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Asume steady-state, one-dimensional heat conduction through the symmetric shape shown. Assuming that there is no internal heat generation, derive an expresion for the thermal conductivity \(k(x)\) for these conditions \(A(x)=(1-x), T(x)=300\left(1-2 x-x^{3}\right)\), and \(q=6000 \mathrm{~W}\), where \(A\) is in square meters, \(T\) in kelvins, and \(x\) in meters.

Short Answer

Expert verified
The thermal conductivity is given by \(k(x) = \frac{20}{(2 + 3x^2)(1-x)}\).

Step by step solution

01

Understand the Governing Equation

For steady-state, one-dimensional heat conduction without internal heat generation, we use Fourier's law given by \[q = -k(x) \frac{dT(x)}{dx}A(x)\] where \(q\) is the heat transfer rate, \(k(x)\) is the thermal conductivity, \(T(x)\) is the temperature, and \(A(x)\) is the cross-sectional area.
02

Differentiate Temperature Function

Given the temperature distribution \(T(x) = 300(1 - 2x - x^3)\), differentiate it with respect to \(x\):\[\frac{dT(x)}{dx} = 300 \left(-2 - 3x^2\right)\]
03

Substitute into Fourier's Law

Substitute \(dT(x)/dx\), \(A(x)\), and \(q\) into Fourier's Law:\[6000 = -k(x) \times 300(-2 - 3x^2) \times (1-x)\] Simplify the equation:\[6000 = k(x) \times 300(2 + 3x^2) \times (1-x)\]
04

Solve for Thermal Conductivity \(k(x)\)

Rearrange the equation to solve for \(k(x)\):\[k(x) = \frac{6000}{300(2 + 3x^2)(1-x)}\]Simplify:\[k(x) = \frac{20}{(2 + 3x^2)(1-x)}\]

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Key Concepts

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

Fourier's Law
Fourier's law is a fundamental principle of heat conduction that helps us understand how heat is transferred through materials. It's particularly useful in solving problems where heat flows steadily from one area to another without changing over time, known as steady-state conditions. According to Fourier's law, the heat transfer rate, denoted as \(q\), is proportional to the negative gradient of temperature \(\frac{dT(x)}{dx}\) and the cross-sectional area \(A(x)\) through which heat is conducted. Mathematically, it is expressed as:
  • \(q = -k(x) \frac{dT(x)}{dx}A(x)\)
Here, \(k(x)\) represents the material's thermal conductivity, which indicates its ability to conduct heat. The negative sign shows that heat flows in the direction of decreasing temperature. If, say, the temperature decreases as you move in the \(x\) direction, heat will naturally flow from the warmer to the cooler side.
Thermal Conductivity
Thermal conductivity, symbolized as \(k(x)\), is a property that quantifies how well a material can conduct heat. It varies with material composition and temperature, so knowing \(k(x)\) is crucial for calculating heat transfer. For instance, in metals, thermal conductivity is high due to the free movement of electrons, whereas in insulators, it is low.
Thermal conductivity helps determine the rate of heat transfer, especially when dealing with varying temperature and cross-sectional area, like in our problem where \(k(x)\) isn't constant. To find \(k(x)\), we rearrange Fourier’s formula like this:
  • \(k(x) = \frac{q}{A(x)\frac{dT(x)}{dx}}\)
This rearrangement allows us to express \(k(x)\) as a function of the given conditions, enabling analysis of how thermal conduction changes with position across a material's length.
Steady-State Heat Transfer
In steady-state heat transfer, the temperature field within a material does not change over time, even if heat is being transferred. This is an ideal assumption that simplifies calculations, as it implies that the amount of heat entering a section equals the amount leaving it. In our exercise, we assume steady-state conditions to apply Fourier's law effectively.
With steady-state, we deal with constant heat flow, making computations more straightforward as the temperature distribution does not fluctuate over time. Each point in the material maintains a constant temperature, meaning any variations depend only on spatial coordinates, such as \(x\) in our one-dimensional scenario.
  • Steady-state holds true if heat sources and sinks remain stable over time.
  • It typically applies to well-insulated systems or those with constant thermal cycles.
Using steady-state assumptions, molecular movement within the material balances out, preventing buildup or depletion of heat, simplifying the analysis and evaluation of thermal properties or system efficiency.

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Most popular questions from this chapter

A spherical shell with inner radius \(r_{1}\) and outer radius \(r_{2}\) has surface temperatures \(T_{1}\) and \(T_{2}\), respectively, where \(T_{1}>T_{2}\). Sketch the temperature distribution on \(T-r\) coontinates assuming steady- state, one-timensional concuction with constant properties. Briefly justify the shape of your curve.

An apparatus for measuring thermal conductivity cmploys an electrical heater sandwiched between two identical samples of diameter \(30 \mathrm{~mm}\) and length \(60 \mathrm{~mm}\), which are presced between plates maintained at a uniform tempenature \(T_{n}=77^{\circ} \mathrm{C}\) by a circulating fluid. \(\mathrm{A}\) conducting grease is placed between all the surfaces to ensure good thermal contact. Differential thermocouples are imbedded in the samples with a spocing of \(15 \mathrm{~mm}\). The lateral sides of the samples are insulated to ensure cee-dimensional heat transfer thrueugh the samples. (a) With two saruples of SS316 in the apparatus, the heater draws \(0.353 \mathrm{~A}\) at \(100 \mathrm{~V}\) and the differential thermocouples indicate \(\Delta T_{1}=\Delta T_{2}=25.0^{\circ} \mathrm{C}\). What is the thermal conductivity of the stainless steel sample material? What is the average tempenture of the samples? Compare your result with the therrmal conductivity value reported for this material in Table A.1. (b) By mistake, an Armeo iron sample is placed in the lower position of the apparatus with one of the SS 316 samples from part (a) in the upper portion. For this situation, the heater draws \(0.601 \mathrm{~A}\) at \(100 \mathrm{~V}\) and the differential thermocouples indicate \(\Delta T_{1}=\) \(\Delta T_{2}=15.0^{\circ} \mathrm{C}\). What are the thermal conductivity and average temperature of the Armco iron sample? (c) What is the advantage in constructing the apparatus with two identical samples sandwiching the heuter rather than with a single heater-sample combination? When would heat leakage out of the lateral surfaces of the samples become significant? Under what conditions would you cxpect \(\Delta T_{1} \neq \Delta T_{2}\) ?

The steady-state temperature distribution in a onedimensional wall of thermal conductivity \(k\) and thickness \(L\) is of the form \(T=a x^{3}+b x^{2}+c x+d\). Derive expressions for the heat generation nate per unit volume in the wall and the heat fluxes at the two wall faces \((x=0, L)\).

A TV advertisement by a well-known insulation manufacturer states: it isn't the thickness of the insulating material that counts, it's the R-value. The ad shows that to ottain an R-value of 19, you need \(18 \mathrm{ft}\) of rock, 15 in. of wood. or just 6 in. of the manufacturer's insulation. Is this advertisement techinically reasonable? If you are like most TV viewers, you don't know the R-value is defined as \(L / k\). where \(L\). \(\mathrm{in}\).) is the thickness of the insulation and \(k\) (Btu * in \(/ \mathrm{hr}-\mathrm{ft}^{2}+{ }^{\mathrm{F}}\) ) is the thermal conductivity of the muterial.

A salt-gradient solar pond is a shallow body of water that consists of three distinct fluid layers and is used to collect solar encrgy. The upper- and lower-most layers are well mixed and serve to maintain the upper and lower surfaces of the central layer at uniform temperatures \(T_{1}\) and \(T_{2}\). where \(T_{2}>T_{1}\). Although there is bulk fluid motion in the mixed layers, there is no such motion in the central layer. Consider conditions for which solar radiation absorption in the central layer provides nonuniform heat generation of the form \(\dot{q}=A e^{-u ;}\), and the lemperature distribution in the central layer is $$ T(x)=-\frac{A}{k a^{2}} e^{-a x}+B x+C $$ The quantities \(A\left(\mathrm{~W} / \mathrm{m}^{3}\right), a(1 / \mathrm{m}), B(\mathrm{~K} / \mathrm{m})\), and \(C(\mathrm{~K})\) are known constants having the prescribed units, and \(k\) is the thermal conductivity, which is also constant. (a) Obtain expressions for the rate at which heat is transferred per unit area from the lower mixed layer to the central layer and from the central layer to the upper mixed layer. (b) Determine whether cunditions are steady or transicnt. (c) Obtain an expression for the rate at which thermal cnergy is generated in the entire central layer, per unit surface area.

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