Question

A small body is placed inside of a spherical stainless steel chamber with a diameter of \(3 \mathrm{~m}\). The chamber is evacuated and has a constant surface temperature of \(500 \mathrm{~K}\). Determine the radiation incident on the surface of the small body inside the chamber if the interior surface of the chamber is \((a)\) coated black and \((b)\) well-polished.

Short answer

Question: Determine the radiation incident on the surface of a small body inside a spherical stainless steel chamber for both cases: (a) the interior surface of the chamber is coated black, and (b) it is well-polished. Answer: (a) For the black-coated interior surface, the radiation incident on the small body is \(\frac{(1) \cdot (\sigma) \cdot (500^4)}{9\pi}\). (b) For the well-polished interior surface, the radiation incident is \(\frac{(\varepsilon) \cdot (\sigma) \cdot (500^4)}{9\pi}\), where \(\varepsilon < 1\) is the emissivity of the well-polished stainless steel.

Step-by-step solution

(Step 1: Stefan-Boltzmann law)

(The Stefan-Boltzmann law states that the rate of radiant energy emitted by a black body is proportional to the fourth power of its temperature, i.e., \(Q = \sigma T^4\), where \(Q\) is the rate of radiant energy, \(\sigma\) is the Stefan-Boltzmann constant, and \(T\) is the temperature in Kelvin. We will consider the chamber as a black body in case \((a)\) and a well-polished body in case \((b)\).)

(Step 2: Data given)

(The diameter of the spherical chamber is given as \(3 \mathrm{~m}\) and its surface temperature is \(500 \mathrm{~K}\). The interior surface area of the chamber can be calculated using the formula \(A = 4\pi r^2\), where \(r\) is the radius of the sphere.)

(Step 3: Calculate interior surface area of the chamber)

(Since diameter is given as \(3 \mathrm{~m}\), the radius \(r = \frac{3}{2} \mathrm{~m}\). Now, we can calculate the surface area: \(A = 4\pi (\frac{3}{2})^2 = 9\pi \mathrm{~m^2}\).)

(Step 4: Calculate the radiation incident on the small body (Case a))

(Utilizing the Stefan-Boltzmann law, \(Q = \sigma T^4\) and considering the chamber as a black body (emissivity, \(\varepsilon = 1\)), we compute the total energy emitted by the chamber per unit time: \(Q = (1) \cdot (\sigma) \cdot (500^4)\). The radiation incident on the small body is equal to the total energy emitted by the chamber per unit time divided by the surface area, i.e., \(\frac{Q}{A} = \frac{(1) \cdot (\sigma) \cdot (500^4)}{9\pi}\). )

(Step 5: Calculate the radiation incident on the small body (Case b))

(Considering the chamber as a well-polished body with emissivity \(\varepsilon < 1\), we can compute the total energy emitted by the chamber per unit time: \(Q = (\varepsilon) \cdot (\sigma) \cdot (500^4)\). The radiation incident on the small body is equal to the total energy emitted by the chamber per unit time divided by the surface area, i.e., \(\frac{Q}{A} = \frac{(\varepsilon) \cdot (\sigma) \cdot (500^4)}{9\pi}\). Note that the actual value of the radiation incident on the small body will depend on the value of \(\varepsilon\) for the well-polished stainless steel chamber.) In conclusion, to determine the radiation incident on the surface of the small body inside the chamber, we calculated the radiation emitted by the chamber (using the Stefan-Boltzmann law) and divided it by the surface area. For case (a), where the interior surface is coated black, the radiation received by the small body is \(\frac{(1) \cdot (\sigma) \cdot (500^4)}{9\pi}\). For case (b), where the interior surface is well-polished, the radiation received is \(\frac{(\varepsilon) \cdot (\sigma) \cdot (500^4)}{9\pi}\), where \(\varepsilon < 1\) is the emissivity of the well-polished stainless steel.

Key concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is a fundamental principle in thermal radiation. It tells us that the energy emitted by a black body is directly proportional to the fourth power of its temperature. This law is expressed mathematically as \(Q = \sigma T^4\), where \(Q\) is the radiant energy emitted per unit time, \(\sigma\) is the Stefan-Boltzmann constant, and \(T\) is the absolute temperature in Kelvin.

In the context of our exercise, this law helps us determine the total energy emitted by the chamber acting as a black body. It gives us a way to calculate how much energy radiates from the chamber's surface due to its temperature. Applying this law becomes especially crucial for understanding energy transfer in thermal systems.

Emissivity

Emissivity is a measure of a material's ability to emit thermal radiation relative to a perfect black body. A perfect black body has an emissivity \(\varepsilon\) of 1, meaning it emits the maximum possible radiation at a given temperature. Real materials have emissivities less than 1, indicating they emit less radiation than a black body.

In our practice scenario, we see two different cases:

• Case (a) involves a chamber coated black, embracing a black body's properties with \(\varepsilon = 1\).
• Case (b) considers a well-polished chamber, leading to an \(\varepsilon\) less than 1.

The emitted radiation energy naturally decreases when the surface is not a perfect emitter, highlighting the significant role of emissivity in determining energy transfer efficiency.

Black Body Radiation

Black body radiation illustrates a theoretical ideal for objects that absorb all incident light and emit thermal radiation perfectly. In this concept, no energy is reflected or transmitted through the object. Instead, all the energy is emitted as radiation, making it a perfect emitter.

Understanding black body radiation is vital for accurate thermal analysis. In many engineering applications, considering an object as a black body simplifies radiation calculations. For the chamber's black interior in case (a), imagining it as a black body helps in accurately calculating the radiant energy it can emit.

Surface Area Calculation

Surface area plays a crucial role in radiation heat transfer because the emitted energy depends on the size of the emitting surface. For spherical objects, the surface area is evaluated using the formula \(A = 4\pi r^2\), where \(r\) is the radius of the sphere.

In this exercise, the chamber is a sphere with a diameter of \(3\ \mathrm{m}\), leading to a radius of \(\frac{3}{2}\ \mathrm{m}\). The surface area calculation then becomes \(A = 4\pi \left(\frac{3}{2}\right)^2 = 9\pi\ \mathrm{m^2}\).

This value is crucial as it allows us to correctly determine the amount of radiation incident on the small body by considering how the energy is spread over the chamber’s surface.