Two parallel, aligned disks, 0.4 m in diameter and separated by 0.1 m, are located in a large room whose walls are maintained at 300 K. One of the disks is maintained at a uniform temperature of 500 K with an emissivity of 0.6., while the backside of the second disk (disc 2) is well insulated. Room is large enough to be considered as a black body. Assuming the discs are diffuse, gray surfaces, answer the following questions: (a) Compute the view factors F12 and F13 [2 points] (b) Draw the radiation network for this problem [2 points] (c) Write an equation for radiation balance at surface 1 [2 points] (d) Write an equation for radiation balance at surface 2 [2 points] (e) Compute the temperature of the second disk. [2 points]

(a) Compute the view factors F12 and F13: The view factor, also known as the configuration factor or shape factor, is a measure of how well one surface "views" another surface in terms of potential radiative exchange. For two parallel, aligned and identical disks of diameter 0.4 m (radius r = 0.2 m) separated by 0.1 m (distance S = 0.1 m), the view factor from disk 1 to disk 2 (F12) can be approximated as:

\[ F_{12} = \frac{1}{1+\left(\frac{S}{r}\right)^2} = \frac{1}{1+\left(\frac{0.1}{0.2}\right)^2} = \frac{1}{1+\frac{1}{4}} = \frac{1}{1.25} = 0.8 \]

The view factor F13 (from disc 1 to the walls of the room), knowing that the room is large compared to the disks and can be considered a black body, is given by:

\[ F_{13} = 1 - F_{12} = 1 - 0.8 = 0.2 \]

(b) Draw the radiation network for this problem: The radiation network represents the radiative exchange between surfaces. It consists of nodes representing the radiating surfaces and arrows representing the heat transfer between these nodes.

``` 500K, ε=0.6 ----> ( F12 )----> ?K, Insulated | | (F13) | v | 300K, black body <------------------ ```

(c) Write an equation for radiation balance at surface 1: The radiation balance at surface 1 (the disk at 500 K) accounts for the energy radiating to both the second disk and the large room walls. If A is the area of the disk and σ is the Stefan-Boltzmann constant:

\[ J_1 A = ε_1 σ T_1^4 A F_{12} + ε_1 σ T_1^4 A F_{13} \]

where J1 is the radiosity representing the total exitant radiative flux from surface 1, and T1 is the temperature of the first disk in Kelvin.

(d) Write an equation for radiation balance at surface 2: The radiation balance at surface 2 must consider the radiative exchange with disk 1 and the insulation on the backside. Since the backside is insulated, there is no net radiative exchange with the environment through that surface. If we consider that the second disk reaches a uniform temperature T2, then the energy arriving at disk 2 from disk 1 should be equal to the energy leaving disk 2:

\[ ε_2 σ T_2^4 A = J_1 A F_{12} \]

Since the problem does not specify the emissivity ε_2 of disk 2, we might assume it's the same as disk 1 (ε_2 = ε_1). However, if it's different, you would need that value to proceed with the calculation.

(e) Compute the temperature of the second disk: To solve for the temperature of the second disk, we would need to derive J1 from the balance at surface 1 and then use that to solve for T2 in the balance at surface 2. This would typically require an iterative solution since the radiosity J1 is a function of both T1 and T2. Without a specified emissivity for disk 2, we can't provide a numeric answer for T2 here.

Extra:

To explain further to a student:

The term "view factor" or "configuration factor" is a concept in radiation heat transfer that assesses how well two surfaces can exchange thermal radiation. Specifically, it measures the proportion of radiation leaving one surface that directly reaches another surface. The emissivity (ε) of a surface reflects how effectively it emits thermal radiation compared to an ideal black body. An emissivity of 1 indicates a perfect black body, while real materials have emissivities less than 1. The Stefan-Boltzmann constant (σ) is a physical constant that appears in the Stefan-Boltzmann law, which 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 constant is roughly equal to 5.67 x 10^-8 W/m^2K^4. To find temperatures in radiative heat transfer problems, we often set up energy balance equations considering emissive power and exchange factors and solve for the unknown quantities. As mentioned in this case, solving for the second disk's temperature could be complex and may require you to assume some properties or use numerical methods.