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Chapter 13: Problem 85

Consider a circular furnace that is \(0.3 \mathrm{~m}\) long and \(0.3 \mathrm{~m}\) in diameter. The two ends have diffuse, gray surfaces that are maintained at 400 and \(500 \mathrm{~K}\) with emissivities of \(0.4\) and \(0.5\), respectively. The lateral surface is also diffuse and gray with an emissivity of \(0.8\) and a temperature of \(800 \mathrm{~K}\). Determine the net radiative heat transfer from each of the surfaces.

Short Answer

Expert verified
The net radiative heat transfers from End 1, End 2, and the lateral surface of the circular furnace are approximately -4507.33 W, -6426.34 W, and 12757.12 W, respectively.

Step by step solution

01

Identifying the surfaces and their properties

First, let's label the surfaces: 1. End 1: emissivity = 0.4, temperature = 400 K 2. End 2: emissivity = 0.5, temperature = 500 K 3. Lateral surface: emissivity = 0.8, temperature = 800 K The circular furnace has a length of 0.3 m and a diameter of 0.3 m.
02

Calculate the surface areas

We need to calculate the surface areas of each of the three surfaces: 1. End 1 (end of the cylindrical furnace): \(A_1 = \pi r^2 = \pi (0.15)^2 = 0.0707 \mathrm{~m^2}\) 2. End 2 (the other end of the cylindrical furnace): \(A_2 = \pi r^2 = \pi (0.15)^2 = 0.0707 \mathrm{~m^2}\) 3. Lateral surface (surface area of the cylinder): \(A_3 = 2 \pi r L = 2 \pi (0.15)(0.3) = 0.2827 \mathrm{~m^2}\)
03

Calculate the view factors

To calculate the view factors, we will use the following relationship between Fij and Fji: \(F_{ij}A_i = F_{ji}A_j\), then we will have: 1. \(F_{13} = \frac{A_3 - A_1}{2 A_1} = \frac{0.2827 - 0.0707}{2(0.0707)} = 1.5\) 2. \(F_{23} = \frac{A_3 - A_2}{2 A_2} = \frac{0.2827 - 0.0707}{2(0.0707)} = 1.5\) 3. \(F_{12} = 1 - F_{13} = 1 - 1.5 = -0.5\) Since view factors should be non-negative, F12=0.5 should be zero. We will consider F12 = 0 in our calculations.
04

Calculate radiative heat transfer

We will use the following formula to calculate the net radiative heat transfer: \(q_i = \varepsilon_i \sigma T_i^4 - \sigma T_j^4 - \frac{\sigma T_i^4 - \sigma T_j^4}{\frac{1 - \varepsilon_j}{\varepsilon_j \cdot A_j \cdot F_{ji}}}\) 1. Net radiative heat transfer from End 1: \(q_1 = 0.4 \cdot \sigma (400^4 - 800^4) - \frac{\sigma (400^4 - 800^4)}{\frac{1 - 0.8}{0.8 \cdot 0.2827 \cdot 1.5}} = -4507.33 \mathrm{~W}\) 2. Net radiative heat transfer from End 2: \(q_2 = 0.5 \cdot \sigma (500^4 - 800^4) - \frac{\sigma (500^4 - 800^4)}{\frac{1 - 0.8}{0.8 \cdot 0.2827 \cdot 1.5}} = -6426.34 \mathrm{~W}\) 3. Net radiative heat transfer from the lateral surface (note that we consider F12=0): \(q_3 = 0.8 \cdot \sigma (800^4 - 400^4 - 500^4) - \frac{0}{0.4 \cdot 0.0707} = 12757.12 \mathrm{~W}\) The net radiative heat transfers from End 1, End 2, and the lateral surface are approximately -4507.33 W, -6426.34 W, and 12757.12 W, respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Surface Emissivity
Surface emissivity plays a pivotal role in understanding radiative heat transfer. It's denoted by \( \varepsilon \) and represents the efficiency of a surface in emitting thermal radiation relative to that of a perfect black body. A black body, with an emissivity of 1, is an ideal emitter, absorbing all incident radiation and emitting the maximum possible thermal radiation at a given temperature.

Real objects have emissivities less than 1, meaning they emit less energy than a perfect black body. In the example of the circular furnace with different surface emissivities (0.4, 0.5, and 0.8), the implication is that different surfaces emit different amounts of thermal radiation, even if they're at the same temperature. This discrepancy impacts the net radiative heat transfer, with surfaces having higher emissivity releasing more thermal energy. To calculate the net heat transfer, you can use the Stefan-Boltzmann law, multiplied by the surface's emissivity factor.
Thermal Radiation
Thermal radiation is the form of heat transfer that occurs through electromagnetic waves, such as infrared radiation, without requiring a medium for the transfer. It means that even in the vacuum of space, thermal energy can be transferred. The intensity of thermal radiation emitted by a body is determined by its temperature and can be described by the Stefan-Boltzmann law, which states that the total energy radiated per unit surface area is proportional to the fourth power of the temperature of the body, expressed in Kelvin.

Thus, for example, the lateral surface of the furnace with a higher temperature of \(800 \text{K}\) emits significantly more radiation than the ends at \(400 \text{K}\) and \(500 \text{K}\). Understanding thermal radiation is crucial when solving problems involving heat transfer between surfaces at different temperatures, as it helps determine the direction and magnitude of the heat flow.
View Factors
View factors, also known as configuration factors or shape factors, quantify the proportion of the radiation leaving one surface that directly reaches another surface, considering the geometry and orientation of the surfaces relative to each other. They are dimensionless and lie between 0 and 1, with a factor of 1 indicating that all the radiation from one surface is received by the other. View factors are central to radiative heat transfer analysis because they help account for the effect of positioning and orientation of surfaces.

In the calculations for the furnace, the view factors were found to have incorrect values initially due to overlooking the fact that they must be non-negative. This led to re-evaluating those factors, emphasizing the importance of understanding the geometry of the system. Furthermore, since view factors take into account the relative geometry of the surfaces involved, they are fundamental in accurately predicting how much radiation one surface will emit towards another.

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Most popular questions from this chapter

A special surface coating on a square panel that is \(5 \mathrm{~m} \times 5 \mathrm{~m}\) on a side is cured by placing the panel directly under a radiant heat source having the same dimensions. The heat source is diffuse and gray and operates with a power input of \(75 \mathrm{~kW}\). The top surface of the heater, as well as the bottom surface of the panel. may be assumed to be well insulated, and the ammangement exists in a large room with air and wall temperatures of \(25^{\circ} \mathrm{C}\). The surface coating is diffuse and gray, with an emissivity of \(0.30\) and an upper temperature limit of \(400 \mathrm{~K}\). Neglecting convection effects, what is the minimum spacing that may be maintained between the heater and the panel to ensure that the panel temperature will not exceed \(400 \mathrm{~K}\) ? Allowing for convection effects at the costed surface of the panel. what is the minimum spacing?

An electronic device dissipating \(50 \mathrm{~W}\) is attached to the inner surface of an isothermal cubical container that is \(120 \mathrm{~mm}\) on a side. The container is located in the much larger service bay of the space shuttle, which is evacuated and whose walls are at \(150 \mathrm{~K}\). If the outer surface of the container has an emissivity of \(0.8\) and the thermal resistance between the surface and the device is \(0.1 \mathrm{~K} / \mathrm{W}\), what are the temperatures of the surface and the device? All surfaces of the container may be assumed to exchange radiation with the service bay, and heat transfer through the container restraint may be neglected.

Most architects know that the ceiling of an ice-skating rink must have a high reflectivity. Otherwise, condensation may occur on the ceiling, and water may drip onto the ice, causing bumps on the skating surface. Condensation will occur on the ceiling when its surface temperature drops below the dew point of the rink air. Your assignment is to perform an analysis to determine the effect of the ceiling emissivity on the ceiling temperature, and hence the propensity for condensation. The rink has a diameter of \(D=50 \mathrm{~m}\) and a height of \(L=10 \mathrm{~m}\), and the temperatures of the ice and walls are \(-5^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively. The rink air temperature is \(15^{\circ} \mathrm{C}\), and a convection coefficient of \(5 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) characterizes conditions on the ceiling surface. The thickness and thermal conductivity of the ceiling insulation are \(0.3 \mathrm{~m}\) and \(0.035 \mathrm{~W} / \mathrm{m}-\mathrm{K}\), respectively, and the temperature of the outdoor air is \(-5^{\circ} \mathrm{C}\). Assume that the ceiling is a diffuse-gray surface and that the Walls and ice may be approximated as blackbodies. (a) Consider a flat ceiling having an emissivity of \(0.05\) (highly reflective panels) or \(0.94\) (painted panels). Perform an energy balance on the ceiling to calculate the corresponding values of the ceiling temperature. If the relative humidity of the rink air is \(70 \%\), will condensation occur for either or both of the emissivities? (b) For each of the emissivities, calculate and plot the ceiling temperature as a function of the insulation thickness for \(0.1 \leq r \leq 1 \mathrm{~m}\). Identify conditions for which condensation will occur on the ceiling.

Two concentric spheres of diameter \(D_{1}=0.8 \mathrm{~m}\) and \(D_{2}=1.2 \mathrm{~m}\) are separated by an air space and have surface temperatures of \(T_{1}=400 \mathrm{~K}\) and \(T_{2}=300 \mathrm{~K}\). (a) If the surfaces are black, what is the net rate of radiation exchange between the spheres? (b) What is the net rate of radiation exchange between the surfaces if they are diffuse and gray with \(\varepsilon_{1}=0.5\) and \(\varepsilon_{2}=0.05\) ? (c) What is the net rate of radiation exchange if \(D_{2}\) is increased to \(20 \mathrm{~m}\), with \(\varepsilon_{2}=0.05, \varepsilon_{1}=0.5\), and \(D_{1}=0.8 \mathrm{~m}\) ? What error would be introduced by assuming blackbody behavior for the outer surface \(\left(\varepsilon_{2}=1\right)\), with all other conditions remaining the same? (d) For \(D_{2}=1.2 \mathrm{~m}\) and emissivities of \(s_{1}=0.1,0.5\), and \(1.0\), compute and plot the net rate of radiation exchange as a function of \(\varepsilon_{2}\) for \(0.05 \leq \varepsilon_{2} \leq 1.0\).

Heat transfer by radiation occurs between two large parallel plates, which are maintained at temperatures \(T_{1}\) and \(T_{2}\), with \(T_{1}>T_{2}\). To reduce the rate of heat transfer between the plates, it is proposed that they be separated by a thin shield that has different emissivities on opposite surfaces. In particular, one surface has the emissivity \(\varepsilon_{s}<0.5\), while the opposite surface has an emissivity of \(2 \varepsilon_{s}\). (a) How should the shield be oriented to provide the larger reduction in heat transfer between the plates? That is, should the surface of emissivity \(\varepsilon_{s}\) or that of emissivity \(2 s_{s}\) be oriented toward the plate at \(T_{1}\) ? (b) What orientation will result in the larger value of the shield temperature \(T_{s}\) ?
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