/*! 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 86 Consider two very large metal pa... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 86

Consider two very large metal parallel plates. The top plate is at a temperature \(T_{t}=400 \mathrm{~K}\) while the bottom plate is at \(T_{b}=300 \mathrm{~K}\). The desired net radiation hear flux between the two plates is \(q^{\prime \prime}=330 \mathrm{~W} / \mathrm{m}^{2}\). (a) If the two surfaces have the same radiative properties, show that the required surface emissavity is \(\varepsilon=0.5\). (b) Metal surfaces at relatively low temperatures tend to have emissivities much less than \(0.5\) (see Table A.11). An engineer proposes to apply a checker pattern, similar to that of Problem \(12.132\), onto each of the metal surfaces so that half of each surface is characterized by the low emissivity of the bare metal and the other half is covered with the high-emissivity paint. If the average of the high and los emissavities is 0.5, will the net radiative heat flux between the surfaces be the desired value?

Short Answer

Expert verified
In this problem, we found that the required surface emissivity for both plates to achieve the desired net radiation heat flux is 0.5. However, using a checker pattern with an average emissivity of 0.5 will not necessarily result in the desired heat flux, as the heat exchange is not uniformly distributed across the surface. The actual heat flux depends on the specific spatial distribution and values of the high and low emissivities.

Step by step solution

01

Set up the problem and identify relevant formulas

We know that the net radiation heat flux between the two plates can be given by the following equation, where \(F_{12}\) is the view factor from plate 1 to plate 2: \[q^{\prime\prime} = \epsilon(蟽T_1^4 - 蟽T_2^4)F_{12}\] In this case, \(蟽\) is the Stefan-Boltzmann constant, \(5.67 脳 10^{-8} Wm^{-2}K^{-4}\), and both plates have the same emissivity, \(\epsilon\). Since the plates are large and parallel, we can assume that the view factor between them is \(F_{12} = 1\). We're given the temperatures of both plates: \(T_t = 400 K\) and \(T_b = 300 K\). We're also given the desired net radiation heat flux, \(q^{\prime\prime} = 330 W/m^2\). Now, we can solve for the required surface emissivity.
02

Calculate the required surface emissivity

First, plug the known values into the equation: \[330 = \epsilon (5.67 脳 10^{-8})((400 K)^4 - (300 K)^4)\] Next, simplify the equation and solve for \(\epsilon\): \[\epsilon = \frac{330}{(5.67 脳 10^{-8})((400)^4 - (300)^4)}\] \[\epsilon = 0.5\] So, the required surface emissivity for both plates is \(\epsilon = 0.5\).
03

Analyze the checker pattern solution

Now, we need to determine if using a checker pattern with half of each surface having the low emissivity of the bare metal and the other half having a high-emissivity paint, such that the average emissivity is 0.5, will result in the desired net radiative heat flux. However, this situation is not as simple as having two uniform surfaces with emissivity 0.5 because the radiative heat exchange is not uniformly distributed across the surface. Instead, the heat flux will be higher in areas with high-emissivity paint and lower in bare metal areas. Therefore, the net radiative heat flux will not be the same as if both surfaces had uniform emissivities of 0.5. In other words, it cannot be guaranteed that the net radiative heat flux between the surfaces will be the desired value if a checker pattern is applied. The actual heat flux will depend on the specific spatial distribution and values of the high and low emissivities. In conclusion, having an average emissivity of 0.5 with a checker pattern will not necessarily result in the desired net radiative heat flux between the surfaces. Instead, the specific emissivities and their spatial distribution will need to be considered to determine the actual heat flux.

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.

Emissivity
Emissivity is a fundamental concept in understanding how surfaces emit thermal radiation. It is a measure of how effectively a material emits energy as thermal radiation compared to a perfect blackbody, which is an idealized surface that emits the maximum amount of radiation possible. The emissivity of a surface is represented as a dimensionless number ranging between 0 and 1.

An emissivity of 1 means the surface is a perfect emitter, while an emissivity of 0 indicates no thermal radiation is emitted. Real-world surfaces have emissivities less than 1, meaning they radiate less energy than a blackbody at the same temperature. For metal surfaces, emissivity is often much lower due to their reflective properties. This makes understanding and manipulating emissivity crucial in designing systems involving radiative heat transfer.

In the context of large metal plates with uniform emissivity, as in the exercise, knowing the precise emissivity is essential for calculating the net radiative heat flux. If both plates have the same radiative properties, the emissivity directly influences the heat transfer between the plates.
Thermal Radiation
Thermal radiation is an important mode of heat transfer and occurs when surfaces emit electromagnetic waves due to their temperature. Unlike conduction or convection, thermal radiation does not require a medium to transfer heat, which means it can occur in a vacuum.

All objects with a temperature above absolute zero emit thermal radiation, but the intensity and range of the emitted radiation depend on the object's temperature and surface emissivity. For instance, hotter objects emit more intense radiation.

In the scenario with two metal plates, thermal radiation is the primary mechanism for heat transfer between them. The radiative heat flux is determined by the temperature difference between the plates and their emissivities. This is critical for designing effective heat exchange systems, especially in applications where physical contact or a medium like air cannot be used for efficient heat transfer.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law is key to calculating the thermal radiation emitted by a surface. It states that the power emitted per unit area of a surface is proportional to the fourth power of the surface's absolute temperature, with the proportionality constant being the Stefan-Boltzmann constant (蟽 = 5.67 脳 10^{-8} Wm^{-2}K^{-4}). The formula is given by \[ E = 蟽T^4 \] ,where E is the power emitted per unit area, T is the absolute temperature, and 蟽 is the Stefan-Boltzmann constant.

When applied to the radiative heat transfer between two plates, as in the exercise, the net radiation heat flux is calculated using the temperatures of both plates and their emissivity. With the net flux described as q'' = 蔚(蟽(T_1^4 - T_2^4)), where 蔚 is the emissivity of the plates. Understanding this relationship helps to determine the necessary emissivity to achieve particular heat transfer outcomes, as in calculating the net radiation heat flux between the plates at 400K and 300K.

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 double-glazed window consists of two panes of glass, each of thickness \(I=6 \mathrm{~mm}\). The inside room temperature is \(T_{t}=20^{\circ} \mathrm{C}\) with \(h_{i}=7.7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the outside temperature is \(T_{o}=-10^{\circ} \mathrm{C}\) with \(h_{a}=25 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\). The gap between the glass sheets is of thickness \(L=5 \mathrm{~mm}\) and is filled with a gas. The glass surfaces may be treated with a low-emissivity coating to reduce their emissivity from \(\varepsilon=0.95\) to \(\varepsilon=0.05\). Determine the heat flux through the window for case 1: \(\varepsilon_{1}=\varepsilon_{2}=0.95\), case \(2: \varepsilon_{1}=\varepsilon_{2}=0.05\), and case 3: \(\varepsilon_{\mathrm{L}}=0.05, \varepsilon_{2}=0.95\). Consider either air of argon of thermal conductivity \(k_{A R}=17.7 \times 10^{-3}\) W/m. \(\mathrm{K}\) to be within the gap. Radiation heat transfer occurring at the external surfaces of the two glass sheets is negligible, as is free convection between the glass sheets.

Consider a circular furnace that is \(0.3 \mathrm{~m}\) long and \(0.3 \mathrm{~m}\) in diameter. The two ends have diffuse, gray surfaces that are maintained at 400 and \(500 \mathrm{~K}\) with emissivities of \(0.4\) and \(0.5\), respectively. The lateral surface is also diffuse and gray with an emissivity of \(0.8\) and a temperature of \(800 \mathrm{~K}\). Determine the net radiative heat transfer from each of the surfaces.

Consider two diffuse surfaces \(A_{1}\) and \(A_{2}\) on the inside of a spherical enclosure of radius \(R\). Using the following methods, derive an expression for the view factor \(F_{12}\) in terms of \(A_{2}\) and \(R\). (a) Find \(F_{12}\) by beginning with the expression \(F_{i j}=\) \(q_{i \rightarrow j} / A_{i} J_{i} .\) (b) Find \(F_{12}\) using the view factor integral, Equation 13.1.

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 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?
See all solutions

Recommended explanations on Physics Textbooks

Electricity and Magnetism

Read Explanation

Electrostatics

Read Explanation

Translational Dynamics

Read Explanation

Modern Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Radiation

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.