Question

Two parallel disks of diameter \(D=3 \mathrm{ft}\) separated by \(L=2 \mathrm{ft}\) are located directly on top of each other. The disks are separated by a radiation shield whose emissivity is \(0.15\). Both disks are black and are maintained at temperatures of \(1200 \mathrm{R}\) and \(700 \mathrm{R}\), respectively. The environment that the disks are in can be considered to be a blackbody at \(540 \mathrm{R}\). Determine the net rate of radiation heat transfer through the shield under steady conditions.

Short answer

Question: Calculate the net rate of radiation heat transfer through the shield under steady conditions between two parallel disks of diameter 3 ft, seperated by a distance of 2 ft, with a shield that has an emissivity of 0.15, temperatures of the disks being 1200R and 700R, and an environment temperature of 540R. Answer: The net rate of radiation heat transfer through the shield under steady conditions is approximately 1981.28 W.

Step-by-step solution

Convert temperatures to SI units

Since our calculations will involve the Stefan-Boltzmann constant that is given in SI units, we need to convert the given temperatures from Rankine to Kelvin: $$ T_1 = 1200\,\text{R} \times \frac{5}{9} = 666.67\,\text{K} \\ T_2 = 700\,\text{R} \times \frac{5}{9} = 388.89\,\text{K} \\ T_3 = 540\,\text{R} \times \frac{5}{9} = 300.00\,\text{K} $$

Calculate the area of the disk

The area of the disk (A) can be calculated using the formula for the area of a circle: $$ A = \pi \times \left(\frac{D}{2}\right)^2 $$ First, convert the diameter from feet to meters: $$ D = 3\,\text{ft} \times 0.3048 = 0.9144\,\text{m} $$ Now, calculate the area: $$ A = \pi \times \left(\frac{0.9144}{2}\right)^2 = 0.659\,\text{m}^2 $$

Calculate heat transfer rates for upper and lower disks

Using the Stefan-Boltzmann Law, we can calculate the heat transfer rates for the upper and lower disks: $$ q_1 = \sigma \mathrm{A} (T_1^4 - T_3^4) \\ q_2 = \sigma \mathrm{A} (T_3^4 - T_2^4) $$ Plugging in the values, we get: $$ q_1 = 5.67 \times 10^{-8}\, \mathrm{W}\text{m}^{-2}\text{K}^{-4} \times 0.659\, \text{m}^2 \times (666.67^4 - 300.00^4) = 2302.36\,\text{W} \\ q_2 = 5.67 \times 10^{-8}\, \mathrm{W}\text{m}^{-2}\text{K}^{-4} \times 0.659\, \text{m}^2 \times (300.00^4 - 388.89^4) = 452.74\,\text{W} $$

Calculate the net heat transfer rate

Now that we have the heat transfer rates, we can use the emissivity and transmittance relations to calculate the net heat transfer rate: $$ q_{net} = \sigma \mathrm{A} [(1-\epsilon)\,q_1 + \epsilon\,q_2] $$ Plugging in the values, we get: $$ q_{net} = 5.67 \times 10^{-8}\, \mathrm{W}\text{m}^{-2}\text{K}^{-4} \times 0.659\, \text{m}^2 \times [(1-0.15)\times 2302.36 + 0.15\times 452.74] = 1981.28\,\text{W} $$ Therefore, the net rate of radiation heat transfer through the shield under steady conditions is approximately \(1981.28\,\text{W}\).

Key concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann Law plays a crucial role in understanding how objects emit thermal radiation. In essence, this law tells us that the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body's absolute temperature. Formulated as an equation, it is presented as:
\[ E = \text{\(\sigma\)} T^4 \]
where \( E \) represents the emissive power, \( \sigma \) is the Stefan-Boltzmann constant (approximately \( 5.67 \times 10^{-8} \) Wm^\(-2\)K^\(-4\)), and \( T \) is the absolute temperature measured in Kelvin. This law is fundamental in calculating the rate of heat transfer via radiation from a surface, which is pertinent in many fields, from astrophysics to engineering.
In applications like the problem at hand, surfaces are often not perfect black bodies; therefore, their emissivity (\( \epsilon \)) must be considered to adjust the emission calculation, which leads us onto our next key concept, emissivity.

Emissivity

Emissivity (represented as \( \epsilon \)) is a measure of how effectively a material emits thermal radiation compared to a perfect black body. It is a dimensionless number ranging between 0 and 1; a black body, which is an ideal emitter, has an emissivity of 1, while any real object would have an emissivity less than 1. The emissivity depends on factors such as the material's surface properties and temperature.
In the context of the exercise solution, the radiation shield's emissivity is given as 0.15. This information is vital because it tells us that the shield is a poor emitter of radiation compared to a black body. The effect of emissivity on the Stefan-Boltzmann law is applied by modifying the formula to include the emissivity coefficient, resulting in a more accurate representation of the heat transfer rate for non-ideal emitters:
\[ E = \epsilon \cdot \text{\(\sigma\)} \cdot T^4 \]
When calculating the net rate of radiation heat transfer between surfaces, such as the two disks separated by a radiation shield in the exercise, the emissivity of the intervening material becomes a key factor in determining the overall rate.

Thermal Radiation

Thermal radiation is a form of heat transfer that does not require a medium; it is the energy emitted by matter as electromagnetic waves due to the vibrations of its molecules. All objects above absolute zero emit thermal radiation. Its characteristics are determined by the surface temperature and properties of the material.
In the example problem, the disks and the shield between them are exchanging energy through thermal radiation. The temperatures of the disks and the environment set up a thermal gradient that drives this radiation. The hotter disk emits more radiation than it receives from the colder disk and the environment, which in turn radiates energy back. This exchange of energy is what we quantify using the concepts of the Stefan-Boltzmann Law and emissivity.
Understanding thermal radiation is crucial in many practical applications, ranging from designing efficient insulation for buildings to creating thermal protection systems for spacecraft. It's also important for calculating the energy exchange in climatic weather models or even in the design of electronic devices to ensure proper heat dissipation.