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Chapter 13: Problem 58

Consider two large, diffuse, gray, parallel surfaces separated by a small distance. If the surface emissivities are \(0.8\), what emissivity should a thin radiation shield have to reduce the radiation heat transfer rate between the two surfaces by a factor of 10 ?

Short Answer

Expert verified
The problem statement is not valid, and it is not possible to determine the emissivity of the shield under the given conditions. This is because the equation we obtain has no unique solution, as it leads to division by zero.

Step by step solution

01

Write the radiation transfer equation for the 2 surfaces before introducing the shield

To determine the radiation heat transfer between two surfaces, we use the equation: \(Q_{1-2} = A_1F_{1-2}\sigma T^4_1 - A_2F_{2-1}\sigma T^4_2\) Since the surfaces are diffuse-gray, we can use their emissivities to modify the equation: \(Q_{1-2} = A_1F_{1-2}\varepsilon_1 \sigma T^4_1 - A_2F_{2-1}\varepsilon_2 \sigma T^4_2\) Given, emissivities of both surfaces (\(\varepsilon_1\) and \(\varepsilon_2\)) are 0.8.
02

Write the radiation transfer equation for the 2 surfaces after introducing the shield

After introducing the radiation shield, the radiation transfer equation becomes: \(Q'_{1-2} = A_1F_{1-2}\varepsilon_1\varepsilon_s \sigma T^4_1 - A_2F_{2-1}\varepsilon_2\varepsilon_s \sigma T^4_2\) Where the unknown \(\varepsilon_s\) is the emissivity of the radiation shield.
03

Set the ratio between the two equations equal to 10

According to the problem statement, the radiation heat transfer rate must be reduced by a factor of 10 after introducing the shield. Therefore: \(\frac{Q'_{1-2}}{Q_{1-2}} = \frac{1}{10}\) Substituting the equations we have: \(\frac{A_1F_{1-2}\varepsilon_1\varepsilon_s \sigma T^4_1 - A_2F_{2-1}\varepsilon_2\varepsilon_s \sigma T^4_2}{A_1F_{1-2}\varepsilon_1 \sigma T^4_1 - A_2F_{2-1}\varepsilon_2 \sigma T^4_2} = \frac{1}{10}\) Since the areas, view factors, temperatures and the value of \(\sigma\) are equal (cancel out) in both before and after the shield introduction, we can simplify the equation to: \(\frac{\varepsilon_1\varepsilon_s - \varepsilon_2\varepsilon_s}{\varepsilon_1 - \varepsilon_2} = \frac{1}{10}\)
04

Solve for the emissivity of the radiation shield

Substituting the values of \(\varepsilon_1\) and \(\varepsilon_2\) (0.8): \(\frac{0.8\varepsilon_s - 0.8\varepsilon_s}{0.8 - 0.8} = \frac{1}{10}\) This equation has no unique solution as 0.8 - 0.8 = 0, leading to division by zero. The problem statement is not valid and it is not possible to determine the emissivity of the shield under these conditions.

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Key Concepts

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

Emissivity
Emissivity is a measure of a material's ability to emit thermal radiation compared to an ideal emitter or black body. A black body, with an emissivity of 1, is the perfect emitter and absorber of radiation. Most real-world surfaces have emissivities between 0 and 1, and they emit less radiation than a black body.
Surfaces with high emissivity emit energy efficiently, appearing more like a black body, while those with low emissivity reflect more energy and emit less. The emissivity of a surface significantly affects the thermal radiation exchanged between it and other surfaces.
In the provided exercise, both surfaces have an emissivity of 0.8, indicating they emit 80% of the thermal radiation that a black body at the same temperature would emit. This is a relatively high emissivity, meaning the surfaces are relatively efficient in radiating energy.
Radiation Shield
Radiation shields are used to reduce the rate of heat transfer between surfaces by reflecting or absorbing heat. When a thin radiation shield is introduced between two surfaces, it modifies the path and effectiveness of radiation exchange between them.
The shield acts as an intermediate surface that alters the effective emissivity between the original surfaces. With a shield in place, the heat transfer reduction is primarily due to the reflection and absorption properties of the shield.
For calculating the required emissivity of a radiation shield, understanding the effective change in heat transfer rate due to the emissivity of the shield is crucial. In the exercise, the objective was to reduce the radiation heat transfer by a factor of 10 with the help of a shield, though the specific solution was not valid due to equal emissivities of the surfaces.
Diffuse Gray Surfaces
Diffuse gray surfaces are an important concept in radiation heat transfer. They assume that the surface radiates energy uniformly in all directions and behaves consistently across the spectrum.
For diffuse gray surfaces, the emissivity is wavelength-independent, making calculations simpler. This means the spectral characteristics of radiation emitted or absorbed by these surfaces do not vary with wavelength.
In the exercise, the two surfaces were assumed to be diffuse gray, facilitating the use of simplified emissivity-based formulas to predict heat transfer. This assumption simplifies complex problems into easier calculations while capturing important radiation behavior. It is particularly useful when precision regarding spectral wavelength distribution is not crucial.

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Most popular questions from this chapter

A wall-mounted natural gas heater uses combustion on a porous catalytic pad to maintain a ceramic plate of emissivity \(\varepsilon_{c}=0.95\) at a uniform temperature of \(T_{c}=1000 \mathrm{~K}\). The ceramic plate is separated from a glass plate by an air gap of thickness \(L=50 \mathrm{~mm}\). The surface of the glass is diffuse, and its spectral transmissivity and absorptivity may be approximated as \(\tau_{\lambda}=0\) and \(\alpha_{\lambda}=1\) for \(0 \leq \lambda \leq 0.4 \mu \mathrm{m}, \tau_{\lambda}=1\) and \(\alpha_{\lambda}=0\) for \(0.4<\lambda \leq 1.6 \mu \mathrm{m}\), and \(\tau_{\lambda}=0\) and \(\alpha_{\lambda}=0.9\) for \(\lambda>1.6 \mu \mathrm{m}\). The exterior sarface of the glass is exposed to quiescent ambient air and large surroundings for which \(T_{m}=T_{a x}=300 \mathrm{~K}\). The height and width of the heater are \(H=W=2 \mathrm{~m}\). (a) What is the total transmissjvity of the glass to irradiation from the ceramic plate? Can the glass be approximated as opaque and gray? (b) For the prescribed conditions, evaluate the glass temperature, \(T_{R}\), and the rate of heat transfer from the heater, \(q_{k}\). (c) A fan may be used to control the convection coefficient \(h_{n}\) at the exterior surface of the glass. Compute and plot \(T_{g}\) and \(q_{\mathrm{b}}\) as a function of \(h_{e}\) for \(10 \leq h_{0} \leq 100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Boiler tubes exposed to the products of coal combustion in a power plant are subject to fouling by the ash (mineral) content of the combustion gas. The ash forms a solid deposit on the tube outer surface, which reduces heat transfer to a pressurized wated'steam mixture flowing through the tubes. Consider a thin-walled boiler tube \(\left(D_{t}=0.05 \mathrm{~m}\right)\) whose surface is maintained at \(T_{t}=600 \mathrm{~K}\) by the boiling process. Combustion gases flowing over the tube at \(T_{-}=1800 \mathrm{~K}\) provide a convection coefficient of \(\bar{h}=100 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\), while radiation from the gas and boiler walls to the tube may be approximated as that originating from large surroundings at \(T_{\text {arr }}=1500 \mathrm{~K}\). (a) If the tube surface is diffuse and gray, with \(\varepsilon_{t}=0.8\), and there is no ash deposit layer, what is the rate of heat transfer per unit length, \(q^{\prime}\), to the boiler tube? (b) If a deposit layer of diameter \(D_{d}=0.06 \mathrm{~m}\) and thermal conductivity \(k=1 \mathrm{~W} / \mathrm{m}\), \(\mathrm{K}\) forms on the tube, what is the deposit surface temperature, \(T_{d} ?\) The depos茂t is diffuse and gray, with \(\varepsilon_{a}=0.9\), and \(T_{m} T_{\ldots}, T_{\text {}}\), and \(T_{\text {sar }}\) remain unchanged. What is the net rate of heat transfer per wnit length, \(y^{\prime}\) ', to the boxler tube? (c) Explore the effect of variations in \(D_{d}\) and \(\bar{h}\) on \(q^{r}\), as well as on relative contributions of convection and radiation to the net heat transfer fate. Represent your results graphically.

A novel infrared recycler has been proposed for reclaiming the millions of kilograms of waste plastics produced by the dismantling and shredding of automotive vehicles following their refirement. To address the problem of sorting mixed plastics into components such as polypropylene and polycarbonate, a washed stream of the mixed plastics is routed to an infrared heating system, where it is dried and stbsequently heated to a temperature for which one of the components begins to soften, while the others remain rigid. The mixed stream is then routed through steel rollers, to which the softened plastic sticks and is removed from the stream. Heating of the stream is then continued to facilitate removal of a second component, and the heating/removal process is repeated until all of the components are separated. Consider the initial drying stage for a system comprised of a cylindrical heater aligned coaxially with a rotating drum of diameter \(D_{d}=1 \mathrm{~m}\). Shortly after entering the drum, wet plastic pellets may be assumed to fully cover the bottom semicylindrical section and to remain at a temperature of \(T_{p}=\) \(325 \mathrm{~K}\) during the drying process. The surface area of the pellets may be assumed to correspond to that of the semicylinder and to have an emissivity of \(\varepsilon_{p}=0.95\). (a) If the flow of dry air through the drum maintains a convection mass transfer coefficient of \(0.024 \mathrm{~m} / \mathrm{s}\) on the surface of the pellets, what is the evaporation rate per unit length of the drum? (b) Neglecting convection heat transfer, determine the temperature \(T_{b}\) that must be maintained by a heater of diameter \(D_{\mathrm{b}}=0.10 \mathrm{~m}\) and emissivity \(\varepsilon_{h}=0.8\) to sustain the foregoing evaporation rate. What is the corresponding value of the temperature \(T_{d}\) for the top surface of the drum? The outer surface of the dram is well insulated, and its length-to-diameter ratio is large. As applied to the top \((d)\) or bottom \((p)\) surface of the drum, the view factor of an infinitely long semicylinder to itself, in the presence of a concentric, coaxial cylinder, may be expressed as $$ \begin{aligned} F_{U}=& 1-\frac{2}{\pi}\left\\{\left[1-\left(D_{d} / D_{d}\right)^{2}\right]^{1 / 2}\right.\\\ &\left.+\left(D_{k} / D_{d}\right) \sin ^{-1}\left(D_{h} / D_{d}\right)\right\\} \end{aligned} $$

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.

A grain dryer consists of a long semicircular duct of radius \(R=1 \mathrm{~m}\). One-half of the base surface consists of an electrically heated plate of emissivity \(z_{p}=0.8\), while the other half supports the grain to be dried, which has an emissivity of \(\varepsilon_{g}=0.9\). In a batch drying process for which the temperature of the grain is \(T_{s}=330 \mathrm{~K}, 2.50 \mathrm{~kg}\) of water are to be removed per meter of duct length over a 1-h period. (a) Neglecting convection heat transfer, determine the required temperature \(T_{p}\) of the heater plate. (b) If the water vapor is swept from the duct by the flow of dry air, what convection mass transfer coefficient \(h_{w}\) must be maintained by the flow? (c) If the air is at \(300 \mathrm{~K}\), is the assumption of negligible convection justified?
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