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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

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 ?

The lower side of a \(400-\mathrm{mm}\)-diameter disk is heated by an electric furnace, while the upper side is exposed to quiescent, ambient air and sumoundings at \(300 \mathrm{~K}\). The radiant furnace (negligible convection) is of circular construction with the bottoen surface \(\left(\alpha_{1}-0.6\right)\) and cylindrical side surface \(\left(\varepsilon_{1}=1.0\right)\) maintained af \(T_{1}=T_{2}=500 \mathrm{~K}\). The surface of the disk facing the radiant furnace is black \(\left(\varepsilon_{d, 1}=1.0\right)\). while the upper surface has an emissivity of \(\varepsilon_{d, 2}=0.8\). Assume the plate and furnace surfaces to be diffuse and gray. (a) Determine the net heat transfer rate to the disk, \(q_{\text {nated, when }} T_{d}=400 \mathrm{~K}\). (b) Plot \(q_{\text {netd as a }}\) a function of the disk temperature for \(300 \leq T_{a} \leq 500 \mathrm{~K}\), with all other conditions remaining the same. What is the steady-state temperature of the disk?

A laboratory oven has a cubical interior chamber \(1 \mathrm{~m}\) on a s盲de with interior sarfaces that are of emissivity \(z-0.85\). Determine the initial rate of radiation heat transfer to the laboratory in which the oven is placed when the oven door is opened. The oven and surroundings temperatures are \(T_{w}=375^{\circ} \mathrm{C}\) and \(T_{\text {sur }}=20^{\circ} \mathrm{C}\), respectively. Treat the oven door as one surface and the remaining five interior furnace walls as another.

Consider the right-circular cylinder of diameter \(D\), length \(L\), and the areas \(A_{1}, A_{2}\), and \(A_{3}\) representing the base, inner, and top surfaces, respectively. (a) Show that the view factor between the base of the cylinder and the inner surface has the form \(F_{12}=2 H\left[\left(1+H^{2}\right)^{1 / 2}-H\right]\), where \(H=L D .\) (b) Show that the view factor for the inner surface to itself has the form \(F_{22}=1+H-\left(1+H^{2}\right)^{1 / 2}\).

2 A composite wall is comprised of two large plates separated by sheets of refractory insulation, as shown in the schematic. In the installation process, the sheets of thickness \(L=50 \mathrm{~mm}\) and thermal conductivity \(k=0.05 \mathrm{~W} / \mathrm{m}+\mathrm{K}\) are separated at 1 -m intervals by gaps of width \(w=10 \mathrm{~mm}\). The hot and cold plates have temperatures and emissivities of \(T_{1}=400^{\circ} \mathrm{C}\). \(\varepsilon_{1}=0.85\) and \(T_{2}=35^{\circ} \mathrm{C}, \varepsilon_{2}=0.5\), respectively. Assume that the plates and insulation are diffuse-gray surfaces. (a) Determine the heat loss by radiation through the gap per unit length of the composite wall (normal to the page). (b) Recognizing that the gaps are located on a 1-m spacing, determine what fraction of the total heat loss through the composite wall is due to transfer by radiation through the insulation gap.
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