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Chapter 2: Problem 4

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 \(T-r\) coordinates assuming steady- state, one-dimensional conduction with constant properties. Briefly justify the shape of your curve.

Short Answer

Expert verified
In this problem, we derive the temperature distribution in a spherical shell with steady-state, one-dimensional conduction and constant properties. The curve starts at point \((r_1, T_1)\) and monotonically decreases until point \((r_2, T_2)\) due to the radial flow of heat from the higher temperature inner surface to the lower temperature outer surface. This is represented by the equation: \[ T(r) = T_1 + \frac{1}{k} \left( \frac{C_{1} (r - r_1)}{r} + \frac{C_{2} (r - r_1)}{r^2} \right) \]

Step by step solution

01

Applying the heat conduction equation

Since we are dealing with steady-state, one-dimensional conduction with constant properties, we can simplify the general heat conduction equation as follows: \[ \frac{d}{dr}\left( k \frac{dT}{dr} \right) = 0 \] where \(k\) is the thermal conductivity of the material. In the case of a spherical coordinate system, this equation becomes: \[ \frac{1}{r^2}\frac{d}{dr}\left( r^2 k\frac{dT}{dr} \right) = 0 \]
02

Integrate the heat conduction equation

We can now integrate the simplified heat conduction equation with respect to radius \(r\): \[\int_{}^{} \frac{1}{r^2} \frac{d}{dr} \left(r^2 k \frac{dT}{dr} \right) dr=0 \] \[ \frac{d}{dr} \left( r^2 k \frac{dT}{dr} \right) = C_1 \] where \(C_{1}\) is the integration constant.
03

Integrating once more

Now, we will integrate the equation again. Integrate both sides with respect to \(r\): \[ \int \frac{d}{dr} \left( r^2 k \frac{dT}{dr} \right) dr = \int C_{1} dr \] \[ r^2 k \frac{dT}{dr} = C_{1} r + C_{2} \] where \(C_{2}\) is another integration constant. Lastly, we need to rewrite the equation in terms of the temperature: \[ \frac{dT}{dr} = \frac{1}{k} \left( \frac{C_{1}}{r} + \frac{C_{2}}{r^2} \right) \]
04

Apply boundary conditions

Next, we apply the boundary conditions that were given: at \(r = r_{1}\), \(T = T_{1}\) and at \(r = r_{2}\), \(T = T_{2}\). We use these conditions to find the integration constants \(C_{1}\) and \(C_{2}\). From the first boundary condition, we can obtain the following equation: \[ \frac{1}{k}\left( \frac{C_{1}}{r_1} + \frac{C_{2}}{r_1^2} \right) = T_{1} \] And from the second boundary condition, we get: \[ \frac{1}{k}\left( \frac{C_{1}}{r_2} + \frac{C_{2}}{r_2^2} \right) = T_{2} \]
05

Solve for the constants

Now, we solve the linear system of equations involving \(C_{1}\) and \(C_{2}\) to find their values. Then, substitute the values of the constants in the expression for temperature distribution. \[ T(r) = T_1 + \frac{1}{k} \left( \frac{C_{1} (r - r_1)}{r} + \frac{C_{2} (r - r_1)}{r^2} \right) \]
06

Sketch the curve and justify its shape

Using the obtained expression for \(T(r)\), we can sketch the temperature distribution on T-r coordinates. The curve should begin at point \((r_1, T_1)\) and decrease monotonically until point \((r_2, T_2)\). The temperature distribution curve is justified by the fact that heat transfer occurs through steady-state, one-dimensional conduction, and the temperature difference between the inner and outer surfaces causes a radial flow of heat from the higher temperature (inner surface) to the lower temperature (outer surface). This results in a gradually decreasing temperature as we move outwards from the inner to the outer surface.

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Key Concepts

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

Spherical Coordinates
When dealing with problems concerning spheres, it's often helpful to use spherical coordinates instead of rectangular or Cartesian coordinates. Imagine a sphere like an onion with multiple layers. Each layer can be described using three coordinates:
  • Radius (r): This length goes from the center of the sphere to the precise point on the shell where you're measuring.
  • Polar Angle (胃): The angle that's formed with respect to a reference axis, typically the vertical axis.
  • Azimuthal Angle (蠁): The angle formed with respect to the x-axis in the plane perpendicular to 胃.
In the context of heat conduction in spherical coordinates, we use the radius as the main parameter due to the radial symmetry of the sphere. This means that instead of using standard x, y, and z axes, we focus on how the temperature changes as we move away from the center of the sphere in layers. The primary factor here is the radial distance, since we are assuming a steady, one-dimensional heat conduction. Only the radial part of the sphere affects the heat flow in our problem.
Temperature Distribution
Temperature distribution in a spherical shell describes how the temperature varies from one surface of the shell to the other. Imagine heat flowing like a gentle wave starting at the hotter inner surface and travelling through the material to reach the cooler outer surface.

The general heat conduction equation in spherical coordinates can be applied here. For simplicity, consider that properties such as thermal conductivity ( k ) are constant. This reduces the equation complexity, meaning the equation is focused only on the radius.

The temperature distribution is governed by the formula: \[ T(r) = T_1 + \frac{1}{k} \left( \frac{C_1 (r - r_1)}{r} + \frac{C_2 (r - r_1)}{r^2} \right) \] This formula tells us how the temperature changes as a function of radius from the inner to outer surface. A typical temperature distribution curve will start at a high temperature and decrease as we move from the inner radius ( r_1 ) at temperature ( T_1 ) to the outer radius ( r_2 ) at temperature ( T_2 ).
Boundary Conditions
Boundary conditions are essential in solving any differential equation problem since they provide the necessary limits to find a unique solution. In our problem, they specify the temperatures at the inner and outer surfaces of the spherical shell. This means they "anchor" the solution, allowing us to find the temperature constant over the entire sphere.

Given the boundary conditions:
  • At radius ( r_1 ), temperature ( T = T_1 ).
  • At radius ( r_2 ), temperature ( T = T_2 ).
These conditions are used to determine the integration constants ( C_1 ) and ( C_2 ) from our integration steps. These constants adjust the temperature curve so that it aligns with the physical reality of the system. This allows us to construct a precise model of how temperature varies within the shell and ensures that our model fits the conditions at both boundaries perfectly. Thus, the role of boundary conditions is to specify the actual situation at the limits of our system's domain.

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Most popular questions from this chapter

A steam pipe is wrapped with insulation of inner and outer radii \(r_{i}\) and \(r_{o}\), respectively. At a particular instant the temperature distribution in the insulation is known to be of the form $$ T(r)=C_{1} \ln \left(\frac{r}{r_{o}}\right)+C_{2} $$ Are conditions steady-state or transient? How do the heat flux and heat rate vary with radius?

Consider a small but known volume of metal that has a large thermal conductivity. (a) Since the thermal conductivity is large, spatial temperature gradients that develop within the metal in response to mild heating are small. Neglecting spatial temperature gradients, derive a differential equation that could be solved for the temperature of the metal versus time \(T(t)\) if the metal is subjected to a fixed surface heat rate \(q\) supplied by an electric heater. (b) A student proposes to identify the unknown metal by comparing measured and predicted thermal responses. Once a match is made, relevant thermophysical properties might be determined, and, in turn, the metal may be identified by comparison to published property data. Will this approach work? Consider aluminum, gold, and silver as the candidate metals.

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 \(L=10 \mathrm{~mm}\) 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 \(q_{o}^{\prime \prime}\) on each of the sample interfaces, and the heat flux is maintained constant for a period of time, \(\Delta t_{o}\). A short time after sudden heating is initiated, the temperature at this interface \(T_{o}\) is related to the heat flux as $$ T_{o}(t)-T_{i}=2 q_{o}^{\prime \prime}\left(\frac{t}{\pi \rho c_{p} k}\right)^{1 / 2} $$ For a particular test run, the electrical heater dissipates \(15.0 \mathrm{~W}\) for a period of \(\Delta t_{o}=120 \mathrm{~s}\), and the temperature at the interface is \(T_{o}(30 \mathrm{~s})=24.57^{\circ} \mathrm{C}\) after \(30 \mathrm{~s}\) of heating. A long time after the heater is deenergized, \(t \geqslant \Delta t_{0}\), the samples reach the uniform temperature of \(T_{o}(\infty)=33.50^{\circ} \mathrm{C}\). 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.1 or A.2, identify the test sample material.

A solid, truncated cone serves as a support for a system that maintains the top (truncated) face of the cone at a temperature \(T_{1}\), while the base of the cone is at a temperature \(T_{2}

Beginning with a differential control volume in the form of a spherical shell, derive the heat diffusion equation for a one-dimensional, spherical, radial coordinate system with internal heat generation. Compare your result with Equation 2.29.
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