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Chapter 3: Problem 41

From Figure \(2.5\) it is evident that, over a wide temperature range, the temperature dependence of the thermal conductivity of many solids may be approximated by a linear expression of the form \(k=k_{o}+a T\), where \(k_{o}\) is a positive constant and \(a\) is a coefficient that may be positive or negative. Obtain an expression for the heat flux across a plane wall whose inner and outer surfaces are maintained at \(T_{0}\) and \(T_{1}\), respectively. Sketch the forms of the temperature distribution corresponding to \(a>0, a=0\), and \(a<0 .\)

Short Answer

Expert verified
The heat flux across a plane wall with varying thermal conductivity is given by: \(q = e^{-\int \frac{aT+k_{o}}{(k_{o}+a T)^{2}} dT - C}\). When the coefficient \(a\) is positive, the temperature distribution is concave, indicating higher heat flux at the hotter side; when \(a = 0\), the temperature distribution is linear, implying constant heat flux; and when \(a\) is negative, the temperature distribution is convex, meaning higher heat flux at the cooler side.

Step by step solution

01

Review the governing equation for heat conduction.

For a plane wall with varying thermal conductivity, we can use Fourier's Law of heat conduction, given by: \(q = -k\frac{dT}{dx}\) where q is the heat flux, k is the thermal conductivity, and \(\frac{dT}{dx}\) is the temperature gradient in the x-direction.
02

Replace thermal conductivity with the linear expression.

Parametrize the thermal conductivity as a function of temperature: \(k = k_{o}+aT\) Now we can replace k in Fourier's Law by the above expression: \(q = -(k_{o}+aT)\frac{dT}{dx}\)
03

Separate the variables and integrate.

We need to find an expression for the heat flux, so we will separate the variables and integrate. \(\frac{dq}{q} = \frac{aT+k_{o}}{(k_{o}+a T)^{2}} dT\) By integrating both sides, we get: \(-\int \frac{dq}{q} = \int \frac{aT+k_{o}}{(k_{o}+a T)^{2}} dT\) Let's now integrate the right side with respect to T: \(-\ln|q| = \int \frac{aT+k_{o}}{(k_{o}+a T)^{2}} dT + C\)
04

Solve for the heat flux, q.

In this step, we will find an expression for q by taking the exponent of both sides and simplifying. \(q = e^{-\int \frac{aT+k_{o}}{(k_{o}+a T)^{2}} dT - C}\) Since we are looking for the heat flux across a plane wall maintained at temperatures \(T_{0}\) and \(T_{1}\), we can substitute these values for the limits of the integral and simplify the expression for q.
05

Analyze the results for different values of a.

We will discuss the temperature distribution corresponding to three cases: \(a>0\), \(a=0\), and \(a<0\). 1. When \(a > 0\), the thermal conductivity increases with temperature. This leads to a higher heat flux at the hotter side and a lower heat flux at the cooler side. The temperature distribution curve will have a concave shape. 2. When \(a = 0\), the thermal conductivity doesn't vary with temperature. This results in a constant heat flux across the wall, and the temperature distribution will be linear. 3. When \(a < 0\), the thermal conductivity decreases with temperature. This leads to a higher heat flux at the cooler side and a lower heat flux at the hotter side. The temperature distribution curve will have a convex shape.

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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 the ability of a material to conduct heat. It plays a crucial role in heat transfer processes. For solids, this property can vary with temperature. In some cases, it can be represented by a linear equation:
  • \(k = k_{o} + aT\)
Here, \(k_{o}\) is a positive constant representing the baseline thermal conductivity, and \(a\) is a temperature coefficient. The sign of \(a\) determines how the conductivity changes:
  • If \(a > 0\), the conductivity increases with temperature.
  • If \(a = 0\), the conductivity remains constant.
  • If \(a < 0\), the conductivity decreases as temperature rises.
Understanding how thermal conductivity behaves is essential for predicting the heat flow through materials, especially when designing efficient heating or cooling systems.
Fourier's Law
Fourier's Law is foundational to the study of heat conduction. It describes how heat moves through materials. In mathematical terms, it is expressed as:
  • \(q = -k \frac{dT}{dx}\)
Where:
  • \(q\) is the heat flux, representing the rate of heat transfer per unit area.
  • \(k\) is the thermal conductivity of the material.
  • \(\frac{dT}{dx}\) is the temperature gradient, or the change in temperature over a certain distance in the material.
This law explains that heat flows from areas of higher temperature to areas of lower temperature, with the rate of this flow being influenced by the thermal conductivity.
In scenarios where \(k\) varies with temperature, such as when using the linear expression \(k = k_{o} + aT\), Fourier's Law adapts. It provides insights into how varying conductivity affects heat transfer, thus influencing everything from material selection to insulation design in engineering applications.
Temperature Distribution
Temperature distribution refers to how temperature varies within a material. When analyzing heat conduction in a plane wall, different scenarios arise based on the sign of the coefficient \(a\) in the thermal conductivity equation:
  • **\(a > 0\):** Thermal conductivity increases with temperature. This scenario leads to a concave temperature distribution curve, skewing towards the colder end.
  • **\(a = 0\):** Thermal conductivity is constant. The temperature distribution is linear, exhibiting a straight line between the two temperatures at the wall's surfaces.
  • **\(a < 0\):** Thermal conductivity decreases with temperature. The temperature distribution curve becomes convex, bending towards the warmer end.
Each shape of the temperature curve reflects how heat is conducted through the material differently:
  • Concave curves suggest more efficient heat transfer at the heat source and less towards the sink.
  • Linear distributions show steady heat transfer across the plane.
  • Convex curves indicate improved transfer at the cooler end and reduced efficiency near the heat source.
Recognizing these patterns aids in understanding how materials respond to thermal stress and helps optimize thermal systems for desired heat flow characteristics.

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Most popular questions from this chapter

To maximize production and minimize pumping costs, crude oil is heated to reduce its viscosity during transportation from a production field. (a) Consider a pipe-in-pipe configuration consisting of concentric steel tubes with an intervening insulating material. The inner tube is used to transport warm crude oil through cold ocean water. The inner steel pipe \(\left(k_{s}=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) has an inside diameter of \(D_{i, 1}=150 \mathrm{~mm}\) and wall thickness \(t_{i}=10 \mathrm{~mm}\) while the outer steel pipe has an inside diameter of \(D_{i, 2}=250 \mathrm{~mm}\) and wall thickness \(t_{o}=t_{i}\). Determine the maximum allowable crude oil temperature to ensure the polyurethane foam insulation \(\left(k_{p}=\right.\) \(0.075 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) between the two pipes does not exceed its maximum service temperature of \(T_{p, \max }=\) \(70^{\circ} \mathrm{C}\). The ocean water is at \(T_{\infty, o}=-5^{\circ} \mathrm{C}\) and provides an external convection heat transfer coefficient of \(h_{o}=500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The convection coefficient associated with the flowing crude oil is \(h_{i}=450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) It is proposed to enhance the performance of the pipe-in-pipe device by replacing a thin \(\left(t_{a}=5 \mathrm{~mm}\right)\) section of polyurethane located at the outside of the inner pipe with an aerogel insulation material \(\left(k_{a}=0.012 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\). Determine the maximum allowable crude oil temperature to ensure maximum polyurethane temperatures are below \(T_{p, \max }=70^{\circ} \mathrm{C}\).

The wall of a spherical tank of \(1-m\) diameter contains an exothermic chemical reaction and is at \(200^{\circ} \mathrm{C}\) when the ambient air temperature is \(25^{\circ} \mathrm{C}\). What thickness of urethane foam is required to reduce the exterior temperature to \(40^{\circ} \mathrm{C}\), assuming the convection coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for both situations? What is the percentage reduction in heat rate achieved by using the insulation?

Copper tubing is joined to a solar collector plate of thickness \(t\), and the working fluid maintains the temperature of the plate above the tubes at \(T_{o}\). There is a uniform net radiation heat flux \(q_{\text {rad }}^{\prime \prime}\) to the top surface of the plate, while the bottom surface is well insulated. The top surface is also exposed to a fluid at \(T_{\infty}\) that provides for a uniform convection coefficient \(h\). (a) Derive the differential equation that governs the temperature distribution \(T(x)\) in the plate. (b) Obtain a solution to the differential equation for appropriate boundary conditions.

A firefighter's protective clothing, referred to as a turnout coat, is typically constructed as an ensemble of three layers separated by air gaps, as shown schematically. The air gaps between the layers are \(1 \mathrm{~mm}\) thick, and heat is transferred by conduction and radiation exchange through the stagnant air. The linearized radiation coefficient for a gap may be approximated as, \(h_{\text {rad }}=\sigma\left(T_{1}+T_{2}\right)\left(T_{1}^{2}+T_{2}^{2}\right) \approx 4 \sigma T_{\text {avg }}^{3}\), where \(T_{\text {avg }}\) represents the average temperature of the surfaces comprising the gap, and the radiation flux across the gap may be expressed as \(q_{\text {rad }}^{\prime \prime}=h_{\text {rad }}\left(T_{1}-T_{2}\right)\). (a) Represent the turnout coat by a thermal circuit, labeling all the thermal resistances. Calculate and tabulate the thermal resistances per unit area \(\left(\mathrm{m}^{2}\right.\). \(\mathrm{K} / \mathrm{W}\) ) for each of the layers, as well as for the conduction and radiation processes in the gaps. Assume that a value of \(T_{\mathrm{avg}}=470 \mathrm{~K}\) may be used to approximate the radiation resistance of both gaps. Comment on the relative magnitudes of the resistances. (b) For a pre-ash-over fire environment in which firefighters often work, the typical radiant heat flux on the fire-side of the turnout coat is \(0.25 \mathrm{~W} / \mathrm{cm}^{2}\). What is the outer surface temperature of the turnout coat if the inner surface temperature is \(66^{\circ} \mathrm{C}\), a condition that would result in burn injury?

A composite wall separates combustion gases at \(2600^{\circ} \mathrm{C}\) from a liquid coolant at \(100^{\circ} \mathrm{C}\), with gas- and liquid-side convection coefficients of 50 and 1000 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The wall is composed of a \(10-\mathrm{mm}\)-thick layer of beryllium oxide on the gas side and a 20 -mm-thick slab of stainless steel (AISI 304) on the liquid side. The contact resistance between the oxide and the steel is \(0.05 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). What is the heat loss per unit surface area of the composite? Sketch the temperature distribution from the gas to the liquid.
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