/*! 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 43 A flat-bottomed hole \(6 \mathrm... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 43

A flat-bottomed hole \(6 \mathrm{~mm}\) in diameter is bored to a depth of \(24 \mathrm{~mm}\) in a diffuse, gray material having an emissivity of \(0.8\) and a uniform temperature of \(1000 \mathrm{~K} .\) (a) Determine the radiant power leaving the opening of the cavity. (b) The effective emissivity \(\varepsilon_{e}\) of a cavity is defined as the ratio of the radiant power leaving the cavity to that from a blackbody having the area of the cavity opening and a temperature of the inner surfaces of the cavity. Calculate the effective emissivity of the cavity described above. (c) If the depth of the hole were increased, would \(\varepsilon_{e}\) increase or decrease? What is the limit of \(s_{\epsilon}\) as the depth increases?

Short Answer

Expert verified
The radiant power leaving the opening of the cavity is \(40.59 \mathrm{W}\). The effective emissivity of the cavity is \(0.8\). When the depth of the hole is increased, the effective emissivity decreases because of the reduced radiant power leaving the cavity. As the depth increases infinitely, the effective emissivity approaches zero.

Step by step solution

01

Part (a): Calculate the radiant power emitted by the material

First, we need to determine the area of the hole. Since it is a circle with a diameter of \(6\) mm, we can calculate the area using the following formula: $$A = \pi r^2$$ Here, \(r\) is the radius which is half of the 6mm diameter, i.e., \(3\) mm or \(0.003\) m: $$A = \pi (0.003)^2 = 2.827 \times 10^{-5} \mathrm{m}^2$$ Next, we'll use the formula we mentioned earlier, substituting \(\varepsilon = 0.8\), \(\sigma = 5.67 \times 10^{-8} \mathrm{W/m^2K^4}\), and \(T = 1000 \mathrm{K}\): $$Q = A \varepsilon \sigma T^4 = (2.827 \times 10^{-5})(0.8)(5.67 \times 10^{-8})(1000)^4 = 40.59 \mathrm{W}$$ So, the radiant power leaving the opening of the cavity is \(40.59\) W.
02

Part (b): Calculate the effective emissivity

Now, we need to determine the power from a blackbody. A blackbody would have an emissivity of \(1\). By substituting this value into the radiation formula, we can find the blackbody power: $$Q_{bb} = A \sigma T^4 = (2.827 \times 10^{-5})(5.67 \times 10^{-8})(1000)^4 = 50.74 \mathrm{W}$$ With the blackbody power, we can now calculate the effective emissivity: $$\varepsilon_e = \frac{Q}{Q_{bb}} = \frac{40.59}{50.74} = 0.8$$ The effective emissivity of the cavity is \(0.8\).
03

Part (c): Change in effective emissivity and its limit

When the depth of the hole is increased, the emitted radiation has to undergo multiple reflections before leaving the hole, reducing the radiant energy emitted compared to a shallower hole. This leads to a decrease in the effective emissivity, as there is less radiant power leaving the cavity compared to the blackbody having the same opening area and temperature. The limit of the effective emissivity as the depth increases is when the hole becomes a perfect cavity, and the emissivity approaches zero. In other words, when the hole's depth becomes infinite, the internal reflections prevent almost all radiation from escaping, resulting in the least amount of power possible leaving the cavity.

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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 how effectively a material emits thermal radiation compared to an ideal blackbody. A perfect blackbody has an emissivity of 1, meaning it emits all the thermal energy it can at a given temperature. Most real-world materials have an emissivity less than 1, which means they emit less radiation compared to a blackbody. For instance, in the problem above, the diffuse gray material of the cavity has an emissivity of 0.8. This implies it emits 80% of the radiation that would be emitted by a blackbody at 1000 K.

Emissivity plays a crucial role in determining the radiant power that a material emits. It depends on several factors including the surface texture and temperature of the emitting material. Emissivity is contextual and can change with the material's surface condition and the wavelength of the emitted energy.
Blackbody radiation
Blackbody radiation is the theoretical concept where an ideal body absorbs all incoming radiation without reflecting any. This body, known as a blackbody, also emits radiation called blackbody radiation that depends solely on its temperature. The radiation emitted from a blackbody is characterized by a specific spectrum and intensity, dictated by Planck's Law. At any temperature, a blackbody emits the maximum possible radiant energy at every wavelength. This makes blackbodies useful benchmarks in determining real materials' emissivities. For example, in the exercise, we calculate the blackbody power for the cavity using the formula for blackbody radiation:\[Q_{bb} = A \sigma T^4\]where \(A\) is the area, \(\sigma\) is the Stefan-Boltzmann constant, and \(T\) is the absolute temperature. Understanding blackbody radiation helps us appreciate the limits of thermal radiation emission.
Radiant power
Radiant power is the rate of energy emission by an object in the form of electromagnetic radiation due to its temperature. In simple terms, it's how much energy is emitted by the surface of an object.To calculate radiant power, the Stefan-Boltzmann law is employed, which is expressed as:\[Q = A \varepsilon \sigma T^4\]Here, \(Q\) is the radiant power, \(A\) is the area of the surface, \(\varepsilon\) is the emissivity of the material, \(\sigma\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4)\), and \(T\) is the temperature in Kelvin. The given problem uses these principles to calculate the radiant power coming from a cavity. Knowing the area and the emissivity of the material helps compute how much radiant energy the opening will emit. Even slight changes in temperature or emissivity can drastically affect the radiant power, emphasizing the sensitivity of thermal radiation.

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Most popular questions from this chapter

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.

Liquid oxygen is stored in a thin-walled, spherical container \(0.8 \mathrm{~m}\) in diameter, which is enclosed within a second thin-walled, spherical container \(1.2 \mathrm{~m}\) in diameter. The opaque, diffuse, gray container surfaces have an emissivity of \(0.05\) and are separated by an evacuated space. If the outer surface is at \(280 \mathrm{~K}\) and the inner surface is at \(95 \mathrm{~K}\), what is the mass rate of oxygen lost due to evaporation? (The latent heat of vaporization of oxygen is \(2.13 \times 10^{5} \mathrm{~J} / \mathrm{kg}\).)

Consider the perpendicular rectangles shown schematically. (a) Determine the shape factor \(F_{12}\). (b) For rectangle widths of \(X=0.5,1.5\), and \(5 \mathrm{~m}\), plot \(F_{12}\) as a function of \(Z_{b}\) for \(0.05 \leq Z_{b} \leq 0.4 \mathrm{~m}\).] Compare your results with the view factor obtained from the two-dimensional relation for perpendicular plates with a common edge (Table 13.1).

Consider the attic of a home located in a hot climate. The floor of the attic is characterized by a width of \(L_{1}=10 \mathrm{~m}\) while the roof makes an angle of \(\theta=30^{\circ}\) from the horizontal direction, as shown in the schematic. The homeowner wishes to reduce the heat load to the home by adhering bright aluminum foil \(\left(\varepsilon_{f}=0.07\right)\) onto the surfaces of the attic space. Prior to installation of the foil, the surfaces are of emissivity \(s_{e}=0.85\). (a) Consider installation on the bottom of the attic roof only. Determine the ratio of the radiation heat transfer after to before the installation of the foil. (b) Determine the ratio of the radiation heat transfer after to before installation if the foil is installed only on the top of the attic floor. (c) Determine the ratio of the radiation heat transfer if the foil is installed on both the roof bottom and the floor top.

An opaque, diffuse, gray ( \(200 \mathrm{~mm} \times 200 \mathrm{~mm})\) plate with an emissivity of \(0.8\) is placed over the opening of a furnace and is known to be at \(400 \mathrm{~K}\) at a certain instant. The bottom of the fumace, having the same dimensions as the plate, is black and operates at \(1000 \mathrm{~K}\). The sidewalls of the fumace are well insulated. The top of the plate is exposed to ambient air with a convection coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and to large surroundings. The air and surroundings are each at \(300 \mathrm{~K}\). (a) Evaluate the net radiative heat transfer to the bottom surface of the plate. (b) If the plate has mass and specific heat of \(2 \mathrm{~kg}\) and \(900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively, what will be the change in temperature of the plate with time, \(d T_{p} / d r\) ? Assume convection to the bottom surface of the plate to be negligible. (c) Extending the analysis of part (b), generate a plot of the change in temperature of the plate with time, \(d T_{\rho} / d l\), as a function of the plate temperature for \(350 \leq T_{p} \leq 900 \mathrm{~K}\) and all other conditions remaining the same. What is the steady-state tem-
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