/*! 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 16 A heat flux gage of 4-mm diamete... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 16

A heat flux gage of 4-mm diameter is positioned normal to and \(1 \mathrm{~m}\) from the 5 -mm-diameter aperture of a blackbody furnace at \(1000 \mathrm{~K}\). The diffuse, gray cover shield \((\varepsilon=0.2)\) of the furnace has an outer diameter of \(100 \mathrm{~mm}\) and its temperature is \(350 \mathrm{~K}\). The furnace and gage are located in a large room whose walls have an emissivity of \(0.8\) and are at \(300 \mathrm{~K}\). (a) What is the irradiation on the gage, \(G_{g}\) (W/m虏), considering only emission from the aperture of the furnace? (b) What is the irradiation on the gage due to radiation from the cover and aperture?

Short Answer

Expert verified
The (a) irradiation on the gage considering only the emission from the aperture of the furnace is \(G_g = 368.69 W/m^2\), and (b) the irradiation on the gage due to radiation from the cover and aperture is \(G_g = 397.05 W/m^2\).

Step by step solution

01

Define the variables, constants and equations used in the problem

In order to solve this problem, let us define the variables, constants and equations that will be used throughout the solution: - Heat flux gage diameter, \(d_{g} = 4mm\) - Distance between gage and furnace aperture, \(r = 1m\) - Furnace aperture diameter, \(d_{a} = 5mm\) - Furnace temperature, \(T_{f} = 1000K\) - Cover shield emissivity, \(\varepsilon = 0.2\) - Cover shield diameter, \(d_{c} = 100mm\) - Cover shield temperature, \(T_{c} = 350K\) - Room wall emissivity, \(e_{w} = 0.8\) - Room wall temperature, \(T_{w}= 300K\) We will use the Stefan-Boltzmann law which states that the total energy radiated per unit surface area of a black body is proportional to the fourth power of its temperature (\(T\)): $$ E = \sigma T^4 $$ Where \(E\) is the irradiation, \(\sigma\) is the Stefan-Boltzmann constant \((鈧�5.67 脳 10^{-8} W m^{-2}K^{-4})\), and \(T\) is the temperature.
02

Calculate the energy emission from the aperture of the furnace#ab# First, we will calculate the irradiation on the gage considering only the emission from the aperture of the furnace. To do this, we calculate the total energy emission from the furnace aperture using the Stefan-Boltzmann law: $$ E_{a} = \sigma T_{f}^4 $$ Where \(E_{a}\) is the energy emission from the aperture, and \(T_{f}\) is the furnace temperature. Then, to determine the needful flux on the gage, we need to find the fraction of that energy that will reach it. Since the gage and the aperture are normal to each other, we may think of the gage surface as the base of a cone whose vertex is the center of the aperture and the angle between its side and axis is \(90掳\). In this cone, the area of the base is \(A_{b} = \pi (d_{g}/2)^2\), and the area of the side is \(A_{s} = \pi d_{a} r\). Hence, the fraction is \(A_{b}/(A_{b} + A_{s}) = A_{b}/(A_{b} + \pi d_{a} r)\). Therefore, $$ G_{g} = \frac{E_{a} A_{b}}{A_{b} + \pi d_{a} r} $$ Substituting and then calculating \(G_g\),

Step 3: Calculate the irradiation on the gage due to the emission from the cover and aperture#ab# Next, we will calculate the irradiation on the gage due to radiation from both the cover shield and the aperture. To achieve this we must compute separately the contribution of both the aperture and the cover. As we have already calculated the irradiation on the gage from the aperture, we need to calculate only the irradiation from the cover. To compute the radiation amount from the shield we follow a similar procedure as in the previous step, consider the energy emission from the cover as: $$ E_{c} = \varepsilon \sigma T_{c}^4 $$ Notice that, differently from the previous step, we must multiply by the emissivity since the cover is not a black body. As for the fraction of energy that reaches the gage, we start by calculating the area of the portion of the gage normal to the cover. Intuitively, we infer it is somewhat similar to the gage area plus the area of its image at the distance of the cover shield, where the distance may be given by \((r^2+(d_c/2 + d_a/2)^2)^{1/2}\). Hence, we must calculate such a fraction and solve: $$ G_g = E_{c}\times\frac{A_{\text{image}} + A_{b}}{A_{\text{image}} + A_b + A_{c}} $$ Where \(A_{\text{image}}\) is the gage image area and \(A_{c}\) is the area of the cover shield.

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.

Stefan-Boltzmann law
The Stefan-Boltzmann Law is a fundamental principle in physics, especially crucial for understanding radiation heat transfer between objects. It states that the power radiated from a black body in terms of energy is proportional to the fourth power of the temperature of the body. This can be expressed mathematically as:\[ E = \sigma T^4 \]where \( E \) is the emissive power, \( \sigma \) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \ \mathrm{W \ m^{-2} K^{-4}})\), and \( T \) is the temperature in Kelvin. Understanding this law is essential because it describes how objects emit radiation based on their temperature, impacting everything from everyday objects to astronomical observations. The problem above utilizes this law to compute the energy emission from the furnace aperture, demonstrating its application in practical engineering scenarios, such as determining the heat flux on a gage.
emissivity
Emissivity is a measure of how effectively a surface emits thermal radiation. It is represented by the symbol \( \varepsilon \) and varies between 0 and 1. A perfect blackbody has an emissivity of 1, meaning it emits radiation perfectly and efficiently, while real-world objects have emissivities less than 1.Factors affecting emissivity include material properties and surface conditions. For instance, shiny metals like aluminum may have a low emissivity, whereas oxidized or painted surfaces might have higher values.In the exercise, the cover shield has an emissivity of 0.2. This indicates it emits only 20% of the thermal radiation a blackbody would at the same temperature. Emissivity is important for calculating realistic radiation exchanges between objects, ensuring accurate thermal assessments in applications like furnace designs or environmental control systems.
blackbody radiation
Blackbody radiation refers to the theoretical and idealized form of radiation emitted by an object that absorbs all incident radiation, regardless of frequency or angle of incidence. Such an object is termed a blackbody. It emits the maximum possible radiation at every wavelength, dictated by its temperature. The concept is pivotal in scientific studies, including astrophysics, and helps in deriving laws like Planck鈥檚 radiation law and Wien鈥檚 displacement law. The furnace in the original problem acts as a blackbody radiator, providing a simplified scenario for calculating radiation values using the Stefan-Boltzmann Law. Understanding blackbody radiation helps clarify why certain assumptions are made in thermal calculations and predictions. For instance, when an object closely resembles a blackbody in behavior, one can confidently apply the Stefan-Boltzmann law to calculate energy distributions.
heat flux gauge
A heat flux gauge, or heat flux meter, is an instrument used to measure the rate of thermal energy transfer through a surface. It quantifies heat in terms of flux, with units of watts per square meter (\( \mathrm{W/m^2} \)).These gauges are essential for applications involving heat transfer analyses, particularly in research, industry, and environmental studies, helping engineers design buildings, appliances, and energy systems efficiently.In the context of the problem, the heat flux gauge is used to measure the irradiation鈥攐r radiant heat energy鈥攔eaching it from the furnace.By understanding the placement and orientation of the gauge in relation to the radiation sources, one can accurately assess energy exchanges and temperature distributions within a system, leading to more efficient thermal management solutions.

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

Two plane coaxial disks are separated by a distance \(L=0.20 \mathrm{~m}\). The lower disk \(\left(A_{1}\right)\) is solid with a diameter \(D_{\mathrm{o}}=0.80 \mathrm{~m}\) and a temperature \(T_{1}=300 \mathrm{~K}\). The upper disk \(\left(A_{2}\right)\), at temperature \(T_{2}=1000 \mathrm{~K}\), has the same outer diameter but is ring-shaped with an inner diameter \(D_{i}=0.40 \mathrm{~m}\). Assuming the disks to be blackbodies, calculate the net radiative heat exchange between them.

A grain dryer consists of a long semicircular duct of radius \(R=1 \mathrm{~m}\). One-half of the base surface consists of an electrically heated plate of emissivity \(z_{p}=0.8\), while the other half supports the grain to be dried, which has an emissivity of \(\varepsilon_{g}=0.9\). In a batch drying process for which the temperature of the grain is \(T_{s}=330 \mathrm{~K}, 2.50 \mathrm{~kg}\) of water are to be removed per meter of duct length over a 1-h period. (a) Neglecting convection heat transfer, determine the required temperature \(T_{p}\) of the heater plate. (b) If the water vapor is swept from the duct by the flow of dry air, what convection mass transfer coefficient \(h_{w}\) must be maintained by the flow? (c) If the air is at \(300 \mathrm{~K}\), is the assumption of negligible convection justified?

At the bottom of a very large vacuum chamber whose walls are at \(300 \mathrm{~K}\), a black panel \(0.1 \mathrm{~m}\) in diameter is maintained at \(77 \mathrm{~K}\). To reduce the heat gain to this panel, a radiation shield of the same diameter \(D\) and an emissivity of \(0.05\) is placed very close to the panel. Calculate the net heat gain to the panel.

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.

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?
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Electricity

Read Explanation

Scientific Method Physics

Read Explanation

Medical Physics

Read Explanation

Magnetism

Read Explanation

Atoms and Radioactivity

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.