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Chapter 2: Problem 3

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.

Short Answer

Expert verified
The temperature decreases non-linearly from \(T_1\) at \(r_1\) to \(T_2\) at \(r_2\) due to spherical conduction dynamics.

Step by step solution

01

Identify Boundary Conditions

The spherical shell has two boundary surfaces: the inner surface at radius \(r_1\) with temperature \(T_1\), and the outer surface at radius \(r_2\) with temperature \(T_2\).
02

Define the Temperature Distribution

Assuming steady-state and one-dimensional conduction, the temperature distribution in the spherical shell follows the form \(T(r)\), where the gradient occurs between the inner and outer surfaces.
03

Analyze Governing Equation for Heat Conduction

For a spherical shell, the heat conduction equation simplifies to \(d/dr (r^2 dT/dr) = 0\). This indicates a linear gradient in \(1/r\).
04

Integrate the Governing Equation

Integrate the heat conduction equation to find \(dT/dr = C_1/r^2\), where \(C_1\) is an integration constant. A second integration will give the temperature distribution \(T(r) = -C_1/r + C_2\).
05

Apply Boundary Conditions

Use the boundary conditions to solve for constants. At \(r = r_1\), \(T = T_1\), and at \(r = r_2\), \(T = T_2\). Substitute these into the integrated equation to solve for \(C_1\) and \(C_2\).
06

Sketch the Temperature Distribution

Based on the equation \(T(r) = -C_1/r + C_2\), the plot of \(T ext{-}r\) will show temperature decreasing with radial distance. It will not be linear but instead form a curve that starts from \(T_1\) at \(r_1\) and decreases to \(T_2\) at \(r_2\), reflecting that temperature decreases with increasing \(r\).
07

Justify the Shape of the Curve

The shape is a result of the inverse relationship with \(r\) in the integrated equation, causing the temperature gradient to decrease faster initially and slower as \(r\) tends toward \(r_2\). This is due to the curvature of the sphere and the nature of heat conduction in spherical coordinates.

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Key Concepts

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

Understanding Temperature Distribution in Spherical Shells
In the context of heat conduction, especially for a spherical shell, temperature distribution refers to how temperature changes between different surfaces of the shell. To grasp this, visualize a shell with two surfaces having temperatures \(T_1\) and \(T_2\), with \(T_1 > T_2\). The distribution is described by the function \(T(r)\), where \(r\) is the radial coordinate.

In steady-state one-dimensional conduction, the heat flowing in from the inner surface equals the heat flowing out to the outer surface. Therefore, the temperature varies smoothly from the higher temperature at \(r_1\) to the lower at \(r_2\).

It forms a continuous curve, influenced by the thermal properties and geometry of the material. Unlike flat surfaces, the spherical shell exhibits a non-linear profile due to its curvature. This is expressed mathematically via the governing heat conduction equation, leading to a distinct thermal pattern.
Exploring Steady-State Conduction
Steady-state conduction refers to a scenario where the temperature at any given point in the material does not change over time. Rather than being affected by transient conditions鈥攍ike initial heat pulses鈥攖he system reaches a state where conditions remain constant.

In our spherical shell, this means the temperature value at each radial position remains the same indefinitely, as long as the external conditions do not change. This constancy is why the term 'steady-state' is used: it implies a balance between the heat entering the system and the heat leaving it.

In one-dimensional conduction through the spherical coordinate system, this balance is described mathematically, indicating that the radial temperature variation remains steady. The system's energy equation, once solved under these assumptions, helps to determine how temperature gradients establish themselves radially.
Importance of Boundary Conditions
Boundary conditions are crucial assumptions in solving differential equations for heat conduction problems. They define known values of the temperature at specific points, usually the surfaces of the material.

In the spherical shell, the boundary conditions are given by the temperatures \(T_1\) at \(r_1\) and \(T_2\) at \(r_2\).

This information is essential for solving the heat conduction equation. By substituting these values, we can determine the constants arising from integration, which tailor the general solution to a specific situation.

These conditions ensure that the temperature distribution aligns with physical reality: they anchor the theoretical model to actual, observed situations, leading to a calculated distribution of temperature over the sphere's thickness.
Understanding Radial Temperature Gradient
The radial temperature gradient refers to the rate of change of temperature with respect to the radial distance in a spherical shell. In essence, it describes how steeply the temperature drops from the inner surface to the outer surface.

For spherical coordinates, the governing equation yields a relationship that shows an inverse dependence on the radius \(r\). The equation derived from steady-state conduction is \(dT/dr = C_1/r^2\).

This gradient is steeper near the inner surface and flatter towards the outer surface. The reason is the geometry of the sphere, which disperses heat over a larger area as the radius increases. As a result, the temperature curve drops sharply at first but levels off as you move outwards. The radial gradient directly affects material performance, indicating regions of rapid temperature change potential, critical for assessing thermal stress.

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Most popular questions from this chapter

A method for determining the thermal conductivity \(k\) and the specific heat \(c_{p}\) of a material is illustrated in the sketch. Initially the two identical samples of diameter \(D=60 \mathrm{~mm}\) and thickness \(I\), \(=10 \mathrm{~mm}\) and the thin heater are at a uniform temperature of \(T_{i}=23.00^{\circ} \mathrm{C}\). while surrounded by an insulating powder. Suddenly the heater is energized to provide a uniform heat flux \(q_{e}^{\prime \prime}\) on each of the sample interfaces, and the heat fiux is maintained constant for a period of time, \(\Delta t_{t r}\) A short time after sudden heating is initiated, the temperature at this interface \(T_{\omega}\) is related to the heat flux as $$ T_{d}(t)-T_{i}=2 e^{\prime \prime}\left(\frac{t}{\pi \alpha c k}\right)^{i / 2} $$ For a particular test run, the electrical heater dissipates 15.0 W for a period of \(\Delta t_{e}=120 \mathrm{~s}\) and the temperature at the interface is \(T,(30 \mathrm{~s})=24.57^{\circ} \mathrm{C}\) after \(30 \mathrm{~s}\) of heating. A long time after the heiter is deenergized, \(t>\Delta t_{\text {e. }}\) the samples reach the uniform temperature of \(T_{d}(\propto)=\) \(33.50^{\circ} \mathrm{C}\). The density of the sample materials, determined by measurement of volume and mass, is \(\rho=\) \(3965 \mathrm{~kg} / \mathrm{m}^{3}\), Determine the specific heat and thermal conductivity of the test material. By looking at values of the thermophysical properties in Table A.I or A.2, identify the test sample material.

Sections of the trans-Alaska pipeline run above the ground and ate supported by vertical steel shafts \((k=25\) W/m \(-\mathrm{K})\) that are \(1 \mathrm{~m}\) long and have a cross-sectional area of \(0.005 \mathrm{~m}^{2}\). Under normal cperating conditions, the temperature variation along the length of a shaft is known to be governed by an expression of the form $$ T=100-150 x+10 x^{2} $$ where \(T\) and \(x\) have units of \({ }^{\circ} \mathrm{C}\) and meters, respectively. Temperature variations are small over the shaft cross section. Evaluate the temperature and conduction heat rate at the shaft-pipeline joint \((x=0)\) and at the shaft-ground interface \((x=1 \mathrm{~m})\). Explain the difference in the heat rates.

A young engineer is asked to design a themal protection barricr for a sensitive clectronic device that might be exposed to irradiation from a high- powered infrarcd laser. Having learmed as a student thut a low thermal conductivity material provides good insulating characteristics, the enginecr specifies use of a nanostructured aerogel, characterized by a thermal conductivity of \(k_{e}=\) 0.0ns W/m - \(\mathrm{K}\), for the protective harrier. The engineer's boss questions the wisdom of selecting the acrogel because it has a low themal conductivity. Consider the sudden laser irradiation of (a) peire aluminum, (b) glass, and (c) aerogel. The laser provides irradiation of \(G=10 \times 10^{6}\) Whm \({ }^{2}\). The absorptivities of the materials are \(a=0.2,0.9\), and \(0.8\) for the aluminum, glass, and acrogel, rerpectively, and the initial ternperature of the barrier is \(T_{i}=300 \mathrm{~K}\). Explain why the boss is concerncd. Hint: All materials experience thetmal expunsion (or contraction), and local stresses that develop within a material are, to a first approximation, proportional io the local temperahure gradicnt.

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 \(\Delta T_{1} \neq \Delta T_{2}\) ?

A spherical shell of inner and outer radii \(r_{\text {c and }} r_{e}\) ts spectively, contains heat-dissipating components, and at a particular instant the temperature distribution in the shell is known to be of the form $$ T(r)=\frac{C_{1}}{r}+C_{2} $$ Are conditions stendy-state or transient? How do the heat flux and heat rate vary with radius?
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