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Chapter 3: Problem 141

An experimental arrangement for measuring the thermal conductivity of solid materials involves the use of two long rods that are equivalent in every respect, except that one is fabricated from a standard material of known thermal conductivity \(k_{\mathrm{A}}\) while the other is fabricated from the material whose thermal conductivity \(k_{\mathrm{B}}\) is desired. Both rods are attached at one end to a heat source of fixed temperature \(T_{b}\), are exposed to a fluid of temperature \(T_{\infty}\), and are instrumented with thermocouples to measure the temperature at a fixed distance \(x_{1}\) from the heat source. If the standard material is aluminum, with \(k_{\mathrm{A}}=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and measurements reveal values of \(T_{\mathrm{A}}=75^{\circ} \mathrm{C}\) and \(T_{\mathrm{B}}=60^{\circ} \mathrm{C}\) at \(x_{1}\) for \(T_{b}=100^{\circ} \mathrm{C}\) and \(T_{\infty}=25^{\circ} \mathrm{C}\), what is the thermal conductivity \(k_{\mathrm{B}}\) of the test material?

Short Answer

Expert verified
The thermal conductivity of the test material is approximately \(285.71 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Step by step solution

01

Heat transfer equation for thermal conduction in each rod

First, let's write down the basic equation for heat transfer through thermal conduction in a solid. The Fourier's Law of heat conduction states that: \[q = -k \frac{dT}{dx}\] Where: - \(q\) is the heat flux (W/m^2), - \(k\) is the thermal conductivity of the material (W/m路K), - \(dT\) is the temperature difference across the material, - \(dx\) is the distance between the points with different temperatures (in this case, \(x_1\)). Now, we need to write down this equation for Rod A and Rod B, considering the corresponding thermal conductivities and temperatures. Rod A: \[q_A = -k_A \frac{T_A - T_\infty}{x_1}\] Rod B: \[q_B = -k_B \frac{T_B - T_\infty}{x_1}\]
02

Solve for heat flux (Q) for Rod A

Since Rod A is made of aluminum, we know its thermal conductivity \(k_A = 200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Using this value, along with the given temperatures \(T_A = 75^{\circ} \mathrm{C}\) and \(T_\infty = 25^{\circ} \mathrm{C}\), we can find the heat flux in Rod A: \[q_A = -200 \frac{75 - 25}{x_1}\] \[q_A = -200 \frac{50}{x_1}\] We will leave the heat flux in this form for now, as it will be useful in the next steps.
03

Write the equation for heat flux (Q) in Rod B

We'll do the same for Rod B, writing down the heat flux equation using the unknown thermal conductivity \(k_B\) and given temperatures \(T_B = 60^{\circ} \mathrm{C}\) and \(T_\infty = 25^{\circ} \mathrm{C}\): \[q_B = -k_B \frac{60 - 25}{x_1}\] \[q_B = -k_B \frac{35}{x_1}\]
04

Set heat flux (Q) for Rod A equal to heat flux (Q) for Rod B and solve for \(k_B\)

Since both rods have the same heat source and are exposed to the same fluid, we can assume that the heat flux through both rods is equal. Therefore, we can equate the heat flux equations for Rod A and Rod B, and solve for the unknown thermal conductivity \(k_B\): \[-200 \frac{50}{x_1} = -k_B \frac{35}{x_1}\] The distance \(x_1\) cancels out, leaving: \[200 \cdot 50 = k_B \cdot 35\] Now, we can solve for the unknown thermal conductivity \(k_B\): \[k_B = \frac{200 \cdot 50}{35}\] \[k_B = \frac{10000}{35}\] \[k_B \approx 285.71 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\] So, the thermal conductivity of the test material is approximately \(285.71 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

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Key Concepts

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

Fourier's Law of Heat Conduction
Fourier's Law of heat conduction is a fundamental principle used to predict the rate at which heat energy is transferred within materials due to a temperature gradient. This law mathematically expresses the relationship stating that the heat flux moving through a material is proportional to the negative gradient of the temperature and the material's intrinsic thermal conductivity.

Imagine holding one end of a metal rod that is being heated at the other end. The heat travels through the rod from the hot end to your hand. Fourier鈥檚 Law is what we use to describe how quickly that heat will travel. The equation form of Fourier's Law is given by: \[q = -k \frac{dT}{dx}\], where \( q \) is the heat flux, \( k \) is the thermal conductivity, \( dT \) is the temperature difference, and \( dx \) is the thickness of the material through which heat is passing.

For our example with the rods, we apply Fourier鈥檚 Law to predict the heat transfer properties of the unknown material by comparing it to a standard material of known properties. This comparison is crucial because it allows us to work out the unknown thermal conductivity, given we have the information about the standard material, the temperature difference, and the heat flux.
Heat Flux
Heat flux is a measure of the rate of thermal energy transfer through a given surface area per unit time. It's typically represented in units of watts per square meter (W/m虏). Essentially, it is the amount of heat that passes through a square meter of material in one second.

We often encounter heat flux in real-life situations such as feeling the warmth of the sun on our skin or noticing how quickly a stovetop heats up a pan. In our exercise, we use the concept of heat flux to equate the rate at which heat energy is passing through both rods. By doing so, despite not knowing the thermal conductivity of the test material, we utilize the heat flux of the known material to derive the conductivity of the unknown one.

To further visualize this, suppose you have two kitchen pans on a stove, one made of aluminum and the other of an unknown material, and you want to find out how well the unknown pan conducts heat. By knowing how quickly the aluminum pan, with a known heat flux, heats up, you can estimate the heat flux of the unknown pan when it's exposed to the same heat source.
Thermal Conduction in Solids
Thermal conduction in solids refers to the process by which thermal energy, or heat, is transported through the material due to the random movements of the atoms and molecules within it. Solids conduct heat at varying rates based on their material properties 鈥� particularly their thermal conductivity.

While metals like copper and aluminum are known for their high thermal conductivity and ability to quickly transfer heat, materials such as wood or ceramic have much lower conductivity and are effective insulators. In our rods example, the heat from the source travels through each solid rod to the point where the temperature is measured. By knowing that the same amount of heat must flow through each rod in the same amount of time, we use the known thermal properties of aluminum to determine those of the test material.

So essentially, by understanding and being able to measure how heat conducts through a familiar material, we've set a standard. This gives us a reference to compare against and uncover the thermal conduction capabilities of the unknown solid material through the experimental setup described in the exercise.

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Most popular questions from this chapter

A device used to measure the surface temperature of an object to within a spatial resolution of approximately \(50 \mathrm{~nm}\) is shown in the schematic. It consists of an extremely sharp-tipped stylus and an extremely small cantilever that is scanned across the surface. The probe tip is of circular cross section and is fabricated of polycrystalline silicon dioxide. The ambient temperature is measured at the pivoted end of the cantilever as \(T_{\infty}=\) \(25^{\circ} \mathrm{C}\), and the device is equipped with a sensor to measure the temperature at the upper end of the sharp tip, \(T_{\text {sen. }}\). The thermal resistance between the sensing probe and the pivoted end is \(R_{t}=5 \times 10^{6} \mathrm{~K} / \mathrm{W}\). (a) Determine the thermal resistance between the surface temperature and the sensing temperature. (b) If the sensing temperature is \(T_{\text {sen }}=28.5^{\circ} \mathrm{C}\), determine the surface temperature. Hint: Although nanoscale heat transfer effects may be important, assume that the conduction occurring in the air adjacent to the probe tip can be described by Fourier's law and the thermal conductivity found in Table A. \(4 .\)

The air inside a chamber at \(T_{\infty, i}=50^{\circ} \mathrm{C}\) is heated convectively with \(h_{i}=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) by a 200 -mm-thick wall having a thermal conductivity of \(4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and a uniform heat generation of \(1000 \mathrm{~W} / \mathrm{m}^{3}\). To prevent any heat generated within the wall from being lost to the outside of the chamber at \(T_{\infty, o}=25^{\circ} \mathrm{C}\) with \(h_{o}=5\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), a very thin electrical strip heater is placed on the outer wall to provide a uniform heat flux, \(q_{\sigma^{\prime}}\) (a) Sketch the temperature distribution in the wall on \(T-x\) coordinates for the condition where no heat generated within the wall is lost to the outside of the chamber. (b) What are the temperatures at the wall boundaries, \(T(0)\) and \(T(L)\), for the conditions of part (a)? (c) Determine the value of \(q_{o}^{\prime \prime}\) that must be supplied by the strip heater so that all heat generated within the wall is transferred to the inside of the chamber. (d) If the heat generation in the wall were switched off while the heat flux to the strip heater remained constant, what would be the steady-state temperature, \(T(0)\), of the outer wall surface?

A technique for measuring convection heat transfer coefficients involves bonding one surface of a thin metallic foil to an insulating material and exposing the other surface to the fluid flow conditions of interest. By passing an electric current through the foil, heat is dissipated uniformly within the foil and the corresponding flux, \(P_{\text {elec }}^{\prime \prime}\), may be inferred from related voltage and current measurements. If the insulation thickness \(L\) and thermal conductivity \(k\) are known and the fluid, foil, and insulation temperatures \(\left(T_{\infty}, T_{s}, T_{b}\right)\) are measured, the convection coefficient may be determined. Consider conditions for which \(T_{\infty}=T_{b}=25^{\circ} \mathrm{C}, P_{\text {elec }}^{\prime \prime}=2000\) \(\mathrm{W} / \mathrm{m}^{2}, L=10 \mathrm{~mm}\), and \(k=0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) With water flow over the surface, the foil temperature measurement yields \(T_{s}=27^{\circ} \mathrm{C}\). Determine the convection coefficient. What error would be incurred by assuming all of the dissipated power to be transferred to the water by convection? (b) If, instead, air flows over the surface and the temperature measurement yields \(T_{s}=125^{\circ} \mathrm{C}\), what is the convection coefficient? The foil has an emissivity of \(0.15\) and is exposed to large surroundings at \(25^{\circ} \mathrm{C}\). What error would be incurred by assuming all of the dissipated power to be transferred to the air by convection? (c) Typically, heat flux gages are operated at a fixed temperature \(\left(T_{s}\right)\), in which case the power dissipation provides a direct measure of the convection coefficient. For \(T_{s}=27^{\circ} \mathrm{C}\), plot \(P_{\text {elec }}^{\prime \prime}\) as a function of \(h_{o}\) for \(10 \leq h_{o} \leq 1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What effect does \(h_{o}\) have on the error associated with neglecting conduction through the insulation?

When raised to very high temperatures, many conventional liquid fuels dissociate into hydrogen and other components. Thus the advantage of a solid oxide fuel cell is that such a device can internally reform readily available liquid fuels into hydrogen that can then be used to produce electrical power in a manner similar to Example 1.5. Consider a portable solid oxide fuel cell, operating at a temperature of \(T_{\mathrm{fc}}=800^{\circ} \mathrm{C}\). The fuel cell is housed within a cylindrical canister of diameter \(D=\) \(75 \mathrm{~mm}\) and length \(L=120 \mathrm{~mm}\). The outer surface of the canister is insulated with a low-thermal-conductivity material. For a particular application, it is desired that the thermal signature of the canister be small, to avoid its detection by infrared sensors. The degree to which the canister can be detected with an infrared sensor may be estimated by equating the radiation heat flux emitted from the exterior surface of the canister (Equation 1.5; \(E_{s}=\varepsilon_{s} \sigma T_{s}^{4}\) ) to the heat flux emitted from an equivalent black surface, \(\left(E_{b}=\sigma T_{b}^{4}\right)\). If the equivalent black surface temperature \(T_{b}\) is near the surroundings temperature, the thermal signature of the canister is too small to be detected-the canister is indistinguishable from the surroundings. (a) Determine the required thickness of insulation to be applied to the cylindrical wall of the canister to ensure that the canister does not become highly visible to an infrared sensor (i.e., \(T_{b}-T_{\text {sur }}<5 \mathrm{~K}\) ). Consider cases where (i) the outer surface is covered with a very thin layer of \(\operatorname{dirt}\left(\varepsilon_{s}=0.90\right)\) and (ii) the outer surface is comprised of a very thin polished aluminum sheet \(\left(\varepsilon_{s}=0.08\right)\). Calculate the required thicknesses for two types of insulating material, calcium silicate \((k=0.09 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and aerogel \((k=0.006 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The temperatures of the surroundings and the ambient are \(T_{\text {sur }}=300 \mathrm{~K}\) and \(T_{\infty}=298 \mathrm{~K}\), respectively. The outer surface is characterized by a convective heat transfer coefficient of \(h=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) Calculate the outer surface temperature of the canister for the four cases (high and low thermal conductivity; high and low surface emissivity). (c) Calculate the heat loss from the cylindrical walls of the canister for the four cases.

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.
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