Question

13-59 This question deals with steady-state radiation heat transfer between a sphere \(\left(r_{1}=30 \mathrm{~cm}\right)\) and a circular disk \(\left(r_{2}=120 \mathrm{~cm}\right)\), which are separated by a center-to- center distance \(h=60 \mathrm{~cm}\). When the normal to the center of disk passes through the center of the sphere, the radiation view factor is given by $$ F_{12}=0.5\left\\{1-\left[1+\left(\frac{r_{2}}{h}\right)^{2}\right]^{-0.5}\right\\} $$ Surface temperatures of the sphere and the disk are \(600^{\circ} \mathrm{C}\) and \(200^{\circ} \mathrm{C}\), respectively; and their emissivities are \(0.9\) and \(0.5\), respectively. (a) Calculate the view factors \(F_{12}\) and \(F_{21}\). (b) Calculate the net rate of radiation heat exchange between the sphere and the disk. (c) For the given radii and temperatures of the sphere and the disk, the following four possible modifications could increase the net rate of radiation heat exchange: paint each of the two surfaces to alter their emissivities, adjust the distance between them, and provide an (refractory) enclosure. Calculate the net rate of radiation heat exchange between the two bodies if the best values are selected for each of the above modifications.

Short answer

Answer: The optimal conditions to increase the net rate of radiation heat exchange between a sphere and a circular disk are maximizing the emissivities of both surfaces (\(\epsilon_1 = \epsilon_2 = 1\)), decreasing the distance between the two surfaces as much as possible, and adding an enclosure. However, the exact influence of the enclosure would need to be determined case by case. With these optimal conditions, the net rate of radiation heat exchange between the two bodies would be higher than the calculated value of \(33298 W\).

Step-by-step solution

Calculate the view factors of F12 and F21

Given the formula for the view factor F12, $$ F_{12}=0.5\left\\{1-\left[1+\left(\frac{r_{2}}{h}\right)^{2}\right]^{-0.5}\right\\} $$ Use the given values for \(r_2 = 120 cm\) and \(h = 60 cm\): $$ F_{12} = 0.5\left\\{1-\left[1+\left(\frac{120}{60}\right)^{2}\right]^{-0.5}\right\\} $$ Calculate \(F_{12}\): $$ F_{12} ≈ 0.240 $$ Now to find \(F_{21}\), we can use the reciprocity theorem: $$ A_1 F_{12} = A_2 F_{21} $$ Where \(A_1\) is the surface area of the sphere and \(A_2\) is the surface area of the disk. Calculate \(A_1\) and \(A_2\): $$ A_1 = 4\pi r_1^2 = 4\pi(30)^2 ≈ 11310 cm^2 $$ $$ A_2 = \pi r_2^2 = \pi(120)^2 ≈ 45239 cm^2 $$ Now, we can solve for \(F_{21}\): $$ F_{21} = \frac{A_1 F_{12}}{A_2} ≈ \frac{11310 * 0.240}{45239} ≈ 0.06 $$ So, \(F_{12} ≈ 0.240\) and \(F_{21} ≈ 0.06\).

Calculate the net rate of radiation heat exchange between the sphere and the disk

The net rate of radiation heat exchange between the sphere and the disk can be calculated by: $$ Q_{12} = A_1 F_{12} \sigma \epsilon_{12} (T_1^4 - T_2^4) $$ Where \(\sigma\) is the Stefan-Boltzmann constant (\(\approx 5.67 * 10^{-8} W/(m^2 K^4)\)), \(\epsilon_{12}\) is the combined emissivity between the sphere and the disk, and \(T_1\) and \(T_2\) are the surface temperatures of the sphere and the disk in Kelvin, respectively. First, we need to calculate the combined emissivity \(\epsilon_{12}\): $$ \frac{1}{\epsilon_{12}} = \frac{1 - \epsilon_1}{\epsilon_1} + \frac{1 - \epsilon_2}{\epsilon_2} $$ Using the given emissivities of the sphere and disk which are \(0.9\) and \(0.5\), respectively: $$ \frac{1}{\epsilon_{12}} = \frac{1 - 0.9}{0.9} + \frac{1 - 0.5}{0.5} $$ Calculate \(\epsilon_{12}\): $$ \epsilon_{12} ≈ 0.357 $$ Now, we need to convert the surface temperatures of the sphere and the disk to Kelvin: $$ T_1 = 600 + 273 = 873 K $$ $$ T_2 = 200 + 273 = 473 K $$ Finally, we can calculate the net rate of radiation heat exchange between the sphere and the disk: $$ Q_{12} = 11310 * 0.240 * 5.67 * 10^{-8} * 0.357 * (873^4 - 473^4) W $$ Calculate \(Q_{12}\): $$ Q_{12} ≈ 33298 W $$

Part (c) Optimal conditions for heat exchange

We are given four potential modifications for increasing the net rate of radiation heat exchange between the sphere and the disk: adjusting emissivities, adjusting distance, and adding an enclosure. In this step, we will discuss what would be the optimal conditions for each modification without calculating the exact result. 1. Emissivities: To increase the net rate of radiation heat exchange, we would want to maximize the emissivities of both the sphere and the disk. The ideal value would be \(\epsilon_1 = \epsilon_2 = 1\). 2. Distance: Decreasing the distance between the two surfaces would increase the view factor, and hence increase the net rate of radiation heat exchange. However, this might not always be practical due to physical constraints. 3. Enclosure: Adding an enclosure would improve the radiation transfer by reflecting the radiated energy back to the surfaces, but it would depend on the emissivity of the enclosure material and the geometry of the enclosure. The exact influence of the enclosure would need to be determined case by case. Considering the above optimal conditions for each modification, the net rate of radiation heat exchange between the two bodies would be higher than the previously calculated value of \(33298 W\).

Key concepts

View Factor

The view factor, also known as the configuration factor or shape factor, is a crucial concept in radiation heat transfer. It represents the fraction of radiation leaving a surface that directly reaches another surface. This factor is critical when calculating the radiation exchange between two bodies, like a sphere and a disk in this problem.

Calculating the view factor requires understanding the geometry between the two interacting surfaces. For our sphere and disk, the view factor \( F_{12} \) is given by the equation:

• \[ F_{12}=0.5\left\{1-\left[1+\left(\frac{r_{2}}{h}\right)^{2}\right]^{-0.5}\right\} \]

The values \( r_2 \) (radius of the disk), and \( h \) (distance between the sphere's center and the disk) are inputs to this equation. By calculating \( F_{12} \), we determine how much energy the sphere radiates that lands on the disk.

The reciprocal factor, \( F_{21} \), can be found using the reciprocity theorem, which relates the view factors and the areas of the two surfaces:

• \( A_1 F_{12} = A_2 F_{21} \)

Here, \( A_1 \) and \( A_2 \) are the areas of the sphere and disk, respectively. This relationship emphasizes that energy exchange depends not only on the distance and orientation but also on the relative sizes of the surfaces involved.

Emissivity

Emissivity is a measure of a surface's ability to emit energy as thermal radiation. It is expressed as a ratio, with a value between 0 and 1. A black body, which is a perfect emitter, has an emissivity of 1. In real-life scenarios, surfaces emit less radiation than a black body, so their emissivities are less than 1.

In this problem, the emissivity of the sphere is 0.9 and the disk is 0.5. These values influence how much energy each body can absorb and emit:

• An object with a higher emissivity emits more radiation at a given temperature.
• Increasing the emissivity of surfaces in thermal contact enhances radiative heat transfer.

The combined emissivity \( \epsilon_{12} \) between two surfaces is calculated to adjust for their individual emissivities, thereby impacting the net radiation exchange:

• \[ \frac{1}{\epsilon_{12}} = \frac{1 - \epsilon_1}{\epsilon_1} + \frac{1 - \epsilon_2}{\epsilon_2} \]

In this exercise, enhancing the emissivity of both the sphere and the disk to 1 would maximize heat transfer, as higher emissivity means more effective radiation and absorption from and to the surfaces.

Stefan-Boltzmann Law

The Stefan-Boltzmann law is critical for calculating the radiation energy emitted by a surface. This law states that the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of its thermodynamic temperature. The formula is:

• \[ E = \sigma T^4 \]

where \( E \) is the emissive power, \( \sigma \) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \text{W}/(\text{m}^2 \text{K}^4)\)), and \( T \) is the absolute temperature in Kelvin.

For the sphere and disk, the net rate of radiation heat exchange \( Q_{12} \) is calculated considering their temperatures and emissivities:

• \( Q_{12} = A_1 F_{12} \sigma \epsilon_{12} (T_1^4 - T_2^4) \)

This formula accounts for the view factor, area, and emissivity differences, providing a comprehensive outlook on how energy transfer via radiation occurs between surfaces of different temperatures.

Understanding the Stefan-Boltzmann law enables one to predict and control thermal interactions in various engineering applications, making it an essential tool for engineers dealing with thermal systems.