/*! 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 141 An experimental arrangement for ... [FREE SOLUTION] | 91Ó°ÊÓ

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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·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’s 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’s 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/m²). 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 – 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 spherical vessel used as a reactor for producing pharmaceuticals has a 10 -mm-thick stainless steel wall \((k=17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and an inner diameter of \(1 \mathrm{~m}\). The exterior surface of the vessel is exposed to ambient air \(\left(T_{\infty}=25^{\circ} \mathrm{C}\right)\) for which a convection coefficient of \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) may be assumed. (a) During steady-state operation, an inner surface temperature of \(50^{\circ} \mathrm{C}\) is maintained by energy generated within the reactor. What is the heat loss from the vessel? (b) If a 20 -mm-thick layer of fiberglass insulation \((k=0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is applied to the exterior of the vessel and the rate of thermal energy generation is unchanged, what is the inner surface temperature of the vessel?

Rows of the thermoelectric modules of Example \(3.13\) are attached to the flat absorber plate of Problem 3.108. The rows of modules are separated by \(L_{\text {sep }}=0.5 \mathrm{~m}\) and the backs of the modules are cooled by water at a temperature of \(T_{w}=40^{\circ} \mathrm{C}\), with \(h=45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the electric power produced by one row of thermoelectric modules connected in series electrically with a load resistance of \(60 \Omega\). Calculate the heat transfer rate to the flowing water. Assume rows of 20 immediately adjacent modules, with the lengths of both the module rows and water tubing to be \(L_{\text {row }}=20 W\) where \(W=54 \mathrm{~mm}\) is the module dimension taken from Example 3.13. Neglect thermal contact resistances and the temperature drop across the tube wall, and assume that the high thermal conductivity tube wall creates a uniform temperature around the tube perimeter. Because of the thermal resistance provided by the thermoelectric modules, it is no longer appropriate to assume that the temperature of the absorber plate directly above a tube is equal to that of the water.

Two stainless steel plates \(10 \mathrm{~mm}\) thick are subjected to a contact pressure of 1 bar under vacuum conditions for which there is an overall temperature drop of \(100^{\circ} \mathrm{C}\) across the plates. What is the heat flux through the plates? What is the temperature drop across the contact plane?

Turbine blades mounted to a rotating disc in a gas turbine engine are exposed to a gas stream that is at \(T_{\infty}=1200^{\circ} \mathrm{C}\) and maintains a convection coefficient of \(h=250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) over the blade. The blades, which are fabricated from Inconel, \(k \approx 20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), have a length of \(L=50 \mathrm{~mm}\). The blade profile has a uniform cross-sectional area of \(A_{c}=6 \times 10^{-4} \mathrm{~m}^{2}\) and a perimeter of \(P=110 \mathrm{~mm}\). A proposed blade- cooling scheme, which involves routing air through the supporting disc, is able to maintain the base of each blade at a temperature of \(T_{b}=300^{\circ} \mathrm{C}\). (a) If the maximum allowable blade temperature is \(1050^{\circ} \mathrm{C}\) and the blade tip may be assumed to be adiabatic, is the proposed cooling scheme satisfactory? (b) For the proposed cooling scheme, what is the rate at which heat is transferred from each blade to the coolant?

A very long rod of \(5-\mathrm{mm}\) diameter and uniform thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is subjected to a heat treatment process. The center, 30 -mm-long portion of the rod within the induction heating coil experiences uniform volumetric heat generation of \(7.5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). The unheated portions of the rod, which protrude from the heating coil on either side, experience convection with the ambient air at \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assume that there is no convection from the surface of the rod within the coil. (a) Calculate the steady-state temperature \(T_{o}\) of the rod at the midpoint of the heated portion in the coil. (b) Calculate the temperature of the rod \(T_{b}\) at the edge of the heated portion.

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