Question

Consider two black coaxial parallel circular disks of equal diameter \(D\) that are spaced apart by a distance \(L\). The top and bottom disks have uniform temperatures of \(500^{\circ} \mathrm{C}\) and \(520^{\circ} \mathrm{C}\), respectively. Determine the radiation heat transfer coefficient \(h_{\text {rad }}\) between the disks if they are spaced apart by \(L=D\).

Short answer

Answer: The radiation heat transfer coefficient between the disks is approximately 4.00 W/m²·K.

Step-by-step solution

Convert temperatures to Kelvin

To work with the temperatures in the Stefan-Boltzmann law, we need to convert the temperatures from Celsius to Kelvin: \(T_1 = 500 + 273.15 = 773.15 \, \text{K}\) \(T_2 = 520 + 273.15 = 793.15 \, \text{K}\)

Calculate the view factor

Since the disks are parallel, coaxial, and have equal diameters, the view factor \(F_{1 \to 2}\) can be determined using the equation for parallel disks of equal diameter: \(F_{1 \to 2} = \frac{1}{2} \left(1 - \frac{1}{\sqrt{1 + \frac{4L^2}{D^2}}} \right)\) Given that \(L = D\), we have: \(F_{1 \to 2} = \frac{1}{2} \left(1 - \frac{1}{\sqrt{1 + \frac{4D^2}{D^2}}} \right) = \frac{1}{2} \left(1 - \frac{1}{\sqrt{1 + 4}} \right) = \frac{1}{2} \left(1 - \frac{1}{\sqrt{5}} \right)\)

Calculate the net radiative heat flux

We can now calculate the net radiative heat flux \(q_{1 \to 2}\) between the disks using the Stefan-Boltzmann law, given as: \(q_{1 \to 2} = F_{1 \to 2} \sigma (T_1^4 - T_2^4)\) where \(\sigma \approx 5.67 \times 10^{-8} \, \text{W} \cdot \text{m}^{-2} \cdot \text{K}^{-4}\) is the Stefan-Boltzmann constant. Substitute the values into the equation: \(q_{1 \to 2} = \frac{1}{2} \left(1 - \frac{1}{\sqrt{5}} \right) \times 5.67 \times 10^{-8} (773.15^4 - 793.15^4)\) Calculate the net radiative heat flux: \(q_{1 \to 2} \approx -80.04 \, \text{W/m}^2\) (negative sign indicates heat transfer is from disk 2 to disk 1)

Calculate the radiation heat transfer coefficient

Finally, we can determine the radiation heat transfer coefficient \(h_{\text{rad}}\) by dividing the net radiative heat flux by the temperature difference between the disks: \(h_{\text{rad}} = \frac{-q_{1 \to 2}}{T_2 - T_1} = \frac{80.04}{793.15 - 773.15} \approx 4.00 \, \text{W/m}^2 \cdot \text{K}\) Thus, the radiation heat transfer coefficient between the disks is approximately \(4.00 \, \text{W/m}^2 \cdot \text{K}\).

Key concepts

Stefan-Boltzmann Law

In the realm of thermal radiation, the Stefan-Boltzmann Law plays a crucial role in determining the radiant flux from a black body. Simply put, 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 absolute temperature. Mathematically, it is expressed as:

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

This law helps in calculating the heat transfer between objects by radiation. When applying this to two objects, like our coaxial disks, we use the temperatures given after converting them from Celsius to Kelvin. Understanding that temperatures must be in Kelvin is crucial as it creates a standard unit of measure for thermal calculations.

View Factors

View factors, also known as configuration factors, are essential in analyzing radiant heat transfer between surfaces. They describe the fraction of radiation leaving one surface that strikes another surface directly. This concept is especially important when dealing with parallel surfaces, such as the coaxial disks in our problem. The view factor between two surfaces, like our disks, takes into account their geometrical alignment and influences the net radiant exchange. For coaxial disks with equal diameter and spacing, the view factor \(F_{1 \to 2}\) can be calculated with:

• \[F_{1 \to 2} = \frac{1}{2} \left(1 - \frac{1}{\sqrt{1 + \frac{4L^2}{D^2}}} \right)\]

Here, \(L\) is the separation between the disks and \(D\) is their diameter. This calculation helps to quantify how much thermal radiation is effectively exchanged between the disks, emphasizing that not all emitted radiation reaches the other disk due to their spatial configuration.

Coaxial Disks

The term coaxial comes into play when two or more structures share a common axis, which is exactly what the coaxial disks in our problem do. These disks are characterized by being parallel, having equal diameters, and being aligned along the same central axis. This alignment facilitates a simplified model for radiation heat transfer. With coaxial disks, their geometry allows for specific view factor calculations as discussed earlier. Since they share an axis and have equal diameters, their alignment simplifies many variables within heat transfer equations. Such geometrical settings are not only common in textbook problems but also crucial in practical applications, like in electronic systems where heat dispersion between components is vital for performance and longevity.

Heat Transfer Coefficient

The heat transfer coefficient \(h\) is an important parameter in characterizing how effectively heat is transferred between surfaces via various modes like conduction, convection, or radiation. In our scenario, \(h_{rad}\) specifically refers to the radiation heat transfer coefficient between the coaxial disks.Radiation heat transfer coefficient can be calculated by dividing the net radiative heat flux by the temperature difference between the two surfaces. The formula goes like:

• \[h_{\text{rad}} = \frac{-q_{1 \to 2}}{T_2 - T_1}\]

Here, \(q_{1 \to 2}\) is the net radiative heat flux obtained through the Stefan-Boltzmann law, and \(T_2 - T_1\) is the temperature difference between the two disks.This coefficient is crucial for understanding how much heat energy is transferred through radiation alone, ignoring other heat transfer modes. In design processes and simulations, using a precise heat transfer coefficient ensures more accurate modeling of thermal responses.