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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

The fire tube of a hot water heater consists of a long circular duct of diameter \(D=0.07 \mathrm{~m}\) and temperature \(T_{x}=385 \mathrm{~K}\), through which combustion gases flow at a temperature of \(T_{\mathrm{m} \text { g }}=900 \mathrm{~K}\). To enhance heat transfer from the gas to the tube, a thin partition is inserted along the midplane of the tube. The gases may be assumed to have the thermophysical properties of air and to be radiatively nonparticipating. (a) With no partition and a gas flow rate of \(\dot{i i}_{\mathrm{e}}=0.05 \mathrm{~kg} / \mathrm{s}\), what is the rate of heat transfer per unit length, \(q^{\prime}\), to the tube? (b) For a gas flow rate of \(\dot{m}_{e}=0.05 \mathrm{~kg} / \mathrm{s}\) and emissivities of \(\varepsilon_{n}=\varepsilon_{p}=0.5\), determine the partition temperature \(T_{F}\) and the total rate of heat transfer \(q\) ' to the tube. (c) For \(\dot{m}_{g}=0.02,0.05\), and \(0.08 \mathrm{~kg} / \mathrm{s}\) and equivalent emissivities \(z_{p}=\varepsilon_{x}=\varepsilon\), compute and plot \(T_{p}\) and \(q^{\prime}\) as a function of \(e\) for \(0.1 \leq \varepsilon \leq 1.0\). For \(\dot{m}_{g}=0.05 \mathrm{~kg} / \mathrm{s}\) and equivalent emissivities, plot the convective and radiative contributions to \(q^{\prime}\) as a function of \(\varepsilon\).

Waste heat recovery from the exhaust (flue) gas of a melting furnace is accomplished by passing the gas through a vertical metallic tube and introducing saturated water (liquid) at the bottom of an annular region around the tube. The tube length and inside diameter are 7 and \(1 \mathrm{~m}\), respectively, and the tube inner surface is black. The gas in the tube is at atmospheric pressure, with \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{v})\) partial pressures of \(0.1\) and \(0.2 \mathrm{~atm}\), respectively, and its mean temperature may be approximated as \(T_{g}=1400 \mathrm{~K}\). The gas flow rate is \(\dot{m}=2 \mathrm{~kg} / \mathrm{s}\). If saturated water is introduced at a pressure of \(2.455 \mathrm{~ b a r s , ~ e s t i m a t e ~ t h e ~ w a t e r ~ f l o w ~ r a t e ~ s i t}\) which there is complete conversion from saturated liquid at the inlet to saturated vapor at the outlet. Thermophysical properties of the gas may be approximated as \(\mu=530 \times 10^{-7} \mathrm{~kg} / \mathrm{s}-\mathrm{m}, \mathrm{k}=0.091 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). and \(P_{r}=0.70\).

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 \(F_{22}=1+H-\left(1+H^{2}\right)^{1 / 2}\).

A cylindrical furnace for heat-treating materials in a spacecraft environment has a \(90-\mathrm{mm}\) diameter and an overall length of \(180 \mathrm{~mm}\). Heating elements in the 135 -mm-long section (1) maintain a refractory lining of \(\varepsilon_{1}=0.8\) at \(800^{\circ} \mathrm{C}\). The linings for the bottom (2) and upper (3) sections are made of the same refractory material, but are insulated.

Boiler tubes exposed to the products of coal combustion in a power plant are subject to fouling by the ash (mineral) content of the combustion gas. The ash forms a solid deposit on the tube outer surface, which reduces heat transfer to a pressurized wated'steam mixture flowing through the tubes. Consider a thin-walled boiler tube \(\left(D_{t}=0.05 \mathrm{~m}\right)\) whose surface is maintained at \(T_{t}=600 \mathrm{~K}\) by the boiling process. Combustion gases flowing over the tube at \(T_{-}=1800 \mathrm{~K}\) provide a convection coefficient of \(\bar{h}=100 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\), while radiation from the gas and boiler walls to the tube may be approximated as that originating from large surroundings at \(T_{\text {arr }}=1500 \mathrm{~K}\). (a) If the tube surface is diffuse and gray, with \(\varepsilon_{t}=0.8\), and there is no ash deposit layer, what is the rate of heat transfer per unit length, \(q^{\prime}\), to the boiler tube? (b) If a deposit layer of diameter \(D_{d}=0.06 \mathrm{~m}\) and thermal conductivity \(k=1 \mathrm{~W} / \mathrm{m}\), \(\mathrm{K}\) forms on the tube, what is the deposit surface temperature, \(T_{d} ?\) The depos茂t is diffuse and gray, with \(\varepsilon_{a}=0.9\), and \(T_{m} T_{\ldots}, T_{\text {}}\), and \(T_{\text {sar }}\) remain unchanged. What is the net rate of heat transfer per wnit length, \(y^{\prime}\) ', to the boxler tube? (c) Explore the effect of variations in \(D_{d}\) and \(\bar{h}\) on \(q^{r}\), as well as on relative contributions of convection and radiation to the net heat transfer fate. Represent your results graphically.
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