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Chapter 13: Problem 40

Consider the very long, inclined black surfaces \(\left(A_{1}, A_{2}\right)\) maintained at uniform temperatures of \(T_{1}=1000 \mathrm{~K}\) and \(T_{2}=800 \mathrm{~K}\).

Short Answer

Expert verified
The heat transfer per unit area between the two inclined black surfaces with temperatures \(T_{1} = 1000 K\) and \(T_{2} = 800 K\) can be calculated using the Stefan-Boltzmann law. The result is approximately \(3.35 脳 10^{4} W/m^2\). The inclination of the surfaces affects the heat transfer by a factor called the view factor, but since we are not given any information on the angle or dimensions, we cannot include this factor.

Step by step solution

01

Identify relevant information

In this problem, we are given the following information: - The surfaces A1 and A2 are "black," which means they have an emissivity of 1. - Surfaces A1 and A2 have temperatures T1 = 1000 K and T2 = 800 K, respectively. - Surfaces are inclined.
02

Understand the Stefan-Boltzmann law

The Stefan-Boltzmann law states that the heat transfer between two black bodies is proportional to the difference in the fourth power of their absolute temperatures. Mathematically, the heat transfer can be represented as: Q = 蟽 * A * (T1^4 - T2^4) where 蟽 is the Stefan-Boltzmann constant (around \(5.67 脳 10^{-8} W m^{-2} K^{-4}\)), A is the surface area of the black bodies, and T1 and T2 are their temperatures, respectively.
03

Calculate the heat transfer

Plug the values of T1 and T2 into the Stefan-Boltzmann equation: Q = 蟽 * A * (1000^4 - 800^4) At this point, we are not given the surface area, A, so the heat transfer per unit area, q, can be calculated: q = 蟽 * (1000^4 - 800^4) q 鈮� \(5.67 脳 10^{-8}\) * (10^12 - 4.096 脳 10^11) q 鈮� \(5.67 脳 10^{-8}\) * (5.904 脳 10^11 W/(m^2)) q 鈮� 3.35 脳 10^4 W/m虏 If the area of the surfaces was provided, you could multiply the q value by the area to get the total heat transfer Q.
04

Consider the effect of inclination

The inclination of the surfaces affects the heat transfer by a factor called the view factor. However, since we are not given any information on the angle or dimensions of the surfaces, we cannot include this factor. With the given information, the heat transfer per unit area between the two surfaces is approximately 3.35 脳 10^4 W/m虏.

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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 fundamental principle in thermodynamics that describes how much energy a black body radiates based on its temperature.
  • Defined by the formula: \(Q = \sigma \times A \times (T_1^4 - T_2^4)\)
  • \(\sigma\) is the Stefan-Boltzmann constant, approximately \(5.67 \times 10^{-8} \text{ W} \text{ m}^{-2} \text{ K}^{-4}\)
  • Ideal for calculating heat transfer between surfaces at different temperatures.
The law is based on calculating the power emitted by a surface per unit area, which is directly proportional to the fourth power of its absolute temperature.
Consider its significance in engineering, astrophysics, and climate science for understanding energy emission and temperature effects.
Black Body Radiation
Black body radiation refers to an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. Such a body also perfectly emits radiation at all frequencies, described by Planck's law. Characteristics of black body radiation include:
  • Perfect absorbers and emitters of radiation.
  • Emission depends solely on its temperature, making them ideal for studying thermal radiation laws.
  • At a given temperature, no body can emit more energy than a black body, making it the standard reference.
Understanding black body radiation helps with analyzing stars' spectra and aids technologies like thermal cameras by providing a model of radiation at different temperatures.
Inclined Surfaces
Inclined surfaces are those positioned at an angle relative to a baseline, which affects various physical processes including heat transfer. When dealing with thermal radiation between inclined surfaces:
  • The inclination affects how radiation is exchanged due to directionality and view factor considerations.
  • The view factor quantifies the proportion of the radiation leaving one surface that directly falls onto another.
  • It is crucial in systems like solar panels and building design to maximize or minimize gains or losses.
Inclination may also influence convective processes, where the angle compared to gravity affects fluid flow, impacting cooling and heating efficiencies.
Thermal Emissivity
Thermal emissivity is a material-specific characteristic that measures how efficiently a surface emits energy as thermal radiation compared to a perfect black body. It can be expressed mathematically as:
  • \(\epsilon = \frac{E_{actual}}{E_{black\ body}}\)
Values range between 0 and 1:
  • An emissivity of 1 corresponds to an ideal black body.
  • Real-world materials have emissivities less than 1, meaning they emit less thermal radiation.
Knowing a material's emissivity is critical in fields such as building ecology, astronomy, and thermal engineering, where accurate temperature readings and energy assessments are necessary.

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Most popular questions from this chapter

Consider a circular furnace that is \(0.3 \mathrm{~m}\) long and \(0.3 \mathrm{~m}\) in diameter. The two ends have diffuse, gray surfaces that are maintained at 400 and \(500 \mathrm{~K}\) with emissivities of \(0.4\) and \(0.5\), respectively. The lateral surface is also diffuse and gray with an emissivity of \(0.8\) and a temperature of \(800 \mathrm{~K}\). Determine the net radiative heat transfer from each of the surfaces.

Consider coaxial, parallel, black disks separated a distance of \(0.20 \mathrm{~m}\). The lower disk of diameter \(0.40 \mathrm{~m}\) is maintained at \(500 \mathrm{~K}\) and the surroundings are at \(300 \mathrm{~K}\). What temperature will the upper disk of diameter \(0.20 \mathrm{~m}\) achieve if electrical power of \(17.5 \mathrm{~W}\) is supplied to the heater on the back side of the disk?

A cylindrical cavity of diameter \(D\) and depth \(L\) is machined in a metal block, and conditions are such that the base and side surfaces of the cavity are maintained at \(T_{1}=1000 \mathrm{~K}\) and \(T_{2}=700 \mathrm{~K}\), respectively. Approximating the surfaces as black, determine the emissive power of the cavity if \(L=20 \mathrm{~mm}\) and \(D=10 \mathrm{~mm}\).

Derive expressions for the view factor \(F_{12}\) associated with the following arrangements. Express your results in terms of \(A_{1}, A_{2}\) and any appropriate hypothetical surface area, as well as the view factor for coaxial parallel disks (Table 13.2, Figure 13.5). (a) A circular disk and a coaxial, ring-shaped disk. (b) A circular disk and a coaxial, right-circular cone. Sketch the variation of \(F_{12}\) with \(\theta\) for \(0 \leq \theta \leq \pi / 2\), and explain the key features.

Consider two large (infinite) parallel planes that are diffuse-gray with temperatures and emissivities of \(T_{1}\), \(\varepsilon_{1}\) and \(T_{2}, \varepsilon_{2}\). Show that the ratio of the radiation transfer rate with multiple shields, \(N\), of emissivity \(\varepsilon_{s}\) to that with no shields, \(N=0\), is $$ \frac{q_{12, N}}{q_{120}}=\frac{\left[1 / \varepsilon_{1}+1 / \varepsilon_{2}-1\right]}{\left[1 / \varepsilon_{1}+1 / s_{2}-1\right]+N\left[2 / s_{s}-1\right]} $$ where \(q_{12, N}\) and \(q_{12,0}\) represent the radiation heat transfer rates for \(N\) shields and no shields, respectively.
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