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

Answer: The emissivity of the enclosure is approximately 0.108.

Step-by-step solution

Calculate the view factor from the enclosure to the sphere

To calculate the view factor, we can use the formula given below for an equilateral triangle with a sphere sitting inside it: \(F_{1 \to 2} = 1 - \frac{\sqrt{3}}{4} \frac{d^2}{L^2}\) where \(F_{1 \to 2}\) is the view factor from the triangle to the sphere, \(d\) is the diameter of the sphere, and \(L\) is the side length of the tetrahedron. Given \(L = D\sqrt{6}\) and \(d = 1 \mathrm{~m}\), we can calculate the view factor as follows: \(F_{1 \to 2} = 1 - \frac{\sqrt{3}}{4}\frac{1^2}{(1\cdot\sqrt{6})^2} = 1 - \frac{\sqrt{3}}{24} \approx 0.936\) So, the view factor from the enclosure to the sphere is approximately \(0.936\).

Calculate the radiative heat transfer between the sphere and the enclosure

The radiative heat transfer between the sphere and the enclosure can be calculated using the radiative heat exchange formula: \(q_{rad} = A_{sphere} \cdot \epsilon_{sphere} \cdot \sigma \cdot (T_{sphere}^4 - T_{enclosure}^4)\) where \(A_{sphere}\) is the surface area of the sphere, \(\epsilon_{sphere}\) is the emissivity of the sphere, \(\sigma\) is the Stefan-Boltzmann constant, \(T_{sphere}\) is the temperature of the sphere, and \(T_{enclosure}\) is the temperature of the enclosure. We are given the following information: - Diameter of the sphere (\(d\)): \(1 \mathrm{~m}\) - Emissivity of the sphere (\(\epsilon_{sphere}\)): \(0.45\) - Sphere temperature (\(T_{sphere}\)): \(600 \mathrm{~K}\) - Enclosure temperature (\(T_{enclosure}\)): \(420 \mathrm{~K}\) Using this information, we can calculate the radiative heat transfer as follows: \(q_{rad} = 4\pi (0.5^2) \cdot 0.45 \cdot 5.67 \times 10^{-8} \cdot (600^4 - 420^4) \approx 2747.4 \mathrm{~W}\)

Calculate the emissivity of the enclosure

We know that heat is generated within the sphere at a rate of \(3100 \mathrm{~W}\). Since heat is conserved, the heat absorbed by the enclosure must be equal to the heat generated within the sphere. We can write the heat balance equation as follows: \(q_{gen} - q_{rad} = A_{enclosure} \cdot \epsilon_{enclosure} \cdot \sigma \cdot (T_{enclosure}^4 - T_{sphere}^4)\) where \(q_{gen}\) is the heat generated within the sphere (\(3100 \mathrm{~W}\)), \(q_{rad}\) is the radiative heat transfer between the sphere and the enclosure (\(2747.4 \mathrm{~W}\)), \(A_{enclosure}\) is the surface area of the enclosure face, and \(\epsilon_{enclosure}\) is the emissivity of the enclosure. We can rearrange the equation to find the emissivity of the enclosure: \(\epsilon_{enclosure} = \frac{q_{gen} - q_{rad}}{A_{enclosure}\cdot\sigma \cdot (T_{enclosure}^4 - T_{sphere}^4)}\) Using the view factor we calculated in step 1, we can determine the surface area of the enclosure: \( A_{enclosure} = \frac{4\pi (0.5^2)}{0.936} \approx 6.695 \mathrm{~m}^2\) Now we can calculate the emissivity of the enclosure: \(\epsilon_{enclosure} = \frac{3100\mathrm{~W} - 2747.4\mathrm{~W}}{6.695\mathrm{~m}^2 \cdot 5.67 \times 10^{-8} \cdot (420^4 - 600^4)} \approx 0.108\) So, the emissivity of the enclosure is approximately \(0.108\).