Question

Two parallel concentric disks, \(20 \mathrm{~cm}\) and \(40 \mathrm{~cm}\) in diameter, are separated by a distance of \(10 \mathrm{~cm}\). The smaller disk \((\varepsilon=0.80)\) is at a temperature of \(300^{\circ} \mathrm{C}\). The larger disk \((\varepsilon=0.60)\) is at a temperature of \(800^{\circ} \mathrm{C}\). (a) Calculate the radiation view factors. (b) Determine the rate of radiation heat exchange between the two disks. (c) Suppose that the space between the two disks is completely surrounded by a reflective surface. Estimate the rate of radiation heat exchange between the two disks.

Short answer

Answer: The rate of radiation heat exchange between the two disks with a perfect reflective surface is 3443 W.

Step-by-step solution

Define the known quantities

The given quantities are: - Diameter of smaller disk, \(D_1 = 20 \mathrm{cm}\) - Diameter of larger disk, \(D_2 = 40 \mathrm{~cm}\) - Distance between disks, \(H = 10 \mathrm{~cm}\) - Emissivity of smaller disk, \(\varepsilon_1 = 0.80\) - Emissivity of larger disk, \(\varepsilon_2 = 0.60\) - Temperature of smaller disk, \(T_1 = 300^{\circ}\mathrm{C} = 300+273.15 = 573.15 \mathrm{K}\) - Temperature of larger disk, \(T_2 = 800^{\circ}\mathrm{C} = 800+273.15 = 1073.15 \mathrm{K}\)

Calculate the radiation view factors

For concentric disks, the radiation view factor (also known as the shape factor) is given by the following formula: $$F_{1 \to 2} = F_{2 \to 1} = \frac{1}{1 + \frac{D_1^2}{4 H^2}}$$ Where \(F_{1 \to 2}\) is the radiation view factor between the smaller and larger disks. Now, substitute the given dimensions into the formula to find the view factor: $$F_{1 \to 2} = F_{2 \to 1} = \frac{1}{1 + \frac{(20 \mathrm{~cm})^2}{4 (10 \mathrm{~cm})^2}} = 0.80$$ So, the radiation view factors between the disks are 0.80 . #b) Determining the rate of radiation heat exchange between the two disks#

Use the Stefan-Boltzmann Law

The formula for the net rate of radiation heat exchange between the disks is given by: $$q_{net} = \varepsilon_1 \sigma F_{1 \to 2} A_1 (T_1^4 - T_2^4)$$ Where \(q_{net}\) is the net radiation heat exchange, \(\sigma = 5.67 \times 10^{-8} \mathrm{W/m^2K^4}\) is the Stefan-Boltzmann constant, and \(A_1 = \pi r_1^2\) is the area of the smaller disk with \(r_1 = \frac{D_1}{2}\). Now, plug in the values and solve for \(q_{net}\): $$q_{net} = 0.80 \times 5.67 \times 10^{-8} \times 0.80 \times \pi \times (0.1 \mathrm{m})^2 (573.15^4 - 1073.15^4) \mathrm{W} = -2583 \mathrm{W}$$ The negative sign indicates heat is radiated from the larger to the smaller disk. Thus, the rate of radiation heat exchange is \(2583 \mathrm{W}\). #c) Estimating the rate of radiation heat exchange with a reflective surface#

Set up equations for heat exchange with the reflective surface

Let the effective emissivity of the surrounding reflective surface be \(\varepsilon_s\). With a reflective surface, the rate of radiation heat exchange is given by: $$q_{net} = \frac{\varepsilon_1 \varepsilon_s}{1 - \varepsilon_1^2 - \varepsilon_s^2 + \varepsilon_1 \varepsilon_s} \sigma F_{1 \to 2} A_1 (T_1^4 - T_2^4)$$ Assuming that the reflective surface is perfect, and the effective emissivity is near zero, \(\varepsilon_s \rightarrow 0\). With this assumption, the equation simplifies to: $$q_{net} = \frac{\varepsilon_1}{1 - \varepsilon_1^2} \sigma F_{1 \to 2} A_1 (T_1^4 - T_2^4)$$

Calculate the new rate of radiation heat exchange

Substitute the values into the simplified equation: $$q_{net} = \frac{0.80}{1 - (0.80)^2} \times 5.67 \times 10^{-8} \times 0.80 \times \pi \times (0.1 \mathrm{m})^2 (573.15^4 - 1073.15^4) \mathrm{W} = -3443 \mathrm{W}$$ The rate of radiation heat exchange between the two disks with a reflective surface is \(3443 \mathrm{W}\).

Key concepts

Radiation View Factors

Understanding radiation view factors is crucial for solving heat exchange problems involving radiation between surfaces. In the context of the exercise mentioned, the radiation view factor—or shape factor—measures how well two surfaces 'see' each other. Given the concentric arrangement of two disks, their view factor can be calculated using the formula
\[\begin{equation}F_{1 \to 2} = F_{2 \to 1} = \frac{1}{1 + \frac{D_1^2}{4 H^2}}\end{equation}\]
where \(D_1\) and \(H\) represent the diameter of the smaller disk and the distance between the two disks. This calculation assumes that all the radiation emitted by one disk is intercepted by the other. For our exercise, this factor turned out to be 0.80, indicating that a significant portion of radiation between the disks is effective. Here's a tip: remember that view factors always lie between 0 and 1, and they help quantify the directional dependence of radiation interactions.

Stefan-Boltzmann Law

When delving into the heat exchange through radiation, the Stefan-Boltzmann Law serves as the foundational equation. This law states that the power radiated per unit area of a black body is proportional to the fourth power of the black body's temperature. Mathematically, it is represented as
\[\begin{equation}j^{\ast} = \varepsilon \sigma T^4\end{equation}\]
where \(j^{\ast}\) is the total power radiated per unit area, \(\varepsilon\) is the emissivity of the material, \(\sigma\) is the Stefan-Boltzmann constant—approximately equal to \(5.67 \times 10^{-8} \mathrm{W/m^2K^4}\)—and \(T\) is the absolute temperature in Kelvins. Emissivity is the ratio of the radiation emitted by a surface to the radiation emitted by a black body at the same temperature and can vary between 0 and 1. The negative sign in our heat exchange calculation indicates the direction of heat flow—from the hotter to the cooler disk.

Emissivity

Emissivity, denoted by \(\varepsilon\), is a measure of how effectively a surface emits thermal radiation compared to a perfect black body. In the provided exercise, emissivity values are given for both disks. The smaller disk has an emissivity of 0.80 and the larger disk has an emissivity of 0.60.

• High emissivity indicates that a surface is an efficient emitter of radiation.
• Materials with low emissivity are poor emitters and tend to reflect most of the radiation.

The role of emissivity is crucial in the rate of heat transfer, as it directly affects the Stefan-Boltzmann Law calculation. When improving this exercise, ensure that the concept of emissivity is clear: it is a property of the material's surface and influences how much energy is radiated away.

Reflective Surface Influence on Heat Transfer

Introducing a reflective surface between radiative bodies significantly influences heat transfer calculations. Reflective surfaces can reduce the net radiation heat exchange between the bodies by reflecting a portion of the radiation away.

• In an ideal case with a perfectly reflective surface, no heat would be transferred between the disks by radiation.
• However, in real-world scenarios, even highly reflective surfaces have some level of emissivity that needs to be considered in the calculations.

In the exercise at hand, by introducing a reflective surface and assuming it to be perfect, we see an increased rate of radiation heat exchange due to multiple reflections between the disks before the radiation is absorbed. This can be quantified using modified equations that factor in the presence of such a surface. This concept can be seemingly counterintuitive, so make sure to explain that in reality, the heat transfer rate will typically decrease rather than increase when introducing real-world reflective barriers.