/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 74 A radiative heater consists of a... [FREE SOLUTION] | 91Ó°ÊÓ

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A radiative heater consists of a bank of ceramic tubes with internal heating elements. The tubes are of diameter \(D=20 \mathrm{~mm}\) and are separated by a distance \(s=50 \mathrm{~mm}\). A reradiating surface is positioned behind the heating tubes as shown in the schematic. Determine the net radiative heat flux to the heated material when the heating tubes \(\left(\varepsilon_{h}=0.87\right)\) are maintained at \(1000 \mathrm{~K}\). The heated material \(\left(\varepsilon_{\mathrm{a}}=0.26\right)\) is at a temperature of \(500 \mathrm{~K}\).

Short Answer

Expert verified
The net radiative heat flux to the heated material is approximately \(121588.17 \mathrm{W/m^{2}}\).

Step by step solution

01

Find the heat emission of the ceramic tubes (heater)

To determine the radiative heat emission from the ceramic tubes, we apply the Stefan-Boltzmann Law, which states that the heat emitted by a black body is proportional to the fourth power of its absolute temperature. For a non-black body, we must include the emissivity term. The formula for the radiative heat emission is given by: \(q = \varepsilon \sigma T^4\) where \(q\) is the heat emission, \(\varepsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} W/m^2K^4)\), and \(T\) is the absolute temperature in Kelvin. For the ceramic tubes (heater), we have \(\varepsilon_{h} = 0.87\) and \(T_{h} = 1000 K\). Plugging these values into the equation, we find the heat emission of the heater: \(q_{h} = (0.87)(5.67 \times 10^{-8})(1000^4) = 147547.97 \mathrm{W/m^{2}}\)
02

Determine the heat absorption by the heated material

The heated material absorbs a fraction of the radiative heat emitted by the heater tubes. To find the heat absorption, we can use a similar formula as in step 1 but with the heated material's emissivity and temperature instead. We have \(\varepsilon_a = 0.26\) and \(T_a = 500K\). Plugging these values into the equation, we get: \(q_{a} = (0.26)(5.67 \times 10^{-8})(500^4) = 9189.135 \mathrm{W/m^{2}}\)
03

Calculate the effective emissivity for radiative exchange

The effective emissivity for radiative exchange (\(\varepsilon_{eff}\)) takes into account the reradiating surface. It can be calculated using the following formula: \(\varepsilon_{eff} = \frac{\varepsilon_h + \varepsilon_a - \varepsilon_h \varepsilon_a}{1 - \varepsilon_h \varepsilon_a}\) Plugging in the emissivities for the heater (\(\varepsilon_h = 0.87\)) and the heated material (\(\varepsilon_a = 0.26\)), we get: \(\varepsilon_{eff} = \frac{0.87 + 0.26 - (0.87)(0.26)}{1 - (0.87)(0.26)} = 0.7937\)
04

Determine the net radiative heat flux to the heated material

Now that we have the effective emissivity, we can calculate the net radiative heat flux to the heated material using the following formula: \(q_{net} = \varepsilon_{eff} \sigma (T_h^4 - T_a^4)\) Plugging in the values for \(\varepsilon_{eff}\), \(\sigma\), \(T_h\), and \(T_a\), we get: \(q_{net} = (0.7937)(5.67 \times 10^{-8})(1000^4 - 500^4) = 121588.17 \mathrm{W/m^{2}}\) So, the net radiative heat flux to the heated material is approximately \(121588.17 \mathrm{W/m^{2}}\).

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Key Concepts

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

Stefan-Boltzmann Law
The Stefan-Boltzmann Law explains how objects emit heat in the form of radiation. It is particularly useful in understanding how heaters, like ceramic tubes, release energy. The law states that the total energy radiated per unit surface area of a black body in unit time is directly proportional to the fourth power of the black body's absolute temperature. However, most real materials aren't perfect black bodies, so we introduce a factor called emissivity. This law can be expressed mathematically as:
  • For a perfect black body: \( q = \sigma T^4 \)
  • For real materials: \( q = \varepsilon \sigma T^4 \)
Here, \(q\) is the heat emission, \(\sigma\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \mathrm{~W/m^2K^4})\), and \(T\) is the absolute temperature in Kelvin. Emissivity (\(\varepsilon\)) adjusts the equation for real-world materials, scaling the potential radiation output.
Emissivity
Emissivity is a measure of a material's ability to emit energy as radiation compared to a perfect black body. It ranges from 0 to 1, where 1 represents a perfect black body, which emits radiation most efficiently. In our example, the ceramic heater has an emissivity of 0.87, meaning it radiates 87% as effectively as a perfect black body.
  • Higher emissivity values mean more efficient radiator.
  • Lower emissivity materials are less effective at radiating heat.
For instance, metals tend to have lower emissivity, making them less efficient radiators. The ceramic heater’s high emissivity makes it effective in releasing energy to heat surrounding areas, such as in the given problem.
Net Radiative Heat Flux
Net radiative heat flux helps us understand the balance of heat transfer between two surfaces. In our scenario, it's about how much heat the ceramic tubes transfer to the heated material. The net flux is computed by considering both the emitted and absorbed radiation, incorporating the temperature difference.The formula is:\[ q_{net} = \varepsilon_{eff} \sigma (T_h^4 - T_a^4) \]Where:
  • \( q_{net} \) is the net radiative heat flux.
  • \( \varepsilon_{eff} \) is the effective emissivity.
  • \( T_h \) is the heater temperature and \( T_a \) is the temperature of the heated material.
In our example, the result is approximately \(121588.17 \mathrm{W/m^2}\), indicating the net energy transferred from the ceramic heater to the material.
Effective Emissivity
Effective emissivity accounts for interactions between different surfaces and materials, providing a more realistic measure of heat exchange. When multiple bodies are involved, especially with a reradiating surface like in our problem, we need this concept.The formula is:\[ \varepsilon_{eff} = \frac{\varepsilon_h + \varepsilon_a - \varepsilon_h \varepsilon_a}{1 - \varepsilon_h \varepsilon_a} \]This includes:
  • \( \varepsilon_h \) - Emissivity of the heater.
  • \( \varepsilon_a \) - Emissivity of the absorbing material.
Using this, we calculated \( \varepsilon_{eff} \) as approximately 0.7937. This shows that the presence of additional surfaces, like reradiators, affects the overall heat transfer efficiency.

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Most popular questions from this chapter

Consider the cavities formed by a cone, cylinder, and sphere having the same opening size \((d)\) and major dimension \((L)\), as shown in the diagram. (a) Find the view factor between the inner surface of each cavity and the opening of the cavity. (b) Find the effective emissivity of each cavity, \(\varepsilon_{e}\), as defined in Problem 13.43, assuming the inner walls are diffuse and gray with an emissivity of \(\varepsilon_{1 N^{-}}\) (c) For each cavity and wall emissivities of \(\varepsilon_{w^{\prime}}=0.5\), \(0.7\), and \(0.9\), plot \(\varepsilon_{e}\) as a function of the major dimension- to-opening size ratio, \(L /\), over a range from 1 to 10 .

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.

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 \(10 \leq h_{0} \leq 100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A radiant oven for drying newsprint consists of a long duct \((L=20 \mathrm{~m})\) of semicincular cross section. The newsprint moves through the oven on a conveyor belt at a velocity of \(V=0.2 \mathrm{~m} / \mathrm{s}\). The newsprint has a water content of \(0.02 \mathrm{~kg} / \mathrm{m}^{2}\) as it enters the oven and is completely dry as it exits. To assure quality, the newsprint must be maintained at room temperature \((300 \mathrm{~K})\) during drying. To aid in maintaining this condition, all system components and the air flowing through the oven have a temperature of \(300 \mathrm{~K}\). The inner sarface of the semicaircular duct, which is of emissivity \(0.8\) and temperature \(T_{1}\), provides the radiant heat required to accomplish the drying. The wet surface of the newsprint can be considered to be black. Air entering the oven has a temperature of \(300 \mathrm{~K}\) and a relative humidity of \(20 \%\). Since the velocity of the air is large, its temperature and relative humidity can be assumed to be constant over the entire duct length. Calculate the required evaporation rate, air velocity \(u_{m}\), and temperature \(T_{1}\) that will ensure steady-state conditions for the process.

Long, cylindrical bars are heat-treated in an infrared oven. The bars, of diameter \(D=50 \mathrm{~mm}\), are placed on an insulated tray and are heated with an overhead infrared panel maintained at temperature \(T_{p}=800 \mathrm{~K}\) with \(\varepsilon_{p}=0.85\). The bars are at \(T_{b}=300 \mathrm{~K}\) and have an emissivity of \(\varepsilon_{b}=0.92\). (a) For a product spacing of \(s=100 \mathrm{~mm}\) and a product length of \(L=1 \mathrm{~m}\), determine the radiation heat flux delivered to the product. Determine the heat flux at the surface of the panel heater. (b) Plot the radiation heat flux experienced by the product and the panel heater radiation heat flux over the range \(50 \mathrm{~mm} \leq s \leq 250 \mathrm{~mm}\).

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