/*! 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 4 A right-circular cone and a righ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A right-circular cone and a right-circular cylinder of the same diameter and length \(\left(A_{2}\right)\) are positioned coaxially at a distance \(L_{0}\) from the circular disk \(\left(A_{1}\right)\) shown schematically. The inner base and lateral surfaces of the cylinder may be treated as a single surface, \(A_{2}\). The hypothetical area corresponding to the opening of the cone and cylinder is identified as \(A_{3}\). (a) Show that, for both arrangements, \(F_{21}=\left(A_{1} / A_{2}\right) F_{13}\) and \(F_{22}=1-\left(A_{3} / A_{2}\right)\), where \(F_{13}\) is the view factor between two coaxial, parallel disks (Table 13.2). (b) For \(L=L_{o}=50 \mathrm{~mm}\) and \(D_{1}=D_{3}=50 \mathrm{~mm}\), calculate \(F_{21}\) and \(F_{22}\) for the conical and cylindrical configurations and compare their relative magnitudes. Explain any similarities and differences. (c) Do the relative magnitudes of \(F_{21}\) and \(F_{22}\) change for the conical and cylindrical configurations as \(L\) increases and all other parameters remain fixed? In the limit of very large \(L\), what do you expect will happen? Sketch the variations of \(F_{21}\) and \(F_{22}\) with \(L\), and explain the key features.

Short Answer

Expert verified
In summary, for a right-circular cone and a right-circular cylinder of the same diameter and length, we get \(F_{21}=\left(A_{1} / A_{2}\right) F_{13}\) and \(F_{22}=1-\left(A_{3} / A_{2}\right)\) for both configurations. For the given parameters, \(F_{21}\) is the same for both configurations, while \(F_{22}\) is 1 for the conical configuration and 0 for the cylindrical configuration. As \(L\) increases, \(F_{21}\) remains the same, \(F_{22}^{cone}\) stays constant at 1, and \(F_{22}^{cylinder}\) approaches 1.

Step by step solution

01

(a) Show \(F_{21}=\left(A_{1} / A_{2}\right) F_{13}\) and \(F_{22}=1-\left(A_{3} / A_{2}\right)\)

We can apply the reciprocal rule for view factors. As we know that \(F_{i,j} A_{i} = F_{j,i} A_{j}\), where \(F_{i,j}\) represents the view factor from surface i to surface j. Let's apply the rule for surfaces 1, 2, and 3. From surface 1 to surface 2: \(F_{21} A_{1} = F_{12} A_{2}\) From surface 1 to surface 3: \(F_{13} A_{1} = F_{31} A_{3}\) From surface 2 to surface 1 and 3 combined (as cylinder inner base and lateral surface can be treated as a single surface): \(F_{22} A_{2} = F_{11} A_{1} + F_{13} A_{3}\) Now we can write \(F_{11}=1-F_{12}\) as view factors add up to 1 from the same surface. Combining all these equations, we get: \(F_{21}=\left(A_{1} / A_{2}\right) F_{13}\) \(F_{22}=1-\left(A_{3} / A_{2}\right)\)
02

(b) Calculate \(F_{21}\) and \(F_{22}\) for conical and cylindrical configurations

For the conical and cylindrical configurations, we have \(L=L_{o}=50 \mathrm{~mm}\) and \(D_{1}=D_{3}=50 \mathrm{~mm}\). For the conical configuration, the area \(A_{3} = 0\) (as there's no opening), so we get: \(F_{22}^{cone}=1\) For the view factor between parallel disks, we have \(F_{13} = 0.158\). Using this value we can find \(F_{21}^{cone}\): \(F_{21}^{cone}=\left(A_{1} / A_{2}\right) F_{13} = (pi*25^2) / (pi*25^2) * 0.158 = 0.158\) For the cylindrical configuration, the area \(A_{3} = pi*D_{3}^{2}/4 = 1963.5 \mathrm{~mm}^2\) (as it's a simple disk), so we get: \(F_{22}^{cylinder}=1-\left(A_{3} / A_{2}\right) = 1-\left(1963.5 / 1963.5\right) = 0\) Now, we can find \(F_{21}^{cylinder}\): \(F_{21}^{cylinder}=\left(A_{1} / A_{2}\right) F_{13} = (pi*25^2) / (pi*25^2) * 0.158 = 0.158\) The magnitudes of \(F_{21}\) are the same for both conical and cylindrical configurations, and \(F_{22}\) is 1 for the conical configuration and 0 for the cylindrical configuration.
03

(c) Variations of \(F_{21}\) and \(F_{22}\) with \(L\)

As we observe the view factors for the conical and cylindrical configurations, we notice that \(F_{21}\) stays the same for both configurations and isn't affected by increasing the value of \(L\). However, when \(L\) increases and all other parameters remain fixed, we would expect the \(F_{22}\) for the cone to remain constant at 1 since the opening remains closed, whereas the \(F_{22}\) for the cylinder will approach 1 as it becomes more difficult for radiation to escape the larger cavity. In the limit of very large \(L\), both \(F_{22}^{cone}\) and \(F_{22}^{cylinder}\) would approach 1, as radiation getting trapped inside both geometries becomes more prominent. As for \(F_{21}\), it stays constant and doesn't change with \(L\). To summarize, the key features of the variation of view factors with L: \(F_{21}\) remains the same for both conical and cylindrical configurations, but \(F_{22}\) for the conical configuration stays constant at 1, while \(F_{22}\) for the cylindrical configuration approaches 1 as \(L\) increases.

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.

Heat Transfer
Heat transfer is a fundamental concept in physics and engineering that deals with the movement of thermal energy from one body or substance to another. This can occur through various mechanisms: conduction, where heat is transferred via direct contact; convection, which involves the movement of fluid; and radiation, which involves the transfer of energy through electromagnetic waves.

Understanding heat transfer is crucial in a variety of applications, from designing efficient heating and cooling systems to analyzing the thermal properties of materials. In our example involving a right-circular cone and cylinder, heat transfer analysis is used to study how thermal energy might radiate between different surfaces—crucial for applications in thermal engineering and energy management.
Radiative Heat Transfer
Radiative heat transfer is one of the modes of heat transfer which occurs through electromagnetic radiation. It involves the transfer of heat in the form of infrared rays and does not require any medium, thus it can occur in a vacuum.

When considering radiative heat transfer between two surfaces, the concept of view factors becomes important. A view factor, sometimes referred to as a configuration factor or a shape factor, is a dimensionless quantity that represents the fraction of radiation leaving surface A that strikes surface B directly.

Calculating View Factors

In our exercise, we use the reciprocity and summation rules for view factors to calculate the proportion of radiative heat transfer between the surfaces of a cone, cylinder, and a disk. These factors are critical in predicting how much heat is exchanged through radiation, and hence, are foundational for designing systems with thermal components.
Reciprocal Rule for View Factors
The reciprocal rule for view factors is a mathematical relationship derived from the conservation of energy and the geometry of radiative heat exchange between surfaces. It states that the product of the view factor from surface 1 to surface 2 and the area of surface 1 is equal to the product of the view factor from surface 2 to surface 1 and the area of surface 2, represented mathematically as: \(F_{12} A_{1} = F_{21} A_{2}\).

This reciprocal relationship greatly simplifies the calculation of view factors in complex geometrical arrangements. In our example, the reciprocal rule is applied to derive that the view factor from the disk to the cylinder or cone is proportional to the view factor between the coaxial, parallel disks. This concept is critical to solving radiative heat transfer problems efficiently and is a vital tool for engineers and scientists who work with systems where radiation plays a significant role in heat exchange.

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 long, cylindrical heating element of \(20-\mathrm{mm}\) diameter operating at \(700 \mathrm{~K}\) in vacuum is located \(40 \mathrm{~mm}\) from an insulated wall of low thermal conductivity. (a) Assuming both the element and the wall are black, estimate the maximum temperature reached by the wall when the surroundings are at \(300 \mathrm{~K}\). (b) Calculate and plot the steady-state wall temperature distribution over the range \(-100 \mathrm{~mm} \leq\) \(x \leq 100 \mathrm{~mm}\).

A row of regularly spaced, cylindrical heating elements (1) is used to cure a surface coating that is applied to a large panel (2) positioned below the elements. A second large panel (3), whose top surface is well insulated, is positioned above the elements. The elements are black and maintained at \(T_{1}=600 \mathrm{~K}\). while the panel has an emissivity of \(\varepsilon_{2}=0.5\) and is maintained at \(T_{2}=400 \mathrm{~K}\). The cavity is filled with a nonparticipating gas and convection heat transfer occurs at surfaces 1 and 2 , with \(\bar{h}_{1}=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(\bar{h}_{2}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (Convection at the insulated panel may be neglected.) (a) Evaluate the mean gas temperature, \(T_{w^{-}}\) (b) What is the rate per unit axial length at which electrical energy must be supplied to each element to maintain its prescribed temperature? (c) What is the rate of heat transfer to a portion of the coated panel that is \(1 \mathrm{~m}\) wide by \(1 \mathrm{~m}\) long?

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.

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?

Determine the steady-state temperatures of two radiation shields placed in the evacuated space between two infinite planes at temperatures of 600 and \(325 \mathrm{~K}\). All the surfaces are diffuse and gray with emissivities of \(0.7\).

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.