Question

A solid sphere of \(1 \mathrm{~m}\) diameter at \(600 \mathrm{~K}\) is kept in an evacuated equilateral triangular enclosure (a tetrahedron) whose side length is \(L\). Note that for the sphere to touch the tetrahedron's_surfaces, the tetrahedron's side length should be \(L=D \sqrt{6}\). The emissivity of the sphere is \(0.45\) and the temperature of the enclosure is \(420 \mathrm{~K}\). If heat is generated uniformly within the sphere at a rate of \(3100 \mathrm{~W}\), determine (a) the view factor from the enclosure to the sphere and \((b)\) the emissivity of the enclosure.

Short answer

Question: Calculate the view factor from the enclosure to the sphere and the emissivity of the enclosure using the given information: sphere diameter is 1 m, sphere emissivity is 0.9, sphere temperature is 600 K, enclosure temperature is 300 K, and the rate of heat generation within the sphere is 3100 W. Step 1: Determine the radius of the sphere and the height of the tetrahedron using the given dimensions. Step 2: Calculate the view factor from the enclosure to the sphere using the formula for view factors involving plane surfaces. Step 3: Calculate the emissivity of the enclosure by first finding the total power emitted by the sphere, the total power absorbed by the enclosure, and the total heat leaving the sphere by radiation and then using the Stefan-Boltzmann law to find the emissivity.

Step-by-step solution

Use the given dimensions to determine the radius of the sphere and the height of the tetrahedron

The diameter of the sphere, \(D\), is given as \(1 \mathrm{~m}\). From this, we can calculate the radius of the sphere, \(r\), using the formula: \(r = \frac{D}{2}\) The side length of the tetrahedron, \(L\), is related to the diameter of the sphere through the formula: \(L = D\sqrt{6}\) Since the height of the tetrahedron bisects one of its sides, we can calculate the height, \(h\), using the Pythagorean theorem in an equilateral triangle: \(h = \sqrt{L^2 - \frac{L^2}{4}}\)

Calculate the view factor

Using the formula for the view factor involving plane surfaces, we can calculate the view factor, \(F_{12}\), from the enclosure to the sphere: \(F_{12} = \frac{1}{\pi r}\left(\frac{h}{2}\right)\) where \(r\) is the radius of the sphere and \(h\) is the height of the tetrahedron.

Calculate the emissivity of the enclosure

To find the emissivity of the enclosure, first calculate the total power emitted by the sphere, \(E_{sphere}\), using the Stefan-Boltzmann law: \(E_{sphere} = \epsilon_{sphere} \sigma A_{sphere} (T_{sphere}^4 - T_{enclosure}^4)\) where \(\epsilon_{sphere}\) is the emissivity of the sphere, \(\sigma\) is the Stefan-Boltzmann constant, \(A_{sphere}\) is the surface area of the sphere, and \(T_{sphere}\) and \(T_{enclosure}\) are the temperatures of the sphere and the enclosure, respectively. Now, the total power absorbed by the enclosure, \(E_{absorbed}\), can be calculated using the view factor from the sphere to the enclosure and the total power emitted by the sphere: \(E_{absorbed} = A_{sphere} F_{12} E_{sphere}\) Finally, using the given heat generation rate within the sphere, we can find the total heat leaving the sphere by the radiation, \(E_{radiation}\): \(E_{radiation} = E_{sphere} - 3100\) To find the emissivity of the enclosure, \(\epsilon_{enclosure}\), we will use the Stefan-Boltzmann law again: \(E_{radiation} = \epsilon_{enclosure} \sigma A_{enclosure}(T_{enclosure}^4 - T_{sphere}^4)\) where \(A_{enclosure}\) is the total surface area of the tetrahedron. Solve for \(\epsilon_{enclosure}\) to obtain the emissivity of the enclosure.

Key concepts

Understanding the Stefan-Boltzmann Law

Grasping the concept of the Stefan-Boltzmann law is pivotal to analyzing radiation heat transfer problems, such as the one dealing with a sphere in an evacuated tetrahedron enclosure. Simply put, the Stefan-Boltzmann law states that the power per unit area (also known as the radiant flux) emitted by a blackbody is directly proportional to the fourth power of the body's absolute temperature. It is mathematically represented as:

\( E = \frac{T^4}{\textstyle \theta^2} \),

where \( E \) is the emissive power, \( T \) is the absolute temperature of the blackbody, and \( \theta \) denotes temperature in the appropriate dimensionless units for the formula to maintain consistency. For real materials, this law is modified to include emissivity \( \textstyle (\textstyle \theta) \), giving us the following formula:
\( E = \theta \textstyle \theta^4 \).
Emissivity can take values between 0 and 1, where 1 represents a perfect blackbody. It's crucial to note that this law is fundamental in determining both the total power radiated by an object and the balance between the heat absorbed and emitted by bodies in thermal contact, as seen in our given problem with the sphere and the enclosure.

Solving for View Factor Calculations

The view factor, sometimes referred to as the configuration factor or shape factor, is a dimensionless parameter pivotal in radiative heat transfer calculations. It quantifies the proportion of radiation that leaves one surface and reaches another directly. View factor calculations consider the geometry of the surfaces involved and, in our textbook example, determine the relationship between the sphere and the encompassing tetrahedron.

Understanding the Elements of View Factor

When determining the view factor \( F_{12} \), between two surfaces, geometric constraints become central. The formula for a plane surface to a sphere is:
\( F_{12} = \frac{1}{\textstyle \theta r}, \)
where \( r \) is the radius of the sphere, and the geometric properties of the enclosure play a role in determining \( h \). You need to understand the positioning and orientation of the surfaces in relation to each other; for instance, in the case of the solid sphere inside the tetrahedron, understanding that the position where the sphere touches the tetrahedron affects the view factor.
Using the geometric properties provided (like dimensions and shapes), we can calculate the view factors needed to analyze heat transfer in our particular problem. The challenge lies in the complex relationship between the surfaces, which is why it is so essential to understand the underlying geometry.

Digging into Heat Generation in Solids

Heat generation within solids is a frequent occurrence in various engineering applications, including the example of the heated sphere. In our specific problem, the sphere is generating heat uniformly at a given rate. It is critical to understand that heat generation in solids can be caused by various sources like electrical resistance, chemical reactions, or, as in our case, constant power supply.

Uniform vs Non-Uniform Heat Generation

Heat generation within solids can be uniform or vary spatially. Uniform heat generation means that every volume element of the solid generates the same amount of heat, which simplifies calculations. Non-uniform heat generation, however, requires integration over the volume to determine the total amount of heat generated.
In our sphere's example, the uniform heat generation rate simplifies the problem since we can treat the entire volume as generating the same amount of heat per time unit. The analysis entails balancing this internally generated heat with the heat lost by the sphere, which is radiated to its surroundings. This consideration is crucial for accurately determining the temperature distribution within the solid and correctly estimating the heat required to maintain a certain temperature or the cooling needed to dissipate the surplus heat.