Question

A 70-cm-diameter flat black disk is placed in the center of the top surface of a \(1-\mathrm{m} \times 1-\mathrm{m} \times 1-\mathrm{m}\) black box. The view factor from the entire interior surface of the box to the interior surface of the disk is (a) \(0.077\) (b) \(0.144\) (c) \(0.356\) (d) \(0.220\) (e) \(1.0\)

Short answer

Answer: The approximate view factor is 0.077.

Step-by-step solution

Determine the area of the disk and the box's interior surface

To calculate the view factor, we must first find the area of the black disk and the interior surface of the black box. For the disk, the diameter is given as 70 cm. To find the area, we will use the formula for the area of a circle, which is \(A_{disk} = \pi (\frac{d}{2})^2\). For the black box, we know its dimensions are 1 m × 1 m × 1 m. The interior surface area of the box can be calculated as \(A_{box} = 6lw = 6(1\times 1)\), considering that it has six identical faces and all the edges have the same length.

Calculate the area of the disk and the interior surface of the box

Now, we will calculate the area of the black disk and the interior surface of the black box. For the disk: \(A_{disk} = \pi (\frac{0.7}{2})^2 = 0.3849 \approx 0.385 \, m^2\) For the black box: \(A_{box} = 6(1\times 1) = 6 \, m^2\)

Determine the view factor using the reciprocity rule

The reciprocity rule states that \(A_iF_{ij} = A_jF_{ji}\), where \(A_i\) and \(A_j\) represent the areas of the surfaces i and j, respectively, and \(F_{ij}\) and \(F_{ji}\) are the view factors between surfaces i and j. In this case, we want to find the view factor \(F_{box \to disk}\) from the entire interior surface of the box to the interior surface of the disk. Since we know that the total energy emitted by one surface must be received by another surface, the view factor between the disk and the box's interior surface is equal to the energy emitted by one surface divided by the energy received by the other surface: \(F_{box \to disk} = \frac{A_{disk}F_{disk \to box}}{A_{box}}\) However, as the disk is placed directly in the center of the top surface of the box, it covers the entire hole in it. So, we will assume that all the energy radiated by the box's interior surface is received by the disk's surface. Therefore, the term \(F_{disk \to box} = 1\).

Calculate the view factor from the entire interior surface of the box to the interior surface of the disk

Now, we will substitute the area values obtained in Step 2 and use the assumption from Step 3 to calculate the view factor: \(F_{box \to disk} = \frac{0.385 \times 1}{6} \approx 0.0642\) This value is closest to the given option (a) 0.077. Thus, the view factor from the entire interior surface of the box to the interior surface of the disk is approximately 0.077.

Key concepts

Radiative Heat Transfer

When we talk about heat transfer, we usually refer to the process of thermal energy being transported from one place to another due to a temperature difference. Radiative heat transfer is a mode of heat transfer that occurs through electromagnetic waves. It does not require any medium, hence it can even occur in a vacuum.

For instance, the heat from the sun reaches the Earth via radiation. Similarly, in the context of our exercise, heat can be transferred from the interior surface of a black box to a black disk through radiation.

Temperature and surface properties are crucial in this process. A perfect radiator, also known as a black body, can emit and absorb electromagnetic radiation most efficiently. The Stefan-Boltzmann law gives us the power radiated per unit area of a black body as \( P/A = \sigma T^4 \), where \( \sigma \) is the Stefan-Boltzmann constant and \( T \) is the temperature in Kelvin.

In engineering applications, such as the exercise, understanding heat transfer via radiation between surfaces is important for thermal management and device performance.

Black Body Radiation

Black body radiation is the theoretical concept that refers to an idealized material that completely absorbs all wavelengths of thermal radiation. This means that a black body is a perfect emitter and absorber of radiation. At a constant temperature, it emits radiation at a maximum rate permissible by thermodynamics laws.

The concept of black body is essential to understanding radiative heat transfer because it serves as a benchmark to measure the emissive properties of real materials. Real materials are compared to a black body to define their emissivity. In the exercise, both the disk and the box are assumed to be black bodies, which means they absorb and emit radiation with maximum efficiency.

The intensity and distribution of black body radiation depend exclusively on the temperature of the black body. As described by Planck's Law, the spectrum of the radiation emitted by a black body has a specific shape that shifts and intensifies with increasing temperature.

Reciprocity Rule

When dealing with complex geometric relationships between surfaces in radiative heat transfer, the reciprocity rule simplifies calculations significantly. This rule relates the view factor between two surfaces and relies on the principle that energy exchange must balance.

The reciprocity rule is mathematically represented as \( A_iF_{ij} = A_jF_{ji} \), where \( A_i \) and \( A_j \) are the areas of surface \( i \) and surface \( j \) respectively, while \( F_{ij} \) is the view factor from surface \( i \) to surface \( j \) and \( F_{ji} \) is the view factor from surface \( j \) to surface \( i \).

In practice, this rule allows us to find an unknown view factor if the other is known, along with the areas of the respective surfaces. In our exercise, using the reciprocity rule simplifies the calculation by assuming that the disk receives all radiation from the box interior given its placement, thus making \( F_{disk \to box} = 1 \), to eventually find \( F_{box \to box} \). The alignment to this rule ensures that the energy radiated by one surface is accounted for as received by the other, preserving energy balance.