conditions, such as temperature or structure. In our exercise, thermal conductivity varies across the thickness of a wall and is described by a linear function, with 'a' and 'b' being constants and 'x' the position within the wall:

  • For materials with uniform composition, 'k' would be a constant value. However, in our case, 'k' changes as 'x' changes, making the wall 'inhomogeneous'. This variation in 'k' greatly affects how we calculate the rate of heat transfer through the material.
Temperature Gradient
The temperature gradient is the rate at which the temperature changes from one point to another within a substance. Picture a snowy hill where the temperature at the top is much lower than at the base. The change in temperature as you move down the hill is akin to the temperature gradient.

In our exercise, the temperature gradient, denoted as
(dT/dx) , is what we are trying to find. It's crucial for determining how quickly heat will migrate through the wall's variable conductivity. Using Fourier's Law, we linked the heat flux 'q''', which is constant throughout the wall, to this gradient. The challenge arises when 'k' is not a simple number but an expression dependent on 'x', complicating our calculations but providing a realistic model for materials whose properties are not uniform.
Heat Transfer
Heat transfer involves the movement of thermal energy from one place to another. There are three main types of heat transfer: conduction, convection, and radiation. Our focus here is on conduction, which is how heat moves through a solid when there is a temperature difference.

Fourier's Law is a fundamental principle predicting how heat flows through materials via conduction. It plainly states that heat moves from regions of higher temperature to regions of lower temperature, with a quantity known as heat flux 'q''', which is affected by the material's thermal conductivity 'k' and the prevailing temperature gradient 'dT/dx'.

In our exercise, we work with a constant 'q''' to find out how the temperature throughout the wall changes. The equations we derived not only help us understand the physics behind heat transfer but also have practical applications in engineering, for instance in designing wall materials that manage heat flow to keep indoors comfortable regardless of outside conditions.

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Most popular questions from this chapter

A square isothermal chip is of width \(w=5 \mathrm{~mm}\) on a side and is mounted in a substrate such that its side and back surfaces are well insulated; the front surface is exposed to the flow of a coolant at \(T_{\infty}=15^{\circ} \mathrm{C}\). From reliability considerations, the chip temperature must not exceed \(T=85^{\circ} \mathrm{C}\). If the coolant is air and the corresponding convection coefficient is \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the maximum allowable chip power? If the coolant is a dielectric liquid for which \(h=3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the maximum allowable power?

A transmission case measures \(W=0.30 \mathrm{~m}\) on a side and receives a power input of \(P_{i}=150 \mathrm{hp}\) from the engine. 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 \(T_{\infty}=50^{\circ} \mathrm{C}\) ?

Pressurized water \(\left(p_{\text {in }}=10\right.\) bar, \(\left.T_{\text {in }}=110^{\circ} \mathrm{C}\right)\) enters the bottom of an \(L=10\)-m-long vertical tube of diameter \(D=100 \mathrm{~mm}\) at a mass flow rate of \(\dot{m}=1.5 \mathrm{~kg} / \mathrm{s}\). The tube is located inside a combustion chamber, resulting in heat transfer to the tube. Superheated steam exits the top of the tube at \(p_{\text {out }}=7\) bar, \(T_{\text {out }}=600^{\circ} \mathrm{C}\). Determine the change in the rate at which the following quantities enter and exit the tube: (a) the combined thermal and flow work, (b) the mechanical energy, and (c) the total energy of the water. Also, (d) determine the heat transfer rate, \(q\). Hint: Relevant properties may be obtained from a thermodynamics text.

During its manufacture, plate glass at \(600^{\circ} \mathrm{C}\) is cooled by passing air over its surface such that the convection heat transfer coefficient is \(h=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent cracking, it is known that the temperature gradient must not exceed \(15^{\circ} \mathrm{C} / \mathrm{mm}\) at any point in the glass during the cooling process. If the thermal conductivity of the glass is \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and its surface emissivity is \(0.8\), what is the lowest temperature of the air that can initially be used for the cooling? Assume that the temperature of the air equals that of the surroundings.

A concrete wall, which has a surface area of \(20 \mathrm{~m}^{2}\) and is \(0.30 \mathrm{~m}\) thick, separates conditioned room air from ambient air. The temperature of the inner surface of the wall is maintained at \(25^{\circ} \mathrm{C}\), and the thermal conductivity of the concrete is \(1 \mathrm{~W} / \mathrm{m}=\mathrm{K}\). (a) Determine the heat loss through the wall for outer surface temperatures ranging from \(-15^{\circ} \mathrm{C}\) to \(38^{\circ} \mathrm{C}\), which correspond to winter and summer extremes, respectively. Display your results graphically. (b) On your graph, also plot the heat loss as a function of the outer surface temperature for wall materials having thermal conductivities of \(0.75\) and \(1.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Explain the family of curves you have obtained.
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