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Chapter 13: Problem 34

An electrically heated sample is maintained at a surface temperature of \(T_{s}=500 \mathrm{~K}\). The sample coating is diffuse but spectrally selective, with the spectral emissivity distribution shown schematically. The sample is irradiated by a furnace located coaxially at a distance of \(L_{\mathrm{s}}=750 \mathrm{~mm}\). The furnace has isothermal walls with an emissivity of \(s_{f}=0.7\) and a uniform temperature of \(T_{f}-3000 \mathrm{~K}\). A radiation detector of area \(A_{d}=8 \times 10^{-5} \mathrm{~m}^{2}\) is positioned at a distance of \(L_{\mathrm{dd}}=1.0 \mathrm{~m}\) from the sample along a direction that is \(45^{\circ}\) from the sample normal. The detector is sensitive to spectral radiant power only in the spectral region from 3 to \(5 \mu \mathrm{m}\). The sample surface experiences convection with a gas for which \(T_{\infty}=300 \mathrm{~K}\) and \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surroundings of the sample mount are large and at a uniform temperature of \(T_{\text {sur }}=300 \mathrm{~K}\). (a) Determine the electrical power, \(P_{c}\), required to maintain the sample at \(T_{s}=500 \mathrm{~K}\). (b) Considering both emission and reflected irradiation from the sample, determine the radiant power that is incident on the detector within the spectral region from 3 to \(5 \mu \mathrm{m}\).

Short Answer

Expert verified
The electrical power required to maintain the sample at \(T_s = 500\mathrm{K}\) is 5.7054 W. The radiant power incident on the detector within the spectral region from 3 to 5 \(\mu\)m is \(2.02\times10^{-6}W\).

Step by step solution

01

Calculate convective heat transfer

For convective heat transfer, we use the formula: \[q_c = hA(T_s - T_\infty)\] We know the convection coefficient, \(h = 20 \frac{\mathrm{W}}{\mathrm{m}^2\cdot\mathrm{K}}\), and the temperature difference between the sample and the gas surrounding it (\(T_s - T_\infty = 500 \mathrm{K} - 300\mathrm{K} = 200\mathrm{K}\)). However, we first need to calculate the area of the sample, \(A\). Notice that we are given the area of the detector, but not that of the sample. To proceed, we will assume that the sample and the detector have the same area (since the problem does not specify otherwise). Therefore, \[A = A_d = 8\times10^{-5}m^2\] Substituting the values, we get: \[q_c = (20\frac{\mathrm{W}}{m^2\cdot K})(8\times10^{-5}m^2)(200K) = 0.32W\]
02

Calculate the radiant heat transfer

For radiant heat transfer, we account for emission and reflection and use the following formula: \[q_r = A\sigma(T_s^4 - T_{sur}^4)\] The Stefan-Boltzmann constant, \(\sigma = 5.67\times10^{-8} \frac{\mathrm{W}}{\mathrm{m}^2\cdot \mathrm{K}^4}\). Plug in the values given for the area, \(A\), and the temperatures, \(T_s\) and \(T_{sur}\). Then calculate the radiant heat transfer: \[q_r = (8\times10^{-5}m^2)(5.67\times10^{-8}\frac{\mathrm{W}}{\mathrm{m}^2\cdot \mathrm{K}^4})(500^4\mathrm{K}^4 - 300^4\mathrm{K}^4) = 5.3854W\]
03

Determine the electrical power required

The net heat transfer from the sample is the sum of convective and radiant heat transfer, i.e., \(q_c + q_r\). The electrical power required to maintain the sample at \(T_s = 500\mathrm{K}\) is equal to the net heat transfer: \[P_c = q_c + q_r = 0.32W + 5.3854W = 5.7054W\] The electrical power required is 5.7054 W.
04

Calculate radiant power in the spectral region from 3 to 5 \(\mu\)m

We will use Plank's law to determine the power emitted by the sample in the given spectral region. In the simplified form for this range, it is as follows: \[p_{emit} = A\sigma_e(T_s^4 - T_{f}^4)\] The emissivity, \(\sigma_e\) for the specified spectral region is not directly given. Since the spectral emissivity distribution is provided schematically, we will estimate the average emissivity in the 3-5 \(\mu\)m range, and call it \(\sigma_{avg}\). Considering the problem statement, it's reasonable to suppose that the emissivity of the sample surface in the 3-5 \(\mu\)m range is close to the emissivity of the furnace walls. Therefore, we will assume \(\sigma_{avg} = s_f = 0.7\). Now we can compute \(p_{emit}\): \[p_{emit} = A\sigma_{avg}(T_s^4 - T_f^4) = (8\times10^{-5}m^2)(0.7)(500^4\mathrm{K}^4 - 3000^4\mathrm{K}^4)(5.67\times10^{-8}\frac{\mathrm{W}}{\mathrm{m}^2\cdot \mathrm{K}^4}) = 0.00899W\]
05

Calculate the reflected power

To calculate the reflected power \(p_{ref}\) from the sample, we will first find the fraction of power absorbed by the sample using spectral distribution. Since we assumed that \(\sigma_{avg} = s_f\) for the spectral region 3-5 \(\mu\)m, the sample will absorb and reflect the same amount of power. Therefore, the reflected power is the same as that emitted in the given spectral region: \[p_{ref} = p_{emit} = 0.00899W\]
06

Determine the radiant power incident on the detector

To find the radiant power incident on the detector, we consider both the emitted and reflected power. However, we also need to factor in the detector's position and angle relative to the sample. The radiant power is distributed uniformly over a hemispherical area centered on the sample and attenuated by the cosine of the angle between the sample normal and the line connecting the sample and detector. Given the angle (45 degrees) and distance between sample and detector (\(L_{dd} = 1.0m\)), the factor of attenuation is \(\cos(45^\circ) = \frac{\sqrt{2}}{2}\). Finally, we can calculate the total radiant power incident on the detector: \[p_{incident} = A_d[(p_{emit} + p_{ref})\frac{\sqrt{2}}{2}] = (8\times10^{-5}m^2)[(0.00899+0.00899)W\frac{\sqrt{2}}{2}] = 2.02\times10^{-6}W\] The radiant power incident on the detector within the spectral region from 3 to 5 \(\mu\)m is \(2.02\times10^{-6}W\).

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Key Concepts

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

Convective Heat Transfer
Convective heat transfer is a mechanism of heat transportation where energy is moved between a surface and a fluid moving over it. It commonly occurs in everyday phenomena, such as the cooling of a hot cup of coffee as the air passes over its surface.

When we want to calculate the heat transferred due to convection, we use the equation \( q_c = h \times A \times (T_s - T_{\infty}) \). In this formula, \( h \) identifies the convective heat transfer coefficient, \( A \) is the area through which heat is being transferred, and \((T_s - T_{\infty})\) is the temperature difference between the surface (\( T_s \)) and the fluid far away from the surface (\( T_{\infty} \)).

This coefficient, \( h \) has units of \( \frac{W}{m^2 \times K} \) and varies depending on the properties of the fluid and the type of flow, whether it be laminar or turbulent. For example, in the provided exercise, the convection coefficient for the gas was given as \( 20 \frac{W}{m^2 \times K} \), signifying that for every square meter of surface area and each degree of temperature difference, 20 watts of heat is transferred per second.

Important factors to consider for accurate calculations include the characteristics of the fluid, such as viscosity and thermal conductivity, and the geometry of the surface. For instance, a flat plate will have a different convection coefficient compared to a cylinder due to the difference in flow patterns around these objects.
Radiant Heat Transfer
Radiant heat transfer, different from convection, does not require a medium to occur. It is the transfer of energy by electromagnetic waves, particularly in the infrared spectrum. Everyday examples include the warmth felt from sunlight or from a fireplace.

In problems related to radiant heat transfer, we utilize the Stefan-Boltzmann law, which is expressed as \[ q_r = A \times \sigma \times (T_s^4 - T_{sur}^4) \]. Here, \( A \) again denotes the area emitting or absorbing the radiation, \( T_s \) and \( T_{sur} \) represent the absolute temperatures (in Kelvin) of the surface and the surroundings, respectively, and \( \sigma \) is the Stefan-Boltzmann constant, which has a value of \( 5.67 \times 10^{-8} \frac{W}{m^2 \times K^4} \).

The fourth power of the temperature in this equation implies that the rate of heat transfer by radiation increases dramatically with temperature鈥攁 hot body at double the temperature does not radiate just double the energy but 16 times more energy. This aspect central to radiant heat transfer was highlighted in the original exercise, where the energy radiated was calculated between an electrically heated sample and its surroundings.
Stefan-Boltzmann Constant
The Stefan-Boltzmann constant (\( \sigma \)) is a critical value in thermodynamics and plays a pivotal role in calculations involving radiant energy transfer. It is a physical constant that describes the total intensity of radiation emitted by a black body per unit surface area.

In detail, the Stefan-Boltzmann constant is derived through the work of Josef Stefan and Ludwig Boltzmann, who studied and formalized the relationship between the temperature of an object and the energy it radiates. An understanding of this constant is crucial for those studying heat transfer because it links the thermal radiation emitted by an object to its temperature raised to the fourth power, as expressed in the Stefan-Boltzmann law (\( q_r = A \sigma (T^4) \)).

With a value of \( 5.67 \times 10^{-8} \frac{W}{m^2 \times K^4} \), \( \sigma \) shows how efficient an ideal emitter, also known as a black body, would be at every temperature. Its relevance extends beyond just theoretical exercises; it is significant in real-world applications such as understanding the heat loss from buildings, estimating the power output of stars, and in designing thermal systems for engineering applications. In the context of the exercise showcased, knowledge of the Stefan-Boltzmann constant is essential to determine the radiant heat transfer between surfaces and their surroundings.

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Most popular questions from this chapter

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 $$ \begin{aligned} F_{U}=& 1-\frac{2}{\pi}\left\\{\left[1-\left(D_{d} / D_{d}\right)^{2}\right]^{1 / 2}\right.\\\ &\left.+\left(D_{k} / D_{d}\right) \sin ^{-1}\left(D_{h} / D_{d}\right)\right\\} \end{aligned} $$

Hot coffee is contained in a cylindrical thermos bottle that is of length \(L=0.3 \mathrm{~m}\) and is lying on its side (horizontally). The coffee containet consists of a glass flask of diameter \(D_{1}=0.07 \mathrm{~m}\), separated from an aluminum housing of diameter \(D_{2}=0.08 \mathrm{~m}\) by air at atmospheric pressure. The outer surface of the flask and the inner surface of the housing are silver coated to provide emissivities of \(\varepsilon_{1}=\varepsilon_{2}=0.25\). If these sarface temperatures are \(T_{1}=75^{\circ} \mathrm{C}\) and \(T_{2}=35^{\circ} \mathrm{C}\). what is the heat loss froen the coffee?

Two convex objects are inside a large vacuum enclosure whose walls are maintained at \(T_{3}=300 \mathrm{~K}\). The objects have the same area, \(0.2 \mathrm{~m}^{2}\), and the same emissivity, 0.2. The view factor from object 1 to object 2 is \(F_{12}=0_{3} 3\). Embedded in object 2 is a heater that generates 400 W. The temperature of object 1 is maintained at \(T_{1}=200 \mathrm{~K}\) by circulating a fluid through channels machined into it. At what rate must heat be supplied (or removed) by the fluid to maintain the desired temperature of object 1? What is the temperatare of object 2 ?

A radiometer views a small target (1) that is being heated by a ring-shaped disk heater (2). The target has an area of \(A_{1}=0.0004 \mathrm{~m}^{2}\), a temperature of \(T_{1}=\) \(500 \mathrm{~K}\), and a diffuse, gray emissivity of \(s_{1}=0.8\). The heater operates at \(T_{2}=1000 \mathrm{~K}\) and has a black surface. The radiometer views the entire sample area with a solid angle of \(\omega=0.0008 \mathrm{sr}\). (a) Write an expression for the radiant power leaving the target which is collected by the radiometer, in terms of the target radiosity \(J_{1}\) and relevant geometric parameters. Leave in symbolic form. (b) Write an expression for the target radiosity \(J_{1}\) in terms of its irradiation, emissive power, and appropriate radiative properties. Leave in symbolic form. (c) Write an expression for the irradiation on the target, \(G_{1}\), due to emission from the heater in terms of the heater emissive power, the heater area, and an appropriate view factor. Use this expression to numerically evaluate \(G_{1}\). (d) Use the foregoing expressions and results to determine the radiant power collected by the radiometer.

The spectral absorptivity of a large diffuse surface is \(\alpha_{\lambda}=0.9\) for \(\lambda<1 \mu \mathrm{m}\) and \(\alpha_{\lambda}=0.3\) for \(\lambda=1 \mu \mathrm{m}\). The botton of the surface is well insulated, while the top may be exposed to one of two different conditions. (a) In case \((a)\) the surface is exposed to the sun, which provides an irradiation of \(G_{5}=1200 \mathrm{~W} / \mathrm{m}^{2}\), and to an airflow for which \(T_{=}=300 \mathrm{~K}\). If the surface temperature is \(T_{s}=320 \mathrm{~K}\), what is the convection coefficient associated with the airflow? (b) In case (b) the surface is shielded from the sun by a large plate and an airflow is maintained between the plate and the surface. The plate is diffuse and gray with an emissivity of \(z_{p}=0.8\). If \(T_{\infty}=300 \mathrm{~K}\) and the convection coefficient is equ盲 valent to the result obtained in part (a), what is the plate temperature \(T_{p}\) that is needed to maintain the surface at \(T_{x}=320 \mathrm{~K}\) ?
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