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Consider two concentric spheres forming an enclosure with diameters of \(12 \mathrm{~cm}\) and \(18 \mathrm{~cm}\) and surface temperatures \(300 \mathrm{~K}\) and \(500 \mathrm{~K}\), respectively. Assuming that the surfaces are black, the net radiation exchange between the two spheres is (a) \(21 \mathrm{~W}\) (b) \(140 \mathrm{~W}\) (c) \(160 \mathrm{~W}\) (d) \(1275 \mathrm{~W}\) (e) \(3084 \mathrm{~W}\)

Short Answer

Expert verified
Question: Calculate the net radiation exchange between two concentric black spheres with diameters of 12 cm and 18 cm and surface temperatures of 300 K and 500 K, respectively. Answer: The net radiation exchange between the two concentric black spheres is approximately 5814.7 W. Unfortunately, none of the given options matches this value, which might indicate a mistake in the provided options or the question itself.

Step by step solution

01

Write down the Stefan-Boltzmann Law

The Stefan-Boltzmann Law states that the power radiated per unit area of a black body is given by: \(P/A = \sigma T^4\) where - \(P\) is the radiated power, - \(A\) is the surface area, - \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{Wm^{-2}K^{-4}}\)), - \(T\) is the temperature of the body in Kelvin.
02

Calculate the surface area of each sphere

To calculate the surface area of each sphere, we can use the formula: \(A = 4\pi r^2\) where \(r\) is the radius of the sphere. The diameters are \(12\mathrm{~cm}\) and \(18\mathrm{~cm}\), so the radii are \(6\mathrm{~cm}\) and \(9\mathrm{~cm}\). Convert these values to meters. \(r_1 = 0.06\mathrm{~m}\) \(r_2 = 0.09\mathrm{~m}\) Now, we can calculate the surface areas: \(A_1 = 4\pi (0.06)^2 = 0.0452\mathrm{~m^2}\) \(A_2 = 4\pi (0.09)^2 = 0.1018\mathrm{~m^2}\)
03

Determine the radiated power of each sphere

By using the Stefan-Boltzmann Law, we can now determine the radiated power of each sphere: \(P_1 = A_1 \sigma (300)^4 = 0.0452 \times 5.67 \times 10^{-8} \times (300)^4 = 304.775\mathrm{~W}\) \(P_2 = A_2 \sigma (500)^4 = 0.1018 \times 5.67 \times 10^{-8} \times (500)^4 = 6119.475\mathrm{~W}\)
04

Calculate the net radiation exchange between the spheres

The net radiation exchange is the difference between the radiated power of the outer sphere and the radiated power of the inner sphere: \(P_\mathrm{net} = P_2 - P_1 = 6119.475 - 304.775 = 5814.7\mathrm{~W}\)
05

Compare to the given options

Comparing our result with the given options, we can see that none of the options matches our calculated value of \(5814.7\mathrm{~W}\). Therefore, it seems that there might be a mistake in the provided options or the question itself.

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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 is a cornerstone of understanding thermal radiation for black bodies. This important principle tells us that the energy emitted by a black body per unit area can be described by the formula: \(P/A = \sigma T^4\). In this equation:
  • \(P\) represents the total power radiated by the surface,
  • \(A\) is the area over which the power is spread,
  • \(\sigma\) is the Stefan-Boltzmann constant, which has a value of \(5.67 \times 10^{-8} \, \mathrm{Wm^{-2}K^{-4}}\),
  • \(T\) is the absolute temperature of the surface measured in Kelvin.
This law exemplifies that the amount of radiation emitted increases dramatically with the temperature, as it is proportional to the fourth power of \(T\). Therefore, small increases in temperature result in large increases in radiated power, emphasizing why managing thermal conditions is crucial in various technologies.
Concentric Spheres
Concentric spheres are a fundamental setup in thermal dynamic problems, involving two or more spheres sharing the same center point but having different radii. This structure is particularly relevant in thermal radiation studies as it simplifies the analysis of heat transfer between surfaces with significant curvature.
In the given problem, two spheres of radii 6 cm and 9 cm are concentric, meaning that one lies entirely inside the other without touching. This setup allows us to apply thermal radiation principles effectively because the heat radiated from one surface (inner sphere) directly corresponds into the larger enclosings.
Concentric arrangements are widely used in many engineering applications including insulation systems and radiation shielding, as they provide a straightforward model for analyzing spherical symmetry in thermal exchanges.
Surface Area Calculation
To understand heat transfer through radiation, calculating the surface area of the objects involved is crucial. When dealing with spheres, the formula to determine the surface area is given by \(A = 4\pi r^2\), where \(r\) is the radius of the sphere.
  • For a 12 cm diameter sphere, the radius is half of that, or 6 cm (converted to 0.06 m for calculation).
  • For an 18 cm diameter sphere, the radius is 9 cm (or 0.09 m).
Plugging these into the formula, we compute:
  • The surface area of the inner sphere as \(A_1 = 4\pi (0.06)^2 = 0.0452\, \mathrm{m^2}\).
  • The surface area of the outer sphere as \(A_2 = 4\pi (0.09)^2 = 0.1018\, \mathrm{m^2}\).
Accurate calculation of surface area ensures precise determination of radiated power, as surface area is a pivotal factor in applying the Stefan-Boltzmann Law.
Black Body Radiation
Black body radiation is a concept used to describe objects that are perfect emitters and absorbers of thermal radiation. A black body absorbs all incident electromagnetic radiation, regardless of the frequency or angle of incidence. This makes it an idealized physical object, though such perfect black bodies do not exist in reality.
In our exercise, we assume that both spheres behave as black bodies. This assumption simplifies our calculations because it means all radiation emitted by the surface and absorbed by the other sphere adhere strictly to the laws of thermodynamics and electromagnetic theory.
This foundational concept is leveraged in practical engineering and physics applications to model thermal radiation. Understanding black body radiation helps in designing systems like thermal cameras and in studying stellar radiation, all fundamentally dependent on similar emission and absorption principles.

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Most popular questions from this chapter

Two gray surfaces that form an enclosure exchange heat with one another by thermal radiation. Surface 1 has a temperature of \(400 \mathrm{~K}\), an area of \(0.2 \mathrm{~m}^{2}\), and a total emissivity of \(0.4\). Surface 2 has a temperature of \(600 \mathrm{~K}\), an area of \(0.3 \mathrm{~m}^{2}\), and a total emissivity of \(0.6\). If the view factor \(F_{12}\) is \(0.3\), the rate of radiation heat transfer between the two surfaces is (a) \(135 \mathrm{~W}\) (b) \(223 \mathrm{~W}\) (c) \(296 \mathrm{~W}\) (d) \(342 \mathrm{~W}\) (e) \(422 \mathrm{~W}\)

\(3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and both natural convection and radiation heat transfer occur at the outer shield surface. To keep the ammonia inside the tube in its liquid state, determine the surrounding temperature that would maintain temperatures of the outer tube and the shield at the specified values. 13-68 PtD A hot liquid is being transported inside a long tube with a diameter of \(25 \mathrm{~mm}\). The hot liquid causes the tube surface temperature to be \(150^{\circ} \mathrm{C}\). To prevent thermal burn hazards, the tube is enclosed with a concentric outer cylindrical cover of \(5 \mathrm{~cm}\) in diameter allowing a vacuumed gap in between the two surfaces. The concentric outer cover has an emissivity of \(0.6\) and the outer surface is exposed to natural convection with a heat transfer coefficient of \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and radiation heat transfer with the surrounding at a temperature of \(20^{\circ} \mathrm{C}\). Determine the necessary emissivity of the inside tube so that the outer cover temperature is below \(45^{\circ} \mathrm{C}\) to prevent thermal burns.

Two long parallel 20 -cm-diameter cylinders are located \(30 \mathrm{~cm}\) apart from each other. Both cylinders are black, and are maintained at temperatures \(425 \mathrm{~K}\) and \(275 \mathrm{~K}\). The surroundings can be treated as a blackbody at \(300 \mathrm{~K}\). For a 1 -m-long section of the cylinders, determine the rates of radiation heat transfer between the cylinders and between the hot cylinder and the surroundings.

What is operative temperature? How is it related to the mean ambient and radiant temperatures? How does it differ from effective temperature?

Consider two infinitely long concentric cylinders with diameters 20 and \(25 \mathrm{~cm}\). The inner surface is maintained at \(700 \mathrm{~K}\) and has an emissivity of \(0.40\), while the outer surface is black. If the rate of radiation heat transfer from the inner surface to the outer surface is \(2400 \mathrm{~W}\) per unit area of the inner surface, the temperature of the outer surface is (a) \(605 \mathrm{~K}\) (b) \(538 \mathrm{~K}\) (c) \(517 \mathrm{~K}\) (d) \(451 \mathrm{~K}\) (e) \(415 \mathrm{~K}\)

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