/*! 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 3 Derive expressions for the view ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Short Answer

Expert verified
(a) For a circular disk and a coaxial, ring-shaped disk, the view factor \(F_{12}\) can be derived using Hottel's crossed-string method by dividing surface A2 into differential elements and calculating their view factors to A1 before integrating over A2. The final expression for \(F_{12}\) is in terms of \(A_1\), \(A_2\), and any appropriate hypothetical surface area from Hottel's method. (b) For a circular disk and a coaxial, right-circular cone, the same method is applied, dividing surface A2 into differential elements and calculating their view factors to A1 before integrating over A2. The final expression for \(F_{12}\) is in terms of \(A_1\), \(A_2\), and any appropriate hypothetical surface area from Hottel's method. To sketch the variation of \(F_{12}\) with angle \(\theta\) for \(0 \leq \theta \leq \frac{\pi}{2}\), observe how the view factors change with the cone's angle and explain the key features of this curve.

Step by step solution

01

Identifying the dimensions and surfaces

Determine the dimensions of the circular disk (radius R1) and the ring-shaped disk (inner radius R2 and outer radius R3). Identify the surfaces A1 (circular disk) and A2 (ring-shaped disk).
02

Dividing A2 into differential elements

To use the Hottel's crossed-string method, we need to divide surface A2 (the ring-shaped disk) into differential elements, each having an area dA.
03

Finding the view factor to the differential element dA

For each differential element in A2, calculate its view factor to A1, defined as F1->dA. Use Table 13.2 and Figure 13.5 for the view factor of coaxial parallel disks as reference.
04

Integrating over A2

Integrate F1->dA over A2 to find the view factor F12. \( F_{12} = \int_{A_2} F_{1 \rightarrow dA}\, dA \)
05

Expressing the result

Express the final result for F12 in terms of A1, A2, and any appropriate hypothetical surface area from Hottel's method. #(b) A circular disk and a coaxial, right-circular cone#
06

Identifying the dimensions and surfaces

Determine the dimensions of the circular disk (radius R1) and the right-circular cone (radius R2 and angle θ). Identify the surfaces A1 (circular disk) and A2 (right-circular cone).
07

Dividing A2 into differential elements

To use the Hottel's crossed-string method, we need to divide surface A2 (the right-circular cone) into differential elements, each having an area dA.
08

Finding the view factor to the differential element dA

For each differential element in A2, calculate its view factor to A1, defined as F1->dA. Use Table 13.2 and Figure 13.5 for the view factor of coaxial parallel disks as reference.
09

Integrating over A2

Integrate F1->dA over A2 to find the view factor F12. \( F_{12} = \int_{A_2} F_{1 \rightarrow dA}\, dA \)
10

Expressing the result

Express the final result for F12 in terms of A1, A2, and any appropriate hypothetical surface area from Hottel's method.
11

Sketching F12 for varying θ

Sketch the variation of F12 with angle θ for 0 ≤ θ ≤ π/2. Explain the key features of this curve. This will show how F12 changes with the cone's angle.

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 91Ó°ÊÓ!

Key Concepts

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

Hottel's Crossed-String Method
Hottel's Crossed-String Method is a valuable tool used in radiation heat transfer to calculate view factors, which are crucial for understanding how radiation is exchanged between surfaces. The core idea is to simplify complex geometrical configurations into manageable mathematical problems by using the strings analogy. Imagine a flat surface split into infinitesimally small differential elements. These elements are conceptual 'strings' that cross over to another surface. The key is to understand that not the entire surface will radiate evenly to every other point on another surface. By calculating radiation exchange for differential elements individually, we can gain an accurate picture of the system's radiation behavior. The method begins by identifying relevant surfaces and dividing the target surface into these differential elements. This approach allows us to establish a relationship between a surface and its corresponding radiation view factor, even when the two surfaces have complex orientations like concentric circles or angles, such as a right-circular cone. The results are then integrated to find total exchanges, considering multiple strings that represent paths of radiant energy traversing between the surfaces. This process is useful in heat transfer analysis, especially when dealing with coaxial geometries, which we will delve into next.
Coaxial Surfaces
Coaxial surfaces are surfaces that share a common axis or center line. In radiation view factor analysis, typical examples include the circular disk and the coaxial ring-shaped disk arrangement or a disk and a coaxial right-circular cone. These configurations are frequently encountered in engineering applications where symmetry around a center axis can simplify calculations and assumptions. The symmetry inherent in coaxial shapes allows for more straightforward mathematical expressions of view factors. For instance, both a disk and a coaxial cone can utilize the symmetry to simplify the view factor calculations, as the math involved can often be reduced to angular integrations or reference from known solutions, like those found in tables such as Table 13.2 in thermal engineering resources. This problem's unique aspect is using the coaxial alignments to easily visualize how radiation flows between surfaces, whether it's from a circular disk to a ring or from a disk to a partial volume like a cone. The perspective provided by coaxial arrangements makes the entire field more intuitive, as we can often support view factor derivation through geometric relationships and predefined mathematical models, which allows deeper insight into the mechanics of radiation in engineering systems.
Heat Transfer Analysis
Heat transfer analysis involving radiation view factors is fundamental in designing systems where the exchange of thermal energy between surfaces plays a critical role. This involves determining how much heat is transferred from one surface to another via radiant energy, which is significant in applications like furnaces, electronic cooling, and climate control systems.In radiation heat transfer, the view factor \( F_{12} \) quantifies the proportion of energy leaving one surface that arrives at another. This can vary based on geometry, orientation, and relative size, requiring precise calculation to optimize heat exchange efficiently. The temperature differences and emissive properties of surfaces often drive these exchanges, making view factors an essential component to be evaluated in every heat transfer problem.The entire process demands an understanding of both the mathematical and physical principles involved, including energy conservation and the Stefan-Boltzmann law, which relates the total energy radiated by a surface to its temperature. In practical scenarios, engineers use these principles to design equipment that ensures energy is utilized effectively and systems are maintained within desired operating temperatures. Understanding and applying the correct view factors can significantly enhance a system's efficiency and reliability, especially when complex configurations like disks and cones are involved.

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

Liquid oxygen is stored in a thin-walled, spherical container \(0.8 \mathrm{~m}\) in diameter, which is enclosed within a second thin-walled, spherical container \(1.2 \mathrm{~m}\) in diameter. The opaque, diffuse, gray container surfaces have an emissivity of \(0.05\) and are separated by an evacuated space. If the outer surface is at \(280 \mathrm{~K}\) and the inner surface is at \(95 \mathrm{~K}\), what is the mass rate of oxygen lost due to evaporation? (The latent heat of vaporization of oxygen is \(2.13 \times 10^{5} \mathrm{~J} / \mathrm{kg}\).)

Heat transfer by radiation occurs between two large parallel plates, which are maintained at temperatures \(T_{1}\) and \(T_{2}\), with \(T_{1}>T_{2}\). To reduce the rate of heat transfer between the plates, it is proposed that they be separated by a thin shield that has different emissivities on opposite surfaces. In particular, one surface has the emissivity \(\varepsilon_{s}<0.5\), while the opposite surface has an emissivity of \(2 \varepsilon_{s}\). (a) How should the shield be oriented to provide the larger reduction in heat transfer between the plates? That is, should the surface of emissivity \(\varepsilon_{s}\) or that of emissivity \(2 s_{s}\) be oriented toward the plate at \(T_{1}\) ? (b) What orientation will result in the larger value of the shield temperature \(T_{s}\) ?

The arrangement shown is to be used to calibrate a heat flux gage. The gage has a black surface that is \(10 \mathrm{~mm}\) in diameter and is maintained at \(17^{\circ} \mathrm{C}\) by means of a water-cooled backing plate. The heater, \(200 \mathrm{~mm}\) in diameter, has a black surface that is maintained at \(800 \mathrm{~K}\) and is located \(0.5 \mathrm{~m}\) from the gage. The surroundings and the air are at \(27^{\circ} \mathrm{C}\) and the convection heat transfer coefficient between the gage and the air is \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the net radiation exchange between the heater and the gage. (b) Determine the net transfer of radiation to the gage per unit area of the gage. (c) What is the net heat transfer rate to the gage per unit area of the gage? (d) If the gage is constructed according to the description of Problem 3.107, what heat flux will it indicate?

Two concentric spheres of diameter \(D_{1}=0.8 \mathrm{~m}\) and \(D_{2}=1.2 \mathrm{~m}\) are separated by an air space and have surface temperatures of \(T_{1}=400 \mathrm{~K}\) and \(T_{2}=300 \mathrm{~K}\). (a) If the surfaces are black, what is the net rate of radiation exchange between the spheres? (b) What is the net rate of radiation exchange between the surfaces if they are diffuse and gray with \(\varepsilon_{1}=0.5\) and \(\varepsilon_{2}=0.05\) ? (c) What is the net rate of radiation exchange if \(D_{2}\) is increased to \(20 \mathrm{~m}\), with \(\varepsilon_{2}=0.05, \varepsilon_{1}=0.5\), and \(D_{1}=0.8 \mathrm{~m}\) ? What error would be introduced by assuming blackbody behavior for the outer surface \(\left(\varepsilon_{2}=1\right)\), with all other conditions remaining the same? (d) For \(D_{2}=1.2 \mathrm{~m}\) and emissivities of \(s_{1}=0.1,0.5\), and \(1.0\), compute and plot the net rate of radiation exchange as a function of \(\varepsilon_{2}\) for \(0.05 \leq \varepsilon_{2} \leq 1.0\).

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?

See all solutions

Recommended explanations on Physics Textbooks

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.