/*! 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 57 Consider two large (infinite) pa... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 57

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.

Short Answer

Expert verified
The ratio of the radiation transfer rate with multiple shields to that with no shields is given by: \[ \frac{q_{12,N}}{q_{12,0}} = \frac{\left[1 / \varepsilon_{1} + 1 / \varepsilon_{2} - 1\right]}{\left[1 / \varepsilon_{1} + 1 / \varepsilon_{2} - 1\right] + N\left[2 / \varepsilon_s - 1\right]} \]

Step by step solution

01

Understand the radiation heat transfer rate with no shields

When no shields are present, the radiation heat transfer rate between two parallel plates is modeled by the following equation: \[ q_{12,0} = \sigma F_{12}\left( T_1^4 - T_2^4 \right) \] where \(\sigma\) is the Stefan-Boltzmann constant, and \(F_{12}\) is the view factor between the two plates, representing the fraction of radiation leaving surface 1 and reaching surface 2.
02

Find the view factor for no shields

For the case with no shields, the view factor \(F_{12}\) can be found using the following equation: \[ F_{12} = \frac{1}{1/\varepsilon_{1} + 1/\varepsilon_{2} - 1} \]
03

Determine the radiation heat transfer rate with multiple shields

For the scenario with \(N\) shields, each shield introduces an additional layer that impacts the fraction of radiation reaching surface 2. The view factor for this case, \(F_{12,N}\), is given by the following equation: \[ F_{12,N} = \frac{1}{1/\varepsilon_{1} + 1/\varepsilon_{2} - 1 + N(\frac{2}{\varepsilon_s} - 1)} \]
04

Calculate the radiation heat transfer rate for the case with multiple shields

Now that we have the view factor for the case with multiple shields, we can determine the radiation heat transfer rate, \(q_{12,N}\), using the same formula as in Step 1, simply substituting \(F_{12,N}\) for \(F_{12}\): \[ q_{12,N} = \sigma F_{12,N}\left( T_1^4 - T_2^4 \right) \]
05

Determine the ratio of the radiation transfer rates

Finally, we divide the radiation heat transfer rate with multiple shields by the radiation heat transfer rate with no shields to find the ratio: \[ \frac{ q_{12,N} }{ q_{12,0} } = \frac{\sigma F_{12,N}\left( T_1^4 - T_2^4 \right)}{\sigma F_{12,0}\left( T_1^4 - T_2^4 \right)} \] The \(T_1^4 - T_2^4\) terms and the Stefan-Boltzmann constant, \(\sigma\), cancel out, resulting in: \[ \frac{q_{12,N}}{q_{12,0}} = \frac{F_{12,N}}{F_{12,0}} = \frac{\left[1 / \varepsilon_{1} + 1 / \varepsilon_{2} - 1\right]}{\left[1 / \varepsilon_{1} + 1 / \varepsilon_{2} - 1\right] + N\left[2 / \varepsilon_s - 1\right]} \] The result is the expression we were asked to show, as requested in the exercise.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Diffuse-Gray Surfaces
Diffuse-gray surfaces are an idealized concept used in the study of radiation heat transfer. These surfaces are characterized by their uniform reflectivity and emissivity in all directions and across all wavelengths. In simpler terms, a diffuse-gray surface is a perfect emitter and scatterer of radiation, without preference for direction or wavelength.

When we say a surface is 'gray,' we mean that its emissivity does not change with wavelength, which is not the case for real materials. Consequently, we can represent complex, actual material properties with a single, average value for emissivity, \(\varepsilon\). This simplification allows for more manageable calculations in engineering problems involving radiative heat transfer and is typically quite accurate for engineering estimates.

For the purpose of textbook exercises and practical engineering calculations, assuming surfaces to be diffuse and gray helps in applying the Stefan-Boltzmann law. This law is only strictly accurate for ideal blackbodies, but by treating surfaces as diffuse-gray, we can make use of the law by adjusting it with an emissivity factor. Thus, such surfaces serve as an essential model in thermodynamics and heat transfer courses.
Stefan-Boltzmann Constant
The Stefan-Boltzmann constant, denoted as \(\sigma\), is a physical constant that plays a crucial role in the domain of thermal radiation. It appears in the famous Stefan-Boltzmann law, which states that the total energy radiated per unit surface area of a black body per unit time is directly proportional to the fourth power of the black body's thermodynamic temperature.

The precise value of the Stefan-Boltzmann constant is \(5.670374419 \times 10^{-8} \text{ W m}^{-2} \text{K}^{-4}\). What this means is that each square meter of a blackbody's surface, at a temperature of one Kelvin, radiates \(5.670374419 \times 10^{-8} \text{ watts}\).

In our textbook example, the constant is used to compare the heat transfer rates with and without shields. The beautiful simplicity of the Stefan-Boltzmann law, when combined with the constant, allows us to estimate the thermal radiation from bodies that are close to ideal black bodies or, as in our case, from diffuse-gray surfaces by using their emissivity.
View Factor
The view factor, also known as the configuration factor or shape factor, is a dimensionless quantity in thermal radiation that represents the fraction of the radiation leaving one surface that directly reaches another specified surface. View factors depend solely on the geometry of the involved surfaces.

To calculate the heat transfer between surfaces, especially in cases without intervening media, the view factor is crucial. The value of a view factor lies between 0 and 1, inclusive. A view factor of 1 means that all the radiation from one surface directly strikes the other surface, while 0 implies none reaches it.

In the context of our textbook problem, the presence of shields alters the view factor, and hence, changes the radiation heat transfer rate between the two planes. View factors come into play in this problem to indicate how adding shields affects the fraction of radiation between the two initial large parallel planes. Understanding how view factors change with different configurations is essential for predicting the efficacy of thermal shields and similar barriers.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A radiant heater, which is used for surface treatment processes, consists of a long cylindrical heating element of diameter \(D_{1}=0.005 \mathrm{~m}\) and emissivity \(\varepsilon_{1}=0.80\). The heater is partially enveloped by a long. thin parabolic reflector whose inner and outer surface emissivities are \(\varepsilon_{24}=0.10\) and \(\varepsilon_{20}=0.80\), respectively. Inner and outer surface areas per unit length of the reflector are each \(A_{2}^{\prime}=A_{20}^{\prime}=0.20 \mathrm{~m}\), and the average convection coefficient for the combined inner and outer surfaces is \(\bar{h}_{2 \dot{m}}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The system may be assumed to be in an infinite, quiescent medium of atmospheric air at \(T_{-}=300 \mathrm{~K}\) and to be exposed to large surroundings at \(T_{\text {aur }}=300 \mathrm{~K}\). (a) Sketch the appropriate radiation circuit, and write expressions for each of the network resistances. (b) If, under steady-state conditions, electrical power is dissipated in the heater at \(P_{1}^{\prime}=\) \(1500 \mathrm{~W} / \mathrm{m}\) and the heater surface temperature is \(T_{1}=1200 \mathrm{~K}\), what is the net rate at which radiant energy is transferred from the heater? (c) What is the net rate at which radiant energy is transferred from the heater to the surroundings? (d) What is the temperature, \(T_{2}\), of the reflector?

Options for thermally shielding the top ceiling of a large furnace include the use of an insulating material of thickness \(L\) and thermal conductivity \(k\), case (a), or an air space of equ盲valent thickness formed by installing a steel sheet above the ceiling, case \((b)\). (a) Develop mathematical models that could be used to assess which of the two approaches is better. In both cases the interior surface is maintained at the same temperature, \(T_{s,}\) and the ambient air and surroundings are at equivalent temperatures \(\left(T_{w}=T_{\text {ue }}\right)\). (b) If \(k=0.090 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L=25 \mathrm{~mm}, h_{a}=25\) W/ \(\mathrm{m}^{2} \cdot \mathrm{K}\), the surfaces are diffuse and gray with \(\varepsilon_{i}=\varepsilon_{o}=0.50, T_{s u}=900 \mathrm{~K}\), and \(T_{w}=T_{\mathrm{ar}}=\) \(300 \mathrm{~K}\), what is the outer surface temperature \(T_{\text {a }}\) and the heat loss per unit surface area associated with each option? (c) For each case, assess the effect of surface radiative properties on the outer surface temperature and the heat loss per unit area for values of

A flat-plate solar collector, consisting of an absorber plate and single cover plate, is inclined at an angle of \(\tau=60^{\circ}\) relative to the horizontal. Consider conditions for which the incident solar radiation is collimated at an angle of \(60^{\circ}\) relative to the horizontal and the solar flux is \(900 \mathrm{~W} / \mathrm{m}^{2}\). The cover plate is perfectly transparent to solar radiation \((\lambda \leq 3\) \(\mu m)\) and is opaque to radiation of larger wavelengths. The cover and absorber plates are diffuse surfaces having the spectral absorptivities shown. The length and width of the absorber and cover plates are much larger than the plate spacing \(L\). What is the rate at which solar radiation is absorbed per unit area of the absorber plate? With the absorber plate well insulated from below and absorber and cover plate temperarures \(T_{a}\) and \(T_{c}\) of \(70^{\circ} \mathrm{C}\) and \(27^{\circ} \mathrm{C}\), respectively, what is the heat loss per unit area of the absorber plate?

Hot coffee is contained in a cylindrical thermos bottle that is of length \(L=0.3 \mathrm{~m}\) and is lying on its side (horizontally). The coffee containet consists of a glass flask of diameter \(D_{1}=0.07 \mathrm{~m}\), separated from an aluminum housing of diameter \(D_{2}=0.08 \mathrm{~m}\) by air at atmospheric pressure. The outer surface of the flask and the inner surface of the housing are silver coated to provide emissivities of \(\varepsilon_{1}=\varepsilon_{2}=0.25\). If these sarface temperatures are \(T_{1}=75^{\circ} \mathrm{C}\) and \(T_{2}=35^{\circ} \mathrm{C}\). what is the heat loss froen the coffee?

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 \(\varepsilon\).
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Circular Motion and Gravitation

Read Explanation

Fundamentals of Physics

Read Explanation

Particle Model of Matter

Read Explanation

Electricity and Magnetism

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.