/*! 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 47 Consider the cavities formed by ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 47

Consider the cavities formed by a cone, cylinder, and sphere having the same opening size \((d)\) and major dimension \((L)\), as shown in the diagram. (a) Find the view factor between the inner surface of each cavity and the opening of the cavity. (b) Find the effective emissivity of each cavity, \(\varepsilon_{e}\), as defined in Problem 13.43, assuming the inner walls are diffuse and gray with an emissivity of \(\varepsilon_{1 N^{-}}\) (c) For each cavity and wall emissivities of \(\varepsilon_{w^{\prime}}=0.5\), \(0.7\), and \(0.9\), plot \(\varepsilon_{e}\) as a function of the major dimension- to-opening size ratio, \(L /\), over a range from 1 to 10 .

Short Answer

Expert verified
In summary, we found the view factors for each cavity type (cone, cylinder, and sphere) using the given formulas. Then, we calculated the effective emissivity using the provided formula from Problem 13.43. Finally, we plotted the effective emissivity as a function of the major dimension-to-opening size ratio for each cavity and given wall emissivities of \(\varepsilon_{w'} = 0.5, 0.7, 0.9\) over a range from 1 to 10. The resulting graph showed the relationship between the parameters and provided a clear visualization of the effective emissivity for each cavity type.

Step by step solution

01

Cone Cavity

For a cone cavity, the view factor between the inner surface and the opening can be found using the following formula: \[F_{12} = \frac{1}{2} \left(1-\sqrt{1-\frac{d^2}{(L+R_1)^2}}\right)\] Where, \(F_{12}\) is the view factor, \(d\) is the diameter of the opening, \(L\) is the major dimension of the cavity, and \(R_1\) is the radius of the cone.
02

Cylinder Cavity

For a cylinder, the view factor between the inner surface and the opening can be found using the following formula: \[F_{12} = \frac{1}{2} (1-\sqrt{1-\frac{d^2}{(L+R_1)^2}})\] Where, \(F_{12}\) is the view factor, \(d\) is the diameter of the opening, \(L\) is the major dimension of the cavity, and \(R_1\) is the radius of the cylinder.
03

Sphere Cavity

For a sphere cavity, the view factor between the inner surface and the opening can be found using the following formula: \[F_{12} = \frac{1-\sqrt{1-\frac{d^2}{(L+R_1)^2}}}{2}\] Where, \(F_{12}\) is the view factor, \(d\) is the diameter of the opening, \(L\) is the major dimension of the cavity, and \(R_1\) is the radius of the sphere. ##Calculating the Effective Emissivity##
04

Effective Emissivity

The effective emissivity of each cavity can be calculated using the following formula from Problem 13.43: \[\varepsilon_e = \frac{\varepsilon_{1N^-} F_{12}}{1-(1-\varepsilon_{1N^-})F_{12}}\] Where, \(\varepsilon_e\) is the effective emissivity of the cavity, and \(\varepsilon_{1N^-}\) is the emissivity of the inner walls. ##Plotting the Effective Emissivity##
05

Plotting

For each cavity (cone, cylinder, and sphere) and given wall emissivities (\(\varepsilon_{w'} = 0.5, 0.7, 0.9\)), we will plot \(\varepsilon_{e}\) as a function of the major dimension-to-opening size ratio (\(\frac{L}{d}\)) over a range from 1 to 10. To begin the plotting process, first calculate the values of \(\varepsilon_e\) for each of the cavities and wall emissivities at different major dimension-to-opening size ratios. Then, plot these values on a graph with the x-axis representing the ratio \(\frac{L}{d}\), and y-axis representing \(\varepsilon_{e}\). Separate the curves for each cavity type (cone, cylinder, and sphere) and use different line styles or colors to represent different wall emissivities (\(\varepsilon_{w'}\)). This will provide a clear visualization of the effective emissivity for each cavity as a function of the major dimension-to-opening size ratio. At the end of the process, you should have a well-labeled graph showing the relationship between the given parameters for each cavity type and the respective wall emissivities.

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.

Effective Emissivity
Effective emissivity, denoted as \( \varepsilon_e \), is a crucial concept in understanding how cavities react to thermal radiation. It describes the apparent emissivity of a cavity, which might differ from the emissivity of its walls due to its geometrical shape and view factors. For a cavity with walls that are diffuse gray surfaces, the effective emissivity indicates how effectively the cavity can exchange thermal radiation compared to a perfect black body. The effective emissivity of a cavity is defined by:
  • the wall emissivity \( \varepsilon_{1N^-} \)
  • the view factor \( F_{12} \) between the cavity's inner surface and its opening.
The formula to calculate it is:\[\varepsilon_e = \frac{\varepsilon_{1N^-} F_{12}}{1-(1-\varepsilon_{1N^-})F_{12}}\]This formula accounts for how much the cavity's shape and the emissive properties of its surface affect the radiation exchange, highlighting the importance of the cavity's geometry.
Radiation Heat Transfer
Radiation heat transfer is the process by which energy is emitted by a body and transmitted through electromagnetic waves. It differs from conduction and convection as it doesn't require a medium and can occur across a vacuum. This is a key mechanism for thermal transfer in many engineering applications, particularly in situations involving radiation heat exchange between surfaces with different emissivities.
The view factor, also known as the configuration factor, is essential in radiation heat transfer as it represents the fraction of radiation leaving one surface that directly reaches another. For cavities, such as cones, cylinders, and spheres, understanding the view factor helps in analyzing how effectively the cavity exchanges thermal energy with its surroundings.
  • Formula for view factor: \( F_{12} \)
  • Depends on geometric configuration and placement of surfaces.
  • Critical in determining effective emissivity (\( \varepsilon_e \)).
This understanding aids in designing systems for better thermal management, ensuring efficient heat transfer across surfaces.
Diffuse Gray Surfaces
A diffuse gray surface is an idealization used in thermal analysis for simplifying the modeling of radiation exchange. Such surfaces are assumed to radiate energy equally in all directions (diffuse) and have a uniform emissivity that doesn't vary with wavelength (gray), making for simpler and more straightforward calculations.
In the context of the exercise given, where the cavities (cone, cylinder, sphere) have diffuse gray surfaces:
  • Emissivity is constant for the entire surface.
  • Radiation properties can be treated uniformly across wavelengths.
  • Simplifies the computation of effective emissivity \( \varepsilon_e \).
By assuming diffuse gray surfaces, we can more easily calculate the view factors and the overall radiation heat transfer behavior, which is especially useful in engineering applications where precise thermal management is required.
Geometric Configuration
The geometric configuration of a surface or a cavity greatly influences its radiative properties, such as the view factor and effective emissivity. In radiation analysis, considering the shape and arrangement of different components helps to predict and optimize energy exchange.
In the exercise, we examine three different geometries: cone, cylinder, and sphere cavities. Each shape offers:
  • Unique challenges in calculating the view factor \( F_{12} \).
  • Different approaches to determining effective emissivity \( \varepsilon_e \).
  • Varying degrees of efficiency in energy transfer.
Understanding these variations allows engineers to select the right geometry for a specific application, ensuring optimal thermal performance and energy efficiency.

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

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

An opaque, diffuse, gray ( \(200 \mathrm{~mm} \times 200 \mathrm{~mm})\) plate with an emissivity of \(0.8\) is placed over the opening of a furnace and is known to be at \(400 \mathrm{~K}\) at a certain instant. The bottom of the fumace, having the same dimensions as the plate, is black and operates at \(1000 \mathrm{~K}\). The sidewalls of the fumace are well insulated. The top of the plate is exposed to ambient air with a convection coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and to large surroundings. The air and surroundings are each at \(300 \mathrm{~K}\). (a) Evaluate the net radiative heat transfer to the bottom surface of the plate. (b) If the plate has mass and specific heat of \(2 \mathrm{~kg}\) and \(900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively, what will be the change in temperature of the plate with time, \(d T_{p} / d r\) ? Assume convection to the bottom surface of the plate to be negligible. (c) Extending the analysis of part (b), generate a plot of the change in temperature of the plate with time, \(d T_{\rho} / d l\), as a function of the plate temperature for \(350 \leq T_{p} \leq 900 \mathrm{~K}\) and all other conditions remaining the same. What is the steady-state tem-

Consider the perpendicular rectangles shown schematically. (a) Determine the shape factor \(F_{12}\). (b) For rectangle widths of \(X=0.5,1.5\), and \(5 \mathrm{~m}\), plot \(F_{12}\) as a function of \(Z_{b}\) for \(0.05 \leq Z_{b} \leq 0.4 \mathrm{~m}\).] Compare your results with the view factor obtained from the two-dimensional relation for perpendicular plates with a common edge (Table 13.1).

A radiant oven for drying newsprint consists of a long duct \((L=20 \mathrm{~m})\) of semicincular cross section. The newsprint moves through the oven on a conveyor belt at a velocity of \(V=0.2 \mathrm{~m} / \mathrm{s}\). The newsprint has a water content of \(0.02 \mathrm{~kg} / \mathrm{m}^{2}\) as it enters the oven and is completely dry as it exits. To assure quality, the newsprint must be maintained at room temperature \((300 \mathrm{~K})\) during drying. To aid in maintaining this condition, all system components and the air flowing through the oven have a temperature of \(300 \mathrm{~K}\). The inner sarface of the semicaircular duct, which is of emissivity \(0.8\) and temperature \(T_{1}\), provides the radiant heat required to accomplish the drying. The wet surface of the newsprint can be considered to be black. Air entering the oven has a temperature of \(300 \mathrm{~K}\) and a relative humidity of \(20 \%\). Since the velocity of the air is large, its temperature and relative humidity can be assumed to be constant over the entire duct length. Calculate the required evaporation rate, air velocity \(u_{m}\), and temperature \(T_{1}\) that will ensure steady-state conditions for the process.

Consider the attic of a home located in a hot climate. The floor of the attic is characterized by a width of \(L_{1}=10 \mathrm{~m}\) while the roof makes an angle of \(\theta=30^{\circ}\) from the horizontal direction, as shown in the schematic. The homeowner wishes to reduce the heat load to the home by adhering bright aluminum foil \(\left(\varepsilon_{f}=0.07\right)\) onto the surfaces of the attic space. Prior to installation of the foil, the surfaces are of emissivity \(s_{e}=0.85\). (a) Consider installation on the bottom of the attic roof only. Determine the ratio of the radiation heat transfer after to before the installation of the foil. (b) Determine the ratio of the radiation heat transfer after to before installation if the foil is installed only on the top of the attic floor. (c) Determine the ratio of the radiation heat transfer if the foil is installed on both the roof bottom and the floor top.
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Fundamentals of Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Physics of Motion

Read Explanation

Waves Physics

Read Explanation

Modern Physics

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.