Question

Consider a vertical 2-m-diameter cylindrical furnace whose surfaces closely approximate black surfaces. The base, top, and side surfaces of the furnace are maintained at \(400 \mathrm{~K}\), \(600 \mathrm{~K}\), and \(900 \mathrm{~K}\), respectively. If the view factor from the base surface to the top surface is \(0.2\), the net radiation heat transfer from the bottom surface is (a) \(-93.6 \mathrm{~kW}\) (b) \(-86.1 \mathrm{~kW}\) (c) \(0 \mathrm{~kW}\) (d) \(86.1 \mathrm{~kW}\) (e) \(93.6 \mathrm{~kW}\)

Short answer

The problem might require a more complex solution involving additional assumptions. As shown in our step-by-step solution, when considering the given view factor and base, top, and side surface temperatures, we find a negative net radiation heat transfer which doesn't match any available options. Discussions with your teacher or consulting a textbook may provide further clarification on this exercise.

Step-by-step solution

Understand the problem and collect information

We have a cylindrical furnace with given temperatures for the base, top, and side surfaces. The view factor from the base surface to the top surface is also given. We need to find the net radiation heat transfer from the base surface to the top surface. Given information: Base temperature T1 = \(400\,\text{K}\) Top temperature T2 = \(600\,\text{K}\) Side surface temperature T3 = \(900\,\text{K}\) View factor F12 = \(0.2\)

Calculate radiosity and emissive power for each surface

First, let's calculate the emissive power \(E_{i}\), where \(i=1,2,3\), for each surface using Stefan-Boltzmann's law which states, \(E = \sigma \cdot T^4\). The Stefan-Boltzmann constant, \(\sigma\), is equal to \(5.67 \times 10^{-8} \,\text{Wm}^{-2} \text{K}^{-4}\). \(E_1 = \sigma \cdot T_1^4 = (5.67 \times 10^{-8})(400^4) = 1829.27 \,\text{W/m}^2\) \(E_2 = \sigma \cdot T_2^4 = (5.67 \times 10^{-8})(600^4) = 12372.57 \,\text{W/m}^2\) \(E_3 = \sigma \cdot T_3^4 = (5.67 \times 10^{-8})(900^4) = 77763.47 \,\text{W/m}^2\) Next, calculate the radiosity, \(J_{i}\) for each surface using the emissive power for black bodies and considering the view factors. \(J_1 = E_1 - (1 - F_{12}) (E_1 - E_2) = 1829.27 - (1 - 0.2) (1829.27 - 12372.57) = 5628.97 \, \text{W/m}^2\) \(J_2 = E_2 - F_{12} (E_2 - E_1) = 12372.57 - 0.2 (12372.57 - 1829.27) = 9417.55\, \text{W/m}^2\) There is no need to calculate \(J_3\) for this problem, as we are finding the net radiation heat transfer between the base and top surfaces, which will not consider the side surface.

Calculate the net radiation heat transfer

Now, calculate the net radiation heat transfer from the base surface (q1) using the radiosity and view factors. \(Q_1 = q_{1} \cdot A_1 = F_{12} \cdot (J_1 - J_2) \cdot A_1\) Since we are interested in the value of the net radiation heat transfer, we will leave out the area of the surface as it will be used to normalize the results. Thus, \(q_1 = F_{12} (J_1 - J_2)\) \(q_1 = 0.2 (5628.97 - 9417.55) = -757.72 \, \text{W/m}^2\) The diameter of the furnace is 2 m, so the area of the base surface is: \(A_1 = \pi (\dfrac{d}{2})^2 = \pi (1^2) = \pi \,\text{m}^2\) Finally, find the net radiation heat transfer \(Q_1\) by multiplying \(q_1\) by the area of the base surface: \(Q_1 = -757.72 \,\text{W/m}^2 \times \pi \,\text{m}^2 = -2381.15\,\text{W}\) Now, we convert the net radiation heat transfer to \(\mathrm{kW}\): \(Q_1 = -2381.15\,\mathrm{W} \times \dfrac{1\,\mathrm{kW}}{1000\,\mathrm{W}} = -2.381\,\mathrm{kW}\) As the answer is not within the given options, let's consider that the question only involves the base and top surfaces, with the view factor \(F_{12}=1.0\) and ignore the sidewalls. In this case, \(q_1 = F_{12} (J_1 - J_2)\) \(q_1 = 1.0 (5628.97 - 9417.55) = -3788.58 \,\text{W/m}^2\) \(Q_1 = -3788.58 \,\text{W/m}^2 \times \pi \,\text{m}^2 = -11893.15\,\text{W}\) \(Q_1 = -11893.15\,\mathrm{W} \times \dfrac{1\,\mathrm{kW}}{1000\,\mathrm{W}} = -11.893\,\mathrm{kW}\) Once again, this result doesn't match any of the answer choices. There might be an error in the given options, or the exercise might require a more complex solution involving additional assumptions. In this case, we recommend consulting a textbook or discussing the problem with your teacher to receive further clarification.

Key concepts

Radiation Heat Transfer

Radiation heat transfer is the process by which energy is emitted or absorbed in the form of radiant energy. This energy is transmitted through electromagnetic waves. Unlike conduction and convection, radiation does not require a medium to transfer heat. This makes it a crucial mechanism through which heat can be transferred, even in the vacuum of space.
In most practical cases, like the cylindrical furnace described, surfaces emit and absorb radiative energy. The efficiency of this transfer depends on several factors including the temperature difference between surfaces and their properties as emitters or absorbers of radiation. Understanding these properties is key in applications like furnaces, solar panels, and heat exchangers, where optimizing heat transfer is essential.

Emissive Power

Emissive power refers to the amount of thermal radiation emitted by a surface per unit area and per unit time. It tells us how much energy in the form of radiation the surface emits. This is dependent on factors such as the surface temperature and its nature as an emitter, which can be classified as a black body, gray body, or real body.
A black body is a theoretical object that perfectly absorbs all incident radiation and re-emits it with maximum efficiency. In everyday materials, emissive power can vary significantly, but for many calculations, the concept of a black body is used as a baseline to simplify complex calculations using ideal conditions. Hence, emissive power concept is crucial for designing systems that involve heating or cooling by radiation.
For the given cylindrical furnace in the exercise, each surface has an emissive power calculated by the formula from the Stefan-Boltzmann law, considering the exact temperatures as critical inputs in the calculations.

Stefan-Boltzmann Law

The Stefan-Boltzmann law is a key principle when discussing black body radiation. The law states that the power emitted per unit area by a black body in thermal equilibrium is directly proportional to the fourth power of the body's absolute temperature. Mathematically, it is expressed as:
\[ E = \,\sigma\, T^4 \]
where \( E \) is the emissive power, \( \sigma \) is the Stefan-Boltzmann constant with a value of \(5.67 \times 10^{-8} \text{ Wm}^{-2} \text{K}^{-4}\), and \( T \) is the absolute temperature of the body.
This law provides foundational understanding in various scientific and engineering fields, predicting and calculating the thermal radiation emitted by objects. For example, in the scenario of the cylindrical furnace, Stefan-Boltzmann law helps calculate the radiation emitted by each surface, crucial for evaluating the net heat exchange.

Black Body Radiation

Black body radiation is emitted by an idealized physical object that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. Because it is a perfect absorber, such an object is often used as a model for understanding and analyzing thermal radiation.
Every real object emits thermal radiation, and black body models provide a baseline for these emissions. Even stars, planets, and other cosmic bodies approximate black body radiation when astronomers and scientists analyze the heat they radiate.
In practical scenarios, like radiative heat transfer in engineering (e.g., the cylindrical furnace), real materials approximate black bodies or gray bodies to simulate and assess efficiencies in emitting and absorbing heat. This simplifies complex calculations into more manageable models, allowing for accurate predictions and optimizations based on theoretical principles.
Understanding black body radiation is crucial not only for solving textbook problems but also for applications in designing efficient thermal systems in real-world settings.